Theoretical Exercises in Probability and Statistics

N. A. Rahman M.A., M.Sc., Ph.D.

Other books on theoretical and applied statistics N. A. Rahman A course in theoretical statistics N. A. Rahman · al exercises in probability Prac t IC . . and statistics The advanced theory of statistics Sir Maurice Kendall, Alan Stuart, and J. Keith Ord Vol. 1. Distribution theory. . Vol. 2. Inference and relatIonshIp . . Vol. 3. Design and analysis, and tIme-series M. Boldrini Scientific truth and statistical method (transL Ruth Kendall) Sir Maurice Kendall Multivariate analysis Time-series Sir Maurice Kendall Rank correlation methods Sir Maurice Kendall Statistical papers of George Udny Yule ed. A. Stuart and Sir Maurice Kendall Studies in the history of statistics and probability (2 voL) ed. E. S. Pearson and Sir Maurice Kendall ed. Sir Maurice Kendall and R. L. Plackett M. H. Quenouille Rapid statistical calculations Characteristic functions E. Lukacs E. Lukacs Developments in characteristic function theory J. A. John and M. H. Quenouille Experiments: design and analysis F. Yates Experimental design: selected papers Sampling methods for censuses and surveys F. Yates Biomathematics Cedric A. B. Smith Combinatorial chance F. N. David and D. E. Barton Exercises in mathematical economics and econometrics J. E. Spencer and R. C. Geary S. Vajda Problems in linear and nonlinear programming Mathematical model building in economics and industry (1st and 2nd series) ed. Sir Maurice Kendall Computer simulation models John Smith The mathematical theory of infectious diseases N. T. J. Bailey N. T. J. Bailey The biomathematics of malaria Estimation of animal abundance and related parameters G. A. F. Seber Statistical epidemiology in veterinary science F. B. Leech and K. C. Sellers Statistical method in biological assay D. J. Finney Physical applications of stationary time-series: with special reference to digital data processing of seismic signals E. A. Robinson Style and vocabulary: numerical studies C. B. Williams For a list of Griffin's "Statistical Monographs & Courses", and the "Biometrika" books distributed by Griffin, see the end of this book.

THEORETICAL EXERCISES IN PROBABILITY and

STATISTICS FOR MATHEMATICS UNDERGRADUATES

With answers and hints on solutions

N. A. RAHMAN M.A.(Alld), M.Sc.(Stat.)(Ca1c.) Ph.D.(Stat. )(Camb.) Senior Lecturer in Mathematical Statistics, University of Leicester

SECOND EDITION Including an extensive supplement bringing the total of exercises to over 580

MACMILLAN PUBLISHING CO., INC. NEW YORK

Copyright © Charles Griffin & Co Ltd 1983

Published in USA by Macmillan Publishing Co., Inc. 866 Third Avenue, New York, N.Y. 10022 Distributed in Canada by Collier Macmillan Canada, Ltd.

All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission in writing from Charles Griffin & Co. Ltd., Charles Griffin House, Crendon Street, High Wycombe, Bucks, England, HP136LE. By arrangement with the originating publisher CHARLES GRIFFIN & COMPANY LIMITED London and High Wycombe First published 1967 Second edition 1983 Library of Congress Catalogue Card Number 83-61216 ISBN 0-02-850760-6

Typeset in Northern Ireland at The Universities Press (Belfast) Ltd. Printed in Great Britain by Pitman Press, Bath.

Foreword Statistical theory is one of those subjects which are best learnt by working through many exercises. But good exercises are not easy to find. Authors of textbooks frequently devote inadequate attention to this side of what ought to be their business and, if they provide exercises at all, offer too many which are trivial. There are some notable exceptions to this generalization, but it is true enough to justify an attempt to provide a separate set of exercises which make an essential point and have been piloted through a class of students to ensure that they are of the right standard. Some years ago I myself published a set of exercises in theoretical statistics, with answers. They have been fairly widely used, to judge by the sales, but some of them were of a rather advanced standard and, in effect, have been replaced by the exercises at the end of chapters of Kendall and Stuart's Advanced Theory of Statistics. In the meantime Dr Rahman has for some years been teaching theoretical statistics at the University of Leicester and has got together a set of exercises for undergraduates. They should be useful to teachers and students from the sixth-form level onwards, and, having been worked over by students for several years, should be almost free of error and ambiguity. I hope that the publication of this set may do something to encourage the study of statistics in all parts of the world. LONDON

1967

M.G.KENDALL

Note by Sir Maurice Kendall In the above Foreword I stressed my view that really good exercises-those detailing statistical, not merely arithmetical, method-are of the highest importance. It is gratifying that the steady demand for this book has justified this view and I welcome this new enlarged edition. 1982 M. G. KENDALL

v

TO

BEAUMONT HALL UNIVERSITY OF LEICESTER

Preface to First Edition This collection of theoretical exercises concentrates only on such aspects of probability and statistics as could be covered, with some selection, in a first year's course for mathematics undergraduates. The choice of topics is largely a personal one, though perhaps not entirely arbitrary, as the selection of problems represents a balanced integration of the basic ideas which would normally be included in an introductory course designed to create a lively interest in the subject as a whole. On the other hand, the mathematical requirements have been deliberately made flexible so as to permit a choice of exercises suitable for anyone of the usual three years of a degree course. It is hoped that the use of these exercises will help to clarify the preliminary ideas of a novel subject, and to provide a sample of the kind of mathematics that is a prerequisite for further study. Over 400 exercises are included in this collection, and these have been broadly classified into the following four chapters: Chapter 1 : Discrete probability and random variables (including the use of simple difference equations and generating functions). Chapter 2 : Continuous random variables (including joint and simple derived distributions). Chapter 3 : Estimation, tests of significance, and inference (including bivariate correlation and regression). Chapter 4 : Characteristic functions (their use in deriving sampling distributions explicitly). Chapter 1 starts with the simplest notions of discrete probability and elementary combinatorial methods. The main stress is on an integration of the ideas of probability with the concepts of random variables and expected values. The general development is based on a consistent use of generating functions, and an attempt has also been made to include a number of exercises on the use of difference equations for the evaluation of probabilities and expected values. The plan of the second chapter is complementary to that of the first in providing an introduction to the concept of continuous variation and related statistical ideas of expectations for joint and simple derived distributions. vii

viii

PREFACE

The first two chapters provide the basis for the major Chapter 3, which deals with a variety of statistical problems. These should give the student an indication of the kind of manipulative mathematics used in statistical theory, and a deepening insight into the principles. The emphasis on estimation is based on the belief that this is perhaps the most significant contribution of modern statistical theory, though, unfortunately, a clear understanding of the ideas of estimation is often not readily acquired by a beginner. The omission of confidence intervals and power functions is due to my conviction that these ideas require a thorough grasp of distribution theory which puts them outside the scope of a first course. On the other hand, the notion of efficiency in terms of relative variance is a useful concept which has the additional merit of being intuitively acceptable to a beginner. Multivariate regression and the general least-squares approach used in the analysis of variance have both been excluded as too specialized for a first course. The last chapter is perhaps of limited interest, since there seems to be an increasing tendency in universities to institute a course in statistical theory before the study of complex functions. However, the characteristic function is of such importance in theoretical statistics that its complete exclusion was, on balance, considered unjustifiable. The purpose of Chapter 4 is to provide a small selection of largely simple examples to illustrate some of the ways in which the characteristic function is used in statistical theory. Of course, these examples could also be solved by other methods, but it is hoped that, as presented, they will give a beginner an insight into the analytical power and simplicity of the method. In this sense the chapter can be regarded as a self-contained unit. In general, the above classification may be regarded as developmental and, in fact, many of these exercises have been selectively used in such a planned sequence with mathematics undergraduates during different stages of the three-year degree. The general impression gathered on the basis of a few years' experience is that students almost invariably found the exercises interesting and helpful in clarifying their understanding of the subject, despite some difficulty in the manipulative mathematics involved. Perhaps this limited experience might be an indicator of a more general acceptability and usefulness of this collection. Answers and some hints on solutions have been provided to assist such students as may be working on their own, without the help of teachers. The exercises have been gleaned from many different sources. My major debt is to the standard research journals, and due acknowledgement is made to published papers from which the ideas for the exercises have been obtained. I am obliged to the publishers and

PREFACE

ix

authors of (1) An Introduction to Probability Theory and its Applications, Vol. I, by W. FELLER (John Wiley, New York, 1952) (2) Introduction to Probability and Random Variables, by G. P. WADSWORTH and J. G. BRYAN (McGraw-Hill, New York, 1960) (3) Probability Theory for Statistical Methods, by F. N. DAVID (Cambridge University Press, 1951) (4) Probability-An Intermediate Text-book, by M. T. L. BIZLEY (Cambridge University Press, 1957) (5) Introduction to Statistical Method, by B. C. BROOKES and W. F. L. DICK (William Heinemann, London, 1958) (6) The Advanced Theory of Statistics, Vol. I, by M. G. KENDALL and A. STUART (Charles Griffin, London, 1958) and to the publishers and trustees of (7) Introduction to Mathematical Probability, by J. V. USPENSKY (McGraw-Hill, New York, 1937) for kindly permitting me to include certain instructive problems from their works, Some exercises have also been obtained from past examination papers of the University of Leicester, and my thanks are due to the university authorities for permission to use them. Finally, a number of exercises have been specially constructed for this collection to illustrate some specific points in theory, but no claim to originality is made. I should like to thank Professor H. E. Daniels, who initially suggested my undertaking this compilation, and whose encouragement has been a major support over several years. My thanks are also due to Mr F. Downton, Reader in Mathematical Statistics at the University of Birmingham, for his careful reading of a part of the manuscript and for suggesting several interesting exercises for this collection. Apart from this, I am also obliged to him for his intangible help in the development of my ideas through a close professional association. To this I would also add my thanks to Professor R. L. Goodstein for the opportunity for experimentation with different groups of students which has given me a first-hand experience of the difficulties facing beginners in the field. Finally, I am much indebted to Messrs Charles Griffin & Co., and their consultant editor, Dr M. G. Kendall, for their helpful co-operation in the publication of this work. I should be grateful if users of this book would point out any errors or obscurities in the formulation of the exercises. LEICESTER 1966 N.A.R.

Preface to Second Edition The continuing demand for this book has occasioned a new edition. This is being done with a slight modification of the title which, I believe, now more clearly reflects the nature of the contents and also specifically distinguishes this book from my other collectionPractical Exercises in Probability and Statistics, Griffin, 1972. The new edition also contains a substantial supplement of 160 new exercises. Of these, the majority were initially formulated as questions set in various undergraduate examinations that were conducted over several years at the University of Leicester and, as such, they reflect a pattern in the development of statistical education which, it is hoped, will interest teachers elsewhere. Many of these exercises were gleaned from published papers, but it is regretted that it has not been possible to identify these sources explicitly. Nevertheless, I should like to record my indebtedness to all whose work has been used in the exercises. Furthermore, not all the exercises in the supplement are devoted to the traditionally defined areas of probability and statistical theory. In fact, in keeping with my individual approach to statistical education at the undergraduate stage, I have included in the supplement a few exercises in ancillary mathematics which experience here has shown to be helpful in stimulating student interest in statistics. Personal preference apart, there is a deeper pedagogical reason for this departure from established practice. There is now, more than ever before, a need to stress the underlying unity of the mathematical sciences at the undergraduate stage in order to avoid specious compartmentalisation. I believe this unity is intellectually necessary to provide balance in the structure of the honours degree in mathematics and to meet the academic challenge to the mathematical sciences that is currently being presented by the spectacular advances in computer technology. I am grateful to users of the earlier edition of the book who pointed out various misprints. These have now been corrected. LEICESTER N .A.R. 14th March, 1980

x

Contents EXERCISES Page

Chapter 1 : Probability and discrete random variables . .

1

Chapter 2 : Continuous random variables

27

Chapter 3 : Estimation, sampling distributions and inference; bivariate correlation and regression .

51

Chapter 4: Characteristic functions

124

List of abbreviations used in references to journals, books and examination papers

141

ANSWERS AND HINTS ON SOLUTIONS

Chapter 1

143

Chapter 2

177

Chapter 3

209

Chapter 4

285 SUPPLEMENT

Additional Exercises 1 to 160

309

Answers and Hints on Solutions

373

xi

1

Probability and discrete random variables

1 Two dice are rolled. Let A be the event that the sum of the points on the faces shown is odd, and B the event that there is at least one 3 shown. Describe A u B; A II B; A - B; and (A II B) u A.

2 For k = 1,2, 3,4, let Nk be the event that N has at least k aces in a bridge deal. Let Sk, Ek, w,. be the analogous events for S, E and Wrespectively. How many aces has W in the events (i) Wi; (ii) N2"S2; (iii) NiISl"B i ; (iv) W2 -W3 ; (v) Ni II Si" Ei II Wi; (vi) N3 II Wi; (vii) (N2 u S2) II E2? 3 Given the three arbitrary events A, Band C, find simpler expressions for (i) (A u B) II (A u B); (ii) (A u B) II (A u B) II (A II B) ; (iii) (A u B) II (B u C); (iv) (A u B) II (A u B) II (A u B). 4 Find the probability that no two of r ( < 10) digits, each chosen at random from 0, 1,2, ... ,9, are equal. 5 Three dice are rolled. What is the probability that there is at least one 6, assuming that the dice are unbiased? Also obtain this probability if the dice are loaded, the probabilities of getting a 6 with them being PI' P2 and P3 respectively. 6 A set of eight cards contains one joker. A and B are two players and A chooses five cards at random, B taking the remaining three cards. What is the probability that A has the joker? A now discards four cards and B two cards. If it is known that the joker has not been discarded, what is the probability that A has the joker? 7 Find the probability that two throws with three unbiased dice each will both show the same configuration if (i) the dice are distinguishable; and (ii) they are not distinguishable. 8 Consider three loaded dice. Die A is certain to show a 3. Die B is twice as likely to show a 2 as to show a 5, and will not show any other number. Die C is twice as likely to show a 4 as a 1, and will show no other number. The three dice are rolled once. Find the probability that (i) A shows a larger number than B; (ii) B shows a larger number than C; (iii) C shows a larger number than A; and (iv) A shows a larger number than B, and B shows a larger number than C. 9 The Security Council numbers eleven members with Great Britain, China, France, Russia and the United States as permanent members. If at a meeting

2

EXERCISES IN PROBABILITY AND STATISTICS

the members sit down at random, find the probability that the British and French delegates are next to each other but that the Russian and American delegates are not, if (i) the delegates sit in a row; .and (ii) the delegates sit round a circular table. 10 In a row of sixteen tiles on a roof, eleven chosen at random are blown off in a gale. Defining a "run" as a sequence of gaps or of tiles, find the probability that the row contains (i) 10 runs; (ii) 9 runs.

11 Two unbiased dice are rolled r times. Find the probability that each of the six results (1, 1), (2,2), ... , (6,6) appears at least once. 12

(i) Eight castles are placed at random on a chess-board. Prove that the probability that none can take another is (8 !)2/64(8)

(ii) If the castles are placed randomly on the chess-board but subject to the condition that none can take another, then prove that the probability that none stands on a square of the white diagonal is 8

L (-1)'/r!

,= 2

(iii) Hence, or otherwise, deduce that if the castles are placed completely at random on the chess-board, then the probability that none can take another and none stands on a square of the white diagonal is (8 (8) !)2. L... ~ (_ 1)'/r.' 64

,= 2

13 Find the probability that a hand of thirteen cards from a standard pack contains the ace and king of at least one suit. 14 For a random distribution of cards at bridge, calculate the probabilities for the following events: (i) A specified player (say S) has at least two aces, irrespective of the other players' hands. (ii) Some one player of the four has at least two aces, irrespective of the other three players' hands. (iii) S and no other player has at least two aces. (iv) Exactly one player of the four has at least two aces. 15 X chooses at random an integer from 1 to m (both inclusive). Without knowing X's choice, Y also chooses an integer from 1 to m. Find the probability that the two numbers chosen do not differ by more than n «m). 16 Suppose that each of n sticks is broken into one long and one short part. The 211 parts are then shuffled and arranged into n pairs from which new sticks are formed. Find the probability that (i) the parts will be joined into their original form; (ii) all long parts are paired with short parts; and (iii) at least one of the original sticks is formed.

PROBABILITY AND DISCRETE RANDOM VARIABLES

3

17 The digits 1, 2, ... , n (n < 10) are written down in a random order such that the n ! arrangements are all equally likely to give a number of n digits. If n == 7, find the probability that the number is divisible by (i) 2; (ii) 4; and (iii) 8. 18 The game of craps is played as follows. In a particular game one person rolls a pair of dice. He wins on the first throw if he scores 7 or 11, and loses on the first throw if he scores 2, 3 or 12. For any other score on the first throw, there are two different ways of continuing the game. In the first method, the player continues to roll the dice until he wins with a 7 or loses with the score obtained on the first throw. In the second method, the player also continues to roll the dice until he loses with a 7 or wins with the score obtained on the first throw. Find the total probability of the player winning a game under each of the two systems of play. 19 An urn contains n white and m black balls; a second urn contains N white and M black balls. A ball is randomly transferred from the first to the second urn, and then from the second to the first urn. If a ball is now selected randomly from the first urn, prove that the probability that it is white is n mN-nM - - +-,------:-;;----n+m (n+m)2(N +M + 1)"

20 Three newspapers A, Band C are published in a certain city, and a survey shows that for the adult population 20% read A, 16% B, and 14% C; 8% read both A and B, 5 % both A and C, 4 % both Band C, and 2 % read all three. If an adult is chosen at random, find the probability that (i) he reads none of these papers; (ii) he reads only one of these papers; and (iii) he reads at least A and B if it is known that he reads at least one paper. 21 For three independent events A, Band C, the probability for A to occur is a; the probability that A, Band C will not occur is b; and the probability that at least one of the three events will not occur is c. If p denotes the probability that C occurs but neither A nor B occur, prove that p satisfies the quadratic equation ap2+[ab-(I-a)(a+c-l)]p+b(l-a)(l-c)

= 0,

and hence deduce that (l-a)2+ab

c>

(I-a)

.

Further, show that the probability of the occurrence of Cis p/(p+ b), and that of B happening is (l-c)(p+b)/ap. In the particular case where a = 0'20, b = 0'42, and c = 0·985, verify that p is either 0·18 or 0·14. 22 Each packet of a detergent called "Sowite" contains a coupon bearing just one of the letters of the name of the product. A set of coupons making the word SOWITE can be exchanged for a free packet of the product. If each packet bought is equally likely to contain any of the six letters, calculate the probability, correct to five decimal places, that a housewife who buys ten packets can get a free packet without exchanging any duplicates with her friends.

4

EXERCISES IN PROBABILITY AND STATISTICS

Another firm introduces a rival detergent called "Osowite" and adopts the same coupon scheme, each packet being equally likely to contain any of the six different letters in the name. Show that the probability that the housewife who buys ten packets of "Osowite" will get a set of seven coupons spelling the name is approximately half the probability in the case of "Sowite". 23 An urn contains n balls, each of different colour, of which one is white. Two independent observers, each with probability of 0·1 of telling the truth, assert that a ball drawn at random from the urn is white. Prove that the probability that the ball is, in fact, white is (n - 1)/(n + 80). Also, show that if n < 20, this probability is less than the probability that at least one of the observers is telling the truth. Resolve this apparent paradox. 24 An unbiased coin is tossed 2n times. Find the probability, p(2n), that there will be an equal number of heads and tails. Prove that p(2n) is a decreasing function of n. 25 A sample of r individuals is taken from a population of n people with replacement. Find the probability U r that m given persons will all be included in the sample. Also, show that if n --+ 00 and r --+ 00 so that r/n --+ p, a constant, then Ur --+ (l-e- P)m. 26 A certain number n of identical balls is distributed among N compartments. Find the probability that one specified compartment contains r balls. Further, show that this probability, Pro satisfies the inequality relation

l)n-r( r)(r- 1I/

l)n-r( r)r-l

2 (n/N)" ( (n/N), ( - - 1-1-

27 A caterpillar x inches long starts to cross at right angles a one-way cycle track T yards wide at a speed of f feet per second. Cycles are passing this particular spot at random intervals but at an average rate of N per second. Assuming that the impress of a cycle tyre on the ground is t inches and that the caterpillar may only be touched to be considered hurt, find the probability that the caterpillar reaches the other side of the track safely. 28 If A and B are two events and the probability P(B) =F 1, prove that P(AIB) = P(A)-P(AB) I-P(B) ,

where B denotes the event complementary to B, and hence deduce that P(AB)

~

P(AHP(B)-l.

Also, show that P(A) > or < p(AIB) according as p(AIB) > or < P(A). 29 As a rather simplified example of weather forecasting at a seaside resort, suppose that the probability that the weather (fine or wet) of one day will be the same as that of the preceding day is a constant p. If, on the basis of past records, it is assumed that the probability of the first of August being fine is 0,

PROBABILITY AND DISCRETE RANDOM VARIABLES

5

a constant, determine On, the probability of it being fine on the following nth day. Show also that as n -+ 00, this probability tends to the limit!, irrespective of the values of P and O. Interpret this result. 30 There are n similarly biased dice such that the probability of obtaining a 6 with each one of them is the same and equal to p, (0 < P < 1). If all the dice are rolled once, show that Pn' the probability that an odd number of 6's is obtained, satisfies the difference equation Pn+(2p-l)Pn-l = p, and hence derive an explicit expression for Pn. 31 Each of n urns contains a white and b black balls which are indistinguishable except for their colour. One randomly selected ball is transferred from the first urn into the second, another one from the second to the third, and so on. Finally, a ball is drawn at random from the nth urn. Find the probability for it to be white when it is known that the first ball transferred was white. 32 Two urns contain, respectively, a white and b black, and b white and a black balls. A series of random drawings is made as follows: (i) Each time, only one ball is drawn and it is returned to the same urn from which it was obtained. (ii) If the ball drawn is white, then the next drawing is made from the first urn, the second urn being used in the contrary event of a black ball drawn. Assuming that the first ball is drawn from the first urn, find the probability that the 11th ball drawn is white, and obtain its limiting value as 11 -+ 00. 33 Of three urns, the first contains a proportion x of white balls and a proportion (1- x) of blue balls, the second contains blue and red balls in the proportions y and (1 - y) respectively, and the third contains red and white balls in the proportions z and (1- z) respectively, none of the given proportions being zero. A ball is drawn randomly from the first urn and then replaced. If it was white, a ball is drawn from the second urn and replaced, but if the first ball drawn was blue, a ball is drawn from the third urn and replaced. The process continues, a ball being drawn each time from the first, second or third urn according as the previous ball was red, white or blue respectively. Every ball is replaced before the next is drawn. If rn , Wn and bn denote respectively the probabilities at the outset that the 11th ball drawn will be red, white or blue, prove that as n -+ 00 these probabilities tend to fixed limits, and find the limiting values. 34 A number n of teams, all always of equal skill, compete annually for a cup which is to be awarded outright to the team winning three years in succession. Find the probability of team X winning the cup outright if it must be won every year and team Y won it for the first time last year. 35 Two players A and B agree to playa game of skill in which their respective probabilities of winning a game are Po and Pb' and the chance of a drawn game is Po. Find the probability of A winning at least (m+ 1) games out of (2m+ 1) played, and obtain its numerical value in the particular case of m = 2 and Po =!. Also, if A and B play a match to be decided as soon as either has won two games, find the probability of A winning the match in n games or less, and show that as n -+ 00, this probability tends to the limit p;(I- Po + 2Pb)/(1- POll.

6

EXERCISES IN PROBABILITY AND STATISTICS

Hence obtain the limiting probability that the match will end ,in a win for either player. If 11 = 10. PII = PI> = i, prove that the probability of the match being finished in ten or less games is 1981/2048.

*

In a game of skill a player has probabilities t, 152 and of scoring 0, 1 and 2 points respectively at each trial, the game terminating on the first realization of a zero score at a trial. Assuming that the trials are independent, prove that the probability of the player obtaining a total score of 11 points is

36

u"

=

I)"

3(3)" 4(-:3 . 13 4 + 39

and that the expectation of his total score is ¥. Also, suppose the rules are changed so that the game does not end on the first realization of a zero score at a trial but the trials continue indefinitely. In this case, show that the probability of the player obtaining a score of exactly 11 points at some stage of play is

37 An urn contains a white and b black balls. After a ball is drawn, it is to be returned to the urn if it is white; but if it is black, it is to be replaced by a white ball from another urn. Show that the probability of drawing a white ball after the foregoing operation has been repeated n times is

p"

b

[

1]".

= 1 - (a + b) 1 - a + b

38 Two urns contain, respectively, a white and b black, and c white and d black balls. One ball is taken from the first urn and transferred into the second, while simultaneously one ball taken from the second urn is transferred into the first. Find the probability, p", of drawing a white ball from the first urn after such an exchange has been repeated n times. Also obtain the limiting value of p" as n -+ 00. Two players A and B start playing a series of games with £a and £b respectively. The stake is £1 on a game, and no game can be drawn. If the probability of A winning any game is a constant, p, find the initial probability of his exhausting the funds of B or his own. Also, show that if the resources of B are unlimited, then (i) A is certain to be ruined if p = !; and (ii) A has an even chance of escaping ruin if p = 2 1 / a/(1 +21/a).

39

40 Two players A and B agree to contest a match consisting of a series of games, the match to be won by the player who first wins three games, with the proviso that if the players win two games each, the match is to continue until it is won by one player winning two games more than his opponent. The probability of A winning any given game is p and the games cannot be dra\vn. (i) Prove that f(p), the initial probability of A winning the match, is given by

PROBABILITY AND DISCRETE RANDOM VARIABLES

7

(ii) Show algebraically that

df > 0

for 0 <- p < 1,

(IX) dp

(fJ) f(P)+f(l-p)

= 1,

and interpret these facts in terms of the data of the problem. (iii) Show that the equation f(P) = p has five real roots, of which three are admissible values of p. Find these three roots and explain their significance. (iv) Calculate the expectation of the number of games to be played for A to win the match. 41 An internal telephone exchange in an office has m (> 2) single connections, one for each room in the establishment; and during a working day all the m connections are equally likely to contact the telephone operator for service. If Pn denotes the probability that in any sequence of n calls to the operator no room contacts the exchange more than three times consecutively, prove that

Pn

=

[(1-IX2)IX~-I_(1-IXI)IX~-1 V(IX I -I(2),

where IXI and IX2 are the roots of the quadratic equation in x m 2x 2 -III(m-l)x-(m-l)

= O.

Find the limiting values of Pn when (i) II -+ 00 and m is finite; (ii) m -+ 00 and II is finite; and interpret their significance. 42 In a lottery m tickets are drawn at a time out of 11 tickets numbered from 1 to 11 (m ~ 11). Find the expectation and variance of the random variable S denoting the sum of the numbers of the III tickets drawn. 43 At an office N letters are to be posted, one to each of N different addresses. and a capricious secretary decides to distribute the letters randomly amongst the N addressed envelopes. If all the N ! arrangements of the letters are equally likely, show that the expectation and the variance of the number of correct postings are both unity. In a similar situation, another slightly less capricious secretary decides to make independently a guess of the correct envelope for each letter to be posted, so that the NN choices are all equally probable. Prove that in this case the expectation and variance of the number of correct postings are 1 and (N -l)/N respectively. 44 A box contains 2" tickets among which (~) bear the number r (r = 0, 1.2•... , n). A group of III tickets is drawn at random from the box, and if the random variable X denotes the sum of the numbers on the tickets drawn. show that E(X)

11111

= 2;

( X)

var

=

mil

4

[1- (2" - 1)] (111-

1) .

45 By considering an example of a discrete sample space, show that the probability distribution of a random variable defined over it is a suitable reordering of the elements of the sample space and their associated probabilities.

EXERCISES IN PROBABILITY AND STATISTICS

8

A probability distribution is defined over the positive integral values from that P(r), the probability for the integer r, is proportional to (;)/(r + 1). Evaluate the proportionality factor and hence prove that the mean and variance of the distribution are

o to n such

respectively. Further, show that this probability distribution can be formally obtained from a finite sampling scheme with replacement in which at least one ball is selected randomly at a time from a total of (n + 1). 46 A population consists of all the positive integers, and the probability of obtaining the integer r from this population is P(r)

= k(1-

Or 1,

(r

= 1,2,3, ... ),

where 0 < 0 < 1.

Determine the constant k, and the mean and mode of this population. Show also that if 0 = 1_(Wln, where n is a positive integer, then the median of the distribution may be considered to be n +1. What is the variance of the distribution? 47 A certain mathematician always carries two match-boxes, which initially contain N match-sticks each, and every time he wants a light, he selects a box at random. Obtain Ur , the probability that when he finds a box empty for the first time the other box contains exactly r match-sticks, and verify that N

L

Ur

=

1.

r=O

Also, prove that the expectation and variance of the number of matches left in the box are p

and

[(2N+2)-(1+p)(2+p)] respectively,

where p == (2N + 1)uo-1.

48 There are n compartments into which identical balls are distributed one by one in such a way that each ball is equally likely to fall into anyone of the n compartments. This process is continued until every compartment has at least one ball. Prove that the probability that every compartment is occupied after t balls have been used is

mto (:)(-If(n-m)t/nt. Hence deduce the probability that exactly t balls are needed for filling all the n compartments, and that the expected number of balls required is n

n

L -. m

m=l

49 A box contains N varieties of objects, the number of objects of each variety being the same. These objects are sampled one at a time with replacement; and if Xr is a random variable which denotes the number of drawings

PROBABILITY AND DISCRETE RANDOM VARIABLES

9

necessary to produce any r different varieties in the sample, find the expectation and variance of X,. Also, for large N, show that

E(X,) '" N 10g[Nj(N -r+ 1)], and var

1 N 10g[Nj(N -r+ 1)].

N(r-l) (X ) , '" N

-r+

50 In the previous example, let the N varieties be identified by being numbered from 1 to N. If X denotes the largest number drawn in n drawings when random sampling with replacement is used, find the probability of

X = k. Hence obtain the mean and variance of X. Also, show that for large N and fixed n, E(X) "" nN n+l

and

nN 2 var(X) "" (n + 1)2(n + 2)'

51 Of a finite population of N animals in a region, Ware caught, marked and released. Members are then caught one by one until w (preassigned) marked animals are obtained, the total number of the animals in the sample being a random variable X. Show that the probability

P(X

=

11)

=

(;=:)(:= 11)/(~)'

for w

~ n~ N-

W +w,

and verify that this represents a proper probability distribution over the given range of variation of the random variable. Hence show that (X)

E

= w(N + 1) W+1

d an

E[X(X

+

1)]

= w(w+ l)(N + I)(N + 2) (W+1)(W+2)'

If a new random variable Y is defined by the relation

Y= X(W+ 1) 1, w then prove that

E(Y) = Nand

var(Y) = (N+l)(N-W)(W-w+1)jw(W+2).

52 A large number (N) of persons are subject to a blood test which can be administered in two alternative ways: (i) Each person is tested separately, so that N tests are required. (ii) The blood samples of k (a factor of N) persons are pooled and analysed together. If the test is negative,-this one test suffices for the k individuals. If the test is positive, each of the k persons must be tested separately, and in all (k+ 1) tests are required for k persons. Assuming that the test responses of the N persons are statistically independent and that the probability (1- p) for a test to be positive is the same for all individuals, find the probability that the test for a pooled sample for k persons will be positive.

\0

EXERCISES IN PROBABILITY AND STATISTICS

If S be the number of tests required for the N persons under plan (ii), prove that the mean and variance of S are

N[l+k(1-pk)]/k

and

Nkpk(l-pk) respectively;

and that the value of k which gives the minimum expected number of tests for the N persons satisfies the equation k 2 + l/pk logp = O.

53 In the simplest type of weather forecasting-rain or no rain in the next 24 hours-suppose the probability of raining is p (> 1), and that a forecaster scores a point if his forecast proves correct and zero otherwise. In making n independent forecasts of this type, a forecaster who has no genuine ability decides to allocate at random r days to a "rain" forecast and the rest to "no rain". Find the expectation of his total score (Sn) for the n days and show that this attains its maximum value for r = n. What is the variance of Sn? Devise an alternative scoring system which would ensure that under random forecasting the expectation and variance of Sn will be 0 and 1 respectively. 54 In the above example, suppose the forecaster has some ability to forecast the weather correctly. Accordingly, if the probability of his forecasting rain for any day is ~, and the conditional probability of raining on a day given that a "rain" forecast has been made is p, find the respective probabilities of the four possible outcomes for any day. Hence, assuming independence of weather conditions for n days, obtain a scoring system such that (i) E(Sn) = 0 for random forecasting; (ii) E(Sn) = n if the forecaster has perfect ability; and (iii) var(Sn) is a minimum when the marginal distribution of forecasting is the same as the marginal distribution of rain on any day, the two events being assumed independent. Verify that with this scoring system E(Sn)

= n~(p -

p)/pq,

(p+q = 1).

55 In a sequence of Bernoulli trials with probability of success (S) p and of failure (F) q, (p + q = 1), find the expected number of trials for the realization of the patterns (i) SSSFS; (ii) SF FS ; and (iii) SSSF. 56 For a sequence of n Bernoulli trials in which the probability of success (S) is p and that of failure (F) is q, (p+q = I), show that Yn, the probability that the pattern SF does not occur in the entire sequence, satisfies the difference equation Yn-Yn-l +pqYn-2

= 0 for n ~ 2.

Hence obtain explicit expressions for Yn when p :F q and when p = q. Also, prove that the expected number of trials necessary for the realization of r consecutive repetitions of the pattern SF is I-p'q' , pq'(1 -pq ) ~ ~4' -1).

57 Random drawings are made from an urn containing b black and w white balls. Each ball drawn is always replaced, and, in addition, c balls of the

II

PROBABILITY AND DISCRETE RANDOM VARIABLES

colour drawn are added to the urn. If P(lI, k) denotes the probability of drawing exactly k black balls in the first n drawings, show that P(n, k) satisfies the recurrence relation P(lI,k)=b

b+(k-l)c w+(n-k-l)c (1)·P(n-l,k-l)+b (1)·P(n-l,k), +w+n- c +w+n- c

where P(n, - I) may be taken to be zero. Hence obtain P(n, n) and P(n, 0). Also, for k < n, verify that p(/I,k)

n) b(b+c)(b+2c)" ·{b+(k-l)c}. w(w+c)(w+2c)" '{w+(n-k-l)c} . (b+w)(b+w+c)(b+w+2c)" '{b+w+(n-l)c}

= (k

satisfies the recurrence relation. Further, if p = b/(b + w), q = (1- p), and y P(Il, k) can be rewritten in the form

=

c/(b + w) (y > - 1), prove that

( k) = (") p(p+y)(p+2y)'" {p+(k-l)y}. q(q+y)(q+2y)'" {q+(/I-k-l)y} PII,

k'

1(1+y)(1+2Y)"'{1+(1I-1)y}

.

1

Finally, if now n -+ 00, p -+ 0, y -+ 0 so that np -+,-t, ny -+ -, then show that p the limiting form of P(n, k) is n(k)

= (,-tP+k-l)(~)AP(_1 k

l+p

)k,

l+p

(0 ~ k < (0),

and hence that, as p -+ 00, ll(k) tends to the Poisson distribution with mean ,-t. 58 Two players A and B alternately roll a pair of unbiased dice. A wins if on a throw he obtains exactly six points before B gets seven points, B winning in the opposite event. If A begins the game, prove that his probability of winning is 30/61, and that the expected number of trials needed for A's win is approximately 6. 59 Under a newly proposed motor insurance policy, the premium is £a in the first year. If no claim is made in the first year, the premium is £Aa in the second year, where (0 < ,-t < 1) and ,-t is fixed. If no claim is made in the first or second years, the premium is £,-t2(X in the third year; and, in general, if no claim is made in any of the first r years (r ~ 1), the premium is £,-tr(X in the (r+ l)th year. If in any year a claim is made, the premium in that year is unaffected, but the next year's premium reverts to £(X, and this year is then treated as if it were the first year of the insurance for the purpose of calculating further reductions. Assuming that the probability that no claim will arise in any year is constant and equal to q, prove that in the nth year (n ~ 2) of the policy, the probabilities that the premium paid is nn-l(X or nn- j-l(X, (1 ~j ~ n-l), are qn-l and (l_q)qn- j - l respectively. Hence calculate the expected amount of the premium payable in the nth year and show that if this mean must always exceed k(X (k > 0), then ,

A>

k+q-l kq .

12

EXERCISES IN PROBABILITY AND STATISTICS

60 A player rolls four unbiased dice, and if S is the random variable denoting the sum of points obtained in a single throw of the dice, prove that the probability P(S = n) is the coefficient of (}n-4 in the expansion of (1- (}6)4/6 4(1_ (})4

for all n in the range (4 ~ n ~ 24). Hence, or otherwise, deduce that (i) P(S = 18) = 5/81; and (ii) E(S) = 14. 61 The probability of obtaining a 6 with a biased die is p, where (0 < p < 1). Three players A, Band C roll this die in order, A starting. The first one to throw a 6 wins. Find the probability of winning for A, Band C. If X is a random variable which takes the value r if the game finishes at the rth throw, determine the probability-generating function of X and hence, or otherwise, evaluate E(X) and var(X). 62 The six faces of an ordinary cubical die are numbered from 1 to 6. If two such unbiased dice are rolled once, find the probability distribution of the random variable X denoting the sum of points obtained. Also, find an appropriate numbering of the twelve faces of two unbiased dice which would ensure that the probability P(X = r) is the same for all r in the range (1 ~ r ~ 12), and show that for such a pair of dice the probabilitygenerating function of X is G((})

= (}(1- (}12)/12(1- ()).

63 From an urn containing (2n+ 1) tickets numbered serially, three tickets are drawn at random without replacement. Prove that the probability that the numbers on them are in arithmetical progression is 3n/(4n 2 -1). Further, by considering the sample space corresponding to the possible realizations of the arithmetical progressions, show that the common difference of the arithmetical progression can be regarded as a discrete random variable X with the probability distribution defined by P(X

=

r)

=

(2n+ 1)-2r

n

2

'

for r = 1,2,3, ... , n.

Hence deduce that the probability-generating function of X is G( (})

=

4(} . [1 _ (2n + 1- (}n)( 1 + (}~l. n(l- (})2 4n

J

64 In a game of skill a player has probability p of winning a point and probability q, (p+q = 1), of losing a point at each attempt. If the trials are independent, find the probability distribution of the random variable X giving the player's total score in n trials. Hence, or otherwise, obtain the mean and variance of X. Also, show that if a new random variable Y is defined by the relation Y= (X+n)/2,

then Y has the Bernoulli distribution with probability of success p in each of n trials. 65 At each independent trial in a game, a player has probability p of winning a point, probability q of losing a point, and probability r for no loss or gain,

PROBABILITY AND DISCRETE RANDOM VARIABLES

13

where (p+q+r = 1). Find the probability-generating function of the random variable X giving the player's total score in n trials, and hence deduce the mean and variance of X. Also, show that the probability for a total score of m (~n) points in n trials is P(X

= m) =

L }=m

(n+m)/2(

n . )(m+n-J) . . . {p/rY (q/ry-m. r". m+n-J J

66 A particular constituency has a total of (L + C + F) registered voters of which L are by conviction Labour supporters, C are Conservative and Fare the floating voters. The probability that a Labour supporter will vote for his party in a by-election is Pl' and the probability is P2 that a Conservative will exercise his vote. The probabilities are P3 and P4 for a flQating voter to vote either Labour or Conservative respectively. Prove that in a straight fight between a Labour and a Conservative candidate, the probability of a tie is given by the coefficient of ()c+ F in the function [(1- Pl)+ Pl()]L [(1- P2)9+ P2f [(1- P3 - P4)() + P3()2 + P4]F. Show also that for the random variable N representing the total number of votes cast in the by-election, the mean and variance are: E(N) = Lpl+CP2+F(P3+P4);

var(N) = LPl(1-Pl)+CP2(1-P2)+F(P3+P4)(1-P3-P4)' 67 In an industrial process individual items are in continuous production, and the probability of finding a defective item on inspection is a constant, p. To ensure a reasonable standard of the outgoing product, 100 per cent inspection is carried out until a sequence of r (preassigned) non-defectives is found after the detection of a defective. After this, 100 per cent inspection is discontinued and only a given fractionf(O
(r+ l)p* -rp* = 1, where p = p* for p = p*.

14

EXERCISES IN PROBABILITY AN!) STATISTIC'S

68 In a sequence of n Bernoulli trials with a constant probability of success p, r successes were observed. If Pn(r) denotes ~he probability of such a sequence, show that Pn(r) satisfies the recurrence relatIOn P(r)=E.(n-r+1).p.(r-l) II q r

forr~

1 (p#q== I-p),

which reduces to the form Pn(r)+Pn(r-l)

= 2Pn + l (r) for

p

= q.

Using the factorial moments of the random variable r, or otherwise, prove that Y2, the coefficient of kurtosis of the distribution of r, is zero whenever p or q = 1(1 + 1/)3).

69 For the binomial distribution defined by r successes in n trials with constant probability of success p, obtain the moment-generating function of r, and hence prove that the mean and variance of rare np and I1pq, (p + q = I), respectively. Also, show that the first incomplete moment of r defined by

" (11) 111(r) = r~r r prqn-r.(r_np) has the value (:) pr q,,-r+ I. r,

where r is a positive integer in the range (0 verify that

~ r ~ n);

and hence independently

E(r) = np.

70 If K(t) be the cumulant-generating function about the origin of the probability distribution resulting from 11 Bernoulli trials, show that .

where z = log(P/q) and p == 1- q is the parameter of the distribution. By expanding the right-hand side in powers of t by Taylor's theorem, prove that K" the rth cumulant of the distribution, is given by dr-1p Kr = n . dz r - I . Hence, or otherwise, obtain the recurrence relation Kr

=

dKr-1

pq.~,

and, in particular, prove that KS =

npq(q-p)(I-12pq).

71 For the discrete Poisson distribution with parameter A., obtain the moment-generating function about the mean of the distribution, and hence

PROBABILITY AND DISCRETE RANDOM VARIABLES

15

verify that the fourth, fifth and sixth central moments are Jl.4 = 3A.2+..t; Jl.s = 1O..t2 +..t; and Jl.6 = 15..t 3 +25..t 2 +..t. 72 A discrete random variable X is defined over the values 0, ± 1, ±2, + 3, ... , it being equally likely to take positive or negative integral values. If the probability that X takes the value r > 0 is pr, (0 < p < 1), prove that the moment-generating function of X about the origin is M(t)

=

1 + 24>(1 + 4>)(24) - 1) sinh 2 t/2 1-44>(1 +4» sinh2 t/2 ,where 4>

=P(X =F 0),

and hence deduce that the variance of X cannot exceed 6. 73 A discrete random variable X, taking all the non-negative integral values, has the probability-generating function 00

G(O)

=

L

9' Pro

r=O

Find the probability-generating function of the random variable Y when (i) Y = X + <x, where <X is a real constant; (ii) Y = {3X, where {3 is a real constant; (iii) Y = <X + {3X, where <X and (3 are real constants; (iv) P(Y = n) = P(X ~ n), where n is a positive integer; (v) P(Y = n) = P(X > n), where n is a positive integer; (vi) P( Y = n) = P(X < n), where n is a positive integer; (vii) P(Y = n) = P(X ~ n). where n is a positive integer; (viii) P( Y = 11) = P(X > 11 + m). where 11 and m are positive integers; (ix) P(Y = n) = P(X ~ n+m), where nand m are positive integers; (x) P(Y = n) = P(X ~ n-m), where nand m are positive integers; (xi) P( Y = n) = P(X > n - m), where nand m are positive integers; (xii) P(Y = n) = P(X ~ n-m), where nand m are positive integers. 74 Colonies of an insect population occur at random in a particular region, and it is observed that the distribution of the number of colonies can be taken to be of the Poisson form with parameter ..t. The number of insects in anyone colony is again a random variable which is independently distributed in a form having a moment-generating function 4>(t). Show that the momentgenerating function of the total number of insects observed is exp[A{4>(t)-l} ]. Hence, or otherwise, prove that if the probability distribution of colony size is logarithmic with parameter p, then the total number of insects observed has a negative binomial distribution with parameters (1- p) and - ..t/log(1- pl. 75 There are t distinct types of cards in a pack of N (fixed) cards, the cards of the same type being indistinguishable. For two packs A and B, the distribution is such that they have aj and bj cards respectively of the ith type, where (0 ~ aj, bj ~ N for all i = 1,2, ... , t). For a given ordering of A and a random distribution of B, a "match" is observed if cards of the same type occur in the same position in the two packs. If v,: denotes the number of ways in which at least r "matches" can occur with one distribution of B. and J¥,. the corresponding number for exactly r

16

EXERCISES IN PROBABILITY AND STATISTICS

matches, prove that

v. == (N-s)! H.!n b ! l

1= 1

where H, is the coefficient of zS in the product

i\ L~o

al! bl ! zk/k! (al-k)! (bl-k)!],

m i denotes the smaller of ai' bl for all i. Hence derive the probability of obtaining exactly r matches, and show that the jth factorial moment of the probability distribution so defined is

and

Use this result to obtain explicitly the mean and variance of the distribution. 76 A random variable Y has a distribution with mean By expressing Y in the form

Il and variance

(12.

[ Y-Il]

Y = Il 1+-Il- ,

show that

if the variations in Yare sufficiently small as compared with its mean value. If Y has a Poisson distribution with a large mean, prove that as a first approximation

var(jY) =

1.

n

For n observations Xl> X2," ., X n , the arithmetic, geometric and harmonic means, denoted by a, g and h respectively, are defined by

1

I

n

1

L-' ni=1xi

- = -

h

If the deviations of the

Xi

from a are small compared with a, and n

ns 2 =

L

(Xi- a)2,

i=1

prove that, approximately, g = a

and hence a-2g+h = O.

[1 -2S;2] ;

h= a

[1-:: l

17

PROBABILITY AND DISCRETE RANDOM VARIABLES

78 A discrete random variable X takes the values -1, 0 and 1 with probb'lities tp, (1- p) and tp respectively. If Xl> X2' ... ,Xn are n independent ~e~lizations of X, find the moment-generating functions of the random variables Yand Z defined by n

Y=

L

n

XI

and

=

Z

1= 1

1= 1

and obtain their limiting values when np and p'" O. Also, prove that P(Y = 0)

=

L xf,

,to G) C;2)(A/2Y.

-t

P(Z

m, a positive number, as

II - t 00

= 0), where A. == p/(I- p),

and r takes even values ~ n. 79 A discrete random variable X takes the values ± 2, ± 1 and 0 with probabilities tp2, p(I- p) and (1- p)2 respectively. Derive the momentgenerating function of Sn, the sum of n independent observations of X. If np = m, where n is sufficiently large so that terms of order n - 2 can be considered negligible, show that in this case the moment-generating function of Sn is approximately

M(t) =

exp{4m(1 +m/n) sinh 2 t/2},

and hence, or otherwise, deduce that var(Sn) = 2m (1 +~). 80 A discrete random variable X has a Poisson distribution with parameter

A., where A. is known to be large. Given a small positive number b, prove that for any realization of X

S~I <Xl

Jx+b

=

[

JA.+b 1+

(-I)'-1(2s-2)! (X-A.)'] 22 • IS !(s-I)!' l+b .

By using this expansion, deduce that

[ r::-:J:.]

E yx+b

(1-

1

(24b-7)

= yA.+b- 8JX+ 128A.3/2

(240b 2 -260b+75) 0(1-7/2) 1024A. 5 / 2 + A ,

(32b 2 -52b+ 17)] (A.--3). [ ~] _ ![1 + (3-8b) 8A. + 32-1. +0 ,

var y X+u - 4

and hence verify that the variance of 81

2

J X +i is ! + O(A. - 2).

Two children, A and B, divide at random a bar of chocolate containing ~ 1) sections, so that A has r sections and B (4n-r), with neither having no chocolate. If A is known to have more than B, show that the expected number of sections in A's possession is 3n, and the variance of his number of sections is n(n -1)/3. Show also that if on three successive occasions A has more than three times as much chocolate as B, it is reasonable to doubt that the bar was broken at random.

4n (n

18

EXERCISES IN PROBABILITY AND STATISTICS

82 A lady declares that by tasting a c,;!p o~ tea made with milk she can discriminate whether the milk or the tea mfuslOn was first added to the cup. In an experiment designed to test her claim, eight cups of tea are made, four by each method, and are presented to ~er for judgme.n~ in ~ random order. Calculate the probability of the cups bemg correctly dIvIded mto two groups of four if such a division is entirely random. If the experiment is repeated ten times 'and a "success" in an individual experiment is scored if the lady corre~tly divides the cups, show that a score of two or more "successes" in ten trials provides evidence, significant at the 1 per cent level, for her claim. The lady considers this definition of "success" too stringent, so an alternative experiment is devised. In this experiment six cups of each kind are presented to her in random order, and she is now regarded as scoring a "success" if she correctly distinguishes five or more of each kind. Calculate the probability of such a "success", and also the probability of three or more successes in ten independent experiments, assuming her claim to be false. Explain the difference, if any, between the stringency of these two procedures for testing the lady's claim. 83

In an unlimited sequence of Bernoulli trials with probability p of success (S) and q of failure (F), two patterns are defined by the outcomes of three consecutive trials as (i) SFF and (ii) FSF. Prove that the expected number of trials needed for r (~ 1) consecutive realizations of the pattern (i) is

and that for the pattern (ii) is

84

A printing machine is capable of printing any ofn characters IX" 1X 2, ••• , IXn and is operated by an electrical impulse, each character being, in theory, produced by a different impulse. In fact, the machine has probability p, independent of any previous behaviour, of printing the character corresponding to the impulse it receives, where (0 < p < 1). If it prints the wrong character, all such characters are, independently, equally likely to be printed. One of the n impulses, chosen at random, was fed into the machine twice and on both occasions the character lXi was printed. Show that the probability that the impulse chosen was the one designed to produce the character lXi is (11- 1)p2/( I - 2p + np2). If the machine had printed lXi on the first occasion and IXj (j 1= i) on the second, determine now the probability that the impulse chosen was that designed to produce lXi' Does it make any difference to this probability if it is merely known that the second character printed was not lXi?

85 Electronic tubes in continuous production at a factory are packed in batches of N (large) units for despatch to customers. It is known from past experience of the manufacturing process that the probability of a random batch containing k defective tubes is Pk> where (0 ~ k ~ m). From a randomly selected batch, a sample of n ( < N) tubes is tested and of these r ~ m are found to be defective. Find the probability that the chosen batch had, in fact, k ~ r defective tubes.

PROBABILITY AND DISCRETE RANDOM VARIABLES

19

If according to a customer's specification of acceptable quality, batches with P(~r) defectives are considered to be definitely bad, prove that the total probability of the selected batch being in fact a bad one is

l_:t:Pk(~=;)/(~)

(N-n) (N)' L Pj. I/'. }-r '/ ' } m

j=,

DiscusS briefly how this result could be used for quality control of the production process. 86 In a big department store there are n separate service counters, and a customer entering the store is equally likely to go to anyone of the counters for service. Find the probabilities that a particular counter X will be the first to have a run of r consecutive customers before any other counter receives s successive customers when it is known that just preceding the count only the last customer went to (i) counter X; and (ii) another counter Y. If /I is finite and r = s large, verify that these two probabilities are both asymptotically equal to l/n, and explain the significance of this limiting value. 87 Housewives shopping at a grocery store are equally likely to ask for anyone of n standard brands of detergents, but due to an advertising campaign for a new detergent X, it is expected that the probability for a housewife to ask for X is a constant P (0 < P < 1), the demand for the standard brands remaining equi-probable. Assuming that all housewives coming to the store buy one and only one brand of the (n + 1) detergents, find the probability that on any day the first r consecutive housewives will purchase X before a run of s shoppers buys packets of anyone of the standard brands. Hence determine the limiting values of this probability for (i) n large and r, s finite; and (ii) /I finite and r. s large. 88 An infinite population consists of two kinds of biological particles which may be likened to white (W) and black (B) beads. In their initial state the beads are found linked together in chains of four in all the five possible distinct colour combinations. The probability that a chain has a white bead is a constant p, and that for a black bead is q (== 1 - pl. Each chain temporarily splits in the middle to form two doublets, and these doublets then randomly recombine to form new chains of four beads each. This dual process of splitting up and recombination continues indefinitely without any other kind of change in the original population of beads. Determine the proportions of the three colour combinations of the doublets (WW), (WB) and (BB) immediately after the first stage of splitting. If now only proportions xo, 2yo and Zo (xo + 2yo + Zo = 1) of the doublets are considered further in the process, because of certain accidental destruction, find X n, 2Yn and Zn' the corresponding proportions after the nth successive stage of splitting, and hence obtain their limiting values as n -+ 00. 89 In an infinite population of two kinds of biological particles, which may be symbolised by white (W) and black (B) balls, the balls exist singly and in pairs of the three colour combinations (WW), (WB) and (BB). Initially. the single (W) and (B) balls are in the proportions Po and qo (Po + qo = 1), whilst the paired balls (WW), (WB), (BB) have the proportions roo 2so, to respectively (/'0+2so+to = 1).

20

EXERCISES IN PROBABILITY AND STATISTICS

All the paired balls undergo a simultaneous disjunction, after which they are equally likely either to combine randomly with the original single balls only to form new pairs of the three types, or to remain single. This second population of paired and single balls repeats the process of separation and combination to form a third population, and this continues indefinitely. Find the proportions of the paired and single balls after the nth stage of the process, and hence deduce their limiting values for n -+ 00. On empirical considerations, the demand for a particular brand of breakfast cereal at a retail store may be considered to be a random variable X having a Poisson distribution with mean A. On the first of each month the retailer arranges to have a stock of n packets of the cereal. Find the expected number of lost sales per month because of inadequate stock, and hence deduce that, as a fraction of the average monthly demand, the expected proportion of lost sales is 90

7t"

= (l-i)[I-F(n-l)]+p(n-l),

where the probability P(X = r) := per) for all r ~ 0, and F(r) is the distribution function of X. Also, prove that the expectation of the monthly proportion of lost sales is n <Xl k! p" '" I-I k~O;:r<' [1-F(n+k)].

Explain the difference between this expected value and

7t".

91 In an infinite population of three kinds of biological particles, which may be identified with white (W), black (B), and red (R) balls, the balls initially exist only in pairs of unlike colours, the proportions for the three possible pairs (WB), (WR), and (BR) being Xo, Yo and Zo respectively (xo+ Yo+zo = 1). The probabilities are p, q and r (0 < p, q, r < 1) respectively that the (W), (B), and (R) balls from another infinite population can come into contact with the corresponding (BR), (WR) and (WB) pairs of the first population. When this happens, the paired balls separate and then these balls are equally likely to combine with balls of the third colour to re-form pairs of unlike colours. On the other hand, if a pair does not contact balls of an unlike colour, it remains unchanged. This three-fold process of association, separation and recombination continues indefinitely. If x,,, y" and z" (x" + y,,+z/; = 1) are the proportions of the (WB), (WR), and (BR) pairs after the nth stage of the process, prove that

x"

= pq/(pq+pr+qr)+BIOIA~

YII

=

+B 2 0 2 Ai

prj(pq + pr+ qr)+ BIA~ + B 2 A.'i;

where (i) BI and B2 are arbitrary constants specified by Xo and Yo; (ii) 01 , O2 are the roots of the quadratic equation

(r-p)02-2(q-r)0+(p-q)

=

0 (p =I- r =I- q); and

(iii) AI' A2 sa tisfy the relations AI +..1.2 = 2-(p+q+r) ..1.1..1.2 = l-(p+q+r)+i(pq+pr+qr).

PROBABILITY AND DISCRETE RANDOM VARIABLES

21

Verify that )'1 and .1.2 are both numerically < 1 and hence, or otherwise, deduce the limiting values of Xn and Yn as n --+ 00. 92 In an infinite population white (W), black (B), red (R), and green (G) balls only occur in the possible unlike pairs (WB), (WR), (WG), (BR), (BG), and (RG) in the proportions Xo, Yo, Zo, Uo, Vo, and Wo respectively, (xo + Yo + Zo + Uo + +v +Wo = 1). The pairs then separa~e, and the individual balls of each pair are °equally likely to form pairs by associating with unlike coloured balls from another separate population. This population of new pairs undergoes the above process again to generate another population of pairs, and in this way the process continues indefinitely. If X n , Yn' Zn' Un' Vn and Wn are the proportions of the six types of pairs after the nth stage of the process, obtain their values explicitly for any n ~ 1, and then deduce that as n --+ 00 these proportions all tend to the limit !. 93 In an infinite population of spherical and cylindrical beads, the probabilities for a spherical bead to be white (A) or black (a) are PI and (1- pd respectively, whilst the corresponding probabilities for the cylindrical beads to be white (B) or black (b) are P2 and (1- P2). The beads are found in linked chains of four, two of each shape irrespective of their colour. Find the proportions for the nine distinct bead combinations of the chains. The chains undergo a temporary disjunction to form separate pairs, each pair consisting of one cylindrical and one spherical bead. Assuming that all such pairs which can be formed from anyone chain are equally likely, obtain the proportions of the four distinct types of pairs in the population. Also, if the pairs then randomly recombine to form complete chains, each of four beads, deduce the proportions of the types of chains at this stage. Supposing that this process of disjunction and recombination continues indefinitely, and the initial proportions in the population of the pairs (AB), (Ab), (aB), and (ab) are X o, Yo, Zo, and wo, (xo + Yo + Zo + Wo = 1), respectively, find the values of these proportions at the nth stage of the process. Hence calculate their limiting values as n --+ 00. 94 An infinite population initially consists of white (W) and black (B) balls in proportions Po and qo, (Po + qo = 1) respectively. These balls randomly combine to form the three possible kinds of pairs irrespective of the ordering of the balls; but immediately on their formation, a proportion A. (0 < A. < 1) of the (WW) pairs and a proportion Il (0 < Il < 1) of the (BB) pairs are destroyed. In the resulting population the pairs then separate, and the individual balls randomly recombine to form the (WW), (WB), and (BB) pairs. Of these, a proportion A. of the (WW) pairs and a proportion Il of the (BB) pairs are again destroyed, and then all the other pairs separate to form a second population of (W) and (B) balls. This population undergoes the same stages as the first one, and the process continues indefinitely. (i) Prove that the proportion of black balls in the successi ve populations remains unchanged if qo = A./(A.+ Il). (ii) If A. = 0 and Il = 1, prove that the proportion of black balls in the successive populations --+ o. (iii) Finally, suppose A = Il = 0, but that the (WB) pairs are completely eliminated immediately on formation at each stage. Show that in this case the limiting value of the proportion of black balls is unity if qo > 1, zero if qo < 1, and t if qo = 1.

22

EXERCISES IN PROBABILITY AND STATISTICS

95 In an infinite population, balls of k different colours, which may be denoted by Ai (i = 1,2, ... , k), occur in unordered pairs Aj Aj (i"# J) in proportions x!J), where 1: j<jx\J) = 1. The pairs Aj Aj then separate, and the individual balls Aj and A j are equally likely to form unlike pairs with balls frolll another infinite population. This new population of pairs undergoes the sallie process to generate a third population, and this continues indefinitely. If x!j) are the proportions of the pairs A j A j after the nth stage of the process obtain their values explicitly, and hence show that as n ~ 00 these proportion~ all tend to the limiting value 2/k(k-1). 96 In an infinite population k kinds of biological particles SI' S2," ., Sk exist in unlike but unordered pairs Sj Sj (I' "# j) in initial proportions xlJ) respectively such that 1: j<j xlJ) = 1. A pair Sj Sj cannot be fertilized by its own kind, but it is equally likely to mate with pairs Sj S, (l "# i, I "# j) and S, S, (l "# t "# i "# j) from another infinite population of similar pairs. The mating of Sj Sj and Sj S, produces progeny Sj S, and Sj S, in equal proportions, whilst the mating of Sj Sj and S, S, produces progeny of the types Sj S" Sj S" S1'S" Sj S" again in equal proportions. The results of all possible random matings lead to the first generation of paired particles Sj Sj in new proportions xlJl (1: j<jxlJ) = 1). This generation is again fertilized in the manner described to give the second generation, and so the generative process continues indefinitely. If xli) are the proportions of Sj Sj in the nth generation, find their values explicitly, and hence prove that as n ~ 00, these proportions all tend to the limiting value 2/k(k-1). Hence, or otherwise, verify that for k = 4, (i "# j "# 1"# t).

97 In the simplest type of weather forecasting-"rain" or "no rain" in the next 24 hours-suppose the probability of raining is p (>t), and that a forecaster scores a point if his forecast proves correct and zero otherwise. In making n independent forecasts of this type, a forecaster, who has no genuine ability, predicts "rain" with probability A. and "no rain" with probability (1- A.). Prove that the probability of the forecast being correct for anyone day is [1-p+(2p-1)A.].

Hence derive the expectation of the total score (Sn) of the forecaster for the n days, and show that this attains its maximum value for A. = 1. Also, prove that var(Sn) = n[p-(2p-1)A.][1-p+(2p-1)A.], and thereby deduce that, for fixed n, this variance is a maximum for A.

=

t.

98 In testing the quality of large batches of a mass-produced article, randomly selected items of the product are examined individually from each batch, the probability of finding a defective item in a batch being a constant p, (0 < p < 1), and that of a non-defective q (= 1- pl. Each batch is inspected separately, and if an inspected item is found to be of acceptable quality, a score of one point is made, while for each defective item encountered the penalty is to subtract a points from the total score, a being a known positive integer. Initially the score for a batch is zero, and as soon as the cumulative score is M, the batch is accepted without further inspection. On the other hand, when the total score is ~ - M for the first time the batch is rejected immediately.

PROBABILITY AND DISCRETE RANDOM VARIABLES

23

If for this inspection scheme, u(x) is the probability that a batch will be ce~ted when the cumulative score is x, where x is taken to be in the equiva-

~;nt positive range (0 < x ~ 2M), prove that u(x) satisfies the difference equation v(x+a+ 1)-v(x+a) = -AV(X), where u(x)

==

and

q-Xv(x)

), == pqa.

Further, assuming that a solution of v(x) is a power series of the form 00

v(x) =

I: ArVr(X), r=O

determine explicitly the series expression for u(x). 99 In a game of skill, a trial can result in anyone of m mutually exclusive results R l' R 2 , ••• , Rm· The probability for Ry to happen is proportional to p", where p is a constant (0 < p < t). The stake for each trial is one shilling, and if Ry occurs at a trial the player receives AV shillings, where (0 < A< 1). If S. denotes the amount received by a player after n independent trials, prove that the probability-generating function of S. is

_ [(1- p)8'<{ 1- (POl)m}]. G(O) -

(1- pIH)(1- pO")

.

Evaluate the cumulant-generating function of S. and hence, or otherwise, show that E(S.) = nA(I- exm + {3); var(S.) = nA 2[{3(1 + {3) - m2ex(1 + ex)], where a==pll1/(1-pll1) and (3==p/(1-p). Also, prove that, irrespective of the values of m and n, the player cannot expect to be on the winning side if A < (1- p). 100 In the sex distribution of families each having a fixed number n of children, it is known that the probability of a child being a boy is a constant p. Assuming that the sex of a child is independent of the sex distribution of the other children in the family, find the probabilities of the following sex distributions in a randomly selected family with n children: (i) All children are of the same sex. (ii) The first k ( ~ n) children are boys and the rest are girls. (iii) The first k ( ~ n) children are of the same sex, and the rest of the other sex. (iv) Any.k (~n) children are boys and the rest girls. (v) There are at least k (~n) boys in the family. (vi) The first s children are boys, and in all there are k (n ~ k ~ s) boys in the family. (vii) The first s children are boys, and there are at least k (n ~ k ~ s) boys in the family. 101 In the study of a developing epidemic, the probability of finding an affected person does not remain constant but depends upon the observed number of the affected. Suppose, then, that the probability of recording another affected person after x have been detected is Px, for x ~ 0, and (0 < Px < 1; Px+qx = 1).

24

EXERCISES IN PROBABILITY AND STATISTICS

For such a model, suppose P(n, x) is the probability of obtaining x affected in a total of n persons. Determine the difference equation for P(n, x) and indicate its appropriate boundary conditions. Further, by considering the transformation x-1 P(n, x) = f(n, x). Ph

n

1=0

show thatf(n, x) satisfies the simpler difference equation

f(n+1,x+1) =f(n, ~)+qx+d(n,x+1). Hence prove that

P(n,x) =

x-1

x

x

1=0

,"'0

j'l<,

n PI' L q:. n (q,-qr

1•

As special cases, deduce the following limiting forms: (i) If p, = A. ,In for all r, then x-1

lim P(n, x) = n""oo

(ii) If P,

=

x

x

n A.i' L e- J""n (Aj-A,)-1. A,.

1=0

,=0

P, a constant, then for finite n

P(n, x) =

(:)px(1- pt- x.

102 If P(n, x) is the probability of exactly x successes in n trials and Px is the probability of success after x previous successes, prove that

P(n+ 1, x+ 1) = PX' P(n, X)+qx+l' P(n, x+ 1), where Px+qx = 1, for (0 ~ x ~ n), and with suitable boundary conditions. Obtain an explicit expression for P(n, x). Further, assuming that Px is a linear function of x such that

Px = p+cx, where P and c are constants, show that the probability-generating function of x is G(t)

= the coefficient of enln ! in the expansion of eO[t +(1- t)eCor P/c•

Hence, or otherwise, derive explicitly the first four factorial moments of x. If c > 0 and as n -+ 00, np -+ ex, nc -+ p, where ex and p are constants, prove that the limiting distribution of x is negative binomial. Determine the cumulant-generating function of this limiting distribution, and hence obtain the first four cumulants. Finally, verify that for p = 0, the negative binomial reduces to a Poisson distribution with mean ex. 103 The probability that a certain type of tree has n flowers is given by = 0, 1,2, .... Each flower has probability j of being pollinated and hence producing fruit, independently from the other flowers on the tree. Each fruit has probability! of being eaten by birds before ripening. Show that the probability of any flower producing a ripe fruit is 1.

(1- p)pn, n

PROBABILITY AND DISCRETE RANDOM VARIABLES

25

A particular tree bears r ripe fruit. Show that the probability that it had initially n flowers is

(~) pn-r(2_py+lj2n+l. Show also that if an orchard of k independent trees produces no ripe fruit, the probability that the total number of flowers was initially 11 is

b~~ ~ p' (2-ptI2"". 104 Contagious distributions are applicable to situations where individuals or items are supposed initially dispersed in randomly scattered groups-such as egg masses in the case of insects or clumps in the case of bacteria-which are subject to chance fluctuation in size. It may be supposed that there occurs subsequently some spatial dispersion from the initial groups, or reduction due to attacks of predators. Such phenomena are usually poorly represented by the simple Poisson model, and a more suitable model is provided by a class of three-parameter contagious distributions defined by the probabilitygenerating function G(O)

=

exp{).I[f(O)-I]},

where 00

f(O) == r(n+l)

L A~(O-INr(I1+k+l)_ k=O

and . 1. 1,..1. 2 ,11 are parameters such that ..1. 1 ,..1. 2 > 0 and 11 ~ o. Verify that G(O) represents a probability-generating function of a discrete random variable X for integral values of X ~ o. Also, prove that the probabilities P(X = r) = ar satisfy the recurrence relations

_ Al ~ 1 (HI) ar+1 - --1· L... -k,.f (O).ar-k

r+

k=O

for r > 1,

•

j<')(O) being the sth derivative of f(O) at 0 = o. Finally, determine the cumulant-generating function of X and hence, or otherwise, obtain the first four cumulants of X. Hence show that, in terms of the first three cumulants K I- K 2 and "3' 11

= 6(K~ +"1"2 -K 1"3 -Ki)f(KI+ 2"1"3 -3,,~).

105 From a standard pack of fifty-two playing cards, a hand of thirteen cards is dealt out at random to a bridge player. Find the probability that, of the four aces in the pack, the player has (i) only one ace; (ii) at least one ace; (iv) only the ace of hearts. (iii) the ace of hearts; (v) Suppose the player asserts that he has an ace. Find the probability that the player has (a) another ace; (b) only another ace. How are these probabilities altered if, alternatively, the player initially declares that he has the ace of hearts?

26

EXERCISES IN PROBABILITY AND STATISTICS

106 If z is an integral variable taking all values from 0 to n, prove that in the usual notation of factorial powers (z+k)(r)-(z+k-l)(r) = r(z+k-l)(r-I), where k and r are fixed positive integers. Hence deduce that

t (z+k-1)(r-1) = ~[(n+k)(r)-(k-l)(r)],

z=o

r

and. in particular, fOI k

r-l,

=

L (Z+k_l)(k) = (+k)(k+1) 11 . n

:=0

k+ 1

Use this result to prove that n

(i)

L X4 = n(I1+1)(2n+l)(3n 2+3n-1)/30; x=l n

(ii)

L L: x= 1 ."<.'C

x(x+l)y2

= l1(n+2)(n 2 -1)(1011 2 +411-3)/180.

2

Continuous random variables

I A continuous random variable X has, for X = x, the probability density function k(l-x) defined in the unit interval (0 ~ X ~ 1). Determine the constant k and then find the mean, median and variance of the population. 2 A continuous random variable X defined over the range (0 ~ X < 00) has, for X = x, the distribution function (l_e- tx2). Derive the probability density function of X and then determine the mean, median and mode of the population. Also, calculate the maximum ordinate of the distribution and the variance of X. 3 Find the constant k and the mean and variance of the popUlation defined by the probability density function

f(X

for 0 ~ X < 00

k(1 +X)-3,

= x) =

and zero otherwise. 4 A continuous random variable X has, for X = x, the distribution function proportional to (IXX fJ - fJx"'), where IX > fJ ~ 1 and 0 ~ X ~ 1. Show that the rth moment of X about the origin is

IXfJI(1X + r)(fJ + r), and hence evaluate the mean and variance of X. If IX = 2P, prove that Iii, the median of the distribution, is given by Iii =

(1- ~rfP

5 For a continuous random variable X defined in the range (a ~ X ~ b), and having, for X = x, a probability density function proportional to g(x), prove that Iii, the median of tlle distribution, satisfies the integral relation

f Iii

f b

2 g(x) dx

=

g(x) dx.

a

a

If X has the probability density function proportional to xl(l +X)3 in the range (0 ~ X < 00), show that the median of the distribution is (1 + j2). Also prove that the mode is 1, and hence deduce the sign of the skewness of the distribution.

6 A continuous random variable X has, for X = x, the distribution function (1 _e-atanX) in the range (0 ~ X ~ n/2), where 0 < IX < l. Find the probability 27

28

EXERCISES IN PROBABILITY AND STATISTICS

density function of X, and hence show that 1110' the mode of the distribution is given by , mo = -!
m, the

median of the distribution, has the value

m= tan- I GIOg2), and that

111,

the mean of X, satisfies the differential equation d2m -d 0( 2+ 111

1 =-. 0(

7 A continuous random variable X defined in the range (0 ::;; X ::;; O() has the distribution function F(X

= x) = 20(x/(a 2 +x 2 ).

Derive the probability density function of X, and hence show that for the distribution of X 3a E(X) = O((1-loge 2) '" 10 and

a2

var(X) = a 2[(n-3)-(1-loge 2)2] '" 20' [n = 3'1416;

IOg102 = 0'3010; 10glOe = 0'4343].

8 The function F(X = x) = xII., (0( > 1) defines the distribution function of a continuous random variable X in the range (0 ::;; X ::;; 1). Determine the first four moments of the distribution of X, and hence calculate the coefficients of skewness and kurtosis. What happens when a = 1? 9 A continuous random variable X has, for X = x, the probability density function proportional to e- X (1 +X)2 in the range (-1 ::;; X < co), and zero otherwise. Derive the cumulant-generating function of X and hence show that ')II and ')12' the coefficients of skewness and kurtosis for the distribution, are 2/:./3 and 2 respectively. Also, show that m, the median of the distribution, satisfies the relation e(m+l)

= 1+(m+2)2.

10 A random variable X is normally distributed with zero mean and unit variance. Find the cumulant-generating function of the random variable Y defined by

Y = O(+PX +')IX 2,

where a,

p and ')I are known constants. E(Y)

Hence, or otherwise, show that

= a+')I and var(Y) = p2+2')12.

11 A ceramic part for an electrical unit is manufactured to order, and as a concession to the manufacturer, a customer ordering N items is prepared to accept at the same price up to aN items where 0( is a given constant> 1.

CONTINUOUS RANDOM VARIABLES

29

However, the manufacturer must supply at least the number ordered. Because f production hazards, the number of items S of approved saleable quality ~ a fraction x (0 ~ x ~ 1) of the initial number I put into the moulds. On the ~asis of past experience of the production process, it can be assumed that x is approximately distributed in the form with probability density f(x) = 495x 8 (1-x)2,

for 0 ~ x ~ 1.

If, in general, the initial numb~r. I is equal to mN (m > (X), determine as a function of m and (X the probablhty that (i) the number of saleable items (S) will be < N; and (ii) the number of saleable items will be > (XN. Hence show that the equation for m in the special case in which the manufacturer desires his risk of not meeting the minimum demand to the full to be half his risk of exceeding the maximum number of saleable items is

mll = 55m 2 (2+(X9)-99(2+(XIO)m+45(2+(X1l). 12 A manufacturer of dolls wishes to plan a special consignment for the Christmas season. On each doll sold he would make a net profit of a shillings, whilst the dolls left unsold in the season would be disposed of in the January sale at a net loss of b shillings per doll. On the other hand, if there is a loss of custom in season due to a lack of supply, the manufacturer expects a loss to the concern because of a goodwill factor which may be considered to be c shillings per doll that cannot be supplied. If the volume of demand v can be approximately regarded as a random variable with a probability density function f(v) in the range (VI ~ V ~ V2), prove that the optimum number of dolls x in order to maximize the expected profit is given by

f x

F(x) = a+c b ,where F(x):= a+ +c

f(v) dv,

VI

and that in the particular case when

f(v) = 1/(V2 - VI)' the appropriate value of x is bVI +(a+c)V2 a+b+c

13 A manufacturer sells an article at a fixed price of £1. If the weight of the article is less than 8 ounces he is unable to sell it and it represents a total loss. The articles have normally distributed weights with mean w ounces and standard deviation one ounce. The cost of production per article is £c, where c = 0·05w+0·30. Determine the mean weight w which will maximize the manufacturer's expected profit. 14 The weights of packets of margarine produced at a factory are nominally a ounces, but actually they are normally distributed with mean x ounces and standard deviation (1, the value of x being adjustable. All packets are weighed, and those weighing less than a ounces are retained, having a value

30

EXERCISES IN PROBABILITY AND STATISTICS

to the firm of V/2. Of those not retained 80 per cent are sold at price V, the remainder being returned to the factory, where their value is again V/2. The average cost of production per packet is x + c, where c is a constant. Show that the value of x which maximizes profit is given by

a+CT[IOge(2~:;2)] t. 15 The cost of operating a hydroelectric plant for a specified period consists of a fixed charge of £c I independent of the amount of electricity generated and a cost of £C2 per kilowatt hour for generation. All the power generated can be sold at a price which is £C3 per kilowatt hour (C2 < C3)' The possible amount of generation is related to the rainfall in the area, and can be considered approximately to be a random variable x having the probability density function (X2 x e - ",x in the range (0 ~ x < (0). Show that the expected profit in the period is 2(C3 - C2)/(X - CI' Also, prove that if initially only £C4 are available as reserves with no borrowing capacity (c 4 < CI), then the probability of going into bankruptcy at the end of the period is

16 The quadratic equation x 2 - ax + b = 0 is known to have two real roots and X2 (XI> X2); but the coefficient b is a positive unknown and can be assumed to have a uniform distribution in the permissible range of variation Find the expectations and variances of x I and X2'

XI

A uniform rod of length 21 is divided at random into two parts of length x and (21- x), these being taken as the semi-axes of an ellipse. Find the expectation of the eccentricity of the ellipse so formed and also its variance.

17

18 A man lives at A and works at C, and is due at work at 8.30 a.m. He always catches the first train from A to B, which is scheduled to arrive at B at 8.15 am. Buses leave B for C every 20 minutes, and the bus which leaves B at 8.20 a.m. is scheduled to arrive outside the factory at C at 8.27 a.m. The train is, on average, a minute late and has a standard deviation of 4 minutes. The bus always leaves on time but is, on average, it minutes late with a standard deviation of 2 minutes. Find the probability for the man to be late for work. The man's employer leaves home in his car at 8.15 a.m. and the time for his journey has mean value 13 minutes with a standard deviation of 3 minutes. Find the probability that (i) both the man and his employer are late for work; and (ii) the employer arrives before the employee. All separate parts of the journeys may be assumed to have a normal distribution. 19 Given the two positive quantities a and b such that a > 3b, and the random variable

g(x) = (3x 2 -ab)/(6x-a-3b)

having a uniform distribution between a and b, prove that the probability for x to be real is 2a/3(a-b).

31

CONTINUOUS RANDOM VARIABLES

If the probability distribution of g(x) is re-defined over the restricted nge which ensures x to be real, verify that the mean and variance of g(x) ra ia and a2/7 '1 y. are 2 respective 20 The two equal sides of an isosceles triangle are of length A. each, and the angle () between them has a probability density function proportional to O(n- O) in the range (0 ~ () ~ n/2), and zero otherwise. Show that Y, the area of the triangle, is distributed in (0 ~ y ~ A. 2/2) with the probability distribution _ )d _ 24 sin- 1(2y/A. 2 )[n-sin- 1(2y/A. 2 )] d f(Y - Y Y - n3' (A.4 _4y2)t . y. Hence, or otherwise, obtain the mean and variance of Y. 21 A straight line of length 21 is randomly divided into two parts which are taken to form the adjacent sides of a rectangle. Find the probability that the area of the rectangle is (i) less than 12/2; (ii) greater than three-fourths the maximum possible area of the rectangle; (iii) less than one-third the area of the circle with diameter equal to the diagonal of the rectangle; and (iv) greater than one-half the area of the circle of (iii). 22 The random variable X has a normal distribution with mean p. > 0 and unit variance. If, for X = x, ao

g(x)

= etx2 . f e- tr2 dt, x

show that the expectation of g(x) is l/p.. 23 A continuous random variable X has the probability density function J(x), for X = x, defined in the doubly infinite range (- 00 < X < (0). (i) If the mean and variance of X are m and u2 respectively, prove that the probability of ~ ku is ~ l/k2, where k is any positive real number. (ii) If f(x) is symmetrical about a point X = a, show that the mean of X is a, and also that the rth moment of X about a is zero for all odd values of r. (iii) For any real number e, show that E{IX is a minimum when e is the median of the distribution of X.

IX - ml

-en

24 If g(X) is a non-negative function of the random variable X and k is a positive constant, show that P{g(X) ~ k} ~ E{g(X)}/k.

If X has mean p., variance u 2 and E[(X given constant e, prove that

P[I-e-A.(cx2 +e2 )t

~ (X;p.y ~

p.)4] = (1 + CX2 )U4 , then, for any

l-e+A.(cx2+e2)t]

~

1-1/,1.2.

25 A straight line of unit length is divided into m parts by (m-l) points chosen at random on it. Find the probability that none of the m parts exceeds

32

EXERCISES IN PROBABILITY AND STATISTICS

I in length, where I ~ t, and interpret the result in the special cases (i) I (ii) m = 2; and (iii) m ..... 00. 26 A continuous random variable X defined in the range (0 has, for X = x, a probability density function proportional to

~

=:::

X <

I. I

~I

X-I. exp{ - (1/2(]'2)(log. x - m)2}, where m and (]'2 are constants. Show that the mean and variance of X ate em +ta 2 and (e a2 -1) e 2m +a 2 respectively. Also obtain the mode of the distribution and the value of the maximum ordinate of the curve. If the mean and variance of X are denoted by (XI and (X2' then show that (]'2 =

log.[(X~~(X2]

m=

and

log.[(X~«(X~+(X2)-t].

27 A continuous random variable X defined in the range (0 ~ X ~ 11 has, for X = x, a probability density function proportional to x 2 (1-x). Find the proportionality factor, and then evaluate M(t), the moment-generating function of X. Show by expansion of M(t) that it can be put in the form

t 2 t2 1 +"53 '1! +"5 . 2! + terms

.

10

ascen

d'

109

powers

0

f

t.

Hence evaluate the mean and variance of X, and check these results by direct calculation of the first two moments. 28 A continuous random variable X defined in the range (- 00 < X < <XlI has finite moments. By considering the expectation of the quantity (X 2 +aX +b)2,

where a and b are arbitrary constants, prove that PI and P2, the Pearsonian coefficients of skewness and kurtosis, satisfy the inequality

P2 where PI == liVJ·d, mean.

P2 ==

~ PI + I,

li4/1i~, and Ii, is the rth moment of

X

about the

29 In a single tossing of x unbiased pennies, find the probability P(x) for obtaining all heads. If x is regarded as a continuous variable in the range (0 ~ x < 00), show that P(x), regarded as a function of x and suitably scaled, determines a probability distribution. Obtain the probability density function of this distribution and calculate its cumulant-generating function. Hence, or otherwise, verify that the coefficients of skewness and kurtosis for this distribution are 2 and 6 respectively. 30 There are n biased pennies such that the probability of obtaining a head with the rth penny (r = 1,2, ... , n) is pi +'x,

where 0 < p < 1 and 0

~

x

~

1.

Find the probability P(x) of obtaining all heads in a single tossing of the n pennies. Show that for fixed nand p, the function P(x), suitably scaled, determines a probability distribution of a random variable X defined in the range (0 ~ X ~ 1).

33

CONTINUOUS RANDOM VARIABLES

lIence obtain the cumulant-generating function of X and verify that E(X)

= p-l_(eP _l)-l

and

var(X)

= P-2- eP(e P_1)-2,

where

31 For a contin~o~s rll:ndo~ variable X, defined in the range (0 ~ X < (0), Ihe probability dlstnbutIon IS such that P(X ~ x)

=

l_e- px2 ,

where

P>

O.

FOnd the median of the distribution. Also, if rn, rno and

(1

denote the mean,

~ode and standard deviation respectively of the distribution, prove that 2rn~ - rn 2 =

(12

and

rno

= rn~.

What is the sign of the skewness of the distribution? 32 A random variable X defined in the range (0 ~ X ~ n12) has, for X = x, a probability density function proportional to e -ax sin x, where ()( > 0 is a

parameter. Calculate the proportionality factor and hence obtain the cumulantgenerating function of X. If 0: is sufficiently small for terms involving 0: 2 to be negligible, show that, as a !irst approximation, the effect of a non-zero ()( in the mean and variance of X is - 0·140: and 0·160: respectively, relative to unity. 33 The probability density function of the random variable X is, for X = x, proportional to 1/(1 +X4), and defined in the doubly infinite range ( _ (jJ < X < (0). Find the proportionality factor and identify the form of the distribution. Also, show that the variance of X is unity. For any given 0: > 0, prove that P(X

~

2 [1

0:)

~3.7.11.15 ... (4r-l)]

= (1+o:4)iB(i,i) 3+ r~i 4r(l+o:4)'(4r+3)r! .

34 From an urn containing N identical balls of which a proportion pare white and the rest black, a random sample of r ( < N) balls is drawn without replacement. Find the probability Yx that (x -1) of these balls are white and the rest black, and hence show that the finite difference i\yx satisfies the relation i\yx _ 2[(NP-X+ l)(r-x+ l)-X(NQ-r+X)] Yx+t (Np-x+ l)(r-x+ l)+x(Nq-r+x) ,

=

(l-p) and Yx+t = i(Yx+Yx+d. where q If x is regarded as a realization of a continuous random variable X, deduce that, as a limiting form, x and Y satisfy the Pearson differential equation 1 dy dx

i

a-x bo +b i x+b 2 x2'

a, bo, b i , b 2 being functions of N, p and r. Integrate this differential equation for the particular case of b i = b 2 = 0, and show that the mean and variance of the resulting probability distribution of X, defined in the range (- 00 < X < (0), are a and bo respectively.

34

EXERCISES IN PROBABILITY AND STATISTICS

35 If F(x) is the distribution function of a normally distributed random variable X with zero mean and unit variance, and
=

~

P(O

~ x),

X

prove that F(x)[l-F(x)] = i_
Hence, by using a series expansion for
e-2X2/1t

[1 +2(7t-3) x 4 _(77t 2 -607t+120)x 6 + ... ]. 37t2 457t 3

36 If the continuous random variable X has a the probability density function

1 r(v+1 )

r

distribution with, for X == x I

forO~X
.e-x.xY ,

where v is a positive integer, and the discrete random variable Y has a Poisson distribution with parameter J.l, prove that P(X ~ J.l)

=

P(Y~

v).

If X is a discrete random variable which has the Bernoulli distribution such that

37

P(X

=

'/I '

r)

= (r)pr qn-r, where p+q =

1,

and Ya Beta variable having, for Y = y, the probability density function 1 . yn-r-l(l_ y)' B(r+ 1, n-r)

in the range (0

~

y

~

1), show that F(r)

=

G(q),

where F(x) and G(y) are the distribution functions of X and Y respectively. 38

For a random variable X having a unit normal distribution prove that 1 00 (_1)'x2r+l (x > 0). P(X ~ x) =!+ ~. (2 1)2r I' V 27t r=O r+ .r .

L

Also, using integration by parts, show that an alternative form is 1

e- tx2

v 27t

X

P(X ~ x) = 1- ~.--.

[

1+

Ln (-1)'(2r)!] 2 + Rn(x), .r.x r

I

2r

r=1

where

Verify that this series expansion for the distribution function of X is asymp· totic by proving that, for any n, the remainder is less in absolute value than the last term taken into account.

CONTINUOUS RANDOM VARIABLES

35

39 Starting from the standard asymptotic expansion for the distribution runctio n of a unit normal random variable X, viz., 1 e- tx2 [ ro (-I)'(2r) '] P(X::; x) ~ 1- r-c.~-. 1+ 2' 2, " y 2n x ,= 1 • X • r .

L

obtain Schlomilch's more rapidly convergent form of the expansion given by P(X ::; x) ~ 1-

1 e- tx2 h:.--[I-Al+A2-5A3+9A4-129As+57A6-9141A7+ ...], y2n x

where, for any positive integer r, A,

==

1/l].1

(x 2 + 2k).

40 A random variable X has the probability density function J(X = x) = ksin 2n xcosx defined in the range (-nI2 ::; X ::; nI2), where k is a function of the positive integer II. Determine k and calculate the probability that (i) X ::; () and IXI ~ () for any given (); and n

(ii)

-6::;

X :e;;

1t

6'

Also, prove that where 12n+ 1 satisfies the recurrence relation

(2n+l)212n+l = 1+(2n)(2n+l)1 2n _ 1, with II = 1. Obtain the distribution of Z

sin 2 X and identify its form.

=

41 If (Xx denotes the distribution function of a normally distributed random variable X with zero mean and unit variance, prove that

(Ill) A(m+r+ 1) = t ,~o (-1)' (m)r A(m+r),

m (i) ,~o (-1)' r

(ii) ,to

m

(-1)'(7) B(m+r)

where for

II

=

! ,to (-I)'C) A(m+r),

a positive integer and () > 0 1

A(n) ==

f (X~

d!X x

and

B(n) ==

o

o

Also, if for positive integral values of nand s 1

D(n, s)

==

Jx'!X~ o

then show that

f(X~ 1

d!Xx ,

!Xlix d!Xx ·

36

EXERCISES IN PROBABILITY AND STATISTICS

(iii)

f

(_I),(m) D(2n+l,m+r)

= 0; and

(-1)'(7)

=!

r

r=O

(iv) Jo

D(2n,m+r+l)

rt (-1)'(7)

D(2n,m+r).

42 If X is a random variable with the probability density function for X = x, find the probability density function of Y = X 2, if (i) f(x) = 2x e- x2, for 0 ~ X < co (ii) f(x) = (1 +x)/2, for -1 ~ X ~ 1 (iii) f(x) = t, for -t ~ X ~ 1.

f(x~

43 If X and Yare two random variables with the joint probability density function f(x, y), for X = x and Y = y, find the probability density function of Z if (i) f(x, y) = 4xye-(X2 +)l2), for 0 ~ X, Y < co and Z = (X2 + y2)t. (ii) f(x, y) = 1, for 0 ~ X, Y ~ 1 and Z = X+ Y ifX+Y<1 =X+Y-l ifX+Y>1. (iii) f(x, y) = ala2 e-(OIX+02)1), for 0 ~ X, Y < co and Z = X+ Y. 44 Each of two independent events A and B can only occur once in the future. The probabilities of their happening in the time-interval (t, t+dt) are proportional to t e-ext . dt and t 2 e- fJt . dt respectively for all t in the range (0 < t < oo~ where oc and p are positive constants. Prove that the probability that the two events occur in the order BA is P3(4oc + P}/(oc + P)4. 45 If the independent random variables X and Y have the probability density functions f(x) and g(y) defined respectively in the intervals (0 ~ X ~ oc) and (0 ~ Y ~ P), (P > oc), find the probability distribution of the random variable Z = X + Y, indicating the discontinuities in this distribution. Hence obtain the distribution of Z in the particular case when both X and Yare uniformly distributed, and determine its limiting form as oc -. p. 46 For the independent random variables X and Y having the probability density functionsf(x) and g(y), both defined in the unit interval (0 ~ X, Y~ find the probability distribution of Z = X + Yin the following cases: (i) f(x) = 1 and g(y) = 3(1- 2y)2 ;

n

(ii) f(x)

= 1r: and

g(y)

=

J-=y ;

2yx 2 l-y (iii) f(x) = 3(1-2x)2 and g(y) = 3(1-2y)2. Also, find the distribution of W = X 2 + y2 when both X and Y have a uniform distribution in the unit interval. 47 The two random variables X and Y have, for X = x and Y = y, the joint probability density function 1 f(x, y) = - 2 2' for 1 ~ X < co; llX ~ Y ~ X. xy

Derive the marginal distributions of X and Y. Further, obtain the conditional distribution of Y for X = x and also that of X given Y = y.

CONTINUOUS RANDOM VARIABLES

37

48 The continuous random variables X and Y have a joint probability density function proportional to y"(x- y)(l-x)P, for 0 ~ X ~ 1;0 ~ Y ~ X, h parameter a being > - 1 and {3 a positive integer. t e Find the proportionality factor, and hence determine the probability distribution of the statistic u = y/Jx. Also, for any given uo, (0 < U o < 1), shoW that 2u"0 +1 x ) _ P(u ~ Uo - B(a+3,{3+1)

(P)

P [(a+2) (a+ l)uo + (a+ l)(a+2)uo+ 2r +S ] x r~o r (-1)' (a+2r+5) (a+2r+4) 2(a+,.+3)(a+2,.+4)(a+2r+5)·

49 If X and Yare independent random variables such that X is uniformly distributed in the range (1 ~ X ~ 3), and Y has the negative exponential distribution in the interval (Y ~ 2), obtain the joint distribution of the random variables z = X /Y and W = X Y. Hence derive the marginal distributions of Z and W, indicating the discontinuities in their probability density functions. 50 A gun is fired at a target and the bullet strikes the target at a point P. The vertical and horizontal distances of P from the bull's-eye are independent normally distributed random variables with zero mean and variance (f2. Show that the probability that the bullet strikes the target at a distance R greater than r from the bull's-eye is e- r2 / 2a " and hence deduce the probability that the bullet hits the target in the annular region (r 1 ~ R ~ /'2)' Also, find the probability that of n independent attempts at hitting the target k (0 ~ k ~ /1) fall in the region (rl ~ R ~ r 2) and the rest outside it. 51 Two independent realizations Xl and X2 are given of a random variable X which, for X = x, has the probability density function

0- 1 e- x/8, for 0

~

X <

00,

0 being a parameter.

Derive the sampling distributions of the statistics

= Xl +X2 and v = X t!(x 1 +X2), and hence prove that u and v are independently distributed. Find the mean and variance of v. Further, derive the distribution of w = X2/Xl' and show that wand u are u

also independently distributed. 52 Three independent observations x h X2' and X3 are given from a univariate normal distribution with mean 111 and variance (f2. Derive the joint sampling distribution of (i) u = Xl-X3; (ii) v = X2 -x 3 ; and (iii) w = Xl +X2+X3-3m; and hence obtain the probability distribution of z = u/v. Also, prove that the mode of the distribution of z is !, and explain the significance of this modal value.

38

EXERCISES IN PROBABILITY AND STATISTICS

53 If Xl> X2' X3' and X4 are independent observations from a univariate normal population with zero mean and unit variance, obtain the sampling distributions of the following statistics:

1X2

(i) u = (x~ + x~)/(x~ + x!) ; (ii) v = (Xl +X2+X3)/[!<XI-X3)2+i(XI-2x2+X3)2+X!]t; and (iii) w = (x~ + x~)/(x~ + x~ + x~). By using the appropriate published tables, determine the constants such that P(u ~

1(1)

= O'OS

and

IXI

and

p(lvl ~ 1(2) = 0·01.

54 The continuous random variables X, Y and Z defined in the range (0 ~ X, Y, Z < (0) have the joint probability distribution with the density function f(X = x, Y = y, Z = z) = (xyz)-t. g(x+y+z). Derive (i) (ii) (iii)

the marginal distributions of the random variables U = X + Y + Z; V = Y/X; and W = Z/(X + Y).

A continuous random variable X has, for X = x, a probability density function f(x) defined in the doubly infinite interval (- 00 < X < (0). If Xl and Xn denote the minimum and maximum of the first n independent observations derived from this population, prove that the probability for the (n+ l)th ~nd (n+2)th independent observations to lie outside the interval (Xl ~ X ~ Xn)

S5

IS

6/(n + l)(n + 2).

56 From a standard pack of playing cards having four aces in fifty-two cards, cards are dealt one by one until an ace appears. Prove that P(X = r~ the probability that exactly r cards are dealt before the first ace turns up, is given by P(X

= r) = (Sl-r)(SO-r)(49-r}/13. 49. SO. S1.

Verify that this represents a proper discrete probability distribution defined over the admissible range of values of the random variable X. Hence determine the mean and variance of X. Show, further, that by a suitable transformation P(X = r) can be considered approximately proportional to (Z3 - z), where z varies continuously in the range (0 ~ z ~ SO). Hence calculate the approximate values of the mean and variance of X. Assuming this approximate form of the probability distribution of X as related with z, determine the probability for X ~ 2S, and thereby prove that if this event happens in two consecutive deals of the pack, it is reasonably certain that the card distribution in the pack was not random. 57 Cards are dealt out one by one from a well-shuffled standard pack until the second ace appears. Prove that the probability that exactly r cards are dealt before the second ace turns up is per)

= r(Sl-r)(SO-r}/13.17.49.S0.

CONTINUOUS RANDOM VARIABLES

39

By considering r to be a realization of a discrete random variable X such that P(X

= r) = p(r),

verify that this defines a proper probability distribution over the possible range of values of r, and hence obtain the mean and variance of X. Also, show that by a suitable transformation p(r) can be considered approximately proportional to (50-z)z(z+ 1), where z varies continuously in the range (1 ~ z ~ 49). Hence calculate the approximate values of the first two moments of X and the percentage errors of approximation. Using this continuous approximation to the distribution of X, determine the probability for X ~ 25, and thereby prove that if this event happens in four consecutive deals of the pack, it is safe to infer the non-randomness of the card distribution in the pack. S8 It is known from empirical considerations that the life of an electronic tube produced by a factory may be regarded as a random variable T having a negative exponential distribution with mean T. In an experiment designed to assess the quality of a large consignment of tubes, a random sample of n tubes is taken and put on the test rack. If t I and t2 are the observed lifetimes of the first two tubes, show that the joint distribution of t I and t 2 is

If the minimum standard of approvable quality of the consignment is such that (t 1+t2 ~ c), where c is a known constant, prove that the probability of obtaining random samples of n worse than the approvable quality is

Find the limiting value of Pn as n -+ significance.

00,

and c and

T

are finite, and explain its

S9 In a factory with It lamps lighted at the same time daily, a bulb is replaced immediately it burns out, the cost of replacing it being a constant y. An alternative method is to change all I! bulbs at the same time periodically with time-period T, but still change a bulb when it burns out. When all the I! bulbs are changed together, the total cost is a + {JI!, where a and {J are constants such that a + {JI! < I'll. If the lifetime T of a bulb is a random variable having a uniform distribution in the interval (0 ~ T ~ k), show that the expected number of burnouts per socket is (T < k).

Hence deduce that the equation for determining the optimum value of which makes the average cost per unit of time the bulb is lit a minimum is eX (I-x) = 1-0,

where x == T/k, and 0 == (a + {J1l)!YIl < 1.

T

40 60

EXERCISES IN PROBABILITY AND STATISTICS

If X and Yare independent unit normal random variables, and V(h, q) l(h, q)

= =

P(O

~

~

P(X

~

X

h; 0

h; Y

~

~

Y

~

qX/h),

qX/h),

and (h) = P(O

~

~

X

h),

q and It being positive constants, then prove that 1 1(11, q) = V(h, q)-tI>(h) + 2n cot- 1(q/Il).

Hence, by deriving Polya's approximation,

V(It, q) '"

tan~;q/h) . [1_e-QhI2Ian-l(Qlh'J,

obtain a suitable approximation for 1(1t, q).

61

For the normal bivariate integral

f

0000

1(It,q) = 21nf

e- t (X 2+).2)dydx,

h Qxl"

prove Nicholson's approximation that

1(It,q)",~.e-th2(1+W2). 2n

1 ], n=1f: 1t-2n(~.~)n-l[ dw w(1+w2) W

where w == q/It. It and q being positive constants. Also, verify that the general term of the series is

1t- 2 n( -2rl (n-1)!

nIl (2~')~V'/22r . w2r+ r=O

I

.(1+w2)"-r.

,

62 If XI and X2 (x I < x 2) denote two ordered observations of a uniformly distributed random variable X defined in the range (0 ~ X ~ 2a), derive the joint probability distribution of Xl +x 2 y= - - - an d

2

X 2 -X l

Z=-2-'

Hence show that for the marginal distribution of y the probability density function is (i) y/a 2 , for 0 ~ y ~ a; and (ii) (2a- y)/a 2, for a ~ y ~ 2a. Also, prove that the marginal probability density function of z is 2(a-z)/a 2 , 63

for 0 ~ z ~ a.

Prove that if A > 0, AB-C 2 > 0, then

2~

f f exp[-1
ifJ

-:;.('I

-rJ)

=

(AB-C 2)-! expt[(Ab 2 +Ba 2 +2Cab)/(AB-C2 )].

41

CONTINUOUS RANDOM VARIABLES

64 The two continuous random variables X and Y have. for X the joint probability density function proportional to

= x. Y = Y.

x2 e - y;C2/ 0 + !x!)a defined in the range (0 ~ Y < 00; - 00 < X < (0). where ex is a parameter > 3 Determine the proportionality factor and the marginal distribution of X. He~ce calculate the mean and variance of X. Also, show that the distribution function of Y is

J 00

e-YXl

P(Y ~ y)

= l-(ex-l)

o

dx

I

( +x)"

•

and verify that all the moments of Y diverge. Use these results to prove that X and Yare uncorrelated but not independently distributed random variables.

65 The joint distribution of the random variables X and Y is defined by a probability density function proportional to yfl(1- xl".

for 0 ~ X ~ 1; 0 ~ Y ~ X.

the parameters ex and P being each > - l. Find the marginal distributions of X and Y. and evaluate their means and variances. Further, determine the conditional distribution of X. given Y = y. and that of Y, given X = x. Hence verify that E(X!Y

= y) = 1 +~:~ l)y;

E(Y!X

=

x)

v ar(X!

Y_ _ (ex + 1)(1- y)2 . - y) - (ex+3)(ex+2)2'

(P+ 1)x 2

= (P+ l)x.

var(Y!X = x) = (P+3)(P+2)2'

P+2 •

and determine the correlation between X and Y. Comment on the limiting forms of these two conditional distributions for Y = 1 and X = 0 respectively.

66 If x is the largest of m independent observations from a uniform distribution in the interval (0, ex), and y the largest observation in a random sample of n from a uniform distribution defined in the range (0. P), prove that the probability distribution of the ratio z = ylx is

__ m_n.,----, ·z11-1 dz -:(m+n)p"

for 0 ~ z ~ p,

'

and

mnpm

dz

-'m+n zm+1'

~or

I'

p

~

~

z<

00.

where p == Plex. Use this distribution to show that the rth moment about the origin of w = zip is I

mil

Jl. = . , (n + r)(m - r)'

Also, determine the probabilities P(z ~ Po > p) and P(z ~ P1 < p) for fixed Po and Pl'

42

EXERCISES IN PROBABILITY AND STATISTICS

67 If x and yare the maximum values obtained from independent samples of 111 and 11 (m ~ 11) observations respectively from a rectangular distribution in the (0, 1) interval, find the sampling distribution of the product z = Xy in the two cases (i) m = nand (ii) m < n. Hence, or otherwise, deduce that for equal sample sizes the statistic v

=

-2nlogz

is distributed as a X2 with 4 dJ., but that for m < n the rth moment about the origin of v is E(v r )

=

G),. qr+ 1) .(I-A:+ 1)/(1_).),

(). ==

m/n).

Use this result to show that for m =F n the distribution of v may be approximated as fJx 2, where X2 has v dJ., the constants fJ and v being evaluated as

fJ =

(1+).2) ),(1+),) and

2(1+).)2 v = (1+).2)'

68 If W1 and W2 are the sdmple ranges obtained from independent samples of size and n2 respectively from the (0, a) rectangular distribution, derive the sampling distribution of the ratio

"l

u= Prove that for any positive integer r E(u r)

= (Ill

WdW2'

n1(n1 -1)n2(n2 -1) +r)(n1 +r-l)(112 -r)(n2 -r-l)

Hence, or otherwise, deduce the limiting form of the distribution of u as both nl and n2 .....

00.

If w is the sample range obtained from a random sample of n observations from a rectangular population in the (0, 1) interval, prove that - (2n -1) log w is distributed approximately as a X2 with 4 degrees of freedom. Hence show that if z" is the product of k independent sample ranges from the above population obtained from samples of sizes n1, n2' ... , n" respectively, then for large samples a first approximation gives that - (2n -1) log z" is distributed as a X2 with 4k dJ., n being the average of the sample sizes nj. Further, prove that an imprOVed approximation can be taken as

69

(2n--l)[1

where kV(nj) ==

{1+4V(n i )}] I 2 2n 2 ogz" - X

L" (nj-n)2. j=

1

70 If Xl' X2"'" X" are the maximum values obtained from independent samples of equal size n from a rectangular population in the range (0, 1~ find the exact distribution of the product v" = Xl X2 ••• X". Hence, or otherwise, prove' that - 2n fog v" is distributed as a X2 with 2k degrees of freedom.

CONTINUOUS RANDOM VARIABLES

43

Also, if the Xj are obtained from large samples of size nj (i = 1, 2, ... , k) ectively, then show that a convenient large-sample approximation for the re sp f. distribution 0 Vk IS

7

-2n [ 1- 2V(n.)] logvk

,...,

2k [ 1- V(n.)] n2' dJ.,

X2 with k

where n is the average of the nj and kV(nj) ==

L (nj-n)2. j=

I

71 Suppose UI is the smallest and VI the largest of nl independent observations from a rectangular population in the interval (0,1); and U 2 is the smallest and V2 the largest observation from another random sample of size n2 from the same population. If it is known that UI ::s:; U2' then for fixed UI prove the rollowing: (i) The conditional joint distribution of VI and U2 is

Hence derive the unconditional distribution of the statistic T= (U2-UI)/(VI-UI),

and verify that for positive integral r E(Tr)

=

(nl -1)r(n2 + 1)r(r+ 1) (111 -r-1)r(n2 +r+ 1)·

(ii) The conditional joint distribution of U2' V2 and VI is n2(n2 -1)(nl -1)

°

(VI-UI)n,-2(V2- U2)"2- 2 (1 -U )n 1 +n 2 -I ° dU2 dV2 dvlo I

(UI ::s:; VI ::s:; 1; UI ::s:; U2 ::s:; V2; U2 ::s:; V2 ::s:; 1).

Hence derive the unconditional distribution of the statistic

u = (V2 -

U2)/(VI - u I ),

and use it to prove that E(U r )

=

n2(n2 -1)(nl -1) (n2+ r)(n2+ r - 1)(nl-r-1)

(iii) The conditional joint distribution of VI and V2 is (nl-1)n 2

o

(VI-UI)n l -2(V 2 -u l )n 2 -1 (1 -UI )n+n I ·dvl dv 2' 1 2

(UI ::s:; VI ::s:; 1; UI ::s:; V2 ::s:; 1).

Hence determine the unconditional distribution of the ratio V

and deduce that

=

(VI - UI)/(V 2 - UI),

44

EXERCISES IN PROBABILITY AND STATISTICS

72 The lifetime x (in hours) of electronic tubes mass-produced by a standard process is a random variable with a probability distribution having the density function 1X2 x e -
A research engineer suggests certain modifications in the production process which would aIter the lifetime distribution by increasing the expected lifetime to 2/P, (P < IX). But because of the cost of introducing the change, the manufacturer does not consider it worthwhile unless the modified process also ensures that the probability of the lifetime of a randomly selected tube being ~ m is increased by a fraction A > O. Prove that the manufacturer's condition for introducing the new process is satisfied if 1 P < 1X--·log(l+A). 111 73 A population of N persons is exposed to the risk of accidents. Initially there is an equal probability for anyone individual to sustain an accident, and it may be assumed that an accident does not result in death, so that the population remains of constant size. In general, as suggested by Greenwood and Yule, suppose that the probability of a person having an accident is altered if he has already met with a previous accident. If then !(t, x) t5t is the probability at time t (>0) that a person who has had x accidents will have another accident in the infinitesimal time-interval t5t, and Vx is the expected number of persons who have had x accidents at time t, prove that

dv x =!(t,x-l).Vx-l-! (t,x).v x , (it where, by definition, v_ 1 = O. From this general differential equation deduce that (i) if!(t, x) = kcp(t), k a constant, then v, = N . (r+ l)th term of a Poisson distribution with parameter kT; and (ii) if !(t, x) = (b + cx)cp(t), band c being positive constants, then Vx = N . (r+ 1)th term of a negative binomial distribution with prob· ability-generating function G(l})

= (l-wt'·/(1-wOt'·,

where

f I

T

=-

cp(u) du

and

w =- (l-ecT).

o

74 A large group of persons is exposed to the risk of accidents over a prolonged period, which may be considered to be divided into a number of time-intervals of equal length. Due to personal and occupational differtmces the accident proneness of the persons varies, and it is known from empirical considerations that if, per time-interval, the mean number of accidents for an individual is A, then the probability that he will have x accidents in a random time-interval is given by a Poisson distribution with mean ,t As suggested

CONTINUOUS RANDOM VARIABLES

45

by Green~ood ~~d Yule, it may be assumed that the probability of a person having a given IS e' nr)" e-cJ. . ..1.,-1 dA., (A. ~ 0). B considering the joint distribution of the random variables x and A., prove t:at the marginal distribution of x has the probability-generating function G(O)

= e'/(1 +e-O)'.

Hence, or otherwise, deduce that corr(x, A.) = (1 +e)-t. Also, if var(xIA.) denotes the variance of the conditional distribution of x for any given A., and var(x) is the variance of the marginal distribution of x, verify that for variation over A. E[ var(xIA)]

= r/e, whereas var(x) = 1'(1 + e)/e 2 •

Explain the difference between these two results as measures of the variability of x. Finally, derive the conditional distribution of A. for given x, and verify that E(A.lx) = (x + r)/(l + c).

75 Accidents occurring randomly in time may be classified according as they give rise to injuries to 1, 2, 3, ... persons. For the distribution of these classified accidents, considered over equal time-intervals, it may be assumed that accidents, each of which involved k injured persons, have a Poisson distribution with mean 'l'k, for all integral values of k ~ 1. (i) Find the probability-generating function of the random variable X denoting the total number of persons injured in anyone time-interval. Hence deduce that if (0 < p < 1),

then X has a negative binomial distribution with probability-generating function

G(z) = (1- p)J./(1_ pz).1., so that, per time-interval, the mean number of persons injured is A.p/(l- p), and the probability that no person is injured is (1- p)J.. (ii) If Y is the random variable denoting the total number of accidents sustained by a person, and Pk is the probability of a particular person receiving injury in an accident involving k persons, derive the probability-generating function of Y. Hence, for

Pk = kp

and

'l'k = A.pk/k,

verify that Y has the Poisson distribution with mean A.pp/(1- pl. 76 A city corporation is responsible for continually providing with lamps a large number N of street-lighting posts. Initially, at time t = 0, a new lamp is inserted in each post with the general principle that as the lamps fail, they are

46

EXERCISES IN PROBABILITY AND STATISTICS

to be replaced by new lamps. In order to minimize maintenance costs two alternative plans are suggested for further procedure: Plan I is that the replacement of the lamps as they fail is continued in. definitely. Plan II is that the replacement as the lamps fail is continued only till time t = T. Then all the N lamps are replaced by new ones. This procedure is repeated indefinitely, all lamps being replaced at nT for all integral values of n ;?; 1. Apart from these regular replacements, irregular replacements are still made whenever lamps fail in the intervals nT < t < (n+ 1)T. For comparing the costs of operation under the two plans, it is known that (i) the labour charge for an irregular replacement is u; (ii) the average labour charge per lamp of a regular replacement under Plan II is v; and (iii) w is the price of a lamp. It may also be assumed that the lifetime distribution of a lamp is f(t) dt,

(0

~

t < (0),

with expected lifetime L > T. Assuming that under Plan I lamps fail randomly in time, so that the probability of a lamp failing in the interval (t, t + dt) is dt/L, calculate C 10 the expected cost of maintenance in the interval (0. T). For Plan II, prove that the expected number of irregular replacements of lamps at a given post in the time-interval (0, T) is IX)

G(T)

==

L

F m(T),

m=l

where F m(T) is the probability that at least m lamps fail at a post in (0, T). Hence calculate C 2, the expected cost of maintenance in (0, T) and verify that

being the ratio of the average total costs of regular and irregular replacement of a lamp. As particular eases of this general result, show that (a) iff(t) = A. e- At , then under all circumstances Plan I is more economical than Plan II; and (b) if f(t) = ..1.2 e- A1 • t, then Plan II is more economical than Plan I, provided that

p

°< p < i(1-e-

2AT ).

77 Electric bulbs used for street lighting have a lifetime distribution with probability density functionf(t) for (0 ~ t ~ To), and zero for t > To. Whenever a bulb fails in a lamp post it is immediately replaced by a new bulb, and this method of replacement is continued indefinitely. Starting with all new bulbs at time t = 0, the process is considered up to a stage t = T, where, in general, (r-l)To < T < rTo for all integral values of r ;?; 1, and F~) (T) is the probability of making at least m replacements in a given lamp post in the interval (0, T).

47

CONTINUOUS RANDOM VARIABLES

Prove that if r = I, then F~:' (T) satisfies the integral relation

F~:' (T)

fF:':~ T

=

I(T-t)f(t) dt;

o

but that for r ~ 2, 1"-(,-I)T o

F:;.'(T) =

1"0

f F~'_I(T-t)f(t)dt+ f F~=P(T-t)f(t)dt o

for m ~ r,

1"-(,-1)1"0

whence the expected number of replacements in the pcriod (0, T) is 00

G,(T) ==

L F~'(T). m=l

Hence determine the probability of a replacement being made in a post during the interval (t, t +dt) regardless of when the bulb was put in. Also, verify as particular cases that if the lifetime distribution of bulbs is uniform in the range (0 ~ t ~ To), then GI(T) = (eT/To_I), if 0 < T< To;

and G:z(T) =

(eT/To_I)_(~ -1)

e(T/To-H,

if To < T< 2 To.

78 Electric bulbs, used individually for street lighting in a large number of posts, have a lifetime distribution with probability density function f(t) for o ~ t < 00; and a bulb is replaced immediately it burns out. If, starting from lime t = 0, the process is observed till t = T, calculate the expected number of replacements in a post during the interval (0, T). Hence deduce g(t) dt, the probability of a bulb being replaced in (t, t + dt) for t < T, irrespective of when the bulb was put in. Next, suppose that at the end of the first interval of time T, all bulbs which were put in the posts before time X < T and have not burned out are replaced by new ones, but the bulbs replaced after time X continue to be used, provided, of course, that they have not burned out. Prove that with such a mixture of old and new bulbs, the probability of a bulb having an expected lifetime >t in the second interval of length Tis x S:z{t)

=

(l-P)Sl(t)+

f

g(T-X)Sl(X)Sl(t+x)dx,

(1" < T),

o

where p is the proportion of bulbs not replaced at time t = T and Sl(t) is the probability that a bulb has a lifetime > t. In the particular case whenf(t) = A. e- A1, verify that S:z(1") = !e-·I.t(1+e-J.X).

79 In a large city a number of street lighting posts are supplied with electric bulbs having a lifetime distributionf(t) dt for (0 ~ t < (0). Initially, at time t = 0, all posts carry new bulbs, and in course of time whenever a bulb burns

48

EXERCISES IN PROBABILITY AND STATISTICS

out it is immediately replaced by a new bulb. In addition, all posts are inspected at regular intervals of time T, so that at time t = liT, (n ~ 1), (i) all bulbs which were replaced in the interval (nT-X, nT), (X < T) and have not burned out by t = nT continue to be used; and ' (ii) all bulbs which were replaced during the interval [(n -1)T, (nT-X)], and have not burned out by t = nTare replaced by new bulbs. Suppose that after the above replacements have been made, Pn is the proportion of bulbs not replaced at t = nT, and that of these a proportion pi have their last replacement before t = (n + 1)T in the interval [(n + 1)T- X, (n + 1)T]. Similarly, of the proportion (1- Pn) of bulbs actually replaced at t = nT, a proportion p have their last replacement before t = (n + 1)T in the interval [(n + 1)T- X, (11 + 1)T]. Prove that Pn satisfies the difference equation

Pn+ I

=

PnP' +(1- Pn)P,

for

11 ~

0,

and, assuming that both p and pi are independent of n, find an explicit expression for Pn. Also, verify that if the difference between p and pi is small then, as a first approximation, P2 '" Pro so that the proportion of replacements at t = nT is effectively stabilized soon after the second T interval. Hence, if g(t) dt, (0 < t < T), is the probability of a replacement being made at a given post in (t, t + dt) irrespective of when the bulb was put in, and S(t) is the probability of a bulb having a lifetime> t, show that x p

=

f

g(T-x.)S(x.)dxI'

o and xx pp'

=

f f g(T-x )g(T+X I -X2)S(xdS(x 2) dX l

o

I

dX2·

0

Evaluate p and pi in the particular case when J(t) = A. 2 e- A1 • t, A. being a positive parameter.

80 If U I < U2 < ... < Un are a set of ordered observations from a population having a probability density function J(x) in the range - 00 < x < 00, derive the joint distribution of the random variables PI' P2' . .. , Pn such that Ut

PI =

•

J J(x)dx

"i+ 1

and

Pi+1 =

-00

J J(x)dx,

for i = 1,2, ... ,n-1.

~

Next, suppose that a random sample of size m is taken from another population, also with a doubly infinite range, and that these m observations are distributed over the (n+ 1) intervals formed by the Uj so that m l observations are un• Find the joint distribution of the mi and Ph and then prove that every random set of realizations of the mi has, for given m and n, the probability 1/(m~").

81 A random variable X has the Laplace distribution with probability density function proportional to e-<xlx-ml for (- 00 < X < (0), m being a positive parameter. If c is any given positive number ~m, prove that

1 E{lx -en = -. [e-<x(m-C)+ex(m_e)]. ex

49

CONTINUOUS RANDOM VARIABLES

Hence deduce that the minimum value of the mean deviation of X is 1/1X, which is attained for c = m. 82 Explain clearly the difference between the standard deviation and mean deviation of a continuous random variable X. In particular, if X has a negative xponential distribution with probability density function proportional to : -«" for X ~ 0, and a and b are positive constants, prove that E(X _a)2 =

1 (1IX )2

-+ 1X2

--a

,

and E{IX-bl} = b+(2e-«b-l)/IX.

Hence deduce that for this distribution the minimum value of the mean deviation is (loge 2) X standard deviation. 83 A dam has a total capacity of M units of water for irrigation, but if there is no rainfall during a month then the dam can only supply IXM units of water, (0 < IX < 1). The amount x of rainfall in a month is a random variable, and as x increases, the supply of water in the dam also increases; and since the capacity of the dam is limited it may be assumed on empirical considerations that this increase dies off exponentially. As a simple model, it may therefore be assumed that if the amount of rainfall in a month is x then the total supply or water in the dam is

where (J is a small positive parameter. If the probability distribution of rainfall in a month is

prove that E(S)

=

M[I-(I-IX)/(l +2{J)t].

Hence, provided P is sufficiently small for p2 to be negligible, show that the actual supply S will be at least equal to its expectation if x ~ (1- P/2). Also, prove that the probability for this to happen is approximately 2[1-(1-P/2)], where (t) is the distribution function of a unit normal variable. 84 A random chord of length Y is drawn across a circle of radius p such that its perpendicular distance from the centre of the circle is px, where x is assumed to be a uniformly distributed random variable in the interval (0, If Yh Y2' ...• Yn are It independent realizations of Y, prove that for their mean y

n

pn

E(y)

=2

,. .,

1·57 p

and

Further, if Ym is the largest of the E(Ym)

=

pA(l1)

and

11

var(y)

=

(32 - 3n 2)p2 12n

p2 ,..., 5n'

observed Yi> prove that

var(Ym)

=[

4n(n+3) (n+ l)(n+2)

2] 2

A (n) . p ,

50

EXER(,[SES IN PROBABILITY AND STATISTIC'S

where _

A(n)

=

n-I

n

r(n+l)

(-1)'

r~o 2'(r+2)· r(n-r)

I{ r (r+2)}l 2 .

Hence. or otherwise. verify that for n = 5 E(Ym) =

7(45n~ 128)p

'" 1·95p

and

var(Ym) '" 0·00695p2.

Also. show that for large n. a first approximation gives var(Ym) '" 4p2/(11 + 1)2(11 + 2)2. 85 A chord of length I is drawn at random across a circle of radius p. Find the expectation of I if (i) the perpendicular distance of the chord from the centre of the circle is PX. where x is uniformly distributed in (0 ~ x ~ 1); (ii) the chord makes an angle with a tangent through one extremity where is uniformly distributed in (0 ~ ~ n/2); (iii) as defined in (ii). has a probability density function proportional to e(n - e) in (0 ~ e ~ n/2). For any given k. (0 ~ k ~ 2). suppose Ph P2 and P3 denote the probabilities P(l ~ kp) in the three cases. Prove that PI and P3 are both ~ P2. Comment on the results obtained.

e.

e

e

e

I

86 If XI and X2 are independent observations from a rectangular distribution in the (0. 1) interval. find the joint distribution of the statistics /I

=

XtX2

and

v = (l-x t )(l-x 2).

Hence determine the conditional distribution of II for given v. and the marginal distribution of v. Also. find directly the marginal distributions of II and v. 87 A random sample of II observations XI. X 2 ••••• Xn is given from a popula· tion having mean m and variance (12. Define s\ the least-squares estimate of (12. and verify that the expected value of S2 is (12. Suppose that the sample average x and S2 are known. but the original sample observations are not available. It is then found that an (n + l)th observation of magnitude ks (k a known constant) was erroneously ignored in the determination of x and S2. If S2 denotes the correct least-squares estimate of (12 obtained from all the (11+ 1) observations. prove that

S

2= 2[ (-11n - 1) + (11+11)" (k--;X )2] . S

88 The tails of a normal distribution with mean III and variance (12 are cut off at a distance ±k(1 (k > 0) from the mean. If (1~ is the variance of the truncated distribution. show that

Also. evaluate the P2 coefficient of the truncated distribution and prove that a sufficient condition for it to be < 3 is that k is >.)3.

3

Estimation, sampling distributions, and inference; bivariate correlation and regression

I A random sample containing an even number n of observations is drawn. The first n/2 observations are realizations of a random variable X such that

P(X = 1) = p;

P(X = 0) = q,

where p+q

= 1.

The second n/2 observations are of another random variable Y such that P( Y = 1)

= q and P( Y = 0) = p.

If r is the number of ones in the sample, show that E(r)

n

="2

and

var(r)

2 A random variable X is such that AT e- A P(X = r) = (

A), r.

l-e

for r

= npq.

= 1,2,3, ...,

;, being an unknown parameter. A random sample of size N contains nr observations having the value r. Show that 00

,1,*

=

L:

r. nr/N

r=2

is an unbiased estimate of A. 3 In a sequence of 11 Bernoulli trials with a probability of success p, (q == 1- p), /' successes were observed. (i) Obtain an unbiased estimate of pq, and find its variance. (ii) Find an unbiased estimate of pq2. (iii) Show that p*(1- p*j2, where p* = 1'/11, is not an unbiased estimate of pq2, but that the bias --+ 0 as 11 --+ 00. 4 In tossing a coin with a probability p for a head, a sequence ofr consecutive heads followed by a tail is known as a run of length r for (r = 0, 1,2, ... ,11), where (0 < p < 1). Calculate the mean and variance of the run length r. Also show that when 11 --+ 00,

E(r)--+(l~p)

and

var(r)--+(1~p)2'

and find an unbiased estimate of this limiting variance. Sl

EXERCISES IN PROBABILITY AND STATISTICS 52 5 In a series of Bernoulli trials with a probability of success p, a sequence 01 successes up to and including the first failure is known as a "turn". If S denotes the total number of successes in n turns, prove that the mean and variance of Sn are

np 1- P

an

np

d

. I respective y.

(1- p)2

Hence prove that an unbiased estimate of var(Sn) is

Sn(n + Sn)/(n + 1), and that a reasonable estimate of p is p* = -Sn- .

n+Sn

6 In a sequence of n Bernoulli trials with probability of success p, (q == 1- p), r successes were observed. It is desired to estimate the ratio p/q, and two estimates are proposed: r

and

Il-r

r

n-r+l

Show that

E[_r] = !!. n-r q var

r] [---=n r

E[ _r n r+ r

[1 +~P,=2 f Il,/q'] '

1[

= 1:

(t -1) L<Xl -,.Il, - ( ,=2 L Il,/q' )2] ; q 00

q ,=2

and

1] = !!.(1_pn) = ~ [1 + (P+A.) f Il,/A.'], q P ,=2 .I\.

var [ ll-r+1

]

(p+A.)2[

= -A.-

<Xl

(t-1)

(<Xl

'~2-A.-,-·Il'- '~2Il'/A.'

)2] ,

where Il, is the tth central moment of p* == r/n, (0 < p < 1), and A. == q+ l/n. 7 From a large lake containing an unknown number N of fish, a random sample of M fish is taken. The fish caught are marked with red spots and released into the lake. After some time, another random sample of n fish is drawn and it is observed that m of them are spotted. Show that P(N, m), the probability that the second sample contains exactly m spotted fish, is given by

By considering the ratio P(N, m)/P(N -1, m) deduce that the maximumlikelihood estimate of N is the largest integer short of nM/m. [It may be assumed that there was no change in the fish popUlation of the lake in the interval between the drawing of the two samples.] 8 For the logarithmico-normal distribution defined by the probability density function

f(X = x) =

~

uj2ic

exp

{-2~2(10g.,X_I1l)2} v

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

53

. the range (0 ~ X < co), show that the maximum-likelihood estimates of In d (12 are /II an 1 n m* = g and (12' = - L (log. Xj- g)2, nj=I

where g is. the natural logarithm of the geometric mean of the random sample observations x l' X2" •• , Xn • 9 A continuous random variable X defined in the range (0 ~ X < co) has, for X == x, the probability density function proportional to x e- x /6, (lJ > 0). Find the mean and variance of X. If a random sample of n observations x .. X2"" , Xn is given from this. population, obtain the maximum-likelihood estimate of the parameter lJ and calculate the variance of the estimate. Show also that 1 n

-'L xf 3n j= 1

is an unbiased estimate of var(X), but that the estimate ti2 has a bias of 0(11- 1), where i is the sample average.

10 A random sample of n observations Xl' X2" .. , Xn is given of a random variable X having, for X = x, a probability density function proportional to x'(I-x) defined in the range (0 ~ X ~ 1), where the parameter a is unknown. Show that a*, the appropriate maximum-likelihood estimate of a, is given by the equation * _ (3g+2)+(g2+4)t a ' - 2g

where g is the natural logarithm of the geometric mean of the sample.

11 A random variable X has the probability density function

{I

}

-1 f(X = x) = _x__ . exp - -2 (log. X - p)2, (1Ji1c 2(1

for X ~

o.

From this population, a random sample of size 11 has mean i and variance Use the method of moments to find estimates of p and (12. Also show that the mean of the distribution is greater than the median.

S2.

12 A continuous random variable X, defined in the range (0 ~ X ~ 1[/2), has, for X = x, a distribution function proportional to (l_e-,.. inX), where IX> O. Find the probability density function of X. Given a random sample of n observations from this distribution, derive the maximum-likelihood equation for a*, the estimate of a. Also, prove that the large-sample variance of a* is

4a 2 sinh 2 a/2 11(4 sinh 2 a/2 _( 2 )' 13 If g(X) is a function of a random variable X having mean m and a finite variance, prove that to a first approximation

= and var[g(X)] = E[g(X)]

g(m), [g'(mW. var(X).

54

EXERC'lSES IN PROBABILITY AND STATISTICS

Extend these results to the case of k correlated variables X I, X 2,· .. ,X Hence deduce that if the X j are observed cell frequencies in a multinoOli~i distribution such that k

LX j=

j

== N (fixed),

I

then var[g(X I· X

where E(X;)

=

2·····

mi ,

X k )]

for i

=

'"

k j~'

( cg ) 2 IIlj

eX j

X,=m, -

( eg ) ]2 i~1 III; ax; X,=III, •

1 [k

N

1,2, ... , k.

14 In a biological experiment the observed frequencies in the four distinct classes AB, Ab. aB, ab were found to be Ill' 11 2 , 11 3 , 114 respectively. On a genetical hypothesis, the corresponding expected proportions are 1<2+8), i(1- lIt i(1- 8), i8, where 8 is an unknown parameter. Determine the maximum. likelihood equation for 8*, the estimate of 8, and show that the large-sample variance of 8* is 28(1- 8)(2 + 8) (1:n; == N). N(I+28) Ap alternative simpler estimate of 8 is {J = (nl-n2-n3+5n4)/2N. Prove that 8 is an unbiased estimate of 8, but that its efficiency relative to the maxi. mum-likelihood estimate is 88(1- 8)(2 + 8) (1 + 28)(1 + 68 - 4( 2 )' Hence deduce that

fJ is fully efficient only when 8 = i-

15 In large-scale sampling of a random variable X having a Poisson distribution with parameter m, it is sometimes convenient to record separately the frequency for X = 0 and to pool the sample frequency for X ~ 1. Assuming that in a total sample of N the frequency observed in the zero class is 11, obtain m*, the maximum-likelihood estimate of 111, and the large-sample variance Jf 111*. Show that the efficiency of 111*, as compared with the usual estimate of III when the data are fully recorded, is

m/(elll-I). Alternatively, if a separate sample count for X = 0 and X = 1 is given, and the frequencies for X ~ 2 only are pooled, prove that in this case the maximum-likelihood estimate m** of 111 has the large-sample variance 11l[1-(1 + m)e- m ] Ne- m (1- m + m 2 -e- m ) '

Hence deduce that for

111

sufficiently small

var(m**) ~ (1-tm) var(m*). 16 The blood-group classification of human beings can be made into four mutually exclusive types 0, A, B, AB with relative frequencies proportional to 1'2, (p2+2p1'), (q2+2qr) and 2pq respectively, where (p+q+1' = 1). In a large sample of N persons tested, the observed frequencies in the four types were found to be /1" n2' n3 and n4'

55

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

Without using the method of maximum likelihood, obtain simple approxite estimates of the proportions p, q and r. Also, determine the large-sample rna ' variances of th e estimates. plant-bre~ing experiment, the observed frequencies of progen~ in ~ ur mutually exclusive classes were ai' a2, a3' a4 (l:aj == N). On a genetical

11 In a

h~pothesiS, the corresponding expected probabilities are Ii. (2 + p), l6 (2 - p), 3 (1-p), k (1 +3p), where p == (1-8)2 and 8 is an unknown parameter.

n Obtain the maximum-likelihood equation for p, and show that the largesample variance of the estimate

p is

.i. .(4- p2)(1_ p)(1 + 3p) 3N

(5+2p-4 p2)

Hence derive the large-sample variance of 0, the maximum-likelihood estimate of 9. Another, simpler method of estimating p, suggested by Immer, is to solve the quadratic equation

a l a4 a 2a3

(2+p)(I+3p) 3(2-p)(l-p)

for p. If p* denotes this estimate, prove that its large-sample variance is the same as that of p.

18 Two varieties of plants were used separately in two similar breeding experiments, and progenies of types A and B were obtained in each case but with different relative frequencies. In the first experiment the observed numbers in the A and B categories were nl and n2' and in the second experiment n3 and n4 respectively. On a biological hypothesis, the corresponding expected probabilities of the A and B progenies in the two experiments are k(4-9), i (4+8); and -h (12+8)(4-8), -h (4+8)2,8 being an unknown parameter. Obtain the maximum-likelihood equation for lJ*, the estimate of 9, and show that the large-sample variance of lJ* is (16 - ( 2 )(12+ lJ)

19 The probability density function of a continuous random variable X is a weighted sum of two normal distributions with means 0 and J.I. > 0, and variances O'I and O'~ (espectively, the weights being in the ratio of 1 : A., where O
m,

J

e- tr2 dt

o

=

A.

J

e- tr2 dt.

0

Hence prove that a sufficient condition for E(X) > mis 0'2 ~ 0'1' Given a random sample of N observations of X, where N is sufficiently large for assuming that the variance of the sample median x is 1/[4N(medial ordinate)2], prove that the condition for the sample mean x and x to be of

56

EXERCISES IN PROBABILITY AND STATISTICS

equal precision is 1- 2 {_I -_ 2 2}]2 _~. (1+),)4 p 2 [pexp(-2"m/O'd+..1.exp !(m J.l) /0'2 - 2 (l+..1.)(1+Ap2)+i,jl2/O'f where p == 0'2/0'1' If J.l = 0, show that this condition reduces to a quartic in p which has two positive roots PI < 1 < P2 such that if P lies outside the interval (Pi> Pl) then var(x) > var(x). 20 A random variable X is uniformly distributed over the interval (O'~l where 0( is an unknown parameter. A sample of 11 independent values of is given, and these observations, arranged in order of increasing magnitude are denoted by XI' X2,' •. ,Xn • Derive the sampling distribution of x" th; rth-order statistic of the sample, and verify that

i

rO(

E(x r) = (11+1)

and

var(x r )

=

r(lI - r + 1)0(2 (11+1)2(11+2)'

Hence prove that (i) Xn as an estimate of 0( is biased. but the bias -+ 0 as 11 -+ 00; and (ii) for 11 odd, both the sample mean and median are unbiased estimates of 0(/2, and the ratio of their variances is (11 + 2)/311. 21 In the detection of a rare human hereditary abnormality which does nOI necessarily show up in the offspring, suppose p is the unknown proportion of abnormal children in families of size k (fixed). To exclude doubtful families, only those of size k are sampled in which there is at least one abnormal child. If, in all, 11 independent families are observed and there are ri abnormal children in the ith family (i = 1, 2, ... ,11), use the method of moments to obtain an estimate of p, and show that the large-sample variance of the estimate is pq(1_qk)2 ,where p+q lIk(l- qk-k pq k I)

=

1.

22 An entomologist wishes to estimate the proportion p with which an unusual form of a certain species of beetle occurs in a particular region. He catches individual specimens of the species until he has obtained exactly r (pre-assigned) of this form, and his total sample size is 11. Show that the random variable X representing the sample size has the probability distribution P(X = 11)

=

C=:)

pr qn-r

for 11

~

r,

(p+q

= 1).

Hence prove that the maximum-likelihood estimate of p is r/lI. Also, verify that lI/r is an unbiased estimate of p -I, but that

23 In a study of an abnormality of blood-cells which affects a proportion p of red corpuscles, counts were made until r (fixed) abnormal cells had been

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

57

probability that exactly n cells were counted to give rded prove that the reCo· I· this quota of abnorma s IS

P(n) = G=~) p" q"-'

~ r,

for n

Show that, for non-negative integral values of IX and function g(IX, P) = -

= 1).

(p+q

(n-IX-p-l)f(n-l) r-IX-l r-l

p,

the expectation of the

is pll.qfl;

and hence deduce, as a particular case, that

p* =

(:= !),

(r> 1)

is an unbiased estimate of p. Also, verify that the mode of the sampling distribution of p* lies in the range

p < p* < P

[1 + r- f-p],

so that the distribution has slight asymmetry. Finally, prove that var(p*) = p2

f if ~(r+S-l),

s= 1

/

S

I

and that

p*2(1_ p*)/(r-l- p*) is an unbiased estimate of this variance. 24 The random variables X and Y have a bivariate normal distribution with parameters (ml, m2; 0"1' 0"2; p) in standard notation. Prove that for positive integral values of rand s /l,s

== E[(X -ml)'(Y-m2Y] =

C1'iO"s2

±(~)(1_p2yI2

j=O

pS-iVjV,+s_j,

J

where Vk is the kth central moment of the univariate unit normal distribution. Given a sample of n observations (Xi> Yi) for (i = 1,2, ... , n) from this bivariate population, two possible estimates

1

Ii = ni~1 n

(XI) Yi

i~1 X~~1 Yi n

and

T2

=

n

are proposed for the parameter A. == mtlm2. Find the expectation of Ii and 12· Hence show that only 12 asymptotically converges to A. as 11 -+ 00, and that for large samples 2 var(12) ,.., -A. [ (VI-PV2) 2 +(1-p 2 )V22] , n

VI

and V2 being the coefficients of variation of X and Y respectively.

58

EXERCISES IN PROBABILITY AND STATISTICS

25 A biologist wishes to determine the effectiveness of a new insecticide when used in varying degrees of strength. It is known from certain empirical con. siderations that p(x), the probability of killing an insect at strength x of the insecticide, in a fixed interval of time, is given by the relation 10ge[ 1 ~~:x)] =

a+x, where r:x is an unknown parameter.

To estimate a, the insecticide is used for a fixed period on three random groups of 11 insects each, and the groups are subjected to different strengths of the insecticide, which, on an appropriately chosen scale, correspond to the values x = - 1, 0, 1. If the total number of insects killed in the three groups is r, show that a*, the maximum-likelihood estimate of r:x, is obtained from the cubic

y3 + (1- 4>)(e + 1 +e- I )y2 + (1- 24>)(e + 1 +e- I )y+(I- 34»

=

0,

= -loge y, 4> =l1/r, and e is the Napierian constant. 26 In an animal-breeding experiment four distinct kinds of progeny were observed with the frequencies and (Ll1j =N). The corresponding where

r:x*

Ill' 112' 113

114

expected proportions on a biological hypothesis are i (2 + p), i (l - p), i (1- pl tp, where p is an unknown parameter. Obtain p*, the maximum-likelihood estimate of p, and verify that its large-sample variance is

2p(1- p)(2 + p) N(l +2p) Show further that an estimate of this variance, obtained by substituting

p* for p, is not an unbiased one, but that its bias relative to unity is approxi· mately -2(5 +3p +6p2 +4p3)

which

.--------:-----0.-------,

N(l +2p)3

27

A

r

--+

°

as N

--+ 00.

variable X has the probability density function 1 ( f( X = x) = -_. e- x /a . ~ ar(p)

)P-I

a'

for

°~ X < ""

00

.

Given 11 independent observations XI' X2" •• ,X n of X, prove that the expecta· tions of the sample arithmetic and geometric means are and

ap

a.

[r(p+~)/ r(P)r

respectively.

Hence deduce that the ratio of the population arithmetic and geometric means is (J

=p

e-(P),

where 4>(P)

d =-[log r(p)]. dp

Also, show that (J*, the maximum-likelihood estimate of (J, is the ratio of the sample arithmetic and geometric means.

ESTIMATION. SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

59

If the parameter a is known and only p is estimated, obtain the large.ample variance of p, the estimate of p, and thereby prove that the large:~mple variance of &, the estimate of (J in this case, is (J2[p -

I -

t/>'(P)J2 /Ilt/>'(P).

28 An experiment results in six independent observations Yr (r = 1,2, ... , 6) such that . (2nl') E(Yr) =- ex cos (2nr) 6"" +P SID 6"" ; var(Yr) = (12. Find the least-squares estimates of ex and estimates has variance (12/3.

p,

and verify that each of these

29 A chord I is drawn at random across a circle of radius p, such that it passes through a given point on the circle and makes an angle (J with the tangent to the circle at the given point. Find the expectation and the variance of /, the length of the chord. Suppose p is not known, and it is desired to obtain its estimate from " independent measurements I" 12 , • •• ,In of chords as drawn above. Show how these measurements can be used to obtain an unbiased estimate of p, and also find the variance of this estimate. 30 A target P moves randomly on the arc of a quadrant of a circle of radius r and centre 0, such that OP makes an angle (J with the horizontal, where (J is known to have a probability density function proportional to (J(n - (J) in the range (0 ~ (J ~ n/2). If PM denotes the perpendicular on the horizontal lhrough 0, show that .::\, the area of the triangle OPM, has the expected value 3r2(n 2+ 4)/8n 3 • If /' is unknown, and a random sample .::\1' .::\2, .•. ,.::\n of 11 values of .::\ is given, obtain an unbiased estimate of r, and hence prove that the estimated area of the quadrant is 2n4.::\0

3(n 2 +4)' where .::\0 is the mean of .::\" .::\2,· .. ,.::\n.

31 The

311

ZI' Z2"'"

Zn'

E(xj)

independent observations x" X2'" . , Xn; YI' Y2'" . ,Yn; and each have the same unknown variance (12, and

=

ml

;

E(Yj)

=

m2 ; E(zj)

=

ml +m 2, for i

= 1,2, ... ,11.

Use the method of least squares to obtain the unbiased estimates of ml and m2' and hence derive the best estimate of (12 based on the total available degrees of freedom. Also, show that the mean square for testing the hypothesis H (1111 = 1112) is 11

2

i(X- ji) ,

where

x and ji are the means of the x and Y observations respectively.

32 The

311

zIt Z2.'''' Zn.

E(x j)

=

independent observations XI' X2" .. ,xn; YI' Y2" .. ,Yn; and each have the same unknown variance (12, and ml

;

E(yj)

=

m2 ; E(zj)

=

m l -m 2,

for i = 1,2, ... ,11.

60

EXERCISES IN PROBABILITY AND STATISTICS

Obtain the least-squares estimates of m, and m2 , and hence derive the best estimate of (12 based on the total available degrees of freedom. If it is proposed to test the hypothesis H (m, = Am 2 ), A being a know proportionality factor, then show that the appropriate generalized t statistic i~

t

=

[(2-A)X+(1-2A)y+(1 +A)Z]jn S[6(A2-A+ 1)]"1'

where t has the t distribution with (3n - 2) dJ., S2 is the least-squares estimate of (12, and X, y, z are the means of the x, y and z observations respectively. 33 The 3n independent observations X"X2""'Xn ; Y"Y2,""Y' ... , Zn, have the same unknown variance (12, and their expectatio~; depend upon three independent parameters Oh O2 , and 03 in such a way thaI

Z h Z 2,

E(Xi) = 0, +0 2 +0 3 , E(Yi)

= -0, +0 2 +03 ,

E(Zi)

= -20 2+0 3 , for i = 1,2, ... , n.

and Use the method of least squares to obtain the unbiased estimates of lJ h 02 and 03 , and also the best estimate of (12 on the available degrees of freedom. Also, show that the mean square for testing the hypothesis H (0, = O2 = 0l) can be put in the form

where Y, == y+z and Y 2 == (2x-3y+3z)/JIT, x,y,z being the means 01 the x, Y and Z observations respectively. 34 The 11 observations x" X2,"" Xn are from a population with mean III and variance (12, and the correlation between any pair of observations is constant and has coefficient p. If

where hand k are unknown constants, is an unbiased estimate of (12, sholl' that n

T=

L (Xi- X)2/(l_p)(Il-I),

n

where /lX ==

i= ,

35

LXi'

i= ,

If y" )'2" .. , Yn are independent observations such that E(Yr)

= rO

and

var(Yr)

= r3 (12,

for r

= 1,2, ... , n,

derive the least-squares estimate of the parameter 0 and obtain the variance of the estimate. Hence show that for large /I this variance is asymptotically (12/(loge n). 36 Three parcels are weighed at a post office singly, in pairs and all together. all weighings being independent and of equal accuracy. These weights are

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE. ETC.

61

(" . k == 0, 1), the suffix 1 denoting the presence of a particular parcel

°

~ijk '~' suffix denoting its absence. and~b~ain explicit expressions for the least-squares estimates of the weights the variances and covariances of the estimates in terms IIe Parcels, giving or " l b' or tthe variance of t h e ongma 0 servatlOns.

7 There are 3n independent observations XI' X2" .. ,X n ; YI' Y2" .. ,Yn; and J z each observation having the same unknown variance ci 2 • The ; I' ;2.1' 'v';I~e~' of the observations are given by: mea

E(Xi)

=

0 1 +20 2+30 3,

E(yJ

=

20 1 +30 2 +03,

and where 01 , O2 , and 03 are unknown parameters. Use the method of least squares to derive estimates of the contrasts (/11- 02 )' (0 2-.0 3) and (0 3 -Od, and he.nce also the unbiased estimate of (f2. H it is deSIred to test the hypothesIs H (0 1 = 021a = ()3Ib), where a and b are known constants, prove that the requisite mean square for testing this hypothesis is

where X,

y, z are the averages 0*

of Xi' Yi and Zi respectively; =

(}'IX+A2Y+';'3 Z). (Af + A~ +A~) ,

and

Al == (l+2a+3b);

A2 == (2+3a+b);

A3 == (3+a+2b).

38 A sample of n independent observations y" Y2" .. ,Yn is given of a normally distributed random variable Y such that

E(Yr) = lX+p.r(xr-x)

and

var(Yr) = (f2

ror (r = 1,2, ... , n), where IX and p are unknown parameters, Xr are values of a non-random variable, and (f2 is an independent unknown parameter. Obtain the least-squares estimates of IX and p, and hence derive the best estimate of (f2. Also, calculate the variance of the estimate of p and deduce the appropriate test statistic for testing the hypothesis H (P = Po). 39 Given a random sample of n pairs of observations (xj, Yi), (i = 1,2, ... , n), show how the method of least squares may be modified for fitting a straight line of the type

IXx+Py+l

=

0

by minimizing the sum of the squares of the perpendicular distances of (x;. Yi) from the straight line. Hence show that the estimates of IX and pare /I

and b respectively, where b = I/(mx - y),

ESTIMATION. SAMPLING

where

111*

INFERENCE. ETC.

63

is the best estimate of m; and

(I +1l2-l)S2 I)

DISTRIBUTIONS,

=(A.Il

I

+n 2)[A.

i

(xr-x?+

r;1

I

(Yr-y)2] +;,1I11I2(X-y)2,

r;1

_ - being the sample averages of the x and Y observations. =1= 1, the varian~e of m* is less than the variance of the arithmetic mean of the (II) + n2) observatIOns.

x. }~ISO, prove that, for A.

43 If a random variable t has Student's £(12) =

t

distribution with

I'

dJ., prove that

C. ~ 2)-

The random variables X and Y have a bivariate normal distribution with ncan s III , Ill., variances 0';, 0'; and correlation p. Given a random sample of " paired ob~ervations (Xi' Yi), for (i = 1,2, ... , IIJl, and a further independent ~l~J11ple of 112 observa~io~s on X only, Y not being recorded, an estimate of /II" is given by the statistic

T= YI +hdx-xd, where XI' YI and hi are the sample means and the regression coefficient of Yon X calculated from the first sample, and x is the mean of the X observations in both samples. Show that T is an unbiased estimate of m)' and that var(T)

=

(0';

nll11+n2

)[1I1+112(I-p2)(111-23)]'

nl-

44 A random sample of N observations is given from a normal population with mean 1/1 and variance 0'2, both parameters being unknown. Suppose S2 is the usual unbiased sample estimate of 0'2, v2 any given positive constant, and /I is the smallest integer satisfying S2

11

~ 1: -

v

and n

N,

~

L

If another independent sample of n observations is taken from the same population, show that by using the two sample means another unbiased estimate of m can be derived whose estimated variance is for large N asymptotically equal to v2 •

45 If X2 has the X2 distribution with

II

E[X/jVJ '"

dJ., show that for large

II

(1- 4~)'

A random sample of n observations is given from a univariate normal population with coefficient of variation A.. Prove that the sample coefficient of variation (sample standard deviation/sample mean) is an asymptotically unbiased estimate of }, with the large-sample variance

A. 2(1 + 2A. 2)

2n 46 A sample of II independent observations Xl' X2" •• , Xn is given from a normal distribution with mean m and variance 0'2, and the statistic d2 is

64

EXERCISES IN PROBABILITY AND STATISTICS

defined as n-1

d2

=

I

(Xi+1- X j)2/2(1I-1).

i= 1

Show that d2 is an unbiased estimate of q2, and by determining its variance prove that its efficiency is 2(11-1)/(311-4) as compared with S2, the usual least-squares estimate whose variance is 2q 4/(n-1).

If Xl' X2" .. ,Xn are 11 random variables such that

47

E(xj)

= m, var(xj) =

for i

q2,

1,2, ... , n,

=

and cov(Xj, x j) = pq2,

for i "# j,

show that (i) var(x)

= [1 +(n-1)p]q2/n, where x is the average of the Xj; (ii) E[1:/= I (Xj- X)2] = (11-1)(1- p)q2; and (iii) - 1/(11-1) ::::.; p ::::.; 1.

48

Given two linear functions n

L1

I

=

n

ajXi

and

L2

=

i= 1

L bjxj, j=l

where the a's and b's are constants and the Xj are random variables such that E(xj)

=

IIlj,

var(xj)

=

cov(x j , Xj) = q2Pii' (i"# J),

and

q2,

find the variances of L1 and L2 and also their covariance. Further, calculate the variance of the function (L 1 - L 2 ), and then show that if all Pii = p, this variance reduces to q2

where

~,

[(1- p) J1 ~; + pet ~,) 2]. 1

== (a, - b,) "# 0 for all r, and hence deduce that n

L

~;

--,--'=-,,1__ ::::.;

LL 49

p ::::.; 1.

0(,0(.

If Xl' X2,"" Xn are n independent random variables such that E(x)

= mj , var(x) =

q2

for j

=

=

I

1,2, ... , n,

prove that the random variables n

n

Y1

=

I

ajXj

Y2

and

j= 1

bjXj,

i= 1

where the a's and b's are constants not all zero, are uncorrelated if n

I

j= 1

ajb j

=

O.

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

Hence, or otherwise, show that the arithmetic mean any deviation (Xj - x) from the mean. Also. prove that var(xj-x) = 0'2(1-1/11), and

cov[(Xj-X), (xj-x)] = -a 2 /n,

50 If Xl> X2" E(xj)

=

••

x is uncorrelated with

for i :F j.

,Xn are random variables such that var(xj)

mjo

65

=

cov(Xj, Xj)

0'2,

=

pa 2 ,

for i :F j

= 1,2, ... , n,

obtain the variance of the linear function n

L =

L

ajX"

j= 1

where the

aj

are constants not all zero.

If the Xj are divided into two distinct groups of (v. +V2 = n) to define two new random variables

L Xj, j=1

and

V2

elements

n

VI

SI =

V1

and

S2 =

LXi> j=vl+1

prove that corr(S1,S2)

=

Also, when n -+ 00 such that value of this correlation.

P[{I+(Vl-l);}~~+(V2-1)p}r. V 1 -+ 00

but

V2

remains finite, find the limiting

51 Of two discrete random variables, X can take only the values ±a and Y the values ±{J, with unequal probabilities for the four possible realizations of the pairs (X, Y). In a random sample of N observations, the frequencies corresponding to (-a, -fJ), (a, -{J), (-a,{J) and (a,{J) were found to be "., 112' n3 and 114 respectively, (Ll1j = N). Prove that the sample productmoment correlation between X and Y is P'3-A1A2)/[(1-AD(l-A~)]t,

. 1.1

. 1.2

where and are orthogonal linear contrasts of the observed relative frequencies associated with the sample means of and Y, and is the contrast Hence show that the sample correlation vanishes orthogonal to both and when

A1

X

. 1.2,

A3

By considering the limit of the relative frequencies in this equation as N -+ 00, deduce that for the joint distribution of X and Y, zero correlation in the population ensures the statistical independence of the random variables.

52 From a finite collection of N balls of which M «N) are white and the rest black, two successive random samples of size n1 and n2 respectively are drawn without replacement. If the random variables X and Y denote the number of white balls in the two samples, prove that P(X = r, Y= s) =

(:1)(:2)(N~~~=:2)/(Z),

and indicate the limits of variation of rand s.

66

EXERCISES IN PROBABILITY AND STATISTICS

By considering the appropriate array distribution, show that E(yIX

=

r)

= "2(M -r)/(N -"1),

and hence that corr(X, Y) = - [(N _

":)~~ _ "2)] t.

Also, find var(ylX = r) and deduce that this variance can never exceed "2(N -"1 - "2) 4(N -"1 -1) . 53 Each of two packs A and B has N cards which are of t different types, the cards of any type being indistinguishable. Pack A has a cards of each type (at = N), and pack B has bj cards of the ith type Cr.b j = N; 0 ~ bj ~ ~ For a random arrangement of the two packs, a "match" is said to OCcur in a specific position if in that position the cards in A and B are of the same type, Suppose that X and Yare random variables associated with any two cards of B, each taking the value 1 or 0 according as a match is or is not observed in the corresponding positions of the cards. By considering separately the two cases when the cards associated with X and Yare or are not of the same type, derive the bivariate distribution of X and Y. Hence prove that for the marginal distribution of X E(X)

=

l/t;

var(X) = (t-1)/t 2 ;

and that where I

(t-l)V(b j ) ==

L j=

(b j -Nt- 1)2,

1

so that the correlation is a maximum when b j = N It. Also, if SN is the random variable denoting the total number of matches realized in a random arrangement of A and B, use the above results to obtain var(SN), and then establish that the maximum value of this variance is N 2 (t-1)/(N -1)t 2 . If X and Yare correlated random variables with correlation p and coefficients of variation v I and V2 respectively, prove that, as a first approxima· tion,

54

where ), is the ratio of the expectations of X and Y. Further, assuming that the joint distribution of X and Y is symmetrical. obtain an approximate expression for the bias in the value of E(X/Y) used in deriving the variance. 55 If x and S2 are the usual sample mean and variance based on a sample of 11 independent observations from a normal population with mean m and variance (12, prove that the correlation between x and Student's t statistic (x - m)/s is

In

(";3Yr(";2) / r("; 1).

ESTIMA TION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

by using the AIso,

67

r function approximation r[(v + 1)/2)]

(

(v/2)tr(v/2) '"

I) 1- 4v

for large v,

veri'fy that for large samples corr(x, t) '"

(1- 4~).

56 From a bivariate normal distribution of the random variables X and Y, ith parameters (m", m" (1", (1" p) in standard notation, a random sample of n wbservations (Xi' Yi) for (i

= 1,2, ... , n) is given. Prove that the sample product-

~oment correlation coefficient r is invariant under linear transformation of the observations. Hence deduce, without obtaining the sampling distribution, that the probability density function of r can only involve the parameter p. If X and Yare transformed to unitary orthogonal variables Wand Z with the sample product-moment correlation coefficient R, then show that r2(1_ R2)

[

R2(I_r2) =

Pf] 2 1+ R(I- p2)t '

where f2 is the ratio of the sample sums of squares of the Wand Z observations.

57 The volume (V) of a duck's egg is known empirically to be proportional 10 (Iength)(breadth)2, where length (X) and breadth (Y) may be regarded approximately as having a bivariate normal distribution with coefficients of variation ),' P for X and Y respectively, and correlation p. If 2 and pare sufficiently small for powers greater than the fourth to be negligible, find the coefficient of variation of V. and verify that as a first approximation its value is {2 2 +4p2+42Pp)t.

Also, determine y, the coefficient of variation of the ratio Y/X, to the same degree of approximation, and hence show that an alternative form of the lirst approximation for the coefficient of variation of V is (32 2 + 6p2 - 2y2)t.

58 For the binomial distribution with constant probability of success p, prove that y" the probability for r successes in n trials, satisfies the finite difference relation

Ay, y,+Y,-l

=

(n+ I)p-r (n+l)p+rO-2p)'

If 0 < p < t, show that this equation leads to a limiting r distribution approximation for the binomial distribution such that the statistic 4[r(l- 2p) + (n + I)p]

(l_2p)2 may approximately be regarded as a X2 variable with dJ.

[ 4{n+l)p(I-P) (1-2p)2 +

I] .

68

EXERCISES IN PROBABILITY AND STATISTICS

59 By using a suitable approximation in the region of integration of tilt double integral

f

" a

[4>(a)]2

= 21nf e- t (X 2+ y2 )dydx, (a>

0),

o0 prove P6lya's approximation that

f fo a

-I-

d ' " 21 (I -e - 2a 2'1t)t . e-tx2 X"

o

Hence show that the sample median obtained from (2v + I) independelll observations of a unit normal variable is approximately normally distributed with zero mean and variance nl(n+4v). Verify that the asymptotic efficiency of the median as compared with the sample mean is 21n.

60 Given a random sample of (2v + 1) observations from a unit normal population, prove that the probability distribution of the sample median x is proportional to [F(x) {1- F(x)} ]" dF(x),

where F(x) is the distribution function of a unit normal random variable. By using the Cadwell approximation

F(z) {1 - F(z)} '" e - (2,,,)%

2 [

2(n - 3) (7n 2 - 60n + 120) I + 3n2 Z4 45n 3 Z6 +

...] ,

show that a scalar transform of the sample median defined by

(4V;nr.

y =

x

has the approximate normalized probability distribution

1

(1 +3k)

fo' e-ty-' (1+ky4)dy, 2n

(-co < y < co),

where k == 2( 7T - 3)v/3(4v + 7T)2. Hence verify that with this approximation for the distribution of x var(x) -

(

7T

7T+4v

)[

8(7T-3)V] 1 + (7T + 4v)2 ,

and, for the kurtosis of x,

16(n-3)"

f2(X) '" (n + 4,,)2 .

61 If x and yare independent random variables such that x has a unit normal distribution, and y is distributed as with n dJ., prove that the ratio

·c

xjn

t=--

.JY

has Student's t distribution with

II

dJ.

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC.

69

Assuming that for sufficiently large n

r{(n+ O/2}

1)

(

1- 411 •

(1I/2)tnn/2) -

derive Bartlett's approximation that (n -t)loge [1 +(t2 /n)] is distributed as -I with 1 d.f.

62 If PI' P2'· .. ,Pk are probabi~ities derived from the r~alizations :X.I' X2.··· ,.Xk of the independent random vanables X 1> X 2.· ..• X k With probabilIty denSity functionsfI(xl),f2(X2) •... •f,.(Xk) such that

f X;

Pi =

for i = 1.2•...• k.

fi(t) dt.

-00

prove that the Pearson statistic k

p=

2: i=

2 loge Pi

1

is distributed as X2 with 2k dJ. 63 A sample of II independent observations of a random variable Y is given. where Y has the probability density function

!(y= y) =

e-

l .e-(Y-fJ)/9

in the interval Y ~ JI. and zero otherwise. Obtain the joint distribution of the largest and smallest sample observations, and hence derive the distribution of the sample range w. Verify that for any given fixed value \1'0 the probability ~

P(w

wo)

= 1_(1_e- wo / 9),,-I.

Explain how this result can be used to test a specific hypothesis about the parameter O.

64 A random variable X has, for X = x. the probability density function f'(x) defined in the interval (ex ~ X ~ {3). If a sample of II independent observa'Iions from this distribution is given. and the (algebraically) smallest and largest of these are denoted by x I and x" respectively, derive the joint distribution of x, and x". If the sample range is defined by R = XII - X I. show that the marginal . distribution of R is

f

x, +R

/l-R

g(R)dR = n(n-1)

!(X1)!(XJ+R)[

2:

f

!(X) dX

r-

2

dxl·dR.

XI

where 0 ~ R ~ {3-ex. Hence derive the distribution of R in the particular case when X is uniformly distributed in the interval (0 ~ X ~ 1). 65 A random sample of (2v+ 2) observations of a uniformly distributed random variable X in the range (0 ~ X ~ 2a) is ordered. the (v + 1)th and (1'+2)th observations being XI and X 2 • (x 2 > XI). Obtain the joint probability distribution of XI +X2

Y=--2

an

d

70

EXERCISES IN PROBABILITY AND STATISTICS

Hence prove that for the marginal distribution of y the probability density function is

y (V)

(v+l) a22Y B(v+ 1, v+ 1)' r~o r (-1)

y-r( l- y)2Y-2r [1 +_ (l- y)2r+l]j (2r+ 1), ti

ti

according as the range of y is (0 ::s;;; y ::s;;; a) or (a Also, verify by actual integration that P(O

::s;;;

y

y

::s;;;

::s;;;

2a).

t. of x and =

::s;;; a)

Starting from the joint distribution s, the sample mean and standard deviation based on a random sample of n observations from a normal population with mean m and variance (l2, prove that the sampling distribution of v, the square of the sample coefficient of variation, defined by

66

v = S2/X 2, is

n+2i)[

e-(n/2Jl)

co

(2n/A.2)i

(n-l\J~or(2i+l)'

r ( -2-

~ r -2-'

[

n

] (2J+3l12

~

n ](n+2J)/2 1+ (n-l)v

for 0 ::s;;; v < 00, where A. == (l/m. Also, obtain the distribution of w

•

(n-l) -n- dv

= six.

67 If s is the sample standard deviation based on n independent observations from a normal population with mean m and variance (l2, prove that E(s) = (lCy, where 1 1] ( 2)t r[(v+ 1)/2J [ Cy

and

== ;

.

r(v/2)

(l2 var(s) '" 2v'

'" 1- 4v + 32v 2

'

(v == n-l).

Assuming that, for v moderately large, s can be regarded as approximately normally distributed with mean (lCy and variance (l2/2v, show that the statistic

v=

Cy.t(t2~2Vr,

t being the standard Student's t statistic with v d.f., has asymptotically the unit normal distribution. Also, determine the central moments of v and so prove that the approximation may be regarded as effectively correct to O(v- 1 ). 68 Assuming Stirling's asymptotic formula for the r(n + 1) '"

(~) n • .j2im,

prove that, for n sufficiently large and h (> 0) fixed, r(n + h) '" nhr(n).

r

function,

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE, ETC.

71

1-1 nce show that if the random variable X has the Beta distribution with prob-

a~lity density function f(X

1

(0 ~ X ~ 1),

= x) = B(P,q)'XP-I(1-X)4- 1,

then for large p and relatively small q, a first approximation gives - 2p log. x as X2 with 2q dJ.

Also prove that as an improved approximation -(2p+q-l) log. x is X2 with 2q dJ.

69 For a continuous random variable X having a Beta distribution in the unit interval with parameters p and q in standard notation, Bartlett has shown that if p is large and q relatively small, then - (2p + q -1) log. x is approximately X2 with 2q dJ. By a suitable transformation of this approximation, prove that

x "'" (

2p-l )X2/24, 2p+2q-1

where X2 has 2q dJ.,

and hence, by setting

v = X2 /(2p+2q-l) and s = q/(2p+2q-l), that

x"'" e - v - s(1 +~s)v + sv 2 •

70 The independent random variables X I. X 2" ••• Xn have Beta distributions in the unit interval (0, I) which, in standard notation, have the parameters (aj' pj ) for j = 1,2, ... , n respectively, where a.j==a.j+I+Pj+l, forj= 1,2, ... ,n-1. Prove that the probability distribution of the product of the n random variables is also a Beta distribution in the same unit interval but with parameters

(a.n , ~

.± p

j ).

)=

I

Hence derive the distribution of g, the geometric mean of the n variables.

7.

If X and Yare normal correlated random variables each with zero mean and unit variance. prove that for positive constants II 1.112' k I' k 2 • P(1I 2 ~ X ~ II I ,k 2 ~ Y~ k l ) = = M(h l , k l , p)+M(h 2 , k 2 , p)- M(1I1' k2' p)- M(h 2 ,

kl' p),

where M(a, b, p) = P(X ;;:: a, Y;;:: b)

Further, if a. and

and

corr(X, Y) = p.

Pare negative, show that M(a., P, p) = t- M(a., - p, - p)-(a.),

where P(O

~

X

~ -a.)

== <1>( -!X),

72

EXERCISES IN PROBABILITY AND STATISTICS

and hence finally deduce that M(a.,p,p) = M(-a., -P,p)+(fJ(-a.)+(fJ(-p).

72 If the random variables X and Y have a bivariate normal distribution with probability density f(X = x, Y= y) =

R

2nO' I 0' 2

1- p

.exp[-2(1~ P2){x:-2P~+Y:}], 0' 0' 0'2 0' 2 1

1

prove that the marginal distribution of X and the conditional distribution of y for given X are both univariate normal. Hence, by considering the probability density contours of the normal bivariate surface, show that the contour 01 smallest area in the (x, y) plane which excludes a fraction P of the probability is given by the equation x2 xy y2 2 - 2p- + 2 = 2(I- p2)log.(I/P). 0'( 0'10'2 0'2 73 For the random variables X and Y having a joint bivariate normal distribution with the probability density function f(X

x, Y

=

=

y) =

2

1

1 . exp [ 2(I_ p 2)(x -2pxy+ y 2nJt- p 2 for

-

00

< X, Y <

21)J 00,

verify that of

02f

op

ex cy'

Further, if two new random variables

eand 11 are defined by the relations

x

,

e= Ifo·e-t2/2dt

and 11 =

Ifo·e-t2/2dt. e

prove that the marginal distributions of both and 11 are uniform in the interval (-! ~ 11 ~ !), and that their common variance is /2' Hence prove that R. the correlation between and 11. satisfies the relation

e.

e

p = 2 sin(n:).

74 A random sample of n observations (Xl. yd. (X2. Y2).' ..• (x", Yn) is given of random variables X and Y having a bivariate normal distribution with the parameters (mI. m2. 0' 1.0'2. p) in standard notation. Prove that Q2

=

~

1) f [(X i-m 0'1

p )i=l

(1

2

_2 P(X i -m 1) (Yi- m2\ + (Yi- m2)21 0'1 0'2 J 0'2

is distributed as X2 with 2n dJ. Hence infer the distribution of

sr. s~ and rs s l

2

R2 = (1(n~l~p) [(~)2 _2P(~)(~)+(~)2]. 0'1 0'1 0'2 0'2 being the second-order sample moments.

ESTIMA TION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

73

If the variances of the random variables X and Y having a joint bivariate 7S al distribution are known to be equal, show that the appropriate estimate norm . I t' . of the population corre a Ion p IS

2rs"s). u- S2+S2' "

y

where s~, s; ~nd rs"sy are the usual second-order moments based on a sample f n observatIOns. o Starting from the known sampling distribution of these moments, derive the joint distribution of u and the random variables

2sxSy v = -2--2' S"

W

+ s).

= (2 s,,+Sy2)/(I 2.

Hence prove that the sampling distribution of u is 1 2)(n-l)/2 ( - p . (1- pu)-(n-l) . (1- u2)(n-3)/2 . du,

B(n;l,t)

(-1 ~ u ~ 1),

and thereby deduce that, in this case, the appropriate Student's for testing the hypothesis of zero correlation is t

t

statistic

= uJn=l with (n - 1) dJ.

Jl-u

2

Also, prove that for p :F 0 _ 2(1_p2)(n-l)/2 00 r{(n+2j+l)/2} p2J+l E(u)- r{(n-l)/2} 'j~O r(j+l) . (n+2j)' 76 Shots are fired at a vertical circular target of unit radius. The distribution of horizontal and vertical deviations from the centre of the target is bivariate normal, with zero means, equal variances (12 and correlation p. Show that the probability of hitting the target is

(1_p2)f

p=

1t'

I

J l-exp +

P

[

{

u}]

2(12(1-p2)

du

'u{p2-(1-u)2}f'

I-p

77 If X and Yare two random variables having a joint bivariate normal distribution with the probability density function 1 f(X = x, Y = y) = 2X 21t(lI(12~

xex p[

1 2 {(X-ml) 2 +(y-m2)2 _2 p(x-m 1) (y-m2)}] , 2(1-p) (II (12 (II (12

obtain the sampling distribution of z=

(X:~I)/(Y:~2),

and hence prove that the mode of the distribution of z is p. 78 If x and yare the realizations of the two independent normally distributed random variables X and Y with zero means and variances (I~ and

74 O"~

EXERCISES IN PROBABILITY AND STATISTICS

respectively, find the probability distribution of the statistic u = x/yo

Hence prove that if Xl> X2, . .. ,Xn and YI' Y2' .. . ,Yn are realizations of X and Y respectively, then the statistics 1

and

w

= -

independ~.

n

L (X;/Yi)

n i=1

have identical distributions. Given a random sample of n observations (Xi' Yi) for (i = 1, 2, ... , n) q the random variables X and Y having a bivariate normal distribution Wilf means mx • my, equal variances 0"2 and correlation p, show that .

79

w

=

(1-p)s;/(1+p)s~

has the F distribution with (n-1, n-1)dJ., where and n

(n-l)s; ==

L (Ui- U)2;

n

(n-1)s~

==

i= 1

i= 1

n

nu ==

L (Vi-ti)2;

n

LUi;

nti

i= 1

== LVi. i= 1

80 Starting from the joint sampling distribution of the three second-order sample moment statistics s;, s: and rsx Sy derived from a bivariate normal dis. tribution with correlation p, obtain the joint distribution of

where 0" x and 0"yare the population standard deviations. Hence, integrating for r from - 1 to + 1 by putting 2 r - u(A.+Jl)-(1-u)(A.-Jl) where A. == 1+v 2 , Jl == pv, - u(A.+Jl)+(1-u)(A.-Jl)'

show that the sampling distribution of V is

81 For two continuous random variables X and Y, Karl Pearson suggested a non-skew probability surface with a density function proportional to X2 xy y2]-n [1+bcl-+2b2--+b32 ' 0"1

0"10"2

0"2

n > 0, for

-00

< X, Y <

00,

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

75

the b's are constants, and (1f, (1~ are the variances of X and Y respec\~h~re Determine the proportionality factor of the joint distribution and derive

II\'~ ~arginal di~tributions of. ~ and Y.

.

Ih~ By eonsidermg the condItIOnal expectatIons of X and Y, prove that for . distribution the product-moment correlation coefficient between X 11115 • ,II1d Y IS lienee show that the joint probability density function of X and Y is

_

21tlTI(12

R'(~) [1+ 1 2 {X:_2P~+y:}]-n, 1_p2 n-2 2(n-2)(1- p) (11 (11(12 (12

where 11 == 3(P~ - ~)/(~2 - 3), P2 being the common value of the kurtosis of the marginal dIstrIbutIOns of X and Y. What happens when P2 -+ 3? 82 The Pearson bivariate distribution of the random variables X and Y is defined by the probability density function f'(X == x, Y = y)

=

1 (n-1) [ 1+ 1 {X2 - - 2 pxy - +y2}]-n - 27C(11(12J1- p2' n - 2 2(n - 2)(1- p2) (1f (11 (12 (1~

-

for (- 00 < X, Y < 00). Obtain the marginal distribution of X, and hence show that the conditional variance of Y given X = x is 2 2 [ 1 {X2}] (12(1-p) 1-(2n-3)' 1- (1f

and find the expectation of this conditional variance. Also, prove that the probability of a random observation falling outside the equi-probability contour x2 xy y2 z - 2 p - + z = 2(n-2)I/1 2(1-p2) (11

(11(12

(12

is 1/(1 + I/1 2)n- I. 83 A form of a bivariate distribution for the random variables X and Y proposed by Karl Pearson is defined by a probability density function proportional to

n > 0, in which (1f and (1~ are the variances of X and Y, and the b's are constants. Show that in this case the equi-probability contours must be ellipses, and obtain the marginal distributions of X and Y, indicating the permissible range of variation of the random variables. Hence prove that the joint probability density function has the form

76

EXERCISES IN PROBABILITY AND STATISTICS

where p is the correlation between X and Y, and n == 3(fJ 2 - 2)/(3 - P ~ being the common value of the coefficient of kurtosis of the marginalll tributions of the random variables. IS. Discuss the limiting forms of the joint distribution when fJ2 -. 2 alld fJ2 .... 3. 84 For the joint distribution of the random variables X and Y with the probability density function 1 ------===x

2:n:ala2~ x

(::~) [1

n > 0,

obtain the marginal distribution of X, and hence show that a;.x, the con. ditional variance of Y, given X = x, lies on the ellipse 2

a},.x 2(1 2 (2n+4) a2 -p) 2n+3

2

x_I + (2n+4)a~ - .

Further, prove that the probability is half that a random observation falls outside the equi-probability contour

85 A farmer, interested in increasing the output of his potato crop, experi· ments with a new fertilizer that is claimed to give an appreciable increase. He uses the new fertilizer on n equal farm plots, and obtains the individual yield figures. From past experience, the farmer knows that his average returns have been m Ibs per plot. Explain the method for analysing the data, stating the assumptions underlying the test of significance for the hypothesis that the new fertilizer has given no different average yield from the past. Suppose the test of significance gives a barely significant result at the 5 per cent level, and the farmer considers that this is not sufficiently convincing evidence for him to introduce the new fertilizer. The farmer then wishes to conduct another similar experiment and desires to know how many experi· mental plots he should take to ensure that differences as large as the one already observed would be significant at the 1 per cent level. State the pro· cedure for obtaining the requisite number on the basis of the available experimental evidence. If the farmer suspects rather large seasonal variations affecting his harvest and also considerable differences in the plot fertility, suggest another experimental procedure and its statistical analysis for testing whether the new fertilizer does, indeed, give a markedly better yield. [It 'Pay be assumed that there is no practical difficulty in dividing each experimental plot into two equal parts.]

86 A laboratory carries out regularly tests for the assessment of the breaking strength of cement mortar briquettes produced in large lots of relatively homogeneous kind; and in an investigation to compare the accuracy of technicians A and B working in the laboratory, two samples of size n 1 and n2

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

77

(vely are obtained from tests conducted by them on randomly selected

rc~pec t~es. Explain how the sample data can be analysed to test the hypothesis I1nqu~ifference between the accuracy of A and B, stating explicitly the assump(If 110 underlying the test of significance. If the observed result is significant lion: I per cent level, show how a 95 per cent confidence interval can be ;II I ~~ed for the parameter measuring the relative accuracy of A and B. ol1l~uggest another experimental procedure and its statistical analysis for paring the accuracy of A and B when it is known that briquettes proll'omed in different lots tend to have different breaking strengths. ( lie 87 In a single tossing of a penny the probabilities of obtaining a head or a . '1 are p and (1- p) respectively; and an experiment consists of tossing the ~lnl1Y twice: Find the probability distribution of the four possible outcomes of Ihe expenment. . .. . If in N independent tnals of this expenment the difference between the bserved relative frequencies of two heads and two tails is I"~ prove that the ~laximum-likelihood estimate p* of pis

p* = (1 + ,1)/2. Show that p* is an unbiased estimate of p, and verify that in this case the usual formula for the large-sample variance of a maximum-likelihood estimate gives the exact variance. of p*. Hence, for ~arge tv., indicate two different ways or lesting the hypotheSIS that the penny IS unbiased. 88 The probabilities of obtaining a head on a single tossing with each of IWO pennies are Pi and P2 respectively, the difference () == (Pi - P2) being an unknown parameter. Find the probabilities of the four possible outcomes of if single tossing of the two pennies. Jf it known that the second penny is unbiased, prove that the maximumlikelihood estimate of () is . ()* = (1- 2,1)/2, where ), is the observed relative frequency with which the first penny turned up tails in N independent tossings of the two pennies. Verify that var({)*) = Pl(l;;Pd. On the other hand, if nothing is known about the second penny, derive an unbiased linear estimate of (j in terms of the observed relative frequencies, and show that the variance of this estimate can never exceed 1/(4N). 89 In a factory producing synthetic yarn, the amount of raw material that can be put into the plant at anyone time is, on a certain quantitative scale, an integral non-random variable x which can take the values x = 1,2, ... , n. The quantity of yarn produced, Yx' depends upon the amount x of the raw material used and an efficiency factor of the plant, so that a linear regression relationship between Yx and x does not hold over all the possible values of x. It is therefore assumed that Yx is a random variable such that E(yx)

= a.X+PX2,

var(yx)

= (12,

and it is desired to estimate the magnitude of the linear parameter a. and the damping parameter p. If a set of sample values Yl, Y2,' .. , YII is obtained by independent experimental runs of the plant using the quantities x = 1, 2, ... , n respectively of die

78

EXERCISES IN PROBABILITY AND STATISTICS

raw material, apply the method of least squares to obtain a* and estimates of a and p, and show that formally

P*, t~

S4(]'2 ( *) var a = (S2 S4 -S~)'

Hence, or otherwise, derive explicit expressions for the two variances.

90 Suppose a and b are unbiased estimates of two parameters respectively, such that var(a) =

var(b) =

A\(]'2;

)'2(],2;

and cov(a, h) =

C(

and p .

A3(]'2,

where AI, A,2 and A,3 are known constants, and S2 is an independent estimate of (]'2. Assuming that a and b are normally distributed and ns 2 /(J2 has a X2 dis. tribution with n degrees of freedom, by considering the function a - pb, Or otherwise, obtain the appropriate 95 per cent confidence limits for the para. metric ratio p == a/po Hence verify that in the particular case when )'3 =: 0, the limits are

~±sto [)'l +A2(~r -A,IA2 s;~~r/ [1-A,2 s;~~J. to being the 5 per cent point of Student's t distribution with n degrees of freedom. 91 Suppose x and yare independent normal variables with zero means and variances (JI and (]'~, and a new random variable is defined by the ratio z = (y+b)/(x+a), a and b being known constants, and a > O. Prove that the probability dis· tribution of z is

1 [ ~ V 27t

1

Q(J~ + (JIbz exp { 2 2 22 2 2)f· -(az-b) /2«(]'2+(J\Z )}+R(z) dz, (J 2 + (J I Z .

.( 2

where R(z) is a positive function of z such that 00

f

R(z) dz

=

p(lwl ~ a/(Jd,

-00

w being a unit normal variable. Hence deduce that if

a is sufficiently large

compared with (]' I> then the random variable g(z)

= (az -

b)/«(J~ + (JIz2)t

is approximately normally distributed with zero mean and unit variance. Use this result to prove that if x and yare correlated normal variables with correlation p, then the appropriate generalization of g(z) is h(z)

=

(az-b)/«(]'~-2p(]'I(]'2Z+(JIZ2)t.

92 Saplings were planted at the corners of a square lattice in a large rectangular plantation, there being MN saplings distributed in N rows each with

TIMATlON, SAMPLING DISTRIBUTIONS, INFERENCE, ETC. ES

79

I nts During the ensuing winter a certain number of saplings were killed

Mp ast ~nd it was assumed, as a null hypothesis, that these deaths occurred ~Y/~ndentIy so that the contiguity of saplings did not affect their chance of

10 ePval To test this hypothesis, a random sample of n saplings (n ~ MN) was su~v~ a~d it was observed that amongst them there were in all d contiguous

ta .e of saplings that had died.

palr~y considering the four possible types of contiguities between two points, rove that, in gener~l, if two ~oints are select.ed ~t random, then the probability ~r their being contIguous pomts of the lattIce IS 2{4MN-3(M+N)+2} P2= MN(MN-l) . Hence determine the expected number of contiguous pairs in the sample of I saplings. Also, assuming that the distribution of d can be approximated ~y a binomial distribution with parameter P2, indicate a method of testing the hypothesis of randomness of deaths amongst the saplings of the plantation. Extend the above analysis to show that there are in all twenty different ways in which the four dis~i~ct contiguous pair~ can be ~ormed into c<;mtiguo US triplets by the addItIon of another contIguous po 111 1. By groupmg these twenty triplets into six types, classified according to the row and column placements of the points, prove that the probability that three random points on the lattice constitute a contiguous triplet is 24{5MN-7(M+N)+9} P3 = MN(MN-l)(MN-2) .

93 For a multinomial population with four distinct classes AI> A 2 , A3 and 114 the corresponding expected proportions are PI' P2, P3 and P4 respectively, where r.Pi = 1. A random sample of nl observations is taken from this population and the observed class frequencies are found to be nll, n1 2, n13 and n14' A second random sample of n2 observations is taken by ignoring the Al class, and the observed frequencies in the A 2, A3 and A4 classes are n22, n23 and 1124 respectively. Finally a third random sample of n3 observations is obtained by ignoring both the A I and A2 classes, and the frequencies in the 113 and A4 classes are n33 and n34 respectively. If the total number of observations in the four classes obtained from these three samples are denoted by r l , r2, r3 and r4 respectively rEri = En} = n), find the expectations of the rio Hence derive estimates of the expected proportions Pi and verify that these are identical with those obtained by the method of maximum likelihood. By using the transformation PI = (1-9 2); P2 = 92(1-9 3);

P3 = 9293(1-94); P4 = 929394

on the joint likelihood of the P's derive the maximum-likelihood estimates 9J and 9% of the 9's. Show that for large samples these estimates are uncorrelated and also obtain their variances. Hence calculate the large-sample variance of a statistic Twhich is a known function of 9!, 9t and 9%. Ifn is fixed, but n l , n2 and n3 may be varied, find the optimum values for the II} which minimize var(T), and then verify that this minimum variance is

O!,

~. [(1-92)JO;I:~I+(1-93)JO;I;~I+J9il-94)·I:~1r

so

EXERCISES IN PROBABILITY AND STATISTICS

94 If a random variable y has a lognormal distribution such that x === loll, is normally distributed with mean and variance (f2, prove that for r > 0 )

e

E(yr)

=

er~ + 1-r 2

a"

and thereby obtain explicit expressions for It and V 2 , the mean and varianCt of y. Given a random sample of 11 observations Yt, Y2,··" Yn from the lognormal population, suppose the corresponding x values XI' X2"'" Xn have a mean i and a sample variance S2 defined by n

ns 2

=

I

(xi-i)2.

i= t

Use the infinite series

where k-t

Ak == (n_l)k-t /

}I (n+2j-3),

to show that E[e rx . hHr 2s2)]

=

E(yr)

and thus deduce that efficient estimates of It and V 2 are

m = eX. h(!S2) and

Finally, by considering an asymptotic expansion of Ak in inverse powers of n, prove that to 0 (n- 2 )

h( ) Z

-

e

.-.2/n

[1. + z2(SZ+3)] 3n 2

'

and hence derive, correct to the same order, large-sample approximations for

m and v2 • 9S In extensive sampling of a multinomial population it occasionally happens that the expected proportion in one of the classes is exceptionally large and, as a measure of sampling economy, it is suggested that only a known fraction of the preponderant class be sampled, whereas the other classes are enumerated in the usual way. In such a situation, suppose the expected proportions in the four classes Ai> A 2 , A3 and A4 are proportional to (2+ 9), (1- 9), (1-9) and 9 respectively, where 9 is a parameter which can theoretically have any value in (0 < 9 < 1). If it is decided to record only a fraction p, (0 < p < n of the At observations but otherwise the sampling is complete, suppose that in such a censored sample of size M the observed frequencies in the four classes are Zi> Z2' Z3 and Z4 respectively (l:Zi = M). Derive the equation for 9*, the maximum-likelihood estimate of 9, and find its large-sample variance. Further, prove that 9* will be as efficient an estimate of 9 as &, the maximum-likelihood estimate of 9 obtained from an

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE, ETC.

lln cens

81

ored sample of N observations, if

M=

(1 +20){2(1 + p)-(l- p)lW

4{2(1+p)+(1+7p)O}

.N =N.h(O,p),say,

and hence ?educe that whatever be the value of the parameter O. as a first approximatIOn for small p 4(1 + p)(17 + 23p)(3 +5p)2 M~ (7+ 13p)(13+19p)2 .N. 96 In a plant-breeding experiment four distinct types of progeny AB, Ab, aB d ab are possible; and in a random sample of N 1 observations, the observed ~n quencies in the four classes were found to be Xl> X2' X3 and X4 respectively, == N 1)' If on a genetical hypothesis the corresponding expected probabi'~~ies are proportional to (2 + 0), (1- 0), (1- 0) and 0, determine the equation for 0, the maxi~um-likeli~ood ~stimate of the parameter 0, and derive the large-sample vanance of thiS estlmate. . . . . Since, in general, the parameter 0 hes 10 the range (0 < 0 < 1), It IS suggested that another convenient method for its estimation is to ignore ~ompletely the preponderant AB class in sampling. Suppose, then, that a total number of N 2 observations are taken from such a truncated distribution, and the observed frequencies in the remaining three classes are obtained as }' Y3 and Y4 respectively (I:Yi = N 2 )· Use the method of maximum likelihood t~'obtain 0*, the estimate of 0 based on this second sample, and prove that for large samples (0 *) _ 0(1 - 0)(2 - 0)2 var 2N .

(fe,

2

Hence show that 0* must have a variance ::f var(O) if N 2 = 0·5174 N l' Further, if 0** denotes the estimate of 0 obtained by using the joint likelihood of the above two independent samples. prove that var(O**)

= (1

20(1- 0)(2- 0)2(2+ 0) + 20)(2 - 0)2 N 1 +4(2+ O)N 2'

Use this result to show that if N 1 + N 2 = N, a constant, then the best allocation of the sample sizes for minimum var(O**) is obtained for N 1 = O. Discuss the significance of the results in the light of efficient estimation of 0 by the method of maximum likelihood. 97 Suppose the ordered observations of a random sample of size /1 from a population having a probability density function f(x) in the range (-00 < x < (0) are denoted by III < "2 < 113 < ... < lin' Further. given a random sample of III from another population. let these »I observations be distributed amongst the (/1 + I) intervals formed by the IIi such that »11 observations are <111; "'i+1 lie between (lIi,lI i +d for (i = 1,2.... 11); and finally IIIn+ I are> lin. where. of course. n+1

m =

L

mi'

i= I

If new random variables are defined by the relations

f

PI =

-

[(J

f

U;+1

Ul

f(x) dx;

Pi+ I

=

Uj

Jj

f(x) dx (i = 1,2.... ,11-1); Pn+ 1 =

f f(x) dx, Un

82

EXERCISES IN PROBABILITY AND STATISTICS

find the joint distribution of PI' pz,· .. , Pn, and use this to prove that for an i ::F j and integral rand s I

n I r!s I E(Pi pj) = (n~r+s; !' Hence, by considering the conditional joint distribution of the ml for fixed p show that h

1

n 98 Suppose An is a random sample of size n from a population having a probability density function f(x) in (- 00 < x < (0), and let the ordered observations of An be denoted by UI < Uz < U3 < ... < Un' Further let B be a sample of m observations from another population, and let these observa: tions be distributed amongst the (n+ 1) intervals formed by the Ui such thaI m l observations of Bm are Un' where ' n+1

L mi=

m.

i= I

If new random variables are defined by the relations

f "I

PI

=

f

"'+1

f(x)dx;

Pi+1

-oc:.

=

f 00

f(x)dx (i = 1,2, ... , n-l); Pn+1 =

IIi

f(x)dx,

lin

find the joint distribution of PI' pz, . .. , Pn and m" mz,· .. , mn+ I' Use this distribution to show that the moment-generating function of the random variables Zi=

1m.) (n+l-~'

for i

I)} .f{p, e-

+ pz e- IJ2/ m + ... +Pn+ I

= 1,2, ... ,n+l is


n ! expC~: O;/(n+

IJ1 / m

e- IJn + I/m}m X

n

X

n dpi>

1= I

where the n-dimensional integration is over the appropriate range of PI'

pz,···, Pn'

Hence, or otherwise, evaluate the mean and variance of the statistic n+1

C z = "t... z~" i= I

and find their limiting values when (i) m is fixed and n -+ 00 ; and (ii) n is fixed and m -+ 00.

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

83

A random sample of n observations from a distribution having a prob-

99.. density function f(x) in the range (- Xl < X < (0) is ordered, and these ablhlY d observations are denoted by UI < U2 < ... < Un' which divide the orde~~ infinite range of x into (n+ 1) intervals. Based on this division, new doud m variables Po, PI' P2,· .. ,Pn are defined such that ran 0

I

Po"" j'J(X)dX;

Pi

=

-00

I <Xl

Ui. I

f(x)dx,

for i

= 1,2, ... ,n-l; Pn =

f(x) dx.

Un

•

a random sample of size m is taken from another population having a ~;~bIY infinite range, and these m observations ~re distributed ove~ the (n+ 1) . lervals formed by the Ui so that mo observatIOns are un, where any mj ~ 0 and IIi' ,+.' h .. d· ·b· f h d f,m. == m. Find t e Jomt Istn utlon 0 t e mi an Pj. 'ro lest the hypothesis that the two samples are in fact from the same population. the statistic n. 2 I

d2

=

I [~- Xi] n

i=O

Xi ==

m

±

mj ,

j=O

is proposed. Show that the joint moment-generating function of the random variables (i = 0, 1,2, ... , n),

is

where k

./, = "L...

p.).

'I'k -

e-t}/m.,

j=O

r=j

ror all integralj and k in the range (0, n), and the multidimensional integration is over the appropriate range of the Pi. Hence determine the mean and variance or el 2 ; and assuming that for large m and n d2 '" PX 2 with v dJ., use the first tlVO moments of d2 to determine p and v in terms of m and n. tOO In sampling inspection of a mass-produced article, the population of the product consists of batches of fixed size N such that the probability that a batch will contain X defectives is P(X). If x is the number of defective items observed in a random sample of size n drawn from one of these batches, find the joint distribution of x and X. Use this distribution to prove that for fixed X. the conditional mixed factorial moment of x and (X - x) is (md(N )(m,) E[x(md. (X -x)(m2)lx] = n -n . x(m,+m2) N(m, +m2)

ror positive integral values of nI. and m2 • Further, assuming that the mean and variance of X are JI and (J2 respectively, prove that (i) var(X -x) = var(x)+

.. cov(x, X-x) (11)

(1- ~)(J2; and

1l(J2 = n(N-n) N(2) . [ (J2_fJ. (fJ.)] I- N = N-var(x).

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE; ETC.

If (J. and

Pincrease in such a

85

way that a./ P == p > 0, show that for large

var(X)-E(X)[1-E(X)/N) ='"

p

N(N -1)p -2 (1 + p)3p +O(P ); and

N(N -1)p var(Y)-E(Y)[1-E(Y)/N) = (1+p)3p +O(P-2). Hence verify that in the limit X and Y have the same binomial distribution ith parameter p/(1 + p), but that, as compared with this limiting form, the ;istribution of X for finite a. and P is leptokurtic, whereas that of Y is platykurtic. 104 It is known that in a large batch containing N items of a mass-produced article the probability for the batch to have X defective items in all is P(X), defined for (0 ~ X ~ N), and such that the mean and variance of X are JI. and (12 respectively. Successive random samples of size n" n2 , • •• are taken from the batch without replacement, and the corresponding numbers of defective items observed are XI' X2' .••• For this inspection scheme, if k

Xk

k

= X-I

and

XI

Nk

= N - I ni , for k

i= I

~

1,

j= I

find the conditional distribution of Xi for given Xi-I' Use this distribution to prove that for integral mi and Vi

E[h x\m')IX]

=

J1 n!md. E[x(Im;))/N(Im

ELDI X\V')IX]

=

III N\V,). E[X(IVd)/N(Iv,).

d;

and

Hence, or otherwise, deduce that (i) E(xJ

= njJl./N

n\2)[u2+~:=~~. (JI.-~)]/N(2)

(ii) var(xj) =

(iii) cov(Xj, X j) = njnj [u 2(iv) E(Xj)

=

(v) var(Xj)

G- ~) ]/N(2),

i t= j

NjJl./N =

N\2)[U2+(~~_~j;. (JI.-~)]/N(2)

(vi) cov(Xj,Xj ) =

NjNj[U2_(JI.-~)]/N(2),

(vii) cov(Xj, Xj) = ni(Nj-l) [U 2+ ( : =

j

2-

and

~~j ) ~-~)]/ N(2),

nN [u G-~)]/N(21, I

(i t=j),

ifi

if i > j

~j.

105 In non-destructive testing of an industrial product, batches consisting of N items each are available for sampling inspection. The number of defectives

86

EXERCISES IN PROBABILITY AND STATISTICS

in any batch is a random variable X with mean Jl and variance u 2 • A rand sample of n items is taken from a batch without replacement, and it is fo"o~. to contain x defectives. Prove that Ill! cov(x,X-x)

(Jl-~)]/N(2).

n(N-n)[u 2 -

=

On the basis of the sample data, it is decided that when corr(x, X - x)

<:::

0

(i) the batch is accepted without further inspection if x > a; (ii) the .r~ma!ning (N - n) items are inspected if x ::::; a, where a is a giver,

posItive mteger < n. This inspection procedure is reversed for corr(x, X - x) > O. If Rb is the proportion of defective items examined on the average in. batch, and Rg is the proportion of non-defective items inspected on th: average, then the efficiency of the sampling inspection scheme is E == Rb-Rg •

Show that for corr(x, X -x) < 0 and a specified a, E=

N(N-n)

a

Jl(N-Jl)'x~o

[(X+l) Jl] n+l Pn+l(x)-N,Pn(x) ,

where Pm(x) is the marginal distribution of x for samples of size m. Assuming that the marginal distributions of both x and X are rectangular, determine the values of n and a which maximize E, and hence prove thaI. for large N, this maximum is -t. Alternatively, if the proportion of items to be inspected is fixed at p, prove that, for large N, the values of 11 and a which maximize E are 11 - JNp/(I-p) and a- Jp(l-p)/N. Finally, suppose that a batch is fully inspected if x > a, but accepted on the basis of the sample if x ::::; a. If, on the average, the initial and after· inspection proportions of defectives in the batch are 0( and [3, and the "aver· age outgoing quality" defined by the ratio [3/(0(+ [3) is fixed at p, prove thaI E is maximized for

IN(Ji-l)

11 -

a-

and

fl'

(A == 0(/[3),

provided N is large compared to P- 1. [It may be assumed that all defective items detected are replaced by non· defectives. ] 106 Suppose U 1, U2 and U3 are the first, second and third quartiles respectively obtained from an ordered sample of N == (4n+3) (n being an integer ~l) observations of a random variable x having a probability density function f(x) in the range ( - 00 < x < (0). Find the joint distribution of the random variables Ui + 1 "1

Pi

=

f - 00

f(x) dx

and

Pi

=

f

f(x) dx

(i = 1,2),

IIi

An independent sample of 111 observations from another population with a doubly infinite range is distributed over the four intervals formed by the Ui such that m 1 observations are < U 1 ; mi lie between (uj, U i + i)' for (i = 1,2);

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

87

are > U3' To test the hypothesis that the two samples are, in fact, and '~he same population a statistic w is defined as rrom

4

w

=

L (4mi-m)2/9m2. i= 1

I'rove that w always lies between (0, 1), and that

E(w) = 4(N +m+ 1)/3m(N +2); and 32 (m-l)(N +m+ l)(N +m+2)(N + l)(N +5) var(w) = 27' m3(N +2)2(N +3)(N +4) ,

. that the distribution of w may be approximated by a suitable Beta distribu-

;:~n. Also, compare the limiting behaviour of the distribution of w when (i) N is fixed and m --+ co ;

(ii) ttl is fixed and N --+ co. Finally, suggest an approximate procedure for using the w statistic when N is large but not necessarily of the form (4n + 3).

107 If u denotes the sample median of a random sample of (2n + 1) observalions from a population having a probability density function f(x) in the range (- CfJ < X < co), determine the marginal distribution of the random variable

f u

p=

f(x)dx.

-00

Suppose an independent sample of III observations is taken from another continuous population with a doubly infinite range, and it is found that ml or these observations are < u. Find the joint distribution of m 1 and p, and hence show that the marginal distribution of 111 1 satisfies the recurrence relation P( ml ) = (ml + l)(n+m-md . P(ml (m-1II1)(n+11I1 + l)

+ 1).

Further, use this distribution to prove that the rth factorial moment of 1111 is E[mlrl]

= 111(') (n +1')(')/(211 + I +1')(').

Hence. or otherwise. determine the mean and variance of mI' Also. if 111 and /I increase indefinitely in such a way that m/ll = A (> 0), prove that the asymplotic variance of the statistic Y = (2m l -m)

I{ m2()·+2)} 1

IS

(3),,+2) -2 2()" + 2)11 + 0(11 ). 108 If

Xl ~ X 2 ~ ••• ~ Xn

denote the ordered observations of a random

~a~ple of size 11 from a rectangular distribution in the interval (0, a), find the

Jomt distribution of the statistics 1I =

!{xn-xd and v = t(xn+xd.

88

EXERCISES IN PROBABILITY AND STATISTICS

Hence derive the marginal distribution of v. Prove that v is an unbia estimate of the population mean, and that its efficiency. as compared W~I the sample mean x of the n random observations, is III (n+ 1)(n+2)

6n

>

1

t' 3 lor n ~ .

Also, show that the standardized random variable W

=

v-E(v) Jvar(v)

----;::=====

the limiting Laplace distribution ~. e-..filwl. dw fO "l ' r ( - 00 < W < 00), whereas the standardized sample mean x has a limitio unit normal distribution. In view of this limiting distributional behaviou: comment on the use of v as an estimate of the mean of the rectangular popu: lation. has, as n --+

00,

109 If x is the sample mean obtained from n independent observations frOIll a rectangular population in the range (0, a), prove that x is an unbiased estimalt of a/2 with variance a l /12n. Alternatively, suppose the sample observations are ordered as Xl ~ xa ~ ••• ~ X n , and a linear function y is defined by n

y=

L CIXi> j=

1

where the Cj are arbitrary constants. Derive the joint distribution of lilt ordered observations and use it to establish that E(x~ x~) I

J

= n! U+s-l)! (i +r+s-l)! . ar+s (n+r+s)!U-1)!(i+s-l)!

for i > j, and r, s positive integers. Hence, or otherwise, find the mean and variance of y, and then deduCt that y is an unbiased linear estimate of a/2 with minimum variance ir Cn = (n + 1)/2n, and Cj = 0 for i "# n. Find the distribution of this best estimate of a/2, and then show that. as n --+ 00. the standardized unit variable defined by t = {y-E(Y)}/Jvar(y) has the limiting distribution e'- 1 dt for (- 00 < t ~ 1). Also, show that for finite n the efficiency of y as compared with the sample mean x is (n + 2)/3.

110 If XI ~ Xl ~ ... ~ Xn are ordered observations from a population having a probability density function (k+ l)x k /a(k+ 1) for 0

~ X ~

a,

where a and k are positive parameters, find the joint distribution of the Hence prove that for i > j and r, s positive integers r s E(x.x.) I

J

=

a+s.n(k+l) {n(k+ l)+s+r}'

and use it to evaluate cov(Xj, x n ).

r(n)rp+k:l)r(i+::~) ,

rU)r(i +k: l)r (n+ ::~)

XI'

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

89

Assuming k to be known, prove that

=

y

I'

I

1I(k+ 1)+ 1 n(k+2) . Xn

h best linear estimate of the population mean, and that its efficiency as eared with the sample mean x of n random observations is

(limP

n(k+ 1)+2 > 1 for n ~ 2. k +3 ,\Iso. prove that the standardized random variable U

has. as /I

-+ 00,

y-E(y)

=

-'-----;====

Jvar(y)

the limiting distribution eu - 1 du for (- 00 < u ~ 1).

III Given an ?rdered samp.l~ of 11 o~servatio~s x I ~ X 2 ~ ••. ~ Xn from a pllpulation. havl~g.a p~obablhty densIty functIOn f(x) for (- 00. < x < 00), lind the jomt dIstributIOn of x. and X" (n ~ t > 5 ~ 1). Use thIS result to derive the joint distribution of the random variables P

=

P(x. ~ x ~ x,)

and

Q = P(x

~

x.).

lienee show that the marginal distribution of P depends only on n and the lIilTerence (t - 5). In particular, if E(P) = a, an assigned number, verify that P has a Beta distribution with parameters (n + l)a and (n + 1)(I-a). Indicate how this result may be used to determine the smallest sample size 11 such that the probability or P lying in a specified interval (b, c) is also at least p. 112 Given a random sample of n observations from a normal population with mean m and variance (12, suppose x and 52 are the usual unbiased estimates of the population parameters. For any given positive number k, a random variable Q is defined by

J

i+lc.

Q -- _1_ (1~' V

~/~

e

-tlx-m)2/a 2

d . x.

i-/CS

Prove that I

E(Q)

=

1 (In-I)' In=1B 2' -2-

f -I

dO (

(2)n/2'

1 +n-l

where t == k[n/(1I+ l)Jt. Hence deduce that, on the average, a proportion

or the population lies between the limits

IX

1)

n+ t X±5t,. ( -n- , where t,. is the percentage point of Student's t distribution with (n-l)dJ. such that the probability P( - t,.

~

t

~

t,.)

=

IX.

90

EXERCISES IN PROBABILITY AND STATISTICS

Also, show that if E(Q)

= ex, then to terms of 0(n- 1) var( Q) = t; .e - IiInn.

113 A number k of similar instruments are used for routine measurellle and the observations made by the ith instrument are xij' U = 1,2, ... , n) :' all (i = 1,2, ... , k) . . Assuming that. the ~jj a.re independent normal vari~bl~: with the same vanance (f2 but with dIffenng means such that E(xd:::: ( find the maximum-likelihood estimates of the and (f2. Show furth~r Ih' for (f2*, the estimate of (f2, at

ej

and

var ((f

2*) _ 2(N - k)(f4 ' N2

k

where N ==

L nj. j=

Hence show that

1

(i) if nj = n, a constant, and k -+ co, then (f2* is an inconsistent estimak of (f2; and (ii) if k is a fixed constant and all nj -+ co, then (f2* is a consistent estimatt of (f2. Comment on these results in the light of R. A. Fisher's claim that "an efficient statistic can in all cases be found by the method of maximum likeli. hood".

114 If YI' Y2" .. ,Yn are n independent observations from a normal popula. tion with unknown mean m and variance (f2, obtain explicitly the maximum. likelihood estimates ji and S2 of m and (f2 respectively. Further, suppose a new random variable v is defined by

v = s2+(ji_m)2. Then, assuming the known form of the joint sampling distribution of ji and S2, prove that (i) E{ v'. (ji- m)2 P + I}

(H) E{,'. (y - m)")

= 0 for fixed r and all integral values of p; and

2 2)P+' ~ (-"--. n

rG+p+r) r(p+t) ( rmr i+p)

for all integral values of p and for all r such that the the right-hand side have meaning.

r

functions on

115 Random samples of size n1 , n2' ... ,nk are taken respectively from ( univariate normal populations having the same mean m but differing varianoo (ff, (f~, ... , (ft. If the sample observations are denoted by xij U = 1,2, ... ,IIi; i = 1,2, ... , k), prove that m*, the maximum-likelihood estimate of m, is obtained as a root of the equation

L k

j=)

(*) njXj-m sf+(xj-m*)2

where Iii

njxj

==

L xij j= I

and

=0 '

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

91

enerally, suppose an estimate In of III is obtained as a solution of the \Iore. g k _ 1I'llIon '\' wj(Xj-m) 1 ~I ' L 2 2 = 0, j=1 Sj +(xj-m) A

A

III

Ihich the weights Wi are constants so chosen ~s to .ensure t~at nl is als.o \~ 'stent estimate of m. Prove that the asymptotIc vanance of 111 as k -4 oc IS

,,,n'

~i (n,.~~)a,,/{.f Wd a;}

vac(.h)

,=1

educe that if I Ien ce d

Wi =

var(m*)

(ni - 2) for all i, then

) = var(ln)+.,Ik Pi (rx. _pl. - (} ,= 1

rxi ==

2.

,=1

1

nda;; Pi == (lt i- 2)/a;;

2; {k.I rxj}2, 1= 1

(} ==

and

JIk rxi jki~/i'

J lhal var(m*) > var(lll), except when the It j are all equal. ~I Further, as another consistent estimate of 111, consider the weighted average

k (kI

.x = I

I1 j

/ljXi

j= 1

j=

•

1

Find the variance of x and hence verify that if the ni are equal to n, then (i) var(x) < var(ln) if the ai are equal; but (ii) for unequal ai' var(x) > var(IfI), provided 11 is sufficiently large.

116 Given a random sample of (2n+ 1) observations from a population having a probability density function f(x) in (- CX) < X < CX)), prove that, as a first approximation, the sample median is normally distributed with mean III and variance 1/[8nf2(m)], m being the population median. Hence deduce that (i) if X is normally distributed then the sample median has an asymptotic efficiency of 0·64 as compared with the average of (211 + 1) observations; (ii) if X has the Cauchy distribution, then the asymptotic efficiency of the sample median as compared with the maximum-likelihood estimate of m based on a sample of size (2n + 1) is 0·81. 117 A random sample of n observations is given from a (he probability density function rxV

nv).e

-ax

,'-1

.X

,

for 0 ~ X <

r

distribution with

CX),

where I' is a known positive constant and rx is a parameter. Prove that in this case the method of moments and the method of maximum likelihood both give the same estimate of rx. Show that this estimate is not unbiased but is asymptotically consistent as n -4 CX). Hence obtain the modified unbiased estimate of rx and, assuming that I1V > 2, verify that its variance is rx 2 /(nv - 2).

118 If Xl' X2" •• ,Xn are independent observations from a univariate normal population with mean III and variance a 2 , prove that the expectation of the statistic 1 n IS a - - , a* = - . - ,I IXi 11 2 n i=1 .x being the sample average.

~

xl

. F.l1-l

92

EXERCISES IN PROBABILITY AND STATISTICS

Also, if Ui

= (Xi -

xl/a for all i, show that, for i #

j,

E{luiUjl} =

=

~[Jn(Il-2)+sin-I(-1 )]. 1t1l n-l

Hence evaluate the exact variance of a* and verify that for large n the efficiency of a* as compared with s, the square root of the maximum-likelihood estimate of a 2 , is (1t - 2)-1. Finally, prove that

11-1 but

119 ffthe random variables x and y have ajoint bivariate normal distribution with parameters mx , my, a';, a;' and p in usual notation, prove that

P{IX-l11xl ~ ly-m).1} =

_ 1

- ~. tan

_1{2A~} 1 _ A. 2

'

Hence deduce that, irrespective of the value of p in its permissible range of variation, this probability is > or < t according as a x is < or > a}" 120 During World War II the markings and serial numbers obtained from captured German equipment were analysed in order to obtain estimates of German war production and capacity. As a somewhat simplified model, let the population of serial numbers pertaining to one type of equipment be 0+1, 0+2, ... ,0+p, where. in general, 0 and p are both unknown, and it is desired to estimate p. Assuming initially that 0 is known, the population may be denoted by the p integers 1,2,3, ... , p. Suppose then the captured equipment provides a random sample of n integers XI' Xl> ... ,Xn, (n < p) from this population, and let g be the largest of the sample integers. If it is further assumed that all samples of n integers have equal probability, find the sampling distribution 01 g for given n and fixed p. Prove that there is only one function of g, f(g) say, which is an unbiased estimate of p and that this function satisfies the difference equation h-I

f(h) = h-

L

v=n

f(v). P{g = vIp = h}/P{g = hIp = h}.

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

93

Next, for any integral k such that n + k > 0, prove that

In P

1)

(g+k) (P+k+ n+k = n+k+ 1

and hence, or otherwise, deduce that for integral r E[(g+r-l)(r)]

(_n_). (p+r): n+r p!

=

r.

Usc this result to p~~ve that = {(n + 1)In}g - 1 is an unbiased estimate of f and obtain an ~xphclt expressIOn for var( T). Further, show that g(g _11)/,,2 IS an unbiased estImate of var(T).

121

(i) In the previous example, if,s denotes the smallest of the sample integers, find the joint distribution of g and s. Hence, or otherwise, determine the joint distribution of g and d = g - s, and then show that the marginal distribution of d is

(p-d).

(~=~)

(:),

for (n-l)

~ d ~ (P-l).

Prove that for any positive integer r

<,) _ n(n-l) (p+r)! [ E (d+r-l) ] - ( ) (n+r- I) n+r p." and

cov(d,g) =

(n-l)(p+l)(p-n) (n+l)2(n+2)

Hence verify that the statistic

d

R=g+n_l- 1 is an unbiased estimate of p, and that its efficiency as compared with Tis (l-n- 2 ). Further, show that the statistics

_ g(g-n) d A - n2-1 an

B _ d(d-n+ 1) -

(n-1)2

are both unbiased estimates of var(R), but as a first approximation for large samples var(A)

= ! var(B).

e,

(ii) If as defined earlier, is also not known, prove that an unbiased estimate of p is given by

S

=

d(n+ 1)_1 n-l

and

2(p-n)(p+ 1) var(S)

= (n-l)(n+2).

Determine an unbiased estimate of this variance as a function of d.

94

EXERCISES IN PROBABILITY AND STATISTICS

122 A discrete random variable X is said to have a rectangular distribut' if X. can take any of the N integr.al values betwee.n l.an? N .with equal pr~Oj abilIty. If a random sample of n IS taken from this dlstnbutlOn, the sall1pl'~ being with replacement, prove that the distribution of the sample range R1~1

P(R) = N- n[(R + It - 2R n+ (R -1)n](N - R), =

for R =f. 0

N-n+t, for R = O.

Prove that for any integer () such that 1

P(R

~

() -1)

=

~

() :s:;; N, the probability

N- n [()n + (N - ()){ ()n -

It}],

(() -

and hence, or otherwise, verify that the stated probabilities for different value of R do define a proper probability distribution. ' Also, for large N, show that (n + I)Rj(n - 1) is an unbiased estimate or tho population range with variance 2(N _1)2 j(n - l)(n + 2). ' 123 A set of n unequal observations Xl' X 2, ... ,Xn is obtained in a till1t sequence. In order to test the serial independence of the Xj' it is proposed to consider their first differences Dj = (X j + 1 -X j ), and a test statistic n-l

D=

LD

j

j= 1

is calculated by scoring unity if Dj is positive and zero otherwise. Suppose no distributional assumption can b~ made about the Xj excepi that, on the hypothesis of randomness of the observations, every permutation of the Xj is equiprobable. Show that

E(D) = (n;

1)

and

var(D)

=

(n~

l

Alternatively, suppose the observations are spaced at equal time-intervals so that

(t

=

1, 2, ... , n),

where a, fJ are parameters, and the x, are independently distributed normal variables each with zero mean and variance u 2 • Prove that in this case

E(D) = (n; var(D)

1)(1 +2ed;

= n{l2+2(e2-ed-3ef}+{l2-4(e2-el)+5ef};

and e1 , e2 are defined by

el ==

~

I

fJla,/2

e- t82 d()

fJla,fi

f

¢(()1, ()2; -t) d()l d()2 + 2

fJla,fi

ff 00

¢(()1, ()2; t) d()1 d()2'

0 0 0

where ¢(()l, ()2; p) is the bivariate normal density function of ()1 and ()2 having zero means, unit variances, and correlation p.

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC.

95

Given a random sample of II paired but unequal observations (X j , lj) 124( '_ 1 2 ,'" n), the first differences of the two series are defined as for ) - , , D, :=: X j + I - X j and dj = lj + I - lj, U = 1, 2, .. , , II - 1). J

, g C. Sellin J

:=:

dPj' a correlation statistic II-I

C

L

=

Cj

j= I

'alculated by scoring C j = 1 if either (D j > 0, dj > 0) or (D j < 0, dj < 0), . , d zero otherwIse . •111 If 00 distributional specification is made about the sample observations, I' (;

I

0

assuming that the two variates X and Yare independent so that the

I~)i pairings of the two series and then ordering the pairs are equally probable, ~r~ve that E(C) = (11-1)/2. Also, show that cov(Cj , C j + d = 158' and hence I

deduce that var(C) = (lIn -13)/36.

125 In the previous example, suppose the pairs (Xj' lj) are a sample from bivariate normal population with zero means, unit variances, and correlation ~" Show that under these distributional assumptions 'I

E(Cj ) = !{l+(2/n) sin-I p}

and

var(Cj )

=

t-{(1/n)sin-1 p}2.

Further, for any three pairs (X I' Yd, (X 2, Y2 ) and (X 3, Y3 ), say, prove that Ihe joint characteristic function of the variates UI

=

(X I -X 2)

U3

VI =' (YI - Y2)

=

(X 3 -X 2)

V3 = (Y3 - Y2)

is qexpi(tlltl+t2VI+t31t3+t4V3)]

=

,~ exp- [tf + t~ + t~ + t~ + (t 1+ t3)2 + (t 2+ t4)2 + 2p(2t I t2 + t I t4 + t2t3 + 2t3t4)]'

By inverting this characteristic function, show that the probability P(1I1 > 0, VI > 0; U3 > 0, V3 > 0)

=

I

1

= 4n 2 [(sin- 1 p)2-(sin-1 p/2)2]+ 4n[sin- 1 p-sin- I p/2]+ constant, and hence deduce that cov(Cj , C j + I)

= 316 -

Finally, use these results to verify that

and

(~sin-I p/2

r

96

EXERCISES IN PROBABILITY AND STATISTICS

126 In the previous example, suppose an alternative scoring systelll i\ adopted such that

Cj

=

I if either (D j > 0, dj > 0) or (D j < 0, dj < 0)

Cj

=

-1

and otherwise.

If now a new correlation statistic be defined as n-I

g

=

L

C/(n -I),

j= 1

prove that E(g)

= ~ sin - 1 p, n

and

/

var(g) = ~

~(lnn~I\; (n~1)2 [(n-I)(~sin-I pr +2(n-2)(~sin-1 p/2f] 11n-I3[1 2 9(n-l)

(2. -sm -I )2] n p

.

Further, if an estimate p* of p is obtained as

p* = sin (ng/2), show that for p = 0, the efficiency of p* as compared with the product. moment correlation coefficient r is for large samples "" 36/11 n 2 • 127 Suppose (Xl' yd, (x 2, Y2),·· . ,(X"' Yn) are random observations from a bivariate normal population such that the differences Ui = (Xi - Yi) have mean JI. and variance (12 for all (j = 1,2, ... , n). Using the signs of the Uj, a statistic n

S

=

L

SJn

i= 1

is defined where the Si are random variables such that Si

= 1 if Uj > 0 and

Sj = 0 otherwise. Prove that E(S)

1

= !+t/>(-r) and var(S) = _{-!-t/>2(or)}, n

where

fo [ r

-r == JI./(1

and

t/>(-r) ==

e-)·2/ 2 dy.

Hence, if an estimate -r* of or is obtained as a solution of the equation t/>(-r*) =

eS;I),

show that for large samples var(-r*) "" 2n. e- t h{!_ t/>2(-r)}. n

BSTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

97

Further, if i denotes the usual estimate of T (sample mean of udsample standard . tion) prove that deVla ' E(i)

and var(i) sO

=

TC C== (n~lrr(n~2)/r(n;1) n,

n

n-l

[(n-l) -C; ] '" 2n T2+2

= n(n_3)+T 2 n-3

for large n,

that the asymptotic efficiency of T* as compared with i is

1 T2+2 1-4q,2(T)'

~.

128 In the administration of a lethal drug to a random group of n animals lhe measurement of interest is x, the time to death of each animal; but for reasons of economy it is not expedient to wait till all the animals die, and the sampling procedure is terminated after a fixed time T. Suppose, then, that in lhe sample of n obser~ations r. of them Xl' X 2, ... , x, were found to be < T, lhere being (n-r) survIvors at tIme T. Assuming that x is normally distributed with unknown mt:,an JI. and variance (12 determine the maximum-likelihood equations for (j and 0, the estimates of (1 ~nd () == (T - JI.)/a respectively. Hence show that IJ _ p(T-x) - pb+(l-p)g(b)

and A _ P [ 2 OA2 t 2 g(O )-(1_p)v 2 .2(V+ )-(v-2)O], A

where p is the observed proportion of deaths within time T; oflhe Xi;

f

x is the average

00

v2 == 4.± (Xi-T)2/r(x-T)2;

,= I

and

g(O) == e-tiJ'j

e- tz2 dz.

Ii

129 Suppose XI' X2" •• , Xn are random observations from a normal populalion with mean J1 and variance a 2 • If it is known that the sampled population is lruneated at a point T so that all Xi are < T, determine the maximum-likelihood equations for estimating the parameters a and 0 == (T - JI.)/a. Hence show lhat for their estimates IJ and 0 (j

= (T-x)/{O+g(O))

and A

g(O)

where

=

2(V2

+02)tV_(V 2

2-

2)0 '

x is the sample mean,

-00

98

EXERCISES IN PROBABILITY AND STATISTICS

130 In quality-control inspection of a mass-produced article, inte centres on estimating the unknown proportion p of defective items in r~ product; but the product is randomly grouped into batches of N items ea . and sampling inspection consists of taking and inspecting a random satn;~ from anyone batch, t Suppose, then, that the product is represented by a finite population of At white (defective) and Mq black (non-defective) balls, where p+q == 1. ~ random batch of ,N balls is taken from this population, and t~en a randotr, sample of n balls IS drawn from the batch, If the number of white balls in t~ batch is X (not known) and that observed in the sample is x, find the jOint probability distribution of the random variables X and x when (i) both the batch and the sample are obtained by sampling with replace, ment; (ii) the batch is obtained by sampling with replacement, and the sampk by sampling without replacement; (iii) the batch is obtained by sampling without replacement, and the samplt by sampling with replacement; and (iv) both the batch and the sample are obtained by sampling without replacement. Hence show that in each of the above sampling schemes p* = x/n is an un, biased estimate of p, but that the corresponding variances of p* are (i)

p: [1 + n ~ 1] ;

"') -pq (III

n

pq; (11") -

n

[1 +-n-l

M-N] -- ' N' M-I '

, -pq~1 -11-1] (IV) -- , n

M-l

131 In an investigation into consumer preference for the type of fuel used for central heating of homes, suppose that in a total population of N houses the proportions using gas, oil, and various kinds of solid fuels are p, q and (1- p - q) respectively, In order to estimate the parametric ratio p/q, a random sample of n houses is taken, the sampling being without replacement, and it is observed that the frequencies of gas and oil users are x and y respectively, Find the sampling distribution of x and y, and then show that E[x/(y+l)]

=

Np [ (N-Nq-l)(nl] p Nq+I' 1N(n) ~q[1-(l-qr]

for largeN,

Also, prove that for positive integral values of rand s

E[x(rl, yes)]

=

l1(r+sl, (Np)(r) , (Nq)(s)/N(r+s),

Hence, or otherwise, deduce that if nq is large then the coefficient of variation of x/tv + J) is approximately

[(N -11)(P+q)]t, (N-J)npq

132 Suppose x and yare random variables having means m" 1112' equal variance (12, and correlation p > 0, and for any given two new random variables II and v are defined by

e

1/ = X

Prove that

cos e+ y sin e;

v

=

- x

sin

e+ y cos e,

eSTIMATION, SAMPLING DISTRIBUTIONS. INFERENCE, ETC.

99

deduce that /-tence ') 0 " corr (u, v) ~ corr (x, y); (I '" M (ii) (x+ y)/./i and (y-x)/y 2 are uncorrelated random variables. 33 Apart from the usual random errors of measurement, an instrument 1 tWO independent and additive sources of bias of unknown magnitude h~ch cannot be totally eliminated by suitable adjustment of the instrument. ~ Iwever, according to the manufacturer's specifications, it is possible to set hO instrument so as to make the biases either positive or negative. Four sets t independent measurements each are made of a quantity by setting the ?strument in the four possible combinations for the biases. Denoting the ~ts by Xj' Yj, Zj and Wj, (i = 1,2, ... , n), it may be assumed that

/n

E(xj) = ex + Pl - P2

E(yj) = ex -

Pl + P2

for all i, where ex is the true magnitude of the quantity measured, Pl, P2 are the unknown instrumental biases, and all the observations have the same variance (12,

Find the least-squares estimates of the parameters, and verify that the expectation of the error sum of squares is (4n - 3)0'2. Hence, assuming that the observations are normally distributed, indicate, without proof, the generalized t statistic for testing the hypothesis that the two instrumental biases arc equal.

134 The lifetime x of electric bulbs produced by a factory is known on empirical considerations to have a distribution ee- Bx dx, for (0 < x < 00), where eis an unknown parameter. Prove that the probability that a randomly selected bulb has a lifetime > a constant T is e - TB. To estimate e, a random sample of n bulbs is tested, and in order to save sampling cost, the actual lifetimes of the bulbs are not observed. Instead, at time T it is noted that r < n bulbs are burnt out, so that there are (n - r) bulbs in the sample having a lifetime > T. Determine the probability distribution of the random variable r and hence show that the maximum-likelihood estimate of e is

1 (n) e = -.log . T lI-r A

Also, prove that the large-sample variance of

eis

var(e) = (e TB -1)/nT 2 , and find an approximate value for the variance of l/e. 135 In destructive life-testing of various kinds of physical equipment like ball bearings, radio tubes, electric light bulbs etc.• it is known that the lifetime x of an individual item has the negative exponential distribution

e-

l .

e- x /B • dx, for x ;;:::: 0,

where 0, the mean lifetime, is an unknown parameter. If, in general, II items are put on a test rack, it is economical to stop experimentation after the first r < II failures have been observed. It is also

100

EXERCISES IN PROBABILITY AND STATISTICS

e

theoretically advantageous for efficient estimation of to use the fact that observations occurred in an ordered manner. Denoting the r observation tb: x I ~ X 2 ~ ... ~ X,. prove that the maximum-likelihood estimate of eis S b!

e= Lt1 xj+(n-r)x, ]/r. By using the transformation YI = Xl;

Yj

=

Xj-Xj_ l • for 2

~j ~

r.

fit.td t~e joi~t distribution of the y·s'. Hence. de?uce that 2rO/e ~as a X2 dit tnbuhon with 2r dJ.• so that the maxlmum-hkehhood formula gives the exa' variance of () as 2Jr. ~ Use these results to prove that (i) 0 is fully efficient as compared with e*. the usual maximum-likelihoOi! estimate of based on a completely enumerated sample of r iternl and (ii) for the random variable x,

e

e

,

E(x,)

=

eL

1/(n-j+ 1);

var(x,) =

j= 1

e2

,

L 1/(n-j+ If j= 1

136 Suppose that YI• Y2 •• •• , y" are n independent observations such thai E(Yv ) = IX+/3IXlv+/32X2," and var(Y.) = u 2 • for (v = 1.2•...• n). where IX. /31' /32' and u 2 are unknown parameters. and Xl' X 2 are non-random ex. planatory variables. If f3t and /3! are the usual least-squares estimates o! /31 and /32 respectively. prove that var(/3n = u 2/(l- r2)S 11; var(/3!) = u 2/(I- ,.2)S 22. where. denoting deviations from the sample means by lower-case symbols, n

Sij == LXiv Xjv.

(i.j = 1.2).

v= I

and r is the sample product-moment correlation between X I and X 2' Next, suppose an independent unbiased estimate b1 of /31 is given such that var(b 1) = ur. and sr is an unbiased estimate of ur. By considering the deviations (Y,.-b,X",). prove that a simple least-squares estimate of /32 is n

A

/32 = (S2y-b 1S 12 )/S22'

Si}' ==

L

YvXiv.

(i = 1.2).

v=l

Show that P2 is an unbiased estimate of /32 and that var(P2) = (u2+r2urS,,)/S22' so that P2 has smaller variance than /3! if ui < u 2/( 1 - ,.2)S 11' Find the expectation of the residual S. S.• n

L (Yv- b 1X tv-P2 X 2Y'

v= I

and use it to verify that an unbiased estimate of u 2 is S2 =

Lt1

(Yv-bIX1v-P2x2Y-(l-r2)sisll}/(n-2).

101

IlSTIMATION, SAMPLING DISTRIBUTIONS. INFERENCE, ETC.

. Jly under assumptions of normal theory, show how this estimate can Flr;:'ro; an approximate large-sample test of significance for any hypothesis oe use about P2' In the distribution of the number of accidents per worker in a factory 137 given period of time, the frequency of workers sustaining one, two or over aaccidents is available, but the number of persons who did not have an rno~~ent cannot be enumerated owing to the factory population fluctuating aeci. g that period. This gives rise to a discrete distribution in which the zero dunn d up is unobserve . gro If it may be assumed that this truncated distribution is Poisson with an known mean A, and that in a sample of N the observed frequency for x .un I'dents per worker is lx, for x ~ I, show that a simple estimate of A based ,ICC f . on the method 0 moments IS

A*

m~_1 m,

=

I

,

where m~ is the rth sample moment about the origin. Prove that for large N E[log(l +A*)] = 10g(1 +A)

(1

~~iZ:)~4\O(N-2),

and. as a first approximation, var(A *) = (1- e - A)(A + 2)/ N. Further, as an alternative estimation procedure, show that the maximumlikelihood estimate 1 is obtained as a solution of

x=

1(l-e- 1)-"

.x being the average of the truncated sample. Hence deduce an approximation

for). when x is large, and verify that for large samples. the efficiency of ..1.* as compared with 1 is A(e A - I)/(}. + 2)(e A - A-I). Discuss the behaviour of this efficiency for variation in I,. 138 The first k (~I) classes of the lower tail of a Poisson distribution with unknown mean A are truncated, and a random sample of N observations is obtained from the normalized distribution. If m~ denotes the rth sample moment about the origin, prove that a simple estimate of A. based on the method of moments is

Also, for large N, show that E{log(;. * + k)}

= 10g(A + k) A{(3A. + k + 1)(Jt', - k + 1)+ 2Ji'.} + O(N - 2) . 2N(}.+k)2 (Ji', -k+ 1)2

and, as a first approximation,

(A*) _ A.{(A.-k+I)(Jt',-k+l)+2Ji'd var, N{Jl', -k + 1)2 ' Where Ji'l is the mean of the truncated distribution.

•

102

EXERCISES IN PROBABILITY AND STATISTICS

Alternatively, if x denotes the sample mean, prove that the maxim likelihood estimate ~ of A. is obtained as a root of the equation Uill

x-~ = I/~o B(k, r+ I). ~r/r! B(k, r+ I) being complete Beta functions in standard notation. Hence verify that for large samples the efficiency of A. * as compared W'I ~ is II. A.(j.t'1 -k + 1)2 {Jl~ -(Jl'1-A.)(Jl~ -k+ 1)}{2Jl'1 +(A.-k+ 1)(j.t'1 -k+ I)}' 139 If X and Yare independent random variables with means m1, m2 a~ variances uf, u~ respectively, prove that the variance of the product XY is (ufu~ + mfu~ + m~uf).

Hence, or otherwise, find the variance of the product &.p, where &. and Pale the usual least-squares estimates of the parameters ex and /3 in the regression equation E(y)

= ex + /3(x -

x),

based on n random observations having equal variance u 2 • Further, if S2 is the least-squares estimate of u 2 , and ns; is the S.S. of the sample values or'the explanatory variable x, show that under normal theory assumptions S2 = s~(s~ +n&2 + np2 s;)/n 2s;

is a biased estimate of var(&.p) and that, relative to I, the bias is

2(: =~)U2/(U2 + nex + ns;/32). 2

140 For a discrete random variable Y having a negative binomial distribu· tion, the point probabilities are P(Y = r)

= (m+;-I)

pr q-(m+r)

for all integers r ~ 0, where m is a known integral parameter, and p is an unknown parameter such that q - p = 1. If this distribution is truncated by the exclusion of the zero class, prove that for the random variable X of the normalized truncated distribution, the sth factorial moment is E[X(SI] = (III+S-1)(sl pS/(I_q-III).

Further, suppose a random sample of size N is taken from this truncated distribution and the sample mean is found to be x. Show that the equation for p, the maximum-likelihood estimate of p, is

P= ~[I-(I +p)-"'], m

and that, for large N, pq(l_ q -",)2

var(p)

=

Nm[I-(q+mp)q

( + 1) m

• ]

eSTIMA.TION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

. ely prove that \l!ern aUv , . E(X2) ,nJ hence .

103

= {l+(m+l)p}E(X),

deduce that a moment estimate of p is

p*

=

(V2 -x)/(m+ l)x,

is the sample second moment about the origin. For large samples, \,here Vl lI:rify that (i) E[log{1 + (m + l)p*}] = I) } __ (m+l)q(l-q-m)(3m p +Sq -l) O(N- 2) 2N ( )2 + ,

== log{ 1 +(m+ p

mmp+q

and (ii) the efficiency of p* as compared with

Pis

(mp+3q-l){I-(mp+q)q

(m+

1)r

141 An infinite biological population consists of three kinds of distinct individuals classified according to the three possible pairs AA, Aa and aa of a doubly inherited character (A, a) with probabilities (1-0)2, 20(1-0) and (/2 respectively, 0 being an unknown parameter. If 0 denotes the maximumlikelihood estimate of 0 obtained from a random sample of size n from the population, prove that the variance of the estimate is exactly 0(1 - 0)/2n. However, it is not always possible to identify correctly the AA and Aa individuals in the population without further time-consuming experimentation. Accordingly, to avoid errors of misclassification, the sample data are classified only in the two distinct groups (AA or Aa) and aa. Show that in this case 0*, the maximum-likelihood estimate of 0, has the large-sampl! variance (1- 02 )/411. Hence deduce the efficiency of 0* as compared with O. and discuss the relative merits of the two methods for estimating O. 142 The probability density function f(x) of a continuous random variable X is proportional to (-00 < X < (0),

where a, a real number, and p ~ 2 are parameters. Prove that the proportionality factor is l/l(p) satisfying the recurrence relation

(p+ 1)(p+2) l(p + 2) = [1 + (p + 2)2] . l(p),

. . with 1(0) == 2 smh 1[/2.

c

Use this result to derive the expectation of 2 logf(x)/ila. 2 , and hence deduce that the large-sample variance of the maximum-likelihood estimate of IX obtained from a random sample of II observations. of X is 1+(p+4)2 lI(p + l)(p + 2)(p + 4)"

Also, show that the sample mean .x is not an unbiased estimate of a but is asymptotically consistent for p --+ XJ.

104

EXERCISES IN PROBABILITY AND STATISTICS

143 The particles emitted by a radioactive source are registered as a funel' of time, and it may be assumed that the probability of a particle being elllit:~ in (t,t+dt) is '\Ii for 0 < t <

(J-l . e- t /9 dt,

00,

where (J is an unknown parameter. To estimate (J, experimental cOunts a made for equal time-intervals of length T, and suppose n. is th~ number ~ particles observed in the time-interval (v-l)T ~ t ~ vT such that 00

I n. == N . •= 1 If X and Y denote the total number of particles observed in the odd and even time-intervals respectively, prove that E(X)

=

e'<E(Y),

(A,

==

T/(J),

so that a simple estimate of (J is

0* = T/log(X/Y). Also, for large N, verify that E(e T /OO )

'"

e'< [1 +(e2.<-1)/2N),

and (J2 (e'<+ 1)2 var(O*) '" - . A,2 .< • N e

Hence deduce that for any given 0, var«(J*) will be a minimum if T approximately, so that var«(J*)

=

2·40

2.270 2

~ ~.

Alternatively, prove that the maximum-likelihood estimate of (J is

6=

T/Iog(_ il ), 11-1

where

Nil ==

f

vn,.,

.=1

and that, in general, the asymptotic efficiency of 0* as compared with () is

( ~)2. 1 +e-'< 144 In a decaying radioactive source the probability of a particle being emitted in time (t, t + dt) is (J-I . e- I / 9 dt, for 0 < t < 00, being an unknown parameter. A simple method for estimating (J is to count the total number of particles emitted after a suitably chosen time-interval T and to compare it with the total number N of counts. Accordingly, if " particles are emitted in (0, T), show that an estimate of 0 is

(J

0*

= -

T/log

(1- ~).

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

\lsO pr

105

ove that for large N

*

02 (eA-I)

var(O ) ~ - . - ' - 2- ,

N

A.

(J.

=

TjO).

deduce that for fixed 0 and N, the variance of the estimate is minimized ilcllrCC 1.6(} whence approximately

(M

-

,

var(6*)

1'54(}2

~ ~.

,\ltcrnatively, prove that the maximum-likelihood estimate of () is

e= -TjIOg(I-_n_), N-n and that the large-sample efficiency of

()*

as compared with

eis

J.e- Ajp-2(l-e- A)2}. 145 In a multinomial population with four distinct classes AI, A 2 , A3 and .1 the corresponding probabilities are proportional to (2 + ()), (1 - ()), (1- ()) ;\I:(i 0, where () is an unknown parameter. A random sample of II observations is taken from this population and the observed frequencies in the four classes arc found to be XI' X2' X3 and X4 respectively. To estimate 0, Fisher suggested livc possible estimates defined as follows: (i) 0, = (XI-X2-X3+x4)/n; (ii) (}2 = (XI- X2- X3+ 5x4)/2n; (iii) () obtained as an appropriate root of the quadratic equation

(}(2+0) (l - (})2

(iv) (v)

XIX4

=

X2 X3;

e, the maximum-likelihood estimate; and ()*,

the minimum X2 estimate obtained by minimizing 4

X2

=

L j;

{xj-E(xjWjE(Xj)'

I

Prove that (a) (), and O2 are both unbiased estimates of 0, and that their exact variances are (1- (}2)jn and (I + 6() _4(2)/4n respectively; (b) 0* is obtained as a positive root of the quartic xf (x~ + x~) xl (2 + 0)2 - (1 _ 0)2 + (}2 = 0;

and

(e) for large n, both {j and 0* are fully efficient as compared with Discuss the relative merits of these five estimates.

e.

146 If X and yare independent normally distributed random variables with zero means and variances I/rx and rx respectively, prove that the information supplied by a pair of (x, y) values for the estimation of a is l/a 2 • Further, given a random sample of n pairs of (x, y) observations (Xi> Yj), (i == 1,2, ... , n), find a, the maximum-likelihood estimate of IX, and determine

106

EXERCISES IN PROBABILITY AND STATISTICS

its sampling distribution. Hence prove that the amount of information o~ tained from a single observation of a is

11(11) ' 11 + 1 ;

a2

so that the maximum-likelihood estimate of a does not extract all the in~ mation contained in the sample. Ot.

147 Two independent random variables x and y have the joint probabil't . function . 1\ denslty ' for x, y

e-(9x+ y /9),

~

0,

o being an unknown parameter. Prove that the information supplied by a Pal of observations of x and y for the estimation of 0 is 2/0 2 , and hence deduct the total information obtained from 11 independent pairs (Xi' YI), ro~ i = 1,2, ... ,11. Further, given the sample observations, show that the maximull\, likelihood estimate of 0 is (Y/X)t,

t =

where X and Yare the sums of the X and y sample values respectively. Determine the joint distribution of X and Y. Use the transformation

X

= u/t and Y = ut

to derive the joint distribution of II and t. Hence, or otherwise, obtain the marginal distribution of t. Show that the amount of information to be expec. ted from a single observation of t is 211 (

211 )

0 2 ' 211+ 1 ' so that the maximum-likelihood estimate does not extract all the information in the sample for the estimation of O. Comment on this result.

148 Show that for a single random observation from the Laplace distribution (-00 < x < 00),

e

the information for the estimation of is unity. Suppose XI' X2,' •• ,Xn , (11 == 2s+ 1 and s > 1) are random observations from the above distribution. Prove that the sample median u is the maximum· likelihood estimate of and deduce the sampling distribution of this estimate. Further, show that

e,

f 1

t'(2-t)" dt

=

2 2•• B(s+ 1, s+ 1),

o

and use it to prove that the amount of information provided by the sample median u for estimating is

e

(s+ 1)(2s+ 1) s -1

[1

J

(2s) ! (s !)2 22• 1 .

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC. /ll!ll ce

how that the amount of information lost in using this estimate of s

I'

1] '" 4[ ~1ts -1]

2(2s+ 1)[(S+ 1)(2s)! s-1 (s!)222s ,0 Ihat

107

as s .....

00,

this loss also

(J

for large s,

~;

Discuss this result.

-+ 00.

49 If x X2"'" Xn are independent observations from a normal distribution me;~ Jl and variance (12, find the joint distribution of the random

I.h

Wit

,'ariables Y1

=

xJn

n

L

and Yi =

for 2 ~ i ~ n,

cijXj,

j=l

. 'suming that the x -+ Y transformation is unitary orthogonal. as Next, by considering the transformation Yi

In .SZi,

=

for 2 ~ i ~ n,

where n

ns 2 ==

L (Xj_X)2

n

and

nx

j= 1

~

L Xj'

j= 1

determine the joint distribution of X, sand

i.

==

Z2, Z3,"

., Zn-1'

Hence, if for

2 n

mA

L1 (xrx)A/n,

==

j=

show that X, sand mvm2"v/2, (v > 2), are independently distributed. Finally, use this result to find the mean and variance of gl = m3m2"3/2

and

g2 = m4m2"2-3.

lit may be assumed that if u and v are correlated unit normal variables with correlation p, then E(U 3V 3) = 3p(3+2p2); E(U 4V4) = 3(3+24p2+8p4).] ISO Particles are emitted randomly from a radioactive source, and it may be assumed that the probability of a particle being emitted in the time-interval (I, I +dt) is 0- 1 • e- I / II • dt, for t ~ 0, () being the unknown decay parameter of the source. Find the probability of a particle being emitted in the interval (0, T), where T is a constant, and hence deduce that the probability of a particle, which is emitted in (0, T), being emitted in (t, t + dt) is

e- I /II dt O(1-e-.1.)'

(.:t == T/O).

If t10 t 2 , •• • , tn are the emission times of particles from independent sources under observation in (0, T), find the equation of 9, the maximum-likelihood

108

EXERCISES IN PROBABILITY AND STATISTICS

estimate of (). Also, show that the information per observation for estimatin () is g 1{ '2 -), } ()2 1A)2 •

(t-:

Further, assuming there are a total of N sources under observation in the ex~r~me":t, prove that the expected information for the whole period per umt ttme IS N [(I-e-),)2 - A,2 e-),] ()3

A,(l-e-),)

.

Hence show that the most efficient value of the time-interval T is approxi. mately 4(). 151 There are k events E l , E 2 , •• • ,Ek , and the probability of the occurrence of Er is Pro (0 < Pr < 1), for (r = 1,2, ... , k). In Ar independent trials the event Er occurred Qr times, (r = 1,2, ... , k). Further, a combined event T is defined as the joint occurrence of E l , E 2 , • •• ,Ek with probability P, and in C indepen. dent trials T occurred c times. Assuming that the events Er are independent, find the maximum-likelihood estimates of Pr and P. Hence, for large samples, obtain a Xl statistic for testing the hypothesis of independence of the events Er • 152 If Xl ~ Xl ~ ••• ~ Xn are ordered observations from a population with probability density functionJ(x) defined for ( - 00 < X < 00), prove that for the sample range w

f

E(w)

=

[1- (Xn -(I-a)"] dx,

where a ==

-00

J(t) dt.

-00

Further, denoting E(w) by E[(w-£O)']

f x

00

£0,

show that the rth central moment of w is

= -(r-I)( -(0)' + +r(r-I)

ff 00

Xn

-00

-co

[1-a:-(I-al)n+(an-al)"](W-£O)r-ldxldxn,

where the probabilities P(x ~ x.)

== al and P(x

~

xn) ==

a..

153 If Xl ~ Xl are two ordered observations from a normal distribution with mean m and variance (fl, find the joint distribution of Xl and Xl' Hence show that the distribution of the difference u = Xl -Xl is

1C . e -u1 /4a1 . du, (u

(fv 1C

~

0).

Further, if Sl is an independent unhiased f"stimate of (T2 with v dJ., show that the significance of u may be tested uy the statistic ul /2s1, which has the" lbstribution with (1, v) dJ. Suppose Xl and Xl are marks obtained by two students in a competitive examination. Discuss how, under suitable assumptions, the above result may be used to select the winning candidate for the award of a prize.

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC'.

109

If ser~~ose

In a large national opinion poll just prior to the budget, N randomly ted persons were questioned about their attitude towards the government. included in the poll, a proportion oc were in favour of the government od the rest against. The budget contained some controversial measures ~ich were expected to affect public opinion, but the government asserted h t its support was, on the whole, unaffected by the budget proposals. t aTo test the government's claim, the same N persons were again asked to . ress their opinion. It was found that of those who earlier supported the cX~ernment only 111 still held the same opinion, and of those who were initially ~oainst the government a number n2 confirmed their earlier view. The rest of ;I;e sampled individuals had changed their attitude. Show how these data can be analysed to test the null hypothesis that there has been no significant 'hift in government support. S Also, as a particular case, verify that for oc = 0'5, the appropriate test s[;ltistic for the null hypothesis is N(I1( -112)2/(11 1 + 112)(N -1I( -112)

having a X2 distribution with a single dJ. 155 In seismological research, the arrival time of a given wave is a normally distributed random variable with mean () and variance O'I. However. a proportion p of the observations are affected by an uncertainty arising in most cases rrom the difficulty of identifying the exact beginning of a phase when microseisl1ls are present. Observations ~ffected in this manner may be assumed to be normally distributed with mean (() + Ji) and variance O'~. Derive the probability distribution of x, the arrival time of a random wave. If from past experience p is known, and as a simplification it is assumed that "( = 0'2 = 1, derive the maximum-likelihood equations for the parameters () and I' on the basis of n random observations Xl' X2'·.·' X n · Also, in standard notation, prove that the elements of the information matrix of the estimates of () and Ji are:

_E[02~;;L]

= 1I[1-pJi2(1-rd];

_E[020010ga,lLJ = np[1-11 (rl-r2)]; and _E[02 10g L] = IIp[rt +Ji2(rt -3r2+2r3)]; 2

0112

where, for integral k,

f J2n 00

r

=

k-

1

-00

y2 2

e--/-dy(1 +A

e-P)'t'

and

A

== (1 -pP ) e- p2 / 2 .

l56 In bacterial counts with a haemacytometer, the distribution of the random variable X denoting the number of bacteria per quadrat is generally known to be of the Poisson form with an unknown mean /I, However, it is ~ometimes difficult to count correctly if there is a large number of bacteria 111 any quadrat; and in order to avoid such non-statistical errors it is convenient to record separately all quadrats having t or less bacteria each, but to pool the rrequency of all other quadrats with more than t bacteria,

110

EXERCISES IN PROBABILITY AND STATISTICS

In a random sample of N quadrats it was found that II, quadrats h bacteria each (0 ~ I' ~ t), and there was a total of IIR quadrats each ~~' more than t bacteria. For given t, derive the equation for fl, the maXill1Ulll 1!~e1ihood estimate of fl., and hence show that the large-sample variance:, fl. IS var(fl) = II/N[P,+P,{(II-t-1)+f1.p,/(l-P,)}], where the probabilities

P(X = t) == p, Also, verify that, for t

and P(X ~ t) == P,.

= 0, var(fl) = (e"-I)/N.

157 A binomial variable X has a distribution defined by the point prob. abilities

P(X = s) = (;)pSqll-S,

for 0

~

X

~

n,

p being the conventional probability of a "success" and (p + q = I J. If thi,

distribution is truncated so as to exclude both the terminal classes, prOVe that the mean of the resulting distribution is greater than lip if p < q. Further, if a random sample of II observations is taken from this truncated distribution and I' successes are observed, prove that p, the maximum. likelihood estimate of p, is obtained from the equation

P=!:..n [I_ + (l-PHpn-I_(l_p)n-I}] (I_pn-I) , A

and that for large n

pq

var(p) = - . II

(l- pn _qn)2 [(1_p"_qn)_lIpq(pn 2+qn 2_pn 2qn 2)],

-::c-:----,-------,-,:---:-'-::----;;-"---'----=-----.---::--:;~~

Alternatively, if for large II, p* = 1'/11 is used as a simple approximate estimate of p, determine the bias of this estimate and verify that exactly

pq [(1- pn _ qn) _Ilpq(pn- 2 +qn- 2 _ pn- 2qn- 2)] (1 n n)2 It -p -q

var(p*) = - .

158 An insect pest lays its eggs on the fruit blossom of mango trees, but not all eggs hatch to become larvae, as some of them are destroyed by weather hazards. The only definite indication of the presence of the pest is in wormed ripe mangoes, which show up when the fruit is sliced. In order to assess the extent of damage caused by the pest in an orchard, samples of n mangoes each were examined from different trees, and records of wormed fruit kept from k trees in whose samples there was at least one damaged mango. The total number of fruit obtained from a tree is large compared with the sample size n, so that the effect of sampling from a finite population may be ignored, and the trees in the orchard may also be treated as independent. If the probability of obtaining a wormed mango from a tree is a parameter p, find the probability of the observed distribution of 1'1>1'2, ... ,r1 wormed mangoes from the k trees. Prove that the equation for p, the maximumlikelihood estimate of p, may be put in the form

p = Po[1 - (1 - p)n],

eSTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

111

P is the observed relative frequency of wormed fruit in the k samples. wheresh~w that for large k Also, pq(l_q")2 var(p) = nk(l" -q -npq"1)' (p+q = 1).

9 for an hereditary abnormality, the probability that a particular child

~5 family is affected depends partly on his or her birth-rank. In order to

~n astigate this effect of birth-order, a complete family (i.e. without any

~~s~arriages or stillbirths) of n children was examined, and it was found

hat r of them were abnormal. t Assuming that there were no mUltiple births in the family so that the n children can be ranked without any ties, suppose the sum of the birth-ranks of fer abnormals is A. Then, for constant II and r, show that, on the null ~~pothesis that birth-order has no effect on the occurrence of the abnormality, the probability of obtaining the sum A is

P[A;n,r] = S[A;n,r]

1(;),

where S[A; n, r] is the number of ways of selecting r of the first n natural numbers such that their sum is A. Prove that the function S[A; II, r) satisfies the following recurrence relations: (i) S[A;n,r] = S[A;n-l,r]+S[A-n;n-l,r-l); (ii) S[A;n,r) = S[r(n+l)-A;n,I']; (iii) S[A:II,I'] = S[!II(Il+1)-A;II,II-r). Hence deduce that A has a probability distribution symmetrical about its mean r(n+ 1)/2. Next. by considering the coefficient of (JA in the Taylor expansion of r

F«(J; n, r) ==

fl (Ji(l- (In-i+ 1)/(1- (Jil, i= 1

prove that the probability-generating function of A is F«(J; n,r)

IG).

Use this result to determine the cumulant-generating function of A, and then derive explicitly the cumulants of A. Hence verify that the second and fourth cumulants of A are K2

= (11+ 1)./12;

K4

= -(n+ 1).{n(n+ 1)-.}/120,

[. == r(lI-r)].

Finally, by applying a correction analogous to Yates's correction for continuity, obtain a suitable normal approximation for testing the null hypothesis of no birth-order effect on the incidence of the abnormality. 160 In the study of an hereditary disease which may develop at different ages, it is often found that there is, on the average, greater similarity between the ages of onset of the disease for a parent and child from a single family than between two affected persons chosen at random from the general population.

112

EXERCISES IN PROBABILITY AND STATISTICS

Suppose that the disease may be produced by either of two different genes, say A and B, it being assumed that there are no families in which tar! the genes occur. Furthermore, in families in which the disease is deterJrtiboi~ by gene A, suppose the age of onset is distributed with mean m 1 and vari hb; af ; and it may also be assumed that there is no correlation between the aganI), onset for parent and child within each family. Similarly, gene B gives a ~S(i age of onset m2 and variance a~. e~. If the genes A and B are indistinguishable in their effects, except as rega age of onset, prove that for such a mixed population the correlation betW::: the ages of onset for parents and children is :

_ [1 +

p -

1tlaf+1t2a~ ]-1 1tl1t2(m1 -m2 ) 2 '

where 1tl and 1t2 are the proportions of the A and B genes i~ the populatior. and 1tl +1t2 = 1. Hence deduce that for al = a2 = a, P = o·s Iflm 1 -m 21~20 As a generalization, suppose there are g different genes A, B, ... , G in proportions 1tl' 1t2, ... ,1tg respectively. Then show that under similar con. ditions and with same notation

p =

[1+ f 1t a;/:[ 1t 1t,(m -m,)2]-I, r

r=1

r

r

r,*'

the second summation extending over all pairs of genes. each pair beinl counted only once. 161 On a certain biological hypothesis, the occurrence of a rare hereditary human disease may be explained as due to the inheritance of either of two gene\ A and B, which are known to occur in the population in the ratio of I:A (>Oi The genes are indistinguishable in their effects, except as regards the age 01 onset of the disease; and, as a simple model, it may be assumed that for such a mixed population, the age of onset is a continuous random variable whose distribution is compounded of two unequally weighted univariate normal distributions such that (i) the weights are in the ratio of 1 : A; (ii) on an appropriate scale, the means are - In and m; and (iii) both the variances are equal to unity. Find the probability density function of the age of onset distribution, and hence deduce that this distribution will be bimodal if, and only if, Ilog AI < 2m(m2 _1)1 - 2 log[m + (m 2- 1)1]. As a generalization, suppose the variances of the compounding distribution are af and a~. Prove that in this case the least possible separation between the means of the compounding distributions for bimodality is

Gfa'" where a 2 is the harmonic mean of

1·840-,

at and a~.

162 An hereditary human abnormality like juvenile amaurotic idiocy is believed to be due to a single recessive gene, and it is therefore usual to find abnormal children in families produced by heterozygous parents who are both themselves normal. If p is the probability that the offspring of two heterozygous parents will be abnormal. then for the estimation of p allowance has to

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC.

113

d for the fact that a number of families are excluded from the observed m8 eause of the absence of abnormal children. Consequently, sampling is Jat3 : ; to families de~ived from normal parents such that each family has at ,:(ln ti e abnormal child. Ica st onpose a total of N families are sampled, there being ns families of size s, SUPI ~ s ~ c. If there are R abnormals in the total sample, prove that \I'here xi;;;um-likelihood estimate lJ of q (== 1 - p) is obtained as the real root (he: mathan unity of the equation Ill:

±

"Iher

R s.n. (1-lJ) = •= 1 (1- ~. , show that the information contained in the sample for the estimation

,\ Iso, I)f

q is c

L s. n.(1- spqS- 1 -

qS)/pq(1 _ q")2.

s= 1

A test of significance indicates that the estimate of p is significantly greater Ihan the Mendelian expectation of p = !; but, as an alternative to the rejection )f the hypothesis, it is suspected that this difference may be due to the fact :hat single abnormal cases are not always reported. If, then, families with at Icast two abnormals only are considered, modify suitably the maximumlikelihood estimation of q, and hence derive the large-sample variance of this estimate. 163 A simple method for comparing the fitnesses of organisms of two different phenotypes, which may be assumed to correspond with two different genotypes A and B, is to compare their viabilities before the age at which they are scored. Then any divergence from Mendelian expectation can be ascribed to differences of viability before this age. Suppose that at the stage of sampling, the A and B organisms in the population are in the ratio of 1 : 0, so that interest centres on the estimation of the unknown parameter 0; and that in a random sample of n organisms, r were observed to be of type B. Prove that the statistic r/(n-r+l) is an asymptotically unbiased estimate of O. Also. by considering a suitable logarithmic expansion of r/(n - r + l), show that for large samples var[r/(n-r+ 1)] =

0(1 + 0)2 [20 ] n I +n+ O(n- 2 ) •

For z > 0, prove that

1 L [/k+l k! n (z+m) ] =-. k=O m=l Z 00

If sampling is confined to samples of size n which have r 'I: n, use the above result to evaluate the expectation of 1/(n - r). Hence verify that, for n sufficiently large so that the probability of r = II is negligible.

E[r/(n _ r)] '" 0

[1 +(1: 0) + (1 + O~~2 +0) + 9(n- 3)] .

Discuss why the estimate r/(n - r + 1) of 0 is better than the maximum-likelihood estimate r/(n- r).

114

EXERCISES IN PROBABILITY AND STATISTICS

164 At a factory mass-producing units of electronic equipment, the assernbl units are tested before being packed for despatch to customers. The probabU~ that a unit will be found to be non-defective is a constant p, (0 <: p "II) In order to estimate the parameter p, inspection records are kept, and il I) observed that a total of Sn non-defective units were inspected up to the st II of finding n defective items in all. a~( Assuming that the production process is under statistical control, det mine the moment-generating function of Sn and hence, or otherwise, pr:~· that Sn(n + Sn)/(n + 1) is an unbiased estimate of var(Sn). Also, find the fi: t four cumulants of z = S,,/n, and use them to prove that for the statist:: p* = z/(l +z)

and (q

=

I-p),

so that p* is an asymptotically consistent estimate of p. 165 It is known that a particular gene with two alleles A and a controll melanism in lepidoptera, and it is assumed that the alleles occur in the popu. lation in proportions (1- (J) and (J, where (J is an unknown parameter. O[ the three types of genotypes AA, Aa and aa produced under random matin~ the darker type AA is the dominant form, and, in general, it is possible to distinguish the three genotypes phenotypically. However, in areas of high industrial pollution, it is occasionally possible to misclassify a heterozygote Aa as an AA. In order, therefore, to allow for this error it is decided to divide the whole sampling region into two strata. Of these two, the first stratum refers to those parts of the sampling region where there is, on some specified standard, no pollution to affect materially the classification of the heterozy· gotes. The second stratum consists of such other areas where the pollution is known to be sufficient to give a probability p (unknown) of misclassifying a heterozygote as a homozygous dominant (AA). From the first stratum a sample of v observations is taken and the fre· quencies of the three genotypes AA, Aa and aa are found to be YI' Y2 and }') respectively. An independent sample of size n is also taken from the second stratum, and the corresponding frequencies of the genotypes, as classified by observation, are XI' X2 and X 3 • Use the data from the two samples to deter· mine the equations for () and p, the maximum-likelihood estimates of (J and p. Also, verify that for large samples A

var(O) =

(J( 1- 0 2 )

2[ I' + (2n + v)(J]

;

var(p) = (1- p)[(1- p) + (3p - 1)(J + (v/n){ 1 + (2p - l)(J)]. 2(J(1-0)[v+(2n+I')(J] , (1-(J)(1-p)! corr(p.O) = (I +W"[O-p)+(3p-l)0+(v/nj{ 1 +(2p-l)(J}p· A

and

ESTIMATION, SAMPLING DISTRIBUTIONS. INFERENCE, ETC.

115

A binomial distribution has (n + 1) classes and an unknown "success" ter p. Suppose the first k (0 ~ k ~ n) classes of this distribution are paraJ11~ed and that a normalized distribution of a random variable x is defined (rllnc~he integral range (k ~ x ~ n), If denotes the rth moment of x about IIVCr rigin prove that (hc 0 ' = r.x+np and = (k-l)r.x+ {I +(n-l)p}"l'

166

"r

"1

\I

here II.

==

"2

~. (kn)pk qn-k+ 1;

p+ q = 1;

"0 == 1 _

"0

k~l (n) pX if-X. x=O

x

1'0 determine "3 and "4' 0\ s S~ppose N independent samples, each of size n, are obtained from this ~cated distribution such that there arefx samples having exactly x successes, ~~ that r. fx = N. If for all integral r, n

T,. ==

L

xr fx/N,

x=k

prove that a simple moment estimate of p is p* = (T2 -kT1 )/[(n-l)T1 -n(k-l)].

By using a suitable logarithmic expansion for large N, find the expectation to O(N- 2), and hence deduce that p* is asymptotically lInbias~d, Show further that the large-sample efficiency of p* as compared with the usual maximum-likelihood estimate is given by

or logf(n-l)p*+k}

dr (p*)

== pq[(n-I)T,-Il(k-Ilf

- [llpq-(r, -Ilp)(" +p -k)][ll(k-I){(k-I)-(1l-3)p} +rd(n-I)(Il-3)p-(lIk- 31l+k + I)}]

In particular, verify that (i) For k = 0,

eff(p*)

=

[1+2(n-l)pq/{1+(n-l)p}2r 1 ,

and its minimum value for variation in p is

-t.

(ii) For k = 1,

eff( *) = (n-l)p(1-q") p [(n-l)-(n-3)q][1-npq" l_ q"]' and that for variation in p the minimum of eff (p*) is attained for a value of q =1= 1 which is the root of the equation np2 qn-2 [(n-l)q"-n(n- 3)q+(n-l)2]

= 2(l-q"f

167 The number of insect eggs found in a nest is known to be a random variable having a Poisson distribution with mean A. But a nest can be recognized as such only if one or more eggs are present. In addition, due to faulty observation, a proportion (0 < () < 1) of the nests having only one egg each are overlooked. Thus, assuming that there are no other observational errors or classification, the distribution actually observed is a truncated Poisson distribution in which the zero class and a proportion of the ones are missing.

e

e

116

EXERCISES IN PROBABILITY AND STATISTICS

Determine the probability distribution of X, the number of eggs pet as observed, and show that nt

E(X) = ..1.(1-0 e- A)/[I-e- A (1 + 0..1.)], and var(X)

=

..1.[1-(..1.+ 1) e - A+ 0 e - A{e - A_ (..1. 2 - A+ I)} ]/[1- e- A(1 + 0),)]2,

Hence deduce that, irrespective of the value of 0,

. 1. 2 e- A var(X) 1 1- (I-e A)2~~I-(A+I)e

A'

168 As continuation of the previous exercise, suppose a sample of N ne' was observed and there were II « N) with just one egg. Use this sample~' formation to prove that the maximum-likelihood estimates of A. and 0 :. obtained from the equations It

e= 1- n1/(N - n)(x* -1),

and

where x* is the weighted average of the sample observations for X ~ 1 Hence deduce that if A is small, then a simple approximation for 1 is 1 = x*-2e- x·/ 3 • Also, derive the large-sample variances of the estimates and verify that (I-0)te- A/2 (I-A-e- A) corr(A,O) = {l-(I+A)e-A}t{(1-e A)(I-Oe A)-Ae A(l-O+AO)}!' A

A

169 The zero class of a Poisson distribution with parameter A is truncated. and a normalized distribution of a random variable X is defined. In a random sample of size N from this distribution, n. observations were found in the cl~\ 00

X = r

for

r ~ 1, and

I

n. = N .

•=1

By considering a linear function of the sample class-frequencies, sholl' that an unbiased estimate of A is given by 00

..1.* =

I

r.n.IN .

• =2

Prove that var(A*) =

~.

[1 +

eA~ 1 J.

and hence deduce that the efficiency of ..1.* as compared with the usual maxi· mum-likelihood estimate 1 is

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC.

117

. by using the same estimating technique, prove that AgaJO, 2112+ N ),* T= N2 ,.111

un

biased estimate of var(A.*), and

A.

I

var(T)

[

..1.(7 - 2A)

= N 3 ' 1 + eA - l

A random sample of size N is obtained from a Poisson distribution 17°1 parameter A. in which the zero class has been truncated. The observed \\111 .' .. cIasses are 111,112,113"'" were h ....frequenCles m t he remammg

, I.I~S

00

L1 IIi = N.

i=

If it is desired to estimate the parametric function

(} == e- A, provc that an unbiased estimate of (J is

(J*

= (Tl - T2 )/N,

where 00

Tl

=

I

and

112r- 1

r= 1

Also, show that the exact variance of (J* is given by var((J*) = (1- e - 2).)/ N. lienee compare theA sampling behaviour of (J* with that of the maximumlikelihood estimate 0, and comment on the usefulness of (J*. 171 An insect lays eggs in colony sites distributed over a region, but the eggs are liable to be eaten by predators, so that it is possible to find a site in which there are no eggs. Suppose (J is the probability of selecting a site which has eggs. and the distribution of x, the number of eggs in it, may be assumed to be a truncated Poisson with parameter A.. Determine the probability distribution of x for given fJ and ).. In a random sample of N sites, there were I1x sites with x eggs, where ~ ~ 0 and :E1l_~ = N. Prove that the maximum-likelihood estimates of (J and i. are given by the equations

(j = 11*/N and x* = }./(l-e- i ), where 00

11* = N-llo

and

II*X* =

L x./lx· x=1

Also. verify that {j and }. are asymptotically uncorrelated and that their large-sample vari:mces are

A. (l-e-il var(A.) = N(J' 1-(..1.+ l)e A

var(O) = (J(1-(J)/N

and

i.'

118

EXERCISES IN PROBABILITY AND STATISTICS

whence

A

A

2A

NO ~ var(,l,) ~ NO'

In a microbiological experiment. a homogeneous group of M gui pigs were randomly divided into k groups, there being Ini animals in th:c group (1: mj = M). Each of the guinea-pigs in the ith group was infee": through respiration by ni Brucella suis micro-organisms, and the numbe It: survivors observed in the group was rj, for all (i = 1,2, ... , k). r~ Assuming that (i) the probability of one micro-organism killing its host is an unknol\ parameter P (small), and r (ii) the micro-organisms act independently of each other. prove that t~e expected proportion of surviving guinea-pigs in the ith grouPI' e- nlP approximately. Determine the sample likelihood of all the survivors in the k experiment.. groups, and deduce that the maximum-likelihood equation for the esthna; P of pis t 172

k

P= L (mi-ri)~;/N(I-e-.'i;;), i= 1

where ~i == niP, and N is the total number of micro-organisms used in tht experiment. Also, prove that for large samples var(p) =

p21tl mj xl (eX; - 1)

-1

(Xl

== niP).

Finally, by maximizing the information obtained from anyone group 01 animals in which n (say) micro-organisms per guinea-pig were used, sho~ that the most economical value of n for fixed p corresponds approximatel) to 20·3 per cent survivors in the group. 173 The dilution method is a standard means for estimating, without direct count, the density of bacteria in liquids like water and milk. The method consists in taking samples from the liquid, incubating each sample in a suitable culture medium, and observing whether any growth of the organisms hal taken place. The estimation of bacterial density by this method is based upon the following assumptions: (i) The organisms are distributed randomly throughout the liquid. (ii) Each sample from the liquid, when incubated in the culture medium. is certain to exhibit growth whenever the sample contains one or more organisms. (iii) The sample volume v is small compared with the total volume I', and the space occupied by an organism is negligible relative to v. If there are a total of b bacteria in V, then prove that the probability 01 the sample being sterile is approximately p = e- cJ ".

where (j == bfV is the bacterial density per unit volume. To estimate (j, samples with k different degrees of dilution (i.e. k different values of v) are taken. Suppose there are ni samples for the ith dilution and the volume of each of these samples is Vi' and that Sj of them are found to be sterile (i = 1,2, ... , k). Determine the maximum-likelihood equation for J, tbe estimate of (j.

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

119

rove that for large samples AIso, P A

1'5415 2

var(b) ~ ~ approximately,

N is the total number of samples used in the whole experiment. "here 4 An experimenter wants to test whether two different methods oflearning 17c equally good. He has two groups of subjects, A and B, unmatched in any .Ir and has both groups learn the same material. Group A learns by one \\J.rhod B by the other. Assume that the two methods really do give different n~~ults ~ith error variances O'f and O'~ respectively. S fl (i) If the total number of subjects in the two groups is n (fixed), prove that the most precise method of estimating the difference is to allocate the number of subjects in the two groups in the ratio of 0'1 : 0'2' (ii) Alternatively, if the experimenter wishes that the variance of the observed difference will be a specified quantity c, and he is free to choose the total number of subjects, prove that the minimum number of subjects will be required if the sample sizes are again in the ratio of 111 : 112'

Evaluate the group sizes in (i) and (ii). (iii) Also, determine, under suitable conditions, the group allocations if the total number of subjects is n (fixed), and the experimenter wishes to ensure that the variance of the observed difference is also a preassigned quantity c. Hence verify that if the two standard deviations are equal to 11, then the group allocations are in the ratio of (l ±).,) : (l =+= 2), where 40'2]-! ), == [ provided IlC ~ 40'2.

1----;;;;- ,

175 As a continuation of the previous exercise, suppose that the costs of making an observation in the groups A and B are in the ratio 1 : 0(, where 1> I. Thus for groups of size 111 and 112 of A and B respectively, the total cost of sampling is proportional to t = /11 +0(/12' Determine the optimum allocations for (i), (ii) and (iii) when t is replaced by 11 in the specifications stated in the earlier exercise. 176 A well-known feature of the classical Poisson distribution is the equality of its mean and variance, but it is observed that in many biological situations the variance is characteristically larger than the mean. The negative binomial distribution gives a good representation of such phenomena, and its point probabilities may be defined by (k+x-l)! px P(X=x)= x!(k-l)!'(I+p)"+x' for

X~O,

where k and p are positive parameters. Prove that the cumulant-generating function is K(t) = - k log[ 1- pte' - I)],

and hence, or otherwise, determine the first four cumulants of X. Suppose that in a random sample of N observations, the observed frequency for X

=

x is lx,

L Ix =

x=o

N.

120

EXERCISES IN PRO R A BIll T Y AND STATISTICS

Prove that the maximum-likelihood estimates p and equations

.x-pk and

=

k are obtained frOIll

t~

0

,

I a.~f(k+x)-log(1 +x/k) = o. x=o where .~ is the sample mean and at is the observed relative frequency for X::. for all t ~ o. Also. determine the elements of the information matrix 01; and k. Hence verify that for large samples ' var(fJ) = p( I + p)2r:N[( I + p)kr - p]: var(k) = k(l +p)/N[(l +p)kr-p]; cov(p.

and

k) = - p(l + p)/ N[(l + p)kt - p],

I""

IXx/(k + X)2 and IXx == E(a x). ..-=0 The maximum-likelihood equation for k is cumbersome, and two simplt, methods for determining an estimate of k have been suggested. (i) The first estimate k* is obtained from the equation

where

r ==

k* log(l +x/k*) = 10g(N/Jo) derived by using the relative frequency of zeros in the observed dis. tribution. Show that for large samples var(k*) (ii)

!! S2

=

(l+p)[(l+p)k+l_(l+p)-kp] N[(1 + p) log(l +p)-pj2

is the usual unbiased estimate of the population variance, then

k. the second estimate of k, is obtained from the equation

k=

.~2/(S2_X).

deduced by using the expectations of .x and Prove that for large samples

S2

as estimating equationl

var(k) = k(l + p)[2(k + 1)(1 + p)-l] Np2

Comment on the possible use of k* and likelihood estimate of k.

k for obtaining the maximum·

177 In plant ecology it is customary to use the method of quadrat samplin, for the estimation of the abundance of a given species in a region. In this method, a square lattice-the quadrat-is dropped at random points in the area, and the number of plants of the given species found in the quadrat is counted. Repeated sampling then gives a frequency distribution of the number of quadrats containing k (~O) plants. In describing such observations mathematically, it is simplest to assumt that an individual plant has no area, and further that the plants are distributed randomly in the region sampled, so that the Poisson distribution is applicable However, it is characteristic of data obtained from quadrat sampling thaI the variance is greater than the mean, due to the clustering of the observations. so that the Poisson model usually gives a bad fit.

ESTIMATION. SAMPLING DISTRIBUTIONS. INFERENCE. ETC.

121

more refined model, it is assumed that clusters of plants are distributed

As

tom and the random variable xdenoting the number of clusters in a

.11 r~n t h;s a poisson distribution with mean /L. Further, let y be the number qual rnats above one observed in a cluster such that the probability 'If p a P(y+ 1) = e-).. }//y!, for y ~ O.

Prove

that the probability of observing k plants in a quadrat is P(k)

=

k e-(I'+n)./Lr.(..1.r)k-r JI r ! (k _ r) ! ,for

= e-I',

for

k ~ 1

k = O.

lI~nce p~r

show.that the cumulant-generating function of the number of plants quadra I IS K(t) = It[exp{ t+ ..1.(e t -1)} -1],

ld use it to evaluate the first four cumulants of the distribution . Suppose that in a ran~om sample of N quadrats there were 110 qu~drats lI"ilh no plants and III wIth one plant each. Use these data to obtam the maximum-likelihood estimates of 11 and ..1., and also evaluate their large-sample ((lvariance matrix. Hence deduce approximately the variance of the sample mcan number of plants per quadrat. Comment on the above estimation procedure. •11

178 Explain clearly the term "best estimate of a parameter" in the theory of least squares. (i) If x and yare two independent unbiased estimates of a parameter () with variances aI and a~ respectively, find the best estimate of () and show that its variance is aIa~/(aI + a~). (ii) Suppose that XI' X2"'" XII are a sequence of random observations such that E(Xi) = I' and var(xi) = ;.ia 2 for all (i = 1,2, ... ,11), where jl, ). and a 2 are independent parameters. By considering a linear function of the Xi' or otherwise, prove that

T=

III i= I

;:-i Xi

/,,-1

L;,i

j= 0

is the best estimate of I" Find the variance of T. Hence prove that for all 11 > 1. var(T) < }.a 2 ; and, for large 11, var(T) '" 0 or (). -1 )a 2 according as ;. < or > 1, respectively. (iii) What is the limit of var(T) as ;. --+ I? 179 The tensile strength of a steel cable produced by a factory has an unknown value JI. and to estimate its magnitude, a series of /I independent assessments X I. X 2, ... ,x" were determined by an investigator A. Since these mcasurements were. as far as possible. made under similar conditions. it may be assumed that they have the same but unknown variance a 2 , independent of I'. Another investigator B made a separate series of v independent measurements .1'1' .1'2" •• ,y,. of the same tensile strength, and these may also be assumed to have variance a 2 • However. the y observations appeared to be consistently larger than the X values. It is suspected that this discrepancy was presumably due to a wrong setting of the measuring instrument by investigator B. so that a constant but unknown bias () (> 0) was introduced in the y measurements.

122

EXERCISES IN PROBABILITY AND STATISTICS

Show that the y values cannot provide any information about Il, but th they may be used to estimate the magnitude of and to derive a more pre/' estimate of (12. Find the estimates and, under the usual assumptions of nil( mality, determine a small-sample test of significance for any specific hYPoth Or. about e. Also, if it is presumed that the wrong setting of the measUring ~I. strument by B might have affected the random errors of measurelllein. indicate how this could be verified on the basis of the sample data availabln! e.

e

180

If X, Y and Z are random variables such that

X=X1+ X2;

Y=X2+X3;

andZ=x3+ x4,

where the Xi are uncorrelated random variables each with mean 111 and varianCt (12, find the product-moment correlations between (X, Y), (Y, Z) and (X 2 Comment on the results obtained. •I Also, determine the correlation between X and Z when (i) Y = c, a constant, and (ii) Y = AZ, where A is a constant. Verify that, as ;, --> 00, the limiting value of this last correlation is unit) 181 Prove that the product-moment correlation between two continuou, random variables must lie between -1 and + 1. . If x is a unit normal variable and another random variable y is defined by the quadratic relation

where <x,

P and yare constants,

prove that

corr(x, y)

= P(P2 + 2y2)-1-.

Discuss the relevance of this correlation as a measure of association between x and y in the light of its limiting values for P = 0 and y = O. 182 The ranking given by a judge to N paintings Alo A 2 , ••• , AN is 1,2, ... ,N A second judge is in general agreement with this ordering, but he divides hi, ranking into k tied groups of 11 1,11 2" " , 11k paintings, where 1:l1i = N. Thul the paintings A l' A 2 , • •• , An, receive the same rank; the next An,H An. + 2.' .. , All. +n, are tied, and so on, till the kth group of Ilk paintings haw the same rank. These k ranks are in an increasing numerical order. If r

L

l1i

= tr ,

for

1.-s:; r .-s:; k,

i= 1

prove that the coefficient of rank correlation between the two rankings is p = [ 3 J2I1ititi-dN(N2-1)r

Verify that for N

= 10, and

111

p =

= 2,

112

=

4,11 3

= 3,

114

= 1,

(46/55)1- '" 0·9145.

183 Explain clearly the difference between a goodness of fit X2 for a frequent) distribution with k classes and a X2 obtained from a 2 x k contingency table Illustrate your answer by suitable examples.

ESTIMATION, SAMPLING DISTRIBUTIONS, INFERENCE, ETC.

123

The observed row. frequencies i.n a 2 x k contingency t~b1e are a" a 2 ,· •• , ak bI' b2' ···, bk• Usmg the notatIOn I1 j == aj+b j, for all (I = 1,2, ... , k) d ,,0 Laj == A, Lbj == B, and A+B == N, prov

e that the X2 for testing independence can be expressed as

X2

=

;;[Jl

af/l1 j -A 2/N).

\Vhal are the degrees of freedom for this X2?

4

Characteristic functions

I If X and Yare two independent normally distributed random variab\ with zero means and unit variances, show that the characteristic function t: the random variable Z = XY is (1 + t 2 )-t. U Also, find the characteristic function of Z if the correlation coefficier' between X and Y is p. Deduce the mean and variance of Z in this case. I 2 The /I independent random variables X 1> X 2.· ..• Xn can each take Ih values + 1 and - 1 with probability t. Determine the characteristic functi~ of X r • and hence the characteristic function of the random variable n

Y=

L

Qr

X r • where

al' Q2.·

•••

an are constants.

r= I

If Q r = rr. use the fact that sin 2 kt = 2 cos kt . sin kt. or otherwise, Iii show that as /I -+ CX) the probability density function of Y tends to (-I~Y~I).

!(Y=y)=t

3 A random variable X is defined over the continuous range (0 ~ X < 1.1 and for x > 0 the probability for X > x is F(x). In addition, there is a fillil( probability e- A for X = 0; and in the range (0 < X < 1). F'(x) = -i.e whilst for X > 1, F'(x) = -), e- AF(x -1). Prove that the characteristic function of X about the origin is

A.-it A. eit - it e A' and hence obtain the mean and variance of X. 4 If the random variables X and Y have a bivariate normal distribution wilh zero means. variances ui. u~. and correlation p. derive the joint characterislk function of X and Y. By using the Fourier Inversion Theorem. obtain an integral expression for the probability mass P in the positive quadrant of the (X, Y) plane. Hence prove that 1 . I I P = -+-smp. 4 2n and deduce the probability masses in the other three quadrants.

5 If X and Yare standardized normally distributed random variables wilh correlation p. prove that the characteristic function of X and Y is cP(t1>t 2) = exp -t(ti+2ptlt2+t~). Given that P(X ~ :x) = a.

P( Y ~

fJ) = h. 124

and

PIX ~

!x.

Y ~ 11)

= c,

CHARACTERISTIC FUNCTIONS

125

, e rJ. and Pin terms of a and b respectively, Also, by using the Fourier ,::I~rn~l~ Theorem on 4>(tl, t 2 ), show that p satisfies the equation 11I\~rslo

j

00

2nc, et (a 2 + p2)

=

~ H,} - l (oc)H,}-I (P) , 'I

" L...,

j ; I)

,

, H (z) is the mth-order Hermite polynomiai in z,

" h~r~

.'

If logical and sociological data often give rise to a kind of heterogeneous /I ,:~on which cannot be approximated by any of the usual probability .. Ir~ liS In such situations, as suggested by Karl Pearson, the popUlation rntl,~'considered as composed of an unequally weighted sum of two standard 11l,1} 'butions, If then a random variable X has a probability distribution which 'h'I{I, sum, in the ratio of 1 : A, (), > 0), of the unit normal distribution and I' 'tl~er normal distribution with mean m and variance (J2, obtain the prob:;::~itY densit,y .ru~ction of X, and prove that the characteristic functiqn of X ,I!lout the OrIgm IS 4>(t) = e - t,2 [1 + A exp{ itm - 2 «(J2 - I)} ]/(1 + A).

tt

,,'nce calculate the first four cumulants of X. Also, if v2 denotes the weighted 'I\~ragc of the variances of the compounding distributions, prove that 1. 2 < var(X) ~ l,2 +!m 2 . 7 Given a random sample of n observations from a population having a probability density function

!(X=x)= for

1

(1 +;,)j2n

-

00

[e-txI+~.e-(X-III)2/2a2],

< X <

(J

and

00

prove that the distribution of the sample average

n

n

(I+A)n~' r~o

A > 0,

x is

[ (n) Ar n2(x - r111/n)2] _ r (l1-r+r0'2)t' exp - 2(I1-r+ra2) .dx

( - 00

<

x<

00).

lienee, or otherwise, calculate the coefficients of skewness and kurtosis for Ihis distribution, and obtain their limiting values when (i) A and (J are constant, and 111 -+ 00; (ii) A and 111 are constant, and (J -+ 00.

X is a weighted 'um of two normal distributions, each with unit variance but means 0 and 1/1 respectively, the weights being in the ratio of 1 : A, (A > 0). Given a random 'ample of 11 observations x I, x 2 , .•. , Xn from this distribution, obtain the joint characteristic function of K The probability density function of a random variable

n

I x; r= 1

n

and

I x,;.jIi. r=1

Invert this characteristic function and, by integration over the range of variation of the sample mean, prove that the characteristic function of the

126

EXERCISES IN PROBABILITY AND STATISTICS

sample sum of squares is

rto C)Ar.exp [(it)rm

2

(1- ;;)}1- 2it)]/(l + A)n(1- 2it)(n-ll/2.

Deduce as a special case that the sample sum of squares from a unit norlll' population has the X2 distribution with (/1-1) dJ. ' 9 Given a random sample of n observations Xl> X2" •• ,Xn from a nOflll. population with mean m and variance a2 , show that the joint characterisl' function of Z j and is I,

zJ

4>(t 1, t 2 ) =

exp{ - tf!2(l- 2it 2 )} . (1- 2it 2 )-t,

where Zj = (xj-m)/a, for all (j = 1,2, ... , n). Hence derive the joint characteristic function of n

n

L zJ

L zlJn;

and

j =1

j =1

and by inverting this characteristic function prove that the sample mean and the error sum of squares are independently distributed. Obtain explicitly the distribution of the sample standard deviation and show that PI' the coefficient of skewness for this distribution, is

l[2A 2 - 2 + (l/v)]/(l- A2)3, where

v == 11-1

and

(~rr(V; 1)/rG)·

y ==

Establish the asymptotic formula

y-

(1- L+3L2)'

and then verify that for large v

1(

1

1)

PI - 2v 1 + 4v + 8v 2

•

10 If X and Y have the joint bivariate normal distribution with parameters (m x , m)" ax, a)" p) in standard notation, evaluate the expectation of

exp(it II e 2 + it 221] 2 + it l2 e1] + it l e + it 2 1]),

the t's being real constants, and ( =

(X -m..}/ax~,

1] =

(Y -my)/ay~.

Hence, given a random sample of n observations from this population, prow that the joint characteristic function of n

XI

=

L

n

ejl../n,

j= I

YI =

L

n

1]j/../n, X 2 =

j=1 n

Y2 =

L

j=1

j=1 n

1]J, and Z =

L

L

j=1

ej1]j

eJ,

127

CHARACTERISTIC FUNCTIONS

~

I (l_p2)n/2 l).-n/2 exp- 21). [(I-2itdd+(l-2itl1)t~ +2(P+it I2 )t l t 2], I).

== [(I-2itIIHI-2itnl-(p+itd 2].

. mally invert this characteristic function, and by integration over the I or pie means show that the joint characteristic function of ,a 111 II

= (n - I )s;/O";(I- p2).

v

=

(n-l)s;!0";(l_p2),

IV = (11-

where s;,

I)rsxs),/O" xO"),(I- p2),

s; and rsxs), are the sample second-order moments. is (1_ p2)ln-I)/2. I). -In-I)/2.

II If S2, s~ and rsxs y are, in standard notation, the second-order moments Icrived frotn a random sample of size II from a normal bivariate population :vilh second-order moment parameters 0";, 0"; and pO" xO")" then it is known that for II, v, IV and 1)., as defined in the previous example,

E[exp(it II II + it 22 v + it 12 w)]

= (1- p2)ln- 1)/2 . I). -In- 1)/2.

Using the result that. for t real, (I - it) - P is the characteristic function of a I' variable with parameter p > 0, prove that the joint probability density funclion of the distribution of II, v, and w is ( I p2)ln-I)/2 -

2n-lfir(Il;1)r(Il;2) .e

-i(u+v-2pw)

(

.

2)'n-4)/2

IIV-W

lIence obtain, in the usual form, the joint distribution of sx,

•

Sy

and r.

12 For a random variable X having the negative exponential distribution wilh probability density function

I(X

= x)

= A. e-).x, for X

~

0,

oblain K(t), the cumulant-generating function of X, and show that the sth clIl11ulant is (s -1) ! ". = -A.-'-' Suppose that the range of X is divided into an infinite number of intervals offinite length h, and the probability distribution of a discrete random variable Y is defined such that

21)h]

(2r+ p [Y =

=

P[rh

~

X

~

(r+ 1)h], for

Prove that the cumulant-generating function of Y is sinh (!A.h) ] [ "h(t) = log sinh Uh(A. - it)} ,

r

=

0,1,2,3, ....

128

EXERCISES IN PROBABILITY AND STATISTICS

and hence that

I

/(h(t) = /«t)+

j=2

.Bj .. (AMi [1- e- jJC(II], .1!.1

where Bj is thejth Bernoulli number. Also, if Kl and K2 are the mean and variance of Y, verify that

>

;(1

and

"I

K2

<

/(2'

~3 For the random variable X having a tr.ia~gular ?istribution in It mterval (- a ~ X ~ a), show that the charactenstIc functIon of X is .

[ Sin(at/2)] 2. (at/2) Suppose that the range of X is divided into 2m intervals each of length. and a probability distribution of a discrete random variable Y is defined su.. that P for (1'

[Y = (2r-2 1)h] = P [Y =

(2r-1)"] 2

= P[(r-l)h

~

= 1,2, ... , m). Prove that the characteristic function of

X ~ rh], Y is


=

22j-3

"2j~'

h2j B 2j'

where B2j is the (2j)th Bernoulli number.

14 In the previous example, suppose that the range of X is divided into (2m + 1) equal intervals of length h, and a probability distribution of anothe, random variable Z is defined by P(Z

= 0) = P [-~

~ X ~ ~].

and P(Z for (1'

= rh) = P(Z = - rh) = P [

(2r-1)h (2r+ 1)h] 2 ~ X ~ 2 '

= 1,2, ... , m). Prove that the characteristic function of Z is "'h(t)

h2

=
and hence, or otherwise, deduce that

h2

Ill.

(,

3h 2 )

Jl2 - 12 ~ - 4a 2 '

where 112 and Jl2 are the variances of X and Z respectively.

\29

CHARACTERISTIC FUNCTIONS Continuous

random variable X has a parabolic distribution with the

< The . func t'Ion bility densIty

I.

.r,lba .

!(X

,hilI!' t

= x) = 3(a 2 -x 2 )/4a 3

for

-a:;;; X:;;; a.

hat the characteristic function of X is 3(sin at - at cos at)/(at)3,

, ify that the variance of X is a2 /S. ,nd ~er ose that the range of X is divided into 2m equal intervals each of ShuPr, and the probability distribution of another random variable Y is

::ng l

,

.!dined by p[Y=

(21'~l)h]= P[Y= _(21'~1)h] = P[(1'-l)lI:;;; X:;;; 1'11],

Ir -- I , 2,. .. , m). Prove that the characteristic function of Y is of the form

hlr

112

3[eoit,(') - eoit2(t)]/(at)2 + 4a 2 . eoit3(t), ,\here tnt)

_

Bji' jk

.

.

L -.,-.. (Ity i=2 J .J OC;

=

(k

= 1,2,3),

Ail == (2a)j+(2 i + l-S)hi,

A. i2 == (22i-2i)aj+(2j-3)hi,

.1Ilt! the Hj are the Bernoulli numbers. Hence prove that

var(Y)

=

112 (

2112)

var(X)+T2 1- Sa 2 > varIX).

16 In the above example, suppose the range of X is divided into (2m+ 1) IIllcrvals each of length II, and another random variable Z is defined by the probability distribution

P(Z

=

0)

= P[ -

~ :;;; X :;;; ~],

and P(Z

(21' - 1)11

= I'll) = P(Z = -I'll) = P [ - - 2 -

~ X ~

(2,. + I)h]

2

'

(I' = I, 2, ... , /11).

Prove that the characteristic function of Z has the same form as that fur }'.

17 For a random variable X having the semi-triangular distribution with 11robability density function

I(X = x) = 2(a-x)/a 2 for 0:;;; X ~ a,

130

EXERCISES IN PROBABILITY AND STATISTICS

prove that the characteristic function of X is 2(cos at-I) 2 sin at-at (ait)2 + ait' at Hence verify that the mean and variance of X are a/3 and a 2 /18 respect"" Further, suppose the range of X is divided into m equal intervals eal~, length h, and another discrete random variable Y is defined by C P [Y=

(2r-l)h] 2 = P[(r-l)h

~

X

~

rh],

for r = L 2, ... , m.

Show that the characteristic function of Y is

2 [Q(at)Q(th) Q2(th/2)

ait

Q.(th) [Q(2at) -1] /Q3(th/2) 1] IIIQ(th/2)+ 2(mt)2 Q(at) ,

sinO where Q(O) == -0-' By a suitable expansion of this characteristic function, prove that h2

E(Y) = E(X)+ 6a'

and var(Y)

=

h 2 ( 1 + ah 22 ) • var(X)+ 36

18 Determine the characteristic function about the origin of a uniform: distributed random variable X defined in the range (0 ~ X ~ 2a). Suppose the range of X is divided into m equal intervals each of lenglh. and that another probability distribution of a random variable Y is definedh,

p[Y= (2r;l)h]= P[(r-l)h

~

X

~

rh],

for

r = 1,2, ... ,m.

If
prove that
=


sin(th/2) th/2 '

and hence determine the relations between the cumulants of X and Y. 19 The discrete random variables X and Y have the probability distribulio: defined by 1 1 P[X = I"h] = - and pry = (21"-1)11] = n n

for (I" = 1,2, ... , n), II > 0, a constant. If
=


Hence determine the relations between the cumulants of X and Y. Also, if II -4 X and II -4 0 such that 1111 -4 A, a finite constant, find limiting forms of (P,,(t) and ",,,(f).

It

131

CHARACTERISTIC FUNCTIONS

If X and Yare two independent unit normal variables, derive the teris tic function of Z = Xl+ yl, and so prove that Z is distributed as ,haJa~ith 2 d.f. .11. further, given the function 20

f(u, v)

= 2~ e-t(U+V) (uv)-t +e[(u- v)(uv-u-v+2) e-(u+v)]

( v ~ 0), show that this can represent a probability density function of ("r "~ndom variables U and V, provided that e is sufficiently small. 1\\O~btain the joint characteristic function of U and V. Hence prove that . V) is distributed as a Xl with 2 dJ., and that the marginal distribution It : ch of U and V is a X2 with 1 dJ., but that U and V are not independently "I f tlarl·buted. Comment on these results in the light of the distributional ,,~ fZ IlI:haviour 0 . 'I The two continuous random variables X and Y have a joint probability j~nsity functi~n, ~nd if g(x) denotes th; probability density function of .t~e arginal distnbuhon of X, show that Ilr(X), the rth moment about the ongm ;::~ the conditional distribution of Y given X = x, satisfies the equation r:1J

Ilr'() x . g() x =~ 2 1t

f

[or(I*I' t 2)] 'r ~ r I ut l

-x

. e-it,Xdt 1, 12=0

where cP(t I' t 2 ) is the joint characteristic function of X and Y. 22 Given a random sample of n observations (xv, Yv), for (v = 1,2, ... , n), from a bivariate normal population with the probability density function

J(X = x, Y = y)

1

1

21ty 1 - p2

2(1- pl)

~. exp-

=

. (X2+ y2-2pxy),

derive the joint characteristic function of the statistics n

n

L

ZI =

x;/2

and

Z2

=

L

y;/2.

v= 1

lienee prove that for fixed lllllllilional distribution of

Z l'

1/; (z I)' the rth moment about the origin of the

Zl'

is given by

II III'1\'

)1(")

== n(n + 2)(n + 4) ... (11 + 2,. - 2) . rr.

Veriry that the mean and variance of the conditional distribution of Z2 are

.1I1l1

var(: 21 z.l

= (1 -

p2)r~(1- p2)+ 2p 2Z1] .

:\Iso. deduce from the above results the correlation of

Z1

and

Z2'

132

EXERCISES IN PROBABILITY AND STATISTICS

23 From a bivariate normal distribution with zero means, variances a2 . and correlation p, a random sample of n pairs (xv, Yv), for (v = 1,2, ... h 0; given, In addition, there are n l unpaired X and n 2 unpaired Y observati~n~1 WI and W 2 denote half the sum of squares of the standardized X and' I observations respectively, obtain the joint characteristic function of w I I an' W2' Hence show that . corr(wl' w 2) = np2/[(n+nl)(n+n2)]t, Further, by inversion ?f,the characteri~t~c funct,ion! pr?ve that 1l~(w.J, It ~o~ent about the ongm of the condltlonal dlstnbutlOn of w2 for fixt: WI' IS given by .

rth

Il~(w.) =

Y2(r)+y(r)

st

C)p2S Ps(w l ),

where y(r) == n(n + 2)(11 +4) "'(n+2r-2)2- r , Y2(r) == l1in 2 +2)(n 2 +4) "'(n 2+2r-2)2- r , and

Jo

Ps(W.) ==

(_l)S+k

(;)~r(l1~nl)/r(l1~nl+k),

In particular, for n = nl = n2, verify that E(w 2 Iw.) = 11

[1 + ~;

(WI

-11)].

24 The two random variables X and Y have a bivariate normal distribuliv with zero means, variances ai, a~, and correlation p, Given a random samp'. of n observations (x I, YI)' (x 2' Y2)' ' , , ,(xno Yn) from this population, derive It. joint characteristic function of the random variables n

WI

=

L x;/2at, v= I

n

W2

=

L YV/a2~; v= I

and by inverting it show that the joint distribution of probability density function

WI

and

where, for positive integral values of N, W) =

g (w N

I'

2

-

_1_ e-tw~ __1_ e-w1 1,,(N/2)- I 'r(N/2)' , '1 '

fo'

Also, prove that for the conditional distributions, var(wt!w 2 ) and

11

=

p4

2+P2(I-p2)(w~-I)-2

1V2

has It,

133

CHARACTERISTIC FUNCTIONS

5 If XI' X 2' ... , X n are independent normal. random. variables, each ~i~h .• ,111. respectively, denve the charactenstlc unltt'lon of the random variable

2 . variance and means ml , 111 2"

.

~~

L X~,

Y=

A= I

,Ind hence show that the distribution of Y is Jl.2r

rr.;

L0 -,.f(Y;Il+2r)dy ,. •

e- p2

(0 ~ Y < 00),

r=

where, for positive integral values of v, "

1

=

!() , v) - 2V/2r(V/2)' e

_ -h'

n

.y

(v/2) - I

and

Jl.2 ==

1L

111~.

A= I

Hnd the mean and variance of Y.

26 The continuous random variables X and Y, with a probability distribu,ion defined in the range (- (/) < X, Y < (/)), have the joint characteristic function (W,' t 2 )· Prove that the characteristic function of X given Y = Y flixed) is

f

I/t(t rly) =

-

2/f oc

x.

e - it 2)' cP(t" t 2) dt

-

':FJ

e - iI2)' cP(O, t 2) dt 2·

if)

27 A continuous random variable X has the probahility density function tlclincd in the range (- r:JJ < X < 00), and the characteristic function cP(t). Suppose a set of (Ill +1l 2 -v) independent observations of X (v < III and 11 2 ) arc randomly divided into three distinct groups of (Il,-v), (1l 2 -v) and v clcments, their sums being SI, S2 and S3 respectively. Find the joint charactcristic function of the random variables Y=SI+S3

and

Z=S2+S3,

and thereby prove that for the conditional distribution of Y, given Z

= z,

v

E(Ylz) = (1l1-v)E(X)+-.z. 112

Ilcnee deduce that the correlation between Y and Z is v/~. Also. show that l'ar()'lz)

= (II, -1')var(X)+~ (1_~)Z2+ 112

112 "",/.

+V(1I 2 -1')

f

,

e- il: {cP(t)}"2-2

(~~r dt

-y

IJ 7.

e- irz {cP(t)j"2 dt.

-x

whcre £(e iIX ) == cP(t). 28 If

oX" X2," .,

x" are independent realizations of a normally distributed

ral~d~m variable X with zero mean and unit variance, find the joint charac-

tcnsltc function of the statistics U =

" L x~ A=I

and

v=

L" A= ,

aAx A,

134

EXERCISES IN PROBABILITY AND STATISTICS

the a being constants not all zero. Hence derive the characteristic funcr of u ~ven v = e, and thus show that the conditional distribution of U f).IOli . \'t~ v = e IS <0 ( _ c52)' e 62 --,-.f(u;n+2r-l)du (0 ~ u < 00), r=O r.

L

where for positive integral v f( u· v) = ,

-

1 20/2r(V/2)·

e- tu

.

U(0/2)-1

,

and n

c5 2 == 82/2

L1 a1·

).=

Comment on the case 8 = O. 29 Given two independent realizations XI and X2 of a random variable X having a Laplace distribution with probability density function

j(X = x) = ! e- 1xl , for -

'XI

< X<

00,

find ('*1' t 2 ), the joint characteristic function of the statistics U=X 1+X2

and

V=Xl-X2.

Hence show that the characteristic function of the conditional distribution of u given v > 0 is

Verify that v2(v+3) var(ulv) = 2+ 3(v+ 1)· 30 If X and Yare approximately normal random variables with the joint cumulant-generating function

( 1, t2 ) = ~ "t L..

(itd r (it 2Y r!s!

"rs·---

r+s=2

11

== !["20(it 1)2 + 2" (itd (it 2) + "02(it 2)2] + I/I(itl> it 2), prove that the characteristic function of X given Y = y is c/>(t I ) _ exp I/I[it1, -(%z)]. exp[!{"20(it d2-z 2/"02}] 1 y exp 1/1[0, -(%y)]. exp( _!Z2/'(02) ,

where z = y-(it 1)"1l. Hence show that, as a first approximation, E(Xly)

= ~.y; var(XIY) = "20-"il/"02.

"02 Also, prove that if linear terms involving the third-order cumulants "21' Ktl and "03 only are retained, then the appropriate approximations are E(Xly)

= ~. y+-21 ["12 "02 ~ "03 "11] (y2 - '(02), "02

"02

135

CHARACTERISTIC FUNCTIONS

"f. var( XI) y = "20--+ "02

l"2."~2-2"12"11"02+"03"f.] 3 ,y, "02

I prove that t~e c~aracteri~tic function of a rando,m variable ,1 'r fin distributIOn In the Ulllt II1terval (0 ~ X ~ I) IS

X having a

ull ilO

(e it -l)/it.

"'cn a random sample of

II observations of X.

obtain the characteristic

~'I\ction of the sample mean X, and by inverting it establish that the prob;~~litY density function of the distribution of x is nn

r( ) '

11

,12 For a

,L_( -1)1, (II)J,(.~-j/nr 1,

)~nx

ran~om var~able

for

0~

x ~ 1.

X having the Laplace distribution with the

probability denSity functlOn 1 J(X = x) = 2a' e-Ixlla,

for

-00

<X <

00,

prove that the characteristic function of X is (l + a 2 t 2 )-I, Hence obtain the characteristic function of the sample mean x based on n independent observations of X. and by inverting it prove that the probability distribution of x is

c-III'
G)

dx

(- 00 <

x<

00),

Verify that this distribution actually integrates out to unity over the entire range of variation of X.

33 For a random variable X, which is uniformly distributed in the range 10 ~ X ~ a), prove that the characteristic function of log X is d t ( 1 + itr I, If x l' X2' ' , , 'X n are n independent observations of X, derive the probability distribution of the statistic n

LI =

L

logxj •

j= •

and hence show that g, the geometric mean of the sample, has the distribution

nn

---n-()' gn-. [log(a/g)]"-

a

r II

1,

dg

(0 ~ g ~ a),

Also, deduce from this distribution that 2nlog(a/g) is distributed as a X2 I'ariable with 2n dJ, 34 For a Laplace distribution of a random variable X defined by the probability density function 1 J(X = x) = 2a,e-lxlla,

prove that the statistic

for-oo < X < 00,

136

EXERCISES IN PROBABILITY AND STATISTICS

based .on a r~ndom sample of size /I, is distri~uted as (~/2/1)X2, ~here X2 i X2 vanable with 2/1 d.~. Hence ?educe. that v IS an unbiased estImate or ;~' parameter (J, and find Its samplIng vanance. l

If X is a random variable having the Laplace distribution with Ihl probability density function I(X = xl = 1. e -lxi, for - 'l.. < X < -f_ ,

35

and Y another independent random variable with the same distribution as \ obtain the joint characteristic function of the random variables . U = X - Y and V = X + Y. Given II independent realizations of X and Y each, derive the distributio ?f ii, the average of the U observations, and also p~ove that this distributio~ IS the same as that of ii, the mean of the V observatIOns. By considering the joint characteristic function of ii and ii, or otherwise establish that ii and ii are uncorrelated but not independently distributed random variables.

36 For a random variable X having a r distribution with parameter p > 0 find .the cha~acteristic function of log X: Hence s~ow that the probabilitj denSity functIOn of u, the sum of the loganthms of /I mdependent observationl of X, is - p+i F. PII

e

-=-.,.-;-::::-:--;-::.

2ni{r(p)l"

Je"" fr(-z)l"dz. J

l -p-i,Y.J

Evaluate the complex integral by using a suitable contour, and then deduce that the sampling distribution of v, the geometric mean of the II observations of X, is

[d

IlV"p-l'f-

1

ll -

" (_I)"r+,,+ I _ _ r(1l) [r(p)}'" r'S-O dZ,,-1 .

37

VII:]

{r(l +Z):"

:=r •

dv

(0 ~ v

< 'X).

If 11 is a positive integer, prove Euler's formula

L( r

(z + ~ )

= /J t -

(2n)(II-II/2 •

liZ.

and hence, as a particular case, derive the

r(z)r(z+t)

r

r(IIZ),

function duplication formula

In r(2z)/22Z-I.

=

The independent random variables X I' X 2, ... , X II all have tions such that the probability density function of X j is

38

I(Xj = Xj) = e-·'tJ xfJ-1/r(p),

for 0

~

r

distribu·

Xj < co, Pj > 0,

where all the Pj are unequal and also do not differ by an integer. Find the characteristic function of the statistic U

=

I

logxj ,

j= I

and then show that the probability density function of u is

J iXJ

"

1

n r (p) j= 1

1 . 2--;' 1tI

e'lZ

-ioo

n r(pj-z)dz. II

j= I

CHARACTERISTIC FUNCTIONS

137

rove that the probability distribution of v = e",n is lienee P n 00 (_1)r+ 2 vn(r+ p.)-I ~. L L r(' I) J _ . r(PA-pj-r). dv [(Pj) j=1 r=O r+ k'¢j

n

fI

(0::::; v < (0).

jd

In the previous example, if Pj = p+(j-I)/n,p > 0, for (j = 1,2, ... ,n), that the characteristic function of u is ,how

39

n-nil

r(np + nit)/r(np).

" ce obtain the probability distribution of the geometric mean v, and verify he~ it is of the same form as the distribution of the arithmetic mean of n I ~ependent observations from a population having the probability density ,n , function

f(X = x) = e- X xP-Ijr(p),

for 0::::; X < 00.

.• , Xn are n independent realizations of a random variable X having zero mean and finite higher moments Pr (r ~ 2), derive the characterislic function of X, the average of the Xj' Hence, or otherwise, deduce that the second and fourth central moments of x are

40 If XI' X2,'

P2

-

n

an

d

P4

+ 3(n -I)pi n3

respectiVely.

Use these results to prove that

where n

(n _1)s2

=

L

(Xj-

X)2.

j= I

41 A continuous random variable X has the probability density function [(xl defined for (- CI) < X < CI), with mean P and variance (52. Suppose Ihat X and S2 are the usual unbiased estimates of P and (52 respectively, based on a random sample of n observations. Let the joint characteristic function of.x and S2 be

('*1> t If.x and

S2

2)

== E[exp(itlx+it 2 s2)].

are known to be independently distributed, prove that

a4J] [at2

=

[I/I(tdn{ x [a4J2] ~

12=0

at2

, 12=0

where 1/1('1:) is the characteristic function of X and deduce that 1/1('1:) satisfies the differential equation

d (dl/l) d'l:

2 1/1 1/1 . d'l:2 -

2

+ U 2 1/1 2 =

4J2(t 2) that

of

S2.

Hence

0,

so that X must be normally distributed. 42 Given that XI' X2,' •• , Xn are n random observations from a normal population with mean m and variance u 2 , find 4J(tI, t 2 ), the joint moment-

138

EXERCISES IN PROBABILITY AND STATISTICS

generating function of the random variables A and B defined by n- I

2a 2A

= (1I-1)c5 2 =

L (x j -xj+d 2,

j= I

and n

2a 2B = IIS2 =

L (Xj_X)2, j= I

X being the sample average. Further, if Mn denotes the usual 11th-order determinant such that cP(tI' t 2) = M;;t,

show that Mn satisfies the difference equation Mn = (l-t2-2td(Mn-l-tiMn-3)+tiMn-4'

for 11 ~ 5.

Hence verify that an alternative explicit expression for Mn is Mn = nil

(211-~-I\(-td\'(I-t2)n-I-\"

.=0

J

l

Use this series representation of cP(t I, t 2) to prove that the jth moment about the origin of the ratio AlB is }-
j

{

1-

L

l l/2

't'r

r=1

-00

j

X

TI

d't'"

r=1

where

so that mj

=

jth moment about the origin of A jth moment about the origin of B'

Hence deduce finally that 2

E(c5

2

Is ) =

211 (11-1)

2

and

var(c5

2

Is ) =

4112 (11- 2) (11-1)3 (11+ 1)'

43 Given a random sample of II observations (Xj' y), li = 1,2, ... , II), from a bivariate normal population with unit variances and correlation P, a correlation statistic C is defined by n

C

=L

C)n,

j= 1

where Cj is a random variable which takes the value + 1 if deviations from the sample means (Xj-x) and (Yj- y) are either both> 0 or both < 0, and zero otherwise. Prove that

CHARACTERISTIC FUNCTIONS

139 further, for any two paired deviations, say (Xl -X), (Yl - ji) and (X2 -x), -) show that the characteristic function Y

/1'2- ,

.

E[exp i{t 1(Xl - x)+ t 2(Yl - ji)+ t 3{X2 -X) +t4(Y2 - ji)}]

1

= exp -2n [(n-l)(tf+t~+ti+t~)-2(tlt3+t2t4)+

+ 2p{{n- l)(tl t2 + t3t4)-{tlt4 + t2t3)}]' and by inverting it prove that

P{(XI- X) > O'(Yl-ji) > 0; '"

~[(Sin-l p)'2 -

(x 2 -x) > 0, (Y2-ji) > O}

[sin -1 p/{n-l)]2] + 4~ [sin -1 p +sin -1 p/{n -1)] + constant,

and hence that cov{Cl> C 2) = constant- [(1/n}sin- 1 p/{n-l)]2. Use these results to verify that var(C)

= H*-(~sin-lpr +

+(n-l){ (~sin-l1/{n_l))2 -(~sin-lp/{n _1))2}]. Finally, if an estimate p* of p is obtained as p

* =S1O . {n{2C-l)} 2 '

prove that for large samples the efficiency of p* as compared with the usual product-moment estimate of p is

4 n2 '

I_p2

(2.

-1

1- -sm

n

)2-+ p

0 asp-+ ±1.

44 Given two random observations Xl and X2 from the Laplace distribution

!e-1x-m1dx

(-oo<x
and a positive constant Jl, (0 ~ Jl < t), prove that the characteristic function of

Y

= (I-Jl)xl +JlX2

is

cp(t) = eirm/{1+t2(I-Jl)2}{1+t2Jl2}. Hence, or otherwise, determine the sampling distribution of y, and then show that the information used by Y for the estimation of m is

140

EXERCI<;ES IN PROBABILITY AND STATISTICS

Also, prove that 00

(i) lim 1= 16 w... t

f o

00

e-2Z

Z2

..

1+2z

and (ii) for sufficiently small /l,

dz

fe-w.dW "" 2exO·2193; = 2e IV 1

List of abbreviatiol~s used in references to

journals, books and examination papers

JOURNALS A/lIwls of Ellgenics (H uman Genetics) Annals of Mathematical Statistics Biometrika Biometrics r Econometrica " 1 Journal of the American Statistical Association I .1: .JI .I(i J ol/rnal of Genetics .IIA Joumal of the Institute of Actuaries .IU,fS Jou/'llal of the London Mathematical Society IRSS Joul'Ilal of the Royal Statistical Society .IRSS(S) JO/ll'l1al of the Royal Statistical Society (Supplement) I/N/~AS Monthly Notices of the Royal Astronomical Society /'CPS Proceedings of the Cambridge Philosophical Society /,UlS Proceedings of the London Mathematical Society /,/~S Proceedings of the Royal Society, London, Series A I'TI~S Philosophical Transactions of the Royal Society, London, Series A S Sankhya II: I.IIS II III

BOOKS (1957): Probability-An Intermediate Text-book by M. T. L. Bizley, Cambridge University Press, 1957 IIROOKES and DICK (1958): ["troduction to Statistical Method by B. C. Brookes and W. F. L. Dick, William Heinemann, London, 1958 ])"VID (1951): Probability Theoryfor Statistical Methods by F. N. David, Cambridge University Press, 1951 I'IILER (1952): An Introduction to Probability Theory and its Applications (Vol. J) by W. Feller, John Wiley, New York, 1952 l'ISIlER (1945): The Design of Experiments by R. A. Fisher, Oliver and Boyd, Edinburgh, 1945 IISIlER (1946): Statistical Methods for Research Workers by R. A. Fisher, Oliver and Boyd, Edinburgh, 1946 III/LEY

141

142

EXERCISES IN PROBABILITY AND STATISTICS

FISHER (1956): Statistical Methods and Scientific Inference by R. A. Fisher, Olive Boyd, Edinburgh, 1956 ran: KENDALL and STUART-I (1958): Thl? Advanced Theory of Statistics (Volume I) by M. G. Kendall A. Stuart, Charles Griffin, London, 1958 an KENDALL and STUART-II (1961): The Advanced Theory of Statistics (Volume II) by M. G. KendalJ anA. Stuart, Charles Griffin, London, 1961 USPENSKY (1937): Introduction to Mathematical Probability by J. V. Uspensky, MeGra Hill, New York, 1937 ~. WADSWORTH and BRYAN (1960): Illtroe/uct iOIl to Probability alld Raile/om Variables by G. P. Wadswofli and J. G. Bryan, McGraw-Hill, New York, 1960 Note To allow for possible changes in pagination, the year refers to the date of edition or PIIOI" used.

EXAMINATION PAPERS

UL

University of Leicester B.A. and B.Sc. examinations.

ANSWERS AND HINTS ON SOLUTIONS Chapter 1

---

:luB = A+BA = All points of A and (1,3); (3,1); (3,3); (5,3); (3,5). An B = (2,3); (3, 2); (4,3); (3,4); (6,3); (3,6). A-8 = (1,.2); (2,1); (1,4); (4, I); (1,6); (6,1); (2,5); (5,2); (4,5); (5,4);

15. 6);j6, 5L

_

(An 8)uA = AB.

Feller (1952). Ex. 6, p. 21. (i) None; (ii) None; (iii) Four; (iv) Two; (v) One; (vi) One; (vii) None. I

Feller (1952). Ex. 9, p. 21. (i) A; (ii) 0; (iii) B u AC; (iv) AB. IO!/IO'.(IO-r)!

Ii

Consider the complementary event. The probabilities are

Brookes and Dick (1958). Ex. 14, p. 89. Initial probability of A having the joker is G)fm = i. This probability " unchanged, because if initially A had the joker, it is certain that the joker is Ililh him.

10

Due to W. Feller. (ii) Three types of configuration are possible: (I) three faces different. initial probability 120/216 (2) two faces same, one different, initial probability 90/216 (3) all three faces same, initial probability 61216. Ihe corresponding probabilities for repetition are 6/216, 3/216, 1/216 so that I~ljllired probability is (720+270+6)/216 2 = 83/3888.

0)

m;

K (i) j; (ii)~; (iii) II

i; (iv)~.

Due to R. L. Plackett. Ii) Number of ways in which British and French delegates are together but the Russian and American delegates are not = 2.1O!-4.9!, and the required probability = 16.9 !/11 ! = 8/55. 1111

Because of circular symmetry the prohability no\v is

2.9!-4.8! 10! 143

7 45'

144 10

EXERCISES IN PROBAB I LlTY A ND STATISTICS

There are II gaps and 5 tiles. Vide Feller (1952), p. 57. The probabilities are: (i)

(ii)

2(~)(:)1 C~) = :2' [e~)(;)+ (30)(:)] /C~) = ~~.

Feller (1952). Ex. 6, p. 76. Let Ai be the event that (i, i) does not occur in the ,. throws. Consid intersections of the Ai' and the probah;\:'" is Cr 11

r. (-

k~O

I)k ( 6 ) ( \ _

k

~ )'. 36

12 (i) Let X be the event that no castle can take another. There are 641~, sample points for the random placement of the eight castles. For th. sample points corresponding to X, there a.re 64 ways of placing th~ first castle, 49 for the second, 36 for the third, ... to 1 for the eighth Hence P(X). (ii) There are now 8! sample points, so that no castle can take another Let Ai be the event that a castle occupies a square of the while diagonal in the ith row (i = \, 2, ... , 8). Then P(A i ) = 7!j8!, P(AiA) = 6!/8!, etc. Consider intersectiolb. and the required probability is 8

L

1-

,~

(_I),-I/ r !

I

(iii) If Y is the event that none of the castles is on a square of the whitt diagonaL the required probability is P(X n Y) = P(YIX)p(XI whence the result, using (i) and (ii).

13 Let Ai be the event that the hand contains the ith ace and king pair. P(Ai) =

-2) //(52) (52\3-2 13'

Consider intersections, and the required probability is

i (4)(52-2k)( k 13-2k

_l)k-1 /

k=1

14

(52). 13

Kendall and Stuart-I (1958). Ex. 7.2, p. 183. (i) PI = 1(ii) P2

Jo (~) C34~k) / (~~) - -

= 1-4!

0·257.

e~) G~) G~) G~) / G~) (~~) G~) G~) - 0·895.

P3 = Pl-3(~)(~~) G~) e~) G~) / G~) G~) e~)(!~) (IV) P4 = 4P3 '" 0·759.

(iii)

-

0·190.

I

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

145

B'zley (1957). Ex. 1.7, p. 35.

I~ Le; x and Y be the numbers chosen by X and Yrespectively, (1 ,.;; x, Y ~ m). btain p(lx - YI ,.;; n), consider the complementary event Ix - YI > n. fo 0 the total number of ways of choosing x and y such that y - x > n is rhen

m-"-l

L

(m-n-x)

=

(m-n-1)(m-n)/2,

x=1

whence p(lx - YI ,.;; n)

= 1- p(lx -

YI > n)

= 1-

(m-n-l)(m-n)

m

2

•

16 Feller(1952). Ex. II, p. 45. (i) 2". 1t!/(2n)!; (ii) (n!)22"/(2n)! (iii) Let the original sticks be numbered 1 to n, and Ai be the event that stick i is reformed. P( .) = (2n - 2)! 1(2n)! AI 2" I 2"'

consider intersections of Ai' and the required probability is

i

(_l)k-

k= I

I(.kn) 2k (2n - 2k)!/(2n)!

17 Bizley (1957). Ex. 2.8, p. 48, and UL (1962). (i) Number even if last digit even: 'I' P ro ba bIity

3.6! 7.

3 7

= - , =-.

(ii) Number divisible by 4 if last two digits are divisible by 4: . P ro ba bT I Ity (b y enumeratIOn)

1O.5!

5 21'

= ~ =

(iii) Number divisible by 8 if last three digits are divisible by 8: 'I' b . Pro ba b Iity ( y enumeratIOn)

22.4!

= ~ =

11

105'

18 Let X denote the score at anyone throw, (2 ~ X ,.;; 12). Then P(X P(X P(X

= 2) = = 4) = = 6) =

P(X P(X P(X

= 12) = -16; = 10) = /2: = 8) = 356:

P(X P(X

= 3) = = 5) =

P(X = 7)

=

P(X = II) = P(X

=

9)

=

is":

!;

k.

Therefore probability of win on first throw = ~, and the probability of contin~ation after first throw = t. Next, let Pj be the probability of winning ultimately after X = j in the first throw, U = 4,5,6,8,9, 10). Then under first system of play, total probability of win 2

=

9+

~ pjP(X J

34

=

j) = 55 '" 0·6182

146

EXERCISES IN PROBABILITY AND STATISTICS

since

(1) (3)r 2 4 ="3 =

P4 =

r~o "6

Ps =

r~o "6 18

00

PiO,

00

(1)(13)r = 5"3= P9,

00

(1)(25)r 6 36 = TI = Ps·

P6 = Jo "6

Under the second system of play, the total probability of win 2

=

9+ ~ (l-Pj)P(X =

244

j)

= 495 '" 0·4929.

J

19

Bizley (1957). Ex. 2.8, p. 48.

Denote a white ball by Wand a black one by B. The four distin possibilities for the first two exchanges are BW, BB, WB, Ww. Hence, ir. denotes the event of drawing a white ball after the exchanges, then peE)

= P(EIB W)P(BW) + P(EIBB)P(BB) +

+ P(EI WB)P(WB) + P(EI WW)P(WW), whence the stated result. 20 peA) = 0'20; PCB) = 0'16; P(C) = 0'14; P(AB) = 0'08; P(AC) == O()I P(BC) = 0·04; P(ABC) = 0·02. (i) 0·65; (ii) 0·22. (iii) Let H be the event that the adult reads at least or,. paper so that P(H) = 0·35. Also, let E be the event that the adult read. at least A and B. peE) = P(AB) = 0·08;

P(H IE) = I.

Then p(EIH) = p(HIE)P(E)/ P(H) = 385'

21

UL (1963). Let PtA) = x; PCB) = y; and P(C) = z. Then

x=a;(l-x)(l-y)(l-z)=b;

I-xyz=c;

andp=z(l-x)(l-y).

Elimination of x, y and z from the expression for b gives the quadratic equatior, in p. Now a > 0 and b(1-a)(l-c) > O. Hence the quadratic has positil( roots if

ab-(1-a)(a+c-l) < 0 or c > Also, z

22

=

p/(p+b) and y

=

(l-a)2+ab (I-a) .

(\-c)(p+h)/ap.

Bizley (1957). Ex. 1.5, p. 32. For Sowite the favourable cases are the partitions of 10 in groupS lik(

1

A;\SWERS AND HINTS ON SOLUTIONS: CHAPTER 04,05'

147

a6)' where Laj = 10 and aj ~ 1 for all i. These cases total

"'~';~! Ib, ~ a,!

1O! (;! + ~; +

coefficicnt of x'" in

. +

~~;r

=

coefficient of x lO in 1O! (e-l)6

=

6 10 - 6. 5 10 + 15.4 10 -20. 3 10 + 15.2 10 - 6 = 16435440,

the required probability

=

16435440/6 10

0·27181.

""

.. h~l1ce osowite, exclude all partitions in which al = 1. These total l'Of

L 1O! /1! tI2 a !,

where

j

= coefficient I

II l~'

lee

~1

aj

Ia + 1 =

and

j

10

of Xl 0 in 10! x( e-' - \)' = 8341200,

the required probability

= 8094240/6 10

""

0·13386.

" H. F. Downton-Private communication. .. Let A and B be the two observers. Denote balls by ai' a2,' .. , an, where 'S the white ball. Then Pta,) = P(a r) = l/n. 1/, I•

10-;

PtA says akiad

= P(B says akiak) =

PtA says aria.)

= P(B says arias) = 10' 11- 1 .

9

1

Iknee

I

P(al A an

_ P(al, A and B say al) d B) say a l - PtA an d B say al )

PtA and B say a1IadP(al) /I

I

PtA and B say allas)P(as )

s= 1

( 1 ) 21 10 II

11-1 - - -+ 1 as 11+80

1 )2 1 9 ) 2 I '1 -+(11-1)(- (--)" (10 II .10 11-1 II

n -+

00.

Probability of at least one of A and B telling the truth 81

19

= 1 - 100 = 100' and 19

81(20-11) >0 100(11+80)

11-1 11+80

---- =

100

for 11 < 20.

As II increases, the possibilities of distillct lies increase, and so the prob· ability of both A and B telling the some lie decreases.

24 Brookes and Dick (1958). Ex. 14. p. 89.

p(2n)

=

(2nn)/2

2"

and

p(2n + 2) = 2n + 1 < 1 p(2n) 2n +2 .

148

EXERCISES IN PROBABILITY AND STATISTICS

25 Feller (1952). Ex. 7 and 8, p. 76. Let Ai be the event that the ith individual is missing from the Hence, considering intersections,

u, =

t

(_1)k

k=O

(m) (1-'~)' -. (1-e- P)m k n

as r -.

Sa

IlIp!:

00,

since in the limit

kP)' -. e(1--;:

kP •

26 Uspensky (1937). Ex. 5, p. 168. _ (;)(N -1f-' _ ( n Nn - N

P, -

)r( 1 _.!.)n-, N . P /r ., I

where ,-1

P ==

n

(1-k/n).

k= 1

The inequality follows from the fact that for 1 ~ k

so that r)(,-I)/2 ( 1-
n

1 For N=n-.oo, P-.1 and ( 1- N

~

r-1

(r )'-1 . 1-2n

)N-' -.e-

1,

so that p,-.e- 1/r!

27 David (1951). Ex., p. 62. Probability of caterpillar being run over by a cycle is (x + t)/36T, so thai the probability of a safe crossing is (

1- x+t )

36T

3NTII

'"

e-N(X+t)/121

.

28

UL (1962). p(AB) = p(AIB)P(B) = p(AIB)[1-P(B)) = P(A)-P(AB), and the staled inequality follows using p(AIB) ~ 1. The second inequality relations are obtained from P(AIB) P(A)

=

P(BIA) PCB)

d an

p(AIB) P(A)

=

p(BIA) _ 1- p(BIA) P(B) - 1- P(B) .

29 0n=p.On-l+(1-p)(1-0n- 1) or (On-t) = (2p-1)(On-l-t), whenCi On = t+(2p_1)"-1(0-t) -. t as n -. 00 since 12p-11 < 1. Thp. forecast for the nth day is as good as a guess for n large. 30 Pn = p(1-Pn-l)+(1 -P)PH-l' Hence PH = t[1-(1-2pf].

wnere Po = O.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

I

149

t

LJ pensky (1937). Ex. 1, p. 91. the urns be numbered 1 to n, and Pr be the probability of drawing a .Le ball from the rth urn. Then \dll te .11

= c:;~ 1)Pn-1 + (a+:+ 1)(l-Pn-d,

Pn

a

b Pn = a+b + (a+b)(a+b+1)".

a I

-+

a+b

PI

as n -+

=

1.

00.

LJspensky (1937). Ex. 2, p. 92 .

.12 If Pn is the required probability, then

a

Pn

=

Pn

="2

C:b)Pn-1 +C!b)(l-Pn-l)'

PI

= a+b'

lienee I [

I 1 + (a-b)n] a +b -+"2

as n

-+ 00.

n Bizley (1957). Ex. 5.8, p. 154. . ' Here IVn+ rn + bn = 1 for all integral n; and IVn =

xrn- l +(1-z)bn- l ,

rn = (1- y)Wn-1 +zbn-l> bn

=

(l-x)r,,_l +ywn- l .

Systematic elimination gives Wn+l +wn+[(l- y)(1-x-z)+xz]wn _ 1 = 1-z+xz, whence

lim Wn

n .... 00

= (1-z+xz)/[3-(x+ y+z)+(xy+ yz+zx)].

'

I\y symmetry, lim rn

n .... co

=

(1- y+ yz)/[3-(x+ y+z)+(xy+ yz+zx)],

lim bn = (1-x+xy)/[3 -(x+ y+z)+(xy+ yz+zx)].

n .... co

.14 Bizley (1957). Ex. 4.4, p. 111. Let the probabilities of X winning the cup outright be (i) x when one of the (n -1) teams has won once and not the year before; (ii) y when one of the (n - 1) teams has won twice in succession; (iii) u when X has won once but not the year before; (iv) v when X has won twice in succession. Then

150

EX ERe I S E SIN PRO B A B I LIT Y AND S T A TI S TI (' S

Hence

35

x

=

II

= (11 + 2)/(11 2 + 11 +

(11+ 1)/(11 2 +11+ 1); I):

y

=

L'

= (211 + 1)/(11 2 + 11 +

11/(11 2 +11+ 1); 1).

The probability of A winning r games out of (2m + 1) is

em,~ 1) p~(I_pym+l-r. Therefore the probability of A winning at least (m + 1) games is

1-

f

(211l+1)p~(I_Pa)2111+I-r.

r;O

r

For m = 2, Pa = i, this probability is 53/512. In the second part, A wins in exactly r (~ 3) games if (i) A wins once in (r - 1) games, (r - 2) games drawn, rth win; (ii) A wins once in (r-l) games, loses once, (r- 3) drawn, rth win. Therefore total probability of A winning in n or less games is

P(A)=p;+p;

L"

(r-l)p'O-2+ P;Pb

r; 3

L"

(r-l)(r-2)po- 3

r; 3

=P;(1-PO+2Pb) [1- "-1+( -1) "1_11(II-l)p;PbP~-2 • npo n Po 1 (1 -Po ) 3 -Po --+

p;( 1 - Po + 2Pb)/( 1 - PO)3

as 11

--+7:..

By symmetry, the limiting value of the probability for B to win is

Pt(l- Po +- 2Pa)/(l- PO)3. The sum of the two limiting probabilities is unity, so that an indefinitch prolonged game (11 --+ CIJ) is certain to end in a win for one player. But f~J n = 10, Pa = Pb = i. Po = t, PIA) = P(B) = 1981/4096,

and the probability of a draw is 67/2048. 36

UL (1963). By considering the possible states which could lead to an exact score of II. II"

=

152.11"_1

+i. 11,,-2,

where

110

= j,

III =~.

The expectation of the total score is 00

L0 11. II".

11==

The difference equation for v" is

v" = j.V"+152. V,,_I +i· V,,-2' 37

where Vo = 1, VI =

i.

Uspensky (1937). Ex. 8, p. 179. If Mr denotes the expected number of white balls after r operations, then

Mr+ I

=

Mr+ 1.

(1- a:rb) ,

where Mo

=

Q.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

Mn = (a+b)-b(l- a~br

and

1

151

Pn = Mn/(a+b).

nsky (1937). Ex. 9, p. 179.

\.~ Us';: and Nr be the expectations of the number of white balls in the Let ~fter r exchanges. Then Mr+Nr = a+c, and Mo = a. Also,

.,,(1

urns

Mr+ 1 = Mr+Nr/(c+d)-Mr/(a+b),

. .h~nce

.

a+e + ad-be a+b+c+d (a+b)(a+b+c+d)

~=

P. == a+ b

(a+c)/(a+b+c+d) as n -+

-+

[1 __I___I_]n a+b c+d

00.

19 Let Un be the probability of A's final win when he has in. Then Un

"f

= pUn+1 + (1- P)Un-l, Un+l-un = Un+l-

where Uo = 0 and Ua+b = I,

C~p)(Un-Un-l)

l_p)n Un = ( -P- Ul'

,u that lienee, using Ua+b

= 1, Un + 1

l_p)n+l]j[ = [ 1- ( 1- (l_p)a+b] -.

P

P

rherefore initial probability of A's win is

_ [ pa_(I_p)a ] b Ua - pa+b_(I_p)a+b .p and of his failure l-ua. For p pI!

=

1,

Ua = a/(a+b) -+ 0 as b -+

00,

and for

40 Bizley (i) (a) (b) (c)

(1957). Ex. 4.6, p. 126. Probability of A's win in 3 games is p3. Probability of A's win in 4 games is 3p 3(1- p). Probability of the first four games leading to the proviso is 6p2(1- p)2. If this happens, A can win if he wins (m+ 2) games out of (2m + 2) and loses the remaining m games, the m wins and m losses being permuted in 2m ways, and the last two games being wins for A. Hence total probability of A's win is 00

!(P) = p3+3 p3(I_p)+6p2(I_p)2

L m=O

whence the result.

2mpm+2(1_p)'",

152

EXERCISES IN PROBABILITY AND STATISTICS

(iii) The roots of f(P) = pare 0, 1, t, 1 ± .,fi12, and for the first values the probability of A winning the match is the same as ththrt: his winning a game. at (.I (iv) The expectation of the number of games played for A's win is 3[(1-2pq)2+4q(1- pq)]/(1-2pq)(1-2pq+3q),

where p+q = 1 and p i= O. From th~ given sequence of n calls, we derive a seque~ce of (n-l) s cesses and fatlures. Define a success at the rth (r > 1) call If the call Co Ut· from the same room as the (r-l)th call. In this derived sequence, the p~' ability of a success is 11m; and if in this sequence there is no run of at lea two successes, it follows that in the original sequence no room makes at lea:: three consecutive calls to the exchange. Let Un be the probability of no run of at least two successes in n trials I the derived sequence. Then o. 41

Un

(1)

m-1) un- l + (m-1) = ( ----;;;----;;;- ;;;

Un -2,

and Pn =

UR - l ,

n > 2,

whence

Pn where Pi =

P2

=

=

m-1) (m-1) ( ----;;;Pn-l + --;;J2 Pn-2'

1, since Uo

=

= 1. Hence the result.

Ul

(i) As m -... 00, 1X1 -... 1, 1X2 -... 0 and Pn -... 1. (ii) As n -... 00, IX'l -... 0, lXi -... 0 and Pn -... O. 42

Uspensky (1937). Ex. 5, p. 168. Let Xi be the random variable denoting the number on the ith ticket drawn. Then m

S=

LXI'

i= 1

But E(Xi) = (n+ 1)/2; E(xf) = (n+ 1)(2n+ 1)/6; and for i i= j E(xjxj)

= =

J L rsln(n-l) = L -r- [n(n+1) ---r n

r"'.

r=ln(n-l)

2

(n+ 1)(3n 2 -n-2)/12(n-l).

Hence var(xi) = (n2-1)j12 Therefore E(S) = m(n2+ 1);

and

COV(Xi'X) = -(n+1)/12.

var(S) = m(n;; 1) [1-

43

:=:].

Montmort's matching problem (1708). Let the envelopes be in the natural order 1,2, ... , N. Corresponding to this there are N! arrangements of the letters. Define random variables X, such that Xr = 1 when envelope in the rth place has correct letter and x, =0 otherwise (r = 1, 2, ... , N). Then N

S=

LX,.

r= 1

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

1

153

Eyidently, E(x,) = l/N; var(x,) = (N-1)/N2; cov(x" x,) = 1/N2(N-1), Hence E(S) = var(S) = 1. " 'I ()ihe second case, the probability of S correct postings is

(~)(~r (1- ~r-s.

[n

. 's a binomial distribution with probability of success 1/N. Hence the Ihls I . l11~an and vanance. lJspensky (1937). Ex. 13, p. 181. .s4 Let Xj be a r~ndo~ variable representing the number on the ticket drawn the ilh occaSion (, = 1,2, ... , m). Then ~

m

X=

L1 Xj'

j=

,\Iso.

E(Xj) = ,Ind for i

±d

,=0

t(n\V2/1 = n/2; E(x1) =

~ j,

r2(~) {(~) -I}

/I

E(xjx) =

,~o

2/1(2/1-1)

n2

n

±

,=0

t2(n)1/2/1 = n(n+1)/4; t ~

rs(~) (~) + ,~s 2/1(2/1-1)

- 4- 4(2/1 - i)' fhcrefore var(Xj) = n/4; cOV(Xj• Xj) = -n/4(2/1-1), whence 11111/2; var(X) =

E(X) =

mn[ 4 1- m-I] 2/1-1 .

45 UL (1962).

Per)

k- I

kC)/r+ ke: = ± = r+ 1) =

=

,=0

± r+

1

(2"+ I -1)/(11 + 1).

r(I1+I)V(2/1+I_1) = (11+1)

lienee E(r)=

,=0

E(r2) =

(11 + 1)/(11+ 1)

:)/(11+ 1) so that

1

-

±\r+

,=0

±

(11)1/(211+1-1)-1

,= 0 , '/

(11-1)2/1 + 1 2"+ I-I .

(11+1),2/(2/1+1_ 1) 1

(11 /I = [ lI(ll+I)L

=1I) -(I1+I)L

I1)] /

II (11 + +L (2"+ 1-1) ,=0 r+ I = [2/1-1(11 2 -11+2)-1]/(2/1+1-1). whence var(r) = (11+1)2/1-1[2/1+1_(11+2)]/(2"+ 1-1)2.

,=1

,

/I

(")

,=0 r

154

EXERCISES IN PROBABILITY AND STATISTICS

There are (~~ Dways of selecting (r + 1) balls out of (n + 1), and th number of ways is (2n+ I_I), excluding the possibility of not selectine to~ Hence the probability is g an!

(~: ~ )/(2 + 1_1) = n

46

UL (1960). k = 0; E(r) P(r)

mode

=

=

= 0- 1 ;

var(r)

P(r).

= (1- 0)/0 2 • < I,

0(1- 0),-1, and since (1- 0)

P(r) is a maximum for r::::: I .

II~

1. n

Sn ==

L

P(r)

,= 1

= 1-{l-Or =

t

if 0 = 1-(W /n •

Similarly, 00

L

P(r) = I-Sn

=

,=n+l

t

if 0 = 1-(W /n.

Hence, by convention, median = n +t. 47 Banach's match-box problem.

u, = CNN-r)mN
N

,=0

,=0

for 0

~ r~

N.

L u, = 2- N. L [coefficient of x N-, in (1 +X/2)2N-'] =

2- N • coefficient of xN

. 10

N ( X)2N-' ,~o x' 1 +"2

= 2- N.coefficient of x N in {l+x/2)2N+I(I-x/2)-1 _ _ 2N ~ (2N + - 2 . L. ,=0

1)

r

.

But

f (2N+l)+ r

,=0

2yl (2N+l) ,=N+I r

= 22N + 1,

so that

To find E(r). evaluate first

E(N -r) =

!.

N-I

L (2N -r)u,+

1

= [(2N + 1)(1-110)- JI]/2,

whence the result. Similarly to obtain E(r 2 ), consider N-2 E[(N-r)(N-r-l)] = L (N-r)(N-r-l)u, ,=0

= [2N(2N + 1)(1-2uo)+(4N +2)(1-lIo)+E(,.2)-(4N +3)/IV4. since Uo = u l . Hence E(,.2) = (2N+3)-3(2N+l)lIo.

1

A!'lSWERS A~D HI~TS ON SOLUTIONS: CHAPTER

i

. Ie (1957). Ex. 4.7, p. 121.

.

..

155

. "

~I~ balls and n compartments the total dlstnbutlve posslblhties are

.Ill

Wit t tal number of favourable possibilities is rhe 0

L: t!

III

ai'

where

I

ai

= t and each

ai

~

1

±m

(n)(_1)m(n_m)l,

:=:coefficientofx'int!(eX -lt=

m=O

e the stated probability. This probability depends upon all favourable .. h:Oc·ncluding those in which the tth ball completes the distribution over ,.I'~S ~ompartments. If f(t) be the probability that all compartments are not ,'~,:pied with t balls, then

±(n)(-lr(I-~)', m n

f(t) = 1.

m=O

d Ihe probability that exactly t balls are required .In I' Ihe expected va ue IS

=

\',I!(t-l)-f(t)}

I f(t) = n ±~.m (n)(_l)m+1 m

1=0

oJ

, I

= f(t-l)- f(t), whence

m= 1

1

1

= nf[l-(I-X)n]dX = nf(1-yn)dY = x

o

l-y

0

±

n/m.

m=l

49 Let Xi be the number of drawings following the selection of the ith variety

up 10 and including the drawing which shows up the (i + l)th variety, (i ~ 1).

rhen

r-1

nr

= 1 + L:

Xi'

i= 1

But 00

E(Xi)

=

I

vp(l-p)·-l

= p-1,

.=1

where p == (N - i)/N and var(xi) = (1- p)/p2. Hence

E(nr) = N

r-1 L: (N -

i) -

1

i=O

=

N

N

f. 1=.\

t- 1 '" N -r+ J

f t- 1 dt = N 10g[N/(N -r+ 1)]. N-r+ 1

Similarly,

r-1

var(nr) = N i~O i(N-i)-2 '" N 2

fN

t- 2 dt-N

N-r+1

=

fN

t- 1dt

N-r+1

N(r-l)/(N -r+ 1)-N 10g[N/(N -r+ 1)].

156 50

EX ERC'ISES IN PROBA B I LIT Y AND STATISTIC'S

The probability of X

~

kin n drawings = (k/NY'; and

Hence E(X)

= ktl

~ N-

kpk

N II

[N + II

f til dt] = nN/(lI+ 1),

1 _

o

and

f

E(X2) -- N- [N + 2-2 II

II

f N

N

til + 1 dt-

o

til

~ nN 2/(n+2),

dt]

0

whence var(X). Bailey, N. T. J. (1951). B, 38, 293. For second sampling, the population has W marked and (N - WI u: marked animals. Therefore

51

pen) = probability of (w - 1) marked and (n- w) unmarked animals in II first (11 - 1) captures x probability of a marked nth capture

C~ J(~=:')

W-w+l . N-n+l '

and the stated expression follows by recombining factorials. To prove lh,' the probabilities add up to unity, it is enough to show that

II,.

= N-IV-w (N-n)(n-l)

SeN, W, w) -

Verify that SeN, W, W+ I)-S(N, that by successive reduction

W-w

w, w) =

w-l

SeN -1, W,

=

(N) W .

w+ 1)-S(N -1, W, wI,'

SeN, W, w+l)-S(N, W, w)=S(w, W, w+l)-S(W, W, w)=o. SeN, W, W+ 1) = SeN, W, 1)

Hence

N-IV

= coefficient of X IV -

1

in

L

(I +X)N-t-I

t=O

52

Dorfman, R. (1943). AMS, 14,436, and UL (1962). Let Xi be the number of tests required for the ith group in plan (ii), Then N/k

S=

L Xi'

where

P(x i

= k+ 1)

=

l_ qk

and

P(Xi = 1)

= qk,

i= 1

Hence E(S) and var(S), using the fact that 53

Xi

are uncorrelated.

Let Xi be a random variable such that Xi = 1 if forecast is correct (F) for ith day =

0 if forecast is incorrect (F) for ith day,

(i = 1. 2, ... ,11).

I

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

",'n

P(Xi

1) == 1-P(Xi

===

/'

(II-I')

157

•

= 0) = p. ;;+q. -n-' and Sn = i~1 Xi'

X. are correlated random variables, so that for i '# j d\ll{he

I

. v.) ::: P(Xi

• I \, ,\ J

::: p. Mp·

== 1, Xj = 1)

c= ~)

+q.

G=~) ]+q. (n~r)[p. (n~ 1) +q. (II~~~ 1)].

. cl! E(S.) == np - (n - I')(P - q) < np for p > q, and var(Sn) = npq. Ji,nDenote rain by R and no rain by R. The four mutually exclusive possibilities (he weather-forecast combination for any day are RF, RF, RF, RF. If :"r 's IV,I IV2' W3 and W4 respectively are associated with these outcomes, then .,tift this == whence

Ind II

.

°

E(S.) = r(wlP+w2q-w3P-w4q)+n(w3P+w4q),

for all /' if W 2 = - w l P/q and

W4 =

1 [ 2 pI' 2 p(n - 1') ] var(X.) = - WI' -+W3' - and I II q q

_ np

var(S.) - - . q

Ihcrerore

.md this == 1 for all I' if WI

w3P/q· This ensures E(X i ) =

cov(X;, Xj)

PI' 2 +-. (WI q

.

.

= 0, (I '# J).

2

W3),

±W 3 • Thus the alternative scoring system is

= (q/np)t; W4 = -"'2 = (p/nq)t. 54 For the four possible outcomes RF, RF, RF, RF, the probabilities are WI

= -

=

2 W3

-

W3

1/1. 11(1- p), p - (1.13 and q - (1.( 1 - 13). For random allocation of {he corresponding probabilities are

IIcnce (i) gives W3 = - W4q/P, W 2 = - w IP/q; (ii) gives W4 = (l-W1P)/q, using (1. = P and 13 (iii) gives var(S.) = n[w1p2/q-(1-W1P)2/p], using the probabilities of the outcomes with when WI = q/p. Hence WI = w4'1 = q/p; W2

(1.

=

I'

days to rain,

= 1;

= 13 = p. Var(S.) is minimum W3 = -1.

55 Define xr to be the expected number of additional trials needed to realize the pattern when the first I' letters are in agreement. Difference equations then give (i)

Xo

= p-l+ q -l p -4; (ii)

Xo

= p-l+ p -2 q -2; (iii) Xo

56 UL (1962). The roots of Z2 - z + pq = 0 are p and q. Hence, for p i= q,

dnd for p

=

q, Y.

=

(n + 1)r·. These results hold for n ~ O.

=

q-l p-3.

158

EXERCISES IN PROBABILITY AND STATISTICS

Let Xm be the expected number of additional trials needed to Co the sequence when a stage has been reached where the first m letters fllpl. pattern are in agreement with the observed sequence, for (0 ~ m ~ 2r).or If Then X2k = 1 + PX2k+ 1 + qxo and x 2k+ 1 = 1 + PXl + qX2k+ 2, where x Hence 2, '" x 2k -a = (xo-a)/(pq)\

whence, putting k

= r,

a

==

[l+(l-pq)x o]/(l-pq),

the result for Xo' The inequality follows since r

Xo

=

I

(pq)-k

and

k= 1

°

~ pq ~

!.

57 Feller (1952). Ex. 33-35, p. 128. The event "exactly k balls in n drawings" can occur if (i) k black balls in (n-1) drawings and white ball in nth drawing· (ii) (k-1) black balls in (n -1) drawings and black ball in nth dra~ing Hence difference equation. 58 Huyghens' problem. UL (1963). Probability of A's winning in a trial is 5/36 and that of B's winning is 6.;, Hence probability of A's winning at the (2r+ l)th turn is /.

G!r G~r (:6)' whence total probability on summation for (0 Expected number of trials for A's win is 61 5 00 (31)r (30)r 30' 36 r~o (2r+ 1) 36 36

59

~ r

=

< (0).

5 6+ 61 '" 6.

UL (1962). Expected premium in nth year is

t- 1 +(1- q)

a[(qA

As n

~ 00,

:t:

(qAt] = a[(l-q)+q"A"-l(l-A)]/(l-q},).

the expected premium tends to limiting minimum value a(l- q)/(l- qA)

> ka,

whence the inequality. 60

UL (1962). The probability-generating function of S is G(6) = (6+6 2 + ... +( 6 )4/6 4 = 64(1-6 6 )4/6 4(1-6)4,

whence the result. E(S) 61

=

G/(l).

Probabilities for the wins of A, Band Care p/(l_q3), pq(\_q3). an:

pq2j{1_q3) respectively.

P(X = r) = pqr- 1 for r ~ 1. The probability-generating function of \ is G(6) = p6/(1-q6), whence E(X) = p-l and var(X) = qp-2. P(X = r) = (r-l}/36, for (2 ~ r ~ 7); P(X = 12-r) = (r+ 1)/36. k' (0 ~ r ~ 4). A simple numbering for one die is (1,2,3,4,5,6) and for It. other (0,0,0, 6, 6, 6). Other possibilities arise from suitable partitions of Gil'

62

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

1

159

UL (19 63 ). . ho ut any restriction the three numbers can be chosen in 11(411 2-1)/3 Wlt,( x is the first number in A.P. with common difference r, then ..IP·.•((211-1) and for any x, r ~ [(2n+l)-x]/2. Direct enumeration , r :~ -; A.P.'s in all. Rearrangement of the A.P.'s according to the magnitude I\~S "common difference gives the distribution of X, whence G(O) by sum; Ihe

~I

111011.

, rhe probability-generating function of X is G(O) = (pO+qO-l)", whence ~ n == 1I(2p - 1land var( Xl = 4npq. Also. ,I.

G(O)

=

±(n~t)p("+11/2q(n-'1/201'

I=-n

,llhat PIX ',\h~rc(-II ~

2

( n~t)

= t) = -2-

p(n+I)/2 q (n-I)/2,

t ~ n) and (n-t)/2 is an integer ~O.

h~ The probability-generating function of X is G(O) .:h~nce E(X) = n(p-q) and var(X) = n[4pq+r(1-r)].

= (pO+qO-l+ r

r,

Again,

,md PIX = m) is the coefficient of om in this expansion. Thus, putting I j -II = m, i.e. A. = n + m - j, the limits for j are obtained from n ~ n + m-j ,lIulli +111- j ~ j. I

M

The probability-generating function of the total electorate is G(O,.) = [(1-PI)+P 10] L[(l-P2)+P2-r] C[ (1-P3-P4)+P3 0 +P4-r ]f' ,

where the coefficient of Orr gives the probability of the Labour and Conserliltive candidates obtaining rand s votes respectively. The required probability lor a tie is the sum of the coefficients of orr for all 0 ~ r ~ min(L + F, C + F). lienee the result by putting -r = 0- 1 in G(O, or). The probability-generating function of N is obtained by putting 0 = or, whence the mean and variance.

or

67 Dodge, H. F. (1943). AMS, 14, 264. r-I

iii P = p

L q' =

l_qr.

1=0 r-I

lIil/J

=L

(t+l)pq'/(l-qr) = [l-qr(l+rp)]/p(1_qr).

1=0 00

(a) g =

L

t(l-P)P I = (l_qr)/q'.

1=0

(b) u

= gh + r =

(1- qr)/pqr.

160

EXERCISES IN PROBABILITY AND STATISTICS

(iii) v = f

- 1 X expected

number of items inspected before getting a dct

1Cehl

00

L

=f- 1

1=

(c) ljJ

tpq,-1 = (fp)-I.

I

= (u + fv)/(u + v) = fl[f + q'(1- f)].

(d) p = p(l -ljJ) = pq'(1- f)/[f + t((l- f)].

(iv) p is maximized for p = p* obtained from (1- f)(1- p*Y 68

= f(r + l)p* - 1]/(1 - p*). whence p*

=

[(r + l)p* -II/r.

UL (1961).

Pn(r) Using factorial moments

=

G)p'qn-,.

= (l-6pq)/npq. whence the result for

')12

69 Frisch, R. (1925). B, 17, 165, and UL (1962). The cumulant-generating function of r is ,,(t) mean and variance.

}'2 :::

O.

= n log(q + pe'), when"

" (n-l) l)] = np [ L _ p,-l q"-,+1 - nL- l ( n _p'qn-, ,=t r

= np[

(:=

1

r

,=t

!)pt-l qn-t+

1] = (:)pt q"-t+ 1.

t.

For t = 0.111(0) = O. whence E(r) = np. 70 Romanovsky, V. (1923). B, 15,410, and UL (1962). 00

,,(t)

=

nlog(q+pe' )

==

L

,= 1

",fir!

and

p = (1+e- Z)-I,

whence d,,(t)

Cit = n/[1 +e-(z+ll] == f(t), say, so that d,,(t) dt

00

L jI'l(O). t'lr!

,=0

where

", = jI,-ll(O) d,-lp

=

n[:;,~ll{l+el (z+ll}l=o dp d

{d,-2 p}

= ndz,-l = n dz ' dp dZ,-2 = or Ie, = pq. d",_ tfdp, since dpldx mines "s.

dp d",-l

dz'~'

= pq. Successive differentiation dele:

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

"I The moment-generating function about the mean

I

161

i. is

• t i,t [ t t t t4 ] exp[A(e -t-l)] = eXPT 1+3+12+60+360+ " ' , 2

2

3

the moments by expansion . .. h~nce _, P(X === r) = P(X = - r) = !pr. Hence P(X = 0) = 1 -
,to

Pr(om-

8')] /0"'(1- 0);

'~~~ p,(om-or)]/ om(1-0);

(xi) [l-omG(O)]j(l-O);

(x) O"'G(O)j(l-O);

(xii) [l-OIll+IG(O)]j(l-O).

("'ncrating functions obtained by determining the probabilities of Y for each ... tlue ~O.

74 Quenouille, M. H. (1949). Bs, 5, 162, and UL (1962). If X is the random variable denoting the number of colonies observed, Ihen P(X = n) = e-).. ;'lIjn !, for n ~ O. Let y. be the random variable denoting the number of insects observed in Ihejth cdlony. Then moment-generating function of lj is
L" 1 lj

S" =

j=

.\IId moment-generating function of SII is [
P(SII = N) =

L

P(S" = NiX = n)P(X = n).

11=0

ihcrcfore moment-generating function of SII 00

=

L N=O 00

=

L 11=0

=

f

11=0

00

etN

L

P(SII

= NiX =

/I) e-).. A."jn !

11=0 ;'"

e-)..,

00

L

etNP(SII = NiX = n)

n'N=O

e-).. i:',[
=

exp[A.{
n •

Ill', has a logarithmic distribution with parameter p, then


rell ui red.

162 75

EXERC'lSES IN PROBABILITY AND STATISTIC'S

Greville, T. N. E. (1941). AMS, 12, 350. Clearly,

Solve these equations for r, (r + I), (r + 2), ... ,N to determine W in t for (r :::;; U :::;; N). Multiplying v.. + u by , crll),

v."

(_I)"(r:.u) and adding

where SM ==

L u=o

M-,

(r+ u)( .M r

)

for M ~ r.

(-1)"

I+U

But SM == 0 for M > rand S, = 1. Hence ~

=

N-, L (-l)" (r+ U) v..+u = LN(-1)s-, (

"= 0

s

s=,

U

_S

)

r

v..

v.,

To determine suppose s matches occur at the kith, k2th, ... ,h positions of the A pack, and that the type grouping for these s card)'. AI' A2 , ••• , At, where t L Ai = S i= I

and 0 :::;; Ai :::;; min(ai' bi' s). Out of ai' Ai cards can be selected in

iII G:)

ways.

Now consider the B pack. The s cards are fixed, as matches occur, but tho other (N - s) can be permuted in (N -s) ! ways. t Il (hi-A j ) ! j=

1

Hence

v.

=

~

(bj-)'i)!] .= 1 (~~)/ I~I

[(N-S)!.TI

1'1

=

Hs·(N-s)!

I.D_tl

bi !

The jth factorial moment is

N

Il(})

IN

= ,~j r(j) ~ ,~o~'

where

N

=

L

s

v..s!

.=}

L (-l)s-'J(r-j)!(s-r)! r=}

V.

's-}

(.)

-_ f... \' ~.', L... " (_l)S-j-1 S-] s=j(s-J). 1=0 I N

-_

0 un Iess s

= j.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 1

163

N

L rUlJ¥. =

lj.j!

and

r"'i

,nd \ariance = 76 consider

[et

]! N 2(N-l).

aibir-N. it1 aibi(ai+ bi)+N2. it1 aibi

jY ~ h

[1 +

L

(Y -/1) -

d for the variance use E(jY) ~

.In

8~2 (Y _/1)2]

to obtain E(J}'),

h, whence

var(jY) ~ E(jY -h)2 ~ var(Y)/4/1. 77 Kendall and Stuart-I (1958). Ex. 2.3, p. 52.

logg ~

g

1" [(x.-a) l(x.-a)2] 10ga+-.2: ~'- -~', whence n,=1

~ ae-s2/2a2 '"

,,-1 "'~. f na

a

[1_(xi-a)+(xi-a)2] a

i=1

whence" '"

a 2 a(1-sl/2a l ).

=

a

a- 1(l+s2/ a2),

a(l-s 2/a 2).

78 The probability-generating function of Y is

G(8) = [tp8+(1-p)+tp8- 1 ]",

and the moment-generating function is

G(e t )

=

[1 + 2p sinh 2 t/2f

-+

exp[2m sinh 2t/2].

rhe probability-generating function of Z is H(8) moment-generating function is

=

[(1- p)+ p8]", and its

H(e t ) = [1 + p(et _1)]" -+ exp[m(et -1)].

Now P(Z = 0)

=

(1- pf

=

(1 + ),)-", so that

G(8) = (1 +;.)-" . .to (;) (.A./2)' 8r (1 +8- 2)'

=

(1 + Je)-"

rt (;)(A/2)' or .t G)

whence the relation between P(Z

=

8- 2.,

0) and P( Y = 0).

79 The moment-generating function of X about the origin is

M(t) = [1 +4p sinh 2t/2{l + p(cosh 2 t/2+ sinh 2t/2)}

r,

164

EXERCISES IN PROBABILITY AND STATISTICS

so that 10gM(t) '"

4m(I+~)sinh2t/2.

Differentiation gives E(Sn) = 0 and var(Sn) = 2m(1 +m/n), these being e results since the neglected terms in log M(t) are of power higher than

/a(.

80 Anscombe, F. J. (1948). B, 35, 246. Consider terms up to [(x-A,)/(A,+b)]6 in the expansion for (X+b)1 obtain E[(x+b)t]. Also, var[(x+b)t]

=

I,

(A, + b)-E2[(X+ b)t].

Use the results of Ex. 71 above.

UL (1960). Since (2n+ 1) 1/(2n-l). Hence

81

~

r

~

(4n-l), each of these (2n-l) cases has probabilill .

4n-1

E(r) = L r/(2n-l) = 3n,

and

r= 2n+ 1 4n- 1

E(r2) = (2n _1)-1

L r2

r=2n+ 1 2n-1

= (2n-l)-1 L (s2+4ns+4n2)

=

n(28n-1)/3,

s= 1

whence var(r). 4n-1

P{r> 3(4n-r)} = per > 311) = L(411-1)-1 r= 3n + 1

= (11-1)/(411-1) < 1. Hence the probability of three such breaks < 1/64 which is significant allh. 2 per cent level. UL (1961), and Fisher (1945), p. 11. In the first experiment the probability of a success is 1/70, and the probability for the lady to establish her claim is

82

lol~i

1- P1(0)- PI (1) = 0'0085, where PI(r) is the probability of r successes in ten trials. In the second experiment, the probability of a success is 37/924 > Ifill The total probability for the lady's claim to be accepted is now I-P2(0)-P2(1)-P2(2) = 0'0063, where P2(r) is the probability of r successes in ten trials. The increase in the probability of an individual success in the seeon.' experiment is more than compensated for by the requirement that the I~dl has to achieve at least three successes instead of at least two as in the III' experiment. 83 (i) If Xm is the expected number of additional trials needed to complell" consecutive realizations when the first ttl letters are in agreement, (0 ::;;; /II ~ .1"

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

Ihell

I

165

tor 0 ~ k ~ r-1

il

X3k

=

1+PX3k+l +qxo; X3k+l

=

l+PXl +q X3k+2;

X3k+2 = l+PXl +qX3k+3,

whellce the difference equation X3k

=

1+(1-pq2)xo+pq 2X 3k+3'

with X3r = O.

rhe inequali~Y foll0'Ys dbyl' noting /hat for (0 ~ p ~ 1), (0 ~ pq2 ~ 4/27), the ximum bemg attame lor p = 3' m3 (ii) By an analogous definition of X m , X3k

=

1+PXO+q X3k+l; X3k+l X3k+2

,0

=

1+PX3k+2+q Xl;

= 1 + pXo+q X3k+ 3

that X3k = (1+pq)+(1-pql)xO+pq 2X 3k+3'

X3r = O.

and

lienee xo. The inequality again follows by noting that r

Xo

r-

1

L (pq2)-1 + L (pq2)-1 + 1.

>

1=1

1=1

H4 Downton, H. F. Private communication. Denote the n impulses by AI> A 2, ... , An. Then for any k, P(A k) = l/n. If f;j denotes the event that IXj is printed twice, then p(EjIA j)

=

P(EkIA j)

p2;

=

1-P)2 ( 11I

(k :F i).

lienee p(AjIEj)

=

P(EjIAj)P(A j)/ P(E j) p(EjIAj)P(A j) p(EjIAj)P(A j)+

L

P(EkIAj)P(A j) k*j = (lI-l)p2/(1-2p+np2).

For the second part, let Eij be the event that the letters succession. Then

IXj

and

IX)

are printed

III

p(Ej'IA k) J

1_P)2 = ( -11I

(i :F j :F k).

lienee P(AjIEj j) =

Finally, let E j • be the event that P(Ej·IA j) = p(1-p);

IX j

(11-

1)p/[(II- 2) + lip ].

and a nOn-lXj are printed in succession. Then

P(Ej·IA j ) =

(,~=n[p+(1I-2)c=n].

lienee p(AjIEj .) = (1I-1)p/[(1I-2)+np].

166

EXERCISES IN PROBABILITY AND STATISTICS

Let Ak be the event that a random batch has exactly k def! . P(A k ) = Pk' Given such a batch, let E, be the event that from a random ~hVt. of n tubes tested, r are found defective. atnp:

85

(~~:)e)/(~),

p(E,IAk) =

for k

~ r and zero for k < r.

Hence ;(E,IAk)P(A k) ,

p(AkIE,) =

L p(E,IA)P(A

j)

j=,

and the required probability is

k=p

86 Let Xk be the probability of the run being completed at X when k CUI tomers have already come there in succession, for 1 ~ k ~ (1'-1), and x '" I Also, let Yk be the probability of the run being completed at X when k CUI tomers have consecutively gone to another counter, 1 ~ k ~ (1'-\), and Y, = O. Hence,

1Xk+l + (n-n- 1)Yl' =;;.

Xk

1~ k

~

/'-2,

1 (n-l) =;;+ -n- Yl;

X,-1

and

1~ k Y,-1

(n-2)

1 =-.x I

+ -/1-

/1

~

/'-2,

YI'

Successive elimination gives 1 Xl

=

YI

=

r=T+ 11

11'- 1 -1 11

,

1

. YI'

and

whence

11'- 1 -1 (/1-1)/1'- I

(/1-

,XI

Xl = (/1,-1+ 11 -2)/(11'-1)

2)

+ 11- -1 . and

Yl =

/1,-1-1

II'

I

'Yl'

(,r 1 -l)/(II'-1).

87 Let Xk be the probability of achieving the run of ,. when k housewivC) have already purchased X. Also, let Yk be the probability for a run of I' of.\ when k housewives have purchased consecutively some other brand. Then Xk

=

P,Xk+l +q'Yl'

X,-I=P+q'Yl'

for 0 ~ k ~ 1'-2, (q=l-p),

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

1

167

p.x, +(n:2)q.Yl.

)'s-, =

llinlination gives

x, = p'-'+(I-p'-')y"

~

Yl = ( ) \\hence

= '[1+ Xo

P

Ii) For large n,

Xo '"

-2

()S ~ + ()S-l ~ +P' {1- ()S-'}' ~

] (1-p'){1-(q/nr 1} (q/n)+(q/n)S 1-2(q/n)s+p'{1-(q/nr'}· 1; (ii) For large rand s,

Xo '"

O.

118 InitiallY, the probabilities are

p(W4) = p4;

P(W 3 B) = 4p3 q ; P(W2B2) = 6p2q2;

P(W B 3 ) = 4pq 3;

P(B 4) = q4.

lIence after the first splitting

P(W2) = p2;

P(B 2) = q2;

P(WB) = 2pq.

rhe proportions at the (n + l)th stage are obtained from the algebraic product

[XiW2)+ 2Yn(WB) + zn(B2)j2, whence

Zn+ 1 = (1- On)2 + j.1n ,

where.1n == (xnzn - Y;);

On == (Xn + Yn)·

Therefore ilild

so Xn

=

O~+mn.1o -+ O~,

Yn

=

00(1-00)-Hr.1o

Zn

= (1-Oo)2+(tr.1 0

as n -+ -+

-+

00

00(1-0 0), as n -+

(1-0 0)2,

as It -+

00

00.

89 Let the proportions of (W2), (WB) and (B2) after the (It-l)th stage be

'._1, 2sn _" tn - 1 respectively. Also, at the same stage, the proportions of (W) ilnd (8) are Pn-l and qn-l. Then

= (rn- 1 +Sn-l)Pn-l; tn = (Sn-l +tn- dqn-l; 2sn = (rn-l +Sn-l)qn-l +(Sn-l +tn-1)Pn-l; rn

and

168

EXERCISES IN PROBABILITY AND STATISTICS

Hence so that Pn = PO+j[1-( _t)n]~O 90 For demand r

~

-+

as n -+ 00.

!
n,

00

E(r-n) =

L (r-n)e- A • Ar/r! = (A-n)[I-F(n-l)]+AP(n-l), r=n

whence 1tn =

(l-n/A)[I-F(n-I)]+P(n-I).

Also, r-n] Pn = E [- - = I - n r

L e - A. A /r . r ! 00

r

r=n

To evaluate the summation use 00

r- 1 '" (r+ 1)-1.

L k !/(r+2)(k),

k=O in factorial notation, and reverse the order of summation. 1tn is the expectation of the proportion of sales lost in the long run, and Pn is the expectation of the proportion of sales lost per month.

91

Xn = (l-r)xn-l +(qYn-l +p zn-l)/2, Yn = (l-q)Yn-l +(rxn-l + PZn- .)/2,

= (l-P)Zn-l +(rxn-l +qYn-.)/2. Hence,usingxn_1+Yn_l+Zn_l = I, Zn

Xn = tp+(l-tp-r)xn-l +!
Y" = (l-tp-q)Y,,-l +!
Y" == Yn-pr/(pq+pr+qr).

and

The equations are solved by using the trial solution Xn = A},n and

Y" = BAn.

92 The difference equations form three pairs: Xn = (I+Xn-l-Wn-l)/6; Yn = (I + Y,,-l -vn- .)/6; Zn = (1 +zn-l -un- .)/6;

= vn = Un =

Wn

(1 +W,,-l -xn - 1 )/6; (1 +vn- 1- Yn-l)f6; (1 +Un-l -z.,,-1)/6.

To solve the first pair, transform to 6Xn = X n- 1- w,,-l

and

6w"

=

w,,-1-Xn- 1,

and

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 1

X ::= " \I~ :::: B).",

where

Xn

-!

w" =

and

Wn

-i·

Hence, using trial solutions

169 Xn

= AAn ,

= [1+(xo-wo)(W- 1 ]/6 Wn = [1 +(wo - x o)(j-)n-l]/6. Xn

Similar results hold for the other proportions, and as n -+

00,

all

-+

f;.

3 The probabilities for the nine types of chains are obtained from the

~lgebraiC product

[piA2 +2Pl(1- pdAa +(1- pl)2a 2] x

x [p~B2 + 2p2(1- P2)Bb +(1- P2)2b 2]. On disjunction of the chains

=

P(AB) P(aB)

=

P1P2;

P(Ab)

(l-pdp2;

P(ab)

= Pl(1- P2); = (1-pd(1-P2)'

nd so the proportions of the reformed chains are also unchanged. In the ;cneral case. the nth stage proportions Xn, Yn' Zn' Wn satisfy

Xn

=

x n- 1 +iAn- 1 ;

Yn = Yn-l -iAn-I;

Zn

zn-l-iAn- 1 ;

=

Wn =

W

n- 1 +iAn-

l•

where 6 n == (Ynzn - Xn wn)· Therefore An = iAn-

Xn Yn

= =

1

= mnA o, whence the results:

xo+Ao[1-mn]; Yo-Ao[1 -(1)"];

Zn Wn

zo-Ao[1-(t)"]; wo+Ao[t -(i)"].

= =

94 (i) The surviving proportions (W2), (WB) and (B2) are (1- ).)p~

1-

AP~ - M~'

2poqo d 1 - ).p~ - Mr an

(1- Il)q~ 1- ).p~ - Ilqr

whence for constancy of qo for successive populations

qo

=

(qo - M~)/(l- AP~ + M~),

so that

qo

=

A/()' + Il)·

(ii) For ;. = 0, Il = I,

ql

=

qo/(l +qo), qn

=

whence

qo/(1 +nqo)

q;; 1 -+

°

=

as

1 +q;;_II' so that n

-+ 00.

(iii) For A = Il = 0,

qn = q~- 1/(1- 2qn-l + 2q~_ d or q;; 1 -1 = (q;;..! 1 _1)2, whence q;; 1 -1 = (qi) 1 _1)2" For limiting behaviour, consider

lOge -qn). qn

170 9S

EXERCISES IN PROBABILITY AND STATISTICS

Denote the (~) dimensional column vector of

xU) by XI; then

xlJ)

by Xo and thai 01

1 XI = 2(k-l)' MX o,

mm

where M is a x symmetric matrix such that (i) leading diagonal elements are all 2; (ii) elements of the cells (pq; ps) are 1, (q "" s); (iii) elements of the cells (pq; rs) are 0, (p "" q "" r "" s); the rows and columns of M being numbered (pq), (p < q) in corresponden Ce with the suffices of the ApAq. Next, if U denotes the x (~) matrix with all elements unity, then

m

MU

=

and

(2k-2)U

M2-4U

=

(k-2)M.

Therefore and, in general,

Mn-(3k-4)Mn-I+2(k-l)(k-2)Mn-2 = 0, so that

Mn = A.n_ I M 2 +lin_ I M. The coefficients A.n and lin are obtained from

lin

=

-

2(k - l)(k - 2)A.n- I ;

A.n = (3k-4)A.n_, -2(k-l)(k-2)A.n- 2 · The elements of the transition matrix [M/2(k-l)]n are thus obtained as:

= 2(1-0"-1)/k(k-l); (pq;ps) = (k+2)(1-0n-I)/2k(k-I)-O(l-0"-2)/k: (pq;pq) = (l-on-I)/(k-l)-20(l-on-2)/k, (pq;rs)

where 0 == (k - 2)/2(k - I). Hence (a)

_

X12 -

2( 1 - on-I) on-I [ " k(k-l) + 2(k-l) j';-I

with similar expressions for

(0)

Xlj

]

,,(0)

+ j;-2 x 2j

-+

2/k(k

1)

-,

as n -+

00,

xli).

96 This follows by using closely the argument of the previous exercise. If Xn for n ~ 0 denotes the (~) column vector of the proportions at the 11th generation. then

XI

=

I 2(k-2)' TX o,

where the transition matrix T = M - 21, 1 is the (~) x (~) identity matrix and M is as defined in the preceding exercise. Then

Tn-(3k-1O)T"-' +2(k 2 -9k+ 16)T"-2 +4(k-2)(k-4)T"-3 = 0, so that

Tn

=

where A., Ii and I' are scalars.

A.n_IT2+lin_,T+Yn_,I,

171

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

rhe clements of the transition matrix [T/2(k - 2)]" are (pq; rs) =

2 48'; 20" k(k -I) - k(k-2) + (k-I)(~-2);

(pq ; pq) = (pq; rs)+

20'; k_ 2

(pq; ps) = (pq; rs)+

0" ()'! k~2 - k~2 ;

==

I rc \lIe

() 1==- (k-4)/2(k-2) and O2

lit) '"

k(k-T)- k(k-2) + (k-l)(k-2) +

--+

2

40';

2/k(k - 1) as

(k - 4)0"

+ k _ 2 2;

-1/(k-2). Hence

20i

0"

(0)

2 X I2

tn - O2 ["

+ k-2

(0)

i;-I Xli

"

(0)]

+ j;-2 X2i

11 -+ 00.

97 UL(1963). Let Xi be a random variable associated with the weather on the ith day, ,u.h that Xi = I, if forecast is correct for ith day Xi =

fhen

P(Xi

=

1)

0,

if forecast is incorrect for ith day.

= 1- P(Xi = 0) = pA. + q(l- A.). i= I

and the

Xi

being independent the stated results follow.

98 Barnard, G. A. (1946). J RSS(S), 8, 1. lI(x) satisfies the difference equation /I(x) = q. u(x + I) + p. /I(x - a), whence equation for vex). Assuming the series for vex) and equating coefficients of powers of A. shows that r (x- ra)(r) vr(x) = (-1) . , . constant A. r.

For the changed range (0 u(x)

~ X ~

= 0 for

2M), the initial score is X ~

0

and

u(2M)

X

= M and

= 1.

lienee

and 00

u(x) =

q2M-X'r~0 (-,1,)'.

(x

ra)(r)!

~!

00

r~o (-A.Y.

(2M

;! ' ra)(r)

x> O.

99 UL (1964). The probability P(R.) = (1- p)p'- 1/(1_ pm). Hence G(O) by summation. Mean and variance obtained simply from the cumulant-generating function. ~s III -. 00, rna -+ 0, so that max E(S") = 11,1,(1- p) < 11, provided that I.

< (I-p).

172

EXERCISES IN PROBABILITY AND STATISTICS

UL (1964). (i) pn+(l_p)n;

100

(ii) pk(l_p)n-k;

(iv) (")pk(I_P),,-k;

(v)

k

(vii)

n-",("-S) L

(vi) (n-S)pkqn_k.

(n)pr(l_pr-r; r

k-s'

p.+r qn-s-r.

r

r=k -s

101

t

r=k

(iii) pk(l_p)n-k+(l_p)"pn-k.,

Woodbury, M. A. (1949). AMS, 20, 311. P(Il+1,x+I)

=

px.P(n.x)+qx+1.P(n.x+l)

with the boundary conditions P(II, x)

= r for

x

< 0 or

>

x

n,

and

P(O,O)

= I.

Also.

n

11-1

P(n,O)

= cfo

and

P(II,n)

=

Pn_ I P(n-I,II-I)

=

PI'

1=0 To solve the transformed difference equation use the generating function 00

G(x, 0)

L

=

0". f(lI, x),

n=x

whence G(x,O)

But P(n,O)

=

= 0(1- Oqx)-I . G(x -1,0) =

Ox. G(O, 0)

/fJ.

(1- Oqr)'

qQ, so that G(O, 0) = (1- Oqo) - I. Therefore G(x, 0) = Ox /

rU

(l - Oqr),

and fen. x) is the coefficient of on in G(x, 0). To obtain this, suppose that

r=O so that

n x

x

L Ar(I-Oqr)-1

==

(I-Oqr)-I,

r=O

x

L Ar j'i'r n (I-Oqj) == r=O This is satisfied for 0 Ar

=

=

I.

I/qo, I/ql,' .. , I/qx' Thus, for 0 = I/q"

q:/n (qr-qJ Hencef(n, x)

j'i'r

=

±q~/n

r=O

j'i'r

(qr-qj)'

102 Rutherford, R. S. G. (1954). AMS, 25, 703. This follows by starting from the expression for P(II, x) obtained in the preceding exercise. Then, for Px = P + ex, P(n,x)

1 x-I (P-+i ) . = -,

n x . 1=0

e

Lx

r=O

(X)

(-1)' . (q-cr)" I

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER I

~

I h~re,o

re the probability-generating function of x is G(t) == =

L"

x=o

t X Pen,

x)

coefficient of

e" n!

eQO[ ( 1-

in

t) + t

e - cO] - pic,

J he factorial moment generating function is obtained by putting t iI(nce 111 I

== ,I,ll

e[(I+c)"-l];

]l(2)

('

== ii, II

173

=

=

(1 + a).

!!.(~+I)[(1+2e)"-2(I+e)"+I]; e (

e(E-lI) (!!.+2) [(1 + 3e)" e e_

3(1 + 2ef + 3(1 +e)"-1];

('

== I'IJ)

e(E+ I) (!!.+ 2) (!!.+ 3) [(1 +4e)"- 4(1 + 3e)" + 6(1 + 2e)" -4(1 + ef + 1]. e c e

('

,-or liP == ex, lie = p, we have lim Pen, x)

n~ '"

=

.n

X-I(a .) e-"'(l-e-py -p+1 . , X •

1=0

.n (a+pl) -e-"'(I-e-P)X , -p- ,

x-I

=

1=0

.

X •

)r·IP .

wilh probability-generating function G(e) = e-"'[1- e(l-e- P lirsl four cumulants of this limiting distribution are

"I = "3 =

The

a(e P-l)/p; "2 = ae P(efJ-l)/p ae P(e P-l)(2e P-l)/p; "4 = ae P{6e P(e P-l)+1}/p.

rhe second expression for the point probability of the negative binomial immediately gives the Poisson distribution for p = O.

103 UL(1964). Let Ar be the event that a tree has r fruits and B" the event that a tree has 1/ flowers. Then

rherefore

p(B"IA r) = C)2-"(l-P)P" =

/Jr

(:)2- x (1-P)P X

C)p,,-r(2 - py+ 1/2"+ I.

If N j is the number of flowers on the ith tree which bore no fruit, then P(N j

Hence

=

II)

= p"(2-p)/2,,+1 and E(e Ni ) = (2-p)/(2-pe).

174

EXERCISES IN PROBABILITY AND STATISTICS

104 Beall, G. and Resica, R. R. (1953). Bs, 9, 354. We may write formally 00

G(e) ==

00

L arer r=O

L ar = 1 r=O

so that

from the given form of G(e). Hence G(e) will represent a probability distribllt' if ar O. Again, G'(e) = A,1 G(e)f'(9). Th~o fore, It _ 1 [d r+1 G(O)] _ A,1 ~ 1 (HI) ar+l - (r+l)! de r+1 8=0 - r+l/;'oklf (O).a r_b

which expresses ar+1 in terms of as (0 ~ s ~ r). Therefore all ar are <0 I' Now

pH 0(0) > 0 for all k.

i

pH 1)(0) = r(n+l).A.~+1

s=O

(-A,2)s. r(s+k+2) . r(s+ 1) r(n+s+k+2)

Thus for n = O,f(H 1)(0) > O. But for n =F 0,

i

pH 1)(0) = nA,~+1

(-~2Y.B(n,s+k+2)

s=O

f

s.

1

= nA,~+ 1 e-

A2%

ZH 1(1_

zt- 1 dz

> O.

o

The cumulant-generating function is ,,(t) = A,1[f(e')-lJ, whence on expansion A,1 A,2 [ 2A,2] "1 = A,1A,2/(n+ 1); "2 = 11+ 1 1 + 11+2 ;

[1 + 11+2 6A,2 6A,~ ] + (11+2)(11+3) ; A,1A,2 [1 14A,2 36A,~ A,1A,2

"3 = 11+ 1 "4 =

11+ 1

The expression for

11

A,~] + 11+2 + (11+2)(11+3) + (11+2)(11+3)(11+4) .

is obtained by eliminating A,2 from the ratios "2/K1 and

"3/"1· 105 UL (1964). (i)

(~)(~~)/( ~~) -

0·43885

~~ )/( ~~) -

0·69618

(ii) 1-( (iii)

G~)/e~) = 0·25000

175

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

13

. (48')/(52) -0.10971 12

(IV)

M(a) (b)

1-4(~~)/[C~)-(~~)] - 0·36964 6(1~)/[(~~)-(1~)]

- 0·30666.

1h~ changed probabilities are

1-(~~) / (~~) - 0·56115

106 UL

(1964). n

n

n

x=!

x=o

x=o

L X4 = L X4 = L

Ii)

(n+4)(S)

[(X

+ 3)<4)-6(x+2)(3)+ 7(x+ 1)(2)-X]

(n+3)(4)

5

6.

4

(n+2)(3)

+ 7 . --3- -

(n+I)(2)

2

\\hcnce result on reduction. o

(ii)

n

x-!

x- !

L L x(x+ l)y2 = L x=l)'=O

x(x + I)

x=l

L

[y(y+ 1)- y]

)'=0

o

=

L

x(x+ 1)[j{x+ l)(3)_!X(2)]

x= 1

0-1

=

L

H(u +4)(S)-¥{U+ 3)(4) +%(U + 2)(3)] u=o (n+4)(6) 11(n+3)(S) 5(n+2)(4) 18

whence result on simplification.

30

+

12

'

'

ANSWERS AND HINTS ON SOLUTIONS

Chapter 2

--

I k::: 2; E(X) = j; var(X) =

is; median =

l-I/.j2.

, The probability density function of X is x e - ",2/2 . var(X~;= (4-n)/2; median = (2 log 2)!; maximum ordmate = e .

• E(X):=: (nI2)!;.

mode

= 1;

k == 2; E(X) = I; var(X) .... 00.

I

~ The distribution function of X is F(x)

a/J

E(X)

=

(ax P - /JxfJ)/(a -

Pl. -since F(1) == 1.

a/J[(a+ 1)2+(/J+ 1)2-1)

= (a+ 1)(/J+ I);

var(X)

= (a+ 1)2(a+2)(/J+ 1)2(/J+2)'

UL (1961).

The probability density function of X isf(x) = 2x/(l +X)3. Formode,!,,(x) = -6(I-x)/(I+x)S < 0 for x =!. The distribution has positive skewness. UL (1962).

h

The probability density function of X is f(x) =

a e-fJ Ian '" • sec2x.

lor the turning values, the roots of f'(x) IIC

Ian x = [I

\Iso. III

= f(x)[2 tan x -a sec 2x) = 0

± (1-a 2 )!)Ja.

!"(x)

= f(x)[2 tan x-a sec2x]2+~(x). sec 2x[1-a tan x),

that f"(x) < 0 if tan x > a-I, whence the mode. Integration by parts gives

f

,,/2

m=

e-fJ'an"'dx,

o

-hence the differential equation. 7 The probability density function of X is 2a(a 2-x 2)/(a 2 +x 2)2. Use the transformation x = a tan 6 for evaluating mean and variance. R The probability density function of X is !(x) = a.xfJ - 1• IIcnce E(X') = a./(a + r). Therefore the central moments are: 177

178

EXERCISES IN PROBABILITY AND STATISTICS

a J1.2 = (a+ l)2(a+2);

J1.3

2a(a - I) (a+ 1)3(a+2)(a+3);

= -

3a(3a 2- a + 2)

J1.4 = (a+ 1)4(~+2)(a+3)(a+4r Hence coefficient of skewness = / 3/2 __ 2(a-l) (a+2)1. YI-J1.3J1.2 (a+3)' a '

coefficient of kurtosis _

2

2

Y2 = (J1.4- 3J1.2)1J1.2 =

For a

6(a 3 -a 2 -6a+2) a(a+3)(a+4) .

I, the distribution becomes rectangular in the (0, 1) interval.

=

9

UL (1962). The probability density function of X is k e- X (1 +X)2, where k == k . evaluated by using the transformation y = (1 +x). 2 The moment-generating function about the origin is

f 00

E(e'X ) = t

f ~

e1x-(1 +Xl. (1 +X)2 dx

=

te- I

-1

e-(1-tlY y2 dy

0

(where 1-/ >II,

The first four cumulants are 2, 3, 6 and 18 respectively. For the median, ~

;e

f

1+~

e-x(l +X)2 dx

=

t or

-I

10

f e-

Z •

Z2

dz = 1,

whence result.

0

UL (1962).

fo J 00

E(e'Y )

=

exp[ -!x 2+t(a+px+yx 2)]dx

oo

=

exp[ta+!t 2p2 /(l- 2yt)] x

fo J 00

x

exp-t[x(1-2yt)t-tP(1-2yt)-tFdx, where 1-2yt>0,

oo

= (1- 2yt)-t exp[ta + t 2P2/2(1- 2yt)]. Expand logarithmically to obtain E(Y) and var(Y). II

Wadsworth and Bryan (1960). Ex. 4.10, p. 102, and UL (1962). If F(x) is the distribution function of x, then

P(S < N) = P(x < l/m) = F(l/m); P(S > aN)

and

= P(x > aIm) = 1- F(a/m), whence result.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER I~

2

lJL (1962).

profit is O(x) = av-b(x-v) if v ~ x, O(x)

fI~nce

= ax-c(v-x) if v> x.

x

1.[IKxl]::::

and

V2

f [(a+b)v-bx]f(v) dv+ f VI

[(a+c)x-cv]f(v)dv

~

f x

:::: (a+c)x-(a+b+c)x

f X

f(v) dv+(a+b)

VI

:)O(X)]

vf(v) dv-c

VI

=

f V2

X

0 gives the result.

1\ Downton, H. F. Private communication . . The proportion of articles of weight < 8 oz. is s - ...

p(w) =

~J'X) e- '/2 dt. t

Iknce expected profit is

0= 0.p+(1-p).1-(0·05w+0·30), .lI1d Ihe roots of dO/dw = 0 are 10·04, 5·96.

Bul

d 2 0= __ 1 .(8_w)e-(S-w)2/ 2 <0 if w=1O·04. _ dw 2 jbr 14 UL (1960).

The proportion of packets < a in weight is

f ajbr a

p(x)

=

_1_.

e-(t-x)'/2a 2

dt.

-00

IIcnce expected profit is

0=

v

V

"2' p+0'8(1- p)V +0·2(1- p)"2-(x+c).

IIcnee

dO

-=0

dx

. ~E. = __1_ and gIves dx O·4V

dp dx

vf(v) dv,

179

180

EXERCISES IN PROBABILITY AND STATISTICS

For a maximum,

d 28 dx 2 15

0-4V ( )212 2 --.[(a-x)e- a -." (1]<0 a3

J2rr

• If a-x
Wadsworth and Bryan (1960). Ex. 12, p. 225, and UL (1961). For given x, expected profit = (c 3 -c2)E(x)-c l , whence result. Also, bankruptcy results if profit ~ - C4' i.e. if x ~ (CI - C4)/(c3 - e2).

16 The roots of the quadratic are XI

= [a+(a 2 -4b)t]/2 and X2 = [a-(a 2 -4b)t]/2;

and b is uniformly distributed in the range (0, a 2 /4). Hence

E(x 2) = a-E(xd = a/6.

a2 a E(xf) = 4+2: E[(a 2 -4b)!]+iE(a 2 -4b) = 17a 2 /24. Hence var(x l )

= var(x 2 ) =

a 2 /72.

For reasons of symmetry, it is enough to consider X ~ I and assume thai has a uniform distribution over the range (0 ~ x ~ I). The semi-axes art uniquely defined and the eccentricity is

17

X

e

=

[41(l-x)/(2/-x)2]~.

To determine E(e) use the transformation I-x = It 2 , whence E(e) = (4 -n) "-' 0'8584, var(e) = 2(2 log 2-1)-(4-n)2 - 0·0356.

and 18

Downton, H. F. Private communication. Probability of man being late = P (train arrives after 8.20) + P (train arrives before 8.20) x P (bus arrives after 8.30) =

1- (f)(1)(f)(0'75) - 0'349,

where (x) is the distribution function of a unit normal variable. Probability of employer being late = 1 - (~) - 0·251. Probability of both being late - 0·088. Probability of employer arriving before man =

P (train before 8.20) x P (bus after car) + P (train arrives after 8.20)

=

<1>(1)[1-<1>( -0'5/j13)]+ 1-<1>(1) '" 0·627.

19 The quadratic 3x 2 -6gx+g(a+3b)-ab = 0

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

2

18]

eal roots if 3g 2 - ~a + 3b~ + ab ~ O. This inequality is satisfi~~ for g < b ~ a/3. Hence x IS real If a/3 ~ g ~ a, whence the probablhty. jod{he redefi,ned distribution of y = g(x) is (3/2a) dy, (a/3 ~ y ~ a), whence and vanance. 11Ca Jl hJS r

'0 UL (1962).

• 'The probability density function of 0 is 120(n - O)/n 3 and y = tl 2 sin O.

f f ,,/2

E(y)

12

= n3

l2 "2sinO.O(n-O)dO = 12l2 /n 3 , and

o

,,/2

E(i)

= n123

l4 "4sin20. O(n-O) dO

= l4(n 2 +3)/8n2 , whence

o var(y) = l4(n 6 + 3n4 -1152)/8n6 •

21 (i) The inequality x(2/-x) ~ [2/2 is satisfied for the regions 0, 1(1-1/j2)

and 1(1 + 1/j2), 2/. Hence probability is (1-1/j2). (ii) The inequality x(2/- x) ~ 3/ 2 /4 is satisfied for

1 2

31 2

-~x~-,

and so the probability is

t.

(iii) The inequality

is satisfied for

o ~ x ~ l[l-(:~:r]

l[l+(:~:r] ~ x ~ 2/,

and

whence required probability is

1-(6-n)t. 6+n 12 Gordon, R. D. (1941). AMS, 12, 115.

rr $(z) denotes the distribution function

of a unit normal variable, then

g(x) = ex2 / 2 [1-«I>(x)]fo. 1I~lIcc

f 00

E[g(x)) =

e _(X-II)2/2 • ex2/ 2 [1_ «I>(x)] dx

-00

f 00

= e- II2 / 2

ellx [1-«I>(x)]dx

-00

= 1/J.l on integration by parts.

182 23

EXERCISES IN PROBABILITY AND STATISTICS

(i) Tchebycheff's inequality. For k > 0 m-ku

00

(12;::,: f(x-m)2f(x)dX+ f (x-m)2f(x)dx -

m+ku

CJJ

or 0"2;:a.

f

P[\x -111\;:a. ku] x (kuf

a

(ii) E(X)

=

f 00

xf(x) dx+

xf(x) dx

a

-00

o

f (y+a)f(y+a)dy+ o

a+

(y+a)f(y+a)dy

0

- 00

=

f 00

f 00

f

yf(y+a)dy+

yf(y+a)dy

=

a.

0

-00

Similarly, 00

E(X-a)'

=

00

f y,/(y+a)dy+ f .v'f(y+a)dy

(-1)'

o

0

= 0 for odd r. (iii) For c >

m, define 00

S(c) ==

f

Ix-clf(x) dx

-00

f f C

=

=

f 00

(c-x)f(x)dx+

(x-c)f(x)dx

- 00

C

m

I

I

m

m

c

00

(c-x)f(x) dx+

-00

(x-c)f(x) dx+2

(c-x)f(x)dx.

Hence C

S(c)-S(m)

=

2

f (c-x)f(x)dx > 0, m

unless c 24

= m. Similarly for c < m.

Downton, H. F. Private communication.

Suppose 0 ~ g(X) < t/J(g). Then

00, 00

E[g(X)]

=

and let the probability density function of g Ix 00

00

f gt/J(g) dg ;::,: f gt/J(g) dg ;::,: k f t/J(g) dg, o

k

k

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

.hence

2

183

the first inequality. For the second, assume

g(X) = [( X;J1r -(l-C)r S'z1ey (1957). Ex. 7.11, p. 219.

:5

I.e: the straight line be divided into mn intervals each of length

.1

e

l/mn. Then tal number of ways for choosing m parts so that they total mn intervals !hChlO coefficient of xmn in (1 +X+X2+ ... +xmn)m

. h is A == m(m+ 1) ... (m+mn-1)/(mn)! ,.hl~or the favourable combinations each of the m parts:::; I, and so we can II

most take mnl intervals for any part. Hence the favourable number of comh ffiClent . f mn'10 0 x (1 +X+X2+ ... +xmn/)m

~Inalions is t e coe

. hich is B == A -m2(m+ 1)(m+2) ... {m+mn(1-/)-2}/{mn(1-1)-1}! Hence ;~c required probability is 1-m(1-l)m-l obtained as the limit of B/A as ~ -+ ro.

26 UL (1958).

The proportionality factor is Ir;lnsrormation y = log x.

l/(1A

which is obtained by using the

f (1fo

00

00

E(X,)=_1_fx,-le-(I08X-mll/2al dx =_1_ =

(1fo 0 exp r(m + r(12/2).

e'Ye-(y- m l l / 2c71

-00

lienee the mean and variance. mode = em- al ;

maximum ordinate =

1

M::.' e- m+ al /2.

(1v 2n

Solution of a 1 = e m+ al/2 and a2 = e 2m + al (e al -1) gives the last results.

27 The proportionality factor is 12. Hence

f ,x 1

E(e'X )

=

12

e

.x 2(1-x)dx

=

12t- 4[e' (t 2 -4t+6)-2(t+3)]

o hy successive integration by parts.

28 Pearson, K. (1929), B, 21, 370, and UL (1961). Ir E(X) = m, then set z = X - m so that

(X2+aX +b)2

=

(z2+Az+B)2,

where A == a+2m, B == m2+am+b. Hence, since E(Z2+Az+B)2 ~ 0,

P2+(J[3; +A/j"i;;y+(1+B/J12)2 ~ Pl +1, where

dy

184

EXERCISES IN PROBABILITY AND STATISTICS

But a and b are arbitrary, so the inequality follows if

I+BIJl.2 =

J7i; +AIJii; = O.

0 and

P(x) = 2-" = exp( -x log 2). Thus x has a negative exponential dj. tribution with density function

29

(log 2) exp( - x log 2),

(0

x < 00).

~

Hence the cumulant-generating function is

= -10g(1- tllog 2), t < log 2. = p"+n(n+ 1),,/2 = exp(n log p - px). Hence the normalized dens K(t)

30 P(x) function of X is

II.

pe- P"/(I-e- P),

and for t <

for O~X
p, E(e'X)

= (l-tlP)-I[I-(e'-I)/(e P-l)].

A logarithmic expansion now gives the mean and variance.

31

UL (1961). The probability density function of X is 2px e- p,,2. Hence

m = (x/4P)t; median = 32

(12

= (4-x)/4P; mo = (2P)-t;

(P- 1 log 2)t; skewness positive.

The probability density function of X is

+(

2 ) e-a.x . sin x (1 -'----:---'----=/2-'

(0

l-a.e "'''

~

X

~

12) x .

For a - t > 0, the cumulant-generating function is

_

K(t) - log

{[ 1+a2 ] [1-(a-t)e-("'-t)"/2]} 1 -ae ",,,/2 1 + (a-t)2

Hence,

1 - ax/2

E(X)

2a

= e"'''/2 -a + 1 +a2 ;

(e",,,/2 -a)(x-ax 2/4)-(1-ax/2)2 2(a 2 -1) var(X) = (ett"/2 _ a)2 + (a2+ W. Therefore for a > 0 and a2 negligible E(X)"" 1-(x-3)a,.., 1-0'14a,

and

[ x 2 -4(x-l)(x-2) ] var(X) '" (x-3) ll+ 4(x-3) .a"" (x-3)(1+0·16cx). 33 If k is the proportionality factor, then 00

kf

-00

00

-~=2kf 1+x4 0

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

,.Ibere

2

185

w:== 1/(1 +X4), so that k = 2/B(i, i)· 1

f:tX):::= 0 and var(X) = E(X2) = B(;, !)'

f

W-

i (1- w)-t dw

= 1.

o

wo

P(X

~ IX) = 2B(~,!)' f w- t (1-w)-tdw,

where Wo ==

(1+1X4)-1,

o

=

2;r!)'

j

C t (1-wot)-tdt,

o II

result, using term-by-term integration after expansion of hence the t

Ii -lI'olr .

\.I A ramous result due to K. Pearson.

'.1

hence

_ [(NP-X+ 1)(r-x+ 1)] Yx+l - Yx (Nq-r+x)x Ayx = Yx+l-Yx

and

Ayx 1 dy - - '" -'-d . Yx+t Y x

For bl = b2 = 0, the normal distribution with mean a and variance bo I'

obtained .

.IS Cadwell,1. H. (1952), B, 39, 207. x

F(x) = t+cI>(x) and

cI>(x) = -1-fe- tx2 dx, o

fo

so that

.11111

lIence

4F(x)[I-F(x)] '"

-2x 2/"

,I

2X4 7x 6 3n 45n

2X2 n

1--+---+ ... 2(n-3)x4 (60-7n2)x6 3n2 + 45n 3 +

'" e

+

'" -2x 2/" e

4 (60-7n 2)x6 ... } 2X 2/,,] [1 + {2(n-3)X 3n2 + 45n + e ,

hence the stated result.

3

...

186 36

EXERCISES IN PROBABILITY AND STATISTICS

A classical result due to E. C. Molina (1915).

= 1- F(/l),

P(X ~ /l)

where

Hence, integrating by parts successively, F(!l)

=

l-e- p

•

•

L

/lr/r!

= I-P(Y ~

v).

r=O

37 F(r) = P(X

= B(

=

~ r) =

.t

qn-r 1 _)'

r+ ,n r

(:)p"qn-.

r

L

.=0

(r) p" qr-·.B(n-r,r-s+l) s

f

f 1

(r)

qn-r . Lr p' qr-. B(r+ 1, n-r) .=0 s

zr-s(1- zr-r-1 dz

o

1

_

-B(

qn-r

I

).

r+ ,n-r

(l-Z)

n-r-1

r

.(p+qz)dz,

o

by reversing the order of summation and integration, q

= =

1 . B(r+ 1, n-r)

f yn-r-1(1_ y)r dy,

y=q(l-z)

o

G(q).

f x

38

P(X

~ x) =

1

1:

1 + fo' 0

e

-x 2 /2

d

x,

and the first result is obtained by term-wise integration of the series fore-"; For the second result,

ANSWERS AND HINTS ON SOLl'TIONS: CHAPTER

',lhe Oce

2

187

successive integrations lead to the series. Finally, oc· (211 + 1) ! e - x2/2 e - (1 2 + 2Ix)/2 IR.(x)1 =

f f

2., fo, n! . 0 (t+x)2n+2 . dt 00

(2n+l)!e- x2 /2 dt (2n)!e- x2 /2 1 < 2n.n!fo '0 (t+x)2n+2= 2n,n!fo'x2n+I'

19 Eolow, E,.R. (1934), AMS, 5, 137. . Use the series (x 2 +2r)-1 = x- 2

00

L (-I)k(2rNx 2k k=O

'valuate AI' A 2 " , , ,A, as power series in inverse powers of x, The stated ::~,~It is obtained by direct substitution in the expression for P(X ~ x).

.ro Direct integration gives k = (2n + 1)/2. (i) P(X ~ 0)

= (1 +sin2n+ 10)/2. p(IXI ~ 0) = 2P(X ~ 0) = 2[ 1- P(X < 0)] = (1- sin2n+ 10). Iii) P( -7t/6 ~ X ~ n/6) = 1t)2n+ I. E(X) = 0 and var(X) = E(X2) as stated, where

f

71/2

12n + 1 ==

xsin 2n +1xdx,

o

whence integration by parts gives the recurrence relation. Z has a Beta distribution. 41 Romanovsky, V. (1933). B. 25. 195. (i) Set

Ux = a~:+ 1(l-axY"

Then

=

fo' I .~

e- x2 / 2 dx.

!a~(l-axY"+ux(!-u~y".

Expand (1 - ax)m and integrate term-wise, whence (i) by noting that

f 00

1 (I 2\111 -x 2 /2 d - 0 fo'-oo U x '4- UxJ e x - .

fo'[

(Jx

(ii) (1.Sx

=

!+u(Jx, where U(Jx =

e- I / 2 dt.

The equality follows by integration of a~(I-ax)ma(Jx

=

!a~(I-ax)m+u(Jx(!-u~y".

(iii) Integrate

(iv) Integrate x2n a~+ 1(I-axY" = tx2n (X~(1-(XxY" +x2n

ux(!-u~y".

188

EXERCISES IN PROBABILITY AND STATISTICS

42 Probability density function of Y is (i) e-",

(0 ~ Y

< (0); (ii) 1/2Jy,

(0 ~ Y ~ 1);

(iii) 1/2Jy, for 0 ~ Y ~ !; and 1/4Jy, for! < Y ~ 9/4.

43

(i) Transform to polars by using x = z cos e, y = z sin 0, whence f probability density function of Z is I

2e-: 2 • Z3, (0 ~ Z < (0). (ii) With the transformation z = x + y and x = x, the region x +y _ x = y = 0 is transformed into the region bounded by x == 0 z~. I and z - x = O. The limits of x are (0 ~ x ~ z), whence the probabil density function of Z in this region is z for (0 ~ Z ~ 1). II, In the region x + y = 1, x = 1, Y = 1 make the transformal • Ii' Z = X +Y -1 and x = x, whence the transformed regIon is x " ' z = O. z = x. The limits of x are (z ~ x ~ 1) for fixed z. Therefore' this region the probability density function of Z is 0- z), ,;" (0 ~ Z ~ 1).

The two densities may be added as there is no discontinuil)' i the range of Z. (iii) The probability density function of Z is

al aZ . [e-aIZ_e-a2Z~, az-a,

(0 ~ Z

< (0),

obtained by using the transformation z = x + y, x = x. 44 The probability density functions of A and Bare (lh e - at and respectively for 0 ~ t < 00. Probability of A not happening in (0, t) is (1 +oct)e-"'. Probability of B happening in (t, t + dt) is !p 3 t Z e -fJI dt. The required probability of occurrence in the order BA is

f

!pJ,le '

IX)

(1 + oct) e- at . !p 3 t Z e- fJI • dt.

o 45 The probability density function of Z is Jf(x)g(z - x) dx. where II" integration is over the range of x for fixed z, and the region is bounded~. x = 0, x = oc. z = x and x = z - p. Therefore the density function of Z is (i) z/oc/3, if 0 ~ Z ~ oc; (ii) 1/13, if oc ~ Z ~ 13; (iii) (oc + 13 - z)/ocp, if 13 ~ Z ~ oc + p. For oc = 13, the distribution of Z is triangular with density function: i' for 0 ~ Z ~ 13, and (2p-z)/pz for 13 ~ Z ~ 213. Baten, W. D. (1934), AMS, 5, 13. Use the method of the preceding exercise. The pro babili ty density funclill!' of Z are as follows: (i) z(4zZ-6z+3), for 0 ~ Z ~ 1; and -4z 3 +18z z -27z+14, for 1 ~ Z ~ 2.

46

") 1 ( 11 4"

1

Iog [11 _+Jz] Jz ' for 0 ~ Z ~ 1; and

4" log

[3-Z+2(2-z)i] 1 ,for z-

1 ~ Z ~ 2.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

2

189

''') i(SzS-40z4 +80z 3 -60z 2 + 15z), for 0 ~ Z ~ 1; and i(_SzS+40z 4 -80z 3 +100z 2 -95z+46), for 1 ~ Z ~ 2. The joint distribution of u = x 2 and v = y2 is

(UI

dudv

C'

4y uv

for 0 ~ U, V ~ 1.

Hence the probability density function of w = u + v is 11:

4'

for 0 ~ W ~ 1;

1

'4 cos-

~7

1

[

-

and

w2 -8W+8] w2 for 1 ~ W '

~

2.

The probability density function of the marginal distribution of X is 1 log x, for 1 ~ X < x

2'

00.

Ihe probability density function of the marginal distribution of Y is

~, 2y

lor fixed X =

for 1 ~ Y <

Y

~

1.

Y is

1 for -:s;;;y:s;;;x.

x

for y ~ X <

00,

l/x 2 y, for l/y ~ X <

~

~

y, the conditional density function of X is

y/x 2 , .md

for 0

x, the conditional density function of 1 2y log x

lor fixed Y =

t,

and

00

00,

if 1 ~ y <

00 .

if 0 ~ y ~ 1.

The proportionality factor of the joint distribution

= (oc + l)(oc + 2)/B(oc + 3, P + 1). Use the transformation u = or II and x as

y/j'X and x = x

to obtain the joint distribu-

111111

(oc+l}(oc+2) «(<<+2)/2 II r: B(oc+3,P+l),U .x .(i-x) (yx-u)dxdu. I hcrerore the marginal distribution of u is

f

(P)( -1)'

=-:(IX:-+_l_}(,--OC_+_2.:....) . u« ,=0 r B(IX+3,P+I)

f 1

[x(<<+ 2,+ 31/2 - ux(<<+ 2'+21/2] dx du

'

u2

.hence the result.

~9 The joint distribution of wand z is e2

dzdw

4' exp{ -(w/z)t} '-z-'

190

EXERCISES IN PROBABILITY AND STATISTICS

and the region of integration lies between the hyperbolas wz ::: 9 and to the right of the line z = IVz, For z fixed in the range H :::; z :::; ~), then (4z :::; W :::; 9/z), and for in the range (0 :::; z :::; t), then (l/z :::; w :::; 9/z). Hence the marginal; 6) . function of z is en.

tw.

'

e2 2z[e- l/Z (1 +z)-(3+z) e- 3/Z ],

for 0:::; z :::;

t

and ] 3 e2 [ ---(3+z) , 2 2z

1

3

for 2:::; z :::; 2'

For w fixed in the range (2 :::; w :::; 6), then (l/w :::; z :::; w/4), and for IV 6). in the range (6 :::; w < (0), then (l/w :::; z :::; 9/w). Hence the marginal den;'. function of w is e2 2

fW

-

l e - I t- dt ,

for 2 :::; w:::; 6

2

and

e2 2

fW

e-1t-ldt,

for6:::;w<00.

w/3

50 If Xl and x 2 are the vertical and horizontal distances of P from the bUll eye, use the transformation Xl = R cos (}; X2 = R sin (}, whence the distrib. tion of R. Therefore

P(R

~

r)

=

e-r2/2a2

P(rl:::; R :::; r2)

and

For n trials with k in the annulus (r 1

C)

:::;

=

e-d/2a2 -e -r~/2al.

R :::; r 2), the probability is

[e-rtJ2a2 _e-r~/2a2Jk[1_e-rV2a2 +e-rl/2a2]n-k.

51 Use the transformation U = joint distribution of u and v as (}2 e- ul9 • u du dv,

+X2 and v =

Xl

where

Xl/(Xl

+X2) to obtain II

0:::; v :::; 1; 0 :::; u < 00.

The variables u and v are independently distributed; v has a rectangul' distribution, and that of u is (}2.

e- uI9 • u du,

(0 :::; u < (0).

E(v) = t; var (v) = l2' Finally, v = 1/(1 + w), so that u and ware also independently distribul\· and the distribution of w is

(1+w)-2dw,

for 0:::; w < 00.

52

Wadsworth and Bryan (1960), Ex. 6.19, p. 172. The variables u, v and w can be expressed in terms of the devialil)' (Xi-m). The joint distribution of u, v and w is

1 3(0' 2n)3

.j21r. . exp[(u 2+v 2-uv)/30'2-w2/60'2]. du dv dw, (- 00

< u, v, w < c£)~

:2

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

t the joint distribution of u and v is ",lha 1 -----==------=~. e - (u 2 + v 2 j3((1j2n)2 • _ II/V -

i·~I.

Uv)/3(12

191

du dv.

and t' = v; note that the joint distribution of v and z is

j3

~

. e -v 2(z2- z -1)/3(12 . lid v v. dz,

(-

00

< v, z <

)

00 ,

61[(1

. ce the distribution of z is

j3

.h~n

dz

2;' (Z_!)2+! .

t.

~refore mode of the distribution is z = Also, z = 1 - t, where ,Ih (Xt- X2 )/(X3 ~X2)' But by symmetry z and t must have the same mode, .h~nce mode ='2'

(illl has the F distribution with (2,2) d.f., i.e. du/(l + U)2, (0 ~

<\

(iiI Put

Zl = (XI Z3

+X 2 +X3)/j3;

Z2 = (XI

= (XI- 2x 2+ X3)/j6;

Z4

U

<

00).

-x3)/j2;

= X4'

The X -+ Z transformation is unitary orthogonal, and each Zj is an independent unit normal variable. Hence v has the t distribution with 3d.f.• i.e. 2 dv

(-oo
7rJ3' (1 + v 2 /3)2 '

liiilll'-t = 1+xV(xi +x~). where 2x~/(xf +x~) has the F distribution with (1,2) dJ. Hence the distribution of w is dw (0 ~ w ~ 1). 2(1- w)t' IXI

=

19'00; (X2 = 5·841.

~ Transform to u = X + y + z, v = y/x and w = z/(x + y). The Jacobian ,·f the transformation is (1 + v)2(1 + W)3/U2, and the joint distribution of U,

1and

Wis

vr.:II. g(u) du. r:. v

dv

v(1 + v)

.

v

Co

dw

w(1 + w)i

,where 0 ~ U, V, W <

00.

lienee U, Vand Ware independently distributed. ~s Wadsworth and Bryan

(1960), Ex. 5.19, p. 140.

The required probability is co

P=n(n-1)

f

00%"

f(Xl) dx l

- 00

1'I\:

f[f Xl

f(X) dX

r-

Xn

2

f(x n)

[l-f f(X)dXfdXn. Xl

XI

the transformation

f XI

U=

-00

f Xn

f(x)dx,

v=

-<X>

f(x)dx,

192

EXERCISES IN PROBABILITY AND STATISTICS

where u

~

v

~

~

1; 0

u

~

1. Hence 1 1

P = n(n-1)

ff

(v-u)"-2[1-(v-u)j2dv.du,

a "

whence result.

56 Carter, W. H. (1933), JIA, 64, 465. P(X

=

r) = er8)(52~J/er2),

(0

~ X ~ 48).

Set N == \3.49. 50. 51. Then

so

48

L

P(X

= r) = N- 1

r=O

L

t(t 2 -1)

1= 1

=

1 50 4N ,~ [(t+ 2)(t+ 1)t(t-1)-(t+ 1)t(t-1)(t-2)]:: I. 50

= N- 1

E(X)

L

(50-t)(t 3 -t)

1= 1

50

= 50-N- 1

L

(t+1)t(t-1)[(t+2)-2]

= 9·6.

1= 1

For the variance, consider so E(50-X)2 = N- 1 L t 3 (t 2 -1) 1=1

50

=

N- 1

L

(t-1)(t+ 1)t[(t+2)(t+3)-5(t+2)+4]

= 17(1,

1= 1

Hence, var(X) = 67·84. For the approximation, let P(X Then

=

= kx(x 2 -1), where (2

r)

~ x~ ~

50

k

J

(x 3 -x)dx

= 1, whence k- 1 = 1561248

2

and

so E(x)

=k

f

(x 4 -x 2 )dx

= 40·0053, so that E(X) ,.., 9·9947.

2

Similarly, E(x 2 ) P(X

= 1667·0024, and var(X),.., var(x) = 66·5744.

~

25) ,.., P(x

~

2S

25) = k

f (x

3-

x) dx

= 0·0623,

2

and the probability for two consecutive outcomes ,.., 0·0039 < 1 %.

ANSWERS AND HINTS ON SOLUTIONS: ('HAPTER

2

193

. rter W. H. (1933), J I A, 64, 465. "

(8 hod

\Id

similar to that of the preceding exercise.

PIX = r) =

(~)t~I)(52~,)/rr2),

(1

~ X ~ 49).

. .; ~ 13. 17 . 50. 49. Then, .

49

~'P(X "" r) = N- I~l (50-t)t(t+1) 1

~I

= N-

1

[H

H]

H

50 I~l t(t+ 1)- I~l t(t+ l)(t+2)+2 I~l t(t+ 1) = 1.

49

E(49-X) =

I

(49-r)P(X = r)

r= 1

49

= N- 1

I

1=

(50-t)t(t+ l)(t-1)

= 28'8, and

E(X)

= 20·2.

1

, rnilarly, 49

£[(49- X)(48- X)] = N- 1

L 1=

(50-t)(t+ l)t(t-l)(t- 2),

I

.h~nee E[(49-X)2]

= 931·2 and varIX) = 101·76. the approximation, put P(X = r) = kx(x+ 1)(50-x), 1'-: x ~ 49) and I· or

f

where

49

k

(50-x)(x2+x)dx = 1

gives

k- I = 540384.

I

Iknee

E(50-X) '" E(x)

= 29'7650, so that E(X)

'" 20·2350.

111o,

E(x 2 ) = 987'1429, whence var(X) '" var(x) = 10H877. P(X

~ 25) '" P(x ~ 25) = k

f

25

(50-x)(x2+x)dx

= 0·3204.

1

Ihcrerore the probability of four consecutive outcomes'" 0·0105 < 2 %. c.x

The joint distribution of t 1 and t2 is

f

el2

P = n(n-l) 2' 1:

., hence result.

o

f

e-It

e-ll/
1

II

e-(n-l)12/
2,

194

EXERCISES IN PROBA B I L ITY AND STATISTICS

59 Wadsworth and Bryan (1960), Ex. 5.19, p. 140. . Consider anyone socket, and suppose in time (~,.r) it has r burno tImes t 1 ::;; t 2 ::;; t 3 ::;; .•• ::;; tr ::;; r. The total probabIlIty for this eVent ~II 1\

r

P(r)

=

r t f

fd~lfd:2fd:3 ... o

'1'2

rr

t

d:r[l-f~tJ

f

I,.

',.- 1

rr+ 1

kr .'-!-kr + 1 .(r+I)! ' by successive integration. Hence
E(r) =

L rP(r) = (etlk_I). r;O

Therefore average cost per unit time is C

=

yn(etlk-l)+a+pn r and

dC

r

..

dt = 0 lor mInImum.

60 Cadwell, J. H. (1951), B, 38, 475. By definition, OC;

/(11, q)

=


2~ f

f e-(x 2+y')/2 dx dy h qxlh

00

h

IX)

00

= -1 f f e -(x'+)")/2 d x d y - 1- f f e -(x'+y2)/2 d x d y 2n

2n

o qxlh

0
=

V(h,q)+2Inf

qxlh

f e-(X'+Y')/2dXdY-2~f f

h '"

'"

o qxlh

0 0

rf)

=

tt/2

V(h,Q)-¥t>(h)+21nf o

whence result.

f e-(X'+Y') /2 dxdy

e- r '12r drde

(a == tan - 1qJh),

7

h qxlh V(h, Q) = 21nf f e-(x'+y2)/2 dx dy. o

0

If the region of integration in the (x, y) plane having area hQJ2 is the sam. as the area of a sector with radius R and angle a, then R 2 = IIq/a. Thercfol( a

V(II, q) =

21n

h sec9

ff o

e- rll2 r dr de,

0

and P6lya's approximation is obtained by approximating the double inlW U

«R

21n

ff o

e -r2/2 r dr de,

0

whence the approximation for /(11, q).

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

2

195

Cadwell, 1. H. (1951), B~ 38, 475. bl put x == r cos e, y = r SIn e. J(h, q)

n/2

= 2~

f e-

th2 sec 28

de,

a

w

ate by parts successively, using [_h2ze- th2 (l+z2)] as the first function Inlegr . 'ach integratIOn. 11 t The alternative expression for the general term is obtained by putting : == I so that (1/w) d/dw = 2d/dt, and then differentiating by Leibnitz's :h~orem.

~2

~,l

Obtained ~y direct integration. For fixed y In (0 ~ Y ~ a), (0 ~ z ~ y); and [or fixed y in (a ~ y ~ 2a), [0 ~ z ~ (2a- y)]. For fixed z, [z ~ y ~ (2a-z)]. Astandard result obtained by using the transformation

~_ C'1+ a , v = '1_(Ca+Ab)

u=

A

Ihe variables

II

AB-C 2 '

'

and v are independently distributed and

Ae + B'12 -

2C~'1- 2a~ - 2b'1

(Ba 2 +Ab 2 +2Cab) (AB-C 2)

(AB-C 2)V2 A'

Au 2 - - - - - ' - - - - -

"hence result. fl.I The proportionality factor k is obtained from

k

f'"

fOO

x 2e - yx 2 dx d y _ k foo

(1+lxlt

-

-000

"~I

k = (IX -

dx

_ 2k foo _d_ x_ - I (1+x)a-'

(1+lxl)a-

-00

0

0/2.

The marginal density function of X is IX-I

(-00 < X < (0),

2(1 + Ixl)a'

IIllllhat of Y is

f 00

1 (IX-)

dx (1 +x)a '

x2 e-

YX2

(0 ~ Y

< (0).

o

!'q = 0; var(X) = 2/(1X-2)(1X-3). All moments of Y diverge since

f 00

E(Y) = (IX-l)

o

dx 2 -+ 00. x (1 +x)a

196

EXERCISES IN PROBABILITY AND STATISTICS

But

since the integrand is an odd function of x. Now E(X) 1 corr(X, Y) = - Jvar(X)' [E(y2)/E2(Y)-1]t' But var( Y) = E( y2) - E2( Y) > 0, so that

E(y2)/E2(Y)-1 > O. Hence corr(X. Y)

=

O.

65 UL (1963). Marginal distribution of X is (l-X)fZ XP+ 1 dx

B(a+ I, P+2) ,

(0 ~ X ~ 1).

Marginal distribution of Y is

1(1- y)fZ+ 1 dy B(a+2, P+ 1) ,

E(X) =

(0

~

Y ~ 1).

P+2 . (ct+ 1)(P+2) ct+P+3' var(X) = (ct+p+3)2(a+p+4);

P+ 1 Y (P+ 1)(ct+2) E(Y) = a+p+3; var( ) = (a+p+3)2(Ct+P+4) Conditional distribution of X for Y = Y is

[(Ct+ l)(l-x)fZ/(l- yt+ 1] dx,

(y ~ X ~ 1).

Conditional distribution of Y for X = x is

[(P + l)//xP+ 1] dy,

(0 ~ Y ~ 1).

(Ct+ l)(P+ 1)]t corr(X, Y) = [(ct+ 2HP+ 2) , given by the product of the two regression coefficients. 66

Murty, V. N. (1955), JASA, 50, 1136. The joint distribution of x and y is (0 ~ x ~ Ct; 0 ~ y ~ fJ).

Transform to polars using the transformation

x/a

= r cos 0, yiP = r sin O.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

2

197

ate out over. r, noting that lolC~ for fixed () 10 (0 :::;; () :::;; n/4), (0:::;; r :::;; sec ()); !~)) tor fixed () in (n/4 :::;; () :::;; n/2), (0:::;; r :::;; cosec ()).

III

I'

P(z :::;; Po > p) = 1P(z:::;;

(_n_) (p/po)m ; m+n

PI < p) = (~)(P/Plr. m+n

Rider. P. R. (1955), J ASA, 50, 1142. h7 Ii' For m = n, the distribution of z is - n2 z"-1 log z dz, (0:::;; z :::;; 1), whence the distribution of v. Iii) For m < n, distribution of z is

(0 :::;; z :::;; 1), whence the distribution of t defined by z

= e- I is (0:::;; t < (0).

Therefore, for v = 2nt,

E(v')

= (2/A)' r(r + 1)(1- A:+ 1)/(1 -

,t).

The X2 approximation is obtained by assuming v - f3x 2 with v dJ., where f3 and v are obtained by equating the first two moments. 68 Rider, P. R. (1951), JASA, 46,502. The joint distribution of WI and W2 is 11 1112(nl-1)(n 2 -1) ,-2 2-2 rx n , +n2 • wi (rx - WI) w'2 (rx - W2) dW I dW2, (0 :::;;

Wio W2 :::;;

rx).

\Iake the transformation wdrx = r sin (), W2/rx = r cos (). Then, for fixed () in 0 ~ n/4), (0 :::;; r :::;; sec ()); and for () in (n/4 :::;; () :::;; n/2), (0 :::;; r :::;; cosec ()). Integration over r gives the distribution of (), whence the distribution of II ~ tan (I is

In ~

n l n 2(nl -1)(n2 -1) (111 + 11 2 )(nl + n2 -1)(n 1 + n2 - 2)

~--~~~--~~~--~X

x [(III-n2)un ,-2_(n l +n 2 -2)u n ,-I]du, for 0:::;; u:::;; 1, .md

for 1 :::;; u <

l-or ". and 112 ~ probability one.

00, E(u)

00.

~ 1 and var(u) ~ O. The limiting form is u

=

1 with

198

EXERCISES IN PROBABILITY AND STATISTICS

69

Rider, P. R. (1953), JASA, 48, 546. Rahman, N. A. (1964), JASA 59 S The distribution of w is n(n -1)wn- 2(1_ w) dw, (0 ~ w ~ 1). Put ~ ,=:' ,57 so that the distribution of u is proportional to e \, e-(n-l)u/n.(1-e- u/n)du =

e-(n-l)U/n.~[1-~+~-"']dU 2 n

2n

6n

'" e -(2n-l)u/2n . ~ du, n whence - (2n -1) log w '" X2 with 4 dJ. Also, from the exact distribution or~. E[(log w)'] =

(-l)'[n'+ 1_(n -1y+ 1 ]r(r+ 1) n'(n -1)' ,

Therefore

(r ~ 1).

1)

(2n-1)2 ( E[-(2n-1)logw] = n(n-1) ",4 1+ 4n 2 ' and var[-(2n-1)10gw]

=

(2n-1)2(2n2-2n+ 1) ( 3) n 2(n-1)2 '" 8 1+ 4n 2 .

Assume that - (2n - 1) log w '" PX 2 with v dJ. Then p and v are obtained by equating the first and second moments of the exact and approximalin~ distributions. Thus

p '"

(1+2~2)

and

v'" 4(1-

4~2)'

whence

Therefore -IOgZk'"

)-1 xt, L (2n l-1)-1 (1 1--2? n, k

i=1

where

xt is an independent X

2

with 4(1-

4~t) dJ.

But

1+4V(ni)] E[-(2i1-1)logzd '" 4k [1+ 4i12 ' var[ -(2i1 -1) log zd

'" 8k [ 1 +

3{ 1 + 4 v(n i )}] ' 4n 2

whence the stated approximation by using the first two moments. 70

Rider, P. R. (1955), JASA, 50, 1142. Rahman, N. A. (1964), JASA, 59, 55; If Xi = e - ytln, then 2Yi is a X2 with 2 d.f. Hence - 2n log Uk is a l with ~(

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

dJ.

2

199

'fherefore the distribution of Vk is

nk

r(k) . 1ft- I ( -log Vk)k-I dVb ~~

ee, for unequal samples.

(0 ~ Vk ~ 1).

k

L x1/ni>

-2logVk =

i= I

l

,\here xl is an independent with ~ dJ. The approximation follows by using rhe lechnique of the precedIng exercise. 71 Hyrenius, H. (1.953), J ASA, 48, 534. . (i) 0 ~ T ~ 1 If U2 ~ VI and T > 1 If U2 > VI' Therefore, for 0 ~ T ~ 1, the conditional distribution of T given UI is

(nl -1)n2 I . H~ I (n2 -1)(1-UIY( _ T)H2-,-1 dT x (l-ulf,+H2 ,=0 r I x

f

(z_u.)H,+H2-,-2dz.

II,

For T> 1, the limits of integration for z for fixed UI and Tare UI ~

Z

~

1 +uI(T-l) T '

whence the second part of the conditional distribution of Tis

(nl -1)n2

H~ I (n2 -1) (-1)' dT . ,=0

r

TH'(nl+r)

The conditional distribution of Tdoes not involve UI explicitly, hence this is the unconditional distribution. (ii) Use the transformation V = (V2 -U2)!(VI -UI), Y = U2 and z = VI' Note that (0 ~ V ~ 1) if (V2 -U2) ~ (VI -utl, and (1 < V < (0) if (V2 - U2) > (VI - UI)' For fixed z and V ~ 1, limits of yare I'I ~ Y ~ 1-(z-utJV; and for fixed V, (UI ~ Z ~ 1). Hence. for (0 ~ V ~ 1). the conditional distribution of V for fixed UI is

n2(n2- 1)(nl-l)

("1+ 112- 1)(III+n2- 2) '

[( nJ + n2- I)V"2- 2 -nl+n2( 2)VH2-I]dV .

For V > 1, the corresponding limits for y and z successively are UI

~

Y

~

1- V(Z-UI)

and

UI

~ Z ~

UI +(1-u l )fV.

whence the conditional distribution of V is

n2(n -1)(n -1) 2 I V-H'dV (n l +n2- 1)(n l+n 2 -2)' . As before. the conditional distribution is free of U \.

200

EXERCISES IN PROBABILITY AND STATISTICS

Use the polar transformation WI = R cos () and W2 = R sin 0 wh after appropriate integration over R, the distribution of V is' ent~

(O~ V~ 1); and

n2(n l -l). Vn 2- l dV,

n l +n2- 1 n2(n l -l). V-n'dV, n l +n2- 1

(1

~ V< 00).

72 UL (1963). The manufacturer's specification is satisfied if for

e- pm > (1 + A.) e- am ,

or

f3 < a

whence the inequality.

73 Irwin, J. O. (1941). Discussion following E. G. Chambers and G. U. Yul (1941). JRSS(S), 7, 89. . At time t, (lV x = Vx - vx + I, and at time t + (it the corresponding differcnl' is v x - I .f(t,x-l)(lt-vx .f(t,x)(it, so that at time t+(it , (lVx Tt =

f

vx-I' (t,x-l)-v x ·f(t,x),

whence the differential equation as (it (i) For f(t, x)

= kljJ(t), we have for

-+

O.

x

dvo

=0

dt = and since at t

=

0,

Vo

= N, therefore

~/ VI

kljJ(t) . vo, Vo

= N e- kT . Again, from

= kljJ(t)[N e- kT -v.J

= e- kT • NkT. In general, if -kT

vr_I=Ne

(kTy-1 . (r-l)!'

then

~r + kljJ(t) . Vr = kljJ(t). Vr - I gives

_ N _kT(kT)' e -,-. r.

Vr -

(ii) A similar method to (i) leads to the negative binomial.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 2

20]

Irwin, J. O. (1941). Discussion following E. G. Chambers and G. U. Yule JRSS(S), 7, 89. '19~)idently, E(xIA) = var(xIA) = A.. The cumulant-generating function of A is v _rlog(1-tlc), so that E(A) = ric and var(A) = rlc 2 • The marginal dis· jll~ " Irlbution of x IS 7~

x . AX c'. e A,-I dA C' (x + r-l) ! Je-XI' (r-l)! =(l+c)x+"x!(r-l)! forx~O.


-C). •

o

lIenCC G«(J) = c'/(l+c-(J)'

and

E(x) = ric, var(x) = r(l+c)/c 2 •

~rom the joint distribution of x and

f co

_

E(XA) -

co

x~o x

e -). . AX +I c'. e - c). • A' - I dA x! . (r-l) !

o c'

co

=

A,

(x+r)!

x~dl +c)x+,+ I' (x-I)! (r-l) ! =

2

r(r+ 1)/c .

lIence corr(x, A) = (1 +c)-t. For variation in A, E).[var(xIA)] = E(A) = ric. Finally, the conditional distribution of A given x is (1 +c)x+'

(x-r-l)"e

-).(1+.) 1x+,-1 d1 .11.

11.,

1:>-; 0 11.""

,

whence x+r E(Alx) = - , l+c

1S Liiders, R. (1934), B, 26, 108. (i) If in any time-interval Vk accidents occur, each of which involves j persons, then the probability of the configuration (VI' V2" .. Vk" .. ) is co

n e-

('lktk/Vk !,

Yk •

k= I

whence co

P(X

= r) =

L n e- Yk • (Yktklvk !, k=1

where the summation is over all non-negative integral values of such that co

L

k. Vk

= r.

k= I

Hence the probability-generating function of X is G(z)

=

,~o z' P(X = r) = exp [- k~1 Yk

(l-z k)]

= (1- p))./(1- pz)). for Yk = Apk/k.

~

202

EXERCISES IN PROBABILITY AND STATISTICS

(ii) Given .the con~guration of. accidents ~Vl' V2" •• ), .the probab'r generatmg function of the accidents sustamed by a given person :slly.

n (OPk + qk)Vk. 00

k=l

Hence the probability-generating function of Y is

L n e-Yk{Yk(OPk+qkWk/Vk!' v k= 1

the summation being over Vk

~

0 such that

00

L

k. Vk

k= 1

= exp

r,

with r -+

Ltl

YkPk(O-I)).

=

00

Campbell, N. R. (1941), JRSS(S), 7, 110.

76

00

C 1 = N(u+w)T/L,

where L =

f tf(t) dt. o

C2

N(u+w)G(T)+N(w+v).

=

To determine G(T), let Fm(T) be the probability of at least m failures al;, given post in a T interval, and F[m)(T) the probability of exactly m failure, at the post. Then and 00

G(T) =

L

00

mF[m)(T) =

L

Fm(T).

m=l

m=l

Also, for t < T,

f I

Fl (t)

=

f(t) dt

o

and T

Fm(T) =

f Fm-

1

(T-t)f(t) dt,

(m> 1).

o

(a) For f(t)

=

Ae-).I, G(T)

(b) For f(t)

= A2

=

AT,

C 2 -C 1 =N(w+v»0.

e-).I. t, G(T)

and

and

= AT -t(1-e-UT), 2

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

-1

2

203

C J11pbeIl, N. R. (1941), JRSS(S), 7,110. interval 0 < T < To, if first replacement is made in (t, t + dt), then In t (III -1) more replacements are made in (T - t), so that

:he

·I~as .

T

T

F!.:)(T) = f F!.:~l(T-t).f(t)dt, o

F\l)(T) = ff(t)dt.

and

0

. h interval (r- 1) To < T < rTo, (r > 1), clearly F~)(T) 1:11 e,( III divide the range into

i ,If r....

= 1 for all r > m.

'

(i) from 0 to T-(r-l) To < To; (ii) from T - (r - 1)To to To; and /iii) from To to T. In (i). if the first replacement is made in (t. t+dt). then since (r-l)To < I ./ < rTo. the contribution to F~)(T) is l' - (r - 1)1'0

J F:~)_

1

(T - t) .f(t) dt.

o

\Imilarly, in (ii) since (r- 2)To < T - t < (r - 1)10, the contribution to F~)(T) is To

f

F~= P(T -

t) .f(t) dt.

T-(r-l)To

I rom (iii). the contribution is evidently zero. 00

L

Gr(T) =

00

m[F~)(T)-F~)+l(T)]

m=l

L

=

F~)(T),

m=l

,lnd the probability of a replacement being made at the post during the Inlerval (t, t+dt) irrespective of when the bulb was put in is

d~~(t). dt. 78 Campbell, N. R. (1941), JRSS(S), 7, 110. tr Fm(T) is the probability that at least m replacements take place at a post In (0. T). then

f f T

Fm(T) =

Fm_l(T-t).f(t)dt,

for m > 1,

and

o T

F 1(T)

= f(t) dt.

o Then the expected number of replacements at the post is

G(T) =

f Fm(T),

m=l

and

dGd(t). dt t

For the second part, there are two possibilities: (i) a bulb replaced at t = T; and (ii) a bulb not replaced at t = T.

=

g(t) dt.

204

EXERCISES IN PROBABI LITY AND STATISTIC'S

For (i), consider a bulb replaced in (x, x + dx), where x < X is th . to t = T. The probability that this replacement is not replaced at t : illlt g(T-x)S\(x) dx. so that I'

f .Y.

P

=

g(T-X)SI(X) dx.

o

Hence the expected lifetime of a bulb replaced at t = T is > t is (I-P)SI(t).

Also for (ii), the probability of a replacement bulb not replaced at having a lifetime > t is x

f

t ::: .,

g(T-X)SI(X)SI(t+x)dx.

o

79

Campbell, N. R. (1941), JRSS(S), 7, 110. where PI = p,

P [1-(p' _p)n] Pn -- (l-p'+p) ,

whence 1

--:----:-:----:-;0 ,..,

1-(p' _ p)2

1.

The integral expression for p is obtained as for P in the preceding exercise. The expression for pp' is obtained by considering two consecutive replace. ments at Xl and X2' where (T-X < Xl < T) and (2T-X < X2 < 2T). Since no change takes place at t = T, the probability for given Xl and X2 is g(T-XI) dXI . S(xd. g(T-x l -X2) dX2' S(x 2 )·

For f(t) = .1 2 e-).I. t. p

and pp'

= (l-e-).X)- ~ . e-).x[1 + e-l),(T-X)],

= [1-e-).X (1 + ~)] [1-e-).X (1 +

_A.:

e-).(2T-X) •

A.:) _A.:

e-).(2T-X)

(l-e- 2ATl]

[1- e- 3).X (1 +lA.X)].

80 The joint distribution of Uj is

n f(uj) dUj, n

n!

j=

I

and that of Zj is n

n!

n dZ

h

1= I

where

f U;

Z,

=

f(x) dx,

-00

and 0 <

Zl

<

Z2

< ... <

Zn

< 1.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER I,

\Ia~e

2

205

the transformation and

PI = ZI

Pj =

for 2 ~j ~ n.

Zj-Zj_1

Ihe distribution of Pi is

n dp II

11

!

;,

i= I

(0 < \~xt,

PI

< 1; 0 < P2 < 1- PI;' .. ; 0 < Pn < 1- PI - P2 - ... - Pn-l)'

put

II

L: Pj

PII+ I = 1 -

j=l

.,1 th at the joint distribution of mi' Pi is m! n! 11+1 n+1

n

n Pi' n d Pj'

. ,i=1

II

IIIi

j=1

l1l i .

i= I

IIcnee successive integration over p", P,,_I

, ...

,P2' PI gives the result.

HI UL (1964).

The proportionality factor is rx/2 obtained from III

k

[f -

<XI

e-a(III-x) dx +

f e- (x-III) dX] = a

C

E[IX -cll

=

~ [f

1.

m

00

III

(c -x) e-a(m-x) dx +

-~

f (x- c) e-a(m-x) dx+ C

f 0()

+

(x-c)e-a(X-m)dX}

III

whence the result. H2 UL (1964). b

E[IX-bll

=

0()

ex [f(b-X)e-axdx+ o

J

(x-b)e-axdx

b

whence stated result. E(X -a)2 is obtained by using the result E(xr)

=

nr+ l)/ex r, for r ~ 1.

HJ UL (1964).

s ~ E(S)

if (I-ex) e- px2 ~ (l-ex)/(l +2P)1-,

'c. ir 2

1

x ~ 2,R log(1 + 2/J),

l

or x ~ (1- P/2).

206

EXERCISES IN PROBABILITY AND STATISTICS

Hence

~

P[S

~ (1- /312)].

E(S)] '" P[x

84 E(y) = pnl2; E(y2) = 8p2/3, whence E(y) and var(y). If x. is the of the sample x values, then the distribution of x. is sl1lallt (0 ~

n(l-xs)"-I dx s ' and

Ym

Xs

~ 1)

2p(1-x;)l.

=

Hence

nl2

f cos 2 () . (1- sin ())"- I d()

E(ym) = 2pn

o

=

n~.,l _ 1 r r(n + 1) r[(r + 1)/2]

Pr~o(

)'r(n-r)'

r(i)

r(r+l) 'r[(r+4)/2],

and the result follows by using the duplication formula

(r~ 1)

2f r

re': 2)

=

.fir: .r(r+ 1).

I

( 2

_

EYm)-4np

2

r _ J(1 x )(1 2

s

_

n- 1

x s)

d

_ n(n + 3) 2 x'-(n+l)(n+2).4P .

o

For large

11.

4p2

var(Ym) '" Jt(n + 2) . var(xs), where var(x s ) is obtained by noting that E(x~) =

85

r(n+l) r(r+ 1)/r(n+r+l), for r

~

1.

UL (1964). (i) PI = P(2p~ ~ kp)

(ii) P2

=

P(2p sin ()

(iii) P3 = P(I

~

kp) =

= (l-k2/4)t; E(I)

1-~ sin - I k12; n

= pnl2.

E(I)

=

4pln.

~ kp) = [1-~sin-1 kl 2]X x

[1+(~sin-IkI2) (1-~sin-IkI2)]; E(I) = 48pln 3.

For 0

~ k ~ 2, ~sin-I kl2 ~ n

1

so that P3

~ P2'

Bartlett, M. S. (1936), PRS, 154, 124. The Jacobian of the transformation u = X IX2, V = (l-xtl(1-X2) is IXI -x21, where (XI -X2)2 = (l-u+v)2-4v ~ O. Combining the contribution~ for XI-X 2 ~ 0 and X2 -XI ~ 0, the joint distribution of u and v is

86

2[(I-u+v)2-4vrt du dv.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

°: : ;

2

207

. ed v in (0 ::::; v ::::; I), u ::::; (1- Jv)2, and so the marginal distribuv is -log v dv: (0. ::::; ~ ::::; 1 ) . . . In of onditional dlstn butlon or II given r IS

,II I1X

The C

2[(I-u+vf-4vrt(-)ogv)-1 du,

0::::; u::::; (1_Jz.)2.

2 c1ly• 1/ ::::: e-(Ox : , w~erle) X2 has 4 dJ., and so the marginal distribution of , -log 1/ du, . '" u "'. . ..... . I'The distributIOn of v IS the same, smce If x IS umform 10 (0,1) so IS (I-x)

I rl

10,1). \1

The sample mean of the (n + 1) observations is (nx + ks)/(n + 1). Hence S2

1 n

= -[(n-l)s2+nx2+k 2s2-(nx+ks)2/(n+ 1)],

,h~nce result.

(32

=

3 -kW + 3) e- k2 / 2/J(k) [l-k e k 2 /2/J(k)j2 ,

Iknee P2 < 3, if J(k) >

3ke- k2 / 2 3-k2

l(k) > 0.

ANSWERS AND HINTS ON SOLUTIONS

Chapter 3

---

Ifr and r2 are the number of ones in the first and second n/2 observations,

~hen /== r1 +r2' and rl' r2 are independent binomial variables. Plackett, R. L. (1953), Bs, 9, 485.

for r ~ 1, whence result.

£(n r ) = NAre-A/(l-e-A)r!

Use factorial moments of a binomial distribution. (i) Unbiased estimate of pq is Q* == r(n - r)/n(n - 1), and var(Q*) = E(Q*2)_p2q2

=

pq[(1-4pq)+2pq/(n-1)]. n

(ii) Unbiased estimate of pq2 is r(n-I')(n-r-l)/n(n-l)(n-2).

(iii) p*(I_p*)2

2r2 n n r

=

r3

---+-, 3 2

£[p*(l- p*)2]

n

= pq2 .

and

(n-l)(n-2) (/1-1) 2 + pq . - 2 - -+ pq2, n II

as n

~

-+ 00.

For finite n, the probability for a run of r heads is (l_p)pr/(l_pn+I). rherefore n

E(r)

=

L t(l- p)p'/(l- pn+ I) '=0

=

p[I-(n+l)pn+npn+l] (l-p)(l-pn+l) -+p/(l-p), as n --+

00.

Similarly £(r2), whence

() _ p(1+p-2pn)+pn+I(l-p)[n2p-(2n-l)] varr (l_p)2(1_pn+l) -+p/(l_p)2, asn-+oo. Also, £(r2)

-+

p(l +p)/(l_p)2, as n --+

00.

lf u = (l.r+ f3r2 is an unbiased estimate of p/(l- p)2, then

(l.p I-p Hence (I. = {3 =

+

{3p(1 + p) P (1_p)2 == (1_p)2'

-i, so that u = r(r + 1)/2. 209

E2(r)

210

EXERCISES IN PROBABILITY AND STATISTICS

5

UL (1961). Let Xi be the number of successes in the ith turn; then random variable. and

Xi

is an indepe d n eOI

But 00

=

E(Xj)

L

vqpV = p/q

and

E(xf)

= p(p+ 1)/q2

(q=l-p).

v=o

Hence E(Sn) and var(Sn). Assuming that rxS n+ f3S; is an unbiased estimate r var(Sn), then rx = n/(n+ 1) and f3 = 1/(n+ 1). 0, Since E(Sn) = np/q, Sn = np*(I- p*). 6

Write _r

n-r

= ~ = e[I+(p*-P)] I-p*

q

p

[1- (p*_p)]-l q

= e [1+(P*-P\! q pq P

f (P*-P)'q-']'

r=2

whence E(_r_). n-r Again. r )

(-

2

n-r

1 L (t-l+2p)(P*-p)'q-'. =P22 [ 1+ 2(P* -p ) +2 q pq p ,=2 C()

]

Hence var(_r_). n-r

E(n_~+I) = JJn_~+I)(;)prqn-r = p(l_pn)/q. Alternatively,

Hence the second expression for expectation. Finally, (

)2

= p2 r n-r+ 1..1.2

[1 + 2(P+..1.)(P*-p) + p..1.

(* )'

1 + p2'~2 P ;p {(P+..1.)[(t+l)p+(t-l)..1.]) C()

which leads to the variance. 7

Feller (1952), Ex (e), p. 37, and UL (1962).

]

ANSWERS ASD HINTS ON SOLl'T10NS: CHAPTER

3

211

arrangement of the binomial coefficients.

,n re

P(N, m)/P(N

-1, m) ~ (1- ~) / (1- ~=';A-

lod this is > or < 1 according as 11/ N < or > (n - m)/(N - M). Hence estimate

"fN. The logarithm of the sample likelihood is n n 1" log L = - "2 log a 2 - i~l log Xi - 2a 2 i~l (log Xi - m)2 + constant, "hence estimates by differentiation with respect to ~

The distribution of X is 0- 2 xe- x /8 dx,

111

and a 2 •

(0:::; X < (0).

I herefore

E(xr) = or

Je-X/8(~)r+

1

= ()' [,(1'+2), so that

d(x/O)

o

E(X) = 20; E(X2)

=

60 2 ; var(X) = 20.

The maximum-likelihood estimate of 0 is 0* = x12. Therefore var( 0*) = var(x) = 02/2n.

*

i'inally,

f Xf]

1 .. E[-3 n ,= 1

20 2 , and

=

E(x 2) = var(x)+E2(x) whence

=

20 2 n

-+ 40 2,

E(~X)2 = 202( 1 + 21n).

10 The probability density function of X is (IX +

1)(1X+2)x a (1-x)

so that 1X>-1.

rhe roots of the maximum-likelihood equation are

* _ (3g+2) ± (g2+4)t IX

~ow -

00

< g :::; 0, and

- 2g

-

IX* -+ -

1 as g -+

-

00.

.

Hence the positive root for

It.

II

whence E(X)

=

eJ.l+O'2/2;

var(X)

Equate the mean and variance with

=

x and

e 2J.1+O'2(e U2 _1). S2 to obtain two simultaneous

212

EXERCISES IN PROBABILITY AND STATISTIC'S

equations which give the moment estimates

11*

X4

! log [ S2'+ xi

=

]

The median of the distribution is e,l < E(X). UL (1962). Since the distribution function must be I for X = n/2, the proporti . factor is (I - e - IX)- I, and the probability density function of X is on~llI. 12

(0 ~ X ~ n/2).

The maximum-likelihood equation is eIX * - I 1 I 1X*-eIX*-1 = T, or IX* = I+T(eIX*-l)'

1 T=-;;

/I

i~lsinxi'

13 By Taylor's Theorem, g(X) - g(m)+(X -m)g'(m), neglecting high •. derivatives, so that E[g(X)] - g(m), and

var[g(X)] - E[g(X) - g(m)f - [g'(mW var(X). The generalization follows by considering an expansion of g(X 1, X 2, .. '. X ' about Xi = mi' for i = 1,2, ... , k. If Xi are cell frequencies with a multinomial distribution, then I

E(X i )

=

mi;

Hence the result. 14

UL (1961). The equation for ()* is nl n2 +n3 n4 2+()* - 1-()* + ()*

= O.

Var(()*) is obtained from the second derivative of the logarithm of the sampk likelihood. Use the results of the preceding exercise to prove that E(O) = ()

var(O) = (l +6()-4()2)/4N.

and

UL (1963). In the first case, P(X = 0) = e- m; P(X Hence the sample likelihood is

15

L =

~

1)

= (1-e- m).

(~)(e-m)n(l_e-rn)N-n,

whence m*

= 10g(N/I1) and var(m*) = (l-e- rn )/N e- rn .

In the second case, P(X

=

0)

= e- rn ;

P(X

=

I)

=

m e- m ;

P(X ~ 2)

= 1-(1 +m)e-

lil

Hence the sample likelihood is L

= no!

111

N ! . (e -1II)no (m e -1II)n, [1- (1 + m) e -lIIt- no -.,. ! (N -no-l1d!

•

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

var ( 111

**)

=

3

m[l- (1 + 11l) e-"'] Ne "'(1-I1l+m 2 -e Ill)

m(e'" -1-11l) =var(I11*).( e '" - 1)(1 -11l+m-e 2 Ill)'

,nee the result on retaining the first power of 111. r* = (nd N )1.

el I):::: Nr2;

I~ ,,1\

2 (.(11\+113)= N(l-p);

1 +11 3')1 -----r;;.

p* = 1- (11

. approximate estimates do not add up to unity. Use Ex. 13 above to prove that

InN

var(r*) =

l-r2 4N;

p(2-p) var(p*) =~;

* q(2-q) var(q) = ~.

17 The maximum-likelihood equation for j) is

~-~-~+~=O 2+/3 2-/3 1-/3 1+3/3

Ihen

p:::: (1-0)2

so that var(O)

=

.

var(j)/4p.

IrQ = ala4/a2a3' then

4

;~1

=

Iltll var[log Q]

64(5+2p-4p2) l/E(a;) = 3N(4- p2)(1_ p)(l + 3p)'

= var(Q)/[E(Q)j2, and var(Q) =

[dd~] 2 p

•

var(p*),

p.=p

'Ihcnce var(p*) = var(j) by appropriate substitution. 18 UL (1963).

The sample likelihood from the two experiments is L=

(~J {l(4-0)}"' {i(4+9)jN,-n. x x

(N1132) {6~ (12 + 9)(4 - O)}"3 {614(4 + 9)2}NZ-II,.

213

214

EXERCISES IN PROBABILITY AND STATISTICS

e*

whence the equation for

is

n,+n 3 N,+2N 2 -n,- 2n3 n3 - 4-0* + 4+0* + 12+0* = O. For large samples, to test the hypothesis R(O

= ( 0)

use

. . 0*-0 0 S.E. (0*) - a umt normal vanable.

19

Pollard, H. S. (1934), AMS, 5, 227. The probability density function of X is

f(x)

=

1 . (1 +A)fo

[~.e-X2/2ai+~.e-(X-p)2/2(1~], (11

(-00 < X <

(12

'1.,

By direct integration, •

/,/1

E(X) = 1+..1.;

2+'2'2 (11 /,(12 + 1l/1

E(x 2 ) =

1+..1.

so that

The median is determined from the equation Iii

f

=

1-

if (12 ~

(11

f(x)dx

-7:

It may be shown that

dE(X)

dm

~ > d/1'

Also, for /1 E(X) > m.

0, E(X)

=

~

var(x)

=

=

m=

O. Hence

~ (11

(12

is a sufficient condition

[Of

27(1 + ..1.)2 [1 A ]2' 4N ~. exp{ -m 2 /2(1i} +~. exp{ -(m- Jl)2/2(1~} (11

Therefore, for Jl

and /1 > O.

(12

= 0, the condition var(x)

=

var(x) reduces to

(1+AP2)(P+).V-~(1+A)3p2 This is a quartic in P, say g(p)

=

=

O.

O. Then g(O) = . 1. 2 > 0, g(oo)

-+ 00

and

g(1) = (1 +,;l(1-n/2) < O.

Therefore the quartic has two positive roots PI and P2' satisfying the statcJ condition for var(x) and var(x). 20

UL (1963). The distribution of Xr is

B

1

rl", /1-1"+

1 )

.00-nx~-1(0(-xr)n-rdx"

(0 :::;;

Xr :::;;

0:),

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

J

3

215

E[(Xr)"] = r(n+ 1)r(k+r) 0( r(r)r(n+k+ 1)'

the mean and variance of X r • = 0("/(k+1), for k ~ 1. This gives Also, E(x) = 0(/2 and var(x) = 0(2/12n.

,·~nce E(Xk)

,Ir n

"" 2v+ 1. the median is x v + 1 so that E(Xv +l) = 0(/2 and

var(xv+Il = 0(2/4(n+2).

, Since families with no abnormal children are excluded, the total proba,I I'S (1-t/'). Therefore . lilY

E(r,) ==

i

ri (k.) pr. q"-r'/(l_q") r,

r.=1

= kp/(1-t/'),

for i

= 1,2, ... , n.

lherefore for the average of the rio E(r) = kp/(l-q"), whence the moment ;,Iimate of p is obtained from

_

kp* r = 1-(l-p*t' \Iso,

k

[k

"-1]

n var(r) = var(r) = 1 pq", 1-~ -q 1-q

var(r) = (ddr*) 2 • var(p*). p pO=p lZ Haldane, J. B. S. (1943), B, 33, 222.

-1) L (n) - (n _

E(n/r) =

00

r

n=r

1

r

prqn-r

= L 00

s=o

(r +s) prqs = p-l. s

Ilul

E(r/n) =

r~. q

q

I: (n =11) f t

n=r

r

q

= r(!!.)rftr- l . q

o

n- 1

dt

o

I: (r+S-1\ tSdt s J

s=o

q

=

r(~r f tr - l (1- t)-r dt o 1

= rp f ur- l [1-q(1-u)r l dll, o

whence the result by term-wise integration.

U

= pt/q(1-t),

216 23

EXERCISES IN PROBAbILITY AND STATISTICS

Finney, D. J. (1949), B, 36, 233. 00

E[g(a,

PH = pll+ 1 qP . coefficient of X'-Il-l =

in

L (q + PXY'-Il-P-t

n=,

p"qP. coefficient of X'-Il- 1 in (q + pXY-Il- P-l .

f

X,

1=0

whence E[g(1,0)]

=

p

and

E[g(2, 0))

=

p2.

Also,

1

= (r-1)p2

f 11- 2[1-q(1- u)r

du,

1

u

=

pt/q(1-t)

o

= p2

f tf/(r + ss - 1) .

• =0

Hence an unbiased estimate of var(p*) is p*2 _ g(2, 0) = p*2 (1- p*)/(r- 1- p*).

For the mode, consider P(n)/P(n-1) = (n-1)q/(n-r),

so for maximum probability r-1

r-1

- - 1 < n-1 < --. p

p

24 Use the transformation

y-m2 and - - = v. (]'2

Then

E[(X :~lr (Y~~2r]= -Q)

ff 00

= 2~

-

a)

00

exp{-!(u 2 +w2)}.u'[w(1-p2)t+pul'dudw,

-00-00

where w = (v-pU)(1_p2)-t, whence the result by term-wise integration.

ANSWERS AND HINTS ON SOLl'T10NS: (,HAPTER 3

Also. since

217

x ( u)(1+-V)-l

-=..1. 1+Y ml

m2

~r~fore

Hence V~k -

1 (2k) 2k . k !

OJ

E(Tl ) = . 1. [ 1 +(V2 - PV1) k~l

!] 1+

..1., as n --+

00.

Hul X and ji have a normal bivariate distribution with correlation P and

.fficients of variation

,(lI:

x]

E[Y

[1

vdJn, v2IJn respectively. Therefore

= . 1. 1 + In(V2 -

OJ (2k)! (V2 )2k-l] PV1)' k~l 2k. k! In

-+

..1., as n -+

00.

\\ a first approximation,

var[~]

mi 2 var(x)+mfm;:4 var(ji)-2mlm;:3 cov(x, y).

=

'11

2~ Let '2 and '3 be the number of insects killed at strengths x = -1,0,1 "spcctively so that' 1 +, 2+, 3 == ,. Hence the joint likelihood of the sample is

1.=

( ")[ '1

1'1[

1 1+e (a liJ X

]"-'1

e-(a-l) 1+e (a 1)

('3n)[ 1+e 1

(n)[ '2

X

]"[e-(a+l)]"-" 1+e (<<+1)

(<<+1)

logarithmic differentiation with respect to .Ihich involves only,.

(X

]'2[

1 1+e- a

]"-'2

e- a 1+e- a

•

gives the estimation equation

26 The estimate p* is obtained from the equation

_n_l__ n2+n3+ n4 = O. 2+p* 1-p* p* I or the bias in the estimate of var(p*) assume that E(p*) '" p. Then

E[2P*(1- P*)(2+ P*)] '" N(1 + 2p*)

var

(P*)

~

(*)~[P(1-P)(2+P)]

+ N var p dp2

(1 + 2p)

,tnd

~[P(1-P)(2+P)] dp2 .I'hence result.

(1+2p)

=

(5+3p+6p2+4p3) (1 +2p)3

218

EXERCISES IN PROBABILITY AND STATISTIC'S

27 Norris, N. (1938), AMS, 9, 214. For t > 0, E(X') = a'r(p+t)/r(p).

= E(X) = ap. Also, E(Xl/n) = a 1/ n r(p + l/n)/r(p),

Therefore for the sample mean X, E(x) and since the

Xj

are independent, therefore for the sample geometric m

ean ~,

E(gn)

= a[r(p+ l/n)/r(p)]n.

The population geometric mean is lim a[r(p+ l/n)/r(p)f = a e~(p)

n .... co

since lim [log r(p+t)-log r(P)] = q,(p). t The logarithm of the sample likelihood is ' .... 0

nx

log L = -np log a-n log r(p)--+n(p-l) log gn' a whence the results on differentiation. 28 UL (1960). Minimize with respect to a and f3

n ==

,tl [y,

-a cosC:r) - f3 sine:r )

r

The estimates a* and f3* are a* = !(YI-Y2-2Y3-Y4+Ys+2Y6)

f3* 29

=

1M(Yl + Y2 - Y4 - Ys). 2",3

I = 2p sin 0; E(I) = 4p/n; var(1) = 2p2(n 2- 8)/n 2. Minimize n

n ==

L [/ -4p/n]2 j

j=1

for variation in p to obtain the estimate p* = ni/4, where of the Ii' var(p

*)

=

(n 2- 8)p2

8n

i is the

.

30 UL (1961). The probability distribution of 0 is [120(n-0)]/n 3 • dO, for 0 ~ and Il = ir2 sin 20, whence E(Il). To estimate r, minimize n

n == L [ll -3r2(n 2 +4)/8n 3 ]2 j

i=1

avera~,

with respect to r.

(J ~ 11/:.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

.11 UL (1961).

Minimize

n

n ==

L [(Xj-ml)2+(yj-m2)2+(zj-ml -m2)2] j=

1

n

=

L j=

',10

[(Xj-X)2+(yj_y)2+(Zj-Z)2]+

1

bta in the estimates

mf

m!

(2x-y+z)j3;

=

= (-x+2y+z)j3.

Ihe residual sum of squares is n

~

[(Xj-X)2 +(Yj- y)2 + (Zj -z)2]+n(x+ y -z)2j3

with (3n- 2) dJ.

1= 1

On the hypothesis H(m. = m2)' the estimate of m. is

IIlf*

=

(x + y + 2z)j6,

'1llhat the sum of squares due to hypothesis is /I [(5- 2-)2 (5- - 2-)2 4(- - -)2] -3n(_ - -)2 , 36 x-yZ + y-X- Z + x+y-z x+y-z

",hence the result. ,\2 UL (1962).

Minimization of n

n == L j=

[(xj-md2+(yj-m2)2+(zj-m1 +m2)2]

1

n

L [(Xj-X)2+(Yj_y)2+(Zj-z)2]+

=

i= •

~i\'es

the estimates

m1

=

(2x+y+z)j3;

m!

=

(x+2y-z)j3.

Ihe residual sum of squares is

wilh (3/1- 2) d.f. The best estimate of m. -},m 2 is U=

Jnd var(u)

(2-A)x+O-2A)y+(1 +A)Z 3

= 2(A 2 - A+ 1)0"2 j3n, whence the t test.

219

220

EXERCISES IN PROBABILITY AND STATISTICS

33 UL (1962). The method same as that of Ex. 31 above. The estimates of (J (J are Or = (x- y)/2; O~ = (x+ y-2z)/6; Ot = (x+ y+z)/3; and the r~iJ and, of squares is Ual \," n

L [(X/-X)2+(y/- y)2+(Zi-Z)2] with 3(n-1)d.f. i= I

On the hypothesis H(OI = O2 = ( 3 ), the estimate of 0 1 (say) is

Or* = (3x + y - z)/l1, whence the sum of squares due to the hypothesis is n[Xf+X~+(XI-X2)2/9],

where XI == (3x+y+10z)/11; X 2 == (-3x+10Y-i-z)/11. Hence the stated result.

34

/I

T= (h+k) L xf+k LX/x), /~}

i= 1

when: E(xf) = 0'2+m2 and E(x/xj) = p0'2+m2. Evaluate hand k using E(T) == 0'2. 35 Due to D. R. Cox and quoted by Kendall and Stuart-II (1961), p, :: The variables y,/rt are independent and have variance 0'2. Hence II. least-squares estimate of 0 is 0*

= [

± y,]/±,=

,= 1

r- 2

r- 1 ,

I

and

36 Downton, H. F. Private communication. The estimates are:

where 4xI

= LXljk; 4X2 = LXj1k; 4X3 = LXjU' hk

hk

hk

Variances of the three estimates are all equal to 30'2/8, and each of the Ihr •. covariances is equal to - 0'2/8.

37 The solution of the normal equations gives for the estimates of 0, (r=/= s),

8f-8~ = z -Kx +y + z);

I'

8~-8t = y-i(x+y+z);

8t-8f = z-1(X+ y + z). The residual sum of squares 3(n-1) d.f.

is

the within-sample sum of squares \\;It

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

II' hen

,\

221

the hypothesis H(8 1 = 82 /a = 83 /b) is true, then E(xj)

;L~

3

= Aiel; E(yJ = A2 e l ; E(Zi) = A3el'

the appropriate sum of squares by minimization.

Minimize

n

I

n

L:

==

[Yr-C(- pr(x r -x)]2

r=1

'n respect to

C(

Pto obtain the estimates _ -(n+l) -) *= y -- . P*(_ Xw-X

and C(

2

II

P*

L:

r(x r - x)(Yr - ji)

r= I

=-n--~~---~~----

"2(

-)2

L... r Xr-X

-

n(n + -)2 4 1)2 . (Xw-X

r= I

.:.~rc.1:

and ji are the averages of Xr and y" and Xw

=

JI rx,jJI r.

tlll'residual sum of squares is

,lIh (11- 2) dJ.

Hence the t statistic for H(P

=

Po).

19 Minimize n

Q

== (C(2 + P2)-1.

L (/XXi + PYi + 1)2

with respect to C( and

i= 1

Ih~

estimates a and b satisfy ax+bji+ 1 = 0, and

lfI

The least-squares estimates of C( and pare

'"d the residual sum of squares is "

L: (Yi -

i =1

II

ji)2 -

P* 2 L: i= 1

(Xi - X)2

with (n - 2) dJ.

p.

222

EXERCISES IN PROBABILITY AND STATISTIC'S

On the hypothesis H(rx

- P()),

=

the relative minimum gives n

L (Xi -

f3* f3**

=

X)2 - n«() - x)y

_.:...i=-"I'-_ _ _ _ __ n

L (Xi -

X)2 + n«() - X)2

i= 1

Hence the sum of squares due to the hypothesis is

=

nf3*2

(~: +()Y itl (Xi-X)2![I1«()-X)2+ itl (Xi-W]

with 1 dJ., whence the required F statistic. 41

Bose, S. S. (1938), S, 3, 339. Sp

p(p+ 1)

P

=

L Yr-Prx--2-· f3;

r= 1

Sq

=

~

Ym-qrx-

'--

q(2n-q+ 1) f3 2 ..

m=n-q+ 1

Minimization of S; + S; gives

_ q(2n-q+1)l:Yr-p(p+1)l:Ym. , (2n-p-q)pq

Q-

b

=

2[pl:Ym-ql:Yr1. (21l-p-q)pq ,

(1

~

r ~ p; n-q+ 1 ~ m ~ n).

Hence the variances of Q and b. Compared with standard least-squares cst I mates the efficiencies of Q and bare

2(2n + 1)/n(n -1) eff(Q)=l[ p+1 1[ p+1 - 1+ +- ~-P 2n-p-q q 2n-p-q

J2

J2'

eff(b) = 3pq(2n- p - q)2. n(n 2 -1)(p+q)

For maximumeff(b), p

= q =

n13.

42 The variables Xr and Yrl.ji have the same variance. For the least-squar" solution minimize

with respect to m. This gives the estimate

m* = (Anlx+n2y)/(Anl +n2),

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

Jod

3

223

Ihe residual sum of squares is nl

"2

r=1

r=1

L (X r-x)2+-1.- 1 L (Yr-ji)2+nln2(X-ji)2/(-1.nl+n2) h( 1I I +1I2-1)dJ. Var(m*) 1/1/11 "" 1110)·

=

.11

-1.a 2/(-1.1I1 +11 2). Hence the t statistic for testing

"hence the inequality.

,Ind

var[bl(X2-XI)]

= E[br(X2- XI)2] = EA(X2 -Xl)2 E,(bilx)] = E[(X 2-X I )2{ a;(1- p2} +p2~; }], (n l -l)sx ax

where (III -1)s~ is the within-sample sum of squares for the n l X observations,

_ 22(nl+n2) "1n2

- p a,

+

(n l +n2)a;(1-p2) E [~(X2-Xl)2] ( ) . x +n2 • Sx2 ' nl n 2 n l -1

"I

( ~)(X2-XI)2 nl +n2

I, a /2

SX

statistic with (II 1- 1) d.f. Hence var(T).

44 Stein, C. (1945), AMS, 16, 243. For any given n, the linear function (ax+bji) is an unbiased estimate of II wilh estimated variance v 2 if a and b satisfy the equations

a+b=1;

a 2 b2 v2 N+-=2".

n

s

(1)

II . " IS chosen as stated, then with a = N /(N + II) and b = n/(N + n), a + b = 1 Ind a2/N +b 2/11 = l/(N +n). Clearly (1) have an intersection since 1/(N +n) ~

224

EXERCISES IN PROBA BILITY AND STATISTICS

that the ellipse a 2 /N +b 2 /11 = V 2 /S 2 is contained in a2 /N +b 2/ 1/(N + 11). II Distributionally, 11, a and b all depend upon S2, v1 and N, of which S2 is a random variable known to be distributed independently of bO~hnl) and y. For given S2, '

V 2 /S 2 , SO

0

a2

b2

N

11

w = (ax+bY-II) ( -+--

)-t

is a normal variable with zero mean and variance (J2. Hence its unconditio . distribution is the same. Therefore w/s has the t distribution with (N _I)~.;I Hence ~

var(ax+by) = v2 (N-l)/(N-3) 45

r

Use Stirling's approximation for the

v2 for large N.

function to prove that

If x and s are the sample mean and standard deviation, and ponding population parameters, then

1) - r (11)/ 1) 2 r (11-2-1) ~ (1- 411'

11 E(s/(J) = ( -2-

is E (~)

t

(S) (1 +;Z)-I ,where

= A -;;

=

AE (~) (J

X

f:

Z

=

In, (J

the corre\.

and

X-In, so that

In - 2r E(z2r)

r=O

Ar(i) A2r(2r)! (11; 1) tr(n; 1) .r~o 2 ! ~ ), (1- L) (1 + I;] A as n 00

nr

--+

r

r

--+ 00.

Similarly,

Hence var(s/x) ~ A2(l +2A 2)/2n. 46

von Neumann, 1. et al. (1941), AMS, 12,153; and UL (1962). Let Zj = Xj-In, for U = 1,2, ... , n). Then 1 n- 1 1 n-I d 2 = 2(n-l)' i~1 (Zi+I-Zj)~ = 2(11-1)' i~l (z;+I+ z;-2z j zj+ I).

whence E(d 2 )

= (J2.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 3

n-I

+4 i~1

zt zt+ 1+ odd power terms

225

]

.

Ihcrefo re

(311- 4)0"4 whence var(d 2 ) = ( I 2 11- ) 41 UL(1962).

(i) var(x)

=

11- 2

,.

var(xi)+ JI

j~i COV(X;,X

0"2 j )]

=

-[I +(I1-I)p] 11

II

L

(ii)

Ltl

(Xi- X)2 =

i= 1

I

(Xi-jl)2_ I1 (X-jlf,

i= 1

\\hcnce

E[J\ (X i -X)2]

=

11 var(Xi)-11var(x)

(iii) Inequality follows because varIx) and .j8 UL

=

(11-1)(I-p)0"2.

E[;t\ (Xi-X)2]
(1961).

var(L I - L 2 ) =

n

L

rx; var(x r)+

r=1

"

L L rxrrxs cov(x r, x s ), r=\r#<s

whence resul t. 49 UL (1961).

The condition for Y\ and Y2 to be uncorrelated is obtained from first principles by evaluating n

cov(Y1, Y2 ) =

0"2

L aibi' i= 1

\lote that 'Iatcd.

(Xi-X)

and

(Xj-x)

are particular linear functions, whence the results

~ UL (1962).

Let S =

s, +S2 so that S 2) = Mvar(S) -

cov(S I>

var(S I) - var(S 2)]

0"2

= 2"[I1{I+(11-1)p}-v l {I+(v l -l)p}-v 2{1+(v 2 -1)p}] = 0"2pV\ V2, whence corr(SI, S2)'

226

EXERCISES IN PROBABILITY AND STA TlSTlCS

The limiting value of the correlation as

VI ---+ 00

is

51 UL (1963). Sample mean of X values Sample mean of Y values

= =

(n2+ n4 -nl -n3)(X/N == A, 1(X. (n3+ n4 -nl -n 2)f3/N == A, 213. (1l 1-n Z-1l3+ n4)/N.

A,3 =

The correlation is obtained by evaluation of the two sums of squares and Ih sum of products. For zero correlation, t

~= and if this is satisfied as N P(X

52

=

-(X,

---+

Y

(nl ~n2) (nl ~n3),

'X),

then

= -13) =

P(X

=

-(X)P(Y

= -{i).

Pearson, K. (1924), B, 16, 172. P(X = r) =

(~)(~l-:) (~)

and P(Y= siX

= r) =

(M:r)(N-~=;I+r)/(N~n}

Hence the result for P(X = r, Y = s) is obtained by recombining the binomial coefficients. The limits are 0::::; r ::::; min (nl' M); 0::::; s ::::; min (n2' M -r) and 0::::; r+s ::::; min (n l +n 2, M). For X = r the conditional distribution of Y is hypergeometric, so Ihal E(YIX

Similarly,

= r) = n2 (M -r)/(N -nIl.

E(XIY = s)

=

nl (M -s)/(N -n2)'

The linear regression coefficients of Y on X and of X on Yare -n2/(N -11 11 and -nl/(N -n2)' whence corr(X, Y).

whence result.

53 Greenwood, J. A. (1938), AMS, 9, 56. Consider any two cards of the B pack. There are two cases. I.. Cards ali of same type. II. Cards are of different types. I and II can occur III

(~)

-(XI

== (X2 ways, respectively.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

227

t- 1 NN-a-I j(N) -1 . IX, 2 t-l. - -N-l P(X = 0, Y= 1) = . I X , )(N) 2 1 N-a P(X = 1, Y= 0) = -.--.IX, N-l j(N)2 1 a-I P(X= I,Y= 1)=-.--.IX, N-l j(N)2 .

rhUS for I,

P(X = 0 Y = 0) = ,

t

.

a

t

t

t

for II,

'

P(X = 1, Y = 1) =

1 = ('

P(X

=

P(X

= 0, Y= 1) =

1, Y= 0)

~. N ~ 1 . 1X2j(~)

P(X=O, Y=O)=

N-a-l N-l .1X2j(N)2

[~. (~~II)+e~2)(N~I)] .1X

2

/G)

[~·(Z=~)+(t~2)(N;~~I)] .1X2/(~)'

Ihe last two probabilities are obtained by considering separately the possibilithat the first card of the A pack mayor may not belong to the type of Ihe second card of the B pack. Hence the joint distribution of X and Y. Finally, let z" Z2,'" , ZN be random variables each associated with one "f the cards of the B pack such that Ii~S

1 if a match occurs at the kth position;

Zk

=

Zk

= 0 if no match occurs at the kth position.

Ihen N

SN =

L Zk'

Zk

has the marginal distribution of X or Y,

k=l

md for i #- j, corr(Zj, Zj) = corr(X, Y). ~ Put X -m, = wand

Y-m2

=

z. Then

(1 +~) (I---=-+ Z22_"') m1 m2 m2 z) (WZ WZ2)] w ",..1. [1+ ( m 1- m2 - ml m2 - m~- mlm~ .

X = w+ml = . 1. Y z+m2

Z2

(1)

Psing the first approximation X/y", ..1.[1 + (w/ml -z/m2)] gives E(X/Y) '" . 1. Ind the approximation for var(X/Y). The approximation for the bias in E(X/Y) is obtained by considering the l1~xt terms in the expansion (1) for (X/Y) and noting that E(WZ2) = 0 for a 'Ymmetrical distribution. Hence E(X/Y)-}, '" ..1.ViV2-PVl)'

228

EXERCISES IN PROBABILITY AND STATISTICS

Pearson, K. (1931), B, 23, 1.

55

Since E(t) = 0, cov(,x, t)

=

E[(x-m)t]

E[(x-m)2fi/s]

=

(11-1)1 J~'r(I1;I)" -2- , by direct integration.

=

r(I1-2) -2

a

But var(t) = (11-1)/(11-3), and varIx) = a 2 /11, whence corr(x, t). With the transformation X

56

= a+bU

r = it! (Xi-X)(Yi-

=

J!

and Y

= a+{W,

jiy[ it! (Xi-X)2. it! (Yi- ji)2 r

(Ui-U)(Vi-Vlj[it! (Ui- U)2. it! (Vi-W]},

III

where (Xi-X) = b(Ui-U); (Yi-ji) = P(Vi-V). Hence the invariance ofr. Put a = m x , b = ax, a = Illy, P = a y , then r is as in (1), and the joint di,. tribution of U and V becomes

1

21t(1_p2)!'ex p

[1

2

2

]

2(1_p2)'(U +V -2puv) dudv,

(-00 < U,V
This only involves p as a parameter, and so the distribution of r can oull involve p. . Next, put z = (u - pv)/(1 - p2yr,

W

=

v,

then U2+V2_2puv = (l_p2)(z2+w 2).

Hence Z and Ware unitary orthogonal random variables. Also, L(Ui-U)2 = (1- p2)L(Zi-zf+ p2L(Wi-iW+2p(1- p2)! L(Zi- Z)(lI'j-lll. L(V i -V)2 = L(Wi-iW;

and L(Ui - U)(Vi - v) = pL(Wi -

wf + (1- p2)! L(Zi -

Z)(Wi - w).

Then so that r

=

[p+RJ(1-p2)!]/[(1-p2)p+p2+2p(1-p2)!Rf]t,

whence the result. Haldane, J. B. S. (1943), B, 33,56. If E(X) = IIIx; E(Y) = my; varIX) = var(Y) = then for the standar.! 4 ized variables X = (X -mx)/ax , Y = (Y -my)/a y, E(x ) = E(y4) = 3; E(x-'yl E(xi) = 3p; E(X2y2) = (1+2p2); and E(X 4y 2) = E(X 2y 4) = 3(1+4p2).

57

a;;

a;,

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

229

These may be deduced from the bivariate moment-generating function l(tf + t~ + 2ptl t2)· tlP"2

E(V) and E(V2) are evaluated directly, whence, using the identity

= E(V2),

(1 +v2)E2(V)

v2 '" (,P +4p2 + 2ex 2 )-2(ex4 + 3p4 +ex 2 {F +ex 2 A2 _ 2A 2P2), ,here ri := 2APP· ~ The results for Y/ X are obtained by expressing Y m <Xl - = -2.. L (- A)' (x' + Pyx'), X my ,=0 JPd

remembering that (2r) ! E(x 2 ,) = - 2' r !

and E(x 2 ,- 1 y)

(2r) !

= p __ .

2' r !

lienee ~

The limit of the finite difference relation gives 1 dy y. dx

2(x-a) bx+c'

where (/:= (n+ l)p-t; b := 1-2p > 0; c := np+t. Ihererore y=

1 -(x+m)/P ( + )a pa.+1 r(ex+l)·e . x m,

(

~ X < -/11....,

)

00,

where {J:= (1-2p)/2 > 0;

ex:=

4(11 + I)p(l- p) (I_2p)2

211p + 1

> 0; /11:= 2(1-2p) > 0,

'!llhat

2(x+m)

fJ \1

2{r(1-2p)+(I1+ l)p} (I_2p)2

hence the result.

~9

[t/>(aW '"

2~

nl2 2ul,!;

ff o

e- r212 .rd,.dO,

0

,Ihieh gives P61ya's approximation. For a standardized normal variable X, P(X ~ x)

= !+t/>(x),

md the distribution of the sample median z is

(21'+1)! (1.!)2

1 [I {J.( )12]\. -=212 d ·fo· 4- 'l'ZJ e . Z.

Iknee the result by using P6lya's approximation.

230

EXERCISES IN PROBABILITY AND STATISTICS

60 Cadwell, J. H. (1952), B, 39, 207. Using the Cadwell approximation, the distribution of the median proportional to [ ] e-(4Vh)x 2/21f. 1 + 2(~:23)V . X4 dx,

. X 1\

whence the distribution of y on normalization. The moments of yare obtai

~n~qilid

f 00

_1_

.j2ic .

2r

y. e

~

d _ (2r)! y - 2r . r !.

-y2/2

-00

61 A particular case of a general result due to M. S. Bartlett. The distributi ~t~ ~ 1 dt (-00 < t < (0). JnB(n/2,!) . (1 + t 2/n)(n+ 1)/2'

Hence the distribution of w = (1 + t 2/n)-1 is

1 ~-2U2 -t B(n/2,!), w (l-w) dw,

(0 ~ w ~ 1),

and the distribution of y = - n log w is proportional to e->'/2. y-t [1-(y/2n)+O(n- 2)]dy. The result follows by approximating the quantity in the square brackets a\ exp( - y/211). This is K. Pearson's well-known PA test. Each Pi is an independcnI variable with a uniform distribution in (0, 1).

62

63 64

UL (1962). Obtained by using the general result of the next exercise. The joint distribution of x I and Xn is Xn

(1

!)2~n!-2) ! ·f(x.) dXI [f f(x)dx x,

(x I ~ Xn ~

13; ex

r-

~ xI ~

2 f(xn)dxn,

13)·

For fixed XI' make the transformation R = Xn-X I, and then for fixed R integrate for x I in (ex ~ x I ~ 13 - R). If X is uniform in (0, 1), then the distribution of R is

B(n

~ 1, 2) . R

n-

2(1- R) dR,

65 The joint distribution of XI and

X2

(0

~

R

~

1).

is

(2v+2)!. (XI)" (1- X2)v. dXI dx 2, (V!)2 2a 2a (2a)2 and the joint distribution of y and z is

_-:::---7"~4( v + 1)

(2a)2l'+2B(v+l,v+l)'

(y _ z)V (2a _ _ z)V d dz y y.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

231

obtain the distribution of y integrate over z, the range being ro (i) (0 ~ z ~ y) if (0 ~ y ~ a); (ii) [0 ~ z ~ (2a- y)] if (a ~ y ~ 2a). rhus for (0 ~ Y ~ a), the probability density function of y is y

4( v + 1) (2a)2.+2B(v+1,v+1)'

f [(a-z)

2

2 •

-(a-y)] dz,

o

Ivhence the result by expansion and term-wise integration. Similarly for the range (a ~ y ~ 2a). Direct integration gives P(O

±(~

~ y ~ a) =

22 .+ I B( 1 1 1)' v+ , v+ r=O v

1

-

- 22'+ l B(v+1,v+1)

. •

I~O

(V)

t (

)(-l).-r/(2V-2r+1)

r

-11 JI )

Z21

dz

o

1

= 22>+

1

1 B(V +

f

1, v+ 1)' (1-Z2)" dz o

B(v+1,!)

1

= 2 2.+ 2 B(v+ 1, v+ 1)

= 2,

using the duplication formula Jnr(2v+2) = 22 .+ 1 r(v+ 1)r(v+!).

xJn

66 Make the transformation = r sin (J and sJn=l = r cos (J to obtain Ihe joint distribution of rand (J as e -nm'/2a 2 - - - - - - - 1 - . cosn- 2 ede. e-r2/2a2. rn - 1 x 2(n-2)/2Jn qnr(n;

xexp

)

(mJn ..sm e) . dr, ~.I

Expansion of the last exponential and term-wise integration over r then gives Ihe distribution of e as

~"

e- n/ 2V2

- - - - . L:.

/it

r(II;1)

j)

(2n)i,2r(n + 2

vJr(j+ 1)

j=O

.cos

Rut v = [n/(n - 1)] cot2e and cot( - e) distribution of e for (0 ~ e ~ n/2) is

2 e- n/ 2v2

In r (

00

L:

-1 ..

11:2

)

J=O

(2njV2)i r(2'

1)'

n-

=

2e sini e de,

-

cot e, so that by pooling, the

r (n+2j) -2- . cosn- e sin

'.}+

lienee the () ...... v transformation gives the result.

2

.

2J

O dO.

232

EXERCISES IN PROBABILITY AND STATISTICS

Jv,

To obtain the distribution of w = consider the distribution separately over the ranges (-n/2 ~ () ~ 0) and (0 ~ () ~ n/2). Hence byOfhO transformation Ic

w=

C: r. 1

cot (),

the distribution of w is

e-n/2V2

co

2 "2r(n+i) (2nIV)' -2-

'L

In r ( n~ 1)

(n-l)t dw

(n-1)w 2

r(j+ 1)

j=O

}(j+2 l/ 2

n

{

{ I+

n

}(n+ jl/ 2

n (n -1)w 2

for 0 ~ w <

\1;,

and

f

Jnr(ll; I)

{ n } (j+2l/2 (1l-1)t (n-l)w 2 -n- dw

n+i) (-IY(2njV2yI2r (-2-

r(j+l)

'j=o

.

{I +

n for -

67

}(n+J1/2

(1l-l)w 2 00

< w ~ O.

Hendricks, W. A. (1936). AMS. 7,210. The joint distribution of x and s is

n~2 [V(V; 1)

r.

exp {-

:2

[(v + l)(x - m)2/2 + v(s - O'C.)2] } dx ds, ( - 00

< s, X <

(0)

so that the joint distribution of t and s is

1 (I'.2)t . exp {I - 0'2 [12t 2s 2+ v(s - O'c.) 2} ] . II s dt ds.

n0'2'

Hence integration with respect to s gives the distribution of t as

(I')t

1 .2 . exp{ ;.

vt2c~/(t2 + 2v)} x 2 ,·e./(t 2 + 2,·)1,2

X [

For fixed

1'.

4vc,. (t2+2v)"-

3'

f e- w2/2 . dw+_2_. exp{ -2v2C~/(t2+21')}] dr. t 2 +2v '

o

the integral is always <~. and the second term in the brackcl'

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

233

. ...0 and of O( V-I e - V). Therefore for large v, the distribution is asymplotically

IS"

2 2 21' ] t . dt, M-. exp{ -2 vt Cy/(t + 2v) } . [ --Y--2

Cy

v 2n

t

+

v

,hence the distribution of v. " By direct integration of the exact distribution of t £Ill'} ==

V-2r) (21"+ 1)/ (V) (1) 1"r( -2r -2- r 2 r 2 '

(r

~ 1)

] t2 3t4 3) v = tc [ 1--+--+0(vy 4v 32v 2 .

fherefore E( v2r + 1)

=

0 and

E(v 2r ) = (2r -1)(2r- 3) ... 3. 1 [1

r(2r+ l)(;r+ 7) + O(v-

241'

3)].

68 A famous result due to M. S. Bartlett. Put x = e - }'. Then the distribution of y is

L

f(p+q) e-Py.yq-l pq f(P) . f(q) . q

p '" f(q). e - P}' yq -

1

[1_~+y2 _ ... ]q-l dy

d y,

2

6

'

(0

~

y < (0)

.. whence th e fi rs t approXimatIOn.

The improved approximation is obtained by considering the probability density function of y as proportional to q-I e -P}' y . e -t(q-I)y .

69 Thomson, D. H. (1947), B, 34,368. From the preceding exercise - (2p + q - 1) log x '"

lIut for large

l

with 2q d.f. so that

11, 2 . log ( -11) - '" - 11-1

211-1

lienee, putting

11

=

(2p +2q -l}/2q,

_. ( X .-

2p-1 )1.2/2q 2p+2q-1

or x'" (1-2st/2.,

234

EXERCISES IN PROBABILITY AND STATISTICS

whence result by using the approximation

e- V

v2 v3 1-v+ 2 -"6'

'"

(0 ~

Un

~ 1).

This is obtained by successive use of the transformations Ur=XIX2".Xr

and

for

X r - l = Xr - l '

r=2,3, ...

,n.

Cadwell, J. H. (1951), B, 38, 475. Results follow by simple manipulation of the double integrals and b using the relation ) a:) a:)

71

ff

f(x,y,p)dxdy =

t,

o -a:)

where f(x, y, p) is the bivariate density function of X and Y. 72 UL (1962). Let the probability density contour be x2

xy

y2

0'1

0'10'2

0'2

2 - 2p - - + 2 = k 2 ,

where k is determined such that JJf(x, y) dx dy the density contour. Use successively the transformations . y X (1) z=--p.-; 0'2

w=

xJ1- p

0'1

(ii) z = R cos e;

2

=

1- P for the region Wilhin

; and

0'1

w

= R sin e.

Then

whence the result. Hotelling, H. and Pabst, M. R. (1936), AMS, 7,29. As x -+ ± 00, ~ -+ ±t, and the normal distribution of x is transformed II' a uniform distribution for ~. So also for 1'/. Therefore R = 12E(e1'/), and

73

dR fa:) fa:) a2f ~1'/ . ox ay dx dy, dp = 12

= 12

-co

-00

CXl

f

f

. of 02f SlDce op = ox ay

00

-00 -00

d~

d1'/ dx dy ,f(x, y) dx dy,

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

235

ntegration by parts with respect to x and y successively,

on "

---it f f

- n2

1- p

2'

00

00

_ 00

_ 00

ex p {-

1 2·[(2-p2)(x2+i)-2PXY1}.dXdY

2(1- p )

6 :::;J4_p2' ~cnce result since for p

= 0, R = O.

•~ Make the transformation Uj

[hen, setting

= Wj

(xj-ml)/O'I;

=

Vj

=

(Yj-m2)/0'2,

(i

=

1,2, ... ,n).

(Vj- pUj)(l- p2)-!, n

Q2

=

L (ul+wl) 1= I

distributed as X2 with 2n dJ., since uj and Wj are independent unit normal ..Iriables. Also, n R2 = Q2 --1- 2 ' (u 2 +iJ 2 -2piW), -p

i'

.Ihere ii and iJ are the averages of the Uj and Vj' The quadratic form of the means is distributed as l with 2 dJ. independently of R2. Hence by Cochran's Ihcorem R2 is distributed as l with (2n - 2) dJ. Alternatively, use characteristic functions. If q,(t) is the characteristic function of R2, then the characteristic function of R2

+ 11(1- p2)- I(U 2+ iJ2 -

2puiJ)

1\ If - 2il) - 1 q,(t) because of the independence of the distributions of the means ,tnd second-order moments. But, directly, the characteristic function of Q2 is 11- 2irrn. Hence q,(t) = (1 - 2it)-n+ I.

75 De Lury, D. B. (1938), AMS, 9, 149. The joint distribution of u, v and W is (n-l)n-I n2n(1- p2)(n-ll/2

nn _2) . exp

{

x (v 2 - u2)(n-4)/2 . wn -

(11-1)W(I- PU)} 2(1- p2) x 2 •

v(l- v2)-! . du dv dw,

where 0 ~ w < co; 0 ~ v ~ 1;

- 1 ~ u ~ 1.

= /'V, so that for u > 0,

(u ~ v ~ 1),

But

II

and for u < 0, ( -

II ~ V ~

1).

Inlegrate the joint distribution for wand v successively, whence the distribution "f II.

236

EXERCISES I N PROBABILITY AND STATISTICS

=

(1

f 1

2)(n-1)/2 00 r( + 2·) -P . L 11 J .p 2j +1 (l-t)(n-3)/2 t (2i+1)/2 11-1.l) j=O r(11-1)r(2j+2) .dr, 2 ,2 0

B(

whence the result by using the r-function duplication formula. Kendall and Stuart-I (1958), Ex. 15.7, p. 367. The region of integration is x 2 + y2 :s;; 1, subject to the condition thaI x 2+ y2 - 2pxy > O. With the transformation

76

x

=

rcosO, y

rsinO,

=

the region is transformed into r2:s;; 1, subject to 1- p sin 20 ~ 0 for all Ipi :s;; 1. Therefore the limits are (0 :s;; r :s;; 1) and (- n/4 :s;; 0 :s;; 7t/4). Henc the result, since u = 1- p sin 20. c Transform to u = (X-m1)/0"1 and v = (y-m 2)/0"2· Then the joint dis. tribution of z = u/v and v is

77

2 n R · exp { 2(1

~p2). V2(1-2 Pz+z 2)} ·Ivl dvdx, ( - 00

< v, z < (0).

Hence the distribution of z is (1- p2)t dz 7t . (z_p)2+(I_p2)'

( - 00

< z < (0).

78 The joint distribution of u and y is _1_. exp 27t0"10"2

{_ty2(U:+~)} .IYI dydu, 0"1 0"2

( - 00

< u, y <

00).

Hence integration over y gives the distribution of u as (-00

< u < (0),

Therefore the distribution of v is the same since n

L1 Xj

n

L1 Yj

and

j=

j=

are independent normal variables with zero means and variances mrf ami 110"~ respectively. For the distribution of w, set tj = xJYj. Then the joint distribution Ilf t h t 2, ... , tn is ( - 00

Next, make the transformation

~i = tj

<

tj

< (0).

for 1 :s;; ; :s;; (/I-I) and n

~n =

Lt j=

j •

1

Use partial fractions to integrate out successively over ~n-hen-2,···,¢,'I'

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

237

(Iblain the distribution of ~n as

A n d~n ~ . (nA)2 + ~;'

( - 00

<

~n

< (0),

,(hence the distribution of w = (1jn)~n' i9 Note that lIj, Vj are independent normal variables with means (mx + m)'), and variances 20'2(1 + p), 20'2(1- pl. Hence ,111,- III) y

(n -1)s~/20'2(1 + p)

(n -1)s~/20'2(1- p)

and

Ir~ independent X2 's each with (n-1) dJ. ~ Finney, D. J. (1938), B, 30, 130. , The joint distribution of P= log(O').sx/O'.xSy) and,. is

(1- p2)(II-I)/2 (n - 2)

(1- ,.2)(n-4)/2 P-pr)n

. (cos h

?_Tr ( - 00

< P<

00;

-

Inlegrate the joint distribution of v and,. over

I'

dP dr,

1 ~ ,. ~ 1), as v 1I,

(0 ~

1I ~

=

etl .

1), using the sug-

~ested transformation.

SI Pearson, K. (1923), B, 15, 231. Direct integration gives the proportionality factor of the joint distribution .1\

(II-l)(b I b 3- b~)t/1[(11 0'2' The marginal distributions of X and Yare

(11-1) (b,bJ-b~)t B( _1 1) [1 (blbJ-b~)x2]-n+t d . b . n 2' 2 + b 30'12 . x, nO'I J (-00 < X < CD);

.tnd (n-l) (blb3-b~)t B( _11)[1 (blb3-b~)y2]-n+t d . b . /I 2, 2 + b 2 . y, n0'2 1 10'2 ( - 00

< Y < (0).

To determine the b's it follows from the equi-density contours of the joint ,\islribution that b2 0'1 I = -. - . y and E(Y X = b, 0'2 lienee hi = b3 since p = b 2/b 1 = -b 2/b J •

I

E(X Y= y)

x)

b2 0'2 = --b .-.X. 3

0'1

Also, from the marginal distribution of X, the evaluation of var(X) gives

1 = b 3 /2(n - 2)(b I b 3- b~). Ihcrefore

hi

= bJ =

1/2(n-2)(I-p2) and

b 2 = -p/2(n-2)(1-p2).

Imally, 11 is evaluated from

P2 == E[(X/0'1)4] I\lr

P2 ~ 3, n --+

00,

= 3(n-2)/(n-3).

and the joint distribution becomes bivariate normal.

238 82

EXERCISES IN PROBABILITY AND STATISTICS

Pearson, K. (1923), B, 15, 231. If f1(X) denotes the marginal density function of X, then

f1(X). var(ylX =

00

x) =

f

p'-

p ::

xr f(x, y) dy,

-00

f1(X) =

(n-1)B(n-!,!) [ x2 nO"l(2n-4)t . ~ + 2(n-2)O"f

and

]-n+ t '

whence the variance by direct integration. The expectation of the conditio. variance of Y for variation in x follows since E(x 2) = O"f. nijl The semi-axes of the equi-probability contour are 0"1",(2n-4)i ad 0"2",[(1- p2)(2n -4)]t. Hence the volume outside the contour is n

f 00

83

f(x, y) . 4n(n - 2)0" 10" 2(1- p2)t '" d",.

'"

Pearson, K. (1923), B, 15, 231. Since n > 0, the probability will be finite if the equi-probability contour, are ellipses. Also the bivariate density should be zero along one of the ellipse, within which the total probability should lie. Hence the joint density funcliu/; must be of the form

f(x,y)

{X2 xy y2}]n 1 = k [1-(1 -p 2)C2 2 - 2p - + 2 ' 0"1 0"10"2 0"2

where c2 is a constant and k is the proportionality factor. Direct integration gives

k = (n+1)/nc 20"10"2(1-p2)t. The marginal distribution of X is

(n+1)B(n+1,!) ~ x 2 )n+1 - - - - - - . 1-22' dx, ncO" 1 c 0"1 with an analogous result for Y. Evaluation of var(X) gives c 2

=

2(n + 2), and n is obtained from

P2 == E[(X/0"1)4] = 3(n+2)/(n+3), so that for n > 0, 2 < P2 < 3. For P2 -+ 2, n -+ and the distribution becomes uniform. For P2 -+ 3, n -+ <Xl and the distribution becomes bivariate normal.

°

84

Pearson, K. (1923), B, 15, 231. Using the notation of the preceding exercise,

E(YIX

=

x)

0"2

= p-x. 0"1

Therefore. if f1 (x) is the marginal density function of X. a

fl(X). var(YIX =

x) =

f -a

(y-p ::

x)

f(x, y) d(y-P ::

x),

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

fl(X)

=

M:

r(n+2)

Y 2n . 0' 1(n + 2)tr(n + i)

3

239

]"+t

[X2 2 ' 2(n + 2)0' 1

. 1

The semi-axes of the equi-probability contour are

0' 1 I/I(2n

+ 4)t

.!~[(I_ p2)(2n + 4)]t, so that the total probability outside the contour is

f

and

1

f(x, y) . 4n(n + 2)0' 10' 2(1- p2)tl/l dl/l = (1- 1/1 2)"+ 1.

I/!

85 VL (1961).

Let x and s be the sample mean and standard deviation obtained from the 'Irst experiment. Then oe == (x-m)/s is known. If the required number of plots is (v+ 1), then oeJv+"1

= 1 % value of Student's t with v d.f.

Jnd so oe

=

1 % value of t with v d.f. - f( ) r::-:1 = v , say. yv+l

Inverse interpolate on the sequence of values N, f(N) obtained from the t !able for varying N d.f. to determine v such that f(v) = oe. The number of plots is the least integer containing (v + 1). To allow for seasonal variations, divide each experimental plot into two equal sub-plots. Allocate randomly one of each pair of sub-plots to the new fertilizer, using the other as a control. Test the hypothesis of no difference by Student's paired t statistic.

86 UL (1961). First part is the standard F statistic for testing the equality of two lariances. For second part, consider pairs of briquettes from the lots, and assign landomly one of each pair to A and B respectively. If (Uj, Vj) are the measurements of A and B on the ith pair, then consider new variables Xi = Uj - Vi; 1',= IIj+Vi, (i= 1,2, ... ,n). Since cOV(Xi,Yi) = var(ui)-var(vi), therefore a test of the equality of the lariances of the measurements of A and B is equivalent to testing that the pairs (x;, Yi) arise from an uncorrelated population. Thus, if R is the sample Nimate of corr(xj, yj), then the appropriate test statistic is t

= RJN-2/Jt-R 2

with (N-2) d.f.,

Vbeing the number of pairs of briquettes used.

~ The probabilities of the four outcomes-HH, HT, TH and TT-are r. PC/. C/p and q2 respectively. (p + q = 1). If the corresponding observed

240

EXERCISES IN PROBABILITY AND STATISTICS

frequencies are n l , n2, n3, n4, then

P*

=

(N +nl -n4)/2N

= (1

+A,)/2;

1

var(p*) = 4N 2 ' var(111 - n4) = pq/2N. To test the hypothesis H(p = t), use (i) z

=

(P*-t)

/(8~r as a unit normal variable.

4

(ii) X2 =

L

(4nj-N)2/4N, as X2 with 3 dJ.

j= I

88 Since PI = P2 + b, the probabilities of the outcomes HH, HT, TH and TT are P2(P2 + b), (P2+ b)(1-P2)' (1-P2- b )P2' (1-P2)(1-P2-b). If tho observed frequencies are nl, n2, n3, n4, then for P2 = t the estimate of cl t easily obtained. " In the second case, consider 4

T=

L ajnJN j=1

as an unbiased estimate of b, the constants aj being determined by E(T) == b. Hence

T= (n2-n3)/N, where

var(T) = 4J(l-(M/N,

and

4J == PI + P2 - 2PIP2' Thus 0

~

USill l'

4J

~

1, whence result.

89 UL (1963). The formal results follow by a direct use of least squares. Explicitly, S2

=

n(n + 1)(2n + 1)/6; S 3

=

n4(n + 1)4/16;

and S4 = n(n+1)(2n+1)(3n2+3n-1)/30. Hence var(IX*)

=

24(2n + 1) (3n 2+ 3n - 1)0'2 . and n(n 2 - 1)(n + 2)(3n 2+ 3n + 2)'

* 120(2n+ 1)0'2 var(fJ ) = n(n2 _ 1)(n + 2)(311 2 + 3n + 2)' 90

Fieller, E. C. (1932), B, 24, 428, and UL (1963). (2

is Student's

(2

= (a - pb)2/S2(A,1 + A, 2p 2- 2pA,3)

statistic with n dJ. The confidence limits are obtained from t~ =

(a-pb)2/s2(A,1 +A,2p2_2pA,3)

giving two roots for p. The stated result follows for A,3

=

O.

.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

241

Geary, R. C. (1930), J RSS, 93, 442. ql put u = x + a and v = y + b. Then the joint distribution of u and z

~.exp{-t[(u-af +(uz-bfJ} .Iuldudz, W' t 0"2

0"1

= vlu

( - 00 < u, z < 00),

0"2

.• hence ~he ~ist~ibu~ion of z by integration over u. But since P(z) dz is a rrobabihty dlstnbuhon, co

f

P(z) dz = 1,

which gives

-00

f

f

a/a,

1

fo·

-a/a,

00

e- w2/ 2 dw+

-00

az-b

R(z) dz

I-p[lwl ~

or

=

1,

where

:J+ f

R(z)dz

(0"~+O"fZ2)i =

w,

00

= 1,

-00

or

f""

R(z) dz

=

p[lwl ~

:J

lienee the result for a sufficiently large compared with 0"1. For correlated lariables. x and y' = y-p(0"2/0"t)x are independent normal variables. Then z

=

y'+(b-')'a) +,)" x+a

where,),

0"2

== p--, 0"1

= z'+,),. ~'hcre

z' satisfies the conditions of z for uncorrelated variables. Therefore az'-(b-')'a) [O"~(1- p2)+ O"fz 2]i l

Ii

approximately a unit normal variable. Hence the result.

92 Todd, H. (1940), J RSS(S), 7, 78. Consider the four vertices of a square. The four kinds of contiguous pairs are: (i) endpoints of a horizontal side of the square; (ii) endpoints of a vertical side of the square; (iii) and (iv) endpoints of the two diagonals of the square. Enumeration then gives P2. Define a triplet by (r, s) where rand s denote the between-row and between(olumn intervals of the triplet. There are 20 distinct types of triplets, grouped .1\ rollows: (1, 1)-4; (2,2)-2; (0,2)-1 ; (2,0)-1 ; (2, 1)-6; (1,2)-6. I he number of ways a triplet (r, s) can occur in the field is (M - r)(N - s).

242

EXERCISES IN PROBABILITY AND STATISTICS

Therefore P3

=

4{5MN-7(M+N)+9}

/(~N).

For testing randomness of deaths with contiguous pairs, use

as a unit normal variable approximately. 93

Daniels, H. E. (1941), JRSS(S), 7, 146.

and n2P4 -PI

= n IP4 +1 -+1

E(r4)

n3P4 -PI-P2

Using these as estimating equations and solving,

*=

rl pf = -;

P2

nl

*

P3

=

r2(nl-rl)

.

nl(nl +n2- r d'

r3(nl-rl)(nl+n2- rl- r2 ) . , nl(nl +n2 -rl)(nl +n2 +n3 - r l -r2)

*-1

P4 -

* * *_

- PI - P2 - P3 -

r4(nl-rd(n l +n2- r ,-r2) . nl(nl +n2 -rd(nl +n2 +n3 - r, -r 2 )

The sample likelihood is 4

L

=

constant x (1- pd- n2 (1- PI - P2)-n 3

n pi',

i= ,

whence the same estimates of the p's. Transform L in terms of the (J's; thel maximization gives nl-rl. 0* _ -0*2 --' 3-

nl

r3+ r4- n 3 r2+ r 3+ r4- n 3

The asymptotic variances are:

var(T) ,...,

L4 (OT)2 ;10* . var(On u 1 6f=6,

1=2

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

243

,fill Ibe minimum variance is obtained by using Lagrange's method on the ',ocl ion ~ ~

F

=

var(T)+A.(n l +n2+n3-n).

Finney, D. J. (1941), JRSS(S), 7,155.

for P > O. E( s 2P) -_ (n -1)(n + 1)(n + p3) ... (11 + 2p- 3)-.(1 2p ,

and

=

=

11

II~oce

E[M!r 2s2)]

e[(n-I)/2n),2a 2

and E[e'XhHr 2s2)]

=

e'~+tr2,,2

E(y').

ence the estimates, their efficiency being inferred from the properties of 2 and s . By logarithmic expansion,

,h I

k(k-l) [1

I gll.k---o 1

_

11

"Ilhill

2(k-2) 2k ---+ 311

2 -6k+5 O( + n 3n 2

-3)] ,

k(2) 1 A.k = 1--+-[3k(4)+ 16k(3)+ 6k(2)] + O(n- 3), n 6n 2

.• here in factorial notation kim) == k(k-l)(k-2) ... (k-m+ 1). Therefore

lienee

9S For an uncensored sample of N observations var(O)

= 20(1-0)(2+0)/N(I+20)

by a straight application of the maximum-likelihood procedure. For a censored sample, the expected proportions in the AI, A 2 , A3 and 14 classes are proportional to p(2 + 0), (1- 0), (1- 0) and 0, with the proporI,onality factor [2(I+p)-(I-p)O]-I. Hence

var(O)

= 0(1-0)(2+0)[2(I+p)-(I-p)O]2. 2M[2(1 + p)+(l + 7p)O]

244

EXERCISES IN PROBABILITY A NO STA TISTICS

Now 1(0 1

,

) P

=

(I+ P) (1+20)(1-..1.10)2 2' 1+ . 1. 20 '

where..1.1 == (l-p)/2(1+p), ..1.2 == (1+7p)/2(I+p). Note that 0 < ..1.1 < t < ..1.2 < 2, and 3..1. 1+..1.2 = 2. Then for fix h(O,p) is maximized for 0", 2(I+p)/(13+19p), whence the inequality. cd

96

The equation for

I'

6 is

and var(O)

= 20( 1 - 0)(2 + 0)/( 1 + 20)N I'

For censored data, the expected proportions in Ab, aB and ab classes.

~1- 0)/(2- 0), (1- 0)/(2- 0) and 0/(2 - 0) respectively. Hence equation r~/I:~ IS

YI+Y2 Y3 N2 1 - 0* + 0* + 2 _ 0* = O. For full efficiency of 0* as compared with

N2

= NI·

0

(1 + 20)(2- 0)2 4(2+0) == NI·g(O),

say.

But g(O) is maximum for 0 = 0'1472, whence the inequality. Var(O·*) I, obtained simply by adding the information obtained from the two samplc~

97 The joint distribution of the

n f(u

Uj

is

n

n!

j)

du j ,

<

( - 00

UI

<

U2

< ... <

Un

<

00).

i= I

Next, put

u, Zj

=

f

for i

f(x) dx,

= 1,2, ... , n,

-00

so that PI = of the pj is

ZI

and Pj

=

Zj- Z j_ 1

for (2 ::;; j ::;; n). Hence the joint distribolioll

n n

n!

j=

dpj,

1

with the limits O
0< P3 < I-PI-P2;"';

0 < Pn < I-Pl-P2-'''-fln

I·

E(P'i pj) is then obtained by integration. For fixed P's the conditional distribution of the mi is multinomial, so lhal E(milp's)

= mpj;

E(mflp's)

= mpi+m(m-l)pf;

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

:d

3

245

E(mjmjlp's) = m(m-l)pjpj.

Dixon, W. J. (1940), AMS, 11, 199. fhe marginal distribution of the p's is as obtained in the preceding exercise, .d the joint distribution of the p's and m's is

.j

n n dpj. j= mj! j=

n+l p.ml

.

n !m !

n

_1_.

1

1

Ir.cexpre~slion for CPt. iS obtained by reversing the order of integration and the

dltinOmla summa IOn.

E(C2) =

[nf 02~] 0 j=1

E(C4) =

()j

and /1'5=0

[t: ~:fL5=0 + L~j o~4:()f L.=o

Ihe integration over p's is carried out after differentiation. Thus

[02f/J]

o()f

)2 +(~_~)p+(m-l)p~] Ii dp. /1'5=0 = n .'J[(_l n+l m n+l m 1

1

J=1

J

= n(n + m + l)/m(n + 1)2(n + 2). lienee

E(C 2) = n(n+m+ l}/m(n+ 1)(n+2). The same technique is used for obtaining E(C4 ), but the task is laborious. II is also tedious to deduce the simple result var(C 2 ) = E(C 4 )_E2(C 2)

4n(m -l)(n + m + l)(n + m + 2) m3 (n + 2)2(n + 3)(n + 4) '/9

Ageneralization of a statistic proposed by A. M. Mood. Vide the reference

"I the preceding exercise,

The joint distribution of the m's and p's is

n _1_, n dpj' j=O mj. j=O n

n !m !

p,!,1

n-l

with the limits for the p's

O
f/J = expLto i()dn] . E[ exp {-

= exp

ita ()iXdm }]

[jto i()dn].E[exp {- ito 'rjmJm }],

,hence the result on multinomial summation. Note that

246

EXERCISES IN PROBABILITY AND STATISTIC'S

Now

where

2cjJ ] [aao~. I

00=0

_- 2"+ i2 ( --1 2i) (m-l) 2 E(PI)+ - - E(PI ), n m n m

PI

==

L PJ i

j'=O

= ~+ (~_ 2i) (i+ 1) + (m-l) n2

m

(i+ l)(i+2) m ' (n+ l)(n + 2)'

n (n+ 1)

Hence E(d 2) = m(n+l)+n 2.

6mn Similarly, E(d4) =

±[a4~]

1=0

aO I

0'0=0

+ 2 L [ a24cjJ 2] i<j

aO I aO j

0'0=0

The evaluation of the right-hand side is straightforward but extremely labo rl ous. It may be deduced that

2 var(d ) For n

=

=

(m+n+ 1){3(2m-n)(2m+n)+2mn(2m+2n-l)} 180nm 3

m, an

(d 2) _ (2n+ 1)(8n+ 7) var 180n2

d

[These results do not agree with those attributed to Mood by Dixon.] 100

Mood, A. M. (1943), AMS, 14,415.

P(X, X) = p(xIX)p(X) =

e)(:=:) (~)

P(X)

.

o ~ x ~ min(n, X).

,

For integral m l and m2,

by hypergeometric summation. Hence var(x) 101

=

n(N -n)J.I. [ It] N(2) . 1- if

n(2)

+ N(2) . (j

2 .

Mood, A. M. (1943), AMS, 14,415. p(xIIX I _ d

=

(XI-l) XI

(N I - l

-XI-l)!

nl-xi

(NI-l), nl

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

247

ince X, = X,_I-X" Ni = N,_l-n,. " p(x,IX i- 1) = p(XIIX,_ l ), 0 ~ Xi ~ min(Xi_l> N i). I~ds

E[ (m)] _ n\m) X,

-

N\"!,\ .

E[x(m)] _ n\m) E[x(m)] 1-1 - N\'~,\' i-2,

nee the result by successive reduction. ,hefor the joint distribution of Xl' X2" .. ,Xk' X p(x", Xk-l' ... , X2' Xl' X) =

== P(xklxk-l,· .. 'Xl> X)P(Xk-ll x k- 2, ... , XhX), .. P(xlIX)P(X) == P(xkIXk-l)P(Xk-lIXk-2)" . P(x21X l)P(X1 IX)P(X). Therefore the final result is obtained by summation over the successive ndilional distributions of Xk, Xk-l,' .. ,Xl and then lastly X . ~ 101 Mood, A. M. (1943), AMS, 14, 415. Use the argument of Ex. 100 to prove that

.

E[x(m l ) (X -x)(m2)lx]

=

n(m I) (N )(m 2 ) - n . x(m , +m2), N(m,+m2 )

.hence cov(x, X - x), var(X - x), var(x). Also 2 cov(X, x) = var(X)+var(x)-var(X -x). Now corr(x, X - x) = 0 if (12«(12 - N f.l + f.l2) = 0, which gives the two possibilities (i) E(X - f.l)2 P(X) = 0, (ii) EX(X - N)P(X) = O. Ii) gives the distribution P(X = f.l) = 1 and P(X '" f.l) = 0; (ii) leads to the distribution P(X = 0) = a constant p, P(X = N) = 1- p, and P(X) = 0 for 1 ~ X ~ (N -1). 103 Mood, A. M. (1943), AMS, 14, 415.

f (\N~ )(a) r=Of (N~ )x coefficient of zr in + z)" = coefficient of ZO in (l+z)" f ( P )z-r r=O N-r coefficient of ZN in +z)« f ( ~ )zN-r = (a+ P). r=O N

.=0

r

r

=

(1

r

(1

=

N

r

Similarly,

~ (a+r)(p+N-r) N (P+N-r) r P = r~o \ P

,;

.

x coefficIent of z' in (l-Z)-«-l

= coefficientofzN in (l-z)-«-l

f (P+N-r)zN-r N-r

r=O

= coefficient of

ZN in (l-z)-«-l

f 1=0

(P+t)ZI t

= coefficient of ZN in (l-z)-«-1I-2 = (a+ P~N +

1).

248

EXERCISES IN PROBABILITY AND STATISTICS

The mth factorial moment of X is E[x(m)] = Ct(m) N(m)/(Ct+ p)(m), and that of Y is E[y(m)] = N(m)(Ct+m)(m)/(Ct+p+m+l)(m).

104 Mood, A. M. (1943), AMS, 14, 415. Clearly, P(X;jX i - 1) = P(x;jx j_ d, and the general results folIow rro Ex. 101 above. (iHvi) are obtained as particular cases. For (vii), n, j

E(x/xjIX)

= E[XxiIX]-

L E[xjxrIX],

(i =1= r),

r= 1

105 Mood, A. M. (1943), AMS, 14, 415. For corr(x, X -x) < 0, the number of items inspected on the average 11<' batch is a ' / =

L Pn(x),

n+(N -n)

x=o

and the average number of defective items inspected per batch is N

B

=

E(x)+

a

L L (X -x)P(X, x).

x=o x=o

Therefore E

=

[f i (X

i

NB-fi/ = N(N-n) -x)P(X,X)_~ P(X)] fi(N-fi) fi(N-fi) X=Ox=o N-n Nx=o n ,

where xto xto

(~=:)P(X,X) = xto (:~~)Pn+l(X).

For corr(x, X - x) > 0, E =

1)

N(N -n) n [(x+ fi] ( _ ). - 1 Pn+1(x)-N·P.(x) . fiN fi x=a+l n+

L

If x has a discrete rectangular distribution, P(x

=

r)

= l/(n+ 1) == P.(x),

(x

= 0, 1, 2, ... , n).

Hence E = 2(N-n)(n-a)(a+l) N(n + l)(n + 2)

if X is also rectangular. Its maximum is attained for a

= (IN +2-3)/2 and

n

=

IN +2-2,

max E = (IN +2- V/2N

-+

t,

as N

so that -+

w.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 3 if ,::::

249

Np, then E is to be maximized subject to pN = n+(N -n)(n-a)/(n+ 1).

, Ihis to express E as a function of n, p and N, and maximize this with tIC '" . eet to varIatIOn In n. '"Pfinally, oX = ! and {J = (N - n)(a + 2)(2)/2N(n + 2)(2). therefore maximize E

= 2(n-a)/A(a+2)

subject to the condition that

(N -n)(a+2)(2) Np N (n+2)(2) = 1-p == T' say.

106 Mathisen, H. C. (1943), AMS, 14, 188. Denote

Yi = mi-i for (i = 1,2,3,4) and C ¢(tl' t 2, t 3, t 4) = E [ex p =

ttl f

4

= i~/r

Then

t i Yi)]

(4N+3)! (N !)4

N

m

'" (PIP2P3P4) dpi dp2 dp3,

,here the integration is over the region 4

n01 ~ 1; 0 ~ P2 ~ 1- PI; 0 ~ P3 ~ 1- PI - P2; and", ==

L Pi e'i. i= I

Ihercfore 4

E(C) 2

E(C )

[024>]

= i~1 Ct~

=

".=0'

and

4 4 [04¢] L -4 + L [ ~ 02 4> 2 ] . i= I Ct 1'.=0 ioFj (li 1'0=0 :l

(tj

i

lienee E(w) and, with some tedious simplification, var(w). Note that max C

Ii) If '"

-+

= 9m 2/16, and so (0

x, N is fixed,

4 E(w) --+ 3(N + 2)

Iii) If N -+

YJ,

32 and

(N+1)(N+5)

var(w) --+ 27' (N +2)2(N +3)(N +4)'

m is fixed, 4 3m

E(w) --+1117

~ w ~ 1).

and

var(w)

32 m-1 --+

27 .~.

Mathisen, H. C. (1943), AMS, 14, 188.

The joint distribution of ml and P is

'£11+

1) ! m ! n+ml (1 )n+m-ml d 1I1!)2"'1!(m-md!'P -p p,

(0

~ ~

1 0

~p~;

~~ml~m. ~)

250

EXERCISES IN PROBABILITY AND STATISTICS

For the rth factorial moment of ml' use the marginal distribution of . the point probabilities Iti. Wilt,

(n:~I) (n:::~l)/

e

n +:+ 1),

(0

~ ml ~ m).

E(ml) = m/2; var(ml) = m(2n+m+2)/4(2n+3), whence the asyrn variance of y. PIOII, 108

Craig, A. T. (1943), AMS, 14, 88.

The marginal distribution of v is (i) n. 2n - 1 • a- n. vn- 1 dv, for 0 ~ v ~ a/2; (ii) n. 2n- 1 • a- n(a_v)n-l dv, for a/2 ~ to ~ a.

E(v

r)-_(~)(~)r[_I __+_r!_ ±(n+k-l)! 2r-k] 22 n+r (n+r)!'k=O k! . ,

so that E(v) = a/2; var(v) = a2/2(n+ l)(n+2). For (- co < w ~ 0), the distribution of w is

n [2(n+l)(n+2)]t·

[1- J(n+l)(n+2) lwlfi ]n-l dw

--+

_1_e-lwlfi dw

fi

By symmetry the same limiting form holds for (0 The characteristic function of x is

~

'

as n --.

'J

w < co).

and that of the standardized variable (x - a/2)J12i1/a is

cP(t)

= e - it/4j3n . [exp{ it(12/n)t} -1]n/[it(12/n)t]n, whence log cP(t)

--+

!(it)2,

as n --+ co.

109 Craig, A. T. (1943), AMS, 14, 88.

For the rectangular distribution in (O,a), E(xr ) = ar/(r+l). The distribution of the ordered observations is

n!

n

". a n dx/o j=

I

whence E(x~ xj) by successive integration. The function y is an unbiased estimate of a/2 if n

L

iC i = (n+ 1)/2,

i= 1

and var(y)

=

=

[ {n}2 a2 2 iC j -(n+ 1) n (i-l)ci+ n i(i+ l)cf(n+l)(n+2) i=2 i=2 i=2

-2 (

L

L

L

n }{ n } {n .L iCi .L (j+ l)cj +2 .~ (i+ l)ci x .~ jCj ,=2 )=2 ,-3 )-2 i -\

} ]

(n + l)a 2 a2 + 2(n+2) '4'

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

,n~

n+l 2

251

n

= -- -

CI

3

L ic,. j=2

{heCCfore for minimum variance , _J)cr+2r

,~r

r-I

n

j= 2

J=r+ I

L (j-l)cj+2(r-l) L

jCj

= (n+l)(r-l), (r

= 2,3, ... ,n).

. unique solution is Cr = 0 for 2 ~ r ~ (n-l), Cn = (n+l)/2n, whence ::: O. The distribution of Xn is na-I(xn/a)n-I dxn, (0 ~ Xn ~ a), so that var(xn) = na 2/(n + 1)2(n + 2).

Jr.C

The distribution of t is

(n:2 )t(1+n

1)-n[

~

,hich -+e,-I . dt, for (- 00

t

]n-I

1 +t/Jn(n+2)

~

1), as n --+

. dt

00.

110 Craig, A. T. (1943), AMS, 14,88.

For any random observation x, E(xr) = (k+ l)ar/(k+r+ I), so that _ (k+ l)a _ (k+ l)a 2 E(x) = k+2 and var(x) = n(k+3)(k+2)2' Ihe distribution of the ordered observations is n!(k+l)n a"(k+ I)

I\

k

n

•

Xj

dx j ,

lienee E(xjxj) by direct integration. Therefore, n(k+ l)a 2 var(xn)= {1+n(k+l)l2{2+n(k+l)}; and

n(k+ l)a E(xn) = l+n(k+l);

r(n)rU+ {1/(k + l)}] n(k+ l)a 2 cov(xJ, xn) = {I +n(k+ 1)}{2+n(k+ I)}' r{j)r[n + {1/(k+ 1)}J , Illr J' 10

be unbiased,

"cr('l+ _1_)/ k+ 1 t...'

r(')- 1+n(k+1) I

n(k+2)'

r[n+{1/(k+l)}] = r(n)

\Iinimization of var(y) subject to the above constraint gives I ~ i ~ (II-I), and ell = {l +n(k+ 1)}/n(k+2). Hence var(y) = (k+ l)a 2/n(k+2)2{2+n(k+ I)}.

0. Cj

= 0 for

Ihe distribution of y is n(k+l) [ n(k+2) ]n(k+l) (n-I)(k+I)+k d an(k+I)' 1+n(k+1)'y . y,

o~ y

1 +n(k+ 1)] ~ [ n(k + 2) . a.

252

EXERCISES IN PROBABILITY AND STATISTICS

Hence, by transformation, the distribution of u is 1

~~----~~~

n(k+ 1)

~~~--~~~~x

[1 + {1/n(k + I)} ]n(k+ 1) [n(k + 1) {2 + n(k+ 1)} 1!-

u ]n(k+I)_1 [n(k+l){2+n(k+1)}]t .du,

x [ 1+~~--~----~~ whence the limiting distribution, as n 111

~

co.

Wilks, S. S. (1941), AMS, 12, 91.

f x.

Let

u=

f ~

f(x) dx

and

v

=

f(x) dx.

x,

-~

Then the joint distribution of u and v is

nn+l) .-1 n-I(1 )1-.-1 d ns)r(t-s)nn-t+ 1)' u v -u-v . udv, the region of variation being bounded by II + V = I, u = v = O. Now P = 1 - u - v and Q = u. Also for fixed P, 0 ~ Q ~ (1- Pl. Hell the distribution of P is l\

1 . pr-I (l_p)n-r dP, B(r,n-r+1) . where r = t - s is an integer 112

~

(0 ~ P ~ 1),

1.

Wilks, S. S. (1941), AMS, 12,91.

J

(x-lII+ko)/a

fo 1

Q=

e -,,2/2 d u.

(x -III-ko)/a

Therefore

E(Qls)

J 00

1

= r-c

e

y2n

-

-v2/2

1 d v. M: y2n

f

v/./ii + k./a

e -,,2/2 d u

vjjn _ ks/a

00

ks/a

fo (,,:S b:(-'C.:l)w'J·dW Hence

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

II I

1 ::::-~.

V 2,.

(1I-1)(n-I)/2

(11-1)

253

exp[-t{(/1-1)+(}2}z2].zn- 1 dzdO,

2(n - 3)/2f _ _

2

-I 0

where 0: = 'I'I! •• 1

3

'J'_

( 11 )'

II' --- -

11+1

,

rl!sull.

pul

=~,

(x-m)/u

var(Q)

s/u

=

= '/, and ct>(z) =

(8Q)2 ~~

~=O

v."

. var(~)+

*-

(OQ) ~ vII

2 ~=O

fry:; e-·I•2 / 2 . dy.

. var(I71.

Neyman, 1. and Scott, E. L. (1948), E, 16, 1. The logarithm of the sample likelihood is 1 k n, 10gL = -N log(ufo)- 2u 2 ' j~l j~1 (Xij_~j)2

11.1

1 k - N log(ufo)- 2u 2 j~1 [(lI j -1)st+llj(Xj- ~y],

n,

n,

where Xj =

L1 xjllj;

(lIj-1)st

=

L

(xij-xY

j= 1

j=

lienee results by application of the standard maximum-likelihood procedure.

114 Neyman, J. and Scott, E. L. (1948), E, 16, 1. Transform the joint distribution of y and S2 to polars so that

.j;t(ji -

ml/u

=

R sin 0

and

s.j;t/u

= R cos O.

Ihen (0 ~ R < 00) and (-n/2 ~ () ~ n/2). Also, for given rand k, r(-)k V y-m

=

/1

U 2r + k R2r+k (2r+k)/2' .

. k(} sm .

lienee the results on integration.

liS Neyman, J. and Scott, E. L. (1948), E, 16, 1. The logarithm of the joint likelihood of the sample observations is k

log L = constant-

k

1

L Il j loguj-t L 2{lljst+lIj(x-m)2},

j= 1

.hence the equations for m* and u7*.

j= 1 Uj

254

EXERCISES IN PROBABILITY AND STATISTICS

The equation for

mmay be formally written as

k

L WiePi(Xi,m) = 0,

where ePj(xj,m) == (x i -m)/[st+(x,-m)2].

i= I

Hence by Taylor's expansion, ePj(Xj, m) ~ ePj(X j, m)+(ln-m)eP;(x j, m),

so that the asymptotic variance of mis E

var(m)

=

[.t

wt ePt(Xj, m)+

,=1

.L. WjWjePi(X j , m)ePj(xj , m)] ,*)

k

L~, wjE{eP;(Xj, mn]

2

The expectations are evaluated directly by using the results of the preccd . to give . var (m A) . In.. exercise 116 (i) L~t the ordered sa?1ple obser~ations be XI ~ X2 ~ .. ,' ::::;. x2n , I " that X n + I IS the sample median. Assummg that X n + I > m, the dlstnbution 1,1 x n + 1 is .

(2n+l)![, (n !)2

4-eP

2

(Xn+l)

]nf(

Xn+1

)d

Xn+l,

where

f

Xn+1

eP(X n+ I)

=

f(x) dx ,.." (Xn + I - m)eP'(m)

=

(Xn+ I - m)f(m).

m

Also for large samples, f(xn + tl ~ f(m). Hence the approximate distribUliol, of X n + l . (ii) The Cauchy distribution has the probability density function 1 1 f(x) = ;'1 +(x-m)2' (-00 < X < 00) and _

[o2 10g f (X)] _ ~ E om 2 - 7t'

00

f

1-(x-m)2 d = {1 +(x_m)2}3' X

I

2'

-00

117 The estimate of ex is ex* = vlx, and the characteristic function of thl distribution of the sample average x is

it )-nv ( 1-. nex Inversion gives the distribution of x as

(nex)"V _nax -nv - 1 dr(nv)' e X x,

(0 ~

x<

00).

Hence E(ex*) = (vnex)'r(nv-r)/r(nv), provided nv-r > O.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

II

3

255

8 fisher, R. A. (1920), MNRAS, 80,758. "

Lei IV ==

L:

Zi = Xi - X.

IZil,

i= 1

n z· is normally distributed with zero mean and variance [(n -1)/n]a 2 • c~rr(zi' Zj) = -1/(n -1), for i =F j. Therefore

rnlto

II,

E(lzil) = a[2(n -1)/nn]t,

E(w 2 )

Ltl

zr +

whence E(w).

i~j IZil x IZjl ]

=

E

=

(n -1)a 2 +n(n-l)a 2 E[luil x lUll].

3ul (IIi' u) have a normal bivariate distribution. Therefore co

E!lUil x IUjl] = 21n

C:2Y f f -

=~.

co

<X)

C:2f f f

IUil x IUjl x

co

-

coco

o

uiujexp{ -(n-l)(ur+ uJ)/2(n-2)} x

0

x cosh(uiuj) dUi dUj

= \ya

~ [{n(n-2)}t+ si n- 1 nn

(_I_)], n-l

polar transformation.

119 Geary, R. C. (1944), B, 33, 123.

e

Lei = x-mx and" = y-my. Denote the joint density function of e and p). Then

~ by f(e,,,,

Pllx-mxl

~ ly-myl1 =

I~I

f f f(e,,,, co

p) de d"

- co -I~I co

=2

ff (f(e,,,, o

~

p)+ f(e,,,,

-

p)] de d"

0

haluate the double integral by a polar transformation.

256

EXERCISES I=" PROBABILITY AND STATISTICS

Goodman, L. A. (1952), JASA, 47, 622. For given p and 11, g will be the largest number in the sample if the rel11" ing (11-1) elements of the sample arise from the first (g-l) integers. There~ln. lOr,

120

p(glll, p)

(~= ~)/(~),

=

(11

~ g ~ pl·

Let T = f(g) be an unbiased estimate of p. Then E(T) = P for all p. (1) If p = n, then g = 11, so that E(Tlp = n) = f(lI) = 11. (2) If p = 11+ 1, then g = 11 or (11+ 1), and so

E[Tlp = 11+ 1] = f(I1). P{g = I1lp = n+ I} + f(l1+ 1). P{g = 11+ lip == IIf\' J

=

(11+ 1) for T to be unbiased

or f(II+1) = [(n+l)-nP{g

= nip = 11+1}]IP{g = 1I+11p = 11+1}.

Hence, more generally,

f(lI+k) = [(II+k)-

f g=1I

(g+k) lI+k

=

~!(:')' P{g = vip =

lI+k}J/P{g = lI+klp = Il+k).

PIli (1I+II+k) 11=0

II

p-u

=

L coefficient of X O in x-

u (l-X)-II-h-I

11=0

=

. coeffiCient of , x P - 1I In 1/(l-X)"+k+2

=

I)

(P+k+ lI+k+1

.

Therefore

E[(g+r-l)(r)]

=

(1I+I'-1)! (/I-I)!

=

f Y=II

(g+I'-I), ~(P) 11+1'-1/ 11 1

(_11_) (p+r) ! .

p!

11+1'

121

Goodman, L. A. (1952), JASA, 47, 622. If g and s are as defined for the sample, then (11- 2) of the sample obscrvOi' tions must lie in the range (g, s). Therefore, the joint distribution of g and .1 I' (11

~

g

~

p; 1 ~ s ~ g -/I + 1)

and that of g and d is (11 ~

g ~ P; 11-1 ~ d ~ g-I).

But for fixed d, (d + 1 ~ g ~ p), whence the marginal distribution of d. p-I E[(d+r-l)(r)] = (p+r)I(d+r-l)(r) d=1I-1

(d =1)/() p -I (d+r)(r+l) (d-')/(l'l _ p-I

11

2

11

d=II-J

II

/I

2

= [(p+r)(lI+r-2)(r) ( p+r-I) -(II+r-1)(r+l) (p+r)]j(/l) . 11+1'-1

11+1'

/I

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 3

>'

257

hence the stated result on simplification. The other results follow as particucases by using the preceding exercise for the moments of g.

IJf

22 Rider, P. R. (1951), JASA, 46, 375. 1 There are Nil ways of selecting a sample of 11 integers. A sample of 11 can , selected from (R+1) consecutive integers 0, (0+1), ... ,(e+R) in (R+lr ~ s But unless and (0 + R) both occur in the sample, the range will not :, 3YR: The number of samples of size 11 not containing either 0 or (0 + R) is and the number not containing both 0 and (0 + R) is (R - 1)". Therefore th; number of samples having a range R is

e

r

(R+ l)1I-2R"+(R-1)" for R #- O. [here are (N - R) subsets of (R + 1) consecutive integers. Hence peR) for R #- O. To sum the probabilities, use 9- I

I

[(R+l)k_Rk] = Ok-1.

R=I

[he mean and variance of R are also obtained by using this method of summation. Thus N-I

I

E(R) = (N-1)-2N-1I

R"

R=I

2 (N-Ir+1 -(N-I)-NII' n+l

(N-1)(1l-1) 11+1

,lIld E(R2) = N-"[N"(N-l)2- 6 2 [

Rtl

R"+1+2N

6 2]

~ (N-l) 1- 11+2+ n+1

Rtl R"]

(N-l)2n(n-l) (n+1)(n+2) .

123 Stuart, A. (1952), JASA, 47, 416.

E(D)

!; var(Dj )

!.

Also, there are six possible permutations of three unequal numbers Xj' X j + I' X j + 2 ; and each permutation gives a pair of values ior Dj and Dj + l . Of these six only one pair gives Dj = Dj + 1 = 1. Hence fWPj+ d = ! and cov(Dj , Dj + I) = - /2' But cov(Dj , D j + r ) = 0, for r > 1. =

=

Alternatively, (X, - X, _ d is a normal variable with mean ~(/ Also, corr{(X, - X, _ d, (X, + I - X,)} = -!. Hence

f1

and variance

P(D,-I = 1) = P(X,-X'_I > 0), ,md

P(D,D'_I = 1) = P(X,-X'_I > 0; X'+I-X, > 0).

124 Stuart, A. (1952), JASA, 47, 416. E(C j ) = 1; var(Cj ) = t· To determine E(CjC j + d, compare two sets of the

'IX. ~ossible permutations of three unequal numbers. Of the thirty-six possiolitlles, ten lead to CjC j + I = 1. Therefore E(CjC j + I) = 15S'

258

125

EXERC'lSES IN PROBABILITY AND STATISTICS

Stuart, A. (1952), JASA, 47, 416.

=

P(C j

1)

=

2P(Xj +,-Xj > 0; lj+,-lj> 0).

To obtain this probability, a simple orthogonal transformation shows that P(Xj+,-Xj > 0; lj+1 -lj > 0) '£00

= 2nR'

JJ

exp {-

o

where

x

2(I~P2)' (x 2+ y2-2pxy) }. dx dy;: / (saYI

0

= (Xj + 1 -Xj }/j2; Y = (lj+1 -lj)/j2.

*

But J = + (l/2n) sin - I P by the medial dichotomy theorem of Shepp d This result is proved by formally inverting the characteristic function and)' to obtain '

:r

0/ ( 21 ) 2 JOO Joo exp{-1(t,+t I 2 2 } 1 1 i)= 2+2pt,t 2) dt,dt 2 = - · _ _ _ . p

n

_ 00

2n

_ 00

JI=P2

Similarly (using particular suffixes 1 and 2 instead of j and j + 1), P(C,C 2 = 1)

= 2P(u 1 > 0, VI > 0; U3 > 0, V3 > 0)

+ 2P(u I >

0, VI > 0; U3 < 0, V3 < 0).

The inversion of the characteristic function gives the stated result fill P(1I 1 > 0,1'1 > 0; 113 > 0, tl3 > 0). In the same way, after sign change U[ I, and (4 in ¢, integration gives P(Ul> 0, VI > 0; 113

1

< 0, V3 < 0) = 4n 2[(sin -1 pf -(sin -1 pj2)2]+

1

+-[sin- 1 p+sin- 1 p/2] + a constant. 4n Hence cov(C" C 2), which still involves an arbitrary constant. This is evalualed by using the condition that for p = 0, cov(C" C 2 ) = l6' (from preccdillf exercise). Chown, L. N. and Moran, P. A. P. (1951). B, 38, 464. Use the results of Ex. 125. For C j = Cj + 1 = 1, the contribution P(CjCj + 1 = 1) is

126

~[(sin-l p)2-(sin-l pj2)2]+~. sin- 1 p+ n

n

a constant.

Also, when C j = C j + I = -1, then CjC j + I = 1, and the contribution this to P(CjC j + 1 = 1) is obtained by changing the sign of p. Therefore P(C j C j + 1

= 1) =

hI

[[(1111

2

2" [(sin- 1 p)2-(sin-l p/2)2] + a constant.

n

Hence E(CjC j + 1), the arbitrary constant being determined by using th .. condition that for p = 1, E(CjCj+d = 1.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

259

The inequality for var(g) is established by proving that

t

(~sin-l P/2) ~H~sin-I P This is deduced simply, since for 0 ~ P ~ 1 sin - 1 p/2 - hin - 1 P ~ O.

127 Cochran, W. G. (1937), JRSS, 100, 69. P(u j > 0)

f

"" = _fol . exp{ -(u j-pf/2I1 Z } dUj = t+rb(.). 2n

11

o

lienee mean and variance of S. Also, var(S)

e::r'=t'

= var[rb('*)] -

var(r*).

Use the independence of Ii and s to obtain the moments of i by noting that for any given r such that n - r - 1 > 0,

Stirling's approximation gives

Cn

,...,

e in so that c; - 1 + 3/2n.

128 Halperin, M. (1952), J ASA, 47, 457.

fo .f

en

P(x

> T)

=

e-%2/ 2 dz,

II

.md the logarithm of the sample likelihood is

1 logL = constant-rlogl1--z ' 211

r

L j=

Or

rOz

11

2

(x j - T)2+_(T-x)-- +

1

'YO

+(n-r) 10g[f e-%2/ z dzl 8

lienee the results stated by standard procedure. 129 Halperin, M. (1952), JASA, 47, 457.

fo'

8

P(x

<

T)

=

f

e-%2/ 2 dz,

-'" Jnd the logarithm of the sample likelihood is 1 logL = constant-nlogl1-?"2' _(1

n

L

11

j= I

8

-

nO (xj-T)z--(x-T)-

n~2 -n 10g[f e-%2/z dz -co

l

260

EXERCISES IN PROBABILITY AND STATISTICS

Hence equations for 1 and whence result since g(O) > o.

e.

Elimination of

a- gives a quadratic in . g(O),

130 Deming, W. E. (1953), JASA, 48, 244, and UL (1964). (i) P(X) =

(~)pXqN-X;

P(xIX) = (:) (~r

(1- ~r-x ;

[O~X~N;

(ii) P(X) =

(~)pXqN-X;

p(xIX) =

O~X~n].

(~)(:=:) / (~); [0 ~ X ~ N; 0 ~ x ~ min(n, X)].

(iii)

P(X) =

(7)(:-~)/ (~);

p(xIX) =

(:)(~r(I-~r-X;

[0 ~ X ~ min(N,Mp); 0 ~ x ~ (iv) P(X) =

(7) (:-~) / (~);

(~)(:=:) / (~);

min(N,Mp); 0 ~ x ~ min(n, x)). Work in terms of factorial moments. First evaluate E[x(r)lx] and then average over X to obtain E[x(r)]. [0

~

p(xIX) =

nJ.

X

~

131 The joint distribution of x and y is

(Nq) (NI1-X-Y - Np -Nq)/ (:), (NP) x y

(0:;;:; x,

Y:;;:;I1;O:;;:;X+Y~Il).

11

Therefore

y n n- ( x ) E [- X] = L L - P(x, y) y+l y=Ox=O y+l

f (_1)(Nq) n~Y(NP-l)(N-Nq-l-NP-l)/(N-I) y x=l x-I n-y-I-x-I n-I n-I( 1 )(Nq)(N-Nq-l)/(N-l) = np Y~O y+l y+l n-y-l n-l np n-I (Nq+ 1) (N -Nq-l) / (N -1) =Nq+l'y~o y+l n-y-l n-l

= np

y=o y+l

= N::l

[(~) - (N-~q-l)]/(~~D, whence the result.

In the same way, E[x(r)y(s)] is obtained by repeated use of the identity

f (M)u (N-M) = (N) n-u

u=O

n

Hence the moments of x and y. To obtain the coefficient of variation of

x/(" + 1) use the following result:

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

261

If X and Yare random variables with coefficients of variation VI> V2' and rrelation p, then, as a first approximation, the coefficient of variation of ,0 . ( 2+ 2_2 )t .\'If IS VI v2 PVI v2 • 31 UL (1964).

I folloWS directly from first principles.

33 UL (1964).

I The estimates are: cr:* = (x+y+z+w)/4; In = (x-y+z-w)/4; R.",(-x+y+z-w)/4; x,y,z,w being the averages of the x, y, z, and w pI • observatIons. . The error sum of squares IS n

M ==

L [(Xj-X)2+(Yj- y)2+(Zj_Z)2 +(Wj-w)2]+n(x+ y- w-z)2/4, j=

I

with (4n-3)d.f. The t statistic for testing the hypothesis H(fJ 1 = fJ 2) is

(x-y)[n(4n-3)/2M]t,

with (4n-3) dJ.

134 UL (1964). The random variable r has a binomial distribution and

P(r) = Hence the estimate

C)(1-e- )'(eT8

T8 )n-r,

(0 ~ r ~ n).

0 and its variance. For large samples,

var(1/0) - 8-4 yare B) = (e TO -

I)j11 (8 2 T)2.

135 Epstein, B. and Sobel, M. (1953), J ASA, 48, 486. XI

Since P(x ~ x r ) = e- xr/8, the joint distribution of the observations ~ Xz ~ ••. ~ Xr is

(_;'\1 't' o-r. n r.

.

.± Xj/O) . [e-

exp (-

n

xr/8 ]n-r ..

J= 1

J= I

dXj'

whence the estimate O. The Yj are independent random variables with joint distribution

it (n-~+1)

e-
(0 ~ Yj

< 00).

Therefore the characteristic function of r

0= L:

(n-j+1)Yj/r

j= 1

is (l-itO/r)-r, so that 2rO/O is a X2 with 2r dJ. Note that Xr

and as the mediately.

=

L

Yj'

j= 1

Yj are independent, the moments of Xr stated in

(ii) follow im-

262

EXERCISES IN PROBABILITY AND STATISTIC'S

136 Durbin, J. (1953), JASA, 48, 799. The estimate P2 is obtained by minimizing n

==

Q

L

CYv-b IX lv-y-P2 X 2v)2

v= I

with respect to y and P2' (y == a+ PIX 1+ P2X 2)' For the estimation of cr 2, the residual sum of squares S: may be expre sscd as n

S; == L

(yv- b IX lv-P2 X 2V)2

v= I

v= I

by using the normal equation for P2' But n

L

(Yv- b I X lv-P2X2v)2 =

v= I n

=

L (Yv-PI

X lv-P2 X 2v)2+(b l

-PI)2S 11 -

v= I n

-2(b l

-

Pd

L xlv(Yv- PIX lv -

v=1

P2 X 2.),

so that

Hence E(S:)

=

(n-2)cr2+(1-r2)cr~Slh

whence the unbiased estimate of cr 2. For large samples, use (P2 - P2)/ S.E. (P2) approximately as a unit normal variable. 137 David, F. N. and Johnson, N. L. (1952), Bs, 8, 275. The moment-generating function of x is E(e tX )

= [exp{ A.(e t -I)} - e- A]/(1- e -A),

whence the first four moments of x about the origin are: Ji.'1

= A./(l-e- A);

Ji.~

= A.(A.+I)/(I-e- A); Ji.3 = A.(A. 2 +3A.+I)/(l-e- A);

Ji.~

= ,1.(,1.3 + 6,1.2 + n+ I)/(l-e- A).

Hence Ji.~ = (1 + A.)Ji.'I, which suggests the moment estimate. A logarithmic expansion gives log(l +,1.*),...,

10g{!!.~) + [m~~Ji.~] _t[m~~Jl~J2 _[111'1 ~Ji.'I] +trl1l 'l ~Ji./I]: \it I Ji.2 Ji.2 Ji.1 Ji.1

I

(I

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

263

<Xl

* Px*-I"IN mr' - 't..." x r Px, - JxI • x~l

E[(m~-JL~)(m~-JL~)] = ![JL~+s-JL~JL~]. Therefore (1) gives E[log(1 + ,1,*)] and var[log(1 + ,1,*)].

138 This generalization follows closely the method of the preceding exercise. k-l

If h(,l.) ==

L:

,1,r/r !, then

r~O

E( IX) e

=

exp{A.(el -1)}-e-J.h(,1,el ) 1-e-J.h(,1,)

Therefore the moments of x about the origin are: JL2 = ..1.(..1.+ 1)+(k+,1,)(JL'I-,1,); JL3 = -(k-1),1,(,1,+k+ 1)+ {P+(k+2),1,+,1,2}JL'I; and JL~ = -(k_1),1,{,1,2 + (k + 4),1,+ (k 2 + k+ 1)} + + {k 3 +(k 2 + 3k + 3),1,+ (k + 5),1, 2+,1, 3}JL'l'

Thus

,1, = <JL2 - kJL'l )/(JL'l - k + 1),

whence the definition of A*. Also, log(,l.*+k)

=

log(,1,+k)+ log [1 +

(k+;)\;'l~k+ 1)] -log [1 + JL7~~j 1J.

The results now follow by the method used in Ex.137.

= a 2 and var(s;) = 2a4 /(n - 2). The results are obtained by a straightforward application of normal regression theory. 139 E(s;)

140 P(Y = 0) = q-m, and so for the truncated distribution

P(X = r) = ( m+r-l) r pr q-(m+r)/(1_ q-m),

for X

~

1.

Therefore E[X(S)] =

(lII+r-1) I I . I' prq-(m+r)/(l_q-m) r~s (r-s) . (m-1) .

L ex

= (m+s-1)(S) pSq-(m+s)

t

I ~O

whence the stated result.

(111+ t+ S-1) (p/q)I/(1_q-m), t

264

EXERCISES IN PROBABILITY AND STATISTICS

The logarithm of the sample likelihood is log L Hence

=

constant + Nx log p-N(m+x) log q-N log(1-q-m).

p and its variance

by noting that

E(,x) = E(X) = mp/(l_q-m).

For the moment estimate p*, log[l +(m+ l)p*] = log[l +(m+ l)p]+ log [1 +

(V2 ;~J1~)

]-log [1 +( ~)]

where J1~ is the sth moment of X about the origin. The large-sample result. for p* are now obtained by applying the method of Ex. 137 above. S UL (1964). Let Xl' X2 and X3 be the observed frequencies in the AA, Aa and aa classes. Then ~ = (X2 + 2x3)/2n, and the exact variance of ~ is obtained by noting that 0 is a linear function of frequencies from a multinomial distribution. 141

A

0* 142

(~:

=

r,

var(O*) '" var(x3)/4n 202.

and

Fisher, R. A. (1922), PT RS, 222, 309.

f

nl2

l(p)

=

e- 8 cosPO dO,

-n/2

and the recurrence relation is obtained on integration by parts twice. the definition of l(p),

_E[a21~!!(X)]

=

[2(P+2)I(P+4)-(P+2)I(P+2)+

Usill~

2:;::~)]1 I(p),

whence the stated result. Also, E(X -IX) = -l/p

and

E(X _1X)2 = l(p-2)_1 l(p) .

Hence the moments of the sample average

x.

143 Peieris, R. (1935), PRS, 149, 467. The probability of a particle being emitted in (v -l)T ::( t ::( vT is vT

fO-le-tIO.dt

= e-(V-I).t(1_e-.t).

(v-I)T

Hence the moments of X and Y, since nv have a multinomial distributioli The approximate variance of X/Yis e.1.(e.1.+1)2/N, remembering that cov(X, Y)

=

-N e-.1./(1+e-.1.)2.

Hence var(O*) and var(O*) ~

(e.1.+ W Ne.1.A,2·

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

265

For fixed N, the right-hand side is minimized for A '" 2'4, whence the , equality for the variance. In A logarithmic expansion of eTIO " = 10g(X/Y) about the means of X and Y TIO") ives E(e . g The maximum-likelihood solution is obtained simply, as the logarithm of the sample likelihood is NT NTv log L = constant+N 10g(l-e- TIO )+e--e-'

144 peieris, R. (1935), PRS, 149,467.

The probability of a particle being emitted in (0, T) is (1- e - T/~ so that

E(N -n)/N = e- TIO , whence the equation for

e*.

Also, var(e*) is minimum when (e).-I)/A 2 is a minimum for e). = 2/(2-A), an equation satisfied for A '" 1·6. The maximum-likelihood solution is direct. 145 For a general discussion of the five estimates vide Fisher (1946), p. 301. The estimates (}l and e2 are linear functions of multinomial frequencies. Hence the unbiasedness of the estimates and also their variances. For

e, let F == log (X1X4) X 2X3 N

Since F is a function of the

= log

[0(2+0)] (1-;- 0)2 .

Xi' therefore for large n

var(F) = 8( 1 + 2e)/ne( 1 - e)(2 + e).

e, so that

But F is also a function of

var(F) '"

(~):=o. var(O).

lienee yare 0) = 2e( 1 - e)(2 + e)/n(1 + 20). This is also the large-sample variance of (j obtained by a straightforward application of the maximum-likelihood procedure. The equation for ()* may be written formally as G(Xl,X 2 ,X 3

,e*)

=

O.

Then var(e*) =

L3 (00*)2 j=

1

OXj

Xi

.var(x + j)

= E(Xi)

where

146 The joint distribution of 1 2n' exp{ -

X

and y is _

!(CXX2

+ y 2cx l)} dx dy,

( - 00

<

X,

y <

CXl).

266

EXERCISES IN PROBABILITY AND STATISTICS

If LI is the likelihood for a single pair (x, y), then

_E[a2~~LI] = a-3E(y2) = a- 2. The n pairs

(Xi'

Yi),

(i

= 1,2, ... , n) give

a=

[.±,= yf / ,=.± xf]t, 1

and a2/a 2has an F

1

distribution with (n, n) dJ. Hence, if of a single observation of a, then log

L2 = constant -

L2 is the IikeIih

Ood

n log a - n 10g(1 + a2/( 2),

whence

E[(al~!L2r] =

:2. C:l).

147 For discussion vide Fisher (1956), p. 163. This follows closely the method of the preceding exercise. The distribution of t is

(e t)-

2 B(n, n)" t+e

2ft

(0 ~ t < 00).

t- I dt,

If L denotes the likelihood of a single observation of t, then alogL = 2n

ae

(!_~)/(!+~)

e·et et· Hence the information for estimating e is alog L)2] = (j2 4n 2[ (t e)-2] = (2n+4n 21)()2· E [( ----ae 1-4E e+t 148 Fisher, R. A. (1934), PRS, 144,285. For a single observation of x, the logarithm of the likelihood is log L 1 = constant -Ix -

el,

so that alog LI

ae according as

X

> or <

±1

e. Therefore

(al~~Llr

=

1.

To obtain the distribution of the sample median, consider deviations Yi = Xi - e. If Y > 0 is the deviation of the sample median, then in a sampk of (2s+ 1) observations (i) s deviations are > Y each with probability t e - Y; and (ii) s deviation) are < Y each with probability 1 - t e - Y. Hence the distribution of the sampk median.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

267

to evaluate the integral

f 1

t S(2-t)Sdt,

o

the transformation v

use •

=

t(2 - t), which reduces the integral to a Beta

rdO~~:~ single observation of II, the logarithm of the likelihood is logL2

= constant-(s+ 1)!u-O!+slog{2-e- 1u - 81 }, se-1u-81 ]2. ( alog L 2)2 _ [ (s+ 1) 2_e-1u-fll

ao

-

For reasons of symmetry it is enough to consider the expectation for the rAoge Z == U - 0 ~ 0 when the distribution of z is

2. (2s+ I)! (1 -z)s+1 (1_1 -Z)S d (S!)2 . "Ie 2e z. The evaluation of E[(O log L2/aO)2] now involves the integrals 1

J

t S+1 (2_£)"-I. dt =

-~+ (s: 1)

o

f 1

tS(2-t)s. dt;

0

and 1

f

1

ts+2(2_£)"-2. dt

s(s-l)

o

r

= - 2(s+ 1) + (s+ l)(s+2) tS(2-t)s. dt. s(s-l)

J0

The expressions on the right-hand sides are obtained by integration by parts and are readily evaluated. Finally, the approximation for the loss of information is obtained by using the r function approximation r(2s+ 1) 2S 2 {r(s+ 1)V '"

1

.jm'

which follows by using Stirling's formula.

149 Fisher, R. A. (1930), PRS, 130, 16. The joint distribution of .x. sand Zi' (2

~ i ~

n-l), is

11-1

n dz;

i=2

x (1 -Z22 Z3 2 -'" SO

that X, s and the

Zi

2 )!'

-Zn-l

are independently distributed. Next. since the x

--+

y

268

EXERCISES IN PROBABILITY AND STATISTICS

transformation is orthogonal, n

Yi

L cij(Xj-x),

=

for 2 ::;; i::;; n

j= 1

so that any (x j - x)/s is a linear function of Y2/S, Y3/S, ••• , Yn/ s, i.e. of Z3, ... ,Zn-l' Therefore X, mi and mom2 o/2 are independently distribul~' Hence E(m~m2oP/2) = E(m~)/E(m2P/2).

To obtain the expectations of the powers of m3 and m4' set Xi - J.I =:: so that (Xi-X) = (ui-ii). Then (ui-ii) are normally distributed rando u, variables with zero mean, variance [(n -1)/n](j2, and III for i #- j.

corr[(ui-ii),(Uj-U)] = -1/(n-1),

Thus

E( 2) = 6(n - 1)(n - 2) 6 . E( m3 ) = O·, m 3 3 ' (j , n E( 2) _ 3(n-1)(3n 3 +23n 2-63n+45) 8 m4 4 •a •

4. E(m4) -- 3(n-1)2 - - (j,

n

n

The moments of m 2 are known from general theory. Hence

E(gl) E(g2)

=

=

0;

var(gl)

-6/(n+ 1); var(g2)

6(n-2)/(11+1)(n+3);

=

2411(n-2)(n-3)/(n+ 1)2(11+3)(11+5).

=

150 Peierls, R. (1935), PRS, 149, 467, and Bartlett, M. S. (1936), PRS, l54. 124. The logarithm of the likelihood of the 11 emission times is log L

=

-

n log 0- n 10g(1- e- T /8 )-l1t/0,

where t is the average of the t i' Therefore the equation for where For

11 =

X

0 is

== T/O.

1. the information per observation is

c2 10g L] -E [ c0 2

=

1 [ ,{2e -A] 02 1-(1_e- A)2 •

Since the probability of observing a particle in (0, T) is (l-e- the expected number observed in the interval is N(1-e- A). Hence the expected information from N sources per unit time. For given Nand 0, this informatioll is maximized for A. '" 4. I .),

151

Fisher, R. A. (1942), AE, 11, 306. The likelihood of the sample is

L=

Ii (Ar)p~ro_ Pr)Ar-ar (C)pcO_ pf-c, a X

r=1

C

r

and on the hypothesis of independence k

P =

n Pro

r= 1

ANSWERS AND HINTS ON SOLVTIOl\S: CHAPTER

3

169

fhcrefore the estimates P~ and p* are given by Arp~-ar

P:

I-

=

c-CP* I-=- p*

= I"~ say.

Ileoee the equation for I., using k

n P:, ,.:=

P* =

I

I'

k

k

fl

(C-A)

TI

(ar+;,)-(C-A)

r= 1

(A r+},) = 0,

r= 1

. hieh has a real root in the range (-as, c), where as is the smallest of the a r • ,\ The goodness of fit X2 is

t

r= 1

(ar-Arp~)2

'2[ t

(C-Cp*)2 _ P*) -).

p:J + C P*(l -

ArP:O -

r= 1

Ar-ar C-C] Ar(ar + I,) + C(c - A) .

Ihis X2 has I dJ. since for any distribution of Or and c, I. = 0 if k

fl

P =1=

Pro

r=1

152 Tippett, L. H. C. (1925), B, 17, 364. .'\:,.

J f J.

E(w) = 11(11- 1)

X,.--

(oc n-octl,,-2 (xn-xtJ doc 1 doc"

X l - - :to

f...

k=O

.

II

J

J

co

k

,,_ 2

-- II., " ( - l) L.. k I ( _ k _ 2) I' .

Xu

II -

k- 2

(XII

d OC II

k (

OC 1

x,.----+-Cf)

xn _

Xl

)

doc 1 . d';x I • d' Xl

Xt--rf.)

Ihcrefore integration with respect to x I gives =

f""

(_l)k

11-2

E(w)

-/1! k;O (k+ 1)! (n-k-2)!'

dV docn U.-.-.dxn' dlX II dX II

where

JIX~

J

XII

U ==

OO

+ 1 dx 1

and

-

x"

-00

Ilr n-2

E(w) =

II!

L

dlX dx

1X"- k - 2 _ " dx II n'"

V=

J CG

( _l)k

k=O (k+I)!(/1-k-l)!' -

(l-oc:- k + 1)1X~+ 1. dx n ,

:/.;

Whence the result by reversing the order of summation and integration. Similarly,

J 00

n-2

E[(w-w)']

(_l)k

= n !k~O k !(n-k-2)!'

1X:- k -

xn -

-00

[lXk+1(_W)'

2 n k+l

rW]

+k+l docn,

270

EXERCISES IN PROBABILITY AND STATISTIC'S

where

f a~+!(Xn-x!-w)'-!dx!. .Tn

W==

-7.)

But

f w. a~-k-2

f

00

00

dan

xn--oo

= -

00

dV dan W'd '-d .dxn =

a"

-r.

x"

f

-00

dW V·-.dx dx" "

and

f 00

a:-! dan

.\ " n - -

= ~.

co

Hence the result. 153 Transform to 1/ and y = . . tegratmg out with respect to y.

X2 -111,

whence the distribution of

1/

on

III

154 Let PI be the true proportion of pre-budget government supporters wi continue to favour the government after the budget. Similarly, let P be II)" . 0 f pre- bd . true proportIon u get government opponents W h0 contmue to2 oppo.,h the government after the budget. . The expected government support after the budget is Nap!

+ N(I-a)(l- P2)'

and on the basis of the government's claim the null hypothesis is

If the hypothesis is true, then the logarithm of the likelihood of the pOll. budget distribution is log L

(I-a) ] = constant + nllog [I - -a- . (1- P2) + +(N -n! - n2) 10g(1- P2)+ n210g P2,

and the equation for the estimate p! is

n2(1-..1.)+ p![..1.(N +n2)-(N -nl)]-N..1.pr = 0,

[A. == (1 - a)/a).

Hence the goodness of fit X2 for the post-budget distribution is [nl - N(I-a)p! + N(I- 2aW [n2 - N(I-a)p!]2 + -=-:-=':----,:---:--:c-=--=_:_:_ N(I- a)(l- p!){ 1- [(1- a)/a](l- p!)} N(I- a)p!(1 - p!)

-,---:-'--:'---=:--.,...,-----'----:-:--;-":":"":'-=-:-,----'--:-:---:-:':"'::'-----:-:-:-

with I dJ. For a =

t, A. =

I, p! = (nl +n2)/N, and the X2 is as stated.

155 Jeffreys, H. (1932), PRS, 137, 78. For O'! = 0'2 = I, the probability density function of Xr is

~. {(I- p) exp[ -t{xr-lW]+ p exp[ -t{xr-O- Jl)2]},

y21t

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

e equations for the estimates fJ, and .J Ih

fJ are

n A n

L (Xr -

0)/(1 + Ar) = 0

L

and

r'" 1

3

Ar(Xr -

fJ - (J.)/( 1 + Ar) = 0,

r= 1

Ar ==

(1 ~ p) . exp[fJ,(xr - fJ - fJ,/2)].

. elements of the information matrix are evaluated by noting that

f fo _

Ir.C

1

00

2

00

f fo 1

y2 e - Y /2 d y 1 + A e I'Y

00

y e- Y

-00

2

/2

J.l('C 1 -

=

'C 2) ;

dy

(I+Ae-l'y)2

= 2J.l('C2-'C3)·

these results are deduced by using integration by parts. 156 Tippett, L. H. C. (1932), PRS, 137,434. I

P,

=

and

e-I'J.l'/t!

P,

L e-l'J.l r/r !

=

r=O

[he likelihood of the sample is N! I L= I • (Pj)"J(1-P,)"R. j-O nR! nj!-

n

n

j=O

lienee the equation for

p. may be written as fJ, =

X/[I-(N~RnJ(:JJ.

where fi,

== N

e- fJ,'/t! il

and

fiR

== N

[1-e-

Ji

±

fJ,j/j!] .

j=O

157 Finney, D. J. (1947-49), AE, 14, 319. The point probabilities for the truncated distribution are P(r) =

C)

pr qn-r/(1_ pn - qn), for 1 ~ r

~ n-1.

lienee E() r ,I

= np(l_ pn-l). E[ ( -1)] = n(n_l)p2(I_pn-2) 1 -pn -q n '

rr

1 -pn -qn

hence var(p*). The maximum-likelihood estimation is straightforward.

'

271

272

EXERC'lSES IN PROBA B I LITY AND ST A TISTI C'S

158 The joint likelihood of the k samples is L=

Ii

j=

I

('.l)pr i q"-r i /O_ q")k,

E(I'j)

and

=

np/(I-q").

, j

159 Haldane, J. B. S. and Smith, C. A. B. (1947-49), AE, 14, 117. (i) is obtained by considering all partitions in which largest aj

~ (11-\ and those in which the largest aj = n. J. (ii) follows by considering the complementary set (n + 1) - a. r . h " lOr . 1, 2, .•. ,1' )WIt (1= r

L [(n+ I)-a;] = I'(n+ I)-A. i= I

For (iii), note that for any ranking of the abnormals, the ranking of th. normals will consist of (n-I') different integers, none >n but summing tt tn(n+ I)-A. II The function F((); n,I') satisfies F(O; n, 1')

=

F(O; n-l, I')+O"F((}; n-1, 1'-1)

whence, on equating coefficients of OA, I(A; n, 1'), the coefficient of OA in F(O; n, 1'), satisfies the same relation as S[A; n, 1']. Also for n = 1, I(A; II rJ = S[A; n, r]. Hence the probability-generating function of A. ' The cumulant-generating function ,,(t) is expanded in terms of Bernoulli numbers as ,,(t) =

-lOge) +tr(n+ 1)t+ sinh (n - j + 1) t ] + log [ 2 .. jt/2 . n-j+1 j~1 (n -; + 1) t smh(jt/2) j r

r

'h

=tl'(I'+I)t+L j=

Note that B2

= !; B4 =

(-1)'"+ I t 2m

L1m. 2 (2 )' m.

I m=

.B2m [(n-j+1)2m-lm].

lo.

160 Harris, H. and Smith, C. A. B. (1947-49), AE, 14, 307. If x and y be the ages of onset for a random pair of parent and child, then E(xIA)

=

E(yIA)

= 111.;

E(x 2IA)

=

E(y 2IA)

= m~+a~;

and E1x.rIA) =

mf·

Similarly, and E(xyIB) = m~.

Hence E(x)

=

E(y)

=

1tlml +1t2m2;

E(.\:2) = E(y2) = 1t.(ar+m~)+1t2(a~+m~);

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

273

I !-larris, H .. ~nd Smi.th. C. A: B. (1947-49), A E, 14, 307. 16 The probabIlity denSIty functIOn of the age of onset x is

fIx)

=

~[e-t(X+m)2+Ae-1-(X-III)2]/(1+A)'

(-co < x
y2n

For Ihe stationary values of x, g(x) == log).+2I11x+log(lIl-x)-log(m+x) = O. Bul g(x) is real only for - III < X < III, and -+ ± oc, according as x -+ += m. lor a multimodal distribution g(x) = 0 must have more than one root in _/II

<X<

III.

Now g'(x) = 0 gives x = ±(m 2 -1 )1-. Hence for r~al roots of g(x) = O. This will be so if min g(x) < 0

III

> 1, there must be three

max g(x) > 0

and

ior x == _(m 2 -1)1- and x = + (m 2 -1)t. Hence the result. For unequal variances, set

~ == !(~+~); (12 2 (1i (1~

15 ==

!(~-~), 2 (1f (1~

so that 1151 < (12.

Therefore for the stationary values of g(x) bx 3 +m(1-2X2-m 2bx+m(1-m2(1-1.) = O.

(1)

Bula cubic aox3+3alx2+3a2x+a3 = 0 has three real roots if

a&a~-6aOala2a3+4aoa~+4a~a3-3afa~ < O.

For (1) this gives

!!!. >

3 . [1 +(30:+ 1)t1 t • (12+(3o:+1)t 2 '

where (X == 15 2 (14, so that 0 ~ 0: ~ L Hence least separation is 2m = (11t)!.

162 Haldane, J. B. S. (1932), JG, 25, 251. Let the number of families of size shaving k abnormals be nok' Then the likelihood of the total sample is L=

IJ c

[

s-I

n I . so'

Il nsk I.

IJ S

{ (

kS) pk qO-k/(1_ qS) }/1

5k ]

,

k-I

k=1

whence the equation for

R

=

.tl ktl

q.

Also,

knob and

E(nok)

=

nsG)pk qs-k/(l_ qS).

274

EXERCISES IN PROBABILITY AND STATISTICS

Hence c

=

E(R)

L s. ns . p/(1 -

qS),

and the information for the estimation of q is

_E[iJ

2

10g

iJt/

L].

If families with at least two abnormals are considered, then the probabT that such a family of size s will have k abnormals is III) (2 :::;

k :::; s).

The maximum-likelihood equation for q*, the estimate of q, is now sn~(l-q*'- I)

c

'~2 [1-sq*s-1 +(s_l)q*S] =

R' l-q*'

where c

R'

S

L L k. n.k .=2k=2

=

The total information for estimating q is c

L s. n~ [(1- q.-t)2 -

(s - Wp2q'- 2 ]/pq(l- spq·-t _ qS)2 .

•= 2

Haldane, J. B. S. (1956), JG, 54, 294.

163

The point probabilities of the distribution of rare P(r)

= (~)er(1 +o)-n, for 0:::; r:::;

n.

Therefore

E[n-r+ r ] 1

f(

-

r=O

=

r )p(r) n-r+ 1

0[1-

:or]

C

--+

0 as n --+

00.

Since r has a binomial distribution, the moments of x obtained in terms of p == 0/(1 + 0) and q == (1- pl. But log(

r

x

n-r+l

[(n+ 1)/n]{I-[nx/(n+ I)])'

r

n-r+

A 0==

A

= n+l . A t -

p(nq+ 1)'

2

==

r/n are readily

so that

1) '" log A o +A t (x-p)+A 2 (x-p)2,

np nq+ 1;

=

where

n2p2_(nq+l)2 2p2(nq+ 1)2

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

3

275

fherefore

[

r n-r+l

~

Ao]

A o[A 1(x-p)+(A 2 +Af/2)(x-p)2j.

nee the mean and variance of ,./(n-,.+ 1). »e The series for I/z is proved by summing the differences

r!(z-l)! r!z! _(r+l)!(z-I)! f 1 , or ~ r (z+r)! (z+r+l)! (z+r+l)!

-

0

n!(z-I)! (z + n)!

Jod noting that

--+

,

as n --+

If we exclude samples for which r

E[~]

=

II-r

1

n- ,

OJ.

then

nil (_1 )(n)prqn-r/(l_pn) n-r f !(n)pll_zqz/(l_pn) r

r=O

z= 1 Z

=

= n,

~

Z

n'k' n - z z/( 1 _ n) k=OZ=l(Z+k+l)!(n-z)!'P q p

L L

= (1- Pn) -

..

n

00

1[

~ n !k ! -(k+ 1) n ~ k~l n ! k ! oj ] t.... •q - P t.... t.... . , , • k=O (n+k+ I)! k=O j=O (k-)+ 1). (n+}).

Therefore, for large n,

[ I]

E-

11-"

164 Let

Xj

~

00

'k'

n..

k~O (n+k+ 1)! .

(1 + O)k+l

E[-I" ] n-r

and

=

nE[_1 n-,.

]-1.

be the number of non-defective units inspected after the (i -1)th

and up to the ith defective unit observed, so that

j=

1

Each Xj is an independent random variable such that P(x j

=

v)

= pYq,

for v

~

o.

Hence the cumulant-generating function of z is ,,(t)

= n log q - n 10g(1- p etln ).

Therefore, for the cumulants of z, KI = 0; "2 = O(1+8)/n; "3 = 0(1 +8)(1 +20)/n 2 ; "4 = 0(n- 3 ); but the fourth central moment /14 = 30 2(l +O)2/n2+0(n- 3 ), (0 == p/q). Hence

E[Sn(n + Sn)/(n+ 1)] = nO(I+0). Finally. p* = z/(l + z) so that 1 )[(z-O) (Z-0)2 (Z-0)3 (Z-O)4 ] p* - p = ( I + 0 I + 0 - 1+ 0 + 1 + 0 - 1+ 0 + . .. , ,Ihcnce the mean and variance of z.

276

EXERCISES IN PROBABILITY AND STATISTICS

165 The logarithm of the joint likelihood of the two samples is log L = constant + (XI + X2 + 2YI + Y2) 10g(l- 0)+

+ (X2 + 2X3 + Y2 + 2Y3) log 0 + XI 10g(l- 0+ 2pO) + x 2 log(l_ pl. Hence for the estimates

p and 6 (2x I +X2)0-X2 . an d 2~(xl +X2) ,

P-

(2X3 + Y2 + 2Y3)- (2v - Y2 - 2Y3) ~ - 2(v + n) ~2 = 0, For the elements of the information matrix,

- E

-E

[ o210gL] (102

2 [0 log L] Op2

-E[~ 10gL] op 00

2n{(I-p)+(3p-l)e} 2nO(I- ( 2) = (l-p){I+(2p-l)e}; =

Hence the covariance matrix of

166

2v

= 0(1- O){ 1+ (2p - 1)0} + 0(1 - 0);

_

2n(1-0)

{I +(2p-l)0},

p and 0.

'3 =

(k - 1)[(k + 1)+ (n - 2)p]ex + [1 + 3(11- 1)p + (n - 1)(11- 2)p2 Jr I ;

'4 =

(k-l)[(k2+k+ 1)+{(n-3)(k+ 1)+3(n-l)}p+(n-2)(n-3)p2)c(-f + [1 + 7(n-l)p+ 6(n-l)(n - 2)p2 + (n -I)(n - 2)(n-3)p3Jr(.

These moments may be obtained by direct summation or, more easily. from the recurrence formulae given by Ayyangar (1934), B, 26, 262. Since E(T,.) = Or> P.= E(T2 -kTd/[(n-l)E(TI )-n(k-l)], whence Ih, equation for p*, For the moments of p*, consider

k+(n-l)p*

=

'2-kA. 'I -A

[I + T2-'21/[1 + Yt -'I], '2 -k]

'I-A

where A == n(k-I)/(n-l), Expectations after a logarithmic expansion or Ihl' right-hand side give the mean and variance. Since d'o/dp = ex'o/pq, the maximum-likelihood equation is p = l/II(T(-il. where & is obtained from ex by substituting p for p. Also, var(p)

= p2q2/N[npq-('1 -np)('1 +p-k)],

by using dex ex -- = ---('1 +p-k). dp pq

167 Cohen Jr., A. C. (19601, JASA, 55,342. The distribution of X is: (I-O)A e-).

P(X

= 0) = 0; P(X = 1) = 1 _ (1 + AO) e -). ;

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

e - 'l,~.' P(X

=

277

for X ~ 2.

r) = [1-(1 +AO)e ).],. !'

,.. mean and variance are

3

obtained by direct summation.

10-

Set

A(A) == ';011'

e-'<[(I-A+A 2 )-e-'<] Ae-). l-e-).(1 +A) and B(A) == l-e-'<(1 + A)"

A(A) > 0, B(A) > 0 and var(X)

A 1-0A(A) 1- e '«1 + A) . [1 + (1- O)B(A)j2'

=

lIence

.nd ~

var(X)

A![I-e-).(1 +,{)].

168 Cohen Jr., A. C. (1960), JASA, 55,342. 'Xl

Let

/Ix

be the number of nests with x eggs each, so that

L

nx

= (N -

n).

x=2

Ihe logarithm of the sample likelihood is logL

= constant+n log(I-0)-AN +Nx 10gA-N 10g[1-e-).(1 +AO)],

.• here x is the sample average, so that Nx = n+(N -n)x*. The equations for land (J are

x -::-l _ e - :i(1 + ~O) - ~ ~ -1 = 0 1- e - ).( 1+ AO)

A

(1)

tnd

_ n + ~

Iolving for

N e-1l = O. l-e-1 (l +AO) AA

(2)

8, (1) and (2) give A

_

0-

~-x(l-e-l) _ N~ e- 1 -n(l-e- 1) A'

A e- ,t(1-x)

-

A ' ·

(N - n),{ e-).

(3)

lienee

x*

=

[

l

-1

]

l 1+1-(1:l)e- 1 '

.hich may be rewritten as e1-l = x*A/(x* - ~). he this t9 eliminate e1 from the first equation of (3) to obtain the stated i(,uit for O.

278

EXERCISES IN PROBABILITY AND STATISTICS

For small A,

1-(1+A)e-). = A. 2 (1-~A+ A. 2_ ... ) '" A. 2 e- 2 )'/3 2 3 4 2'

l

whence the approximation for For large samples, var

(X) _

~ [l-(1+A)e-).][1-(1+A.O)e-).].

[(1_e- A)2_A.2 e-A]

- N'

,

(0) = (1-0)[1-(1 +AO) e- A][(1-e-).)(1-(} e-).)-A(1-(}+A.O)e- AI var NAe A[(1-e )')2_A 2 e-).] ----=; and

0) _ (1-0)(1-A.-e-).)[1-(1+AO)e-).]

A

cov(A.,

N[(1-e-A)2 _A2 e A]

-

Plackett, R. L. (1953), Bs, 9, 485.

169

00

Let A* =

L a,n,/N. Then A* is an unbiased estimate of A if

,= 1

which determines the a,. var(A*)

= -1[00 L r2p,_ N ,=2

{OO rp, }2] , ,=2

L

where p, == (

;'"

A

e -1)r!

and

_ A (l-e-)')2 var(A) = N' 1 _ (A + 1) e -).' 1 Write

T ==

-2

N

whence eft' (A *).

10

L b,n, ,=2

so that

But 00

'L

L b,p, = 2P2+ Lrp,

,= 2

=

A+,F/(e).-I),

,=2

00

L b; p, =

CfJ

12p2 +

,=2

L

,= 2

r2p,

A. + 6,1,2 /(e). - 1)+ A2 e)./(e). -1).

=

Hence var(T).

170

Plackett, R. L. (1953), Bs, 9, 485.

P(X

= s) = AS/(e). -1)s! == Ps' say, for s ;;:: 1.

00

E(T1 ) = N

L

,= I

00

P2,-1

and

E(T2 )

=

N

L

,= I

P2,'

;

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER 3 ~ence

E((}*)

279

= e- A• =

var(T.)

NL~, P2r-l- (~/2r-lr].

2)= NL~, P2r- C~I P2r) 2].

var(T

co

cov(T1• T2) = -N

and

co

L L P2r-lP2s'

whence var((}*).

r = I s= I

i1I Cohen, Jr., A. C. (1960), Bs, 16,446. The distribution of x is: PIx = 0) = (I-());

PIx

= r) = o;.:/(eA-l), for r

~ I.

[he estimation and the variances of the estimates are direct. Set

I/I(l) == (1-e- A)2/[l-(1 +l)e- A]. rhen lim I/I(l) = 1 and

A-+co

lim I/I(l) = 2.

A-+O

Also

d log I/I(l)

d}.

e- 2A[(2 + l)(e A + 1)- 4 eA] (l-e A)[l-(t +l)e A]'

lienee t//(l) ~ 0 for l ~ O. Therefore the inequality for var(l). 171 Peto, S. (1953), Bs, 9, 320. The probability that ni organisms administered to a guinea-pig in the ith group do not effect a kill is (1- p)n, ...... e-n,p. Iherefore the likelihood for the whole experiment is

L

=

Ii (~i)(e-n,pY'(l_e-nfpr·-r'. r,

i= 1

lienee Pand its variance. The information obtained from n organisms per animal in a group of 1/1 guinea-pigs is In =

mn 2 e- np/(1-e- np ),

dod for variation in n, this is maximized for a value of n given by "p::: 2(1- e - np ), with root np = 1· 5936. The most economical expected proportion of survivors is e -np . . . 0·203. 173 Cochran, W. G. (1950), Bs, 6, 105.

280

EXERCISES IN PROBABILITY AND STATISTICS

The equation for

J is

J=

Jl (ni-s/)~i(l-e-il)-1 ~tl

v~!c5) = 1/~1 n/x~(e,xI-l)-I, A

(~i == Jv/)

nivi>

"

1

1

~-

".... N'

(Xi

== c5v/)

1·54

(X2)

=--

N

max~l

e -

since

x 2 j(e,x-l)

is maximum for

X

= 1·5936.

174 First part is Query 78, Bs (1950), 6, 167.

Let the group means be x and ji based on nl and n2 observations res . tively. so that n = n t +n2 and var(x- ji) = (1Un. +(1~/n2' pc\. For (i) minimize (12 (12 ---.!.+ __ 2_ with respect to n1 .

nl

n-nl

Hence n 1 = n(1d«11 +(12) and n2 For (ii) minimize

n=

= MJ«11 +(12)'

en 2+n (12_(12) 1 1 22 1 with respect to nl' enl- (11

Hence nl = (11«11 +(12)je; n2 = (12«11 + (12)je. For (iii) solve the simultaneous equations and

(1~

(1~

-+-=e. nl n2

This gives the quadratic in n1 en~-(en+(1~-(1~)nl +(1~n

= O.

A necessary condition for it to have positive roots is that

en + (1~ - (1~ > 0, and the roots are real if

(en + (1~ - (1~)2 - 4ne(1~ = [en - «1~ + (1~W - 4(1~(1~ ~ O. For (11

= (12 = (1, nt

=

~ [1± (1- ~:~r}

n2 =

~ [1+ (1- ~:~r].

175 The argument is similar to that of the preceding exercise with t = III tall) instead of n = n 1 + n2' For (i),

nl = t(1.1«11 +(12~); n2 = t(12j~«11 +(12~)'

For (ii), nl = (11«11 +(12~)je; n2 = (12«11 +(12~)je~. For (iii), if et+(1~-IX(1~ > 0 and [et-«1~+IX(1~W-41X(1~(1~ ~ O. then

n1 =

~ [{I + (1~~;(1~}± {I

«11

+::~)2r {I

«11

-::~)2r].

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

n2 == ;rt [{

1- ut ~trtu~} +{I

(Ul +

3

281

::ja)2f{1- ::ja)2fl (u 1-

16 Bliss, C. I. (1953), Bs, 9, 176, and the accompanying note by R. A. Fisher. I

The probability-generating function of X is <Xl

L

Ox. P(X = x) = (1 +p_pO)-k.

x=o

lienee ,,(t). The first four cumulants of X are:

"1 = kp; "2 = kp(1 + p); "3 = kp(1 + p)(1 + 2p); "4 = kp(1 + p)(1 + 2p)(1 + 3p). The logarithm of the sample likelihood is log L = constant + Nx log p - N(k + x) log(1 + p) + <Xl

+

L Ix log[k(k+l) ... (k+x-l)],

x=o

\,hence the equations for p and t The covariance matrix follows directly. (i) Set p~

= Ix/N, for all x. Then the equation for k* is

~ * - (p*)-I/k 1+ ~ k* . ;/::\ xpx 0

O

,

which may be written formally as

k*

= F(P~,

pT, p~, .. .).

Note that E(P~)

= Px ==

P(X

= x);

var(p~)

= Px(1- Px)/ N ;

and (x 1= y).

Hence var(k*). (ii) Set E(x)

= kp == J.t, and E(s2) = kp(1 + p) == It2' Then _

J.t[I+~r

[2

].

k= ( (s - I(2) - (x - II) J.t2-J.t ) 1 + (It2 - It) Hence, using a logarithmic expansion,

var[Iogk] '" (It 2 - It)-2 [(2J.t:: Ilf var(x)+ var(s2) 2(2 It;l- It) COV(.X,S2)], where varIx) = It2IN; var(s2) '" (It4-lt~)IN; cOV(X,S2) Ilr == E(X - Il)r. Hence var(i<).

= It3IN: and

282

EXER('JSES IN PROBABILITY AND STATISTICS

177 Thomas, M. (1949), B, 36, 18.

=

Probability of no plants

probability of no clusters

=

e- p.

Ie

P(k)

=

L L P (r clusters having exl> ex2,' .. ,ex, plants),

,= 1

"

,

L exj = k and exj > 0 for all j;

where

j= 1

and t,h,e second summation is over all possible sets of ex's satisfying the staled conditions, Thus Ie -p, , 1 P(k) = ~e-A' ,=1 r! " j=l (exj-1)!

A.1e-'L n --

L Ie

=

-p,

L ~e-A', A.1e-'rle-'/(k-r)!, r!

, =1

by considering the multinomial expansion of (Xl + X2 + .,. + X,)k-, where each Xj = 1. The probability-generating function of the number of plants is 00

E(OIe)

= e- P +

L Ole P(k) = e- P[exp{JlOeA(II-1)}], whence ,,(t), k=1

The first four cumulants are:

"1 = Jl(1+A.); "2 = Jl(1+3A.+A. 2); "3 = Jl(1+7A.+6A. 2+,P); "4 = Jl(l + 15A.+25A.2+ 10A.3 +A.4 ). The likelihood of the observed sample is

N'.

L=

, e- pno [Jl e -(p+A)]n, [l-e- P - Jle -(p+A)]N-no-n"

no! nl ! (N -nO-nl)! Hence the equations for the maximum-likelihood estimates

va..tll) <\f<

eP - 1

A

= -N' - ' var(A.) = A

and

cov(A., {1)

1 and jl are

Jl e(H p) + (Jl- WeP -1 ' NJl2'

=

1+(1I-1)eP Jl

An estimate of the population mean is m

';v

= jl(1 + 1). Hence

var(m) ,.., (1 + A.)2 var(jl) + Jl2 var(l) + 2Jl(1 + A.) cov(jl, 1)

1

= N[eP (Jl-A.-2)2 _(A.+2)2 + Jle(Hp)]. 178 (i) If 0* = ax+by is the best estimate of 0, minimize var(O*) subjcd to the condition a + b = 1 to determine the constants a and b. n

(ii) Let T

=

L ;= I

ajXj'

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

it

r

Minimize F == var(T) - () i."

iii =

i

283

11 by Lagrange's method. Thus

iii -

(i. - 1)/(i." - I).

An(A -1)0- 2 < A0-2 var(T) = . 1."-1 For large

3

whence for all n > 1.

11,

An (A -1)/(An-1) '" 0 for A < I, and", (A -1) for A > 1. (iii) For A -+ I, var(T) -+ 0- 2 /n.

li9 UL (1964). Least-squares estimation gives J.L* rtl 2 is S2 = Il)

Lt,

(Xi-X)2+

=

it,

x and

0* = ji-x, and the estimate

(Yi- ji)2 ]/(n+v-2).

lest H(O = ( 0 ),

t [0 lest

H[var(x i)

F . Ir its

=

=

ji-x-Oo (

s

- nV)1n+I'

.h

WIt

(n+v-2)dJ.

var(Yi)), use the F statistic

I\"

v-I"

= n-l' i~l (Xi- X)2 i~' (Yi-ji)2

with (n-l,v-l)d.f.,

reciprocal to ensure F > 1.

= corr(Y, Z) = t, and corr(X, Z) = O. (i) If Y = c, then X3 = c-x 2, and so X = x, +X2, Z = C-X 2+X4'

ISO eorr(X, Y)

-to

Hence carr (X. ZIY = c) = lii) If Y = i.Z. then.\2 = (i.-I).\3+i'.\-l-' so that

X

=

Xl+(A-l)X3+AX4,

and

Z = X3+X4'

Hence

2A-l

corr(X, ZIY

= AZ) = 2(A2 -A + 1)1-'

181 UL(1964). E(y)

= (X+y, E(y_(X)2

=

p2+3y2,

and cov(x,y) = E(xy) =

p.

lienee corr(x, y).

182 UL (1964).

The sum of ranks for the two rankings is N(N + 1)/2. The sum of squares for the first ranking is N(N 2 -1)/12. The sum of squares for the second ranking is k-l

=i[(t2+2tf+tk )+

k

L t;ti+l(t i+ -ti)]-N(N+1)2/4 =! L 1

i=1

njtjtj_l'

j=2

This is also the sum of the products of the two rankings. Hem:e p.

284

EXERCISES IN PROBABILITY AND STATISTICS

183 UL (1964). The result is obtained by simplification of the X2 in standard form . E(aj) = AnJN; E(b j) = BnJN. There are (k-l)dJ. The result is due to BrWl\h and Snedecor. and,

ANSWERS AND HINTS ON SOLUTIONS

Chapter 4

E(e iIZ )

ff 2~ f f 'Xl

,::1)

-Y]

-co

C1J

00

-!'fJ

-:0

= 21n =

exp{itxy-!(x 2+ y2)} dx dy

exp[-t{(x_ity)2+(1+t2)y2}]dxdy = (l+t 2)-t.

For correlated variables, itxy

2(I~P2).(x2-2Pxy+y2) = -2(1~P2)[X-{P+(I-P2)it}y]2-t[1- 2pit+ (1- p2)t 2]y2.

lienee integration gives

E(e iIZ )

=

[1-2pit+(I-p2)t2rt.

A logarithmic expansion gives for the first two cumulants of Z and K2 = (1 + p2).

Ki =

2 The characteristic function of ajXj is cos(ajt), and that of Y is n


n cos(ai)·

j= 1

2n sin(t/2n)
that
sin t t/2n sin t (/2n) --+ - , as n --+ t sm t t

= -. .

00.

This is also the characteristic function of a uniform distribution in (-1, 1).

J Tanner, J. C. (1951), B, 38, 383. P(X

P(X ~ x)

= 0) = e-). and

= I-F(x) for x>

The density function of X isf(x) = -dF/dx. Therefore I/!(t) = E(e iIX )

f

l-E

=

!~ [e

il £ P(X

= e)+

f 00

e i1x f(x) dx+

1 +.

f.

285

e i1x f(x) dX]

o.

P

286

EXERC'lSES IN PROBABILITY AND STATISTICS 1-1:

=

~~~ [eil< P(X =

e)+Ae-.t

f

L

eitx dx+

<

=

lim rei/< P(X

<~O

=

f eit,~,

Ae-.t F(x-1) dX]

1+<

e)+ A ~-.t {e it (I-<)_e il<}+A e-.t It

f 00

eit(I+},) F(y) d ,] J

<

=

lim [ei/
=

r.~O

e)+ A~-.t{eit(I-<)_eil<}+Aeit-.t It

f

1 -<

eit}'F(y)dy+

L

+Aeit-.t

f

eit}'F(Y)dY] ,

1+<

Hence, integrating the two integrals by parts and then taking the tirol'ls. ' e -+ 0, Aeit-.t A e-.t ¢(t) = - ,- , ¢(t)+e-.t--,-, as F(O) = P(X > 0) = (l-e-,t)

,,~

It

It

'

whence the result. For the first two cumulants Kl =

(e.t-A-1)/A;

K2

= (e U -2Ae.t-1)/A2.

4 Sheppard, W. F. (1898), PTRS, 192, 101, and PRS, 62, 170.

But 1 2 [u 2 + v2 - 2pu- 2(1- p2)it lalu - 2(1- p2)it2a2v] = 2(1-p ) 1 2 [{u-pv-(1-p2)itlad2+(1-p2){v-pitlal-it20'2}2H 2(1- p )

+ Wit I al)2 + (it 2a2)2 + 2p(it I a d(it 2( 2)]'

Hence E(e it,X +it2Y)

=

exp t{(itlad 2+ (it 2a2)2 + 2p(it l ad(it 2a 2)} == ¢(t 1,t2), say.

Therefore

P=

II (2~r J J

exp(-itlx-it2y)·(tt>t2)dtldt2dxdy,

00

-00-00

or, reversing the order of integration,

A~SWERS A~D HINTS ON SOLUTIONS: CHAPTER

ff 00

2

oP

4n Op =

4

287

00

(1'1(1'2 _." _."

2n 4>(tI,t 2 )dt l dt2 = ~'

hence result, since P = i for p = o. • BY symmetry, the total probability in the negative quadrant is the same, consequently in each of the mixed quadrants it is Ind 1 1 . -1

4- 2n Sin

p.

S The density function of X and Y is f(x,y) =

4~2

ff 00

00

-')')

-tX)

-co

-Q'J

exp(-itlx-it2y)·4>(tl,t2)dtldt2

fherefore (_P)J[ 1 foo

P(X~(X, Y~P)=.L -.,)=0 J. 00

f f 00

X

21n

-2 n

a

f exp{-!
dx

-00

00

dy

p

exp{ -!
+2it2y)}·t~ dt1'

-00

But

)J

1 foo {I 2 }' .( a 1 fOO 1 2 27t exp -WI +2itlX) . t{ dt l = i) ax . 2n exp( -2tl - itlx) dtl -00

-00

by the Inversion Theorem. Hence

1 foo

2n

CI

dx

foo -to

. exp{ -!
ii foo (O)i tx2 fo ax edx a:

=iJ[- ~.(:(Xr-l e- ta2] =(-i'YHj_l((X).~e-ta2. Similarly for the other double integral, whence the result.

288 6

EXERCISES IN PROBABILITY AND STATISTICS

UL (1963). The probability density function of X is

J2n [e- +~. exp {--212 (x-m)2}1· 2n (J (J tx2

1 (1 + A)

(-00 < X < cr:).

The characteristic function of X is E(e iIX ) = e- 12 / 2 [1 +A eiIA]/(l +A),

where A == m +-!(t)

(itA)

(itA)2

(itA)3

= -tt 2 +log [1+[J(itA) {1+-2- + 6 -+2"4+'" ([J == 1

}]

,

~J,

whence the power series expansion in (it)'. The first four cumulants of X are

"1 =

"2 =

1+[J«(J2-1)+[J(1-[J)m 2 ; K3 = [J(l- [J)m[3«(J2 -1) + (1- 2[J)m 2]; "4 = [J(1- [J)[3(q2 _1)2 + 6(1- 2[J)m 2(q2 -1)+(1- 6[J + 6[J2)m4 ]. [Jm;

Finally, so that var(X)

= v2+ [J(1- p)m 2, whence

the inequality as 0 < [J(l- P) ~ !.

7 Using the result of the preceding exercise, the characteristic function of the sample mean x is [ e - 12/2n2 + Aexp {i~m _

= (I+A)-n

t;:: }]"/(1

+ A)"

£ (n)Arexp [_t(n-r~rq2) {t2_ 2itrmn 2}]' r n n-r+m

r=O

But

1 2n

fCio

[

l(n-r+rq2) { 2 n2 t

exp - itx -2

2itrmn }] n-r+rrr2

whence the distribution of x by the Inversion Theorem. If "r is the rth cumulant of X, then, again using the preceding exercise. the coefficients of skewness and kurtosis of the distribution of x are K~

I't = nK~

[J2(1- [J)2m 2[3«(J2 -1)+(1- 2[J)m 2j2 = n[1 + [J(q2 -1)+ [J(1- p)m2j3 ; and

[J(1- [J)[3«(J2 _1)2 +6(1-2p)m 2«(J2 -1)+ (1- 6[J+6p2)m 4 ] 1'2 = n"~ = n[1 + [J(q2 -1)+ [J(1- fJ)m 2F "4

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

rhcrerore (i) for A and

(J

constant and m --.

constant and

In

289

00,

yr ---. (I-A)2/IlA ; (ii) for A and

4

Y2 ---. (l-4A+A 2)/IlA.

yr --. 0; Y2 ---. 3/nA.

00,

(J ---.

Use the identity

(1-2it2) [ -'----=..:.. X 2

m+it 1 J2 + 2(1-2it2)

+ [(it 1)2 + 2it lnH 2it2rn2 ]/2( 1- 2i( 2) to prove that (.

.

E[exp,t 1x+,t 2x

2)]

=

exp{-d/2(1-2it2)}

(I+A)(I-2it2)t

Hence the joint characteristic function of n

Yl

,1,( 'l'

t \lt2

) =

=

n

xr/Jn = xJn r=l I

and

Y2

I

=

exp{-tf/2(1-2it2)} ~ (n)Ar (1+At(I-2it 2t/ 2 'r~O r exp

r=l

x~ is

[it l rm/Jn +it2m2rJ 1-2it2

.

Inversion of this characteristic function gives the joint density function 01 Y. and Y2 · But the within-sample S.S. is Z = Y2 - Yr. Hence the joint density function of Z and Y1 is

GJ f f 00

g(Z, Yt)

=

00

2

-00

-

exp{ - it 1 Y1 - it 2(Z +

ym .J/!(tl, t2) dtl dt 2.

~

Integration over tl gives

f 00

Iz 1":)

~

,

I

1 ~ (n)Ar = (2n)t(l +At' r~O ,. 2

(1-2it 2 )-(n-l l/2 x

it2rm2 (1- 2i(2) { 2') 2 Y1 -

xexp [ -it 2 (Z+Ytl+(1

-

It2

r:.

rm} 2

.

yn(1-2lt 2 )

dt 2 •

Finally, integrate out over Y1 to obtain the density function of Z as

IItA)-n __ 21t

.

"n

L., r= 0

(n) r

'r A.

rm

r)] dt

2 ( foo (1_2't)-(n- l l/2 [ it -'tZ+ I 2 • exp I 2 (1 2 2' ) 1-- It2 n

2·

-00

Hence, reversing the order of summation and integration, and using the Inversion Theorem gives the characteristic function of Z. The special case follows when either m = 0 or A = O. The error sum of squares is n

(n-l)s2 ==

I

r= 1

(X r -X)2,

290

EXERCISES IN PROBABILITY AND STATISTICS

(n-1)s2 _ ~

so that

".2

-

L.-

2

Zj -

j='

v

(~

L.-

/ ;:)2

Zj

n ,

V

j='

where each Zj is a unit normal variable. Put A = 0 in the preceding e . to ?b.tain th~ joint characteristic function of Z j and zJ. Hence the joint c~erCI\~ tensttc functton of arac· II II X, = L Zj/jn and X 2 = L zJ is j='

j='

(1- 2it 2 )-1I/2 . exp{ - tU2(1- 2it 2)}.

The Inversion Theorem now gives the joint density function of

X,

ff «>

'7J

-Ct.)

-00

(2~r

and

y=x 2 -Xf=(n-l)s2/u 2 as

(1-2it2)-1I/2 x

00

fo'

f

e- il2Y dt 2 2n -00 (1-2it 2 )11I-1)/2'

1 e-txt _1 = __

.

l1ence X, is a unit normal variable and Y is independently distributed as a X with V dJ. /2 r(v+r)jr V E(xr) = 2r -22' so that E(X 3 ) = (v+ 1)E(X). Hence p,. The approximation for y is obtained by using Stirling's formula, whence lhl' approximation for p,. 10 The joint distribution of e and '7 is

(1_p2)t 2n . exp-!(e+ '72-2pe'7) . de d'7,

(- 00 < e, '7 < 00)

and so

E[exp(it 11 e 2 + it 22 '7 2+ it 12 e'7 + it, e + it 2'7)] = (1-:/)t .

exp[2~ . {(1- 2it 22 )(it,)2 + (1- 2it 11)(it2)2 + 2(p + itd(it,)(iI2111

This is evaluated by a straight application of Aitken's integral (vide Chapler ~. Ex. 63). Hence the joint characteristic function of X" Y" X 2, Y2 and 1 Invert this characteristic function to obtain formally the joint distribution (II X" Y" u = x 2-xf, v = Y2 - Yf, and w = Z-X,Y,. Integration over II and t2 can again be carried out by an application of Aitken's integral, M' that the joint distribution has the density function (1- 2)"/2

(2:)4

«>

ll1~

III -')/2.

exp{ -i(ut11 +Vt22 +wtd} x

-00

Hence the result after an integration over X, and Y, for ( -

00

< X" Y, < (/.1

A NSWERS A ND HINTS ON SOLUTIONS: CHAPTER

4

291

Using the joint characteristic function, the probability density function of II . . II ,I,

and w IS

(I_p2)(n-l)/2

X

-(2n)3

)( f 00

[exp{-i(utll+vt22+wtI2)} (l-2it 22 )-(n-I)/2(},-2it 1 d-(n-l)/2]x

-<X)

X dIll dt22 dt l2 ,

. here ,l. == 1 - (p + it (2)2/(1- 2it 22 ) is to be regarded a constant for integration ::ver til' Therefore the joint density may be written as

.

(1_ p 2)(n-l)/2 e- tll . u(n-3)/2 f'~' 1 . e- u22v (1-2it 22 )-(n-l)/2 x (2n)2(n-I)/2 -<Xl

r(I1;)

00

1 x2n

f

d2

exp [U(P+it -----'-----

2(1- 2it 22 )

-00

Since II > 0, the integration over t 12 is simple and gives the density function

{-t(u - 2pw +~) }.u(n-4)/2 2(n-l/l2for(ll; 1)

(1- p2)(n-l)/2 exp

X

00

x2~ f

(l-2it22)-(n-2)/2exp{-it22(V-

~:2)}

dt 22 ,

which for ut' - w2 > 0 finally gives (l - p2)(n -1)/2 exp{ -!(u + v - 2pw)}. (uv _

W 2)(n- 4)/2

2n-IJn r(n; 1)r(Il;2) The (II, V, W) -4 (Sx, s)" r) transformation is straightforward, but the standard form of the constant of the joint distribution is obtained by using the duplicalion formula 2n -

3

r(Il-2)r(~) 2 2

_

r::. 1(n-2). - ....;n.

12 Elderton, W. P. (1933), B, 25, 179. For the continuous distribution of X, E(e iIX ) discrete random variable Y, PIY,= h/2)

=

(1- it/.,1,)-I. For the

= (I-e- Ah ); P[Y = (2r+ 1)h/2] = e-rhA(l_e-Ah), for,.

Hence the probability-generating function of Y is E(OY) = Oh/2(l-e- Ah )/(l-O" e- hA ),

~

1.

292

EXERCISES IN PROBABILITY AND STATISTICS

and the characteristic function is _ (l_e- lh) e irh / 2

CPh(t)-

_

1 -e irh-hl

_~)-I

(

1 ,I.

-

whence the result for K( /) and K,,((). K1

=

~

"I -

sinh(V.h)/(-V.h)

' Sill . h{-!(' '/11'1 - 2 A-It) '( 12h().-i/)}'

[I

1 I]

zit coth(z,l,.It) - i. '

V-It.

and the inequality follows since 0 < tanh(-V-h) <

K2

=

Also,

K2 + .12[1-!J.2lt 2 cosech 2(ti.h)], I.

and sinh(tAIt) >

13

V.It,

whence K2 < K2'

Kupperman, M. (1952), B, 39, 429. The probability density function of X is

(a+x)/a 2, for -a ~ X ~ 0 (a-x)/a 2, for 0 ~ X ~ a.

and

Hence cp(t) by integration. For the discrete distribution, 111ft = a, and

P[(r-l)h ~ X ~ rhl

=

h{2a-(2r-l)h}/2a 2, for 1 ~ r ~

Ill.

Therefore

cph(t)

=

I~[e-ita

a =

(2r-l)heith(2r-ll/2+eita 2

h2[e-ita.~.!!..a

=

f r= 1

f

(2r-l)he- ith (2r- l l/2]

i: eith(2r-l)/2_eita.~ iJ f

10t,=1

2

r= 1

e- ith(2,-ll/2]

10t,=1

h [ .O{l_eita } . O{l_e- ita }] 2a2 e - Ita ot sin(tlt/2) + ella at sin(tlt/2) ,

whence the result. Therefore

cph(t)

=

tlt/2 ] 3 [Sin tit] cp(t) [sin(th/2) th'

and a logarithmic expansion gives the relation between K2j and K2j'

14

Kupperman, M. (1952), B, 39, 429. P(Z = 0) =

h(4a-h)

whence the result.

4

a

2

'

and

P(Z = rh) =

h(a-rlt) a

2'

for 1 ~ r ~

Ill.

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

4

Bxpansion of '" h(t) gives h2 (it)2 K2 = + - + the coefficient of -2' in

"2

12

.

•bich, on differentiation, is found to be -

h4 /32a 2.

IS Kupperman, M. (1952), B, 39, 429. a

E(e i'X )

= 2~3

f

1

(a 2- x 2) cos tx dx

=~

o

.h'oee result . \;

P [ y=

f

(1- u2) cos atu du,

0

(2,.-1)11] 311 2 = 4a 3 1!(3a 2 -h 2)-h 2,.(r-l)].

Ihererore the characteristic function of Y is

~ [eilh(2r-ll/2 + e - ilh(2r- 1 l/2] . ~ [!(3a 2- 112) - 11 2r(,. - I)] ~

40 3

" I

--

1I(3a 2-h2) [eiIIl12(eilllth_l) e ilhI2(e-illllh_I)]_ 'h + 'h 4a 3 e"-l e-II-I 311 3 ---3

16a

LIII [(2r-1)2 _1][ei,h(2r-ll/2 +e- ilh(2r-1 l/2] r= 1

11(120 2-112) sinmth + -311 (0)2 ~ Ie illl(2r-ll/2 +e -ilh(2r-l l/ 2] L 160 3 4a 3 ot r= 1 • sin(thI2) -

h(12a 2-h 2)

=

16a 3

.

sin at 3h(0)2[ sinat] sin(tIl/2) + 4a 3 ot sin(thI2)'

\\"hich may be reduced to the form h2 [sin at thl 2] 3 [Sin 2at sin til at (thI2)3] 4a 2 ----;;t. sin(th/2) - (at)2 ~. ~ . sin at . sin 3(th/2) + 3 [Sin at sin 2 til (thI2)5 ] + (at)2 ~. (tll)2 . sin 5 (tIl12) .

lienee the result by using the expansion in Bernoulli numbers. 16 Kupperman, M. (1952), B, 39, 429. 1I(12a 2 -112) 3h 3 ,.2 P(Z = rh) = 16a 3 4a 3 '

for,. ~ O.

Ihcrerore the characteristic function of Z is 1,112(/2_112) [III ] 3h3 III " - 3 - 1 + L (e ithr +e- ithr ) - - 3 L ,.2(e i1hr +e- itlrr ) 16a r=1 40 r=1 2 2 h(12a -h ) [ eilh(eillllll-l) e-i'h(e-illl'h_I)] =

16a

3

1+

'h

e"-I

+

. e""-l

(2)2

lienee the slaled result.

3h 4a ct

+--3" -..;-

+

LIII r= 1

(e i1hr +e- illrr).

293

294

EXERCISES IN PROBABILITY AND STATISTICS

Kupperman, M. (1952), B, 39, 429. The characteristic function of X is obtained by direct integration a d be rewritten as ' n rna:. 17

_2_ [Sin 2at . ~ _ 1] +~ [Sin at -1] (ait)2 2at sm at art at

==

~[et/11(tl_l] +~ [et/12(t l _ 1] , (Glt)2 Glt r/!l(t) ==

I: Bj~j~:t)j(22j_l); J.J.

where and

j=2 00

r/! 2(t) ==

L: )=2

B .(2a)j (it)j J

• .,

the B j being Bernoulli numbers.

'

J.J.

Hence the mean and variance of X.

P[Y _ (2r-l)h]

__ 2h_h 2(2r-l), a a2

2

for 1 ~ r ~

Ill,

so that the characteristic function of Y is

f eith(2r- l/2[2h _h 2(2r2-1)] l

a

r= 1

a

= 2h.eith/2. eit.hm _l_ 211 (~.~) ;, eith(2r-l)/2 e'th -1

a

a2 i at r~1

'

whence the result. The mean and variance are deduced by considering Ihl' expansion of log QUJ) in terms of the series involving Bernoulli numbers.

18

Kupperman, M. (1952), B, 39, 429. E(e itX ) = eita[(sin at)/at], so that the cumulant-generating function of X II log [e

ita

(Sin at)] _ ~ - - = L.. at r= 1

(it)' "r'-,' r .

The characteristic function of Y is

!. ki1eith(r+tl = eiat [Sin at] [ k ,=0

at

. tl1/2 ]. sm(tl1/2)

Therefore, if Kr denote the cumulants of Y, '.0

(it)'

00

_

(it)'

00

(it)i Bi hi

L: "r·-, = L ",.-.-, + L: -.-, .-. , r=1 r. r=1 I. j=2J· J the Bj being Bernoulli numbers. Hence

19

"2r+ 1

=

K2r+ 1; "2r

=

h2r K2r+ B 2r.];,

r ~ 1.

Kupperman, M. (1952), B, 39, 429. The characteristic function of X about the origin is

eith(n + 11/2 sin(ntl1/2) th/2 . nth ''2 . sin(th l '2)'

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

E(X) = (n + 1)h/2,

4

295

whence
Similarly, the characteristic function of Y about the origin is

einth sin(nth)

n' sin(th) , '0 that E(Y) = nh. Hence I/Ih(t). Logarithmic expansions of
"2r+l(X) = "2r+l(Y)

For h ~ 0, n ->

CfJ

such that nh .I,. ( )

'Ph

"2 (X) = B2r.(n2r_l)h2r; 2r

= 0;

t

->

r

A,

~ sin(At/2)

At/2'

./, ( ) sin(At) 'I' h t --+ ----;:t. 10 James, G. S. (1952), PCPS, 48, 443, and UL (1963).

1(11, v)

1 . e -t(u+v) . (uv)-t [1 + e. 2n(uv)t(u - v)(uv -11- v+ 2) e-t(u+v»). 2n

= -

Thus the coefficient of e is bounded for all u, v ~ 0, so that f(u, v) {: 0 for e lufficiently small. Also, since the second term in f(u, v) is anti-symmetrical in U, I'.

coco

f f f(u, v)dudv = o

1.

0

As

Ihcrerore

E[

(. . ) 1 2eit 1t 2{t2- t d exp Itl u -lt 2v) = (1- 2it l)t (1 - 2it2)t + (1- it 1)3 (1- it 2)3'

Hence the characteristic functions of U, V and (U + V). 21 Wicksell, S. D. (1933), B, 25, 121. If p(Ylx) is the probability density function of Y given X = x, then

t 2 )

f

= eit1x g(x) dx

J

eit2Y p(Ylx) dy

296

EXERCISES I N PROBABILITY AN D STA T1STlCS

so that

[or ¢(t~, t 2)]

at

2

= ;r

fe

il "

g(X) Jl~(X) dx.

12 =0

Hence the stated result by using the Inversion Theorem on the fUnction g(x) Jl~(x). Wicksell, S. D. (1933), B, 25, 121. Since

22

1 2(l_p2) [x 2{1-(I- p2)itd + y2{1-(1-p2)it 2} -2pxy]

= _{l-(I- p 2)it d 2(1_p2) therefore

[xpy ]2_y2{(1-itl)(l-it2)-p2(itd(il2)} . 1-(1_p2)it. 2{1-(l_p2)it.} -,

E[exp(it lX2 /2 + it2y2/2)]

= [(1- it d(1- it2)- p2(it l)(it 2)]-i,

whence the joint characteristic function of z. and Z2' Use of the result of the preceding exercise gives 11~(Z d by noting that for N >0

and

(it dO e - il,%, dt 1 (1- it .)N/2 23

Wicksell, S. D. (1933), B, 25, 121. By the preceding evaluation of the joint characteristic function, that of n+n2

'I+nl

w. =

L

xf/20"f

and

j= •

W2

=

L

YJ/20"~

is

j= •

¢(t., t 2) = [(1-itd(l-it 2)-p2(it.)(it 2)]-n/2. (1-it.)-n,/2(1_it 2)-n,fl. Hence by a logarithmic expansion the bivariate cumulants and corr(w., \\'2)

Krs

are obtained.

= KII/(K02 K 20)i.

The conditional moments 11~(W.) are evaluated in the same way as in Ex. 22 above. 24

Wicksell, S. D. (1933), B, 25, 121. To evaluate E[exp(it.x 2+it 2y)], set u _

I 2 2(1- p)

= x/O". and v = Y/0"2' Then

[(~)2+(L)2_2P(X)(L)]+it.X2+it2Y 0". 0"2 0". 0" 2 2(

1 2 [A(u-pv/A)2+B(v-C/B)2_c 2/B], I-p )

where A == 1-2(1- p2)it.O"I; B == 1- p2/A; C == (1- p2)it 20"2'

A !':SWERS AND HINTS ON SOLUTIONS: CHAPTER

4

297

Hence

and so the joint characteristic function of WI and W2 is 4>(t l ,t 2) = (l-itd- n/ 2

exp[Wt2)2{1+;~~;JJ

.

00

= (l-it l )-n/2 ei

L (!p2),(it 2)2r(itd(1-it

(i 12J2

l

)-r/r!

r=O

Ajoint inversion now gives the distribution of WI and W 2 . The variances are obtained by the differentiation of 4>(t l , t 2 ) as in Ex. 22.

Zs Put z), 2 • 1:.+ 111), ) IS

=

~

X), -m), for (1

A.

~ n).

Then the characteristic function of

(I - 2it)-t exp[mi{it)(l- 2itj- I],

dnd that of Y is 4>(t) = (1-2it)-n/2 exp[2112(it)(I-2iW 1 J =

e _,,2 (1- 2iW n/2 exp[Jl2(1- 2iW I J

=

e-,,2

2r

00

L ;(1_2it)-
r=O

whence, on inversion, the distribution of Y. A logarithmic expansion of 4>(t) gives

E(Y)

=

n+2Jl 2 ;

var(Y)

=

211 + 8Jl2.

26 Bartlett, M. S. (1938), JLMS, 13, 62. Let f(x, y) be the joint density function of X and Y, so that

=

f(x, y)

g(xly) h(y)

in terms of the conditional density of X and the marginal density of Y. Then, x,

f

ljJ(t Ily) =

e it1x g(xIY) dx,

-00

and 00

4>(t" t 2)

=

f

ljJ(tlly) h(y) e- it2Y dy.

-00

Therefore, by the Inversion Theorem,

f 00

ljJ(tlIY) h(y)

=

21n:-

e- il2Y 4>(t l , t 2) dt2

-oc,

Jnd

f 00

h(y)

=

2~

-00

,hence the result.

e- it2 ), 4>(0, t 2 ) dt 2 ,

298

EXERCISES IN PROBABILITY AND STATISTICS

27 Kenney, J. F. (1939), AMS, 10, 70.

Since E(e ilX ) = cp(t), therefore the joint characteristic function of Y a Z is nd I/I(t1>t 2 ) = E(eillY+ilZZ) = [CP(tl +t2)]V[cp(td]nl-V[cp(t2)]nz-v.

Further, if h(z) is the marginal density function of Z, then by using the Inv . Theorem, er· SlOn

f e-ilZ'[~. ~I/I] 00

E(Ylz) h(z)

1 = -2 1t

I

ut 1

11:

0

dt 2,

-00

and

.,

E(y2Iz) h(z)

J

= 21 1t

e- i1z '

-00

[~. ~2fJ I uti

dt 2· 11=0

These equations lead to E(Ylz) and var(Ylz), noting that E(X') = [;,.

o'CP~d] ott

I

1,=0

and

f i1z' 00

[cp(t 2)]n

=

e

h(z) dz.

-00

The correlation between Y and Z is obtained from the linear regression equations expressing E(Ylz) and E(Zly). 28

The joint characteristic function of u and v is CP(tl, t 2)

= E[exp(it 1 u+it 2v)] = (1-2it 1 )-nI 2 exp[ -At~/2(1-2itl)], n

where

A

=I

ai.

'<=1

Hence, using Ex. 26 above, the conditional characteristic function or given v = 0 is I/I(tllv = 8) = (1-2it 1)-(n-lI/2. =

ell-

exp[~: {1-(1-2it d}]

f (-(j2Y(1_2it r!

l r(n+2,-l l1 2,

,=0 whence the result on inversion. 29

Bartlett, M. S. (1938), JLMS, 13,62. E(e ilX )

= (1 + t 2) -1.

Therefore the joint characteristic function of u and v is

CP(tl, t 2)

= E[exp(it l u+it2V)] = E[exp i(tl +t2 )xtl. E[exp i(t l -t 2 )X2] =

[1+(tt+t 2)2]-l[I+(tl-t 2)2r 1.

Hence evaluate cp(t dv) by Ex. 26 above.

1/

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

4

299

for v > 0, evaluate the integral 00

f

eit2V cf>(t 1, t 2) dt2

-00

by using a semicircular contour in the upper half of the complex plane.

fherelevant poles are at t2 = i±t1. Hence the numerator in cf>(t1Iv) is ne - v [t 1 cos vt 1 + sin vt 1 ] t 1(1 +tn

2'

.

The denominator is ne- v(1+r)/2. obtained as the limit for t 1 --+O of the numerator. 30 Bartlett, M. S. (1938), JLMS, 13,62. Use the result of Ex. 26 above.

f

00

Je- it2Y

00

cf>(t1' t 2) dt2

=

-00

J

exp{ -it 2Y+K(t 1, t2)}' dt2

-00

00

eoP[it,,-(oloy)). exp[ - it 2y -!(K20t i + 2K 11t 1t2 + K02t~)] . dt2

-00 00

=

f

eoP[it!. -(016y)) . e-tK2ott

exp['t 1( K02 t 2 - I 2Y -2 2 + 2K11 t 1 t 2 )] • dt 2

-00

whence cf>(t 1IY)· 31 The characteristic function of the sample mean

[eit;;/:

x is

I]",

and so by the Inversion Theorem the probability density function of x is f(x)

= -

1

Joo

2n

itl e- itx [e- ."-1till

1]" dt.

-00

The integrand is analytic everywhere. and so the path of integration may be changed to the contour r consisting of the real axis from - 00 to - p, the small semicircle of radius p and centre origin, and the real axis from p to 00. Thus itl f(x) = -1 e- itx - .dt 2n 1till

f

[e "_I]"

r

=

.(11) . feit[(jf,,'-X) dt.

(-1)". " (-1)) j

2n

j~O

(i/Il)"

r

t"

300

EXERCISES IN PROBABILITY AND STATISTICS

But ei~Z

f -Z"

dz

= 0, for 0( > 0,

r

2nO("-1 - i" . (II _ 1)!,

for 0( <

o.

Hence the value of I(x). 32 00

E(e itX )

= 2~

f

f 00

eitx-Ixl/a dx

=~

e -xla . cos xt dx

0

-00

= 1-(12t 2 . E(e itX ). Hence the characteristic function of x is (1 + (12 t2/11 2)-II. Therefore the prob. bility density function of x is ,t·

-00

-00

The integral is evaluated by using a semicircular contour with a pole Ilf order II at t = in/(1, whence the distribution of x. Direct integration gives

f 00

g(x) dx

=

-00

II - 1 (II +,. - 1) 1 L -1 2"+r-1 r=O II II-I (1 +O()"+r-I r~o -2-

= coefficient of 0("- I in

1+0()"-1 211-1 (211-1) ] 2 [(- O(r/2 211 - 1 2 r=O 1

L

= coefficient of 0("-1 in

1 = 2 [1- .,211I· -

LI

II -

r=O

(211 - 1)]

.

.

L rl JJ

5=0

' whence the result.

I

33

Kullback, S. (1934), AMS, 5, 264. The characteristic function of u is aitllO + it)-II, and the probability density function of u is

f

~

h

_ 1 I(u) - 2n

exp( - itu + itll log a) dt __1_ (l + it)" - 2ni"

f

eitA dt , (t - i)"

-00

-00

where A. == II log a - u ~ O. The integral is evaluated by using a semicircular contour. whence The distributions of g and 2n log(a/g) follow by transformation. 34

E(eitIXI) =

(l-(1it)- I, so that the characteristic function of v is (1-(1it/n)-1I

whence the result.

= {1-2it(:n)} -n,

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

~ ECeiIX) I. . Hence

=

(1

4

301

+ (2)- I.

[exp i(t I U + 12 V)] = E[exp i{(t 1+ t 2)X + (t 1- t 2)Y}] = [I +(tl +t 2)2r 1[1 +(tI-t 2)2r I. rherefore the characteristic function of u and (Wl>t 2 )

= [1+(tl:t2rrn

v is

[1+el:~2rrn.

rhUS the characteristic functions (W 1> 0) and cP(O, t 2) of u and v are of the ..Jdle form, but cP(t I> t 2 ) 1= cP(t I> 0) cP(O, t 2 )· Expansion of 10gcP(tl> t 2 ) shows that the coefficient oftlt2 is zero. The probability density function of u is

-00

-00

rhe integral is evaluated by using a semicircular contour. Hence _

ne- n1iil

2n-1 (2n+r-1)!

g(lI) = (2 II _1)1' .

L

r = ()

-1(" _'_1)1' , • _II 1 .

[nluWn-r-1 _ ,)2n+r • (-r:o
J6 Kullback, S. (1934), AMS, 5, 264. E(eitIOgX)

= r(p+it)jr(p).

Therefore the probability density function of u is 00

g(u)

=

-1 2n

f e -,'U. [rIp + it)/r(p)]n dt

-00

ePU f-p+ioo = 2ni[np)]n' eUz [n _·z)]n dz. -p-i'X)

But nz)r(1-:) = nisin nz, so that

f

- p+ ioo

e PU( -nl" 1 g(u) = .[r(p)]n 2ni

eUz dz .' {n1 + z)}" smn nz

-p~ioo

Evaluate the iptegral using the contour Iz/ = m+t, m a positive integer, and the line x = - p. The poles of the integrand are at z = r with residue

[d

n- I ] n-n( _1)nr eUZ (n -1) ! dzn- I . {n1 + z)}"

z=r •

Hence

ePU 00 (_1)nr+n+ 1 [d n- I eUz ] I n g(u) = [r(p)]n' r~o (n-1)! dz - ' {n1 +zW z=; Whence the distribution of v.

302 37

EXERCISES I N PROBABILITY AND ST A TISTI CS

Let

'IY r(Z+~)I/lIt-"Z(21t)(n-Il/2nIlZ). V

R(z) =

,=0

II

Then R(z) = R(z + I) so that R(z) = !~~) R(z). But for large z, Stirling's formula gives r(z+1) = (21tZ)t(;r+ O(Z-I). Hence log R(z + 1) = O(Z-I) so that lim log R(z+ 1) = O. • -+ 00

The duplication formula is obtained for 38

II =

2.

Kullback, S. (1934), AMS, 5, 264. Since E(eil)ogXj) = nPj+itvnPj), the probability density function of 1/ is

•

i 'X

feu. Ii r(pj-Z)' dz.

n 1 .~ r(pj) 2m -ioo

fl

j=1

j= 1

The integral is evaluated by using the contour Izl = m+t, where m is an integer. and the imaginary axis. For fixed j, the poles of r(pj-z) are at z = Pj+r ror r = 0,1,2, .... The residue at z = Pj+r is (-1)'+ 1 e U('+ Pj) nr+ 1) . IJj r(Pk-pj-r).

1

Hence g(u)

=

n

00

·.L L fl r(pj) )= 1 ,=0

(_1)'+2 eu(r+Pj) r('+ 1)

n

I

.

fl. r(Pk- pj-r), k'¢)

j= 1

whence the distribution of v = eu/n• 39

Kullback, S. (1934), AMS, 5,264. Since Pj = p+(j-I)/n, by Ex. 37 above n

fl

r(pj) = nt -np(21t)(n-ll/2nllp).

j= 1

Therefore the characteristic function of u is n -nilnnp + nit)/nnp),

and the probability density function of u is

()- f

~ g u - 21t

00

e

-ilu n-nilnllp+nit) d . nnp) . t

-00

-np+ioo

=

1 - (-)'-2' nr np m eUPn"P

f (n eu/nyn-z)dz.

-np-ir

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

4

303

fb e integral is of the type -a+ ico

I ==

~. Jwz f'( -z) dz. 2m -a-ioo

"/0 and w real, which is evaluated by using the contour Izl = m+t, m a ~sitive integer, and the line x = -a. The residue at the pole z = r is 1~\V)'/1tnr+ 1), so that I = e- W • Hence the distribution of v = e"/n is 1 - - . e-m'(nv)nP-'n du,

(0 ~ v < 00).

r(np)

If

Xl'

X2" .. , Xn are random observations from a n

parameter p, then

I

has the

Xj

r

r

distribution with

distribution with parameter np. Hence the

j= ,

distribution of the sample mean i. ~O

Fisher, R. A. (1929), PLMS, 30,199, and UL (1964). .

E(e" x )

(it)'

7J

=

1+

I 2 Ilr'-" r.

r=

"I

that the cumulant-generating function of i is

I [1 + ~ Ilrn

n og

L...

r'

r=2

(it)'] ,

r.

.

By expansion

-) = -11-2 "2 (X

n

an d

Again, n

(n-l)s2 =

I

xJ-ni 2, so that E[(n-l}s2] = (n-1)11-2'

j= 1

Also, n

(n - 1)2s4 -_ "L... X4j + j= 1

"L...

X2X2 j k-

j*k

so that _ (n-l)2 (n-l)(n2-2n+3) 2 E[(n- 1)24] s - ---·11-4+ ·11-2'

n

n

41 Lukacs, E. (1942), AMS, 13,91. If cf>, (t ,) is the characteristic function of i, then 4> 1 (t d == 4>(t h 0) == [I/!(t dn)]n ;

and. hy definition. 4> 2(t 2) == 4>(0. t 2).

whence var(s2).

304

EXERCISES IN PROBABILITY AND STATISTICS

The independence of x and S2 means that (p(t I' t 2) == ¢I(t .)¢2(t 2) Or

But

a¢ = I.

-? t2

f

S2

. S .exp (.1t 1X+1t 2 2). TIn f( X,. ) dXv>

,.= 1

f(x) being the density function of X. Therefore

[ata¢]

= i["'(ttfn))"-1 feiIOC/n.x2f(X)dX-

2 12=0

_;["'(t.ln)]n-2 [feiIIXln.Xf(X)dxr Hence the differential equation by putting ttfn = t. The solution

follows since ",(0) = 1 and ""(0) = ill.

42 Williams, J. D. (1941), AMS, 12, 239, and McGregor, J. R. (1960), B,47, 111. The variables Yj = (xj-m)/a are unit normal, and n-l

A

=!

L (Yj-Yj+l)2; j= 1

n

B

=!

L(yj-ji)2. j= 1

Therefore

But n

-!

L YJ+tlA+t2B j= 1

where a == 1-tl -[(n-1)/n]t2; b == tl -t2/n; c == 1-2tl -[(n-1}/n]t 2; d == /2/11 so that a+b+(n-2)d = 1: 2b+c+(n-3)d = 1. M n has the leading diagonal (a, c, c, . .. c, a), bordered by diagonals whose c\.cments are all h's, and the remaining elements are d's. Row operations gIve Mn = t~-1 Lln' where Lln is the nth-order determinant:

ANSWERS AND HINTS ON SOLUTIONS: CHAPTER

1 1

1

1 1 1

()

1 0

0 0 0

1

()

1

0 0 0

0 0

1

()

0 0 0

0 0 0

0

1

0

.......

0 0 0 0

()

4

305

1

1 1 (0+ 1) ,

.here 8 == (c-d)/(b-d); ()+ 1 == (a-d)/(b-d). Let D.-l denote the (n -1)th-order determinant obtained from An by xcluding the first row and column of An. Also, let C j denote the jth-order ~elerminant formed by the first j rows and columns of Dn _ 1, U < n - 1). Then Mn=t~-l(Dn_l-An_l)' C 1 =(}, C 2 =(}2-1, and for j>2 II

(.::: (}C j - 1 -C j J

2•

Hence

BUI

and /).n = Dn- 1 -An Therefore

1•

Assuming that An = AIZn+A2Z-n, the constants Al = - z/(1- Z2); A2 = z/(1- Z2). Hence An = An- 2 +S n- 1 , where S. == z'+z-', whence

Mn = t~Mn_2+~-lSn_l' Bul z, Z -

1

are the roots of Z2 - (}z + 1 = O. Therefore Sn-(}Sn-l +Sn-2 = 0,

and the difference equation for Mn follows. j

To obtain the expression for mj' put t2 =

f 00

= [ 0i(t 1;t 2)] ot 1 1,=0

A j e/2B •

-00

L

,= 1

t,

in

fIk=l -1-.e-tY~dYk' fo

and integrate for ( - ex) < t. ~ 0). Also, for the given form of (t 1> t 2)

[ oj(t~,t2)] ot 1

= [dj(tt O)] I, =0

dtl

. (1_t 2 )-tln + 2 j-l). I,

=0

306

EXERCISES [I>< PROBABILITY A;,\;[) STATISTIC'S

43 Cochran, W. G. (1937), JRSS, 100,69. Let uj = xj-x and Vj = Yj- y. Then E(uj) = E(vj) = 0; var(u) = var(vj ) = (n -1)/n; and corr(u j , Vj) = p.

Also, since uj and Vj have a joint bivariate normal distribution, P(u j > 0, Vj > 0)

=

P(uj < 0, Vj < 0)

=

!+(1/2n) sin-I p

by Sheppard's medial dichotomy theorem. Hence E(C) = t+(1/n)sin- 1 p and var(Cj ) = !-[(1/n) sin-l p]2. Note that, by symmetry, P(1I 1

> 0,

VI

> 0; U2 > 0, V2 > 0) = P(UI < 0,

VI

< 0; U2 < 0, v2 < 0).

To obtain P(UI > 0, VI > 0; U 2 < 0, V2 < 0), change the sign of t3 and I . the characteristic function. Then inversion and integration now give 4 III P(u l > 0,

VI

> 0; U2 < 0, V2 < 0)

1

1

= 4n 2[(sin- 1 p)2-{sin-1 p/(n-1W]+ 4n[sin- 1 p-sin- l p/(n-l)]+conslalll = P(u l < 0, VI < 0; u2 > 0, V2 > 0), by symmetry.

Hence cov(C I, C 2 ) = constant - [(lIn) sin -I p/(n -1)F whence var(C), the constant being evaluated since var(C) = for p = 1. For large n,

°

var( C) - :n [1 -

(~ sin - I P ) 2].

which gives the large-sample efficiency of p*. 44 Bartlett, M. S. (1936), PRS, 154, 124. E[eit(X-m)]

= (1+t 2)-I, and y-m = (1-JL)(xl-m)+JL(x 2-m).

Therefore the characteristic function of (y-m) is E[ exp{ it(1- f/)(x 1- m)}] . E[exp{ itJL(X2 - m)}]

= [1 + t 2(1- JL)2] - I [1 + t 2p2]- I 1

= 1-2JL

[

(1- JL)2 1+t2(1-JL)2

JL2] 1+t2JL2'

But (1 + t 2 a 2 )-1 is the characteristic function of the distribution 1 - e- 1zl /C1 dz 2a' ,

( - 00

<

Z

< (0).

Hence the distribution of (y - m) is 1

2(1- 2JL)

[(1- JL) exp{ -ly-ml/(l- JL)} - JL exp{ -ly-mI/JL}] d(y-m). ( - 00

< y < (0).

Al"SWERS AND HINTS ON SOLUTIONS: CHAPTER

4

307

fherefore the logarithm of the likelihood is log L = constant + log[(l- JI) exp{ -ly- ml/(1- JI)} - JI exp{ -1.1'-1111//1]]. ,0

that

1

[1

L

] ,,_ . -+y~ 2 +(y_l)2 00 y' foo exp-[{(r+2){3-(r+l)lX}z].dz, (1- 2JL)(I- JL) IX P r= 0 o

IP> IX). whence the result.

SUPPLEMENT TO THEORETICAL EXERCISES IN PROBABILITY AND STATISTICS

Supplement Finite differences and summation of series Sum the following series by the method of differences: (i)

t

L

r(r + 4)(s - l)(s + 2)

r= 1 s
(ii)

L" L r(r + 3)s(s -1) r=1 s=S;;r n-l

(iii)

L L r(r+2)(r+4)s(s-1)(s+2)

r= 1 s=S;;r

(iv)

t

L (r+1)(r+2)(r+4)s(s-1)(s-3)

r= 1 s=5;;r n+l

(v)

L L (r+2)(r+6)(s-2)(s-4).

r=1

s
Sum the following series by suitable methods: (i)

L" (r -

1)2(r - 2)2(r + 3)

r=1

(ij)

L" r. 2'/(r + l)(r + 2) r= I

(iii)

L'" (r 2+3r-6)/r! r=l

(iv)

L'" (r3+S r2+6)x

r•

r=O

Sum the following finite binomial series: (i)

'f (r+n 1)r(r + 2)x

r

r=O

(")

" r r(r -l)(r + 3)x r • (ii) r~ Given that the rth term of the series

3,4,7,14,27,48,79, ... is a polynomial in r, find the general expression for the rth term. Hence derive a particular value for this term and sum the series to It terms by using Ihis value. 311

312 5

EXERCISES IN PROBABILITY AND STATISTICS

By using Euler's method of summation, sum the following series' 9 25 9 r2 (i) 1+2+-+2+-+-+" .+-+... 4 16 8 2,-1 '

and

00 1 27 125 27 r3 (11) -+1+-+2+-+-+" .+-+ ....

4

16

64

16

6. For a series of terms Uh Wise, that (i)

U I1 +1

= (1 + A)"Uh

(ii) AlI1 u I =

2'+1

U2, U3, ••• , UI1 ,

•••

prove by induction, Or other. .

and

I (-1y(I1l)U r

,=0

II1

+ 1 _"

~her.e, in standard difference notation, A111 +1 U, = AII1 (U,+1 - u,), for all posi.

tIve mtegral values of 111. Also, by considering the differences of the series

where U,

= A-1u,+1 - A-1u"

show that 11 U,=A-l[(1+A)I1_1]Ul='~1 11 (") (iii) '~1 r A'-IU I · 7

A function f(x) is tabulated for a sequence of values x, x + h, x + 2h, ... , x + rh,

If it is known that A2f(x) = 0 + {3x, where

(r;;;. 3).

and (3 are constants, prove that

0

f(x) = ao+ alx + a2x2+ a3 x3

where ao and al are arbitrary constants, but a2 = (0 - {3h)/2h 2 and

a3 = (3/6h 2.

8 For a function f(x) tabulated at the points x, x + h, x + 2h, ... , define the finite difference operation A'f(x) for positive integral values of r. Hence prove that

(.) A2tan-1 x=tan-1[

-2h 2(x+h) ] {(x+hf+1}2_h 2{(x+h)2_1}

I

(ii) [A2 +2(1-cosh)(1+A)]sinx=O, and ( 00') [All -1 III A- 1

]A/3

X

= [(A/3h -1)11 -1 A/3 h - 2

where A and {3 are constants and 9

It

]A/3

X

'

is a positive integer.

If A is the standard finite difference operator with the argument interval

SUPPLEMENT

:: (7 0)

313

operating on the function f(x), define arf(x),

for

r?3l,

lIenee prove that, for h sufficiently small so that terms involving h 3 are ·legligible, (i) a2 10g(1+x)=-h 2 /(1+xf (ii) [a2 + h 2 (1 + a)] cos x = 0, and (iii) a2ex2 = ex2 • 2h 2(1 + 2x 2). (0 (i)

If u" the rth term of a series, is such that Ur == fer) - fer + 1), prove that the sum of the first 11 terms of the series is

f(l) - f(1l + 1).

Use this result to find the sum to 11 terms of the series for which 1 (r ?31). (3r - 2)(3r + 1)(3r + 4) Hence deduce that the sum to infinity of the series is

214.

(ii) Suppose Ur == arx r are such that

a r + aa r_\ + {3a r-2 = 0,

for r?3 2

and a, {3 are constants. If S denotes the sum of the series uo, Ul> U2, ... , then by considering the product (1 + ax + {3x2)S prove that S=ao+(a\+aao)x 1 +ax+ (3x 2 Hence, as a particular case, determine the sum to infinity of the series 1-8x +28x 2-80x 3+ . .. , and indicate the range of x for which the summation is valid.

°

11 (i) For a> and Il small such that terms involving show that a root of the equation is a (1-!1l log ex). (ij) Prove that for any r>

11 2

are negligible,

°

1 r(r+2)

2r+l 2r+3 2r(r+ 1) 2(r+ 1)(r+2) .

By using this identity, or otherwise, find the sum to Il terms of a series whose rth term is 1/r(r+2). What is the limiting sum as Il ~ oo? (iii) For any two numbers a and b such that a > b > and a - b is small compared with a, prove that the value of the function

°

g(a, b) = a;::2

-log(~)

lies between (a-b)3/6a 3 and (a-b)3(3a-2b)/6a 3b.

314

EXERCISES IN PROBABILITY AND STATISTICS

12 (i) If f(r)

= X-I /

(x: r) where

is any positive integer, prove that

X

f(r-1)-f(r)=r- 1 /(x:r),

for r;;;'1.

Hence show that

so that, as n ~ 00, the sum of the series tends to X-I. (ii) If X is a sufficiently large number such that terms involving · 'ble, prove t h at negI19l (

(iii) If

Ixl < 1 and

X

x-4 , tlrt'

-I)! = 1-~+ 1 1 1 2x 2 - 2x3'

x+1

the rth term of an infinite series is

(-xY Ur =

prove that

r(r+ 1)

for r;;;.l,

L u =l- (1_+_x) 10g(1+x). 00

r

X

r=1

13 (i) If the rth term (r ;;;'1) of an infinite series is 1 +2+2 2 +, .. +2 r - 1 U = r (r+ I)! prove that 00

LU

r

=!(e-1f.

r=1

(ii) Prove that for any positive integer

"f (r-1n ) (3r

r

r=1

)

n;;;'

=

1

4"+1-1. n+1

(iii) If x is sufficiently large for terms involving x- 4 to be negligihk. prove that 2 1 1 (x + X)2 [Iog(x + 1) -log x] = 1- 24x2 + 24x 3 . 1

(iv) If x is sufficiently small for terms involving that

XS

to be negligible, prow

14 (i) The function f(r) = 1/(1 + r+ r2) is defined for all non-negative val· ues of r. Prove that f(r-1)-f(r)=1

2r 2 4' +r +r

forr;;;'1.

315

SUPPLEMENT

Hence show that for any positive integer S ==

i r=

"

2,. 1

11

11(11

1 + ,.2 + 1'4

+ 1)

n 2 + 11 + 1 '

so that lim S"

=

1.

.. +--GO

(ii) If

Ixl
~1

=

X

-

1 (1- x)( 1 - 2x) .

(iii) If 0 is a small number such that 03 and higher powers of 0 are negligible, prove that an approximate value of the positive root of the equation eX +x = 1+0

is

x = to(I-iO).

IS (i) If the rth (1';;;0 0) term of an infinite series is Ur =

41'3+211'2+30,.+ 11 (r+3)!

prove that '"

L U = 4(e-l). r

r=O

Hence show that if a discrete random variable X has the probability distribution with point-probabilities proportional to u" then

P(X;;;o2) = 48e-l03. 48(e-l) (ii) If a and (3 (a;;;o (3) are any two positive integers, prove that

at k=O

(a-k)= (a+l). (3 (3 + 1

(iii) If n is a positive integer and

Ixl < 1, prove that the coefficient of xr (n+~-l).

in the expansion of (1-x)-" is

t k=O

k

(a + -1)((3 +a k

a

Hence show that

=k -1) (2a + (3-1), =

k

a

where a and (3 are any two given positive integers.

16 (i) If a is a positive constant and

"

r~o (-1)'

11

a positive integer, prove that

(n) (1+ra)/(1+na)'=O. I'

(ii) If u" the rth term (1';;;01) of a series, is

16

316

EXERCISES IN PROBABILITY AND STATISTICS

evaluate the sum of the first /l terms of the series, and hence . that the sum to infinity is unity venfy (iii) Given that f(r)==or 2(,.-1)2(2r-1)(r 2 -r-1), prove that r6 = tr2 + 1[f(r + 1) - fer)].

Hence show that for any non-negative integer r II

L r6 = /l(/1 + 1)(211 + 1)(3/1 4 + 6/1 3 -

311 + 1)/42.

r=1

17 (i) For any two positive integers 11 and s (s < /1), prove that L (r+s)(s+J) = -1- (11 +s+ 1)(s+2). r=\ s+2 II

Hence, if Sr denotes the sum of the first r natural numbers, that

ShOll'

II

L SrSII-r = l~o(11 + 3)(5), r=\

in standard factorial notation, 11 (;;.1) being an integer. (ii) Prove that for any non-negative integer 11 II (11)/ L (r+2)= 11 . 211+1 + 1 . r=O r (11 + 1)(11 + 2)

18 Prove that (i) the average of the first 11 (;;.1) natural numbers is t(11 + 1); and II

(ii)

L [r-!(11+1)]2=-b11(112-1). r=1

(iii) If r is a discrete random variable which takes the values 1,2, ... , /I with probabilities 11- 1 , then by using the exact expressions for E(Jrl and var(..r;.), prove that [1- 2V(I1) ], [ t Jr]2 =11 2(11+1) 2 11(11 + 1) r=1

where "

V(I1) ==0

L (Jr-O)2

11

and

r=1

0=

L 0-/11. r=\

19 Explain clearly the difference method in the summation of finite and infinite series. (i) If U r = 3r4 (r -1)4 - 4r 3 (r -1)3 for all integral values of ,. ;;.1, prO"1.' that II

Lr

7

= -b11 2(11 + 1)2 + {4UII+\

r=\

( oo) If II

Ur =

2r+ 1

r 2( r+1 )2 '

evaluate the sum to 11 terms of the series deduce that its sum to infinity is unity.

U 1 , U2, U 3 , . . . ,

and henel.'

SUPPLEMENT

317

(iii) By using the identity

.L .L. II

XjXj

1 [(

=-2

1=1/<1

.L )2 -.L x~ II

II

Xj

1=1

]

,

1=1

prove that the sum of the products in pairs of the first n odd integers is in(n-1)(3n 2 -n-1). !O (0 Define the finite differences of a function [(x) tabulated for equispaced values of the argument x. If, for x = r, the value of [(x) is [(r) = sin«(J + rq,), where (J and q, are constants, then prove that for any non-negative integer n L1"[(r) = (2 sin(q,/2))" . sin[(J +!n('1T + q,) + r
Uo, Uh U2, .•• , are the successive values of a function corresponding to those of an integral variable r ~ 0, then prove that for any non-negative integer n,

L1"uo=

~ r=O

(n)(-l)"-rUr. r

Hence, or otherwise, show that if ur =(-lY/(r+1), then Llluo=(-1)1(2"+ 1 -1)/(n + 1).

(iii) Find the sum to n terms of the series whose rth term is

ur =(r2 +r-1)/(r+2)!, for r~1.

Also, verify that the sum to infinity of the series is

!.

ProbabDity and statistical theory 11 At a university in Ruritania the probability of a first-year student passing the June examination is t and that of receiving a res it in the following September :&. The probability that a resit student will pass the September examination is~. Prove that the probability of a first-year student ~ntering the second year at the university is ~. If the probability of a second-year student ultimately getting a degree is ;~, determine the initial probability of a student receiving a degree at the end of three years. Hence evaluate the probability that of n randomly selected first-year entrants to the university (i) exactly k students will receive a degree; (ii) at least k students will receive a degree; and (iii) at least k students will not receive a degree, for O::s:; k ::s:; n. 11 At a game of poker, a player is dealt out a hand of five cards from a Itandard pack of 52 cards. Find the probability that the player has (i) a flush, that is, all cards of the same suit; (ii) at least one spade; (iii) at least one spade, but not a flush; and (iv) a flush, but not of spades.

318

EXERCISES IN PROBABILITY AND STATISTICS

Hence determine the probability that the player has (v) a flush when it is known that he has at least one spade' and (vi) a flush of spades when it is known that he has a flush. '

23 An electrician buys a box containing N fuses of which a number '
1)]':

and (b) the conditional probability that a second sample of v fuses from till' N - v remaining ones is wholly good is

r~l

(;)

(~ ~~v) / (~

1_(N:n)/(~ 24 A bag contains 6n (n;:. 1) identical tickets which are numbered serially from 0 to 6n -1; and an experiment consists of drawing a ticket randoml\' from the bag and noting its number, This experiment is repeated three timc~, the drawings on the second and third occasions being made after tl1l" previously drawn ticket had been replaced in the bag, If S denotes the slim of the numbers on the three tickets drawn, prove that d ( ') P(S=6 )=(3n+2)(6n-l) , 1 n 216n 3 ,an

(ii) E(S) =~(6n -1). Also, show that if the sampling is without replacement of the drawn, then 3n P(S = 6n) = (6n -1)(6n -2)

tickct~

25 An urn contains m + n balls which are identical apart from their colour: m of them are black and the rest white. A ball is drawn at random froIll the urn and its colour observed. The ball is then returned to the urn and. in addition, r balls of the colour drawn are added to the urn. A ball is 11011 drawn at random from the urn. Find the probability that (i) both balls drawn are white; (ii) both balls drawn are black; (iii) one ball drawn is black and the other white, irrespective of the ortle' in which they are drawn;

SUPPLEMENT

319 (iv) the second ball draw.n is whit~; ~n? (v) the first ball drawn IS black, If It IS known that the second ball is white. I\lso determine the expectation of the number of white balls drawn in Ihe tWO trials. '6 An urn contains 2N balls which are identical except for their colour, : If the balls being white and the rest black. If r «N) randomly selected

:ans from the urn are all found to be of the same colour, determine the ~obability that the colour of the balls drawn is white. r In another experiment k «N - r) black balls are first removed from the ~rn, and then, from the remaining 2N - k balls, r balls are selected ranJomly. If these selected balls are known to be of the same colour, show that Ihe probability that the colour of the balls drawn is white is 1 p = 1 +(N-rYk)/N(k)'

If N is large compared with rand k, so that the terms involving N- 2 are negligible, prove that approximately p=

~ [ 1 + 2k~ J.

17 In "The Final Problem", a short story by Sir Arthur Conan Doyle, Sherlock Holmes is pursued by his old enemy Professor Moriarty with murder in mind. It is understood that if Moriarty catches Holmes then the latter must die. Holmes takes a train at London bound for Dover from where he can escape to the continent. Moriarty charters a train and follows dose behind. There is only one stop before Dover-Canterbury. If both Holmes and Moriarty choose to alight at the same station, then Holmes meets his end. Secondly, if Holmes chooses Dover and Moriarty Canterbury, then Holmes makes good his escape. Lastly, if Holmes gets off at Canterbury and Moriarty goes to Dover, then the pursuit is a draw since Holmes, though temporarily safe, is still within the reach of Moriarty. If the independent probabilities of Holmes and Moriarty getting off at Canterbury are x and y respectively, determine the probabilities of the three distinct outcomes of the pursuit. If the probability of Holmes meeting his death in the encounter is a (>0), and this is also the probability of his escape, then prove that y2-3ay+a =0. Hence deduce that in this particular case a must satisfy the inequality 28 (i) A standard pack of playing cards consists of four colour suits, each suit having 13 distinct cards. The 13 cards of a suit include four court cards-an ace, a king, a queen and a jack. From the pack, a hand of 13 cards is dealt out randomly to a player. Find Per), the probability that the hand contains exactly r court cards, 0:;:; r:;:; 13. Hence show that, for r ;;'1, per) satisfies the difference equation P( )_(17-r)(14-r)p(

r -

r(23+r)

P(O) = 36(13)/52(13\

).

r-1,

and

in standard factorial notation.

320

EXERCISES IN PROBABILITY AND STATISTICS

(ii) Two players A and B playa game of chance as follows. From the pack of 52 cards, 13 cards are dealt out to A and then another 13 cards to B, the player having more court cards being the Winner f the game. The game ends in a draw if both A and B have no COU~t cards or the same number of them. Prove that the probability fo the game to be drawn is r

t (16)(16-k)( 36 )(23+k)/(52)(39) k k 13-k 13-k 13 13'

k=O

Also, in particular, if it is known that A has all court cards, then show that the probability of B having no court cards is ~ approximately. Verify that the percentage error in this approximation is;'!7·

29 In a game of chance, a player repeatedly rolls k (> 1) unbiased cubical dice till at least one six shows up. If the event "at least one six" OCcurs on the rth throw of the dice, the player wins £2 1- kr (r;;;:.l). Prove that the probability of the player winning on the rth throw is (~)k(r-l)[l-(~)kJ.

Hence deduce that his expected reward is l-(~)k

£ 2k -

1

[1-(&YJ'

so that, on the average, the player can never be on the winning side if the non-returnable stake to play the game is £2 1- k • Also, show that the expected reward is a decreasing function of k.

30 An urn contains n (> 3) balls, either white or black, but otherwise identical; each ball has a probability p of being white and q of being black (p + q = 1). A random sample of k «n) balls is drawn from the urn. Prove that the probability of the sample containing only one white ball is k

"-t+ (nr-11

k)prq"-r.

r=1

Furthermore, if it is known that the urn contains one black and one white ball and the remaining n - 2 balls are equally likely to be either white or black, prove that the probability of the sample of k balls containing only one white ball is ,,-k+l (11 -

k

n(I1-1)2,,-2'

r~1

k)

r(n-r) r-1 .

31 A man plays a game of chance in which, at each turn, he rolls a pair of unbiased cubical dice whose faces are numbered from 1 to 6. He continues until the first "successful" turn, a successful turn being one in which at least one of the dice shows 6. If this is his 11th turn (11 ;;;:.1), he receives a reward of £a" (a < 1). This terminates the game and the stake for playing it is 50 pence. Prove that the player's expected reward is £l1a/(36-25a). Hence deduce that, on the average, the player will be on the winning side if a> ~~. 32 In a game of poker, five cards out of a standard pack of 52 are dealt out at random to a player. There are four suits in a pack and the 13 cards in a

SUPPLEMENT

321

suit are regarded to be in an ascending order in which the ace is taken to be Ihe lowest card and also the highest. Any five cards (not necessarily of the same suit) which are in an unbroken sequence are said to form a run. I-Iowever, if the five cards in the run belong to the same suit, then the sequence is known as a running flush. Find the probability that the player has a run but not a running flush. Also, show that there is a 4 per cent increase in this probability if it is known Ihat the player has the jack of hearts. What is the explanation for this increase? 32a There are It alternatives in a multiple choice test, and the probability of an examinee knowing the right answer is p. However, if he knows the right answer, then it is not absolutely certain that he will give the correct response because of the emotional tensions of the testing situation. It may therefore be assumed that the probability of the examinee giving the correct response when he knows the right answer is 1 - a where a > 0 is small but not negligible. Alternatively, if the examinee does not know the right answer Ihen he randomly selects one of the It altet:natives. (i) Prove that the probability of the examinee knowing the right answer when he, in fact, gives the correct response is Itp(1-a)

l+[n(l-a)-l]p·

Also, determine the probability of the examinee not knowing the right answer when, in fact, he gives an incorrect response. (ii) Evaluate these probabilities on the assumption that when the examinee does not know the right answer, he knows that m < It of the alternatives are incorrect so that he makes a random choice from the remaining It - m alternatives. (iii) Find the limiting values of the probabilities in (i) and (ii) as It ~ 00, and comment on the results obtained. 33 As a financial inducement to Ruritanians for changing their habits of smoking, the government presented two separate proposals in parliament. The first proposal referred to a decrease in the excise duty on cigars and lobacco, and the second called for an increase in the duty on cigarettes. Two free votes were taken in parliament with the following results. The ratio of Ihe number of M.P.s voting for a decrease in duty on cigars and tobacco to Ihe number voting against it was one and a half times the ratio in the vote on an increase in the duty on cigarettes. Again, of those M.P.s who voted in favour of a reduction in the duty on cigars and tobacco there was a majority of 165 in favour of the increase of duty on cigarettes; and of those who voted against the decrease in duty on cigars and tobacco there was a majority of 135 in favour of an increase of duty on cigarettes. Finally, if 110 of those who voted for both government proposals had, in fact, joined those who voted against them, the number of the former would have been twice Ihat of the latter. Assuming that there were no abstentions in the votings, find (i) the total number of M.P.s attending the meeting; (ii) the numbers voting for each proposal; (iii) the numbers voting for and against both proposals; and (iv) the numbers voting for one and against the other proposal.

322

EXERCISES IN PROBABILITY AND STATISTICS

34 Each of two numbers x and y is selected randomly and independ from the integers 1, 2, 3, 4 and 5. Show by suitable enumeration tha~ntly probability that x2 + y2 is divisible by 5 is is. Further, prove that this is ~he the probability for 2X2+3y2 to be divisible by 5, and that the probab~l'sll that both x 2+ y2 and 2x 2+ 3y2 are divisible by 5 is ,Js. Illy Extend these results to the case when x and yare chosen frollJ h integers 1 to 5N, where N is any integer ~1, to show that the ab t e probabilities are independent of N and so also applicable when x and y ~\'e selected from the set of all integers. dre

35 There are k similar urns numbered from 1 to k, each containing IJJ + 1 (11 > k) balls which are identical apart from their colour. The balls in the urn are all white, but in the rth urn (1'::;; r'::;; k -1) there are m + r black 'I ; 11 - r white balls. An urn is selected randomly and two balls are drawn fl:(;l( it. If both the balls are found to be white, determine the condition~; probabilities that the kth urn was selected, on the assumption that tI;~ second drawing was made (i) after replacement of the first ball drawn; and (ii) without replacement of the first ball drawn. Explain very briefly why the conditional probability in (i) should be less than that in (ii).

kl:

36 There are six unbiased dice Dl> D 2 , • •• , D 6 , the six faces of D j beinl: numbered (i, i, i, 4,5,6) or (i, i, i, 1,2,3) according as i.::;; 3 or i > 3. If one Ilf the dice is selected randomly and rolled twice, then prove that the probahility of obtaining a double-six is -h. Further, given that a double-six has been observed, prove that thl' probability of the selected die being D6 is i. How is this probability altercd if it is known that D3 is a biased die such that, though it is equally likely 10 show a three or a number greater than three, the probability of it showing a six is twice that of obtaining a four or a five? 37 On receiving an external stimulus, a certain kind of biological particle gives rise to a progeny of r (0.::;; r'::;; n) particles with probability (~)p'q" " where p and q are positive parameters such that p + q = 1. The original particle then dies. The particles, if any, constituting the progeny act independently of each other and are subjected to another external stimulus, till' behaviour of these progeny particles being the same as that of the original particle. If Xl and X 2 are random variables denoting the number of particles in the progenies generated after the first and second stimulus respectively, prove that the joint probability-generating function of Xl and X2 is E(OflO~2) == G(Ol> O2) = [q + pOl(q + p02)/I]/I.

Hence, or otherwise, show that (i) P(X2 > 0) = 1- q/l (1 + pq/l-l)/I; (ii) E(Xl + X 2) = np(1 + np);

and

(iii) var(Xl + X 2) = I1pq[np + (1 + np f].

38 A system can be in anyone of sixteen states which may be represented as the equal cells of a square divided by lines parallel to its sides. Initially the system is equally likely to be in anyone of the states. The sixteen states are of three types conveniently denoted as A, Band C, and their distribu-

SUPPLEMENT • J1

11 0

323

is as given in the diagram below. A B B A

B C C B

B C C B

A B B A

After initial placement, the system is moved n times. At each move, the system is transferred to a neighbouring cell in any direction horizontally, "ertically or diagonally, all such moves being equally likely. \ For n;?!: 1, suppose all' {311 and 'YII respectively are the probabilities that after It moves the system is in an A-type, B-type or C-type cell. Prove that Ihese probabilities satisfy the following difference equations: a" =!{311-1 +!'YII-l

{311 = ~all_1 +~{311-1 +!'YII-l 'Y" =1all -l +~{311-1 +i'YlI-l'

Show that the solution of these difference equations is all =!-i{3",

where

10 [ 13 (-9)"] .

{3"=21 1- 90 40

Hence deduce the probabilities that after n moves the system will be in a specific A-type, B-type or C-type cell and verify that, as n ~ 00, these probabilities are in the ratio 3: 5 : 8.

39 Two players A and B contest a series of games. At each trial the probabilities of A and B winning a game are p and q respectively (pi 0, qrO), and there is a non-zero probability r for it to be drawn. It is agreed Ihat the winner of the series will be the player who is the first to win four more games than his opponent. Prove that, for pi q but irrespective of the value of r, (i) the initial probability of A winning the series is 1 1 + (q/p)4 '

and (ij) the probability of B winning the series when he is exactly 4) games behind A is

11

(0 ~ n ~

1-(p/q)4-11 1- (p/q)8

Hence deduce the values of the probabilities (i) and (ii) when a game cannot be drawn, the other conditions for the contest being unaffected.

40 A self-service store has a stock of nt + n undated one pint bottles of milk of which n are fresh and 111 a day old, and it may be assumed that these

324

EXERCISES IN PROBABILITY AND STATISTICS

are placed randomly on the service counter of the store. If a cu selects k (k < Ilt and n) bottles of milk, find the probability that she ~O~lcr fresh milk bottles. as
In

f (v) (In n+~ -:-k k)j(1lt n+ n), n-v-k>O, r

r=O

r

and that she gets no fresh milk is

f (v) (111 + nn~ v- k)j (111 +

r=O

r

r

11).

11

Hence, or otherwise, prove that the conditional probability that the secor I customer obtains k bottles of fresh milk when it is known that at least one :'r her bottles is fresh is I

f

r=O

(v)(I11+~-:-k)j[(m+n)_

nr k

r

n

f

r=O

(v)(m+n~v-k)]. r

nr

41 In a game of chance, a player can score one, two or three points at " trial with constant probabilities q, pq and p2 respectively, where p + q = 1. in repeated trials of the game, Un denotes the probability that the player< total score will be n points at some stage, then show that Un satisfies a third-order linear difference equation with the initial conditions Uo = 1 and Ul = U2 = q. Hence deduce that

Ii

1 [1+(-P)n+l{~Sinn71'/3-(1+P)COSn71'/3}], for 11;;;:0, (1+p+p2) ,,3 42 A finite population consists of the N integers 1,2,3, ... ,N, where N is unknown. A random sample of n integers is taken from this population without replacement, and it is observed that x is the largest integer in the sample. If X is the random variable denoting the largest integer in samples of n obtained from the population, then prove that the probability distrihution of X is given by the set of equations Un

=

P(X=x)= e=~)j(~,

for

n~X~N.

Use this probability distribution to show that for any integer r the rth factorial moment is E[(X + r-1)
Hence verify tliat

(n+r).N. T=(n+1)X 1 n

is an unbiased estimate of N, and that

1)(N -n) n(n+2) .

( ) _ (N +

var T -

SUPPLEMENT

325

[Note: It may be assumed that

f

x="

(X+k)= (N+k+1) l1+k l1+k+1'

ror all integer values of k ~ -1.] 43 A common method of making a saving for the future is by means of periodic payments, usua!ly at equispaced intervals of time. These payments accumulate compound Interest at a fixed rate, and the whole amount is payable after a stipulated number of years. An investor makes an initial payment of £R, and ag.r~es to pay every subsequent year increasing amounts R,2R, 3R, .... In addItIon to these payments, the investor has probabilities pand 1 - p of contributing £2R or £R towards the annuity at the beginning of each year after the first. Prove that the expected amount accumulated after 11 years is

R

SIt =2 [{I +(1 + p)p}{(l + p)" -1}+ p2(1 + p)" - np], p

where the annual percentage rate of interest is lOOp (0 < P < 1). Hence show that, irrespective of the value of p and for 11 large compared with IlP, SIt
Finally, if under the above conditions the investor desires a total return of A, deduce that

1

n >-log[Ap2/R(1 + pf]. p

44 In The Wind on the Moon by Eric Linklater (Macmillan; 1944), the Palfrey sisters, Dinah and Dorinda, have in all the sum of 16 shillings and 11 pence to pay for the unusual services of the barristers Hobson and Jobson. In offering the sum to the barristers, Dinah thinks that it would be rather difficult for them to divide the money equally between them. However, Mr Jobson thinks otherwise and declares "I see there are twice as many half-crowns as shillings and the number of half-pennies is the same as the number of shillings and half-crowns added together, including the very shining bob with the head of Queen Victoria on it. Then if you add to that (that is, the total number of half-crowns and halfpennies) the total number of two-shilling pieces, the total is one and a half times the number of pennies. Now let me see ... " "There are two single shillings", said Mr Hobson, "and if you multiply the number of half-crowns by that you will get the exact number of pennies. It's perfectly easy, Jobson. We divide the 16 shillings and 11 pence into two exactly equal halves, by taking 11 coins each. There are no complaints and nothing left over." Determine the individual face values of the 22 coins offered in payment to the barristers. Hence show that if the 22 coins are randomly divided into two groups of 11 each, then the probability that the resulting division of money will be equal is 200/4199-0·0476. It may be assumed that the symmetrical distribution of coins is the only solution for equal division of the money.

326

EXERCISES IN PROBABILITY AND STATISTICS

Also, show that the same distribution of the face values of the coi . ns IS obtained even if the number of shillings is not known. [Note: For those unfamiliar with the pre-decimal British monetary syste shilling or a bob was worth 12 old pence, a two-shilling piece 24 old P:' a IICr and a half-crown 30 old pence.]

45 (i) Define (x), the distribution function of the unit normal variable X and then show that for any x > 1 . x

1 Jexp[1 -1) +<1> (x-2+ 1) -2 (t2-1)2J dt= (X-2-1.

2J(2'Tr)

-x

(ii) The parliament in Ruritania consists of 2N + 1 members. On . non-party motion, a free vote is proposed. A pressure group consiS\~ of 11 members who agree amongst themselves that they will all VOir in the same way, but it may be assumed that the remaining members will vote independently, and each is equally likely to vote for o~ against the motion. Assuming that there were no abstentions at the actual voting, determine the probability that more than half the total votes will be cast in the same way as those of the pressure group. Jr N is large compared with 11, use a suitable approximation to show that, for fixed N, this probability is an increasing function of II. Further, assuming that the effective range of a unit normal variable is from -3 to 3, prove that it is practically certain that the decision on the motion will go the same way as the voting of the pressure group if 11 ;;;, ~[J(8N + 13) - 3].

46 A population of wild animals consists of N members, where N is a large but unknown number, and a biologist is interested in obtaining an estimate of N. He takes a random sample of W animals, marks and then releases them. After this, he starts capturing the animals randomly and noting whether they are marked or not. He continues this sampling without replacement till this second sample contains exactly w (fixed and preassigned) marked animals. The sampling is discontinued and it is observed thai the total sample size is n. If X is a random variable denoting the size of the sample, prove that P(X=I1)=

(~=:)(:=~)/(~),

for

w~X~N-W+w.

By considering P(X = n) as a function g(N) of the unknown parameter N, show that g(N) g(N-1)

(N -11)(N - W) N(N- W-I1+W)'

Hence deduce that the maximum-likelihood estimate of N is the largest integer contained in IlW/w. Also, assuming that E(X) = w(N + l)/(W + 1), evaluate a simple unbiased estimate of N. [Note: It may be assumed that no change in the animal population occurs in between the taking of the two samples.]

SUPPLEMENT

327 47 A player has a constant probability p (>1) of winning in an independent trial of a .game of c?ance. The sta.k~ for each trial is a shillings. If the player wins a trIal he receives (a + x) shIllIngs (x> 0), but he loses his stake money ~J1 the contrary event. The player decides to play this game with the proviso :hat he would stop play immediately after his first failure. If S is the random variable denoting the net gain of the player at the end of his play, prove that px

E(S)=--a; q

px 2

var(S)=-2 , q

p+q=1.

Also, show that the player can expect to gain at least a shillings at the end of play if x> a/po

48 Small electrical fuses are mass-produced and then packed into large boxes by an automatic filling machine. The number of fuses put into a box by the machine cannot be controlled completely; and in order to ensure that no customer is a loser, the filling machine is so adjusted that it puts at least n [uses in a box. It may be assumed on empirical considerations that the actual number of fuses in a box is a discrete random variable X having the geometric distribution with point-probabilities P(X = x) = (1- p)px-n,

for

X~

n,

where p is a positive parameter such that 0 < P < 1. Given a randomly selected box, the probability that any particular fuse selected at random from the box is usable is a constant 0, where 0 < 0 < 1. If a box contains exactly x fuses, prove that the probability that it contains exactly r usable fuses is P(r I X = x) =

(:)0'(1- oy-r,

for

0:0;;; r:O;;; x,

and write down the expected number of usable fuses in a box containing exactly x fuses. Hence, or otherwise, show that the expected number of usable fuses in a randomly selected box (whatever the number of fuses it contains) is E(r) = nO + pO/(l- p).

49 A manufacturer of a breakfast cereal introduces a coupon scheme as a sales promotion campaign. A set of coupons consists of It (~2) different ones and one coupon is placed in every packet of the cereal. A complete set of coupons can be exchanged for a stock of spring bulbs. A housewife buys one packet of the cereal every week. Assuming that it is equally likely for her to find anyone of the n coupons in a packet, prove that the probability that she will have to wait r (~1) weeks before obtaining the (x + l)th coupon when she already has x different coupons is P(r I x) =

(1-~) (~r-l,

for

1:0;;; x:O;;; n-1.

Hence show that, for any given x, the expectation of r is It/(Il- x), and the expected total number of weeks required for the completion of one set is

328

EXERCISES IN PROBABILITY AND STATISTICS

50 (i) If X is a random variable having the Poisson distribution with /L, and a, (3 and k are non-negative integers such that 0 < k s: l11ean prove that '" Q <:" I~, 13

L r(klp(X = r) = /L kP(a -

k os; X os; (3 - k).

r=a

(ii) A photographic dealer normally sells a standard roll of colou 1'1 for 42 new pence, but in order to encourage sales he deck[ ,1111 allow a discount to customers who purchase more than one rOI~~ III time. To a customer who buys r (2 os; r os; 5) rolls at a time the
where a = P(Oos; X os; 5); b=P(1OS;XOS;4); c=P(Oos;Xos;3). Henn' deduce that var(Z) = 36(1- ~~+e~~b + /Lc)

[6(\~a:_~ /Lbr

What is the average return per roll of film sold under the schellle of the dealer?

51 The attendance records of workers at a large factory kept for a IOIl~ time show that the probability of a worker being absent for r (;;;'0) days in a year was approximately given by the Poisson distribution with mean IJ.. The absence of a worker causes dislocation in the production schedule which results in a loss to the factory; and it was estimated that, on the average, if a worker was absent for r days in a year the net loss to the factory was £a/'(r+ 1) e-13r, where a and (3 are positive parameters and (3 is small. Prove that the expected loss (in pounds) to the factory due to absenteeism of a worker in a year is a/L(/L e- 13 + 2)exp[ -(/L + (3) + /L e- 13 ], Hence show that if /L and (3 are such that 1 < /L < el3, then the expected loss must be less than 3a.

52 The Poisson distribution is occasionally referred to as a distribution describing the occurrence of rare events. Explain this remark. A commercial secretarial organization charges Sa pence for cutting a foolscap stencil. The price includes the cost of the stencil and the remuneration of the typist, and the residual profit of the organization is a fifth of the charge made, provided the typist makes no typographical errors. However. if an error is made, then the profit of the organization is diminished by the additional cost of correction, On empirical considerations, it is known thaI the cost in pence of correcting r errors on a stencil is 2r(3r + 2)/(r+ 1), for r;;;' 1, It may be assumed that, on the average, the typist makes /L errors per

SUPPLEMENT

329

.Ieocil where /L is.lsmaldl. Shuggesht a sUhitable model for the distribution of ors on a stenci an t en s ow t at the expected cost in pence of :rtrrecting the errors made on a stencil is .0 2[3/L -1 + /L -1(1-e- lL )]. Ileoce show that if the organization wishes to obtain, on the average, a oftt of A.a pence per stencil, where A. is a constant such that 0 < A. < 1, and \ assumed to be sufficiently small for terms involving /L 2 to be negligible, ;hell. as a first approximation, the organization will attain its objective if 5/L

a=I_A. .

<3 A stationery store plans an advance order of a standard line of next :ear's calendars which are usually sold in the pre-Christmas season. The order can be placed for any number of calendars, and the cost per calendar 10 the store is two shillings and sixpence. The store expects to sell the 'alelldars in the pre-Christmas season at a (1 < a < 2) times the cost price, ~ut the residual stock will have to be disposed of in the January sale at a fraction (3 (0 < (3 < 1) of the cost price. The past experience of the store ,hoWS that the pre-Christmas sales of this line of calendars are Nx, where N ~ a positive integer and x is a random variable having a Poisson distribution lI'ith mean /L and distribution function F(x). If the store orders Ny calendars, show that the expected profit in pence of the store is y G(Y)=30N[(a-(3)

,,~u (x

y)p(x)+(a-l)YJ,

II'here P(x)=e-ILIL"/x! Hence, by considering the behaviour of the finite difference G(y + 1)G(y), prove that the optimum size of the stock which the store must order to maximize its expected profit is Nyu, where Yu is the smallest integral value of y for which a-I a-(3

F(y)~--.

54 A government wishes to control immigration by restricting the entry of foreigners to an annual intake of a persons. (It may be assumed that the immigrants enter the country at the commencement of each calendar year.) Suppose initially, at the start of the first calendar year, the total population of the country is a constant N. However, due to the natural changes in the population, it is presumed that the actual number of persons in the country at the end of the year will be Nx, where x is a random variable such that its probability density function is A. e-A(x-l),

for x ~ 1 and A. > 1 is a parameter.

Accordingly, at the beginning of the second calendar year the total population, including the annual quota of immigrants, is another random variable }'2=a+Nx.

Prove that E(Y2)=a+N(I+A.)/A.. In general, suppose that the expected population, including the immigrant quota, at the beginning of the rth (r ~ 3) year is 1 + A.) E(Yr-l)' E(Yr) = a + ( -A.-

330

EXERCISES IN PROBABILITY AND STATISTICS

Use this equation to verify that the expected population at the beginni the (k + l)th year is ng of a[

(1 + "f - AkJ

Ak -

1

+N

(1 + A)k A .

Hence deduce that, if the annual immigrant intake a = N/ A, the popular will attain IJ. times its initial size at the beginning of the (k + l)th year whe101J . , re . fi es the equation k satIs 2(1 + A)k = (IJ. + l)A k. If A is large, show that an approximate solution of this equation is

1)

IJ. + . k =(l+A)log ( -2-

55 The annual profit of a steel company depends upon the demand x and the two scaled parameters a and (3 representing the capacity of the plant and the selling price of steel respectively. It is known from past experience that, for given a (>1) and (3 «1), the profit increases with rising demand However, the effect of an increase in price is to decrease demand due to th~ competition of imported steel; consequently, for given a, the total realizable profit tends to decrease somewhat because of the additional cost of holding unsold stock. Accordingly, it is suggested on empirical evidence that a reasonable approximation of this dampening effect of price increase on profit is a factor exp[-(3(a - x)], which proportionately reduces a parabolic increase in profit with rising demand. It may therefore be assumed that, on a suitable scale, the profit of the company is T= aX(l + (3x)

e-!Ha-x).

Hence, assuming that the demand is appropriately scaled so that x is a normally distributed random variable with zero mean and variance (3, prove that the expected profit is E(T) = a(32«(33 + 2)exp[-(3(a - (32/2)]. Finally, suppose (3 is fixed by a retail price agreement amongst the steel companies, and a company has the freedom to determine the production capacity of its plant for maximizing the expected profit. Show that under such a restrictive price arrangement the company can obtain maximum profit when a = 1/(3.

56 A factory produces corned beef for export which is despatched in large consignments of 12 oz. containers to customers overseas. However, due to a recent failure in the sterilization of the product, the government has imposed the restriction that a consignment should only be allowed to pass for export if a random sample of 100 containers from it is found to be completely free from infection. It is suspected that, despite improvement in the sterilization process at the factory, there is an a (small) per cent rate of infection in the consignments offered for inspection. If it may be assumed that the size of the consignments is large compared with the sample size so that the effect of sampling from a finite population is negligible, prove that the probability of a random consignment being passed as free of infection is approximately

SUPPLEMENT

331 Hence deduce that the probability that of a batch of ten consignments least three will be rejected on inspection is

_fl.

:1

1-

L 2

w=O

(10) (l-e-"')W(e-"')lO-

w•

W

If, over a period of time, this probability is assessed to be 1· 5 per cent, verifY that the rate of infection still present in the consignments is approximately five per cent. [Note: e- o.os = 0.95 correct to two decimal places.]

57 A printing house proposes to pay £a as a standard charge for proofreading a book irrespective of the number of errors detected by the proof-reader. If no errors are detected in the proofs, then the proof-reader is paid another £ e-Aa, where A> 0 is a parameter. On the other hand, if the proof-reader detects errors in the proofs, then he is paid £pr-1a if he detects r errors for 1::;;; r
P(x = 0) = l-e- A ; P(x=r)=(I-e- A )e- r>-,

for l::;;;r
Assuming that the proof-reader detects the errors present in the proofs, shoW that the expected amount payable to him is e-A(I-e-A)(2-p e- A)] a + I-pe - A '

[1

and hence deduce that this amount can never exceed 7a/4.

58 A retailer obtains his supplies of a small machine component from the manufacturer in large batches, each batch consisting of M components. Although the production is under statistical control, there is a proportion e of defective components in each batch such that e may be regarded approximately as a continuous random variable with the probability density function 60(1- 0) in the range 0::;;; e::;;; 1. The retailer sells each component for one shilling but, to attract custom, he agrees to refund a shillings (1 < a < 2) to a customer for a defective component supplied. To protect himself under such a guarantee system, the retailer takes a sample of N (fixed and «M) components from each batch supplied to him by the manufacturer for destructive testing, and refuses to accept any batch containing n or more defective components. For any given 0=0, find the conditional probability of the retailer accepting the batch after inspection, and then show that his expected net income in shillings from such a batch is g(n, 0) = (M - N)(I-aO)

~t~ (~W;(I- O)N-x.

Hence deduce that the expected net income of the retailer per batch for variation in e is g(n) =

(M - N)(n + 1)(2) 2(N+4)(4) [6(N+2)(N +4-a)-2(2n + 1){(N + l)a + N +4}+

+ 3an(n + 1)].

332

EXERCISES IN PROBABILITY AND STATISTICS

Finally, assuming that N is sufficiently large for terms involving N-I be negligible but 11/N is finite, show that g(n) is a maximum for 11::: N;o approximately. Q

59 A factory produces packets of margarine and the nominal weight of th contents of each packet is one lb. However, due to random fluctuations .c the automatic weighing machine, the actual weight of the contents of I~l packet is a normally distributCi:d random variable x with mean IL a ~ variance u 2 (IL »u). It is also known from the past experience of the facton( y that . the cost in shillings of producing a packet containing x lb of margari,r~ IS

0: + (3x - (x 2 -

( 2)

e --v(x-/-L)/u,

where 0:, (3 and 'Yare small positive parameters. Prove that the expectcd cost of a packet of margarine is a

+ (3/1- - (/1- - 'YU f e _~-V2.

An economic crisis forces the government to devalue its currency and consequently, there is an increase in the import price of edible oils used ir; the production of the margarine. To offset a part of the increase thc manufacturer makes certain small economies in his fixed costs and also establishes checks to lessen waste. As a result of these measures, thc parameters of the cost function are expected to become 0:' «0:), (3' (>~) and 'Y. The manufacturer aims that, on the average, there will be no increasc in his cost of production; and he plans to achieve this by suitable alterations in the parameters of the weight distribution. Assuming that the new weight distribution is normal with mean /1-' and variance U,2, and /1-' is so chosen that 0: + (3/1- = a' + (3' /1-', prove that there will be no change in the expected cost of production if /1- -/1-' = 'Y(u-u'). Hence show that the production of packets produced by the new system which are less than the nominal weight is ['Y{-(O: - 0:') + «(3' - (3)/1- + (3'(1-/1-)J (0: - a') - «(3' - (3)/1- + 'Y(3' u ' where (z) is the distribution function of a unit normal variable z. 60 It is known from the past experience of a publishing house that a small proportion of typographical errors remains undetected by the usual methods of proof-reading. In order to improve upon the existing procedure, the publishing house agrees to an additional proof-reading of a book with the proposal that the proof-reader will be paid (i) £0·25 for a page on which he detects no error; (ii) £1·00 for a page on which he detects a single error; and (iii) £1·25 for a page on which he detects more than one error. The probability that a word on a page is typographically incorrect is a small number a and, on an average, there are 11 words on a page. If /I is sufficiently large compared with 0: but 110: = /1-, a finite number, give without proof the probability that there are exactly r (~O) errors on a page. Hence,

SUPPLEMENT

333

assuming that the proof-reader detects the errors present, show that the eXpected payment per page to the proof-reader is £~[5 -

(4+ IL) e- IL ].

The proof-reader is unwilling to accept the suggested rates of payment because he realizes that it is unlikely that many pages will contain one or ptO re errors per page. As a counter-proposal, the proof-reader asks for an arrangement which would give him an expected payment of £1·00 per page by raising the amount £0·25 in (i) when the amounts £1·00 and £1·25 in (ii) and (iii) remain fixed. Prove that the increased payment that the publishing house should make for each page on which the proof-reader finds no errors, in order to meet his demand is £M5 + IL -e- IL ].

61 A manufacturer of a duplicating machine rents out the machine to cuslomers on an annual basis. The annual rental for a machine is fa, a fixed charge, plus an amount which depends upon the number of copies made in a year. Accordingly, if x copies are made on a machine in a year, then the lolal rental payable (in pounds sterling) is T(x) === a

+ bx(1-tx),

where b is another constant. If it is known from empirical considerations that x can be regarded as a continuous random variable with the probability density function (A + 3)(A + 2)x (1+X)A+4 ,forx;;a.O,

where A > 1 is a parameter, prove that for r;;a. 0 E(x r ) = f(r+2)f(A - r+2)/f(A +2).

Hence deduce that E(T) =

a

+ (2A -1)b

A(A+1)'

The manufacturer wishes to so arrange the rental of the machines that, on the average, he will obtain a rental income of at least £Aa per machine. Show that the manufacturer will attain his objective if

a

2A-1

b

A(A -1)'

-<---::-2

62 In a certain welfare state, a person pays an annual contribution for social security based on his income x which is greater than a, the minimum individual income in the population; the amount of the contribution is A(x - a), where A is a parameter such that 0 < A< 1. The probability distribution of the incomes of individuals liable to pay the social security contribution is log-normal with the probability density function _1_(x-a)-lexp-_1_[log(x-a)-lLf, for x>a, u.,fi;,. 2u 2 Where u and IL are parameters in standard notation. Prove that, on an average, the contribution paid by an individual is A exp(1L + !u2 ).

334

EXERCISES IN PROBABILITY AND STATISTICS

As a concession to income earners, the government proposes that 0 I persons with incomes greater than (3 (>a) should be required to pay ~ y social security contribution at the rate of A(X - (3). Show that the aVer he contribution now payable is A times age [1-<1>(8 - 0')] e fL +¥>"2 - ({3 - a )[1- <1>(8)], where 8 == [log({3 - a) -IL ]/0' and <1>0 is the distribution function of th standardized normal variable. Hence deduce that the relative change in the average social security contribution is e <1>(8 - 0') + [1- <1>(8)] exp 0'(8 -!O').

63 An owner of a large Edwardian mansion has the choice between tw methods of heating the residence. He can either continue to use solid fuel i:: the existing fire-places or replace them by thermostatically controlled gas fires. It is estimated that the annual consumption of solid fuel would be N tons at a cost of £a per ton. On the other hand, the initial cost of installing t~e gas fires is substantial and it m~y be discounted as an annual cost of £(l Smce the gas fires are thermostatIcally controlled, the consumption of gas would depend upon the daily temperature distribution as recorded within the house. For anyone day, the gas consumption in therms may be regarded as a random variable X with a probability density function proportional to e- Ax2 . X3 for X>O, and A > 0 is a parameter. Prove that the average consumption of gas per day is N w/ A therms. Hence, assuming that the price of gas is £ 'Y per therm and that, on the average, there are 300 days in a year when heating is at all required, determine the expected annual cost of heating the house by gas. Also, show that, on the average, solid fuel heating will be more economical as long as 'Y

aN-{3

J>.

22SJ";'

->---

64 A refinery, located at a seaport, has a capacity of processing N gallons of crude oil during a week. The refinery is supplied with crude oil by tankers which arrive at random intervals of time; and if the refinery receives no fresh crude oil during a week then it is scheduled to process only aN gallons, where 0 < a < 1. The amount of crude oil received by the refinery during a week is a random variable X, and as X increases the scheduled processing capacity of the refinery also increases. However, since the capacity of the refinery is limited, it may be assumed on empirical considerations that this increase dies off exponentially. It may therefore be assumed that if the amount of crude oil received by the refinery in a week is X, then the scheduled processing amount is T = N[l-(l-a) e- 13x ], where (3 «1) is a small positive parameter. If the probability density function of the probability distribution of X is f(x)

=

x

e-~x2

for X;;:. 0,

prove that E(T) = N - N(l-a)[l-J2;. e!13 2{3{1-({3)}],

335

SUPPLEMENT

where. ct>(.) is th~ distri?ution function ~f a unit normal variable. Hence, aSSumIng that (3 IS sufficiently small ~or (3 to be negligible, show that in any one week the amount of crude oIl scheduled for processing cannot be greater than the expected amount if

0 1 log (7T X";-I3· 1- s e12) 2"

•

6S A continuous random variable X has the probability density function proportional to x"'-I/(l+(3x"')2",

for X;;oO,

where a and (3 are positive parameters and n > 1. Prove that if (2/1-1)a - r > 0, then the rth moment of X about the origin is given by

E(X') = 2n-l (3'/""

B(~+

12n -1-~) a

a'

in standard notation for the beta function. Also, show that the stationary values of the probability density function of X are the roots of the equation a '" ,..x

=

a-I (2n-l)a+l .

Hence deduce that for a> 1 the probability distribution of X has a mode at the point [

a-I

]

1/",

X= {(2n-l)a+l}(3

.

66 Explain, very briefly, the concept of conditional probability. A continuous random variable X has the beta distribution of the first kind with the probability density function 1 - - - x l1l - l (l-x) B(m,~

,

for 0..; X..; 1

,

where 111 is a positive parameter. Given a a positive number less than unity, prove that P(X ..;a) = a "'(111 + l-l11a).

Given another positive number (3 < a, prove that . ((3)111 111 + 1 - m(3 P(X"; (3 I X..; a) = - . 1 . a 111+ -l11a

67 A continuous random variable X has the probability density function ![1 + h(l- x 2 )], for -l..;X..; 1, where A is a parameter having some value in the interval (-1,1). Prove that for any non-negative integer r

E(X')=.!. [_1_{1+(-lY+2}+ 2A {I + (-lY+3}]. 2 r+ 1 (r+2)(r+4)

336

EXERCISES IN PROBABILITY AND STATISTICS

Hence verify that 2A

var(X) =

E(X) = 15 ;

75-4A 2 225

Also, prove that the distribution function of X is

( 21[A4

A

1 2)] F x)=- l--+x+-x 2( 1-2:x

2

.

Hence show that the median of the probability distribution of X is a rc' I root of the equation a

A(x 2 -1f-4x = 0, and that this root is > or <0 according as A> or <0.

68 A continuous random variable X has the probability density function proportional to x 2 /(1 + x 4 ), defined in the range -00 < X < 00. If k denotes the proportionality factor, prove that k (2r+3 1-2r) E(X2 r)=2 B - 4 - ' - 4 - '

. provided 1-2r>0,

in standard notation of the beta function. Hence deduce that k = .f2/7r and that none of the even moments of X exists. Also, show that the turning values of the probability density function of X are the roots of the equation

X4=1. Hence prove that the probability distribution of X is bimodal and determinc its modes. Finally, show that the points of inflexion of the probability distribution of X are the real roots of the equation

3x 8 -12x 4 + 1 = 0, and determine these roots explicitly. [Note: It may be assumed that f(t)f(l- t) = 1T/sin 1Tt.]

69 A continuous random variable X has the probability density function ex(3x",-l f(x) = (1 + (3x"'f' for X;;;'O, where ex and (3 are positive parameters. Prove that if ex > r, then the rth moment of X about the origin is given by E(xr) =_l_B(l +..!:. (3r/'" ex '

1-..!:.)

a'

in standard notation for the beta function. Also, for a> 1, show that the mode of the probability distribution of X is at the point [ ex -1 ]11'" x= (3(a+1) .

SUPPLEMENT

337

Finally, prove that the median of the probability distribution of X is at the point

flen ce , for ex> 1, infer the sign of the skewness of the probability distribution of X. 10 (i) For any two positive numbers ex and v, the incomplete gamma function is a ra(v)=r;v)

J

e-xx,,-l dx.

o

If

v> 1, then show by suitable integration by parts that e-aex,,-t ra(v-1)- ra(v) =

r(v)

(ii) For any r;;;:: 0 and 0> 0, prove that II

[(r, 0)=

Je-

4x2 x r dX=2(r-1)/2rA(r~ 1)r(r~

1),

o

where A =!02. (iii) A continuous random variable Z has a non-normal probability distribution with the probability density function 1 [Z4_6z2+3] fu e- tz2 1+ n ,for -oo
P(Z:SO; 0) = <1>(0) + e-!1I2 8 (3- ( 2 )

,

nfu

where <1>(.) denotes the distribution function of a unit normal variable and 8 is a positive constant.

71 A continuous random variable X has the probability density function porportional to (ex + Ix!) exp- ~ lxi, defined for -oo<X <00, where ex and ~ are positive parameters. Evaluate the proportionality factor and then verify that for any integer r > 0 E(X2r )

= (ex~ + 2r+ 1)(2r)! (ex~ + 1)~2r

Hence deduce that 'Y2, the coefficient of kurtosis of X, is 24

'Y2 = 3 -

(ex~ + 3)2 '

so that, irrespective of the values of ex and ~, l< 'Y2 < 3. 72 A continuous random variable X has the probability density function proportional to (ex+x4)e- X2 defined for -oo<X1) is a

338

EXERCISES IN PROBABILITY AND STATISTICS

posItIve parameter. Determine the proportionality factor and verify thai E(X) = O. Also, show that the (2r)th central moment of X is 1L2r

(X) = [4a + (2r + 3)(2r + 1)] r(r + !) (4a + 3)-/;

Hence deduce that X has unit variance if a =~. Prove that in this cas mode of the distribution of X is at the origin and the maximum ordina~ th~ the distribution of X is 3/(4";;). Also, show that the coefficient of kurtC (!f of the distribution of X is OSI, I'z(X) = -i. Compare these values with the corresponding ones for a unit normal distribution and comment on the differences observed.

73 The distribution function of a continuous random variable X having th . "logistic" distribution is defined by the equation ~ P(X:os; x) = (1 + e- x )-l,

for -00 < X < 00.

Evaluate the probability density function of X. By using the transformation y = (1 +e-X)-l, or otherwise, prove that the moment-generating function of X about the origin is Itl< 1. Hence deduce that 1'1 and 1'2, the coefficients of skewness and kurtosis of X are 0 and 6/5 respectively. . Also, show that (i) the semi-interquartile range of X is loge 3; and (ii) the points of inflexion of the probability density function of X arc
74 A continuous random variable X has the probability density function proportional to x 4 /(l + x 2 )(1 + X4), for -00< X < 00. Prove that the proportionality factor is 2/7T. Show that the turning values of the distribution of X are the roots of the equation X6

-x 2 -2=O.

Hence deduce that the distribution of X is bimodal and, assuming that the modes lie in the neighbourhood of the points x = prove that a more accurate approximation for the modal values is x = ± 35/23.

±J3i}'

75 A fluid flows continuously at a constant rate of Q molecules per minute into and out of a reactor. The reactor is said to be "well stirred" if the molecules inside it are always mixed in such a manner that the probability that any given molecule will leave the reactor during an infinitesimal time-interval Sx is approximately Sx/(3, where (3 == V/Q, V being the volullle of molecules inside the reactor. It is then known that the time for which a molecule remains in the reactor is a random variable X having a negative exponential distribution with mean (3. However, if two well-stirred reactors, each containing V molecules, are connected in parallel, then the entering fluid splits between them. If the total flow rate is 2Q molecules per-minute and a fraction a (O
SUPPLEMENT

339

loes through one of the~~ rea~to~s a?d the remaining 1 - a through the ~ther, th~n t~e probablhty. ~hstnbut.lOn of X is approximately hyper~%ponentIaI, with the probablhty density function f(x)

= 2;2 [e- 2aX/l3+

C:o'Ye-20 - a

)x/13

J,

O~X <00.

prove .that in this case the moment-generating function of X about the origin IS 2[2O'(1-O')-{1-2O'(1-O')}{3t]

Mo () t = ---'--------'-------'--'-~ (20' - (3t)[2(1- a) - (3t] . Hence deduce the mean and variance of X. Also, show that, irrespective of the value of a in its permissible range of variation, E(X) = (3

var(X) ~ {32.

and

16 Evaluate the moment-generating function about the mean of a unit normal variable X. Hence verify that the (2r)th central moment of X is (2r)! I-'- (X) - - -

2r

- 2r. r! '

for r;30.

Suppose Z is a continuous random variable with the probability density function proportional to e-!z2[1+O'z(z+1)],

for -oo
where a is a positive constant. Prove that the odd and even moments of Z about the origin are ,

(Z)

a

(2r+2)!

1-'-2r+1

=1+O'·2 r+ I .(r+1)!'

1-'-2r(Z) =

1+(2r+1)O' (2r)! 1 . -2r" +0' . r.

and for r;3 O.

Hence determine the mean and variance of Z. Also, show that both E(Z) and var(Z) are increasing functions of a but that, as a ~ 00, their limiting values are 1 and 2 respectively. 77 For a continuous random variable X having the negative exponential distribution with probability density function A e-AX,

for X;30,

obtain the cumulant-generating function of X. Here deduce that the rth cumulant of X is (r-1)!

Kr(X) = -A-r-

, for r;31.

Suppose e- Ay is regarded as proportional to the point-probabilities of a discrete random variable Y which takes all the non-negative integral values 0,1,2, .... Show that the probability-generating function of Y is 0(8) =E(8 Y ) = (1-e- A)/(1- 8 e- A).

340

EXERCISES IN PROBABILITY AND STATISTICS

Hence determine the cumulant-generating function of Y and verify thaI KI(Y)=1/(e A -1);

and

K2(Y)=e A/(e A -1)2.

Finally, if >.. is sufficiently small for>.. 2 to be negligible, prove thaI

KI(X)-Kl(Y)=~ (1-~)

and

K2(X)-K2(Y)=A

78 A standard pack of playing cards has four aces and 48 other card, l pack is well shuffled and then cards are dealt out one by one face up~. hl, until the second ace appears. If X is a discrete random variable denotin number of cards dealt out before the appearance of the second ace pg. Ill' that ' 111\\.

clrt

P(X = r) =

(~)C~81)(523_ r) /

(5r2) , for 1 ~X ~49.

Also show that P(X = r) satisfies the recurrence relation P(X=r+l)=

(r+ 1)(49- r) ( ) .P(X=r), for 1~r~48. r 51- r

Hence verify that P(X = r) is an increasing function of r only if r"", 16. Explain, very briefly, how the above recurrence relation can be lIsed obtain the numerical value of P(X ~ 5).

10

79 Explain, very briefly, the concept of unbiasedness of an estimate in Ihl' theory of estimation. A continuous random variable X has the probability density function x A f( x) = _1_ 6>..4 e- / .X3, for X

~ 0,

where>.. is a positive parameter. Prove that E(X) = 4>..

and

var(X) = 4>.. 2.

If Xl> X2, •.. ,X" are 11 independent observations of X, show thaI maximum-likelihood estimate of >.. is

llll'

A=~i,

where i is the average of the Xi' Hence deduce that an unbiased estimate of >.. 2 is ni 2 /4(411

+ 1),

whereas the estimate i 2 /16 has a bias of O(n- I ). 80 A continuous random variable X has a normal distribution such lhal both the mean and variance of X are equal to 0, unknown. If XI' Xb ','" x" are independent observations of X, show that the equation of 8, lhl' maximum-likelihood estimate of 0, is e(e+1)=i 2+s 2,

where n

ni=Lxj j~l

n

and

ns 2 =L(xj-if. i~l

SUPPLEMENT

341

~eoce deduce that

8 = ~({1 + 4(.XZ + S2)}! -1] ]lIII that, for large samples,

20 2 var(O) = n(28+ 1)' A

11 Explain,. ver'! briefly, the concept of unbiasedness in the theory of tatistical estimation . . A continuous random variable X has the probability density function

1

f( x) = -3 x 2 e- x/ fJ for X;a.O. , 20

prove that for any r;a. 0

E(xr) = !orf(r+ 3). Hence verify that E(X)=30;

Given that XI> X2, ••• , X" are independent realizations of X, prove that the maximum-likelihood estimate of 8 is 8=li,

Ilhere i is the average of the n observations. Also, determine var(O). Two functions 1 " L1 = x~ and ~=li2 4n i=1

L

Jre suggested as possible estimates of var(X). Show that Ll is an unbiased of var(X) but

~Iimate

\0

that L2 is asymptotically unbiased.

82 An infinite population consists of white and black balls, and it is known that p, the proportion of white balls in the population, is rather small. In order to obtain a good estimate of p, an experimenter decides to sample the balls one at a time till he obtains exactly r (fixed in advance) white balls in the sample. (i) If X is a random variable denoting the size of the sample which includes exactly r white balls, prove that P(X=n)= (;=Dprq,,-r,

for X;a.r;

q==l-p.

(ii) Prove that the probability-generating function of X is E(OX) = (pO)'(1-qOrr.

Hence, or otherwise, deduce that E(X)= rIp.

342

EXERCISES IN PROBABILITY AND STATISTICS

(iii) Given r and the sample realization n of X, prove that the ma . )(llllll01_ likelihood estimate of p is r

A

p=-, 11

and that for large samples

p2q

var(p)=-. r

(iv) Also, verify that

E(:=~)=P, so that

p is

not an unbiased estimate of p.

83 In the distribution of the number of police prosecutions per motorist' · over a given . . d 0 f time, ' . 'III a Iarge cIty peno t he f requency 0 f motorists havin prosec~tions is 11" for r ~ 1 ~nd .2::::"=1 nr == N, b~t the number of motori~t: ~ho dId not ?ave a pro~ecutIo~ IS unknow~ owmg to the motorist popul
p,*=Lr.I1.1N r=2

is an unbiased estimate of p, and that var(p,*) =E: [1 +-p,-]. N eIJ.-1 Also, verify that an unbiased estimate of var(p, *) is

2n2+Np,* T2 =

N2

Give any reasons which in your opinion make the estimate p, * preferable to the maximum-likelihood estimate ,1. [Note: It may be assumed that for large samples

p,(1-e-IJ.)2 ] var(p,) = N{l-(p, + 1) e-IJ.}· A

84 Explain briefly the concept of an estimate of a population parameter, and distinguish clearly between unbiased and consistent estimates. Illustrate your answer by suitable examples. The probability density function of the Cauchy distribution is 1 -. 7T

1 , for -co> X
where p, is a parameter of the distribution. If Xl' X2> ••• ,x.. are independent observations of the random variable X, determine the equation of ,1, the maximum-likelihood estimate of p,. Also, prove that the large-sample variance of ,1 is 2/n.

SUPPLEMENT

343

I~ A continuous random. variable X has. ~ normal ~istributi~n with mean (J

,od variance k(J, where k IS a known posItIve quantIty but (J IS an unknown Jd aOleter. If Xl> ~2' •.. ,XII are n independent observations of X, show that ;'Jf equation for (J, the maximum-likelihood estimate of (J, is ·he e(e+k)=.e+s 2,

"

ni=Lxj

and

"

ns 2 =L(xj-if.

i=1

i=1

~ence deduce that

e =![{k 2+4(i 2+s 2)}L k], ad that, for large samples, J

2k(J2 var«(J) = n(2(J + k) A

What is the limiting value of var(O) if, for fixed n, k ~ 0, (J ~ 00, such ,hat k6 ~ /-L, a finite number?

86 Explain, very briefly, the concept of the best estimate of a parameter in ,he theory of linear estimation. Suppose that Xl' X2, ... ,XII are n independent observations such that (J2 E(x,,) = /-L; var(xv ) = , for 1 ~ v ~ n, n-v+1 Il'here /-L and (J2 are independent parameters. Determine /-L *, the leastsquares estimate of /-L, and show that 2(J2 var(/-L *) = - - n(n + 1) Another linear function suggested as an estimate of /-L is

6 T= (

n n+

1)(2

11

n+

1) L (n-v+1) 2xv. v=l

Prove that T is an unbiased estimate of /-L but that, as compared with #.t *,

2(2n+ 1)2 Eff(T) = 9n(n + 1)

8

9

for large n.

87 Explain clearly the meaning of linear regression and state the assumptions underlying the standard tests of significance in linear regression analysis. Suppose that Yl> Y2' ... ,YI1 are independent observations such that E(yJ = a

+ (3(x" - i) + "(x,, - if;

where x" are values of a non-random variable, i = I~=1 xv/n, and a, (3, " (1-0) and (J2 are independent parameters. Prove that, in general, the linear functions 11

a*=

L

yJn

v=l

are biased estimates of a and {3 and find their expectations explicitly. What happens if the x" are equispaced?

344

EXERCISES IN PROBABILITY AND STATISTICS

Also, for general Xv, determine a linear function of a* and (3* expectation is independent of 'Y and find its variance. wh()~t

88 Suppose that Yl> Y2, ... , Y.. are n independent observations such thaI E(y.,) = a + (3(xv - i),

var(y.,) = 0'2,

for v = 1, 2, ... , n,

where the Xv are values of a non-random explanatory variable with av . i, and a, {3 and 0'2 are independent parameters. If a* and {3* a er,lgt standard least-squares estimates of a and (3 respectively, then t;e tht residual is defined as e /Jlh rv = yv-a*-{3*(xv-i),

for v = 1, 2, ... , n.

Prove that the rv satisfy the two linear constraints n

L

,.

rv = 0

L

and

1'=1

(xv - i)rv = O.

1'=1

Also, prove that for any v

n-1 var(r.,) = 0'2 [ n

E(r.,) = 0; where

(x., -

X

i)2]

.

.

L

X=

(x.,-if;

v=l

and

i

var(r.,) = (n -

2)0'2.

v=l

89 Suppose (Xi> Yi), for i = 1, 2, ... , n are independent paired observations such that each Yi is normally distributed with E(Yi) = a

+ (3(Xi -

i)

and

var(Yi) =

0'2,

where a, {3 and 0'2 are the parameters in standard linear regression analysis and i is the average of the Xi> the values of the explanatory variable. If n'; and {3* are the usual least-squares estimates of a and {3 respectively, find (J *, the best estimate of the parametric function (J = a - {3i. Prove that corr«(J*, (3*) =

i.Jn/

(t1 xfY.

Also, determine the value of this correlation coefficient when, in particular, the Xi (i = 1, 2, ... , n) are the first n natural numbers, and then show that, as n ~ co, the limiting value of the correlation coefficient is -!J3. 90 Explain briefly the concepts of unbiasedness and efficiency in the theory of estimation. It is known that, in biological investigations carried out over prolonged periods of time, the experimenters often obtain quite accurate information concerning the coefficient of variation (standard deviation/mean) of the quantitative characteristic under study. This information can be used to plan further experimentation. Since large random variability is a typical feature of biological material, a knowledge of the coefficient of variation can be

SUPPLEMENT

345

Jsed 10 ~btain an asymptotically unbiased es~imate of a parameter, which .as a van~nce smaller than that of ~h~ best lInear unbiased estimate. , In particular: suppose a ~haracten~tIc Y. (su~h as the length of the ears of ~aiZe of a particular type) IS under Investigation, where

E( Y) = IL

and var( Y) =

0'2,

and 0'2 being independent parameters. It may be assumed that v == 0'/ IL is rnown accurately from previous similar experimentation. If Yl> Y2, ... , YI1 JlC independent observations of Y, an estimate of IL is defined by the linear [Odelion i=l

~here

C is so chosen that E(T-IL)2 is a minImum. Prove that C =

1/(11 +v 2 ). Hence show that T has a smaller variance than the sample

Jverage

y, but that E(T) = IL

(1+:2)-1

so that T is an asymptotically unbiased estimate of IL. Also, prove that the tlliciency of T as compared with y for the estimation of IL is (1+

:2y.

••• ,x,. are n independent observations with mean and variance 0'2. Another independent sample of n correlated observalions Yl> Y2, ... ,YI1 is given such that for all i = 1,2, ... , n

91 Suppose that Xl> X2, ~

E(Yi)=IL;

var(Yi) = 0'2;

and

corr(YhYj)=P,

(i=/=i).

Prove that x and y, the means of the x and Y observations respectively, are unbiased estimates of IL and determine their variances. If T== ax + (3y is a linear function of the sample means, find a and (3 such that E(T) = IL and var(T) is a minimum. Verify that this minimum var(T) is 0'2 - [1 +(n -1)p]/[2+(n -1)p].

n

Also, show that (n _1)2p2 ] var[(x + y)/2] = [ 1 + 4{1 + (n -1)p} . min var(T).

Hence, or otherwise, deduce that for n> 1 and P=/= 0 var[(x + y)/2] > min var(T) and comment on this result. 91 If Xl> X2, ••• ,Xm are independent observations of a random variable X having a normal distribution with mean IL and variance 0'2, find the least-squares estimates of IL and 0'2. Hence indicate, without proof, how an exact test of significance based on Student's distribution may be used to test any null hypothesis about IL.

346

EXERCISES IN PROBABILITY AND STATISTICS

Given the 11 sample observations, an experimenter wishes to obt . interval estimate for an additional independent but unknown observat~tn an X. Show how the argument leading to Student's distribution may be IOn or ified to derive the required 100(1-1/) per cent (0 < 1/ < 1) confi:odinterval for the (11 + l)th observation as encc i -t(1/;

1l-1)S~ (11: 1)~x ~i + t(1/; 1l-1)S~ (11: 1),

where x is the unknown observation, i and S2 are the sample mean v~ri~nce. of th.e Xi and t( 1/; 11 - 1) is the 1001/ per cent point of Stude~~~ dlstnbutlOn with d.f. .

11-1

93 Explain clearly, without proof, how a confidence interval may b. obtained for the regression coefficient in linear regression analysis, and Sht c. the underlying assumptions of the analysis. (C In an investigation the distinc~ values of t.he explanatory variable x arc Xl> X2, ••. ,Xn and the correspondmg observations of the dependent variabl' yare Yl, Y2, ... ,Yn, it being assumed that the standard linear regressi{)~ holds between X and y, so that E(Yi) = a + (3 (Xi - i) var(yJ = (J'2, for i = 1,2, ... , 11, and i is the average of the Xi' The experimenter wishes to make k (preassigned and ~ 1) further independent observations of the dependent variable, each observation corresponding to the same given value Xo 1= Xi (i = 1,2, ... , of x. If Yk is the average of these k observations (yet to be made), prove that the 95 per ccnt confidence interval for Yk obtained on the basis of the known (Xi. Vi) i = 1, 2, ... ,11 is

11)

[1 1

A

a+(3(xo-i)-tO.05 xs 'k+-;;+

(X o -i)2J!

X

~

[1 1 (xo-xf]\ X •

~Yk~a+(3(xo-i)+tO.05xS 'k+-;;+ A

where a, ~ and S2 are the usual least-squares estimates of a, (3 and (12 respectively, X=L:t~l (Xi -i)2, and to'05 is the 5 per cent point of Student's distribution with 2 d.f. Suggest reasons, if any, which might make the above confidence interval inappropriate in a practical situation.

11 -

94 (i) Explain clearly the difference between a poillt estimate and an illterval estimate of a population parameter in the theory of statistical inference. (ii) Suppose that Xl> X2, ••. ,Xn are 11 independent observations from a normal population with mean /.L and variance (J'2. If i and S2 are the sample mean and sample variance respectively, indicate, without proof, how these sample quantities may be used to obtain a confidence interval for /.L. An experimenter plans to make two further independent observations Xn+l> Xn+2 from the above normal population. If A is a known constant (0 < A < 1), prove that the 99 per cent confidence interval for the linear

347

SUPPLEMENT

[unction

~,

X-tOoOIXS

[1-+1-2,\(1-,\) ]~ :S;;L:s;;x+toooIXs[1-+1-2'\(1-'\) J" 2

11

1 1 '

Mre tOoOI is the one per cent point of Student's distribution with 11 -1 d.fo Also, show that, irrespective of the value of ,\ in its permissible range, ,he above confidence interval for L must be greater than the interval W

11 +2)~ x±too01Xs ( 2;- .

9S If

XI' X2,' •• , X are independent observations of a normally distributed rariable X with mean IL and variance u 2 , find the least-squares estimates of ~ and u 2 • Hence indicate, without proof, how an exact test of significance based on Student's distribution may be used to test any null hypothesis about IL· Given the 11 sample observations, an experimenter wishes to obtain another 11 independent observations of X. If Xo is the mean of these 11 observations (yet to be made), prove by a suitable extension of the argument leading to Student's distribution that the 100(1- T)) per cent (0 < T) < 1) confidence interval for the difference x - Xo is Il

where x and S2 are the sample mean and variance of the already observed 11 values of X, and t( T); 11 - 1) is the lOOT) per cent point of Student's distribution with 11 -1 dJ.

96 In 11 Bernoulli trials with a constant probability of success p, w successes were observed. Show that for the probability distribution of the random variable w the third central moment is 1L3(W)=l1pq(q-p),

where p+q=1.

Prove that an unbiased estimate of 1L3( w) is

T

=

113 p*(1- p*)(1-

2p*)

--''-----'------=.----'-

(11 -1)(11 - 2)

,

p* being the observed relative frequency of successes. Also, show that for large 11 var(T)-npq(1-6pqf.

97 In a plant-breeding experiment, the observed frequencies of progeny in the four mutually exclusive classes AI> A 2 , A 3 , A4 were Ill> 112, 113, 114 respectively (I I1j == N). On a genetical hypothesis the corresponding probabilities for the four classes are -&(2+0), -&(2-0), -&(1-0), i6(1+30) respectively, Where e is an unknown parameter. Derive the maximum-likelihood equation for e, and show that the

EXERCISES IN PROBABILITY AND STATISTICS 348 large-sample variance of the estimate 8 is 2 ( A) _ 4(4- ( )(1- 6)(1 + 36) var 6 3N(5 + 26 _ 4(2)

Further, suppose there was some error in the classification of the A A3 progeny, though the Al and A4 plants were classified correctly. PO~I and observed frequencies in the A2 and A3 classes and then derive the equ t~hc for 6*, the maximum-likelihood estimate of 8. Verify that for large sam;I~~1l var( 8*) 4(3 - 28)(5 + 28 - 4(2) var(6) (1-8)(2-8)(29+328)' and hence show that var( 8*) > var( 8). 9d~ . In a Plantf-breeding ebxp~rimdent the observed fdrequencies .of the fOli r

Istmct types 0 progeny 0 tame were al> a2, a3 an a4 respectively whe . On a genetical hypothesis the expected proportions in the fo~c classes are 1(2 - 8), i(1 + 8), i8 and i(1- 8), where 8 is an unknowl~ parameter such that 0 < 8 < 1. Find the equation for 8, the usual maximum. likelihood estimate of 8. Alternatively, a simply calculable linear unbiased estimate 8* may he derived by equating the linear function

L aj == N.

X==al-a2-3a3+3a4

to its expectation. Show that explicitly 8* = (a2+2a3-a4)/N, and hence derive the exact variance of 8*. Also, prove that, irrespective of the true value of 8 1 (*) ~3 -~var8 2N 4N·

99 A multinomial distribution has k (;;a.2) distinct classes and in a random sample of N observations from the distribution, the observed frequencies in the k classes were at> a2, ... , ak respectively. The corresponding expected frequencies were ml> m2, ... , mk (Lk=1 ai = L~=1 mi = N). Assuming that the mi's are all functions of an unknown parameter 8, prove that the equation for 8, the maximum-likelihood estimate of 8, is

t

[ai . dmiJ =0. i=1 mi d8 6=6 Also, show that for the linear function X==t a;.dm i i=1 mi d8

t

l.. (dmi)2. i=1 mi d8 In a particular breeding experiment with a variety of Papaver rhoeas. there were four distinct classes with the expected frequencies var(X) =

E(X)=O;

N

N

N

N

2

4 (3-28+8 2), 4 8(2-8)'4 8(2-8), 4 (1-8).

349

SUPPLEMENT

I'erify that in this case 2N[1 + 2(1- 6)2] var(X)= 6(2-6)[2+(1-6)2]"

100 The serially correlated observations E(Xj) = 0, corr(Xb If X and

Xi+l)

var(Xj) = u = p,

2,

Xl> X2, •••

,x..

are such that

for i = 1, 2, ... ,n;

corr(xb Xi+k) = 0, for k;;:;. 2.

S2 denote the mean and variance of the sample observations, therl

rrove that u2 (i) var(x) = - [1 + 2p(n -l)/n]; n (ii) E(S2) = u 2(1-2p/n); and (iii) -~:s;;p:s;; 1.

Also, if another linear function of the observations is defined as

2

n

xw = n ( n+ 1) 1'=1 L vx", Ihen verify that 2u 2 var(xw ) = 3n(n+ 1) [2n(1 +2p)+(1-4p)].

Hence deduce that, as n ~ 00, the limiting efficiency of x'v as compared with is i.

j

101 If X and Yare two correlated random variables having finite means 3nd variances, define cov(X, Y). (i) Assuming that X and Yare positively correlated and var(X);;:;. var(Y), prove that cov(X, Y):s;;var(X). (ii) Prove that as a first approximation

(Y X) _ cov(X, Y) cov\X' - E(X) . Hence deduce that to the same order of approximation

Y)]

(D E(Y) [ cov(X, E\X)=E(X) 1- E (X)E(Y) . 102 Define the product-moment correlation coefficient between two continuous random variables having finite, non-zero variances, and prove that the coefficient must lie between -1 and + 1. Suppose Xl> X2, ••• ,x.. are random variables such that E(Xj) = p.

x

var(xi) = u 2 , and corr(Xj, Xj) = p, for it- j

II is the mean of the Xi and Yi (Yi =

Xj -

(i = 1, 2, ... , n).

x) are the deviations of the

Xi

from

350

EXERCISES IN PROBABILITY AND STATISTICS

i, prove that (F2 (i) var(i) = - [1 +(n -1)p]; n (11") var (Yi) = (n -1)(1- p)(F2 ,

for i

n

(iii) corr(Yi> Yj) =

-~1' n-

for

i1= j =

=

1,2, , , , , n;

and

1, 2" , " n,

103 If Xl> X 2 " " , XN are correlated random variables with finite m ' and variances, prove that edlJ\

va{~ Xi] = i~ var(Xi)+ i~ j~i cov(Xi, Xj), Suppose that an unbiased coin is tossed n times and the number of tim' the sequence a head followed by a tail (HT) is observed. If Zi is a randoC\ variable which takes the value unity if the sequence HT OCcurs at tI:l~ (i -1)th and ith trials considered together and zero otherwise, express SC the random variable denoting the total number of times HT occurs in the ';; trials, in terms of the Zi' Hence show that (i) E(SII) =i(n -1); and (ii) var(SIt) = 'Mn + 1),

104 (i) If X and Yare two random variables with finite variances, prove that the random variables X + Y and X - Yare correlated unless var(X) = var(Y), (ii) Given that Xl> X2, ••• , XII are serially correlated observations such that and var(xi) = (F2, for i = 1, 2, ... , n, where p and (F2 are parameters (-1 < P < 1 and (F > 0), prove that the variance of i, the average of the Xi is

[1

(F2 + p _ 2p(1- pll)] n I-p n(1-p)2 .

lOS (i) If Xl> X 2 and X3 are uncorrelated random variables, prove that cov(Xl + X 2 , X 2+ X 3 ) = var(X2), (ii) The serially correlated random variables that for any s ~ 1

XI> X2, • , , , XII' , , •

are such

E(xs ) = 0;

\p\ S2 and S3 are defined as n+r It S2=

LXv; v=r+l

S3=

L

v=n+l

Xv·

SUPPLEMENT

351

prove that

a2pll-r+l(1_ pr)2

COV(SI,S3)= ~ertce

(l-p)2

deduce that if COV(Sb S2) = COV(Sb S3), then 2pll pr =

1 + pll·

,06 Prove that the product-moment correlation coefficient between any I\VO

random variables with finite non-zero variances must lie between -1

JU d

+1.

The random variables Xl> X2, X3, .•• are uncorrelated and each has zero l1Iean and constant variance a 2 • Given any integer n > 1, another sequence 01 random variables Yt> Y2, Y3, ... is defined by the equations 1 1 11 -

Yi=-

n

L Xi+;,

fori=I,2,3, ....

;=0

II r is another non-negative integer, prove that for any fixed i ifr;?;n

n -r n

if r
107 Prove that the moment-generating function of a Poisson variable X with finite mean p, is

E(e 'X ) = exp p,(e' -1). If Xl> X2> ••• , XII are random observations of X, then prove that the sample mean i has the moment-generating function

exp[np,(e l/ll -1)]. Hence deduce that the cumulant-generating function of the standardized random variable has the series expansion

so that, as n ~ 00, the probability distribution of z tends to normality. 108 Prove that the probability-generating function of a Poisson variable X with mean A is E(8 = exp A(8 -1). X )

If XI and X 2 are two independent Poisson variables with means AI and

"-2 respectively, show that the joint probability-generating function of Xl and ¥=X I +X2 is E(8i". 8n = exp[A 1 (8 1 82 -1)+ A2(82 -1)].

352

EXERCISES IN PROBABILITY AND STATISTICS

Hence prove that the marginal distribution of Y is POisson W'th Al + A2 but that the conditional distribution of Xl given Y == j is 1

e)(l: P)'(l !py-r,

P(XI = r I Y = j) =

Ilh:"u

for 0:S;;X 1 :S;;j,

where P==Al/A2' Discuss briefly how this conditional distribution may be used to t ' equality of the means of Xl and X 2 , CSllhl'

109 A discrete random variable X denotes the number of successe' ' sequence of n Bernoulli trials with probability P for a success in an';~ 1, a Prove that the probability-generating function of X is n.11

t

E(OX) = (q + pO)",

where P + q = 1.

If Xl and X2 are i~depend~nt, random variables respectively denoting tl1l' number of successes m two dIstmct sets of nl and n2 Bernoulli trials w'lI success param.eters PI and P2, prove that the joint probability-general)') 1 , , ( Ill! functIon of Xl and Y = Xl + X 2 IS O( 01> ( 2) == E( O~· , On = (ql + PI 01 (2)n'(q2 + P2 ( 2)"2,

(PI + ql = 1; P2 +q2 == I\.

Hence show by suitable differentiation of 0(01) ( 2 ) that (i) P(Y=r)=

i

v=O

(nl)( ]1

r

~2

v

)Plq'i·-vp2-vq22-r+v,

for 0:S;;Y:S;;n 1 +11 2;

and

(nl)( n2 ) , (ii) P(Xl=sl Y=r)= r s

r-s p

L (n1)( r-]1 n2 )p v

v=O

]1

where p == Plq2/P2ql> and it is assumed that in both (i) and (ii) the binomial coefficients are zero for inadmissible values of r> nl and r> n2' Discuss very briefly how the conditional probability distribution (ii) may be used to provide a test of significance for the null hypothesis H(PI = P2)'

110 A discrete random variable X has the probability distribution defined by the system of equations P(X = r) = Pn

for integral values of X;;. 0,

Define the probability-generating function of X and show how this functioll may be used to evaluate the factorial moments of X. Conversely, if /Ll/) is the jth factorial moment of X about the origin, prove that 1

00

(-I)'

L -,-#L(r+,)' r, ,=0 s,

Pr = --,

Use this result to verify that if, in particular, #Lej) = (n + j _l)(j)pi,

where n, a positive integer, and P (0 < P < 1) are parameters, then Pr= ( n+r-1) r p r(1+p)-
SUPPLEMENT

353 1 In sampling from a binomial population with success parameter k I~pen~ent samples are obtaine?, there bei?g ni observati(;lns in th/ith pie (I == 1,2, ... , k). The relatIve frequencies of successes In the samples ~ pt, p~, ... ,pt. Prove that the linear functions x= .¢

itl

niPf/N and

y=

Jl

(tl

pf/k

ni ;';:N)

both unbiased estimates of p, but that var(x) = pq/N and var(y) = pq/kno (p +q = 1)

here no is the harmonic mean of the ni' Determine the moment-generating functions of the standardized random ,ariables w = (x - P)(N/pq)! and z = (y - p )(kno/pq)~. I

~ence deduce that if the ni are large then, as a first approximation, both w lnd z are unit normal random variables. Also, if the differences between the .1 are small, show that var(x) -1- v(nJ var(y) ii 2 ' «here I

k

kv(ni);';:

i=l

L (ni -

k

iif,

kii;,;:

i=l

L ni'

112 A finite population consists of the integers 1, 2, ... ,N. A random lample of n «N) integers is taken from the population without replacement. If X and Y are the random variables denoting the largest and smallest integers respectively in the sample, then prove that the joint distribution of X and Y is given by the system of equations

where n~X~N, and 1~ Y~X-n+1. Use this distribution to obtain the joint distribution of X and Z = X - Y. Hence show that the marginal distributions of X and Z are given by the systems of equations

C=!)/(~,

for

n~X~N;

P(Z=Z)=(N-Z)(:=~)/(~,

for

n-1~Z~N-1.

P(X=x)=

and

113 The number of motorists in a large city having one, two, three, ... prosecutions over a given period of time is available, but the motorists who did not have a prosecution cannot easily be enumerated owing to the motorist population of the city fluctuating during that period. This gives rise to a discrete distribution in which the zero group is unobserved. An observed distribution of this kind can usually be well represented by a truncated Poisson distribution with parameter 11-, that is, a Poisson distribution in which the probability of the zero class is ignored. Show that for

354

EXERCISES IN PROBABILITY AND STATISTICS

such a modified Poisson distribution the variance is never greater th mean, and that, in fact, the variance is equal to the mean multipliedan thc the probability of a prosecuted motorist having at least two convict' by /l. IOns. 114 A sample of 2 v + 1 independent observations is taken of a r variable X having a probability density function [(x) defined in thcan~I()lll -oo<X <00. Suppose the observations are ordered and let Xl denotr Xv+l an d X2v+l' If another random sample of 2r observations of X is taken find I probability that, for given Xl> Xv+l and X2v+l> these 2r observ~tions ~ Ic equally distributed in the intervals (Xl> Xv+l) and (xv+l> X 2v + I ). Hence ShMl' that the total probability of the second sample observations being equ'(;~~ distributed between the half-range below and above the median of the ~ ~l~ . samp Ie IS v(2v+l) B(v+r,v+r) p = (2r+ l)(v+ r)(2v+2r+ 1)' B(v, v) B(r+ 1, r+ 1)'

115 If X is a unit normal variable, prove that the distribution function of X is 1 1 (-1),x 2r +! (x)=-+- I , for x;;:oO. 00

2 .J2;r=o(2r+l)2r .r!

Prove that the moment-generating function of E(e tIX1) =

2 e!t

2

•

IXI

is

(t).

Hence obtain an expansion of the cumulant-generating function and veriry that the first two cumulants of IXI are KI

= (2/7r)!;

K2

= (7r - 2)/7r.

116 A continuous random variable X has the probability distribution with probability density function n(n-l) a"

~-~ x,,-2(a -x),

for

O~X~a,

where a >0 and n, an integer >1, are parameters. If Xl and X2 arc independent observations of X, then show that the probability density function of the distribution of the ratio u = XI/X2 is n(n-l) n-2[ 11 - ( n - 1)] 2(211 -1) . u u,

f

n(n-l) ( ) u-"[n-(n-l)/u], 22n-l

for

or

0

1 u~ ;

~

and

Use this distribution to verify that for (3 > 1 P(u

l~u
r/{3].

~ (3) = 1- 2(211~~){3"-1 [1- (n: 1

117 Two continuous random variables X and Y have a joint probability

SUPPLEMENT Sity function proportional to ~~ e- Y(1-x)y"'(l-x)"'+13,

JlI

355

forO~X~l;

O~Y
d zero otherwise, where a > 0 and (3 > 3 are parameters. prove that for positive numbers rand s «(3) E(xrys) = (3 B(r+ 1, (3 -s)f(a +s+ l)/f(a + 1).

Also, show that the marginal distribution of X has the probability density function (3(l-x)13-l, forO~X~l;

Jod that of Y (3f(a + (3 + 1) y-13- 1r (a + (3 + 1) f(a + 1) Y ,

for 0 ~ Y <00,

where, in standard notation, the incomplete gamma function Y

r

(m)=_l_ fe-Itm-l dt

f(m)

Y

(m >0).

o

118 A point P is selected at random on OA, a fixed radius of a circle with centre 0 and radius a, such that the distance OP = X is a random variable having a distribution with probability density function f(x) = 12x 2(a -x)/a 4 , for O~X~a. Determine Y, the length of the chord drawn through P and perpendicular to OA, and verify that E( Y) = (15'77" - 32)a

10 Also, show that the probability distribution of Y has the probability density function

4!4 [2a -(4a 2- y2)!]y(4a 2 - y2)l, for

O~ Y~2a.

Hence, or otherwise, prove that the probability that the length of the chord (Y) is less than the radius of the circle is

43-24J3

16 119 A random variable X has the Laplace distribution with the probability density function f(x)

1

= 20- e-lxI/CT, for -oo<X <00.

Prove that the characteristic function of X is E(ei'CT ) = (1 + t 2 0- 2 )-I.

356

EXERCISES IN PROBABILITY AND STATISTICS

Given that YI and Y2 are independent observations of a random variabl Y having the Laplace distribution with zero mean and variance two, prove that the characteristic function of e z = (l-tJ.)YI + tJ.Y2 is E(e iIZ ) = [1 + t2(1-tJ.\2Il + t2tJ.2] ,

,+

where tJ. is a positive constant such that O:so; w s Hence deduce that the sampling distribution of z has the probability density function 1 [(1-11.) e-lzI/O-IL)_1I. e-lzlllL] for -oo
120 Given that Xl, X2, ... are observations obtained in a temporal order such that E(xj) = mj

for j ~ 1 and var(xj )

=

a constant,

0'2,

the autocorrelations between the Xj are defined as Pk

1

= P-k = 2

0'

E[(Xj - mj)(Xj+k - mj+k)]

for k ~ 1.

If the Xj satisfy the linear relation Xj+2 = aXj+l + (3Xj + Bj+2'

where a, (3 are parameters such that var(Xj) remains constant for all j and the B'S are independent random variables with zero mean and constant variance, show that (i) Pk+2=apk+I+(3Pk for k~-l; and (ii) var(B) = (1 + (3)[(1- (3)2-a 2]0'2/(1- (3). Also prove that Pk=

(3(1- (3)(A~-I- A~-l) + a(A~ - An (1-(3)(A 1 -A2) fork~O,

where Al and A2 are the distinct roots of the quadratic equation z2-az(3 =0.

121 Explain briefly (without proof) the relation between the normal and

x2

distributions. If X and Yare independent unit normal variables, prove that the random variables U and V defined by the equations y x ,2 U=_l_ e-! dt V = -i e -112dt 2

..n;

J

,

..n;

J

are independent random variables, each having the rectangular distribution in the (0, 1) interval. If U and V are independent rectangular variables defined in the interval (0,1), show that the random variables lW and Z defined by the equations W = (-2 loge U)~ cos 2'ITV,

Z

= (-2Io~

U)lsin 2'ITV

SUPPLEMENT

357

are independent unit normal variables. Also, show how these results may be used to construct from k + 2 (k ;;:.1) independent realizations of a rectangular random variable defined in the (0, 1) interval a function which has the X2 distribution with 2k + 1 d.f.

122 A continuous random variable X has the semi-triangular distribution

with the probability density function 2x/a 2 defined in the range 0 ~ X ~ a. If XI and x" are the smallest and the largest of n independent observations of X respectively, find the joint distribution of Xl and Xn" Use this distribution to show that the joint distribution of U

= Xl + x"

and

v = x" - Xl

bas the probability density function

!n(n-1)a- 21 (uv)"-2(u 2-v 2), for

O~v~a;

v~u~2a-v.

Hence prove that the marginal distribution of v has the probability density function

nv"- 2 ---:----::-[(n -1)(20. - v)"+1 - (n + 1)v 2(2a - V)"-1 + 2V"+1] for 0 ~ v ~ a; 2(n+ 1)0. 211

and that of u (n + 1)a 2n

for

O~u~a

'

and

123 Two related components in an electronic unit are such that the failure of one tends to affect the life-time of the other. If X and Yare continuous random variables denoting the individual life-times of these components, then a possible model for representing their joint distribution has the probability density function

f(x, y) = atf3 2 exp[-132y-(al +13l-13~x], for O~X~ Y = 1310.2 exp[-a2x - (0.1 + 131 - (2)y, for 0 ~ Y ~ X, where the unrestricted range of both X and Y is from 0 to 00, and al> 0.2, 131> 132 are positive parameters such that Prove that the probability that X ~ Y is 0.1/(0.1 + 131). Hence derive the joint distribution of X and Y subject to the condition X ~ Y, and then show that in this case (i) the marginal distribution of X has the probability density function (0.1 + 131) e-<<x,+f3,)x, for 0 ~ X < 00; (ii) the conditional distribution of Y given X density function

= X

and has the probability

358

EXERCISES IN PROBABILITY AND STATISTICS

Also, if it is known that the life-time of the first component is less Ih that of the second and that the life-time of the former is X == x, prove I ~11 the probability that the life-time of the second component is at least equ ~ilt but ,not greater than twice that of the first is a 10

1-e-132X •

124 A sample of 2 v + 1 (v;;. 1) independent observations is taken of . random variable X having a probability d~nsity function f(x) defined in Ih~ range -oo<X <00. Suppose these observatIons are ordered and let Xl denote the smallest, X2v+1 the largest, and X v +l the median observation of th. C sample. Determine the joint distribution of Xl' X v +l and X2v+!' If another independent random sample of 4r (r;;' 1) observations of X . taken, find the probability that, for given Xl> X v +l and X2v+h these ~~ observations are equally distributed in the intervals (-00, Xl), (Xl> x )r (Xv+l> X2v+l) and (X2v+l> (0). Hence prove that the total probability that;il~ second sample observations are equally distributed between the four inlervals determined by the smallest, the largest and the median observations of the first sample is v 2B(2r+ v+ 1, r+ v)B(r+ 1, 3r+2v+ 1)B(r+ 1, r+ 11) (4r+ 3)(3)B(v + 1, v + 1)B(r+ 1, r + 1)B(2r +2, r+ 1)B(3r+ 3, r+ 1) .

p=----~~----------------~--~------~----~---

125 If X is a normal variable with zero mean and variance u 2 , and II y2 is an independent random variable having the X2 distribution with II cU., determine the probability distribution of the ratio Xlu Z=Y'

Y;;.O.

Further, suppose U is a unit normal variable independent of X and Y, and c is a known constant. Then, by considering the expectation over the joint distribution of X and Y, prove that Ex.y[(X +cY)] = PX,y,u(U ~X +cY) = P[Z ~cl.J1 +u 2 ], where <1>(,) is the distribution function of a unit normal variable , and PX,y,u(-) is the probability of the event in brackets under the joint distribution of X, Y and U. Hence, assuming that u is also known, discuss very briefly how Ex,y[(X +cY)] may be evaluated by using the standard statistical tables.

126 If X is a discrete random variable having the Poisson distribution wilh mean p" prove that the moment-generating function of X about the origin is E(etX ) = exp p,(e t -1). Hence verify that the fourth central moment of X is P,4(X) = p, + 3p, 2. If Xl> X2, ••. ,Xn are independent realizations of X, show that the moment-generating function of their sample average x about the origin is E(e tX ) = exp np,(e t/n -1). Hence prove that

T

x

= - [(3x

n

+ 1)n -3]

SUPPLEMENT

359

, an unbiased estimate of I-'-iX) and that 's

var(T)=; [(61-'-+1)2+ 1~I-'-J.

IZ7 A random variable X has a uniform distribution in the interval (0, a). Ux\ and X2 are independent realizations of X, determine the joint probabil~y distribution of ~ence derive the marginal distribution of v and then verify that for any positive constant (3"; O! 1[ (3 2 P(v;;;'(3)=Z

1-;].

Also, prove that 3lvl is an unbiased estimate of a and that the variance of • • 1 2 Ihis estimate IS 2'0! •

128 (i) The probability that an event X will not be realized in the first n trials is u". Prove that the expected number of trials required for the realization of X is 00

L u..

where

Uo

= 1.

r=O

(ii) In a sequence of n Bernoulli trials, the probability of a success (S) is p and that of a failure (F) q (p + q = 1). If Yn is the probability that the pattern FSSS will not be realized in the n trials, prove that Yn

=

t (n -r r) (_ qp

3)(n-rl/3,

r=n/4

3 where, for any integer n;;;' 4, the summation over r is for r = nand those values of n - r which are divisible by 3. Also, show that the expected number of trials required for the realization of the pattern FSSS is 1/qp 3. Verify that for n = 15, the probability that the pattern FSSS will be realized at least once in the sequence of trials is 4 qp 3(3 - 9qp 3+ 5q2 p 6).

129 Explain briefly the concept of the statistical independence of two continuous random variables, and illustrate your answer by indicating (without proof) the behaviour of the joint sampling distribution of the mean and variance of a sample of N (> 1) random observations from a normal population. A random variable X is normally distributed with mean I-'- and variance 0'2. If Xl> X2 are the sample means and sf, s~ the sample variances obtained respectively from two independent sets of n observations of X, prove that the function

is an, unbiased estimate of I-'- and that ( )

var w =

(n + 1)0'2

2n 2

360

EXERCISES IN PROBABILITY AND STATISTICS

130 Explain very briefly, without proof, the relation between the X2 d' distributions. an I· If X~ and X~ are independent random variables each having th 2 distribution with n dJ., determine the joint distribution of the ran~ X variables Rand e, where om X~= R cos2

e and

X~= R sin2

e.

Use this joint distribution to verify that the random variable X2_X2

W=_I __2=cot2e 2XIX2 has the probability distribution with probability density function 1

~

(

1:)' (1 + w1

2)
for -00< W <00.

B 2'2

Hence show that if F has the F distribution with n, n d.f., then

t= In 2

[JF-~] JF

has Student's distribution with n dJ. and, for F> 1, 2t 2 2t [ 1+t2]~ . F=1+-+n In n

[Note: For p>O, r(2p)r(!) = 2 2p - 1r(p)r(p +!).]

131 State briefly (without proof) the relation between the unit normal and X2 distributions. (i) If a continuous random variable X has the Laplace distribution with probability density function f(x) =~ e- 1xl , for -oo<X <00, prove that the characteristic function of X is E(e itX ) = (1 + t2)-I. (ii) If Y1 and Y2 are independent unit normal variables, determine the joint probability distribution of the random variables Ul=~(Yl+Y2)

and

U2=~(YI-Y2)'

(iii) By using the above results, prove that, if Z1> Z2, Z3 and Z4 are

independent unit normal variables, then the random variable

W=ZI Z 2+ Z 3 Z 4 may be expressed as half the difference of two independent X2 ,s each having 2 dJ. Also, show that the probability distribution of W is the same as that of X.

132 Cards are dealt out one by one from a standard pack of 52 cards of which four are aces and the rest non-aces. If X is the random variable denoting the number of cards which precede the appearance of the third

SUPPLEMENT

361 and Y another random variable ~enoting the number of cards which plCn up after the appearance of the thIrd ace and before that of the f h (.t, prove that ourt (.t,

P(X = r) =

(~)C~82)(5:_r) /

er2) , for

2~X~SO,

and P(Y=s I X= r) = (50s-r)(51_1r_J / (51s-r), for

O~ Y~50-r.

Hence derive the unconditional probability distribution of Y. Also, show by considerations of random sampling that E(Y)= ~8. [Note: For

n+p;;;'k+1,

L~=l

(r+p)(k+1)={(n+1+p)(k+2)_(1+p)(k+2)}/

(k+ 2).] 133 A continuous random variable X has the probability density function f(x) defined in the interval -oo<X <00 and the mean of X is fL, an unknown parameter. In principle, the average of n independent realizations of X is an unbiased estimate of fL. However, an occasional sample will, due to the

sampling process, contain one or more observations from the upper tail of the distribution of X. When this occurs, and the sample size is small, the sample average will tend to exceed fL by a considerable amount. A "better" estimate can be obtained by using a procedure which reduces the effect of large sample observations. Suppose a random variable Z is defined such that Z = X if X < t and Z == t otherwise, where t is a specified quantity. Show that E(Z) = PfL, + qt

and var(Z) = p[u~+q(t- fL,)2], where fL, and u~ respectively are the mean and variance of the probability distribution of X truncated on the right at X = t (that is all values of X> t are ignored and the distribution is normalized) and

,

P=

J

q = 1-p.

f(x) dx,

-00

Hence determine the mean and variance of the average of n independent observations of X in which values >t are replaced by t individually. 134 If

Xl> X2, ••• ,

XII are equi-correlated random variables such that var(xJ=u7, fori=1,2, ... ,n;

and corr(Xi> Xj) = P a constant, for i =/= j, evaluate the variance of X, the arithmetic mean of the Xi' Hence deduce that

1

362

EXERCISES IN PROBABII,ITY AND STATISTICS

Also, if Yh Y2, ... 'YII are the deviations of the Xi from show that, for any given i,

L

cov(x, Yi) = -1 [ (1 - p )af + pai II n j=1

aj

x

(Yi == Xi -

x),

] - var(x).

Hence prove that a necessary and sufficient condition for the x. (._ ~, 2, ... , n) to have equal variance is that the Yi should be uncorrelated W'ith

x.

135 Three discrete random variables Xl> X 2 and X3 have the joint probability-generating function E«(JX, (JX2(JX3) = ~ 1 2 3 211

[1 +!!.211 {I + (J2(1211+ (J3)1I}1I]1I

where n is a positive integer. Use this probability-generating function to prove that

+ 1)11-1 2 11 (11+1) ;

n 2(211

.

(I) P(X2 = 1) =

211(11-1) (ii) P(X1 = 11 X 2 = 1) = (211 + 1)11-1 ; and ... [211 (11+1) + (211 + 1)11 ]11 (m) P(X3 = 0) = 2 11 (11 +11+1) 2

Also, if Z

= Xl + X 2 + X 3 , then show that E(Z) = n(n 2 + ;n +4)

2

136 If Xl and X2 are independent observations of a normal variable X having mean /L and variance a 2 , prove that U=X1-X2

and

V=X1+X2

are independent and normally distributed random variables with means 0, 2/L respectively and the same variance 2a 2 • Use the probability distribution of u to verify that E(lui) =

2a .[;'

and hence deduce the expectation of the range of Xl and X2. Further, given three independent observations of the random variable X, verify that the sum of the moduli of the differences of the observations taken in pairs is equal to twice the sample range R. Hence deduce that for a sample of three independent observations of X E(R)=

~.

137 Three players A, Band C agree to playa series of games with a biased coin which has a probability p (0 < P < 1) for a showing a head. In each game, each of the three players tosses the coin once, a success being

SUPPLEMENT

363

recorded if the toss leads to a head. The series is won by the first player to obtain a success in a game in which no other player obtains a success. If just 1\\,0 players obtain successes in the same game they continue to play without lite other player who is deemed to have lost the series. If all three players obtain successes in the same game they all continue to play. Determine the probabilities of the possible outcomes of the first game played, from the point of view of a particular player (say A). Hence deduce lit at Un' the probability that A wins the series on the nth game, satisfies the difference equation Un

= (1-3pq)un_l+ 2p3q2(1-2pq)n-2,

Finally, if cf>(z) =

Lr= 1 urz r is the generating function cf>(z) =

SO

forn~2,

q=l-p.

for u'" verify that

pq2z[1-(1-2p)z] [1-(1-2pq)z][1-(1-3pq)z] ,

that cf>(1) =~. Explain very briefly the significance of this value of cf>(1).

138 Two gamblers A and B agree to play a game of chance with initial capital of £a and £b respectively, the stakes at each trial being £1 on the occurrence or non-occurrence of an event E. If E occurs at a trial then B pays A £1, whereas if E fails to occur at the trial, then A pays B £1. The lrials of the game are independent, and the constant probabilities of the occurrence or non-occurrence of E at a trial are p and q respectively, where p+q = 1. A player wins when the capital of his opponent is exhausted. If Un (0 < n < a + b) is the expected number of further trials required for A to win when his capital is in, prove that Un satisfies the difference equation Un = 1 +pUn+1 +qu,.-I'

Hence show that the expected duration of play for A to win is a q_p

(a+b)(qa_pa)pb (q _ p )(qa+b _ pa+b)

if pi= q,

and ab if p = q. 139 A finite population consists of the N elements Xl> X 2 , Ihe mean and variance of the population are defined as

_

1

X=-

1 S2=_-

N

LX;

and

N -1

Nj=l

••.

,XN , and

N

L (Xj-Xf j=1

respectively. Suppose a random sample of n «N) observations is taken without replacement from the population. The sample mean x is defined as the average of the elements of the population included in the sample, but an alternative formal definition of x is 1 N i=-

L ZjXj,

nj=l

where Z b Z2, ..• ,ZN are indicator random variables associated with the individual elements of the population sampled such that Zj is either 1 or 0 according as X j is or is not included in the selected sample. Use this

364

EXERCISES IN PROBABILITY AND STATISTICS

alternative definition of i to prove that (i) E(i) = X;

and

..) (_) (N - n)S2 (11 var x =

nN

140 A continuous random variable X has the distribution function F( defined in the .ra.nge O:s;;; X.~ a. I~ x! < ~2 <: .. < XN are ordered reaIizatio~~ of X, find the Jomt probabIlIty dIstrIbutIon of Xr and x. (s> r). In particul . if X has a uniform distribution in the interval (0, a), show that probabil~' density function of the joint distribution of u = xrla and v = x.la is I y NI (r-l)! (s-r~I)! (N-s)! u r- 1 (v-u)·-r-l(l-v)N-.,

O:s;;;u:S;;;l;

v;:ou.

Hence deduce the marginal distribution of u and the conditional distribution of v given u. Also, given u and some constant Va such that u < Va < 1, prove that P(V > va) = 1 - BA (N - s + 1, s - r)

in standard notation of the incomplete B-function and where A"" (va- u)/(I-u).

141 Given that X and Yare independent negative exponentially distributed random variables in the range O:S;;; X, Y < 00 and with means II n I and I1n2 respectively, prove that

P(X;::;'Y)=~. nl +n2

Determine the joint probability distribution of X and Y given that X;:o Y. Use this probability distribution to derive the joint probability distribution of U = X - Y and Y, given that X;::;. Y. Hence deduce the probability distribution of U given that X;::;. Y. Also, show that the unconditional distribution of IUI has the probability density function

142 In a quantitative study of the spatial distribution of a plant population in an area, a square region was divided into n quadrats of equal size, and the number of plants observed in the region was s. For given s, the distribution of plants inside the square is such that a plant is equally likely to be found in anyone of the quadrats independently of the other plants. Determine the probability that a specified quadrat contains no plants. Furthermore, if X is a random variable denoting the number of empty quadrats in the square region, prove that E(X) =

and

n( 1-~r

SUPPLEMENT

II

365 Also, obtain approximate expressions for E(X) and var(X) when sand ~rJJ in such a way that sIn = A, a constant.

1.3 A continuous random variable X has a uniform distribution in the

'olerval -a";;; X,,;;; a. If X,t+l is the median of a sample of 2n + 1 independent X n +!. Use this probability distribution to prove that, for any non-negative integer r,

~bservations of X, determine the probability distribution of r aT(2n+2)f{!(r+ 1)} E(Ix,t +1 I ) = 22n + 1r(n + l)r{n +!(r+ 3)} .

Hence, in particular, verify that E(lxn+!i) = (2nn+ 1 )a/22n + 1 •

Also, show that for large n a first approximation gives E(lxn+li) = a/~. [Note: It may be assumed that for large m > 0

r(m + 1) = .J21Tm(mle)"'.]

144 Suppose that Xl> X2, ••• ,Xn are serially correlated observations each with zero mean and variance u 2 , and corr(Xj, Xi+k) = P for all i arid k = 1,2 such that 1,,;;; i ,,;;; i + k ,,;;; n and zero otherwise. If x is the average of the Xi> prove that . u2 [ 2(2n - 3)P] (1) var(x) = -;; 1 + n ; and (ii)

E[t

(Xi -

X)2 ]

=

u 2[ n

-1

2(2nn- 3)p

J.

Hence show that as n ~ 00 p ~ -1.

145 Suppose Xl < X2 < ... < x.. are ordered observations of a continuous random variable X having a negative exponential distribution with the probability density function a- I e-(X- I3)f""

X~(3,

where a and (3 are positive parameters. By using the transformation UI = Ur

n(xi - (3)

= (n -

r+ 1)(x,- X,-l),

2,,;;; r";;; n,

or otherwise, prove that the u's are independent and identically distributed random variables such that, for any j, 2u/a has the X 2 distribution with 2 d.f. Hence determine the probability distribution of the ratio n(n -l)(xl - (3)

Comment very briefly how this result may be used to test a specific null hypothesis about the parameter (3.

366

EXERCISES IN PROBABILITY AND STATISTICS

146 State the additive property of the X2 distribution. Suppose X and ~ a~e two indep,endent r~ndom variables such that 2ex and 2 Y/O are both dlstnbuted as X2 s each with .2 d.f., where 0 is a Positive parameter. If Xl> X 2 , ••• , XII are random observations of X, and Yb Y2 .. those of Y, determine the joint probability distribution of the sum; . , y" n

It

L Xi

U =

and

L Yi·

v=

i=1

i=1

Hence, by considering the transformation

u=w/z

v=wz,

and

or otherwise, determine the joint distribution of wand z. Finally, use this joint distribution to deduce that the marginal distribu_ tion of z has the probability density function 2f(2n) (OZ-I+O-I Z)-2I1 Z -l {f(nW '

and that

for O:s;;z
,

E(z) = (2n ~ 1)0 [f(n -t)/f(n)]2.

147 Suppose Xl is the smallest and XII the largest of n random observations of a random variable X having the probability density function I(x) defined in the range -oo<X <00. Find the joint probability distribution of Xl and x . If another random sample of r observations of X is obtained, determine t1~~ conditional probability, given XI and XII' that these observations all lie outside the range (XI' x,,). Hence show that the total probability of the second sample observations lying outside the range of the first sample is A

P = n(n -1)

JI(Xl) dX J [ f/(X) dx 1

r-

X

2 [

1- f/(X) dx T/(x,,) dX II

= f(n + l)f(r + 2)/f(n + r + 1). Also, if r is small and n large, prove that P=

1)]

.

f(r+2) [ r(r+ nr l-~ approximately.

148 Prove that the probability-generating function of a Poisson variable X with mean A is E(OX) = e A (9-1). If Xl and X 2 are two independent Poisson variables with means A1 and A2 respectively, show that the joint probability-generating function of XI and Y=X1 +X2 is E(O;'!.Oi)=exp[Al(OI02- 1)+A2(02- 1)].

Hence prove that the marginal distribution of Y is Poisson with mean Al + A2 but that the conditional distribution of Xl given Y = j is P(X1 = r I Y= j) = where p=AdA 2 •

e)(l :P)'(l !py-r,

for O:s;;XI :S;;j,

SUPPLEMENT

367

Discuss briefly how this conditional distribution may be used to test the equality of the means of Xl and X 2 • 149 Two continuous random variables X and Y have a joint distribution with the probability density function f(x, y) = n(n -1)a-1I211-lxll-2, defined in the triangular region bounded by the lines x = y, x + Y= a, and ,( == O. Find the marginal distributions of both X and Y. Also show that the conditional distributon of Y for given X = x has the probability density function

fey IX =x) = (a -2x)-1,

for x";; Y";;a -x.

Hence verify that var(Y I X = x) = (2x - a)2/12.

E( Y I X = x) =!a ;

150 If (x) denotes the distribution function of a unit normal variable X, and P(O,,;;X";;x)=I{!(x),

then prove that (x )[1- (x)] = ~ - I{!2(X).

Also, by integrating a series expansion of the integrand of I{!(x) , show that 4 x 2 [ 1- x 2+7x ] I{!2(X) = 271' 90 - ....

3

Hence deduce that approximately 4<1>(x)[1- (x)] = e- 2X2 /'7T[ 1 + 2( 71'3~;)X4].

when x is small.

151 A continuous random variable X has the rectangular distribution in the interval (0, a). Prove that P(X";;!a)=!. If Y is another continuous random variable such that the joint distribution of X and Y has the probability density function

f(x, y) =

(

1

a a-x

)'

for 0,,;; Y,,;; a - X; 0,,;; X ,,;; a,

prove that P(Y";;!a) = pea - X - Y ";;!a) =!(1 + log., 2).

In an experiment a particle Xo with energy Eo is subjected to radiation. There is a probability p that Xo will not split, and a probability 1- p that it will split randomly into two particles Xl and X2 with energies EI and Eo- EI respectively such that EI is uniformly distributed in (0, Eo). In a second experiment the unsplit particle Xo or the particles Xl and X 2 are again subjected to radiation under the same conditions as in the first

368

EXERCISES IN PROBABILITY AND STATISTICS

experiment. Prove that the probability that all the particles result in f the two experiments are less than tEo in energy is g rOIl} t(l- p )[1 +!(1- p )(1 + loge 2)2].

152 If X is a continuous random variable having the Laplace distrib t' . h pro b a b'l' . f unction . Wit I Ity d enslty

U

f(x)=texp-ix-8i,

IOn

for-oo<X
prove that the moment-generating function of X about the origin is E(e tX )

=

e t8 (1- t 2 )-1.

Hence show that, if x is the sample mean of n random observations of X then the standardized random variable '

x-8 z=--

mn

has the moment-generating function (1- t 2 /2n)-n. Further, verify that the coefficients of skewness and kurtosis of the probability distribution of z are 1'1 = 0 and 1'2 = 3/n respectively, and thai for large n, z is approximately a unit normal variable. Alternatively assuming that an approximating non-normal distribution representing th~ behaviour of z has the probability density function proportional to e-!z2(1 + az 2+ (3z4),

determine the normalizing factor of this probability distribution. Also, if this approximating distribution has the same first four cumulants as the exact distribution of z, prove that 6

a = -6{3 = - 8n + 3 .

153 A collection of nl unbiased dice, each of which is coloured red, and another collection of n2 unbiased dice, each of which is coloured white, arc rolled together once. The numbers shown uppermost on the nl red dice total Xl and those on the n2 white dice X2' The collection of white dice is rolled once again and the sum of points obtained is X3' Assuming that each of the nl + n2 dice is numbered from 1 to 6, show that the product-moment correlation between XI + X2 and Xl + X3 is nl/(nl + n2)' Also, determine this correlation when it is given that X2 + X3 = c, a constant. Find the values of these two correlation coefficients when (i) n l = n2; (ii) nl ~ 00 and n2 remains finite; and (iii) n2 ~ 00 and nl remains finite. Explain briefly the meaning of these limiting values.

154 A continuous random variable X has a probability density function proportional to (1 + X 4 )-1 defined in the range -oo<X <00. Determine the proportionality factor and show that the random variable Z =(1 + X4)-1

has the beta distribution of the first kind with parameters ~ and standard notation.

!,

in

SUPPLEMENT

369 If Y is another independent random variable having the same probability J~tribution

as X, prove that the ratio

W=X/Y has the Cauchy distribution with the probability density function

1 7T(1 + w2)' for -00< W <00. Verify that this is also the distribution of W if X and Yare independent unit normal variables. [Note: It may be assumed that f(t)f(l- t) = 7T/sin 7Tt.]

ISS A continuous random variable X has the semi-triangular distribution with the probability density function 2x/ a 2 defined in the range O:s:;; X:s:;; a. If x. is the largest of n random observations of X, determine the probability distribution of XII. Hence show that A

_

a -

(2n+1) 2n x,.

is an unbiased estimate of a with variance a 2 /4n(n + 1). Also, derive the sampling distribution of the standardized random variable

a-a

z=--s.d. (a)

and then prove that, as n ~ 00, this distribution has the limiting form with probability density function

e- l

for -00< z:s:;; 1.

156 A savings bank permits a client to withdraw money not more than once a week. It is observed from the past experience of the bank that the probabilities of a client withdrawing money in the week preceding and the

one following Christmas are PI and P2 respectively, and the probability of a client using the facility in both weeks is P12, where O P2). The bank has a total of n clients, who may be assumed to be operating their accounts independently, and Xl, X 2 are discrete random variables denoting the total number of withdrawals made in the week before and after Christmas respectively. Prove that the joint probability-generating function of Xl and X 2 is

E(Oi". 0~2) = [1 + Pl(OI -1)+ P2(02-1)+ pdOI -1)(02 -1)]". Further, assuming that al2 PI2=-, n where at> a2, and al2 are finite constants, determine the limiting form of the above probability-generating function as n ~ 00. Show that for the limiting joint distribution al2 corr(Xt> X 2 ) = ~. vala2 Hence verify that in this limiting case if Xl and X 2 are uncorrelated then they are also independently distributed random variables.

370

EXERCISES IN PROBABILITY AND STATISTICS

157 Two continuous random variables X and Y have the joint distribur with the probability density function IOn 0: - 2 yO< (y ) no: + 1) . (1 + x)13 . exp - 1 + x '

(3 -

defined for O:s;; X, Y < 00, where 0: (> -1) and (3 (>0) are parameters sUch that (3 > 0: + 2. Prove that (i) the marginal distribution of Y has the probability density function «(3 -0: -2)g(y) no: + l)y13-o<-l '

o:s;; Y
where Y

g(y) ==

Je- w 13 w

2

dw;

o

(ii) the conditional distribution of X given Y = y has the probability density function

(Y )

y13- 1 g(y)(l+x)13

exp - - -

l+x '

and

...

(Ill) E(X I Y = y)

e- Yy13- 1

y

= «(3 -2)g(y) + (3 _2-1.

Also, if two new random variables U and V are defined by the relations

1

Y U= l+X

and

V=l+X'

show that U and V are independently distributed and determine their probability distributions explicitly.

158 A continuous random variable X has the exponential distribution with the probability density function 0:- 1

exp -(x-(3)/o:,

for

X~(3,

where 0: and (3 are positive parameters. Suppose that Xl> X2, •.• ,XII are /I independent observations of X. Given that Xl is the smallest of the /I observations, prove that the distribution of Xl has the probability density function no:- l exp -n(xI - (3)/0:, for Xl ~ (3. Hence prove that

Also show that, for given Xl> the conditional distribution of the probability density function o:-lexp-(Xj-XI)/O:,

forxj~xl'

Xj

(if 1) has

SUPPLEMENT

371

Ose this probability distribution to show that

E[~ (xi -

XI)] = (n -l)ex,

and hence verify that

1

L"

ex*=-(xi-Xl) 11 -1 i=2

and

~*=XI

~

respectively.

are unbiased estimates of ex and

159 Given a random sample of n observations Xl' X2,' .. , X" from a unit normal distribution, two statistics t and ware defined as

x~

t=-s-

and

" / Vi~Xf, ,-" W=i~Xi

where x and s are the sample mean and standard deviation respectively. Prove that (i) -JIi ~ w ~ JIi; and (ii) t 2 = (n -1)w 2

n-w 2 Hence deduce that t is a monotonic function of w. Starting with the joint distribution of x and s, use the polar transformation

x~= r sin 6,

sJn -1 = r cos 6

to prove that the probability density function of the distribution of u = wlJli is 1

BH, ~(n -1)}

.

2 !(,,-3)

(1- u )

,

Hence show that the probability distribution of v = Il.Jn -1 has the probability density function 1

2

BH, ~(n-l)}" (1 +v)

_!"

,

-oo
[Note: It may be assumed that the probability density function of the joint distribution of x and s is (n -1)!(,,-J)~

2!("-2)rmf{~(n -1)} exp[-Mni 2+(n -1)s2}]s,,-2.

]

160 In the study of accuracy of shooting at the centre of a two-dimensional target, the horizontal and vertical distances of the point of impact from the centre are known to be two random variables Xl and X 2 respectively having a joint bivariate normal distribution with zero means and second-order central moments uf, u~ and PUIU2 in standard notation. The random variable R defined by

372

EXERCISES IN PROBABILITY AND STATISTICS

is known as the radial error and is taken to be an appropriate measure of t accuracy of marksmanship. Prove that if a rotational transformation is ma~c such that C YI = Xl sin 6 + X 2 cos 6, where _1(-2 Pcr l cr2 ) 6 =It 2 an 2 2' crl-cr2

then the random variables YI and Y2 are independently distributed with zero means and var(YI ) == cr~, = t(cr~ + cr~) +t[(cr~- cr~)2+4(pcrlcr2)2p; var( Y2) == cr~2 = ~(cr~ + cr~) -

mcr~ - cr~)2 + 4(pcr1cr2)2J!.

Hence, or otherwise, show that the moment-generating function about the origin of R 2 is E(e'R2) = [(1- 2tcr~,)(1- 2tcr~2)]-~

and then verify that the jth cumulant of R2 is Kj(R 2 )

= 2j-l(cr~~ + cr~~

. (j -i)!

ANSWERS AND HINTS ON SOLUTIONS Supplement

--------------------------------------------------Finite differences and summation of series

L" r(r+4)[l(r+1)(3)-2r]

=

r=1 n

11

=t L (r+4)(5)-2 L [(r+3)<4)-2(r+1)(2)] n

-2

L [(r+2)(3)+(r+1)<2)-3r] r=\

= h(n + 5)<6) - 2[k(n + 4)(5) -~(n + 2)(3) +!(n + 3)(4) +Hn +2)(3) -~(n + 1)(2)] =-k(n + 1)(2)[5n 4 + 34n 3 -14n 2 -331n -144].

FOrll=1, sum=-10. r

II

L

(ii) Sum =

r(r + 3)

r=1

L

S(2)

.=1

II

L r(r+3) .1(r+1)(3)

=

r=\

=l L"

[(r+3)(5)-2(r+2)(4)-2(r+1)<3)]

r=\

= HHn +4)<6) -hn + 3)(5)-!(n +2)(4)] = -k(n + 2)(4l[5n 2 + 23n + 9].

For

11

= 2,

sum = 20. n-l

(iii) Sum= =

L

r

r(r+2)(r+4)

L (s-1)s(s+2)

r=\

.=2

'1-1

r-l

r=1

1=\

L r(r+2)(r+4) L [(t-2)<3)+(t+1)(2)]

11-1

=L

r(r+2)(r+4)[!(r+2)<4)+Hr+ 1)(3)]

r=\

373

374

EXERCISES IN PROBABILITY AND STATISTICS 11-1

L [(r + 5)(7) -

=!

6(r + 4)(6) + 3(r + 3)(5) + 3(r + 2)(4)]

r=1 11-1

+i L

[(r+4)(6)-3(r+3)(5)-3(r+2)(4)]

r=1

= Uk(n + 5)(8)-~(n + 4)(7) +Hn + 3)(6)+~(n +2)(5)] +Ht(n + 4)(7) -!(n + 3)(6) -~(n + 2)(5)] = 4Ao(n + 2)(5)[15n 3+ 100n 2+ 125n -144]. For n = 3, sum = 384. (iv) Sum=

n

r

r=1

0=1

L (r+1)(r+2)(r+4) L

[(S+1)(3)-4s(2)]

n

=

L (r+1)(r+2)(r+4)[Hr+2)(4)-~(r+1)(3)] r=1 n

L [(r + 5)(7) -

=!

5(r + 4)(6) + 2(r + 3)(5) + 2(r + 2)(4)]

r=1 n

L [(r+4)(6)-2(r+3)(5)-2(r+2)(4)]

-~

r=1

= Ut(n +6)(8)-~(n + 5)(7)+i(n +4)(6)+~(n + 3)(5)] -~[Hn + 5)(7)-i(n +4)(6)-Hn + 3)(5)] = 1O~80(n + 3)(5)[315n 3+ 1005n 2- 4850n -8936]. For n=2, sum =-144.

J: (r+2)(r+6)[~ (S+1)(2)-7:~

(v) Sum =

n+l

=

L

S+8r]

(r+2)(r+6)[i(r+ 1)(3)-~r(2)+8r]

r=1 ,.+1

,.+1

r=1

r=1

=i L [(r+3)(5)+3(r+2)(4)]_~ L

[(r+2)(4)+5(r+1)<3)+5r(2)]

n+l

+8

L

[(r+2)(3)+5(r+ 1)(2)+5r]

r=1

= Hi(n + 5)(6)+~(n +4)(5)]-UHn +4)(5)+Hn + 3)(4)+i(n +2)(J)] + 8[Hn + 4)(4) +i(n +3)(3)+HI1 +2)(2)]

= 12(n +2)(2)[4n 4+ 12n 3-235n 2+411n +6048]. For n = 0, sum = 168. n-2

2 (i) Sum =

L

t2(t + 1)2(t + 5)

1=1 n-2

=

L

[(t+4)(5)-3(t+3)(4)-6(t+2)(3)+6(t+ 1)(2)]

1=1

=i(n + 3)(6)-~(n +2)(5) -~(n + 1)<4)+2n(3) = ron(3)[5n3+ 12n 2-44n +9]. For

11

= 3, sum = 24.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

375

I"

[f(r + 1) - f(r)] r=1 = f(11 + 1) - f(1), where f(r) == 2 r/(r+ 1).

(ii) Sum =

for

It

= 1, sum =1.

(iii) Sum=

'" 1 L -[r(r-1)+4r-6] r=1 r! '"

=

I

1/t!+4

1=0

'"

I

1/t!-6

1=0

'"

L 1/r!

r=1

=6-e. (iv) If Q r denotes the coefficient of xr (r ~ 0), then Aao = 6; A2 a o = 16; A3ao=6; Akao=O for k~4. Hence by Euler's formula the required sum

2(3-6x+ llx 2 -5x 3 ) (1-X)4

for

Ixl < 1.

(i) Sum="f [(r+1)r+(r+1)-1]( n )xr r=O r +1

1" (n) x' L (n -1) xr - - I r x t

(11-2) ,,-I x' + 11 1=0 t r=O

I

,,-2

=

n(n -1)x

=

1 n(I1-1)x(1 + X),,-2+ 11(1 + X),,-I_- [(1 + x)" -1]. x

=

n(2)x 2(1 + X),,-3[(11 + 3)x + 5].

1=1

4 Difference the given terms of the series. Then note that A3u,. = 2, where II, (r~ 1) denotes the rth term of the series. If u,. is a cubic in r then A3 u,. = 2 for all r ~ 1. Then Ur

=

(1 + Ay-I u1

= Ul + =

e~ 1)AUI + (r; 1)A2 U1+

e;

1)A3U1

1(r 3 - 3r2+ 5r + 6).

If A3 ur 1= 2 for all r ~ 1 then u; differs from the above value by a polynomial in r which is zero for r = 1, 2, ... ,7. Hence the possible values of Ur are

t{r 3 - 3r2+ 5r+ 6) + (r-l)(7)P(r) where P(r) denotes a polynomial in r.

376

EXERCISES IN PROBABILITY AND STATISTICS

Assuming that Ll 3 ur = 2 for all r, we have

f

r~1

For

11

Ur

=![{"(11+1)}2_ 11 (1+1)(2"+1)+511(11+1) ] 3 2 2 2 +611 1 = 12"(11 3 -211 2 +511+32).

= 1, sum=3.

5 (i) The rth term of the series can be written as arx r where ar = 2r2

and

x =!,

for r~O.

Observe that Llkao = 0 for k ~ 3. Hence the sum of the series

o

=--+ I-x

2x 4x 2 +---::(l-x)2 (l-x)3

= 12. (ii) Here the rth term of the series is arx r where ar = !r3 and x = t. Also Llkao=O for k~4. Hence the sum is 13. 6

UL (1961). (i) Since Llu" = U,,+1 -

Un

we have U n +l =

=

(1 + Ll)u" (1 + Ll)2U"-1 = ...

= (I+Ll)"ul. (ii) Assume that the result holds for m -1 so that

Hence Ll"'UI =

L

",-I

(-1)'

(m -1)Llu",_ro

where Llu,,,-r = U",+I-r -

r

r=O

Um - r •

The result for m is seen to hold by combination of binomial coefficients n

~

A-I

'- Ur = L.1

(iii)

U,,+1 -

A-I L.1

Ul

r=1

= Ll- 1[(1 =

7

+ Ll)n U1 ]- Ll- t Ut

[(1 + Ll)n -1]Ll- t u 1

UL (1961). Assuming the form of f(x), show that Ll2f(x) = 2h2a2+6h3a3+6h3a3x.

If this is

== a + (3x, then

whence the result.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

8 UL (1962). (i) Put tan-I(x+h)=y and tan- t x=z. Then h

tan(y - z) = -:----:-:-:;-----(x+ h)2- h(x+ h)+ 1 SO that

h ]. h)2 h( x+ x+h)+1 Repeating this argument gives the answer for a 2 tan-I x. II.

L.l

tan - I x = y - z = tan - I [ (

(ii) a sin x = 2 cos(x + h/2) sin h/2, and

a 2 sin x = -4 sin2(h/2) sin(x + It) = -2(I-cos h). (1 +a) sin x, whence the result. (iii) aAl3x = (A13 h -1)A13 and, in general, a rA I3x = (A I3h -1)'A I3x. X

Hence, on summation, (A 13h -1)"-1

II-I

L a Al3x r~O r

=-'------:-:---

Al3h -2

.

Also,

'fl arAl3x = ["f a r] . Al3x = a"a-I-1 . AI3X, r~O

r~O

whence the result. 9 UL (1963). (i) alog(1 + x) = 10g[1 +_h_]

l+x

so that 11.21 L.l

og

(1

+x

)=1

og

[(1+X)(I+X+2h)] (l+x+h?

=IOg[I-(I~Xr{l+ I~J-2] - log [1 - (1 __ (_h l+x

~ x r]

)2 +0(h

3 ).

(ii) a 2 COS x = COS(X +2h) + COS x -2 cos (X + h) = COS(X + h) cos h -sin(x + h) sin h + cos x - 2 cos(x + h) = (cos h - 2) cos (x + h) -sin x sin h cos h + cos X cos 2 h = 2(cos h -1) cos(x + h).

377

378

EXERCISES IN PROBABILITY AND STATISTICS

The result stated now follows by noting that cos(x + It) = (1 + A) cos x 1-cos h = 2 sin2 h/2- h2/2 for small h. and (iii) Ae 2= eX'(e2hx+h2 -1),

and A2ex2 = A[e'(e2hx+h2 -1)] = e'(e4hx+4h2_2e2hx+h2+ 1)_eX2 . 2h 2 . (1 +2X2)+ O(h 3 ).

10 UL (1965). (i) The general result follows on addition and for the particular series note that 1 Ur = f(r) - f(r + 1), where f(r) = 6(3 2)( . r-

3r+1)

(ij) For r = 2 and 3 the recurrence relation gives

28 - 8a + (3 = 0;

-80+28a -8(3

=

0

whence a = (3 = 4. The sum of the series is (1-4x)/(1+2x)2 for Ixl
~-exp(-!n log 0') -1-!n log 0'. 0'

(ii) Here L~'~ 1 Ur

Ur

= f(r) -

= f(1) -

f(r+ 1) where f(r) f(n + 1) ~ i as n ~ 00.

= (2r+ 1)/2r(r+ 1). Hence

(iii) Put x = a-b. Then

g(a,b)=2xa

[1+(1-~rl]+IOg(1-~)

=:: [(!-~)+(!-t)~+(!-t)(~r

+··1

Hence g(a, b»

x3

6a 3

and

whence the result.

12 UL (1975). (i) The expression for f(r -1) - f(r) follows directly. Then the sum of the series to n terms is f(O) - f(n) ~ l/x as n ~ 00. (ii) Set A = [(x -1)/(x + 1)]!. Then log A = ~ [IOg( 1-~) -log( 1 +~) ]

1

1

= --;- 3x 3 + O(x

-5)

.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

379

Hence

1 1 1 -1--+--X 2X2 2x 3 ' (iii) Note that

Ur

13 UL (1973). (i) Note that

Ur

(-xY 1 (-xy+ 1 =--+-. 1 whence r x r+ ~ 1 L Ur = -log(l + x) - - [log(l + x) - x]. r=1 X

= (2 r -l)/(r+ 1)! whence

~

L

r=1

~

ur=t

~

L 2r+1/(r+1)!- L

r=1

r=1

~

=t (ii)

on expansion.

n

1/(r+1)!

~

L 2$/s!- L

$=2

$=2

l/s!,

whence the answer.

(n

r 1 ,,+1 ,,+1 ( L _ )(3- ) =--1 L + 1) 3r, whence the answer on summar=1 r 1 r n + r=1 r tion of the binomial expansion.

(iii) The given expression can be rewritten as x(l + l/x)! 10g(1 + l/x) and the answer is obtained by expansion and retaining terms up to x- 3 • (iv) The given expression = e- 2X '(e2x +e- 2x ) 2x4 ] = 2 e -2X'[1 + 2 x 2 +3+'"

whence the answer on the expansion of e- 2x '.

14 UL (1972). (i) The expression for f(r -1) - f(r) follows on substitution and then S" = f(O) - f(n). (ii) If S denotes the sum of the series then differencing shows that (1- x)S = 1 +2x +(2x)Z+(2x)3+ ... =(1-2x)-1 if 12xl<1. (iii) We have 1 + e = 1 +2x +tx 2+ix 3 + . . " whence, as a first approximation x = te. Now assume that a second approximation is x = te + Ae 2 where A is determined from the equation 1 + e = te + Ae 2 + e~6+M2 Expansion of the exponential and equating the coefficients of gives A = --h.

15 UL (1971). (i) Note that Ur

=

4(r + 3)(3) - 3(r + 3)(2) + (r + 3) + 2 (r+3)!

e2

380

EXERCISES IN PROBABILITY AND STATISTICS

whence the required sum. It now follows that Ur

forX~O

P(X=r)=r(e_1)'

and so

P(X ~ 2) = 1-P(X ~ 1)

= 1 uo+ Ul 4(e -1)

(ii)

L

"'-(3 (

k) = ",-(3 aL coefficient of x(3 in (1 + (3

k=O

X)",-k

k=O ",-(3

=

coefficient of x(3 in

L

(1 + X)",-k

k=O

= coefficient of x(3 in

L'" (1 + x)i j=13

. . (l+x)",+1 (l+x)(3 = coefficIent of x(3 In - - - ------'x x (1 +X)"'+1 = coefficient of x 13 in -x,whence the answer .

(iii)

t (a + kk - 1) ((3 +aa=kk - 1)

k=O

0:>

=

L

coefficient of

Xk

k =0

=

in (1- x)-'" x coefficient of

coefficient of x'" in (1- x)-",-(3,

16 UL (1970). (i) The expression =

t

l+l1a

r

( -1

l+na

)r-l

x 1 +l1a

in (1- xf II

whence the result.

(11)(~)r +na(~)

r=O

X",-k

t

r=1

(n-1) r-1

'

= 0 on summation. (ii) Here u r =f(r)-f(r+1), where f(r) = 4Ir 2(r+ 1f. Hence

L" U = f(l) r

f(n

+ 1).

r=1

(iii) Here f(r+ 1) - f(r) = 14r6 -2r2 whence the expression for r6. Hence f1

fI

L r6=9 L r2+1~[f(n+1)-f(1)], and the answer follows on reduction.

17 UL (1969). (i) Note that (r+s+1)(s+2)-(r+s)(s+2)=(s+2).(r+s)(s+1), whence the

result on summation. In particular, for s = 0, Sr = !r(r+ 1)

and

S,,-r = !(n - r)(n - r + 1).

It now follows that S.5,,-r =;\[(n + 3)(2)(r+ 1)<2) -2(n + 3)(r+ 2)(3) +(r+ 3)<4)].

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

381

Hence II

L SrSII-r =;t[(11 +3)(2)xi(n +2)(3)-2(n+3)Xi(n+3)(4)+!(n+4)(S)] r=1

and the required result follows on reduction. II (11) (ii) Note that r~o r x r+1 =(1+x)II+1_(1+x)lI.

The required result is now obtained by integrating both sides with respect to x from 0 to 1.

18 UL (1968).

(i) The sum of the first n natural numbers is !(n + 1)<2), whence the average.

(ii)

11

II

r=1

r=1

L [r-!(n+1)]2= L [r 2-(n+1)r+;t(n+1)2] (2n + 1)(n + 1)(2) (n + 1)(n + 1)(2) 1( 1)2 6 2 +4 n + n,

whence the result on reduction. (iii) By

definition E( Jr) Hn + 1) - (J2. Hence

Ltl

= L~I=l Jr/n = (J

Jr

and

var( Jr) = E(r) - (J2 =

r

= n 2(J2 =!n 2 (n + 1)- n 2 var(.Jr) =n 2(n+1) 2

[1- l1(n+1) 2V(n) J.

19 UL (1967). (i) Note that ur+l-ur=24r7-8r3, whence 11

n

n

r=1

r=1

L r =1 L r3+2~ L (u r+

1 - U r)

7

r=1

="31 [n(n2+ 1)J2 +2
whence the result.

(ii) Here U r = f(r)- f(r+ 1), where fer) = 1/r2. Therefore II

L ur =f(1)-f(n+1). r=1

(iii) If x takes the values 2r -1 for 1 ~ r ~ 11 and s> r, then the required sum is II

L (2r-1) L (2s-1) r=1

s>r

II =2"1 [{r~1II (2r-1) }2 - r~1 (2r-1)2 J

whence the result on reduction.

382

EXERCISES IN PROBABILITY AND STATISTICS

20 UL (1966). (i) We have Llf(r) =sin[O+(r+ l)ebJ-sin(O+rcp)

= 2 cos[ 0 + eb/2 + reb J sin( eb/2) = 2 sin(eb/2) . sin[O +!( 7r + eb) + reb J. Hence

Ll 2f(r) = 2 sin( eb/2)Ll sinE 0 +!( 7r + eb) + reb J = [2sin(ebf2)f. sin[O+(7r+eb)+ rebJ. The general result now follows by induction. (ii) By definition U r = (1 + Ll)Ur-1 = (1 + Ll)'uo. Therefore

t r=O

(n)(_l),,-r ur = r =

t r=O

(n)(-l),,-r(1 + Ll)'uo r

Lto (;)(-l),,-r(l+LlY].

Uo

For ur =(-lY/(r+1),

Ll"uo=.to (;)<-1)"/(r+1)

t

=(-1)". (n+1) n+1 r=O r+1 (-1)"(2,,+1-1) n+1 ( ...) H 111

_ (r+2)(2)-2(r+2)+ 1

ere

Ur -

=

(r+2)!

[:! -(r: 1)!]- [(r: I)! -(r:2)!]

whence, on summation,

" r [ r~1 u = 1

1 ] [1 1] (n+1)! - 2-(n+2)! .

Probability and statistical theory 21 UL (1965). The total probability of a student entering the second year is 3 5 4 7 4+32·5=8; and the probability of a first-year student getting a degree is

7 20 5 8·21="6·

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT ~e/lce

n)(5)k(1)"-k P(i)= ( k "6 "6 ; P(iii) =

.. -_ L. f P(ll)

383

(n)(5)S(1)"-S --

s=k

S

6

6

st (;)(~)"(~r-s.

zZ UL (1966). Let A be the event that the player has a flush, and B the event that he has at least one spade. (i) P(A) =

(~)(~)

1

(552 ) = 163:60 = 0·0020.

(39)/(52) 5 = 7411 9520 = 0·7785.

(ii) P(B) = 1- 5

(iii) P(BA) = P(B) - P(AB)

= 1-

(39)1 (52) _ (13)1 (52) = 49355488 63441280 5 5 5 5

0.7780.

99 (iv) P(AB) = P(A) - P(AB) = 66640 = 0·0015. 31416 (v) P(A I B) = P(AB)/P(B) = 49 386 904 0·0006. (vi) P(B I A) = P(AB)/P(A) = 1. Z3 UL (1967). P(i)= P(iii) =

1(~

(N:n)/(~

P{ii) = (:)

(:)(~~:)/(~

P{iv) = 1- (N:n)/(~.

(a) Let X be the event that there is at least one bad fuse in the selection of P, and Y the event that all the v fuses selected are bad. Then the required

probability is P(Y I X) = p(YnX)/p(X) = P(Y)/P(X),

whence the result. (b) Let Z be the event that all the· fuses in the second selection of v are good, and Tr the event that there are exactly r bad fuses in the first sample of P, where l:so; r:SO; v. The required probability is P(Z I X) = P(Z n X)/P(X)

where X =

. LT

r=1

Therefore p(znx) =

f P(Z n Tr) f P(Z I Tr)P(T.}. =

r=1

r=1

The required result now follows by noting that

r•

384

EXERCISES IN PROBABILITY AND STATISTICS

24 UL (1968). (i) With replacement sampling the total number of sample points ar" . from the numbers on the three tickets is (611)3. The numbelSlllg sample points leading to the realization of S = 611 is the coefficie~t or 0 611 in the expansion of or H(O)

=

(1- ( 611 )3/(1- 8?

The required coefficient is 18n 2 +9n-2=(3n+2)(6n-1) whe the answer. ' nee (ii) The probability-generating function of S is H(0)/216n 3 and E(S) H'(1)/21611 3 • When sampling is without replacement then the total number of sampl . points is 611(611 -1)(6n -2), and the number of sample points leading to th~' realization of S=611 is also less than 18n 2 +9n-2. From this we have te exclude the sample points corresponding to the integral solutions of x + y :' z = 611 whe~e (a) x = y = z and (b) y = z1= x so that?' +2y = 6n. The solution correspondmg to (a) IS x = Y = z = 2n and the solutions corresponding to (h) are obtained for 1:so; y:so; 3n. The solutions for (b) are 3n -1 in number and hence the total number of sample points are 3(3n -1). Therefore the excluded sample points corresponding to (a) and (b) are 9n -2, whence the answer. 0

25 UL (1969). (i) Suppose Xi is the event that the ith ball drawn is white, i == 1,2. Then . n(n+r) P(I) = P(X1n X 2 ) = P(X2 1 X\)P(X1) = ( ) ( . m +11 m+l1+r)

(ii) Similarly or by symmetry P(ii)= (iii) Suppose

m(m+r) (m+n)(m+l1+r)

Yi is the event that the ith ball drawn is black,

i == 1,2.

Then P(iii) = P(Y2 1X1)P(X,)+ P(X2 1 Y1)P(Y,) 2mn (m + n)(m + n + r) . (iv) P(iv) = P(X2 ) = P(X2 1X,)P(X1) + P(X2 1 Y1)P( Y1) n =-m +11 (v) P(v) = P(Y1 I X 2 ) = P(Y1n X 2 )/P(X2 ) = P(X2 1 Y1)P( Y 1)/P(X2 ) m

m+n+r

Expectation of white balls drawn is 2nl(m + n).

26 UL (1970). Suppose X is ·the event that all r balls are of the same colour, Y tl!l' event that all are black and Z the event that all are white. Then X == Y+Z.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

"here in the first experiment P(Y) = P(Z) =

(J /

385

(2:).

/'he required probability is P(Z I X) = P(Z n X)/P(X) = P(Z)/P(X) =!.

In the second experiment

/'he required probability is

(J

1

Ignoring terms involving N- 2 , this probability is approximately =

[1 + 1- k{2r+ (k -1)}/2~-1 1-k(k-1)/2N - J

=~[1_k(~~1)][1 k(k;~-1)r1

-~ (1+2k~). 17 UL (1979). P(Holmes dies) = xy + (1- x )(1- y) = a P(Holmes escapes) = (1- x)y = a P(Draw) = x (1- y) = 1 - 2a.

(1) (2)

(3)

From (2) and (3), x = y + 1- 3a whence on elimination of x from (1) we obtain the stated quadratic in y. The condition that the quadratic has real roots gives for a> 0, a ;;a;. §. The inequality now follows since from (3) 1- 2a;;a;. o. 28 UL (1971). (i) By the hypergeometric distribution

and

The recurrence relation follows by taking the ratio P(r)/P(r-1). (ii) Let Xk be the event that A has k court cards in his hand (o:s;; k :s;; 13) and Y the event that the game ends in a draw. Then 13

P(Y) =

L P(Y IXk)P(Xk), k=O

386

EXERCISES IN PROBABILITY AND STATISTICS

where P(Xk ) = P(k)

as in (i)

and P(YIXk )=

C6;k)(:~~~)/G~),

=0, for

for

0~k~8

9~k~16.

If it is known that A has all court cards, then the probability of B hav· no court cards is Ing

The percentage error in the approximation is

2 200 7 703x100=~ 200 7· 703

29 UL (1972). The probability that at least one six occurs for the first time on the rth throw of the k dice is

For given k the expected reward is (in £)

Ek =

f 21_kr(~)k(r-1)[1_ (~)k] r=1

=

6

6

1 1-(i)k 1 2 k- 1 • 1-U2)k <2 k- 1 •

Hence, on the average, the player can never be on the winning side if the non-returnable stake to play the game is £2 1 - k • We have

Ek 1-(i)k 1-(fi)k+1 E k+1 =2 ·1-(i)k+1· 1-(f2)k 1-(i)k >2· 1 _(i)k+1 _12 [ (i){1-(i)k-1}] -11 1 + 1_(~)k+1 12 >11> 1, whence Ek > E k + 1 • UL (1973). Suppose Xr (0 ~ r ~ n) is the event that there are r white and n - r black balls in the urn, and Y the event that in the sample of k balls drawn, there is

30

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

387

oply one white ball. Then Il-k+l

P(Y) =

L

P(Y I Xr)P(Xr),

P(Xr) = (;)prqll-r

apd P(YI Xr)=

G)(:~~)/(;)

for

1~r~n-k+1

=0 for r=O.

Ip the second case P(Xo) =0; 31 UL (1975). The probability that the player gets far at the rth trial is (~~y-l. ~~. Hence the expected gain of the player is

r~l 00

(25)r-l 11 36

11a

. 36' a r = 36-25a .

The player, on the average, will be on the winning side if the expected gain >! which gives a >~. 32 UL (1977). Irrespective of the suit of the cards, the following ten realizations of five cards lead to a run: (A, K, Q, J, 10), «K, Q, J, 10,9), ... , (5,4,3,2, A). Since there are four suits and in all 52 cards, the probability of a run is

and that of a running flush is

whence the probability that the player has a run but not a running flush is 40. 255Wn = 81> say. If it is known that the player has the jack of hearts, then the three corresponding probabilities are

The required percentage change is

The initial probability of a run is 128 51.49.13

PI'

388

EXERCISES IN PROBABILITY AND STATISTICS

say and the initial probability of a running flush is

1 51.49.26 = P2, say. When the player has the jack of hearts, these probabilities are chan d ~~PI and ~P2 respectively. Hence the percentage change is ge to

[ ~(Pl- P2)

1J x 100 = 4.

P1-P2

32a

UL (1966). (i) Let X be the event that the examinee knows the right answer and Y

the event that he gives the correct response. Then P(X)=p; Hence

P(X)= 1-p;

P( Y \ X) = 1 - a ;

' P(Y \ X) '= 1/11.

P( Y) = P( Y \ X)P(X) + P(Y \ X)P(X) n(1-a)p+ 1-p n

Then P(X \ Y) = P(Y \ x)P(x)/P(Y) np(1-a) ~1 np(l-a)+ 1-p

as n ~OO.

Also P(X \ Y) = P(X n Y) = 1- P(X) - P( Y) + P(X n Y) P(Y) 1-P(Y) 1-P(X)-P(Y)+ P(X \ Y)P(Y) 1-P(Y) (n-1)(1-p) ~ 1-p <1 npa +(n-1)(1-p) ap +l-p

as n ~OO. (ii) We now have P(Y IX) = 1/(n - m) instead of lIn. Hence

P(X \ Y) = (

(n-m)p(l-a) ) (1 ) 1 ~1 n-11l p -a + -p

.

as n ~ 00 and m fimte

and P(X\ Y)=

(n-m-1)(1-p) ~ 1-p <1 (n-l1l)pa +(n-I1l-1)(1-p) ap + 1-p

as n ~ 00 and m finite. 33

UL (1967). Let A denote the proposal "decrease duty on cigars and tobacco" and B the proposal "increase duty on cigarettes". _ If N(X) denotes the number of M.P.'s voting for a proposal X and N(X) those against X, then we set N(A) =x;

N(B)

= y;

N(A nB)=z;

N(A

nB)=w.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

389

f\lso assu~e that there are 11 M.P.'s in all. Then the conditions of the problem give x 3 y

,l\lso,

I1-X=2:'I1-Y

(1)

z + w + 135 = y w+z-165=I1-Y z-1l0=2(w+llO)

(2) (3)

(4)

N(A nB) =N(A)-N(A nB) = x- z N(A

n B) = N(B) -

N(A

n B) = y- z.

Hence, using the fact that A n B, A n B, A n B and A n B are mutually exclusive and exhaustive events, we have z + (y - z) + (x - z) + w = 11

(5)

Elimination of x, y, wand z from (1)-(5) gives the quadratic in 11 as 11 2-97511+67500=0 with roots n = 900 or 75. For 11 = 900 x=675;

y=600;

N(A

n B) = 420; n B) = 255;

Thus N(A

z=420;

w=45.

N(A nB) =45 N(A

n B) = 180.

34 UL (1968). There are 25 (x, y) sample points. Divisibility by 5 of x 2 + y2 is ensured for the following 9 points.

(1,2), (1, 3), (2,1), (2, 4), (3,1), (3,4), (4, 2), (4, 3), (5,5). Hence the stated probability. Divisiblity by 5 of 2x 2+ 3y2 is ensured for the following 9 points: (1,1), (1,4), (2, 2), (2, 3), (3, 2), (3, 3), (4,1), (4, 4), (5, 5), whence the probability. The point (5, S) is the only one common to the two sets. When 1:5; x, y:5; 5N, then the sample points giving divisibility of x 2 + y2 by 5 are for x = 1, y =2+5r; 3+Sr 0:5;r:5;N-1 for x =2, y=l+Sr; 1+4r for x =3, y = 1+5r; 1+4r for x =4, y =2+5r; 3+5r for x =S, y =5+5r. This pattern repeats itself N times as x varies from 1 to 5N. Hence there are 9N2 sample points out of a total of 25N2 which now give divisibility of x2 + y2 by 5. The probability is independent of N. A similar argument applies for the divisibility of 2X2+3y2 by 5. There are ~ common sample points (S + Sr, 5 + Sr) which give the divisibility of both X2+y2 and 2X2+3y2 by 5.

390

EXERCISES IN PROBABILITY AND STATISTICS

35

UL (1969). Suppose Hr is the event that the rth urn was selected and X the e that the two balls drawn are white. Then vent k

P(X) =

L P(X I Hr)P(Hr)

where P(Hr) = 11k.

r=l

n-r)2 (i) Here P(X I Hr) = (- for 1 ~r~k-l m+n =1 for r = k. Hence P(X) =~

[en ~nr{ n 2(k-l)- nk(k-l)+ k(k -1~(2k-l)}+ 1] for l~r~k-l

(ii) Here P(XIHr)= (n-r)(n-r-l) (m+ n)(m+ n -1) =1 Hence

for r= k.

P(Hk I X)

(n+m)(n+m-l) 2 (k -1)[n - nk +!k(2k -1)]- n(k -1) +4k(k -1) + (n + m)(n + 111 -I) .

36 UL (1966). Let X be the event that the two consecutive throws of the die selectcd show a double-six. Then 6

P(X) =

L P(X I D )P(D j

j)

j=l

where P(Dj ) =!, the probability of selecting D j and P(X I D j ) = 2 for i = 1, 2, 3;

m

=0 =(4)2

for for

i = 4, 5;

i=6

Hence P(X) = is and P(D6 1 X) = P(X I D 6 )P(D6 )/P(X) =~

If D3 is a biased die with probability x of it showing a four or a five, thcn the probability of it showing a six is 2x. Hence 4x = 4 or x = k, and P(X I D 3)= (~f. We now obtain P(X) = ii4 and P(D6 1 X) = ~~.

37

UL (1969). n

G(Ol> O2 ) =

nr

L L O;:O~P(XI = r, X 2 = s) r=Os=O nr

It

=

L Orp(X = r) L 0~P(X2 = s IXl = r) 1

r=O

=

s=O

L 0;: (n) prqlt-r L It

r=O

Itr

r

s=O

= [q + pOI(q + p 02t]lt.

O~ (nr) pSq'''-S

s

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

391

(i) The probability-generating function of the marginal distribution of X 2 is G(l, ( 2) = [q + p(q + p( 2)"]"

whence P(X2> 0) = 1- P(X2 = 0) = 1- G(1, 0). (ii) and (iii) If Y = Xl + X 2 , the probability-generating function of Y is G(O) = [q + pO(q + pO)"]

whence, on differentiation, E(Y) = G'(l)

and var(Y) = G"(1) + G ' (1)[l- G ' (l)].

38 UL (1969). Let X", Y,,, Z.. be the events that after n moves the system is in an A-type, B-type or C-type cell. Then for n ~ 1 X,,-l + Y,,-l + Z,,-l = n, the sample space. Hence P(X,,) = a,,_IP(X.. I X.. - I) + ~"-IP(X" I Y,,-l) + 'Y.. -IP(X" I Z.. _I) and similar equations for P(Y,.) and P(Z..). Symmetry considerations now show that P(X" I X,,-l) = 0, P(X" I Y,.-I) =!, P(X,. I Z .. -l) =i,

P(Y,.I X,,-l) =i, P(Y,.I Y,,-l) =t, P(Y,. I Z,,-l) =!,

P(Z" I X .. - I ) =t P(Z,. I Y,.-l) =t P(Z" I Z,,-l) =i.

Hence the difference equations a" = !~"-I +i'Y"-1 ~Il = iall-l +t~Il-1 +!'YIl-I 'Y" =~a"-l +t~"-l +i'Y,,-1 with the conditions a,,+~,,+'YIl=a"-I+~"-I+'YIl-l=1. Elimination of an and 'Yn gives the following non-homogeneous equation 70= 120~" +27~"_1. The transformation ~" = x,. +¥t reduces this to 40x,. +9x,,_1 =0.

Hence 10

(9)11

~Il = 21 + k - 40

for n ~ 1

where k is arbitrary. For n = 0, ao = 'Yo = t ~o =! and so ~l = flo. Hence using earlier equations,

k = -li9. Therefore,

all =!-*~.. and 'Yll =!-a~". By symmetry, the probabilities for specific A-type, B-type and C-type cells are aa", i~" and h". As n ~ 00, these probabilities tend to i4, i.., /4.

392

EXERCISES IN PROBABILITY AND STATISTICS

39 UL (1967). (i) Let Xo be the initial probability of A winning the series; this i' . I. the probability of A to win whenever the scores of the two PIS d So 'h ' 'h are Ieve,I If XII IS t e pro b a b'l' I Ity 0 f A winning t e series when ayers he .. n games ahead of B (-4~n~4) then the forward diffe" IS , , lenee equation IS for-4
XII=PXII+I+qxlI_l+rx",

x4=1,

X-4==O,

whence PX,t+2-(P+q)XII +1 +qxlI

= O.

The roots of the auxiliary equation are 1 and q/p and so XII =

IX

+ {3(q/p )11,

where the arbitrary constants IX and (3 are determined by using the conditions X4 = 1 and X-4 = 0, Hence p8_ p4-lI q ll+4 XII =

and the stated result for

Xo

8

P -q

8

follows by putting

n =

0,

(ii) This probability is obtained by putting -n for n and interchanging II and q in the expression for XII'

40 UL (1967), The probability that all the k bottles of the first customer are fresh is

Let Xr be the event that r fresh and v - r a day old bottles are sold oul before the selection of the second customer, and Y be the event that she selects k fresh bottles, then v

P(Y) =

L P(Y I Xr)P(Xr) r=O

where

P(Xr) = P(Y I Xr) =

(;)C':') j (m: n) (n k- r)j (m +kn-v),

for r = 0, 1,2, , , , , v.

The stated result is obtained by recombining the binomial coefficients. Again, if Z denotes the event that the second customer gets no fresh milk, then v

P(Z) =

L P(Z I Xr)P(Xr) r=O

where P(Z I X r )=

(m +;-v)j (111 +;-v).

Finally, let W be the event that at least one of the k bottles of the

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

393

second customer is fresh. Then P(YI W)=p(ynW)/p(w) P(Y)

1 - P(Z)' whence the stated result.

41 UL (1966). The difference equation for Un is Un = p2 . U,.-3 + pq . U,.-2 + q . u,.-\> for n ;a. 3 with Uo = 1, Ul = q, U2 = pq + q2 = q. The roots of the auxiliary equation are 1, (-!p)(1 ± i.J3) whence the general solution Un =A+B[(-p)n(cos

~71" +i sin n371")]+c[(-p)"(cos n371" -j sin n371")]

where A, Band C are arbitrary constants. To determine the constants use the initial conditions. This gives B=-JL (l+p)iv'3-q. 2.fj· 1 + P + p2 '

A== _ _ I_

1 + p + p2

C = _JL (1 + p)j.fj+q 2.fj . 1 + P + p2

The stated result follows on reduction.

42 UL (1966). The total number of samples of size n is (~). In any sample of size n, x will be the largest integer if the other n - 1 integers in the sample are drawn from the x-I integers less than x. Hence the probability distribution of X. The assumed equality may be proved as folIows.

xt (=:~)= xt C~~) =

Nf' (t+n+k) 1=0

t

N-n

=

L

coefficient of

ZO

in z-l(1 + z)'+n+k

1=0

= coefficient of

N-n

ZO

in (1 + z)n+k

L {(1 + Z)/Z}I

1=0

= coefficient of ZO in (1+z)n+k[(1+z)N-n+l. z-(N-n)_z] = coefficient of zN-n in (1 + Z)N+k+l

= (N+k+l)= (N+k+l). N-n n+k+l For k = -1, we obtain the sum of the probabilities P(X = x). The expression for E[(X + r -1)<')] is also obtained by the use of the equality whence the required values of E(T) and var(T).

43 UL (1966). The difference equation for Sn is Sn+l =(1 +p)Sn +nR +(I-p)R +p. 2R, and So= R. To solve this difference equation, make the transformation Sn = Xn + a + (3n,

394

EXERCISES IN PROBABILITY AND STATISTICS

where a and (3 are so chosen that the equation is transformed into x,,+1-(1+p)x,. =0 It then follows that

a=-

R(2+p) P

This leads to the stated result for 5" on reduction. To obtain the inequality, observe that R

5" =2"[(1 + p)"{1 +(1 + p)p +p2}-{1 +(1 +p)p+ np}] p

and since 0
<;

(1+ p )n+2=RC;Py(1 +p)"- RC;PYenp

The inequality for n follows on taking logarithms.

44 Suppose there are x shillings (2x half-crowns, 3x halfpennies), y two-shilling pieces and z pennies. Then 5x + y = ~z. If x = 2, then z = 8. Hence, working in halfpenny units, the equation for y is 6 X 1 + 8 X 2 + 2 X 24 + 48 y + 4 X 60 = 406 whence y =2. For equal division of money the probability is

(~)(!)(~)(D(~) / (~~) = ~10909 -0,0476. If x is unknown, then z =2X2; 5x+y =3x 2. Hence, in terms of halfpenny units, the equation for x is 3x X 1 +2x 2 x2+24x+48(3x 2 -5x)+2x x60 =406

which reduces to (x - 2)(148x + 203) = O. Therefore x = 2.

4S UL (1970). (i) Make the transformation z = !(t -1). Then the given integral is 1 (X- l112 1 2 1 (X- l)/2 I 2 1 (-X- l)/2 _h2 dz e J 2Z J2; e- dz = J2; e-'Z dz - J2; .

J

J

(-x-1)/2

whence the result in terms of ct>(.). (ii) There are 2N + 1- n members who are not in the pressure group. If r

is a random variable denoting the number of these members who vote with the pressure group, then r is a binomial variable such that E(r) = !(2N + 1- n); var(r) =i(2N + 1- n). The required probability is P(r~N+ 1-n) = 1-P(r:!S;N-n) =1-

L

N-"

r=O

-ct>(-

(2N + 1- n)(1)2N+l-" r 2

n

)

\.J2N+1-n '

by using the normal approximation for the binomial distribution.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

395

For fixed N this probability is an increasing function of n and if we assume that <1>(3) -1, then P(r ~ N + 1 - n) - 1 if n2~9(2N +1- n),

whence the inequality for n.

46 UL (1967). P(X = n) = Probability that there are w -1 marked animals in the first n -1 members of the second sample x Probability of selecting a marked animal on the nth occasion from a total of N - n + 1 animals of which W - w + 1 are marked

. N-n+1 '

which reduces to the stated result on recombining the binomial coefficients. On reduction (N - n)(N - W) N(N- W-n+w)

g(N) g(N-1)

=

(1- ~ / (1- ~=:).

The stated result follows by considering g(N)/g(N -1) > l. · f · (W + 1)n 1. · d An unblase estImate 0 N IS W

47 UL (1965). The probability that the player loses for the first time at the (r+ 1)th trial is prq and his net gain is rx - a. Hence QO

E(S) =

L (rx-a)prq r=O QO

= -a + pqx L

rpr-1

r=1

d = -a +pqx. dp [(1-p)-1],

whence result.

In the same way, QO

E(S2)=X2q

L {r(r-1)+r}pr-2axp/q+a 2 r=O

Hence var(S). The inequality E(S) ~ a leads to x> a/p since 2q < l.

48 UL (1975). P(r, X = x) =

(:}n1- 6)x-r . (1- p )px-n

396

EXERCISES IN PROBAB ILITY AND STATISTICS

and so

L'"

= (1- p)e

Xpx-n

x=n

Lx

(

r=1

x-

1) or-l(l- oy-r

r-1

'"

L

=(1-p)O

Xpx-n

x=n

'"

=(l-p)6

L (n+t)p'

,=0

'"

=nO+(l-p)O

L tp',

,=0

whence the result on summation.

49 UL (1972). When the housewife has x different coupons, the probability of her obtaining a new coupon is 1- x/n and that of a duplicate COupon is X/II. Hence P(r I x) and E(r I x) =

(1-~) n

n n-x

= --

f

r=1

r(~)r-l n

. on summation.

Hence the expected total number of weeks is "-1

11-1

x=1

,=1

L n!(n-x)=n L t-

50

1•

UL (1971). ~

(i)

L

~

r(k)p(X = r) = f.L k

L e-lLf.Lr-k/(r- k)!

r=a ~-k

=f.Lk

L

e-ILf.LS/s!,

whence the result.

s=Ot-k

(ii) If Y denotes the number of rolls of film sold, then

P(X= r) P(Y= r)= 1-P(X=0)' for Y::::.1. Therefore the probability distribution of Z is P(Z = 0) = P( Y = 1) P(Z=r)=P(Y=r), for 2.:;;Y.:;;5 P(Z = 6) = P(Y::::. 6). Hence 5

rP(X=r)

'"

P(X=r)

E(Z) = r~21-P(X=0) +6 r~61-P(X=0)

1 _ [f.LP(1 ':;;X':;;4)+6{1- P(O':;; X.:;; 5)}], 1-e jJ.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

397

",hence the result. Similarly, 1 E(Z(2) = --_- [/-L 2p(0 ~ X 1-e IL

~ 3) + 30{1- P(O ~ X ~ 5)}]

",hence var(Z). The average return per roll of film sold is (in pence)

42- 6(1-a)+/-Lb. 1-e- 1L

Sl UL (1970). Expected loss =

I e- IL /-L:r. . ar(r+ 1) e-/3r

r=O

-ar e IL =a e- r~1 (r _1)! /-L,[(r-1)+2] 00

=a e- IL e- 2/3/-L2 [

-/3(r-2) r-2 -/3(r-1) r-l] /-L +2e-/3/-L L e /-L , r=2 (r-2)! r=1 (r-1)! <X>

<X>

Le

whence the answer on summation. Since 1- /-L < 0 and /-L e-/3 < 1, we have Expected loss < 3a e 1 -

1L

< 3a.

S2 UL (1969). Assume a Poisson model for the distribution of errors with mean /-L. Then the expected cost of correcting the errors made on a stencil is

I e-IL/-Lr 2r(3r+2) r=O r! . r+1 =2e- I /-L'[3r(r+1)-(r+1)-1] =

1L

(r+1)! r r 1 r+l] 3 L _/-L__ L ~+- L _/-L_ r=l(r-1)! r=or! /-Lr=o(r+1)!'

r=O

=2e- 1L

<X>

[

<X>

whence the result on summation. The residual profit is a -2[3/-L -1

+;

<X>

(l-e- IL )]

and this will be equal to '\a if

(1-'\)a+2 2

3 /-L+1 (1-e-) IL /-L =1+~/-L+i/-L2- ... ,

----=

whence the answer by retaining only the terms in /-L.

53 UL (1968). Suppose Ny calendars are ordered. For 0 ~ x ~ y, the profit is PI = Nx x 30a + N(y -x) x 30~ -Ny x30

398

EXERCISES IN PROBABILITY AND STATISTICS

and for x > y the profit is P2=Ny x30a-Ny x 30

Hence y

00

I

G(y)=30N

L

[(a-{3)x-(1-{3)y]P(x)+30N

X~O

(a-1)yP(x)

x~y+,l

whence the result on using the relation y

00

L

P(x) = 1-

x~y+l

I

P(x).

x~o

Direct substitution shows that G(y + 1) - G(y) = 30N[a -1- (a - (3)F(y)]. Maximum profit is obtained for the smallest value of integer y which makes aG(y)
E(Y2) =

a

f

+ N '\x

e-A(x-l)

dx

1 00

f

=a+N '\(u+1)e- AU du, o

whence the result. Therefore the expected population at the start of the third year is a + (,\ : 1) [a +

N('\ : 1) ]

=

a[ 1 + (,\ : 1) ] +

N('\ : 1

r

Hence the expected population at the start of the (k + 1)th year is a[ 1+ (,\ :1)+ (,\

:1y

+ ... + (,\ :1r-l]+N('\ :1r =

The equation 2(,\ + 1)k

a

[(,\

+ 1)k _,\ ,\ k-l

k] + N (~)k ,\ .

= (/L + 1),\ k follows directly. Then

10g(/L; 1) = -k 10g[1- ,\ ~ 1] ~ ,\ ~ 1 for'\ large. 55

UL (1965). 00

E(T)=

~f x(1+{3x)exp{-~-{3(a-X)}dx

v27T{3 =

-00

2{3

{1

}

-0<11+111 3 fOO a ~2 x(1 + (3x) exp - - (x - {32f dx. 27T{3 2{3

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

399

The integration is completed by using the substitution z = x - (32 a d th . 0 f th i 'Integra.I n en the propertIes e norma Differentiate E(T) logarithmically with respect to a to maximize E(T).

S6 UL (1966).

For small a, (1 - 1~O) 100 ~ e -a. Hence the probability that out of ten

consignments exactly w will be rejected is

Therefore the equation for determining a is

To solve this put z = 1-e-a , where z is small. The terms containing z and Z2 vanish and we have approximately 1-120z 3 ~ 0·985 whence a

=

or z

= 0'05,

0·05.

57 UL (1966). Expected amount payable to proof-reader is 00

T=a+e->-a.P(x=O)+

I

pr-1

a .p(x=r)

r=l

whence the result. O<e->-O,

and

O<e->-(l-e->-):S;;~.

Therefore

whence the inequality by using p e->- :S;;!. 58 UL (1967). The probability of finding < n defectives in the sample of N is

This is also the probability of accepting the batch after inspection, which now contains only M - N components. For each component sold, the expected net income is 1 - a6. Therefore g(n, 6) = (M - N)(l- (6)

~t~ (~6X(1- 6)N-X

400

EXERCISES IN PROBABILITY AND STATISTICS

whence 1

J

g(n) = E0[g(n, 0)] = (M - N) (1- aO)

~~: (~OX(l- O)N-X . 66(1- 0) dO

o 1

= 6(M - N)

=6(M-N)

"of (N\ J(1- aO)Ox+1(1- 0)N-x+1 dO

X~O x)

o

:~: (~[B(X+2, N-x+2)-aB(x+3, N-x+2)J

6(M-N) .. -1

L

)(4) (x + l)(N - x + l)(N + 4 - ax - 2a), on reduction N+4 X~O 6(M-N) .. = (N+4)(4) Z~1 z(N-z+2)(N+4-a-az),

= (

whence the result on summation over z. For ~ = n/ N, we have g(n)-

(M-N)e 2 [6-4(1+a)~+3ae].

It now follows that g(n) is maximized for

a~ =

1.

59 UL (1968). The expected cost of a packet before devaluation is 00

a

+ {3p, -

~

(TV27T

J(X 2_(T2) exp[-(x -p,)2/2(T2_'Y(X-p,)/(T] dx _00

The stated result now follows by putting u = z + l' and then integrating over u by using .the properties of the normal integral. The expected cost after devaluation is a' + {3' p,' - e!"Y'( p,' - 1'17')2

whence, under the given conditions, , a+{3p,-a' p, = {3' ;

,

(a -a')-({3 -{3')p, +'Y{3'(T

17=

'Y{3'

Under the modified system the proportion of underweight packets is ct>[(l-p,')/(T'] whence the stated result.

60

UL (1972). The expected payment per page is (in £)

ix e- IL + 1 x p, e- IL +ix [1-(1 + p,) e- IL ], whence the answer. If £0 is the new payment to the proof-reader for every page on which he

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

401

detects no errors, then 9 x e - JL + 1 x,.., e- JL +lX[l-(l +,..,) e- JL ] = 1

whence

9 =i(S+,..,-e- JL ).

61 UL (1973).

co

dx f (1x'+l +X)H4

E(x') =(,\ +2)('\ +3)

o

=(,\+2)(,\+3)B(r+2,'\-r+2), if '\+2>r whence the result. Using the particular results for E(x) and E(x 2 ) gives the value of E(T).

62 UL (1973). To evaluate E(x - a) use the transformation y = log(x - a) where -00< y
f

E(x - (3) =_1_

u../2;, 10&(/3-0<)

[(a - (3)+eY ]

e-(Y-JL)2 /2q 2

dy

= -«(3 -a)[1-(8)]+eJL+~'[1-(8-u)], by reduction of the two components of the integrand into normal integrals.

63 UL (1977). The probability density function of X is k e-,\x 2x 3, where the proportionality constant k = 2,\ 2 is obtained from the equation co

f

k e- Ax2 x 3 dx = 1. o

The integral is readily evaluated in terms of the gamma function by using the transformation z = x.Ji... The average consumption of gas per day is co

f

co

2,\ 2 e-,\x x 4 dx = ~ 2

fe-"u~ du,

JA. 0

o

where u = '\x 2 ,

whence the result. The required inequality follows by noting that the annual open fire cost is aN and the annual cost with gas fires is

3 /7T (3 + 300-y . 4 Vi .

64 UL (1977).

co

E(T) = N - N(l - a) f x . e-tx2-/3x dx o

=N -

N(1- a) e!/3 2

f /3

(z - (3) e-!z2 dz,

where z = x + (3.

402

EXERCISES IN PROBABILITY AND STATISTICS

Note that

I e-~z2

I

00

z

00

dz =

e-~(32;

e-!z2 dz =.J2";[I-({3)],

(3

(3

whence E(T). For (33 negligible, (3 1 I I I > dt--+-. 1 (3 ({3)=-+- e-,t

2fu o

2fu

Hence E(T) - N - N(I- a) [ 1-

~N -

~ e!(32 . {3 ~; (1- (3 ~;) ]

N(I-a)[ 1-~e~(32

J,

since 0 ~ {3.J2/7T~ 1 and (3.J2/7T(1- (3.J2/7T)- ~~. The stated inequality now follows directly by using the upper limit of E(T) in the inequality T~E(T).

65

UL (1977). If k is the proportionality factor, then

I 00

E(X') = k

X",+,-1

(1 + {3x",)2n dx.

o

The integral is reduced to a beta function of the second kind by using the substitution z = {3x"'. Hence for r = 0, k = (2n -1)a{3. Logarithmic differentiation of the probability density function gives the equation (a -1)(1 + (3x"')-2na{3x'" =0

for the stationary values of X. For the mode the second logarithmic derivative of the probability density function of X is <0.

66 UL (1975).

P(X~a)= B(m, 1 2)

'" IX

Ill -

1

(I-X)dX,

o

whence the result on integration. P(X ~ (3) is obtained by symmetry. Hence P(X~{3 X~a)

P(X~{31 X~a) = '

P(X~a)

67

P(X"':::{3) ~

P(X~a)'

UL (1973). 1

E(X') =! I x'[1 + Ax(l- x 2 )] dx, -1

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

403

I\'hence the result on termwise integration, x

F(x) =

J![1 + AX(1- X2)] dx, -1

whence the stated result. The median is obtained as a root of the equation F(x) = o.

68 UL (1973).

00

E(X2r )=k

x2r+2 k (2r+3 1-2r) J 1+x4dx=2"B -4-'-4-'

if 1-2r>0,

the integral being evaluated by using the substitution u = (1 + X 4)-1, where 0';;; u:;:;;; 1.

Differentiate logarithmically to obtain the modes and the points of inflexion. The modes are at x = ±1 and the points of inflexion are at x::: ±(2 + "'lJ)i; ±(2 -../¥)i.

69 UL (1972).

00

E(xr) = a(3

J(l~;:~)2 dx o

and the integral is evaluated by using the substitution z = (3x"', where O
70 UL (1971). (ii) Use the substitution u = !x 2 , where 0:;:;;; u :;:;;; A. (iii) Note that for fixed r l(r, 6) ~ 2(r-1)/2[[(r+ 1)/2] as 6 ~ 00. Then II

P(Z:;:;;;6>0)=<1>(6)+

~J exp[-!z2(z4-6z 2+3)]dz

nv27T 1 = <1>(6) + r.::- [1(4, 00) - 61(2,00) + 31(0,00)] nv27T 1 + r.::- [1(4,6)-61(2,6)+31(0,6)], nv27T -00

whence the result on reduction. UL (1970). Since the probability density function of X is a function of Ixl and the distribution is symmetrical, all odd central moments of X and E(X) vanish. If k denotes the proportionality factor, then

71

00

E(~r) =

Jx2r(a + Ixl} dx 2k[a J e- x 2r dx + J e-l3xx2r+1 dx J. k

e- 13txl

-~

=

~

~

13x

o

0

whence the stated result on integration using gamma functions. For r = 0, E(X2r ) = 1 gives k = (32/2(a(3 + 1).

404

EXERCISES IN PROBABILITY AND STATISTICS

72

UL (1969). The probability density function of X is an even function of x and . E(X) = O. Hence So 00

iJ.2r(X) = E(X2r ) = k

J

(a + X 4 )X 2r e- x2 dx

00

J

= 2k e- (a + X 4 )X 2r dx, X2

o

where k is the proportionality factor. The integral is evaluated by the use or the substitution x 2 = z. For r = 0, E(X2r ) = 1 gives k = 4/(4a + 3)~. Logarithmic differentiation shows that for a> 1, the mode of X is at the origin. For a =~, var(X) = 1, 1'2(X) = -~ and maximum ordinate is 3/4/;. For a unit normal variable Z, var(Z) = 1, 1'2(Z) = 0 and maximum ordinate is 1/~ which is less than the maximum ordinate of X. But since 1'2(X) < 1'2(Z) , the frequency curve of X is to be "defined" as platykurtic. 73

UL (1968). The probability density function of X is e- x/(1 +e- X )2.

1

=

Jyt(l- y)-t dy o

=B(I+t, I-f),

Itl<1.

E(e tX )

The moments are obtained by expansion of as a power series in /, whence 1'1 and 1'2' (i) P(X:::; x) = ~ and P(X:::; x) = i give the first and third quartiles as -log., 3 and log., 3. Hence the S.I.Q.R. (ii) Logarithmic differentiation leads to the equation e- 2x -4e-x +l=0 for the points of inflexion. 74

UL (1967). If k is the proportionality factor, then

since the distribution is symmetrical about the origin. Hence 00

00

k[J~J~+ l+x2 l+x4

J dX]=1 00

2

x l+x4

.

0 0 0

The second and third integrals cancel out as can be seen by putting x == 1/11 in the third integral. Hence k = 2/7r.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

405

Logarithmic differentiation leads to the stated equation for the turning The equation can be written as (v2+~v+i)(v-~)=k where v=x 2.

~alues.

rhus v -~. The modal values are approximately x = ±~, since the second derivative is <0. The improved approximation is obtained by setting v = 8 + ~ where 8 is small. This gives (8 +~)3_(8 +~)-2 = 0, whence neglecting 1;2 and 83 we have 8 =;k,. Hence the improved approximation for the modal values.

7S UL (1967). 00

J[exp[-2ax/f3+tx]+ (1-a)2 exp[-2(1-a)x/f3+tx] ] dx

2a 2

E(e IX )=13

~

o 00

= 2a 2 J[exP[-Z(2a - f3t)] + o

e

:ayeXP[-z{2(1-a)-f3t}]] dz

where z = x/f3 and 2a - f3t > 0, 2(1- a) - f3t > 0. The stated result follows on integration. To obtain the moments, note that log Mo(t) = log[ 1- 1 ;~~(~:~) f3t]-log[ 1- :~]-log[ 1- 2(:~a)] =f3t+f3

2[1-2a(1-a)] t 2 2a(1-a) 2+""

on expansion,

whence the E(X) and var(X). The lower limit for var(X) is obtained by noting that a(1-a)~~. 76 UL (1966). The moment-generating function of X is E(etX ) = ell" whence the moments. The proportionality constant k for the probability density function of Z is obtained from the equation 00

k

Je-!z,[1+az(z+1)]dz=1,

whence

k

=

1 .Jh(1+a) .

The vth moment of Z about the origin is

J 00

E(ZV)

1 fu(1+a)

= IL~(Z).

e -iZ2 (z 1'+ az 1'+ 1 + az 1'+2) dz.

-00

406

EXERCISES IN PROBABILITY AND STATISTICS

For v = 2r + 1 and v = 2r, the required moments are obtained by using the integral expressions for the moments of the unit normal variable X. In particular, E(Z) = a/(1 +a) and var(Z) = 2-1/(1 +af. 77

UL (1965). 00

J

E(e'X) = A e-(A-')x dx = (1- t/A)-I. o

The cumulants of X are obtained by the expansion of -log(1- t/A). The point-probabilities of Yare c e- Ay where 00

c

L e-A'=1

or c=1-e-A.

• =0

Hence 00

G(6)=(1-e- A)

L (6e-A)' =(1-e-A)/(1-6 e-A) . • =0

Therefore the cumulant-generating function of Y is log(1-e- A) -log(1-e- A e') whence the first two cumulants on expansion. Finally, e A -1-A Kl(X)-Kl(Y) = A(eA -1); K2(X) - KiY) =

(e A - 1f - A2 e A A2(eA _1)2

The stated approximations are now obtained by expansion. 78

UL (1972). P(X = r) = Probability that there is one ace in the first r cards x (r+ l)th card dealt is second ace

The recurrence relation is obtained by considering the ratio P(X = r + 1)/P(X = r). It then follows that P(X = r + 1) > P(X = r) if (49-r)(r+1»r(51-r) or if r<4j. 79 UL (1977). For r>O,

whence the mean and variance of X. The likelihood of the n observations is

" X~ ( 1)" exp (1" -i i~ ) Il

L = 6A 4

Xi

•

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT SO

407

that nx

L 3 log Xj, II

log L = constant - 4n log A- - +

A

j=t

whence the estimate A on differentiation with respect to A. We have E(x) = 4A; var(x) = 4A 2/n = E(x 2) - E2(X). Therefore E(x 2) = 16A 2(1 + 1/4n). Hence an unbiased estimate of A2 is nx 2/4(4n + 1). Also, E(x 2/16) = A2(1 + 1/4n).

80 UL (1973). The likelihood of the sample observations is

( 1)11 .exp--1 L (xj-Of II

L= - SO

that

~2wO

20

j

=t

n 1 log L = constant-llog 0 - 20 [ns 2 + n(x - Of].

Differentiation with respect to 0 leads to the stated equation for 0. The quadratic in 0 has one positive r
81 UL (1972). Use the transformation z = x/O to evaluate E(xr) by a straight application of the gamma function. The logarithm of the sample likelihood is logL=constant-3nlogO+2 whence the stated value of

0.

II

nx

j=t

0

L logxj--

Var(O) = var(X)/9n.

E(L t ) = ~ 4n

r

E(xD = !E(X2)

j=t

and E(L2) = 1E(x2) = 1[var(x) + E 2 (x)], whence the stated results. 82 UL (1971). (i) P(X = n) = Probability that there are exactly r - 1 white balls in the first n - 1 balls sampled x Probability that the nth ball is white =

(ii)

E(OX) =

(nr-1 -l)pr-l(l_ p )II-r X p. i: Oll(nr-1 -l)prqll_r

lI=r

=

=

r

(E.)r (n ~ l)(qO)1I q lI=r r 1

~ (r+s-1) (pO)' s~o s (qO)S = (pO)'(l- qo)-r.

408

EXERCISES IN PROBABILITY AND STATISTICS

(iii) The log likelihood of the sample is log L

=

constant + r log p + (n - r) 10g(1- p)

whence p and var(p) are obtained by a straight application of the method of maximum likelihood.

I

(iv) E(r-1)=pr (n-2)ql1_r n=r r-2 n -1

(r+S-2) qS = pr(l_q)-r+l = p.

~ = pr l.J

S

s=O

83 UL (1969). The point-probabilities of the truncated Poisson distribution are

so that E(nr )

=

NP(X = r)

and

1 <X> E(IA- *) = - L rE(nr ) N r=2

=

IA-, on summation.

A simple argument for the derivation of var(1A- *) is as follows. Let Y 1 , Y 2 , ••• , YN be random variables denoting respectively the number of prosecutions each motorist has. Then Yj are independent random variables, each having the same distribution as that of X. Next, define new random variables Zj such that Zj = Yj if Yj;;.2 and Zj = 0 otherwise. Then N 1 NIA-* = j~ Zj and var(1A-*) = N var(Zj), where

1 <X> rlA- r E(Zj) =-IL- L -, = IAe -1 r=2 r. and

Hence var(1A- *) as stated and var(,l) = var(1A- *)

(elLelL-1)[1_ (_1A-_)2] elL - 1 ~ 1

as IA- ~ 00.

Note that IA- * is more easily calculated than ,1 and for large IA-, IA- * may be preferred to ,1 as a quick estimate of IA-.

84 UL (1966). The log likelihood of the sample observations is 11

log L

=

-n log '7T -

L 10g[1 + (Xj -lA-f], j=1

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

409

whence the equation for (l. is

n 2' the integral being readily evaluated by the use of the transformation x -IL = tan 0, -7T/2 ~ 0 ~ 7T/2.

8S UL (1977). The log likelihood of the sample observations is

n n 1 n log L = --log(27Tk) --log 0 - (~- 0)2 2 2 2kO i =1

L

whence the equation for 6 on differentiation. Since of the quadratic equation gives the value of 6. For the variance, observe that n 2 -2 d 2 logL n d02 202- k0 3 (s +x)

n

=20 2-

6> 0,

the positive root

1" k0 3 i~ x'f.

For the limiting variance put k = ILIO, whence 21L2 1 IL var(O) = - . 2 k 2 -+ - as n IL+ n A

k -+ o.

86 UL (1973). The least-squares estimate of IL is obtained by minimizing

n = L"

(n - v + 1)(x -IL f

with respect to IL,

v=l

whence IL*=

2 " 1) (n-v+1)x". n n + v=I

L

(

Clearly,

4

"

var(1L *) = 2( 1)2 L (n - v + 1)2 var(Xv) n n+ v=1 and 36 " 1)2(2 1f (n - v + 1)4 var(Xv), var(T) = 2( n n+ n+ v=l

L

410

EXERCISES IN PROBABILITY AND STATISTICS

whence

2

2a var(IL*)= n(n+l);

87

9a 2 var(T) = (211 + If .

UL (1972).

1 E(a*) = -

n

E«(3*) =

/I

n v=!

v=l

X)2;

t

1 . (Xv - x)E(yJ (Xv-xf v=!

I v=l

=

l'

/I

L E(yJ = a +- L (Xv -

(3 + l'

JI

(Xv - X)3 /

JI

(Xv - X)2

If Xv are equispaced, then I~=! (Xv-X)3=O. Hence

E«(3*) = (3;

E(a*) f a.

If we set X k == I~=l (Xv - X)k for k = 2 and 3, then

E(a*) = a +I'X2 /n;

E«(3*) = (3 +I'X3 /X2

whence

and

88 UL (1971). Observe that rv

=

Yv -a*- (3*(xv -x)

_ [ 1 (xv - X)2]

- Yv 1---

X

n

-

[1

" (xv - x)(Xj - X)] ~ Yj -+ . j"ov n X

Hence E(rJ = E(yJ - E(a*) - (xv - x)E«(3*) = 0 and

- xf]2 2 () a 2[1 - 1n (xv X- X)2]2 + j"ov,,[1-n + (Xv - x)(Xj a X

var rv

~

=

=

a

2[(n-l)2 n -1 (xv - X)4 - - +--+-'---'------,----'---n

+ (x v;2xf +

ttl

2(x -x) {

~X

n

2(n -1)(Xv -

nX

X2

(Xj -xf-(xv -xf}

j~! (Xj-x)-(Xv-x) /I

=a2 [n:l_ (xv;xf].

}]

xf

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

411

89 UL (1970). From general theory 0* = a* - x(3 *, whence E(e*) = a - x(3

and

-2)

1 var(e*) = u 2 (-+~ , 11

where

X

n

X =

I

(Xi _x)2.

i~l

Hence cov(O*, (3*) = E[(O* - 0)«(3* - (3)] = E[(a* - a)«(3* - (3)]- xE«(3* - (3)2

XU 2 X· When the Xi are the

11

natural numbers, then n

I

x=!(11+1);

X~ = kl1(11 + 1)(211 + 1),

i=l

whence

* * - [ 3(n + 1)

J!

corr (e ,(3 ) - - 2(2n + 1) . 90

UL (1967). E(T- /.Lf = C 2 x E[Jl (Yi - /.L)2+ 2(I1C; 1)/.L Jl (Yi -IL)+ (nC~!)21L 2J =

nC 2u 2+(nC-1)21L2.

Hence the stationary value is C = l/(n + v 2 ).

91 UL (1965). Clearly, E(x) = E(y) = IL; u2 var(x) =-;.

n

var(y) =

u2

-

n

[1 + (n -l)p].

Now T is an unbiased estimate of IL if a + (3

=

1. Also

u2 var(T) = - [(1- (3)2+ (32{1 +(n -l)p}].

n

Hence for minimum var(T) u=

1+(n-1)p . 2+(n-1)p'

(3 =

1

-,------

2+(n -l)p .

Since the Yi are equi-correlated, -[l/(n -1)] os:; p os:; 1, so that (n-1fp2 4{1+(n-1)p}

-,,-'----,--'--'--,- ~ 0

.

Hence for Pt=O. varG(x+y)]>minvar(T). Note that x and y have unequal variances and so the best estimate of IL is T.

412 92

EXERCISES IN PROBABILITY AND STATISTICS

UL (1968).

If x is the unknown (n + 1)th independent observation of X, then x - i . normally distributed with E(x - x) = 0 and var(x - x) = (n + 1)0-2In. Henc~s

irrespective of the magnitude of fJ.

'

x-x

has Student's distribution with n -1 d.f. whence the stated confidence interval for x.

96 UL (1969). Under normality assumptions of the y observations, it is clear that Yk is normally distributed with mean a + /3(xo - x) and var(Yk) = 0-2/k. Therefore Yk - & -/3(xo - x) is normally distributed with mean zero and variance 1 1 (Xo-X)2] . 0- 2 [ k+-;;+ X . Hence the ratIo A

Yk -& - t3(xo-x) 1 1 (xo- X)2]i s [ -+-+..:......:'----~ k n X has Student's distribution with n - 2 d.f. This now leads to the stated confidence interval for Yk.

94 UL (1971). Here x - L is normally distributed with mean zero and variance 0-2[(1In)+1-2A(1-A)]. Hence the ratio

x-L

S[~.+ 1-2A(1- A)] has Student's distribution with n -1 d.f. This leads to the stated confidence interval for L. The lower limit is obtained by noting that 0,..;;; A(1- A) ~!.

9S UL (1975). Observe that x - Xo is normally distributed with mean zero and variance 20-2/n. Hence the ratio x-Xo

s../2frt has Student's distribution with n -1 d.f. This leads to the stated confidence interval.

96 UL (1966). The rth factorial moment of w is E[w(r)]= n(r)pr, for

r~O.

Therefore fJ.3( w) = E[( w - np )3], whence the result on taking expectations after expansion. Since fJ.3(w)=n(p-3p2+2 p3), an unbiased estimate of this is _ [W 3W(2) W(3)] T- --~+2---cJ) , n

n

n

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

413

which gives the answer on reduction. Observe that E(p*) = p; var(p*) = p(1- p )/n, and for large samples dT)2 var(T) = ( d * p

x var(p*)

p*=p

whence the stated result. 1)7

UL (1968). The log-likelihood of the sample observations is

log L = constant+ nl log(2 + 6) + n210g(2- 6) + n3 log(1- 6) + n410g(1 + 36) whence, on differentiation, the equation for 6 is n3 + 3n4 =0. 1-6 1+36 Also, var(6) is readily obtained from d 2 10g L/d6 2. With misclassification, the log-likelihood is

nl 2+6 A

n2

_

A

_

A

A

2-6

log L = constant + nllog(2+ 6) + (n2 + n3) log(3 -26) + n410g(1 + 36) whence the equation for 6* is nl 2(n2+ n 3) + 3n4 =0 2+6* 3-26* 1+36* . For large samples var( 6*) > var( 6) if 4(3 - 26)(5 + 26 - 4( 2) > (2 - 36+ ( 2)(29 + 36) which reduces to (6+2)(36+1»0.

98 UL (1969). The log-likelihood of the observed sample is log L = constant+ al log(2- 6) + a210g(1 + 6) + a310g 6 + a410g(1- 8) whence, on differentiation, the equation for 6 is _~+~+ a3_~_0

2-6 1+6

E(X) = N(1-26)

8

1-8- .

and var(6*) =

1 + 26(1- 6) 2N '

obtained by using the properties of the multinomial distribution. The limits for var(8) follow since 0~8(1-8)~~.

99 UL (1970). Since I~=l mi = N, a constant, therefore n=l dmJd6 = O. The probability of obtaining the observed sample is L

=

kN!

IT

i=l

.

Ii (mi)

a,

.1 i=l N a,.

and the equation for 6 is obtained by logarithmic differentiation.

414

EXERCISES IN PROBABILITY AND STATISTICS

Note that E(X) = 0, var(ai) = mi(1- m;/N) and cov(ai> a j ) == -m.rn/N

i =1= j. Hence var(X) is obtained by using the fact that X is a linear functio~ of

the ai.

100 UL (1966). (i) Clearly, var(x) = :2

[tl var(Xj) + 2 ~t: cov(Xj, Xj+l) J,

whence the stated result. (ii) E(S2) =

~ E[ f. 1

n

i=1

1 [ =---:-E n

=

n

1

(Xj -

X)2]

L x7-nx 2 11

]

i=1

~ 1 [.f. E(x7) 1=1

nE(x 2)]

[.f.

~ 1 var(Xj) - n var(x)], since E(Xi) = E(x) 1=1 n whence the result. =

(iii) This follows by noting that var(x);;:. 0 for all nand E(S2);;:. 0 for all n ;;:. 2. Note that var(xw )

=

[f.

2( 4 )2 i 2 var(Xj) + 2 L ij cov(Xj, ~)] n n+1 i=1 i
=

2(

n n+1

[

)2

11

11-1

L f+2p i=1 L i(i+1) i=1

]

,

whence the result on summation. Eff(iw) = var(x)/var(xw ) 3(n+1) 1+2p(n-1)/n 3 ~ as n~oo. 4n ·1+2p+(1-4p)/2n 4 101 UL (1967). (i) Since var(X - Y) = var(X)+var(Y)-2 cov(X, Y);;:'O, therefore cov(X, Y) .::;;~[ var(X) +var(Y)], whence the stated result. (ii) Suppose that E(X) = 6 1 and E(Y) = 62. By definition, cov(f , X) = E(Y) - E(f)E(X) ( Y - ( 2) + 62 ] 62- 6 1 • E [ (X - ( 1)+ 61 X - 6 1 Y - 62 (X - ( 1)(Y - ( 2 )] - 62-62 . E [1 --6-1-+~6 162 ' whence the result on taking expectations. =

102 UL (1968). Assume that for two random variables X and Y, var(X) = O"f;

var(Y) = O"~;

cov(X, Y) =

pO"10"2.

415

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

Then var(X ± Y) = (0"1-0"2f+2(1 ±p)0"10"2;;:'0 for all 0"1 and 0"2' Hence for 0',==0"2,1±p;;:'0.

1 [" (i) var(i) = n2 i~1 var(Xi)+ i~ COV(Xi, Xj) ] , whence the result on substitution. (ii) Here Yi = Xi - i = (1-1.)Xi -1.

n

L Xj'

n j""i

Therefore var(Yi)=

(1-~r var(x;)+(~)\n-1)Var(Xj)

whence the result on substitution. (iii) Put Zi = Xi - /.L and

z =1. i n

i~1

Zi' Then

Expand and take expectation over the z's whence the answer.

103 UL (1969). Here S" = Ill=2 Zi where P(Z, = 1) =i;

P(Z, =0) =~ for 2~r~ n.

Therefore E(Zr) = t cov(Z" Z,+l) =

--16;

var(Z,) = /i;; cov(Z;, Zj) = 0

for

Ii - il ;;:,2.

Hence E(S,,) and var(S,,).

104 UL (1970). (i) If E(X) = 81 and E( Y) = 82, then cov(X + Y, X - Y) =E[{(X - 8,)+(Y -( 2 )}{(X -( 1)-(Y -82 = E[(X - 8 1f- (Y - ( 2)2] = var(X) - yare Y), whence the result. (ii) By definition,

m

whence the result on summation of the double series.

416

EXERCISES IN PROBABILITY AND STATISTICS

lOS UL (1971). (i) Let E(X;) = 8i, i = 1, 2, 3. Then, by definition, cov(X1+ X 2 , X 2 + X 3 ) = E[{(X1- 81) + (X2 - 82 )}{(X2 - 82 ) + (X3 - 83 )}] = E[(X2 - 82)2], since the X; are uncorrelated =var(X2 ). (ii) Here E(Si) = 0, i = 1, 2, 3. Hence cov(Sl> S2) = =

r

"

r

"

L L E(x.Xt) .=11=r+1 L L

.=11=r+1

L L r

=u 2

cov(x., Xt) "

pl-.,

.=11=r+1 whence the result on summation. Cov(Sl> S3) is obtained in a similar manner.

106 UL (1973). Since E(Yi) = 0, we have 1 i+,,-1 i+r+n-1 cov(Yi> Yi+r) =2 L L E(x.Xt).

n

s=i

t=i+r

If r;;:;' n, then E(x.Xt) = 0 for all permissible values of sand t (s 1= t). Therefore cov(Yi> Yi+r) = O. If r:OO;;;n-1, let n-r-1=k, where k;;:;.O. Then 1 j=n-1 cov(Yi> Yi+r) = 2 E(xr+j), the cross-product terms being zero n j=n-k-1

L

107 UL (1966). For the Poisson distribution P(X = x) = e-"',A,x/X! whence QO

E(e'X ) = e-'"

L (IL e1nx! = exp lL(e' -1). x=o

It then follows that

Hence E(e 'Z ) = e-',/i;; . E(e'x Jn7j;.) = e-',/i;; . exp[nlL(e'/,/i;; -1)]. Logarithmic expansion gives the result for K(t).

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

108

417

UL (1967). 00

L (A8Y/r! = e

E(8 X ) = e- A

A (9-t).

r=O

Therefore G(8l> ( 2) == E(8J' . 8n = E[(8l82)x']E(8~2) = exp[A l (8 l 82 -1) + A2(82 -1)].

The probability-generating function of Y is G(1, ( 2 ) whence P(Y = j). The probability-generating function of the conditional distribution of Xl given Y= j is G(8 l I Y = j) = coefficient of 8~ in G(8 h = (1 + p8 l )i/(1 + p )i.

(

2)/P(Y = j).

On expansion the coefficient of 81 gives P(XI = r I Y = j). 109 UL (1967). E(8 X )

" (n). (p8)iq"-i = (q + p8)".

=L

i=O

]

Hence G (81) ( 2) = E[(0 1 (2)X,] . E[ 0~2] =

(ql + PI 81( 2)"'(q2 + P2 ( 2)"2.

The probability-generating function of Y is G(l, ( 2 ) whence P(Y = r), and the probability-generating function of Xl given Y = r is G(OI I Y = r) =

[arG(O~, ( 2)] a0 2

jP(Y = r), 9 2 =0

whence, on expansion, P(X1 = slY = r). 110 UL (1967). The probability-generating function of X is 00

G(O) =

L Pror, r=O

and the factorial moment-generating function of X about the origin is Ho(a) == G(1 + a) r=O

f t r(s) a;s. = f a; f s.

=

Pr

r=O

s=O

s=O

r=s

Prr(s)

418

EXERCISES IN PROBABILITY AND STATISTICS

Again, putting () = 1 + a, we have co

co

«()-1)S

r=O

0=0

s.

L Pr()r = L - ,=

f.L(s)

t

f

f.L~0)

co

S. ()k

(S)()k(_l)S-k k=O k co (-1y-k

co

()k

co

0=0 =

k~O k! S~k f.L(s) (s-k)!

=

L -k' L -.,- f.L(k+j) k=O 'j=O J.

(-1)j

Equate coefficients of ()' on both sides to get Pro The particular result follows on summation of the series.

111 UL (1966). Here E(ri) = P1;P; var(ri) = P1;pq; cov(rj, rj ) = 0 if j. It then follows that E(x)=E(y)=p and var(x) and var(y) are also readily obtained. Note that E(e'X ) = (q + P el/N)N whence E(e 'W ) = e-P1/
n (q + P e l/",k)", i=1 k

E(e 'Y ) = and

E(e 'Z ) = e-P1/
n k

(1 + P el/n,kuo)".

i=1

Logarithmic expansion of E(e"Y) and E(e 'Z ) now leads to asymptotic normality. The approximation for var(x)/var(y) is obtained by noting that

t ! t ~ [1 _(ni ~ ii) + (ni ~ ii)2 _... ] -~ [1 + V~~i)J. =

i=1 P1;

i=1 n

n

n

112 UL (1966). The sample of n unrestricted values can be selected in (~) ways. Next, if

x and yare the largest and smallest integers in the sample, then the choice of the remaining n - 2 is restricted to x - y -1 numbers in all. Hence

where, evidently, n~X~N and for given X, ¥;;;;.1 and X-¥-1;;;;'n-2, whence the stated conditional limits for ¥. Clearly, for fixed X, n -1 ~ Z ~ X -1 so that the joint distribution of X and Z is

P(X=x,Z=z)=

(:=~)/(~

where

n~X~N.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

419

Hence the marginal distribution of X is

P(X=x)=

L

x-I

(

z=,.-1

z- 1)j(~ n-2 n

where

xf'

Xf z=,.-1

(Z-1)= (U+n-2) n -2 ,,=0 U x-n

=

L

coefficient of 0" in (1 + 0),,+,.-2

,,=0

=

{1 O}" 0 {1 + O}" in (1 + 0),.-2 L - -

x-,. L coefficient of 00

,,=0 =

in (1 + 0),,-2 -+X-,.

coefficient of 00

,,=0

0

= coefficient of 00 in (1 + O)X-l • O-(x-,.l

=\x-n (X-1)= (X-1). n-1 Again, for fixed Z = z, we have z + 1 :s;;; X :s;;; N, whence the marginal distribution of Z is P(Z = z) =

f (Zn =21)j(N\n)

x=z+1

=(N-Z)(:=~)j(~, with the stated limits for Z.

113 UL (1966). The probability that a prosecuted motorist has x prosecutions is e-.... '.c.x/(1-e-.... ). x!,

1:s;;;x
It now follows that the rth factorial moment of x is 00

e

1-e ....

(rl

r

L~

-....

E[x(rl ] =

x=1

x!

f.Lr

-1-e-..... Hence E(x) = _1£_ .

1-e-.... '

Therefore var(x) -1 1£ e-.... E(x) - -1-e-.... ' which is P(x ;;;:.2) for the truncated distribution.

420

EXERCISES IN PROBABILITY AND STATISTICS

114 UL (1966). Consider first the joint distribution of Xl> x,,+l and X Zv + 1 and then the conditional probability of the 2r observations. It then follows that p

=

(2v+1)! . (2r) {(v-1)!F r

Xv + 1

J J J[ Jf(x)dx ]v+r-t 00

-00

00

Xl

00

Xl

X v +l

X,,+l

To evaluate the triple integral make the transformation Xl

U

=

X"+l

Jf(x) dx;

J

v=

X2v+l

f(x)dx;

W=

f

f(x) dx

where O
115 UL (1966). By definition, «I>(x) =

k 1 2

X

f e-!u 1

f

J2;

0

2

du

_1

2

=-+-- L.e "u du z)' f L'" (-!u --du x

1 1 =-+2 J2;

r!

r=O

o

whence the result on integration after changing the order of summation and integration

f J2; '"

E[etIX1 ]=_1_

=

(2)! f'" 1T

etlxl-!x2

t

1

eX-"X

2

dx

dx

o =

(!)! j

e!t 2 e -!<X-t)2 dx o

=

2 e!t 2 «1>(t)

Logarithmic expansion gives the cumulants.

116 UL (1966). Set

Vt

= Xl/a and

Vz = xz/a.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

421

The probability density function of the joint distribution of VI and V2 is n 2(n -1f(VIV2)n-2(1_v l )(1- V2), O::s;;; VI> V2::S;;; l. To evaluate the distribution of u, put VI = UV2, V2 = V2 and consider the cases O
117 UL (1966). If k is the proportionality factor, we have

ff I

E(xrys) = k

00

e- y (l-x)x ry",+s(1- x)"'+~ dx dy

o

0

I

=

00

k f xr(1- X)~-'-l dx f e- y (l-x){(1- x)y}",+s(1- x) dy o

=

0

kB(r+ 1, (3 -s)f(a +s+ 1).

For r = s = 0, E(xrys) = 1 whence k = (3tr(a + 1). The probability density function of the distribution of X is 00

{3

f{a+1)

f

(1- X)~-l e - Y(I-X){y(1- x)}"'(1- x) dy = (3(1- X)~-l

,

for O::s;;;X::s;;;1.

o

The probability density function of the distribution of Y is

f(:+

1

1) . e-Yy'" f e YX (1o

x)"'+~ dx.

This may be put in the required form by use of the substitution y(1- x) = t.

118 UL (1967). Clearly,

Hence by substitution in the probability density function of X, the stated distribution of Y is obtained. E(Y) = 24 a4

f'"x 2(a-x)(a 2-x 2)idx. o

The integral is evaluated simply by using the substitution x = a sin O.

'"

P(Y::S;;;a) =~ f[2a(4a2_y2)i-(4a2_y2)]y dy. 4a o

Evaluate the integral by using the substitution y = 2a sin O.

422

EXERCISES IN PROBABILITY AND STATISTICS

119 UL (1967). co

E(e iIX ) =

J

J:...

eilX-lxl/a

20'

[J

dx

J

00

=

~

00

e-(1+ila)"

o

E(e i1z )

du +

e-(1-ila)"

du ]

0

1 1 + t 20'2' = E[eil(1-.... )y']E[eit....y2] 1 1 1 + t 2(1_1J.)2 . 1 + t 21J. 2 1 [(1-1J.)2 =1- 2 1J. 1+t 2(1-1J.)2

IJ. 2

]

1+t21J.2·

The sampling distribution of z now follows directly by the use of the Inversion Theorem.

120 UL (1967). (i) From the definition we have cOV(Xj_k, Xj+2) = a cov(Xj+l> Xj-k) + (3 cov(Xj, Xj-k) so that Pk+2 = apk+I + (3Pk' (ij) Similarly, var(Xj+2) = a 2 var(Xj+l) + (32 var(Xj) + 2a(3 cov(Xj, Xj+I) +var(ej+2) so that var(e) = 0'2(1- a 2- (32)-2a(30'2pl> where from (i) for k = -1, PI = apo + (3P-l> Po = 1, PI = P-I' Hence PI = a/(1- (3) and var(e) follows. (iii) The general solution of the difference equation in (i) is Pk = AA~ + BAt

with Po= 1,

PI = a/(1-a)

Use of the initial conditions now gives

A

=

-A2+ a /(1-(3). Al - A2 '

whence Pk'

121 UL (1967). Note that 1 1 z v=-tan- 21T W It then follows that a(u, v)

--=

a(w, z)

1 [I( 2 2)] --exp -2" W +z 21T

423

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

whence Wand Z are independent N(O, 1) variables. Therefore W+Z2= - 2 loge U is distributed as x2 with 2 d.f. If Ul> U2, ... ,Uk+2 are independent realizations of U, then n=1-2Ioge Uj is distributed as x 2 with 2k d.f. Also (-2 loge Uk+l)~cos(21TUk+2) is an independent N(O,1) variable. Hence w2+ L~= I - 2 loge u; is distributed as X2 with 2k + 1 d.f. 122 UL (1967). The probability density function of the distribution of XI and x,. is

4n(n-l) 2 211-2 ,,::: ""'">211 (X I1 -Xl) XIX,,, O-"Xl-.,a;x,,"'-XI· a Hence the probability density function of the distribution of

n(n-1)

----'-.........,,--'- (UV)"-2(U 2 - v 2) 2a 2n "

0 ~ V ~ a' v ~ U ~ 2a - v

For the marginal distribution of v, integrate over v~u<2a

U

and v is

U

•

with the limits

-v.

The marginal distribution of u is obtained in two parts. For 0 ~ u ~ a, integrate over v in the range 0 ~ v ~ u, and for a ~ u ~ 2a integrate over v in the range 0 ~ v ~ 2a - u. 123

UL (1967). y

co

P(X~Y)= J Jf(x,Y)dXdY=~ al+~l o

0

so that the normalized joint distribution of X and Y subject to the condition that X ~ Y has the probability density function ~2(al

+ ~l) exp[ -~2Y -(al + ~I -

~2)X],

O~X ~ Y; O~ Y <00.

(i) The marginal distribution of X has the probability density function co

~ial + ~l) exp[ -(al + ~I - ~2)X] J e-/32

Y

dy = (al + ~l) e-
x

O~X
Oi) The required density function is ~2(al + ~l) exp[ -~2Y - (al + ~l - ~2)x]/(al + ~l) e-
The conditional probability is P(x

~ Y ~ 2x) =

2x

f ~2

e-/3,tY-x) dy = 1-e-132x.

x

UL (1968). Consider first the joint distribution of Xl> Xv+l and X2v+l and then .t~e conditional probability of the 4r observations. Then the required probabIlity

124

424

EXERCISES IN PROBABILITY AND STATISTICS

is (2v+1)! (4r)! p = {(v-1)!Y(r!)4

Joo Joo Joo [JXI

f(x) dx

-00

XI

X ... +l

X 2 ... + 1

X [

-00

]r[ XJ"+I

]r+V-l f(x) dx

XI

00

J f(x) dxr+V-l[ Jf(x) dx Jf(X1) dxd(x +l) dX +d(X v

v

2v

+1) dx 2v +1•

To evaluate the triple integral make the transformation

XI

u=

Jf(x)dx;

X 2 ... + 1

X ... +l

v=

J f(x)dx;

Jf(x)dx,

w=

O
Integrate successively over wand v by using the substitutions 1- w = (1- v)t and 1- v = (1- u)z. The integrand in u is easily evaluated as a beta function, whence the stated result.

125 UL (1968). The probability density function of the joint distribution of X and nY2/2 is _1_ e-x2/2a2x _1_ e-ny2/2(ny2/2)n/2-I, -oo<x <00;

ufu

0:0;:;

y <00

r(n/2)

Make the transformation x/u=rcosO; y.vn=rsinO, where O:O;:;r
1 (. 0),,-1 B(nI2, !)' sm . Clearly z n d.£.

=.vn cot 0 whence it follows that Z has Students' distribution with 00

EX,Y[cP(X+CY)]=.l;

00

x+cy

J J[ J -00

0

e- U2/2 dU]t(x,y)dXdY

-00

where f(x, y) is the joint density function of X and Y =Ex.y[pu{U:O;:;x+cy I X=x, Y= y}] = px.y,u(U:o;:;X+cY)

=Px.y.u[y~:O;:;~J. But since (U - X)/(l + ( 2 )! is N(O, 1) and independently distributed of Y, it follows that Z and (U-X)/(1+u 2 )! have the same distribution.

126 UL (1968). The cumulant-generating function of X is 00

K(t) = f.L(e' -1) = f.L

L trlr! r=1

425

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

so that

Kr

= /.t for

r;;'

1. Hence /.tiX) =

K4

+ 3K~ = /.t + 3/.t 2 •

Next,

n

E(e'X) = " E[e'x/,,] i=1

= [exp l1-(e'/" -1)]" = exp n/.t(e'/" -1). Therefore the cumulant-generating function of x is n/.t(e'/" -1), whence E(x) = /.t,

_

var(x)=~,

/.t

3/.t 2

n

n

/.tiX)="3+-2 .

n

E(T) can be readily verified and to obtain var(T), it can first be shown that n-3)]2 911-2 18/.t [T- E(T)]2 = 9(x -11-)4+ [ 6/.t + ( -n(x - /.t)2+ 7 -----;- (x - /.t)2

+ 6 [ 611- +

(n : 3) ] (x - /.t? - 6; [ 6 /.t + (n : 3) ] (x - 11-),

whence the result on taking expectations.

127 UL (1968). Observe that the region of integration In the (u, v) plane has the boundaries u ± v = 0 and U ± v = 2a. Hence (i) for fixed v in 0 < v < a, v ~ U ~ 2a - v ; (ii) for fixed v in -a < v < 0, -v ~ U ~ 2a + v. It then follows on suitable integration that the probability density function of v is a-v -2- for a

O~v
a+v -2- for a

-a~v::;;;O.

P(v;;./3) is obtained directly by integration and so also E(lv \) and E(lvI 2) = E(v 2 ).

128 UL (1968). (i) Clearly, Ur-l - Ur is the probability that X is realized at the nth trial. Therefore the required expected number of trials is 00

L r(Ur-1 r=1

00

ur )

=L

,=0

Ut·

(ii) Here Y,,-1 - y" = probability that the pattern is realized for the first time at the nth trial

= y,,_ 4 X qp 3 whence the difference equation for y", n;;' 4, with the initial conditions YO=Yl=Y2=Y3=1. The generating function for this difference equation is

and y" is the coefficient of zit in the expansion of
426

EXERCISES IN PROBABILITY AND STATISTICS

The expected number of trials required for the realization of X is 00

L

Yr = ep(1) = l/qp 3.

r=O

129 UL (1968). Here Xl> X2, sf and s~ are all independently distributed and F=sVs~ has the F distribution with n -1, n -1 d.f. Now Xl + FX 2 w= 1+F

and

E(wIF)=p.,

whence E( w) = p.. Also, 1 var(w IF) = (1 + F)2 var[(xt + FX2) I F]

u 2 1+F2 =-;;-. (1 + F)2· Hence

where

J 00

{ F} EF (1+F)2 =

1 p"-1)/2 dF (n-1 n-1) (1+F)"+1 B -2-'-2- 0

n-1 4n . 130 UL (1968). The distribution of W is obtained directly by using the transformation suggested. It then follows that w = t/.Jii, where t has Student's distribution with n d.f. Also, W

=!(XI/X2- X2/XI) =~

(.JF - Jp).

where F has the F distribution with n, n d.f. Hence

whence

np2-(4t2+2n)F+n =0. The positive root of this quadratic gives the required answer since for F> 1, t>O.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

427

131 UL (1968).

f 00

(i) E(e itX ) =!

eitx-Ixl dx

00

=

fe

-x

cos tx dx

o = 1- t 2 E(e itX ).

(ii) U j and U 2 are independent N(O,!) variables. (iii) We can write

W = a[(Zl + Z2)2 - (Zl - Z2f + (Z3 + z4f - (Z3 - Z4f] where !(Zl ± Z2) and !(Z3 ± Z4) are independent N(0,1) variables. Hence W = !(xi - x~) where xi and x~ are independent X2 variables each with 2 d.f. Finally,

f fexp[itw -!xi-!x~] dxi dx~

E(e itW ) =!

o =

0

ff

exp[-(1- it)vj - (1- it)V2] dVl dV2

o

0

132 UL (1968). P(X = r) = Probability that the first r cards include any two aces and r - 2 non-aces x Probability that the (r + 1)th card is an ace

(~)(,~82)

2

(5 2) r

. 52-r'

for 2,,;;; X,,;;; 50.

P(¥ = s ! X = r) = Probability that s non-aces turn up after the third ace x Probability that the (r + s + 2)th card is the fourth ace

Hence

P(X=r, ¥=s)=P(X=r)P(¥=s!X=r). 12r(r-1) 52(4)

EXERCISES IN PROBABILITY AND STATISTICS

428 Therefore

50-.

P(¥=S)=

r~

12r(r-l) 52(4)

--'-------;-:-;--- • f or 0 :0;;;; ¥

4(51-s)(3) 52(4)

:0;;;;

48 •

133 UL (1968). I

E(Z) =

00

Jxf(x) dx + Jtf(x) dx = PILI +qt, I I

E(Z2)=

00

Jx 2f(x)dx+ Jt2f(X)dX=P(0'~+IL~)+qt2, I

whence var(Z).

134 UL (1968). Here var(i) = \

n

[ti=1 O'~ + i""jL PO'iO'j]

1 [ (l-p) L " O'~+P (" =2 L O'i n i=1 i=1 The inequality now follows since var(i);;;: O. Set

~

- IL = Zi and let

)2] .

z be the mean of the Zi' Then

cov(i, Yi) = E[Z(Zi - z)]

1" =E [ Zi'L Zj ] -E(Z2) n i=1 = AO'~ + BO'i - C, where I-p

A=->O, n

P

n

B=- L O'j' n j=1

C=var(i»O.

Hence if 0'1 = 0'2 = ... = 0'", then cov(i, Yi) = O. Conversely, if cov(i, Yi)=O, then AO'~+BO'i-C=O. Since A>O and C>O, there is one positive root O'i = 0' say. But A, B and C are independent of i, so that the conditions cov(i, Yi) = 0 give the equations AO'~+ BO'i - C = 0

for i = 1, 2, ... , n,

which have the solution 0'1 = 0'2 = ... = 0'".

135 UL (1969). Let G(Oh O2 , ( 3 ) denote the joint probability-generating function of Xl> X 2 and X 3 • (i) P(X2 = 1) = coefficient of O2 in G(I, O2 , 1). (ii) P(X1 = 1, X 2 = 1) = coefficient of 0 1 02 in G(Ol, O2 • 1), whence

P(X1 = 11 X 2 = 1) = P(X1 = 1, X 2 = 1)/P(X2 = 1)

42~

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

(iii) P(X3 = 0) = G(1, 1,0). The probability-generating E(Z) = [G'(6, 6, 6)]9=1'

function

of

Z

is

G(6, 6, 6)

an«

136 UL (1969). E(e tX ) = e lLt +!o-2,

and

E(e ut ,+vt 2 ) = E[exp[(t1 + t2)x1 + (t 2- t1)X2]]

= eo-2t~ . exp[2/Lt2+ u2t~], whence the result by the Inversion Theorem.

J 1 J =u-/; e00

E(lui)=_12u-/;

lule- u2/ 4

O-

2

du

-00

00

u

u 2/4cr2

2u du = -/;.

o

If Xl> X 2 , X3 are three independent observations of X, then

IX2 - XII + IX3- X 2 1+ IX3- XII = 2R, where for i l' j E(IXi

-

Xj D= 2u/.Jrr.

137 UL (1969). Let X, Y and Z be the events that A, Band C respectively have success in a game. Then for A to win, the eight possible outcomes of the fin game may be classified as follows. E I : XYZ, XYZ-game reverts to initial state P(E I )

=

p3+ q3= 1-3pq

E 2 : XYZ, XYZ, XYZ-A loses P(E2)

=

pq(1 +q)

E3: XYZ-A wins; P(E3) = pq2 E 4 : XYZ-game undecided between A and C; P(E4 ) = p2q Es: XYZ-game undecided between A and B; P(Es) = p2q. Next, if S.. denotes the event that A wins the series on the nth game, the for n~2 5

P(S..) =

L P(S.. I Ei )P(E;) i=1

where, by definition, P(S.. I E 1) = Un-I;

P(S.. I E 2 ) = P(S.. I E 3 ) = 0;

P(S.. I E 4 ) = P(S.. I Es) = (1- 2pq) .. -2pq. Hence the stated difference equation for 00

L n=2

Z.. -1u., = (1-3pq)

u.,.

It then follows that

00

L n=2

00

Z ..

-IUn_1 +2 p3q 2

L (1-2pq) .. n=2

-2

z n-1

430

EXERCISES IN PROBABILITY AND STATISTICS

or

1 2 p 3q 2 z 2 -
138 UL (1969). When A has £n another trial of E must be made. If this trial results in the occurrence of E, then the expected number of further trials needed for A's win is u,.+l, whereas if this trial results in the non-occurrence of E, then the expected number of further trials needed for A's win is Un-l. Hence Un = 1 + Pu,.+l + qU,.-l,

where Uo = 0 and Ua+b = O.

The difference equation is reduced to the homogeneous form Vn = pVn+l + qVn-l by the substitution u,. = Vn +an where a = 1!(q - p). Hence Un =_n_+ kl(q!p)n + k2 q-p where, by using the boundary conditions the constants (a+b)pa+b k l =-k2=-( q-p )( q a+b -p a+b)' (P=l=q)· The expected duration of play for A to win is Ua • When p = q =!, the difference equation for u,. becomes 2u,. = 2+ U,.+l + Un-l· This is reduced to the homogeneous form 2wn = Wn+l + Wn-l by the substitution u,. = Wn + (3n2, where (3 = -1. Hence the solution is now un =(a+b)n-n 2 and ua=ab.

139 UL (1969). Clearly P(Zj

=

n

0) = 1-N .

Hence E(z.) =!!:... var(z.) = n(N - n) . I N' I N2' 1 N _ (i) E(x) =- L XjE(zj) = X nj=l (ii) var(x) =

:2 [tl XT

var(Zj)+

j~j Xj~ cov(Zi> Zj) J.

which reduces to the given answer.

140 UL (1969). By using the multinomial distribution, the probability density function of

431

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

the joint distribution of Xr and Xs is N! (r-1)! (s-r-1)! (N-s)! I! I! [F(xr)]r-l[F(xs)-F(xr)]S-r-l[I-F(xs )]N-S dF(x ) dF(x )

s x -r- - dXr

dxs

For the uniform distribution F(x) = x/a, whence the distribution of u and v. For fixed u, u:s;; v:s;; 1, so that the probability density function of the marginal distribution of u is 1

N! ------:------:----:--,(r-l)! (s - r-l)! (N -s)!

J r-l( U

V - U

)s-r-l(1 - v ) d v = --:-----'-----...:...-u r- 1(1- u)N-r B(r, N - r+ 1)

" by using the substitution v - u = z(l- u). The conditional distribution of v given u has the probability density function B(N - s + 1, s - r)(I- u)N-r , P(v > vo> u) is evaluated by using the substitution v - u = z(l- u) for

integration over v.

141 UL (1969). x

00

P(X~ Y) = JJnl n2 e-(n,x+n 2 )dy dx Y

o

0

n2 nl +n2

.

.

- - - on mtegratIOn. Hence for X ~ Y, the probability density function of the conditional joint distribution of X and Y is nl(nl + n2) exp- (nix + n2Y) = nl(nl + n2) exp-[nl(x - y)+ (nl + n2)Y], X~

Y;

O:s;; Y <00.

lf we now set u = x - Y and v = y, then it readily follows that U and V are

independently distributed. Hence, for U ~ 0, the conditional distribution of U has the density function 00

J

nl(nl + n2) e- n," e-(n,+n2)u dv = nl e-n,u. o

Again, by symmetry the conditional distribution of Y - X given Y - X ~ 0 has the density function n2 e- n2". Also, P(X ~ Y) = n2/(nl + n2) and P(X:s;; y) = nl/(nl + n2)' Hence the unconditional distribution of IUI has the density function

432

EXERCISES IN PROBABILITY AND STATISTICS

142 UL (1969). Let Xi be a random variable such that Xi = 1 if the ith quadrat is empty = 0 if the ith quadrat is not empty. Then

and 11

X=

L X;.

i=1 It is now easily shown from first principles that

cov(x;, Xj) =

(1-~)" _(1-~) 2.

Hence the stated expressions for E(X) and var(X). Under the given limiting conditions E(X)- n e->';

var(X) - n e->'(1-e:->').

143 UL (1969). Observe that whether X +1 >0 or I1

X,,+1

<0,

1 ( 1-7 X~+I) .

P(X~x..+l)P(X.;;;x..+l)=4

Hence the probability density function of X,.+1 is (2n + 1)! 1 ( X~+I)" n!n!1!(2a)·2211 1-7 '

-a';;;x..+l';;;a.

Therefore

J 1

r = (2n + 1)! a (n !f2211 + 1

o

t(r-1)12(1- t)" dt where X~+1 = t ' a2

whence the stated result.

144 UL (1969). 1 [" II (i) var(x) = n2 i~l var(x;)+ i~ j~i cov(X;, Xj) ] 1

["-1

n-2

]

= n2 nu 2+2 i~ cov(X;, X;+1)+2 i~ cov(X;, Xi+2) ,

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

433

whence the result on reduction. (ii)

E[~ (Xj-X)2]=E[~ XT-nx 2] =

n var(Xj) - n var(x), since E(Xj) = E(x) = 0,

whence the result on reduction. (iii) Since var(x) ~ 0, it follows that p ~ -n/2(2n - 3) for n ~ 2. Hence as n~oo, p~-l.

145 UL (1969). The Jacobian of the x ~ u transformation is n! Also, observe that n

UI = n(xl-m

and

n

L Uj = L (Xi -Xl)'

i=2 i=2 It then follows that the Uj are independent and identically distributed random variables such that for any j"2u/a has the X2 distribution with 2 d.f. The ratio n(n -l)(XI - ml'f.r=2 (Xj - Xl) has the F distribution with 2, 2(n-1) d.f.

146 UL (1966). Clearly, both 20u and 2v/0 are independently distributed as X2'S each with 2n d.f., whence the joint distribution of u and v. Next, the Jacobian of the (u, v) ~ (w, z) transformation is 2w/z. Hence the probability density function of the joint distribution is 2 {f(nWexp[-(Ow/z+wz/0)].w2n-l.z-t, O:os;w
E(z)=

2

B(n, n)

f(Oz-l+zO-I)-2n dz, o

and the integral is readily evaluated 'I'J = 0 2 (Z2+ ( 2), where O:OS;'I'J:OS; 1.

by

using the substitution

147 UL (1966). The joint probability density function of the distribution of Xl and Xn is

,

1! 1! ~~-2)!

[fxf(x) dx ]n-2f(XI)f(Xn),

-00< Xl <00; XI:OS;Xn <00.

Also, given Xl and x'" the probability that the r observations of the second sample all lie outside the range (xt> Xn) is

Hence the integral expression for P. To evaluate the double integral use the

434

EXERCISES IN PROBABILITY AND STATISTICS

transformation

",

J

u=

v=

f(x) dx,

r

f(x) dx.

Then 1

1

JJ

P = n(n -1) du (v - u)"-2[1- (v - u)]r dv. o

lC

The integration over v is carried out by using the substitution z = v - u. To obtain the approximation for P, note that for large n

(n:!r)!=n-r(1-~)(1-r:1) ... (1-~)

[1-'!' i t] n-r[ 1- r(;: 1)].

- n- r =

n

1=1

148 UL (1967). co

E(8 X ) =

e- A

L (A8)'/r! = eMII-n. r=O

Then G(81) ( 2)=E(8i'. 8n = E([8182)X,] . E(8~2) = eXp[Al(8 1 82-1)+ A2(82 -1)]. The probability-generating function of Y is G(1, ( 2) whence P(Y = j) = e-(A ,+A 2>(A) + A2)i/j!, 0 ~ Y ~oo. The coefficient of 8~ in G(81) ( 2) is e-(A ,+A 2>(A2 + AI ( 1)i/ j!, so that the probability-generating function of the conditional distribution of Xl given Y= j is G(8 1I Y = j) = e-(A ,+A 2>(A2+.~181)i/P(Y = j) J. = (1 + p( 1)i(1 + p)i On expansion, the coefficient of 8 1 gives P(X1 = r I Y = j).

149 UL (1966). Unconditionally, 0 ~ X ~ a/2 and 0 ~ Y ~ a. For fixed X, X ~ Y ~ a - X. Hence on integrating the joint distribution with respect to Y, the probability density function of the marginal distribution of X is n(n -1)a-"2"-lx"-2(a -2x),

O~X ~a/2.

For Y in the interval 0 ~ Y ~ a/2, 0 ~ X ~ Y and for Y in the interval

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

al2::o.;; y::o.;; a, 0::0.;; X ::0.;; a - Y. Hence, on integration

435

h

forms of the probability density function of Yare t e corresponding two (i) na-"2,,-1 y"-I, 0::0.;; y::o.;; al2 (ii) na-"2,,-I(a - y)"-\ al2::o.;; y::o.;; a. The probability density function of the conditional distrib t' X=x is (a-2x)-\ for x::o.;;Y::o.;;a-x. Hence U Ion of Y givl'lI (a - xy+l- x r + 1 E(yrIX=x)= (a-2x)(r+1) ,

r;::O

whence E(Y I X = x) and var(Y I X = x).

150 UL (1966). ~x)=t+"'.(x) a~d (x)[1-(x)]=!-",2(x). On expansion of the mtegrand m the detinmg mtegral for "'(x) and term-wise integration . obtain ' we

and

Hence 2X2 2X4 7x 6 4<1>(x)[1-(x)]-1-- + - - - + ...

7T

-e

37T

457T

-2X2hr[1 + 2( 7T - 3) 37T 2

x 4 -'"

]

.

151 UL (1967). Here P(Y ::o.;;ta) = 1- P(Y~ta) -1-

'f' ' "'f-x o

~a

dy dx a(a - x)

=t(1-log 2) =P(a - X - y::o.;;ta),

by symmetry.

After the first experiment there are two possibilities: (i) probability p that Xo does not split; (ij) probability 1- p that Xo splits into two particles XI and X 2 with energies EI and Eo-EI respectively. Now P(E I ::0.;; tEo) =t. Also, if EI ::0.;; tEo, then any further splitting or not of XI must lead to particles or particle with energy 1!Eo. Therefore the probability of Xl giving particles with energy ::o.;;tEo after the second experiment is t(1- p). Next, the probability that X 2 has energy ~tEo is ! and the probability that in the second experiment X 2 splits into two further particles X3 and X 4 is 1- p. If the energies of these particles are E2 and Eo - EI - E 2 , then it is required that E 2 ::o.;;tEo and E o -E I -E2 ::o.;;tEo simultaneously.

436

EXERCISES IN PROBABILITY AND STATISTICS

Hence the total probability required is

!(1-p)+!(1- P)2[P(E2 ~!Eo) xP(Eo- El - E2~!Eo)] where P(E2~!Eo) =P(Eo-

E 1 - E2~!Eo) =!(1 + log 2).

152 UL (1967). 00

E(eIX)=~

J

exp[tx-\x-O\]dx

00

=~

J

exp[t(u+O)-\u\]du 00

=! e 18

00

[J e-(I+I)u du + Je-(I-I)u dU]' o

=

\t\ < 1

0

e '8 (1- t 2)-I,

whence E(X) = 0; var(X) = 2; E(i) = 0; var(i) = 2/n. The moment-generating function of i is

( t2)-"

Mx(t)=E(eIX)=e'8 1- n 2

,

whence

t 2 )-" . -_ ( 1- 2n

Hence, for large n, E(e 'Z ) ~ ell'. For the approximating distribution the proportionality factor is 1/[.J2;(1 + a + 313)] and

( ) _1+3a+1513. varz1 +a+ 3 13 ' ( ) _ 3+15a + 10513 fL4 z 1+a+313 .

E(z) = 0;

For the exact distribution of z, E(z) = 0;

var(z) = 1;

Kiz) =3/n.

Hence

6

a =-613=---. 8n+3

153 UL (1967). If y denotes the number of points obtained uppermost on anyone die, then 6

6

E(y) =

L i.s=!; 0=1

E(y2) =

L i. S2 =9j; 0=1

var(y) =~~.

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

Hence

7nl E(XI)=T;

7 E(X2) = E(X3) = 2" n2.

35nl var ( Xl) =1:2 ;

35n2 var(x2) = var(x3) = 1:2 .

437

It then follows that

var(XI + X2) = var(xi + X3) = H(nl + n2); COV(XI + X2, Xl + X3) = E[(XI + X2)(X2 + X3)]- E(XI + x2)E(XI + X3) = E(x~) + E(XI)E(X2) + E(XI)E(X3) + E(X2)E(X3) -

?(nl + n2)2

=~~nl'

Hence corr(XI + X2, Xl + X3) = ~ . nl+n2

Given

X2 + X3

= c, then

X3

var(xi +x31

= c-

X2'

X2+X3 =

It then follows that

c) =var(xI- X2+C)

= var(xI) + var(x2)' Similarly, it may be shown that COV(XI + X2, Xl + x31

X2 + X3 =

c) = ~~(nl -

n2)'

In the special cases (i), (ii), (iii), corr(xi + X2' Xl + X3) =!. 1,0 and corr(XI + X2, Xl + x31 X2 + X3 = c) = 0, 1, -1.

154 UL (1967). The probability density function of X is k(1 + X 4 )-1 where k = ..fi/7I". The probability density function of Z is 1 _1 _:I - - Z 4(1- z) 4 for O:os;; Z:OS;; 1. 7I".fi. ' The probability density function of the joint distribution of Wand Y is 2 Iyl 71"2 . (1 + w4y4)(1 + y4) whence the probability density function of W is

438

EXERCISES IN PROBABILITY AND STATISTICS

If X and Yare independent N(O, 1) variables, then the joint distribution of Y and W = XI Y has the probability density function

1

27T exp[ _!y2(1 + w2 )] .\y\

whence the probability density function of W is ~

2~

!J ~

Jexp[_!y2(1 + w

2 )]

\y\ dy

=

exp[ _!y2(1 + w2 )] • y dy

o

1

155 UL (1967). The distribution of x,. has the probability density function

I (X2)"-l 2x

n (11-1)! I!

o~'

0;

whence for any r> 0

J

'" E(xr) =2n - X211 + r- l dx 11 0211 11 11

(

2n = -) or. 2n + r

o

It then follows that n0 2

var(xl1 ) = (11 + 1)(211 + 1)2· Hence & and var(&) as stated. Note that the limits for z are

y

-2[n(11 + 1)]~"'; z"'; (n: 1

and the probability density function of z is

J

211-1 2n ( 2110 )211-1 [ Z no 1 211 0 2n + 1 1 + 2{n(n + I)}! . 2n + 1 . [11(n + I)]!

J

2 Z 1 )-~ --i>e'-1 1 -)211[ 1+ (1 - • 11-1 ( 1+2n + 1 2{n(n + I)}> n for -00< z < 1. =

as n --i> 00,

156 UL (1967). Probability of a withdrawal only in the week before Christmas is PI - P12; Probability of a withdrawal only in the week after Christmas is P2 - P12; Probability of no withdrawals in these two weeks is 1- PI - P2 + P12. Hence the probability-generating function of the withdrawals of a single client in the two weeks is (1- PI - P2+ pd+(Pl- P12)81 +(P2- pd 02+ PIP2 0 1 02 = 1 + Pl(OI -1) + P2(02-1) + PliOI -1)(02 -1),

whence E(O~'. 0~2)= G(Ol, O2 ).

ANSWERS AND HINTS ON SOLUTIONS: SUPPLEMENT

439

Under the stated conditions the limiting form of G(O 1> 0)' 2 IS exp[al(OI -1) + a2(02 -1)+ aliOI -1)(02 -1)]. The moment-generating function about the origin of Xl and X is G(' , whence the cumulants on logarithmic expansion. For corr(X 2X ) == eO" eh,) . f unction . fX I and 2 ,tc moment-generatIng 0 X 2'IS exp[a l (e"-l)1>+a2(e"-1)] whence independence of Xl and X 2 • '

157 UL (1969). (i) The probability-density function of the marginal distribution of Y is =

(y )

(3-a-2 cxI 1 r( a + 1) . y -(l-+-x-)-=-/3 exp - -l-+-x dx. o

To integrate over x for fixed y use the substitution w = y/(l + x), whence the stated result. (ii) This follows readily by using P(X I Y = y) = P(X = x, Y = y)/ p(Y=y).

I =

•.•

y/3-1

(111) E(XI Y=y)= g(y)

x e- y /(1+x) (l+x)/3 dx.

o

This integral is evaluated by using the substitution z = y/(l + x). The Jacobian of the (X, Y) ~ (U, V) transformation is v- 3 , whence the probability density function of the joint distribution of U and V is

1

---e-uu cx (Q-a-2)v/3-cx-3

r(a + 1)

.

tJ

O,.;;U
,

,

whence independence and marginal distributions.

158 UL (1969). P(X ~ Xl) =

~

I =

e-(x-/3)/cx

dx

= e-(x,-/3)/CX.

The probability distribution of Xl is now obtained by using the multinomial distribution. E(XI) follows directly on integration. Since Xi ~ Xi> therefore for fixed Xl the probability density function of the conditional distribution of Xi is

1

_ e-(x,-/3)/cx/e -(x,-/3)/cx

a

= _1 e-(x,-x,)/cx. a

Hence E(a* I Xl) = a = E(a*). E«(3*) follows on taking expectations.

159 UL (1969).

(i) The limits for ware obtained from I:'=l (xi-i)2~O which may be rewritten as n - W2~O.

440

EXERCISES IN PROBABILITY AND STATISTICS

(ii) Note that

w=

nx [en _1)S2 + nx2]~

=-:------:---;;------;;-::;-

whence the relation between t 2 and w2 • The sign of t is the same as that of w. Also, n - w 2 ;;;. O. Hence as w2 ~ n, t 2 ~ 00. Therefore t is a monotonic function of w. Note that u = wl..fii = vl(1 + v 2 )! and the suggested transformations lead to the stated distribution of v.

160 UL (1969). Here E(Y1) = E(Y2 ) = O. var(Y1) = !( Y 2) = !( Y2 are uncorrelated if P
-I

2

1

2·

For this value of 0, Y1 and Y2 are independent normal variables with zero ' means and variances