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Stoclmstics Cfkfclfllfs; Practical fafrtdffclcn, Richazd Dllrrette ./1.

Chass Expfmsinn,Multiple Wcincr-ln Integrals and Victor Perez-Abreu

Probability and Stochascs Series

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White Noise Dfsfrfsllfion Theonol,Htti-l-lsiung Kuo Topics n Cnntempnt.y Probabiliiy, J. Lattrie Snell

Applications, Chz-isti.artHoudre

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Acquidng Mtor Cover Designer Marketing Manager Associate Marketing Manager Msistant Managing Editor, EDP Pre-press Mmmfactuling Assistant

Library of Congress Cataloginge-Hblication

Data

Catalog record is available 9om the Library of Conp'ess

This book is th: Mincarnation of my qrstriook Bwiti gp Motion and Martingqles in na lysis whih was published by Wadslbiti 144984. For mpre tharza decardeI. ave used Chapters 1, 2, 8, :4d.-9 f (ila'f ook.to give flrst year grarduate probr courses'' to graduate ytqdents Fho have conplitd.the ability course ancl Fere intere.ted i learning more about procsqes that in spac: and time. Taking the advice from biology that torm folcontinuolzsly stochwstic lowsfunction '' I have taken that material on s#ochastic kntegratiohj diferensialequations, Brownian motion and its relation tp p>rtial diferential book (Chaptys 1-5). To this I have >ddd equations tq be the core pf other practically important topics: on dimensipnql diffuppp, semigrqups @nd geperators, Harri, chaipsi and peak convergence.. I have struggled Fith this materi>l for almst twepty years. I zpF Shipk that J Jmderstand most of it, ( ' sp to help you mmster it in less time, t have tmed to pxplain iti ms simply and clearly as I can. My students? motivations for learning this material have been diveyse: some have wanted to apply idemsfrom probability to analysis or diflbrential geometry, others have gone on to do zesearch on difusiop prcsses or stochmstic partial of thes ideas diferential equatins, some have been inte#ested in applications prpblemK ih operations reseyrch. My motiyation for writing to fnance, or to this book, like that for Pvobabilily Teorg and fzcpvled, wws to simplify my life as a teacher b# bringin: together in one plae useful material tht is scattered iu a variety of souicesk tfif you copy from e book that i plagiarism, but ' An old joke says that scholaiship.'' boks that is from From that viewpoint this is tez if you' copy contributors various subjects are (a)stochatic main book. the for Its cholarly and equations: and lkeda artd diferential Chung Williams integration (1900), Wataabe (1981), Revuz and Yor Karatzmq and Shreye (1991), Protter (1990), Stroock and Varadhan (1970); (1991),Rogers and Williams (1987), (b)partil equations: Folland Diedman Port diffrential (1976), (1964), (1975), and Stohe and Karlin and Zhao dimnsional diflksions: Chung (1978), (1995);(c) one and Yor semi-groups and Taylor (198.1)) Dynkin Revuz generators: (1965), (d) and Billlngsley weak Ethier Kurtz (1991); (e) convergence: (1986), (1968), If you bought a1l those books you would spend Stroock and Varadhan (1979). more than $1000 but for a fraction of that cost you can have this book, the intellectualequivalent of the ginzu knife. ,

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This book contains information obtained from authendc and highly regarded sources. Replinted material is quoted with pennission, and sources are indicated. A wide variety of refemnces are listed. Reasonable effortq have beert made to publislt reliabl data and information, but the autlzor atd the publisher cmnnot assume responsibility forthe validity of all materials or for the comsequence,sof theiruse. Neithertls booknoruypartmaybereproduced rtmnsmitted in any fonn orby anymeans.electronic or mechanicak including photocopying, microfllming, nd iecording, or by any information stomge or retrieval systenx witlmut prior permission in writing hp4p tlle publisher. CRC P=s, lnc.'s consent does not extend to copying for geneml distribution, for promodon: for creating new works, or for resale. Specitk pennission mstb obtained in wtiting from CRC Press for such copying. Direct all inquirie.sto C'RCPrss, lnc., 2000 Corpomte Blvd.. N.W., Boca katon,Florida 33431. g

@ 1996 by CRC Pmss, lnc. No clnim to original U.S. Government works Intemadonal Standard Book Number 0-8493-8071-5 Plinted in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acide,e paper

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vi

Preface

Shutting off the laugh-track and turning on the violins, the road from this book's frst publication in 1984 to its rebirth in 1996 hmsbeen a long and winding second half of the 80's I accumulated long list an embarrassingly one. In the edition. Some time at the beginning of the 90's I talked of typos from the flrst to the editor who brought my first three books in t o the world John Kimmel about preparing a second edition. llowever, after the work was done, te second edition was personally killed by 2ill Roberts, the President of Brooks/cole. At the end of 1992 I entered into contract with Wayne Yuhasz at CRC Press produce this book. In the first few months of 1993, June Meyerman typed to of book into TeX. In the Fall Semester of 1993 I taught a course from the most this material and began to organize it into the current form. By the summr of 1904 I.thought I was almost done. At this point 1 had the good (and'bad) fortune of having Nora Guertler, a student from Lyon, visit for two months. When she was through making an average of six corrections per pagej it was clear that the book was faz from flnished. During the 1994-95 academic year most of my tine was devoted to preparing the second edition of my qrst year graduate textbook Probability: Tct)r.q and Eamples. After that experience my brain cells could not bear to work on another book for another several months, but toward the end pf 1995 they decided is now or never.'' The delightftll Cornell tradition of a long winter break, which for me stretched from early December to late January, provided just enough time to finally fnish the book. I am grateful to my students who have read various versions of this book and also made numerous comments: Don Allers, Hassan Allouba, Robert Battig, Marty Hill, Min-jeong Kang, Susan Lee, Gang Ma, and Nikhil Shah. Earlier in the process, before 1 started writing, Heike Dengler, David Lando and I spent and Jacod and Shiryaev (1987), an enterprise a semester reading Protter (1090) ( which contributed greatly to my education. The ancient history of the revision process hms unfortunately been lostk At the time of the proposed second edition, 1 transferred a number of lists of typos to my copy of the book, but 1 have no record of the people who supplkd the lists. 1 remember getting a number of corrections from Mike Brennan and Ruth Williams, and it is impossible to forget the story of Robin Pemantle whq took Bvownian Motion, Martingales and Analysis as his only math book for a yearr long trek through the South Seas and later showed me his fully annotatvd copy. However, I must apologize to others whose contributions were recordd but and 1'11have something whose names were lost. Flame me at rtdllcornell.edu to say about you in the next edition. ,

,

About

the Author

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Rick Durrff

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Ttick Durrett received his Ph.Dt in Opeitions research from Stanford University in 1976. He taught i ihe Mathematics Deprtment at UCLA for nine years before becoming a Professor of Matheniatics at Cornell University. He ws a Sloan Eellow' 1981-83, Guggenheim Fellpw 1988-89, and spoke at the ! International Congress of Math in Kyoto 1990. of > gradlpt tltbppk, Prqbqbility: Teorg and Dmrett is the The Eqentials of Jiropllip. 'zcmpled, and an tudergpdu>ty He hms one, ujjjj co-aythorj an' d en 19 students taj jj written ( almost 100 papers > 9 u. comylete their Ph.D.?s. under his dtxciippr lds recept rrsearc fjcuses n the spatial mpdl: it? vayiolp problerns in zlor. appllcations of stochuti '

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Calculus:

Stochastic 1.

Introduction

A Practical

5. Stochastic Diferential Equations 5.1 5.2 5.3 5.4 5.5 5.6

Brownin.n Motion 1.1 1.2 1.3 1.4

Deflnition and Construction 1 Markov Propexjy, Blumenthal's 0-1 Law 7 Stopping Times, Stropg Markov Property 18 First Formulms 26 ; '

E

,

2 Stochastic lntegraon 33 2.1 Integin ds i P !!Qitble Pr'eje Lc>l Mitikl Integrators: Continus 2.2 .

t.3

.

.

6. One Dimensional Difusions

;

.

.

.

.

'

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37

wa.tt

'

'

'

i

Reculzence and Transience 95 Occupation Times 100 Exit Times 105 Change of Time, Lvy's Theorem 111 Burkholder Davis Gundy lnequalities 116 Martingales Adapted to Brownian Filtrations

4. Partial Diferential Equations

B.

4.4 4.5 4.6 C. 4.7 4.8 4.9

Parabolic Equations The Heat Equation 126 The Inhomogeneous Equation 130 The Feynman-lfac Formula 137 Elliptic Equations The Dirichlet Problem 143 Poisson's Equation 151 Equation 156 The Shrdinger Applications to Brownian Motion Exit Distributions for the Ball 164 Occupation Times for the Ball 167 Laplace Transforzns, Azcsine Law 170

'

Constructiou 211 Feller's Test 214 Recurrence and Qansience 219 Green's Ftmctions 222 Boundary Behavior i29 Applications to Higher bimensions234

7. Difusions as Markov Processes 7.1 7.2 7.3 7.4 7.5

8.

3. Brownirm Motion, 11

A. 4.1 4.2 4.3

.. .

and

E

3.1 3.2 3.3 3.4 3.5 3.6

6.1 6.2 6.3 6.4 6.5 6.6

E ('

Covariahe Prcse 42 Bded 52 Myitinzle Ih t 4 tity 59 2.5 The Kit-Watab 2.6 Integration w.r.t. Local Maztingales 63 2.7 Change of Variables, lt's Formula 68 2.8 Integration w.r.t. Semimartingales 70 2.9 Associative Law 74 2'.10Ftmctions of Several Semimaztingales 76 Chapter Summax.y 79 2.11 Meyer-rfnaka Formula, Local Thne 82 2.12 Girsanov's Formula 90 Variane

2.4 Integration

Examples 177 It's Apmoach 183 Extension 190 Weak Solutions 196 Change of Memsure 202 Change of Time 207

119

Semigroups and Generators 245 Examples 250 Aansition Probabilities 255 Harris Chains 258 Convergce Theorems 268

s7eakConvergence 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

ln Metric Spaces 271 Prokhorov's Theorems 276 The Space C 282 Skorohod's Existence Theorem for SDE 285 Donsker's Theorem 287 The Space D 293 Convergence to DiAsions 296 Examples 305

Solutions to Exercises 311 References 335 lndex 339

1

Motioh

Btownian

''''''' 1.1. Defnltlop

Constructlon

and

yyytj (yyjytyjjoj k( sjju tvaut, . . (. . . .wit . aftei . birtlz of a . lili w itl b ible redtiin: th deqittn of a to have fun with our new arrival. We begin by starting with point d-dimensional Brownian motion to th>t 9 a one a geneqal itikt tw sttie t , (1r >nd (tiifl il s ioal BlWi rfttitt titi: t 0 (t * 2) 1 re part of tmi . .

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jt

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(1.1) Tzmnslatlon hwarlnnce. (.Df Bz, k 0Jis independent of Bz the same distributioh as a Brownia motio with Bz 0. =

of coordlnate (1.2) Independence tj V g jwyj jgyllj fnotions startig Brownian dimensional 0) are independent one

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(a) If to < z < independent. (b). If s, t

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inrement lkmsindependent increments. () says th>t (a) says tat Bs + t) W(s)haq a normal distribution witlt mean vectpr 0 >nd variance t. The reader slkould note that above we have sometimes (c) is self-explanatory. -

Capter

2

1 fz.owan

we will contoirme in what

For each 0

the defnition that will be useful myuy times is:

#r,4z,...xatXl

wzitten Bs and sometimes follows.

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a practice

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To be precise, the two families of random variables have the same flnite dimepsional distributions, i.e., if sz < < sg then .

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ln view of (1.2)it suces to prove this foi one dimensional motion. To check this when n q.z: 1, we note that 1/2 times a pormal Fith me>n 0 and s ls a ppifnt withmean 0 and variance si. The resvlt torn $ 1 toltows vayiqpce from indpendent increments. '

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The flrst question that must be addressed in any treatment of Brownian motion is, tsls there a process with these pxoperties?'' The answer is tYesy'' of course, or this book would not xist. For pedagogical remsops we will pursue an approac h that leds to a dead end and tlken retreat a little t ieitify the diculty. In view of our deqnition we can restrict our attention to a one dimensionalBrownian motion starting from a flked z e R. We ould take z 0 but will ot foi remsons that become clear in the remark afte (1.5). =

x

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R) and Fo be the (1.4) Theorem. Let Do (functions td : (0,x) generatd by the fnite dlmenslonal sets (w: ) E A for 1 K %n) df E ,J2: For each z E R, there is @ unique probability memsqre vr on SO that < ta 1 and when 0 < tl v r (td: (0)= z)

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=

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x x .Aa-,) pw-aykkkia-u.'

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its flnite dimensional distributions are

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by

on Rn

. 0, but By translatzon mvariance, we can without loss pf generality mqsume z of and normalslwith indpendnt the >11this yay. iq that cwse is sum man 0 azd ha,N with tJ-1 normal and istribution 0 vari>nces tj mean p, h+1 tn With th consistency of f.d.d,'s verifed we get our flrst yayiance t.+1 j-l. construction of Brownian motion:

Bt is continuous.

,-.+

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3

0,

=

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Construction

and

and A e the Borel subsets of R. From the formula above it is easy to see that fr flxed' z th fkmil: JI is a onsistent set of qnite dimeional distributios sn-t) then (f.d.d.' )j that isj if (s1, t srz-tl C (tlj t t twl and tj / (t

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Dennition

Section 1.1

Motion

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wher:

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-

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Thij follow: from a generalization of Kolmogorov's extension theorem. We will not bo ther with te details since at this point we are at the dead end referrd to above. If C (td: t td(t) is continuous) theq C / So, that is, C is not a measurable set. The easiest way of proving C Fo is to do --+

=

6 So if and only if there is a sequence of times t1, tz , J-feld T xR k so that e (0,((J e T?/1,2,...) (theinfnite product a1l words, = Bj. In events in Fo depend on only : @(t1),(f2), .4 countablymauy coordinates.

Exerclse x)

.4

1.1. and > B

...

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...)

.

Capter

4

1

Prownan Mbton

The above problem is easy to solve. Let Q2 (m2-n: m, n k 0) be the (tJ : Qa dyadlc rationals. R) and Fq is the c-fleld generated by the If fl flnite dimensional sets, then enumerating and applying the rationals 1, q,, Kolmgorov ' s ex t ens ion theorem shows that we can construct a probability v z) : tJ(0) 1 and (+)in (1.4)holds when the on (f1gFq) so that vrf x) we will show: ii e Qz. To extend the defnition from Q2 to g0, ::zq

=

-+

.

.

.

.

=

,

=

(1.5) Theorem. Let T < (x) and z E R. vo mssignj probability tJ : Qa R that are uniformly continuous on Q2n (0,T).

with

sl1+G (1.6) Theorem. Suppose that Epxj .X)# % ff lt alp then with probability one thete is a contant Cw) 7<

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Proof Let < a/#, p > 0, Q ((,./) (p(aY(./2-n),X(2-n)) Gn S j and t/P(IYl > c) S E lFl# we have .f

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... .

P.

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i

KK

'

=

,

-+

S J'f 2n2n9

PG') where enough

=

so

.

p)(1 + a > 0. To that

(1

pyj

-

-

(1.7) Lemma.

Let

(, jj

-

complete

=

S Xl

-

rl-/

2-'T).

-

(1.6) follows

Proof of (1.5) By (1.1)and (1.3),we can without loss of generality supyose Bz 0 and prove the result for T 1. In this case, part (b)of the deqnltion and the scaling xelation (1.3)imply =

PH'

.E'n(If

/,14)

=

fnlf-,I4

=

Ct

-

s)2

where C = .E'1/114< x. From the last observation we get the desirqd uniform continuity by using a result due to Kolmogorov. In this proof, we do not use the independent increments property of Brownian motion; the only thing we

N

xxGn

=

with 1

rl

-

s

we

have

2-(1-8)X

(1.7):

from

G)

S

)S

On #w'

Q2n (0,1)

(x7

=

-

euily

we can pick p small show

for q, r E

E

p(dz).%(.)

2n9J

Since

ffz-''

=

p). Since y < alp, the prf ow we will

=

=

<

QJ.

2n2n9 1 so

;

(1+

.

Pxq, Xr)

is

E f

(2n92-n)-#-/+1+G

.4)

.P,(.A)

E

-

X(2-n))#

.

3 2(1-n)-//(1

.z4

=

.

E

j

Lp-f--va

Now the number of

%sumption.

'

for

<

(,j)

al1

)z-n )

-

(J)E.Ik

by our

j < 2n, 0

%

)2-r)'T

)2-n)-Jh.E(.Y(j2-'n),

-

0. If

(,J)EJk

.

=

0

:

p>

e Q2n (0,1)

al1 q, r

-

.

Our construction guarntees that Bt @) vt has the right Enite dimensional distributions for t of paths and a sinple limiting argument Continuity e Q2. E x). shows that this is true when t 6 g0, th Ay mentioned earlier the generalization to d > 1 is straightforwd i cooxdinates ate independent. In this gehera tit y C (cntinuos tp (0co)' 14 Rdl and c is the c-field generated by the coordinate rriaps t (). The rder should note that the result of our construction is one set of random variable Bt @)= (t), and a family of probability memsures P=, z G Rd, so that under z) .Pa, Bt is a Brownian motion with P=Bz tlie 1. It is enough to constrct Brownian motion starting from an initial point z, sihce if we want a Browniaxt motion starting from an initial memsure y (i.e.,have PgBz p(X)) we e simply set =

=

=

-+

.

where a, so that

-

,

one to paths

5

In Sectipn 2.11 we will need this result when the metr p; so, we will go ahead and prove it in

use is the moment coudition. Xj take values in a space S that generality.

-+

> Remark. It will take quite a bit of work to prove (1.5).Befoye takipg pn t/ tmsk$ we will attend to the lwst memsure theoretic detail: we tidy tlkings up y = (t2 : movingour proability measures to C, c) where (continuous (0,x) R) and is the J-field generated by the coozdinate maps t w(), To do this, we observe that the map # that takes a uniformly continuous point jzi fl to its and we set extension in C is meyurable, ;

and Construction

Denition

Section 1.1

PGn'

n=N

) S ff

S

nN

rl-/ for This shows pxg, .X'rl < I rl pxq, Xr) S C(Y)I rl-f for q, r e (0,1). .1:

-

-

gg-N

2-n

% :(a)

=

1

-

2-

and implies that we have

-

Proof of so that

(1.7)

Let q, r E

Q2f3 (0,1) with 0 <

2-(m+1)(1-p) < r

-

r

-

2-(1-'?)X. q<

2-m(1=p) q<

Pick rlz k N

6

Chapter 1 Brownian Motion

and

write

Secon

.72

=

...m

.

j

<

so j

.

.

,

f) <

-

(a)

+

.

'

+ 2

.

.

-

.

-

.

.

(

.

-

K p(X(2-m),XU2-'n))

,

i

No* 0 .

<;

r

q <

-

On Il

( .. . .

it follows from the triangle inequality that

P(Xn)

(x)

(b) where Cv

=

1/41

2--/)

-

>

1. Repeating the lmst corriputatiop .

(c) combining (a)-(c)gives

'-tn

pxr

# (X q Xr)

S

,

since 2-m(1-0) < 21-njr.(1.6) and (1.5)

-J2

,

.

x'.i2 )) s cv2

3Cv2--O(1-0)

i

=

n P

1/11<

I. This

completes

-

,

the proof of

l(1.7)and

flft

=

-

(1.6)with

(1.8) Theorem. any Y < 1/2.

(1.2)implies m I2m Cmlt

p

=

2m

-

hence of

and

(:

m

=

-

1 and

Brownian paths aye lllder

Cm

then letting

m

-.+

f(tnz2-n)

=

7 for

aicl

l1se'

n

X.

=.*

(1.9) Theorem. With pmbability one, Brownian paths are not Lipschitz continuous (andhence not diferentiable)it any point.

It

e (0,1) so

=

lft

that

B,

-

IK Clt

41 whep

-

Fk,n

=

Gn =

max

( at

B

k+ j n

-

lemst one 'Fk,n is

B

k+ j z

K 5C/n)

-

resu

:

j

=

0, 1, 2

3

.

(:z.)-1/2

.

?k

(). Noticing

-.+ .a

.

th fmof. istd are not

ts

I::I

of 3 thai if k $ Hlder continuous

B,

-

;k$

t

B (t(?'rz1)2-n). -

-

s.

Compute

''

'

,

.

,,,2 -

a

EmsaaA2m,s

-+

t

a.s.

we consider t? Freedman

lt is true

.

.

. .

lhtuitively

y

1

.

1.2. Markov : '

-

,

.g

+!.r

partttions H1 a sequence of c with mesh 0. See llz c (1970)p.42-46. The true quadratic Var ia tion , deqned as the sup over all partitions, is x for Brownian motion.

The last

-+

-n

-+

the Borel-cantelli lemma to conclqde that

Remek.

Let ( : there is an s sl K 3/n). For 1 %k % zz 2 let

-

El mK2n

gives:

(x7

.

-

'

(0,1).

q at any point of

E

It is easy to show:

Proof

T! zjy

=

1.3. Let Am,a

2

-,z

2

with exponent

continuous

(

-.+

.E'1/112r''

=

s rj

1ya

The next zesult is more evidence that Bt

(3

Exercise slm where

n

10C

.

1.2. Show by considering k incremhts then with probability t, Brownian p>ths

with exporent

+

-

.p(.rt)

s

Exerclse 1/2 + 1/

The scaling relation

So using

If1/aIS

izic

'

3

-,z

.

exptvzi/z) 1. Letting n x show, q all emsing d cotpletes hows /(.A r. ) t f6r i r n

:

. '

i 3Cv 2(l-0hjr .

shows

'

.

5c' a n 5C 3

n P

=m


(1.2)gives

S P(Gn) S

)'--!(2-())'y s 5-!(2-'-') s cvz-rn =l

:

p(x#,x(2-n))

#(r'-2)/al

-

Using l.n c: Gu and the scaling relation

.

.

I#(a-a)/nf(r&-2)/nl < If(r'-a)/r'#11 + Ilz -

.

:' .

< scjn,.

E

.-z

,

((2rn9)2-m)T ..

is

=

.

.

.

.

=

.

.ye

,

.

2-() z-m 2-(1) < r(f) and m < (1) < r < (:)r 2m P and it follows tltt on Vx q=

where m < r(1) 2-m( 1- %)

-r(z)

+ 2

we claim

1, which We claim that C Gn To prov Shis we consider s the worst possibility. In this cmse we want to conc j u (j e tjjat n-2,a 0 is the worst case, but even then For this we observe that j -s

-r(1)

r

Property Blumqnthal's 0-1 raw 7

Marov

1.2

j

Property,

:

theMarkov

Blumenthal's

prrty

0-1 Law

g

.

say

tgiven the present state, B' any other information about Ayhat happened before time s is irrelevant for predicting what happens after tirn.e s.'' ,

.

8

Property Blumenthal's p-1 faw

Secton 1.,2 Mrov

Chapter 1 Prowzkan Motion

Since Brownian motion is translation invariant, the Markov property more simply stated as:

can

be

Bs t k 0 is a Brownian motion that is indepenNf s k 0 then #f+, dent of what happened before time s.''

by

(#,tJ)()

tJ(s

=

+ t)

-

f0r t

,

This should not be surprising: the independent increments property ((a)in the deinition) implies that if sl S < a then %sm s and 0 < tl .:2

.

(#t:+,

-

Bz

.w+,

,

.

.

.

=

.

.

Bsj is independent of

-

.

.

.

(.Gz ,

.

.

The major obstacle in proving the Markov property for Browni>n motion then is to introduce te measure theory necessary to state and prove the result. The flrst step in doing this is to explain what we mean by ttwhat happened before time s.'' The technical name for what we are about to deqne is a flltrat 1 h a f n y term for an increasing collection of flelds F , i e if s < t thz F a c F f. Since we want Ba e $a,' i.e., B, is meadurale with respc' t t Fa', tlte flrst thing that comes to mind is ,

,

F:

tztfr

=

:

r

,

.

it.is convenient y+

to

-+

=

y

o

=

fzh

Likewise, if F4)

0,

J(patJ(t))

=

,

F

.

.

o

.tJf

p,

=

J((s

))

+

=

)

fBs-vt

then

a)

fB,+t

=

.

,

.

.

B,+tn

)

.,

,

K s)

replace

(:.1) The Markov then for al1 z E Rd

lf s k 0 and F is bounded

property.

f'zty

where the right hand side

F: by

njxay-jo

=

.

C, for s k 0

k0

o

%1F,+)

=

and

py)

=

C measurable

En.Y :

For technical reasons,

-+

In words, we cut of the part of the path before time s and then shift the path so that time s becomes time 0. To prepare for the next result, we note that if R is C memsurable then F o 0, is a function of the future after time s. ' :C J(tJ(t)). ln th case To see this, consider the simple example F(tJ)

Bsvj

.

%: C

s,'' it is convenient to deqne the shift transformations

9

EvY evaluate j at

.p

f

=

. .

().

Exylnnation. In words, this says that the conditional expectation of y o % given &+is just the expected value ot F for a Brownian motiop starting at B, ttgiven the present state B, any other ipforpatn To explain why this implies kbout wht happened before time s , i irilevant for? predictihg w jy t ju p jjyjjs after time s,'' we begin by recalling (see(1.1)in Chytei 5 of Duirett (199$)) .E'(Zlf). Applying this with that if c: F and .E'(gl.f') e f then .E'(Zl7) c(l,) = = F+ aud .F we have 4 .

The felds F,+ are nicer because they nt>,Jl+

=

are right

nf>, (nu>t.T-:)

=

continuous.

nuyyaF:

=

That is,

,

Ft

=

peek at the future,'' i.e,, In the Fs+ allow us an ttinflnitesimal e Ss'f if 6 it is in Ss*+zfor any > 0. If Bt is a Brownian motion and f is any memsurable function with fuj > wh.en > 0 the random variable words

.

.

.

E

:

.

.E'.(y lat'i'-o es1.,:-a+)

'tz

Bt limsup t ta

-

#a)/J(t

s)

-

=

o

0s 11,)

If we recall that X = .E'(ZlF) is our best guess at Z giyen the information in F tin the sense of minimizing E LZ X)2 over X E .4F, see (1.4)in Chapter 4 of then we see that our best guess at F o % given B, is the same Durrett (1995)) ..F9- is irrelevant fpr best guess given Fa+, i.e., any other information in as our redicting o % p -

is memsurable with respect to Fs+ but not F(. Exercises 2.9 and 2.10 onsider what happens when w take fn) lg log(1/u).' HoWever, V-uand fn) will see in (2.6),there are no interesting examples of sets that are in F+! ms we , but not in S:. The two J-flelds are the same (u to null sets). To state the Markoy property we need some notation. Firstj recall that we have a family of measures .P., z e Rd, on C, C) so that under Pr, .8t(tJ) () is a Brownian motion with Bz z. In order to de:ne happens after time =

=

'tz

'

'

.

Proof

By the deqnition of conditional

expectation,

what we need to show is

=

=

Gwhat

(MP)

f.(F

o

#, ; W) = &(.Eb.F;

d)

for a1l

.z4

6 F,+

Capter

10

1 Brownan Afoton

Section 1.2

We will begin by proving the result for a special clmss of F's and a special clss of W's. Then we will use the theorem and the monotone clmss theorem to extend to the gerteral cmse. Suppose 'm

lf.I y,,ttmll lKmsn

'i'(-)

=

.,1.

.

.

=

,

extend

the

E=pB,+

clmss of

,

#); .A) for a11 f.d. sets .A in

.A's to al1 of Fo ?,, a+

,

(2.2) The g' theorem. Let be a collection of subsets fl and is closed under intersection (i.e.,if B e Jt then collection satisfy of that sets a (i) if B E f and D B then W B E

'utl

'tzt

FXs

.4

of fl that contains

nB e

.4).

Let

f be

.4,

-

.

(ii) If d,z

.

=

11

we use

,4

.

.

To

%; .A) =

Blumenthal's 0-1 rw

.,

-+

.

o

-

where 0 < :1 < < ri and the fm : Rd R are bndd d fiisurable. Lt (tJ and let 0 < < 1, let 0 < sl : g(s/) < sk S s + E Aj 1 i j i k where Aj E 'Itd for 1 K j K k. We will call these .A's the flnlte dlmensional or f.d. sets in F*a+! Prom the definztioz of Brownian motion it follows that if 0 < < dezi of sity then the given Buzj Bu is by joint . z .

(MP-1) E.Y

-

Markov Property

If

.4

and

e

then the

C

dn

1 .A, then

J-feld

generated

E

.

by

.

d

t7'(.4)

.4,

.

'l-lz

.

.

,

.!%lfu,

=

g1,

.

.

.

Bue

.

.

.

pz)

=

z

=

f=1 where yz

z.

=

(MP-2) E.Y

yij (:w-1,

puf-uf-,

See (2.1)in the Appendix

Proof

..

.

.

''

''

o

#,;

.4)

Eg,pBaw;,

= ''

.

of

.

'

.

Durrey,t ,

#);

-)

(1995).

for

all

.. .

.

..

G Lh

.

..('

.

.

.

.

.

.

ikoof Fix F and 1et be the collection of sets X fpr whith the desired equality i? tiu. A simple subtraction shows that (i)in (1,2)hold, While the zootne shows that (ii)holds. If wr let t be t pllction pf jt nite nvergence theorem (i.4) dimensional sets in Sso..s then we hve shw F,%s / c: d whih is the desired conclusion. (:I

From this it follows that

jt

Er

z

giBui) 'I7 =1

dy puz-unlpa, !/1).1(g1)

=

.

.

dpz puz-uz-z

.

Applying this result with the ui given by th obvlou chois for the gi We have .E'.(F

o

p,;.)

=

1,

k 1x,(l,,)

T1

Ec

.

.

dzlpaztz, z1)

..

.

Wd

,

s +

t1,

.

.

.

+ tn

and

(MP-3) ErY

o

#a;.A)

w

F1'ym(s+,m)

*y)

m=1

,

. dpp,+?,-,.(zk,

p) py,

&)

dpl ;hz-?&(p, pz).f1(p1)

k gi Bui Using the formula for E= 1-Ifal

.

.

.

) again

dyn pta-ta-zlp,:-z, we

have

0); d) for a1t W E Fi >

0,

all we need

.

.

dyn pta-ta-z

0 in

convergence

theorem shows

-+

ezh, ) Using (MP-2) result.

-+

(pn-1,ynjfn (pn)

is bounded and memsurable. Using the dominated 0 and zs that if z then ynlfrtyuj

to do is let

J1(p1) dy1 pta-tztgz, p2)J2(p2)

=

..4.

where w(g,)

EzpBs

Proof Since St C F,*+s for any (MP-2). It is easy to see that

-+

=

=

dzk p,k-,p-z(zk-l;zk)

.

ykz

sk , s +

.

J=1

=

,

(a.??j)

laa

.

.

Our next step is

(pz-1,pz).z(gz)

and

=

J

dplpfz-hlzh,

the bounded

convergence

p1)/(p1)

-+

ez,

0)

theoyem now gives the desired 1Zl

Section 1.2 Afarov

1 Brownian Motion

Capter

12

(MP-3) shows that (MP) holds for F Fllzm
.fm&(m))

th

when

=

'-erallounded

memsurable

are willuse

'

11. we

,4

(i) If

.4

e

(ii) If 0 S

.4,

to

Proof

Exercise

and

bouqded function, then

A 11, a

f

E 7.

'

with

al1 the bounded functions on f th>t are memsurable

(1.4)of

Chapter 5

Durrett

of

' '

.

(i)

'.

LlBt

2.1. Let 0

< s <

If

.

.f

:

Rd

.E'.(J(ft)l?,) 2.2. Take fz) z lude that Bij and BgiBi: are

Exercise

=

. .

.

.!%(S

> 1+

)

=

j

.'&j

=

so

we siuc

if i

#j

# j.

calculations

ip Exercise 2.1 to

=

and

=

pltz, p)Pp(% >

Pnt

S f)

=

j

h(r, Brj

dr1

pt(0, p)J5(% > 1

-

t) dp

:

Rd

j

a

hs + :

Bn4 du

'tz,

R x Rd

-+

R are bounded

.'%(

(T(fz-,)

other

and measurable

-a

R and

-+

#7s-

turn our attention now to

(.('-' (s+

exp

applications tttriviality''

.%4

u,

dn)

j

f the Markov property below, that hms surprising consequences.

.

%1.'F'a+ ) = En.Y

o

C F:

that it follows (see,e.g., (1.1)in Chapter 5 of Durrett (1995)) o

pa1F,+)

=

Eckr

o

pa1J7)

1.4. =

0J.

From the last (2.6) Theorem.

equation

it is a short step to

If Z E C is bounded then for all ErzLFt,

t) g =

=

R is bounded

tr,Bv) dr

(-ftr,m4 j Ljo ) exp

ErY in Sectin

f

t. lf

Ek (Y

with

zizj

<

drj

En.1Bt-s4

=

j'

=

0

The xeader will see many

0) ard let L supt K 1 : Bt Exercise 2.4. Let Tn infts > 0 : B, L is for left or lmst. Use the Markov property at time 0 < < 1 to conclude =

= eXP

. ..

of the Markov proprt

Exerclse 2.3. Let Tn infts > 0 B. 0J 1et R inft > 1 : Bt R is for right or return. Use the Markov property at time 1 to get :

=

E

R is bounded and measurable

-+

martingales

The next two exercises prepare for

.

.

.

f(z)

and

'

.

.

'

The next seven exercises give tyjical applications Perhas the simplest example is

(2.5)

E=

.

.

.

(2.4)

Exercise 2.7. Let 0 < s and measurable then

xfor

.

COnC

drj

in Chapter 4.

-->

+ Ss(a)

(1b95).

.4

Exercise

(Jftr,

Brj

0

(2.1)from (2.3),fix in

' '

If lt : R x Rd

.

J(.1).

See

..

2.6. Let 0 < s <

Er

J1. E F,+ and 1t 7 = the collection of bunded which (MP) holds. 7 clearly is a vector ' space satisfying (ii): functions ' ' . . . . . of sts of the foiin ((J : ulj E Aj, 1 K J S nl Skh be the collctih Let ) .4 shows that if W At E 7 d The specil caae treated aboke e then lx W' . . clusion follchksfror (2.3). holdi and th desired t Thi shows .

E

e 1i.

lx

fn E 7

Then 14 contains

To get

=

The next two exercises prepare for calculations

.4

resp

Exercise 2.5. Let G be an open set and let T inft : Bt / G). Let ff be a closed subset of G and suppose that for al1 z e ff we have #.IT > 1, Bt E ff ) k a, then for all integers n k 1 and z E J'f Wehave P=T > n, Bt e ff ) k an .

be a collection of subsets qf fl that (2.3) Monotone class theornm. Let tmder intersection (i.e.,if .A,B G thep .An B E #). contains f apd is closed Let 7f be a vector space of real valued functions on f/ satipfying

ect

Property Blumenthal's 0-1 faw

=

0J.

Proof when

By .the monotone

clmss

z

=

=

k 0 and

z

e Rd,

Erz6Fsot.

theorem,

(2.3),it

'I'IJm(l(tm))

m=1

s

suces

to prove the result

S'

14

Section 1.2

Chapter 1 Brownian Motion

lh this cmse Z = .XIF o 0aj where and the Jm are bounded and measurable. expectation (see,e.g., X e F: and F E C, so using a property of conditional and the Markov property (2.1)gives (1.3) in Chapter 4 of Durrtt (1995)) f.(Z1Fa+) and the proof is

=

XEnLY

p, 1.T-,+ )

o

e F:

XEB.Y

=

If we 1et Z E F1' then (2.6)implies Z ErzkF:) t are h e same up to null sets. At flrst jlance this The fun starts when we take s = 0 in (2.6)to get =

,

0-1 law. lf

e Ft

.

then for

(iv)

.&,

Another typical application

=

is trivial

=

= >Q(1xIJ7) 1.4 = E (14 1.&+)

>.

.&(.)

=

This shows that the indicator function and it follows that .P.(.) E (0,1)

1.A

is a.s.

equal

.s.

to the number #.(X) I::I

Proof

J%('r

inft

K t) k .Pc(#

metric about 0. Letting t

t

.&l'r

k 0 : Bt >

0)

=

> 0) then

1/2

since

.&('r

=

0)

=

1.

the normal distribution is sym-

0 we conclude =

0)

=

S t) k 1/2

limatr tt0

so it follows from

(2.7)that

#otr

=

0)

(2.9) Theorem.

lf Tc = inftt

>

0 : Bt

=

immediately starting frpm 0, it this forces: Bt is continuous,

0) then .45(Tp

=

0)

E'

.

.

.

.

01/2. The 1aw of the

=

=

=

21/2 when

x J)c a.s., so with probability of oider 1/2 at 0.

.

=

1.

for

a

the

complexity

psitive

of

amount

Of course, on top of your name it will write everybody else's name, ms well as al1 the work of Shakespeare, everal pornographic novel, and' a lot of nonsense. precie statement Thinking of the fuction g' as ttr ijnture we can mak follows: ms

(2.10)Theorem. Let g : (0,1) let E > 0 and 1et trt t 0. Then fi sup 0s8s1 Vrn

D

-+

ft)

ttlf you run Brwnian mtion ih two dimensihs of time, it will write your name.''

Rd be a

-+

Botnj

1.

Once Brownian motion must hit (0,x) must also hit (-x, 0) immediately. Since t

'

shows that c (1995))

limsuptlclttl/l/z are not Holder cpntipuou

=

=

when

Comlc Rellef. There is a wonderful way pf expyssing Brownian paths that t learned tromWilfridKendallz

.

,

=

.

Remark. Let 7G@)be the set of times at Fhich the path u) e C is Hlder jy, continuous of order y. (1.6)shows tht #t7v s (0,(R))) = 1 gor j < Exercise 1.2 shows that Pt7G = 9) = 1 for @ > 1/2. The last exerclse shows #t e 7z/al = 0 for each t, but B. Davis (1983) hms shown P(7f1/2 # 9) 1.

t

In words, the last result says that the germ fleld, Fo+ ls triv j a j sus yesplt is very useful in studying the local behavioz of Brownian paths. Until further notice we will restrict our attention to one dimensiortal Brownian motion. lf r

l

' .

=

.

(2.8) Theorem.

(t

one Brownian (/.(./:,)

=

.

(2.7)is

of

In the next exercise we will see that c cx) iterated logarithm (seeSection 7.9 of Durrett 1oglog(1/t))1/2. T(f)

.zz

:

.

.

xerclqej.40. showthat paths

the fact that (ii)(2.6),(iii) e proof using(i) apd if is trivial E (aYlf) EX gives

under

. .

.

=

e Rd,

.$t,

.,4

'

Exercise 2.8. If c < b then with probability one there is a local maximum of Bt in (c, ). Since with probability one this holds for all rational c < b, the local maxima of Brownian motion are dense.

.!5

,

(0' 11

Pr(X) C

the Markov property you can prove ' .

=

a11 z

0-1 Tzaw 15

Exercise 2.9. Let ft) be a function with ftj > 0 for a11 t > 0. Use (2.7)to conclude that limsuptc #(t)//(t) c a.s. where c E (0,x) is a constant.

J-flelds so the two conclusion exciting. i not

6 F:

(2.8)and (2.9)with ..

IZI

complete.

(2.7) Blumenthal's

Combining

Markov Property Blumenthal's

goj

=

<

6

Propf In view of blinenthal's 0-1 law, this if we cn slkoWthat can prov we

(*)

Jan

lf(#) pspsl sup

-

continuous

function

with

#()

=

0,

surely,

almost

p(p)l < e

for ipinitely many n

(2.t), and the staling >

0 for any e > 0

relation

(1.3),

16

.

17

(x7 to the behavior ms (2.11) allows us to relate the behavior of Bt as t Blumenthal's with leads Combining this idea law 0-1 0. to a very seful t result. Let F; O'CB, : s k ) the future after time t the tall c-fleld T nfkn

This was easy for me to believe, but not so easy for me to prove when students asked me to f11in the details. The flrst step is to treat the dead man's signature g(z) H 0. .

0-1 fzaw

Afarov Property Blumenthal's

Section 1.2

Chapter 1 Pzownfan Motion

,->

-->

,

=

=

.6

Exerclse

Show that if e > 0 and t

2.11.

.l% sup i

sup

tlsas

< cxn thn

1.B,l <

6

>

(2:12)Theorem.

0

If

ln doing this you may fnd Exercise 2.5 helpful. In (5.4)of Chapter 5 we C@nthe rader ;nd get the gyneral result from this one by change of msure. a simple direct proof of @)? ,

'

#.()

#.= (.4: c B with FIs (X+U Vr ) :4

0 or #.4.A)

H

t.

Notice that this is stronger than the conclusion of Blumenthal's 0-1 (tJ : tJ(0) E BI show that for .A in the germ law (2.7).The examples J-fleld F1 the value of .&(.A) may depend on z.

=

.4

=

of B is the same ms the gerin c-field for X, it Proof Since the tail J-fleld follows that .J%(.) E (0,1) To improve this to tlze conclusion given observe that E FL,so lx can be written as ls o 0. Applying the Markov property, .

,4

(2.1),gives

.P.(.A)

t)

=

F,

H

,

With opr discussion of Blumeuthal's 0-1 1aw omplete, the distinction between F+> ad F: is no longer importnt, so we Fill inke on iul irprovemet E in our c'-felds aud remove the superscripts. Let

=

#.4A)

E I then either

.,1.

Remark.

witl

'

=

=

.

%Fsm

E

0)

.E'a(.G(1s o p1lF1))

=

.E'.(1s

=

/(2v)-d/2

o

=

expt-lp

-

zl2/2).Py(#) dy

Taking z = 0 we see that if #(44.A) 0 then PvBj 0 for a.e. y with respect again shows and using #.(.) = 0 for all z. the formula Lebesgue to measure, and observe f%(.Ac) 1 f%(.A) that dc T 0, so the last handle E the case To al1 IZI for result implies 0 z. =

=

Sa are the null sets FJ are the completed c-lelds tor #.. Since we do not want the flltration to depend on the initial stat we tak the intersection of This technicality will be mentioned at on ppint in all the completed J-felds. otherwise section be ignored. but can the next and

EzEnz 1s)

=

=

=

.&(.c)

=

.

of Bt ms i 0. y psing a (2.7)concerns the behavior about result beh>yior the information get this ms s+ &., to --+

trickwe

qll use

(2.11) Theorem. lf Bt is a Brownian motion starting at 0 then so is the process defned by Xz 0 and Xt = t #(1/t) for t > 0. =.

Proof By (1.2)it suces to prove the result in op dipension. We begin by observingthat the strong law of large numbers implies X3 0 ms t 0, so X hmscontinuous paths and we only have to check tht X has tlkeright f.d.d.'s. to show that (i) if By the second deqnition of Brownian motion, it suces rimltirit norwl ditribution .X'(tn)) hms a < ta then (I(t1), 0 < tl < obvious) and is with mean 0 (ii)if s < t the -+

-+

...

...

(which

ECX'XL)

=

s.E'(#(1/s)#(1/t))

=

s

C1

The next result is a typical application closerelativ of the one for (2.8).

(2.13)' nn (#t =

of

(2.12).The

Let B j be a on tltmensional Brownian for some k nJ. Then /.4.) = 1 for all

eorem:

j

argument

here is

motion and let

a

.4

=

.

ln words, one dimensional Brownian motion is recurrent. It will return to 0 often,'' i.e., there is a sequence of times @r& 0. We 1 cxnso that Bt have to be careful with the interpretation of the phrmse in quotes since starting from 0, Bt Fill return to 0 inqnitely many time: by time e > 0. Ginfnitely

=

.h/zt

Proof We begin by noting that under #., with mean zlvi and variance 1, so if we use PrBt

<

0)

=

PrBtlvi

<

0)

x =

hms a normal distribution denote to a standard normal, Px

<

-xIVi)

'

18

Chapter 1

Section 1.3 Stopping Times, Strong Afarov Property

Motion rz'ownlan

0) then the lmst 1/2. If we let Tc = inftt : Bt and limt-.x P.Bt < 0) continums paths and implies that for al1 z > 0 zesult the fact that Brownian are =

(2.9)(orthe

Symmetry implies that the lmst result holds for z < 0, while property) covers the lmst case z = 0. Combining the fact that #.IT: < x) k 1/2 for all property shows that PzBt

=

0 for some t

k

n)

Erpnn

=

(Tn <

(:xa it follows that .&(.A) Letting n .&(Sf = 0 i.o.) H 1.

k 1/2

Markov

the Markov

z with

k 1/2

(x))) E ' . ..

. . . -+

=

ut

.4

(2.12)implies

E T so

(T

Proof

lf Tn is a

D

E1

of stopping

sequence

times

Ta

and

1 T then T is

a

K @) = nn(Ts %tJ.

'' ..

(3.4) Theorem. stoppingtiz.

lf ff is a closed set and T

inft

=

k 0 : Bt

'

6 ff J then T is a

Proof Let D(z, r) fy : Iz pl < r), let Gn = u(.D(z, 1/n) : z e ff ), and inf (t k 0 : Bt 4 dn). Since Gn is open, it follows from (3.1)that T let Ta is a stopping tipe. I clpirp that as n 1 x, T,z 1 T. To prove this notice that proke T % lirri Tn we can supjse T k Tn fo# a1l iz, so liin Ta % T. yhat Ta 1 t < &. Sinte #(Ts) e G,z fr al1 (i and #(Ta) Bt), it foltws that j Bj e ff gg :':c:

'

.

UntTa < t).

..'

(3.3) Theorem. stopping time.

k 1/2

=)

P/I% <

(T < f)

Proof

=

-

=

ro

-+

1.3. Stpplg

Times,

M/kv

Strong

Proprty

tj

.

.'

'

g

.

tlme if for al1 We call a random variable S taking values in (0,x) a stppplng SL. of Bt think t) LS life < bring this denitioh To 0 E to as giving the tk price of a stock and S ms the time we choose to sell it. Then the decision to sell before time t should be inemsurable with resfect to the information known at time In-the lmst de:nition we have made a choice between (S < t) and (S K t). This makes a big diference in discrete time but none in continuous time (fora right continuous fltration hj : '

1

.

E

K ) C Ft then (5'< f) Uats K t 1/n) St. If LS < t) e Fj then LS K @J nat,s < t + 1/n) e &. If (S

=

E

-

=

TV second relies below show that

is incremsiftg. The flrst conclusion requires only that t right continuous. that is the fact Ft t (3.2)and (3.3) on when checking something is a stoppinz time it is nice to knw denfl itions are equivalent .%

-.+

-+

'

tht '

'

the two '

.

(3.1) Theorem.

stopping time.

If G is an

open

set and

T

=

inftt k 0 : Bt 6

Bt is coritinuou (T < t) Since G is open and t rational all so (T < @) e &. Here, wherethe union is over =

-.+

,

so we end up with a countable

union.

tJ
Exerclse that Tx

infft

Gk

. . tirfte

:

.

'

.

' :

'

(3.2) Theorem. stopping time.

If Tn is

a sequence

times

and

Ts

t

T then T is

a

.

j

:

e JI.Jis

Bt

i.e.,

a stopping

.

.

.'

.. . . .

.

.

.

:

.

.

.

.

.

.

Show

union of closed sets.

a cotable

time.

Exercise 3.2. Let S be a stopping time and let Sg (z) the largest inte/er % z. That is,

=

((2nSj + 1)/2n

where

=

s4 .

.

=

.

:

'

(m+ 1)2-n it rzj-n .

.

k

.

,-

,

'

.

. .

.'

'

.'

.

'

.

s < (m+ '

L

E

.

1)2=n

E

'

tirn

>ftr S (i.e.,> S). From In words, we stop at' the first time of the form the verbal description it should be clear that Sn is a stopping time. Prov that it is. mints, T), SVT Exercise 3.3. If S and T are stopping times, then SAT maxts', T), and S + T are also stopping times. In particular, if t k 0, then S A t, S V t, and S + t are stopping times. =

of stopping

.

.

.

3.1. Suppose .A is an F,

=

a

we need to use E1

.

Remark. As the reader might guess the hltj f Brel et /, /x = inf (t : Bt E .A), is a stopping time. However, thij tprnj pltt to be a dicult result to prlke d is ot true tes th fltratip is tiri/leted as we did at the end of the lmst section. Hunt was the fiist to plove this. The reoer cn flnd a discussion of this result in Sectipp 10 of Chapter 1 of Blumenthal and Getoor (1968) or in Chapter 3 6f Dllc h t.i e an ii Meyer (1978)We will not about result here since (3.1)and (3.4),or in a pinch thr next result, that worry will be adequate for all the hitting times we will considf

.

Proof

the rationals

GI then T is

(.

.

=

20

Section 1.3 Stopping Times, Stpng Alhrkov Property

Chapter 1 Brovnian Motion

Ta sup T,z, inf n

limsu

,

T,z

,

liminf Ta n

a

n

Exerclse

times. Show that

of stopping

3.4. Let Ta be a sequence

Exercise

R

times

are stopping

Our next goal is to state and prove the strong Markov property. To do this, we need to generalize two deinitions from Section 1.2. Given a nonnegative random variable S(tJ) we define the random qhift os which Quts ofl- the part of before Sulj and then shifts the path so that time Swj becomes time t.'

(pstd)(t) =

t

tJ(,$'(td)

+ t)

h

on on

3.6. Let S be a stopping

(,$'< x) si x) (x$,

tj

On

.

o,j

xc

is

a stopping

=

Two properties

Proo

f

lf

pl

of Fs that will be useful

If S

KT

E .Fs then

(3.6) Theorem.

If Ts

.

tT

Show that

.

.4

Exercise 3.T. Let S and T be stopping times. (i) LS < t), (S > ), LS = t) are in Fs. (ii) LS < T), LS > T), and (S T) are in Fs

(3.5) Theorem.

lkol Herp L is an extra point we add to C to covr the cmse in which th convztio that all Some authors like to adbpt the ath ts hifted away. f ttions hake J() = 0 to tak car of the second cmse. H6weker, We will j.t sually explicitly restrict our attention to (S < XJ so that the second a jj ojthe defnition will not come into play.

S

(

=

let A : Fs

and

time

21

time

(andin .T).

below

are:

thenFs

are stopping

times

n (T < )

(.4n ( s t)) n (T s ) e h.

=

are stopping

C F1'.

times then Sr

=

nnF(Ta).

Proof (3.5)implies F(Tn) D Fw for all n. To prove the other indusion, let n.T1Tn). Since n (T',z < ) e h and Tn t T, it follows that n (T < .t) eC F so .A Sv E) l e .4

.

The second quantity Ss, ttthe information known at time S,'' i a mor ubtle. We culd have deflned

y, so by analogy

=

g(sw(,+r),t

nroc

tittte

k ())

.

The last result is obvious from the

and Exerciss 3.2 and verbal deinition.

=

ne>()cBths-vzj,

(.

something

that

.

-+

t k 0) wor

with.

The next result goes in the opposite frz is increasing and f c'(fr,). n -+

=

to prove

=

The deflnition we will now give is less transparent but easier to Fs

allow us

3.7

Kxercise 3.8. Bs C Ss, i.e., the value of Bs is memsurable with respect to the information known at time S! To prov this lt Sh ((2nS) + 1)/2n be the (x) and stopping times deflned in Exercise 3.2. Show BSn) E Fsv. then 1et n use (3.6).

we could set

Fs

'

:

.An LS K t) E St for all t

k 0)

Jt that lies in In words, this makes the remspnable demand that the part of af time available information with measurable respect to the (S K ) should be and but in the choice made cmse of < between K t as t a @. Again we have useful that and the know is two it to makes disezkence stopping times, this no equivalnt. deinitious are

Exerclse 3.5. When Ft is right continuous, if we replace (5'K t) by (S < ). Eor practice with the definition

of

Fs do

the deflnition

of

Ss is

unchanged

direction from

=

(3.6).Here frj 1

meas

:7Nerc iS@ 3 9 Let S < (x) and Trj be stopping times and suppoqe th>t Tri 1 (x) as n 1 x. Show that Sshn 1 Fs ms n 1 x. .

.

We are now ready to state the strong Markov property, the Markov property holds at stopping times. (3.7) Strong Mazrkov property. Let (s,tJ) measurable. If S is a stopping time then for E=Ys

o

pslFs)

=

Ens4Ys

-+

F(s,

) be bounded

a1l z E Rd

on

which says that

(5'<

x)

and

T xC

,'.hk:,' -)t' '''!i;7'!i;p''' .,'

',' -' y' kq'q' iyjpjd ygd jjyygg,;-jjjyg;;d .::1::,-* kkd ''-j'

p' :1!.5.1.,,* gjl.ttttt)tyjd yjjj,d gjjgy.y)jgr.yjjj.d ggjj,d ll'-tpd rttd ('' t'y'y' jjd lljjj!;d jgjjjjy;jjd jjyj'j' j'11:::,,* jjjj,d jltd j)';j).ii1(jj' j)' j'.rd jjyd )' ;;' ')1(* (' (j)' 1)* )5*jtjjjj1* ))p))))..' )1*-)* yjgjrjd /')'r';$' (')t' llttlj)d i-.'LL' 1* ''' tjd t'p'

rjd rrjd rjjd'...) W. '

.

ltyyljjjyjjij .

:'.

:

' (E. . . E;j i . E. ! El,r)tt .

.i. )qt('. ......!p'hp ;;.g.)jgt.jyy,pg .yj y.,yyy'jr-.r,jy..y.jyy'yyy .

22

Chapter 1 Pzowzlan Moton

section1.3

-

-

Stopping Times, Strong Markov Property

23

.. is bounded and continuous. ''''t,; . .lj....ijjil, Having assembled the necessary inzredients we can now complete the proof. t)))qj . impli-es Applylng special ) .A the '''$''(7.'1:: j case tet e ss. sinces s s,,, (a.5) :! (ijj. e s(sa). r . ;( E.,l.tli.jijl

-

.

.

where the right-hand

)

side is py,

evaluated

Evj'

=

at y

=

Bsj,

t

=

S.

E

.

.

.

.

.

.g.

Remat'k. ln most applications the function that we apply to the shifted path will not depend on s but this flexibility is important in Example 3.3. The verbal description of this equation is much like that of the ordinary Markov property:

'E .:

. E ' '

.rqlr'ljtlytj ('. . yjy -

--'

-

!j i(Il

.. tlt'plftll)jlt ilt) ( yr ..i(!.y):;)rjt). jj)' . !!

)j ytiytttj

EE .

.''

(3.7)proved

. that

above to Sn and observing

(& <

)..t.)(jj;. .t'.t)l) tjjjtl)

x7) gives .

LS <

=

.

.

.

s

=

E

.;iE.y).)yjy

x)

. .

-

.

tthe conditional expectation of F o os given Ss* is just the expected value of Fs for a Brownian motion starting at Bs

qf

-

.. .-.

. Now

as n

.

.

tyj a

-+

.

.

.

x,

o

n fs

.g.

ps a ;

t S,

Sn

< x))

.n(,$k) -+

%)

BLS),

w(.s(&),%)

pya

Fs

.

.

n fs

;

E

.

.

.

< x))

pBsj,

-.+

.

.

.-

ErpBsnj,

=

.

.

.

5')

and

.ltl

Proof We flrst prove the result under the mssumption that there is a sequence of times ts 1 x, so that PZIS < x) E P=S ta). ln this cmse we simply break' things down according to the value of S, >pply the Markov property and E Fs then put the pieces back together. If we let Zn 'Ia @)and =

=

,1.

=

f.(Fs

o

0s

J1

n (S

< x))

=

V E=Zu

n

=1

,z)

#a ; X n (S

:.=:

o

-+

,

'''7'''l't )t,.))),

,

'

t lte b oun ded convergence theorem implles that (3.7)holds when J hks the given in (+). form ' '. j ttt. To complet, the pwof now we use the monotone clasp theorem, (2.3).Let k the colletion of bouded functions for which (2.7)holds. cleprly7-/ is i jl(ll'i . t))q'. and inciemse t a if ys e a vector space that satisnes are nonnegative (ii) t'.tll! chet i collection of sets of bounded y', then '. j;qrjjj!yl'tjl let be the now, yy (i) e '.y!.).t... the foim (u : (a(t/) e Gjj where Gj is an open set. If G is open the function ))!))i l (:111... j :,!pI. decjeasing limit of the continuous fupctionsfktz) ( )) ) (1 k disttz, G))+ )j)'tt'k('. . .j!!)()*411*:1 pt disttz,c) lx : .H. This here is the distance IE/iIII from then G, if .A t z so ; e '. ). . shows aud the desired conclusion holds from .j. (i) cl ollows (2.3). s

,E'.)t#,)t) ' -

-

''

:

E-(E

',tittjtt) ''itlt -

E -' ;jE .E ;. . ;. .'E.t)'tt) .

tnl)

os

o

'1Eil)

.

-zf

,y),;

.

.

.

''./

. EE :.E -

'

.

i

.tt.

'ro

.4

''l:''pp)j

Now if .A e Ss, A n (S (. n (5' K ts)) that the property follows the Markov from it o =

o

ysa

Elitl))

=

(.4(3 (5' K tn-1J)E above -

.F(in)?

E E

(:

EE ' E.i

. .

.

=

-1t:::;;,,'

'''''i7')t '''?ttt

suin is

-

,

.

-

' -- -

E-

-

-

.'

' .Ej.lEj ...).Etjt)t. .r(

EEi(.l.;.,jjlt't..jqj.',Eg. ..

=

E.E'.(.Eb(f-)zk;.

n fs

n=1

=

ta))

=

.G(.Eb(s)'Fs;.n

fs

< x))

.)..)yj y.y. yjjg

=

-+

;

'

.-

.

.

i.- .

.

n

Fa(=)

hst

E 'E

qE. . .

...

.

:

.. E;.Itg gy ' ' jj )! ( . . . .. .. '.;'...l(lli)ILfj.. tttl (. ' '. . '. . ' .!.. . lt'ty ..

,

'E E . !'

-

. .

! : E :

'

. .

-

.

' -. E

.

.

w(z,s)

=

EZYS

=

hsj

dplptz

:'

(z,pl).f(p1)

'Er'..'i-'...(ytj

.. . j..E. . .(

.

.

dps pta-za-,

(pn-1,pn)J'(pn)

>

o

as()tz

c)

>

.c

.

..

.

.

..y

E ...

..

.- .

E.

.

tz,(j

-

'

.

: . .. .

- : . -.-

E

EEq E EEE E 'E -E

!. . . .t'E(/,.k)( -

E: .

E . . . . . .. . . y. . y . .

E. .- . . - E. E. E . - . E . . .q

. .. - -- . . .

''k-'j

-

.

.

.

'?.l

=

.. . - .

q. E: ...

. .

.

.

..

..

,,.

E E. . E;)))t)y)'. E: . . E: E

=

-

.

. : ..

'itt').. .

.

E

g , tE)'tlE'.i.klE,

:.

E-

. .

. -

t, t :.

.

-

. : : E. E.E Eq.... . . );E),.

.

.

,g(td)

-.

.

E... .... -. . E bijji,, .: . . . : i E . :: . .: ; y.. (. . . E i. . .y . EE ,l,EE . .. -.. . q.. E -

i .. .

.

. .

..

l'll''i

-. . . -. . E E 'E

,

-

.

- .

. . . : y - E. . y. . . y .y . .

.

E::. ..

y.-i),),.

-

..)iyt.)(j

.

to op, cjsx,)

;::jI:;,.

.

. .. . q) Ey

q

that

=

' ji. . . . . Taking the expected value of the lmst equation we see that . .' jt) y .-( .: . ))' ( j.-. .''jiilittty.t o p y., '(..' () tl'ttttj op,> () sr some rational t) 0 ; (.. ;y ... ; )jiki f'tj'jj($-. . .q. .. y.....y . ' . ( r . . . E.. .. i '.'.'').l:jy!:[email protected]('.j;.(j; .J(tJ) rq.rE, .. jh';;j(jjjgy';lll'r, with probability > it follows that From this ne, if a point u e (t : ',iE,t..t,,. r'i'l)r'i tz) ! js rational isolated the left thre is that < t on so a 1!)/)((1 st(w) c) (, n (i.e., @ . . (C;'.'.;il.)r.;@i:)'. .....E)y.. q. = j) then it is a decremsing limit of points in J@). This shows . that tlte .1...j y ( . y (. ..b3jl . . z(w) . ': pl i'). closed set z(w)has no isolated points and hence must be uncountable. For the tit).t jmst and step ; ! ( ) Hewitt Stromberg ) 72. ( pyg see .'.7771''1 . (1969), rr'. ( ry t ' ;ll!:t')pjr. !.i.. If we let lz()I denote the tebesguememsure of then Fubini's theorem . '.-'.. ri..j!:.k jy.(.... y.y.j(.jy.. .iyjpjjt'j. ..yEl/tt)lt))y, . . ) . )( j....y..y y.yy y ; Ltn . jji... j jy..j.... .. .-.....';;Lj,j$-q;3.;qL-1L.L 4(::)1 4(::), i!r .-..)... 41:) 4(:4 dr''..'lk 44:4 ..;!!iEiI, rqEilr''-'ljl p(.-.. rj,jlll (2)) 1(:21) 11 (zliIIII 11 1(:). -.-qyq!)'tri((t$r'. (;'..i;. i. .r . ) . ' . .(.y.j.y(.j..y.r.q.,,.,y.,. (.. . .ttljjijjy/yty )j ljtjj gy,jyjgyjgj ; j . ..yy j.yj jty'jjyy. jj.y. j.y..jyyy.j g y jjr y. j.yjy...y.yjy. . ........yj.(yyj.j'. j j(.jy qjgj..(.jy. ..gy .j.yy..j yrj. '...(.((1y)(j).)t'tjjy'jltjlphtr (r. .(y.yj.y.y......jjy...y..jj . ....y.yy.y(y yg. j.. jy yygj .y ! jjj ,Et

From this

q

E'

.

...','i!kjjll

and induction it follows

.

.

E f't))y E

y)E.

E

'tt

is continuous.

.

.

p

E

-

j jy . . ' yjjj

.-+

..

E-

.

.

..

.

=

E.. q EE)t))l)).j. . q! . . . . .

'

and continuous.

.

.

.

are real valued,

.

E'.

-

bounded < ts and A, where 0 < l < fn If f is botmded and continuous then the dominated convergence theorem implie that dylhz, ylfyj z .

. ..

=

=

q

-

.q.yy

.

.

.

=

.

- -

'

m=1

.

.

.

=

E. s'''jjjjl'. . . . . . ..'.iljtljtyj )qyj(1 j(' ..' .'.

1-fJm(a'(m))

..

.p.(a

-

::

=

. ..

-.

t'[email protected]

=

y ..

.

.

',''':)it) sxnmple3.1. zeros o Brownlan motlon. considerone dimensional Brow. q)jtij nian motion, let Rt 0) and let T'n inf (tz> 0 : Bu infttz > t : Bu 0). trytjyilypj., and x) strong implies Bt) :i k;. < the Markov 1, 0 ; . .. ) so xow(2.1a) ' rttyty lir: )1.)! . ..propeity and .. ..... . imply . . .. .tt'ytlll. . . (2.9) '.!'f)j) l -

+ 1)/2n where (z) the To prove the result in general we 1et Sa ((2n5') largest integer % z. In Exercise 3.2 you showed that Sa is a stopping time. To (:x) we restrict be able to 1et n our attention to F's of the form

.

..

.

...

.

....jj .....,...jjyjy.

E

).E.y...-E.....t.q. ....

-

'

E-

'. E . E- .. . - - . . . .. E ''l lEi!i'j!')itt)@.:...(...'...E....('E.y..(.'...';.',E. . .. . E..iEjlEji.) .), .

. .

--

. .

-

q,

...

E

ig

-

E

i

y

. y

-

-

y .y y y . .y . --

.

qE.. y. ..- g E-

!-

( .- ilj ! E.

-. .

-E

qjjjjyy

.

-

yy . y .

! y.y . . E.y. E q . q - y tid-r'jlqjj.. y.yy.-.y. y...... y ; y.;. g yy -

. -

-

.-

!q

ii

.

-

.,.

.

-

.

j .

,

-

j ggj. yy .

. g

,,.

.-illiiiir':;:,.

.-;:ii!:7

dt:;i!r

.-ilr:::'':,r:-

'i,,.-,-

,,

::::::::::

:::::::::.

.

-

::::::::::

24

Secton 1.d

Chapter 1 Brownian Motion

So Z@) is

a set of measure

In

zelo.

.

..

g

.

=

u(z)

(3.8)

.J%(Tc< t)

2#(,(#f

=

> a)

Intultlve proof We observe that if Bs hits a at some time s strong Markov property implies that Bt .5(L) is independent pened befpre time Tc. The symmetry of the normal distribution c) 0 then imply =

Intultion Since D(z, J) C G, Bt at Bs. When Bs = p, the probability hOW B %0t to y. Proof

of exiting

F

G in

,4

E

To prove the desired formla,

erty, (3.7),to

qrst exiting Dk, J) is nyj independept of

1(s c' Ek)

.

'

Markok propr

strozk

Will apply the

we

=

,

exit G without

cannot

(3.9)

.&(T.

.

.

.

. . . ..

.

'

J%(T

(

<

=

o

pslfs)

ltsz:x) the previous display, we have

Using the deinition of

u,

u(z)

=

.G1(s.Ex)

= E=Ez:

o

.Eb(s)1(s.ux)

0s

=

(1(s.Ex)

=

ltswex)

,

ps IFs)

taking expected

and

)'

,

vlue

=

=

E=

j

the function F that we time S.

allow

< jj

Bs

=

a op

which proves Ekercise then

(3.10)

(()

Bt

> c)

have c)

2.J%(#j >

=

.F, the

=

(

=

< f, u2(t

:

.'. ..

.

< x)

and

#nIL

<

. cl with. inf ,

XJ

=

Markov property on LS <

w(c,s)

1/2 if

=

t, Bt

s) > a

-

=

on LS <

vBs,Sj

=

..

inf (s < t : B,

strong

(3.9).To

othrrwise

1(sf>c)

=

fntys o pslls)

ds + uBs)

shows why we want

) we

(: (T. <

1 if s

=

that if we 1et S

If we 1et wlz,s)

(

to Our third apply to the shifted path to depend on the stopping application

so

Fstps)

No

gB'j

.

of

3.10. Let G, T, D(z, 6j and S be s above, but now suppose Exercse = all y for that EyT < x e G. Let g be a bounded function aud let u .E',I :(0,) dst. Show that for z E G

u(z)

.

ErktBst

S

(I)

2J%(T

=

5',()

We do this

o ps)

f.(1(s.ex) o

nBsj

(#t >

1

j..&(Tc < t)

=

Proof To make the intuitive proof rigorous we only have to prove this from the strong Markov property (3.7),we let extract

=

=

,

.

=

=

)

> a)

t Bt

<

Multiplying by 2, then using g

..

To check that this leads to the right formula, we observe that since D(z, J) C G, cos = Bv and ltsv.uxl o we have Bv os 1(swEx). In words, o and the shifted path eswmust exit G at the same place. #.IT < x) 1, it follows that Since S % T and we have supposed strong Markov property now gives 1. Using the P.IS < ) .G(1(sv.Ex)

< t then the of what hapand PzBt =

-

EruBs)

=

princlple. Exnmple 3.3. Refetion Let Bt be a one dimensiopal Brownian motion, 1et a > 0 and 1et L = inf ( : Bt cJ. Then

=

=

-

25

=

.

inftf : Bt / Gj, Exnmple 3.2. Let G be an open z 6 G, let T of G, ad 1et x) 0G, the boundary and suppose PSCT < 1. Let d (7 P.CBP chosen claim D(z, J) = = that > .A). that if 1et be so u(z) 0 1 E we ; zl infff D(z, J)), and : Bt J) S < then let k 0 fy lp / cG set,

Stoppng Tmes, Srong Markov Property

> c)

=

s <

.E(1/2

(Tc

. ..

.

<

. .

0

=

cxl

then

t)

implies

XJ (:r'.<

t)

=

t, so ; Tc < @)

(3.9).

3.11.

Generalize the proof

#OIT.

<

t, u

<

Bt

<

1?)

=

of

(3.9)to

J%(2c

-

conclude

n < Bt < 2c

that if 'u) -

u <

1)

i

a

Capter

26

To explain our interest in P t)(M

-p)

(u,

f%tMf Diferentiating

> c,

with

(3.10)ilet

Bt

> a, u <

Letting the interval

Section 1.d

Motion

1 frownan

< r)

Bt

z)

=

:ve

f%(#

=

to c now

respect

we

'

Pn (Mf

11)

c

=

,

Bt

z

=

)

17

-

z)

-

k 2a

<

-

Our next step is to 1et t s/lr + into one over t e (0,s). dt = weflrst rewrite the integral as

it ms

1 =

-slr

u)

2%

c

'/r

(aa-mjayay

-

2% a

e

#c(f, S

s)

c'=

We have two interrelated aims in this section: the first to understand the bec) for a one dimensional Brownian havior of the hitting tims T. inf (1: Bt second study motion #; the the behavior of Brownian motion in the upper to R,d = : 0). We begin with T. and observe that the > half space .I zd (z E principle reflection (3.8)implies =

=

.

P n(Tc

t)

<

::c?

215(#t

> tz)

2

=

J

.

Here, and until further notice, we are dealing with a one dimensional Brownian motion. To flnd the probability density of Tc , we change variables z = 1/2c/s1/2 1 dz = -tI2a(2s3I2ds to get 0

(4.1)

Jb(Tu < t)

=

)

t

=

0

d. (-t1/2c/2s3/2j

(2*)-1/2exp(-c2/2s)

2

(:

A's

:3

-1/2

)

a expt-c

2

/ :s) ds

=

=

supt

K1 in Ekercis

Cr

%s) .POIf.

.

. . .

(

1 =

.. .

.

:

(1

u.)

-

x

t-

,

az

?n u -

..

au

to intuitih,

=

-

Q

> 1

s) dz

-

zr

=

2 1

= =

0

Co

-

-

g' 1

1-,

'm

1-,

co

1-a

:$)-1/2 r s/tr

(sr

z exp

+

s)

(2:rr3)-1/2exp(-z2/2r)

(-z2(r + dv

r.

N

=

'

-

(t(1 -

A'

0

t))-1/2

jbr 0 < t < 1

.

is symmetric about 1/2 and blows up near 0 and 1. This is one of two arcsine laws associated with Brownian motion. We will encountr the other one in Section 4.9. The computation for R is much emsier and is left to the Teader. ( (

.

'

.

:

.

. .

Exercise

'

..

: .

Show that the probability

4.1.

,.

density for R is given by

:

P 0 LR = 1 + t) Notation. PT t) =

.

.

.

' . .

=

=

1/(*1/241

+

j))

for t k 0

ln the last two displays and in what follows we will J(t) as short hand for T hms density functio ft). .

.

t

.

Fl

drdp

s)/2rs)

d dr

often

write

.

'

' ' '

' As our next appllcation of (4.1)we will compute the distribqtiop of Bv where r = inftt : Bt / S'J and 11 (z : zd > 0J and, of coprse, Bt is a d dimensional Brownian motion. Since the exit time depepds only on the lmst coordinate, it is independent of the flrst d 1 coordinat aztd we cart compute the distribution of Bp by w'riting for z, 0 E Rd-l ) p 6 R .

*

(sr3)-1/2 ()

to get

the density function of L

-

exp(-z2/2s) (2xs)-1/2

z

.

arcslnjvsp

=

p,(0, z).&(%

=

112

=

N.

Using the lmst formula we can compute the distribution of L Bt = 0J and R inft k 1 : Bt 0J,completing work we startd 2.3 and 2.4. By (2.5)if 0 < s < 1 then =

.!actz, t) =

pith t

,

V

2

-

dv

t))-1/2 dt

(t(1

()

The reader should note that, contrary the lmst 0 before time 1,

.

dz

and thep again

'm c

'

(2>)-1/2exp(-z2/2f)

above

-

';r

1.4. Flrst Forrnulas

s

(r + s)2 '

-

-(

1/2

s) a

*%

1 =

-

r s

1-,

27

to convert the intejral over r E (1 s, x) emsier + slz#r, so to make the calculations

(r +

Changing variables ms indicated

.z;

=

t:r

1 -

a /2f

-(2c-/)

s)

=

get the joint density r(,yy

13

.h

see that 2c

=

B, to rwrite

maxcs/i

=

f%(2c

=

shrink to z

Mt

First Formulas

P (r,&)'B

r

=

o 0)) '

dspn,vtr

=

'S)(2=f)-(d-1)/2e-Ir-#IQ/2J =

0 =

# (2T)d/2

-(ja-#j2+pa)/2a

-(d+2)/2

0

ds s

e

by

(4.1)with

c

p. Changing variables

=

Jc -(Iz ).

#

(2T)d/2

so ve

Section 1.4 Flrst Forznulas

Motion

Chapter 1 lrownan

28

s

-

#1a + !/a ) dt 2t2

-

pj2 + y2)j2t giyes

(lz

=

(a+2)/2

2f pl2 +

lz

-

p2

By induction it follows that n

n-f

Pr,vlBr

(#,0))

=

=

( lz

pjajz

.j.

vqdjgj gd/a

r(a)

=

.L*pc-le-Vdg and :

Proof

.'

The first

c

k 0Jhas

T:

o

-

Tc)

T5

=

-

Tc

'

then

a < ,

(3.7)and

the strong Markov property

so if 1 is bounded and measurable, lation invariance imply

ll'b

0v.

,

...

independent increments.

stationary

is to notice that if 0 <

step

Ez

.'.'

. .

Under f%, (L,

(4.4) Theorem.

1.f.)

=

=

Ez (J(T) E T(T) J

o =

to

.E'(T(A Za)lG) -

G =

:

aud that for each bounded mea-

.E'.f(Z-,)

.

and meeurable. g

n

E

czz'z nL

(4.6)

So consulting Section 2.7 of Durrett (1995) we see that Tu hms a stable 1aw with index t:e = 1/2 and skewness parameter tc 1 (since Tc k 0). of the Cauchy distribution in the hitting 10To explain the appearance cj of a cations, let n = inft : Bt1 component (whereBI is the second application motion) observe and dimensional another Brownian of that the two strong Markov property implies =

z-,) I-fyftzt; i

(4.7) Theorem. increments.

=

+l

f =1

f(r,),

=

s

k 0J has

independent

stationary

(4.7).

(1.3)and

an obvious symmetry

c d sc,

( 8)

4

c, Jc

-c,

,

.

imply

we can conclude that % hms so again consulting Section 2.7 of Durrett (1995) the symmetric stable distributiou with index t:v 1 and hence must be Cauchy. To give a difect derivation of the list fact let isj E (exp(pC,)). Then (4.7) and (4.8)imply

'

(

E

vevett

=

::1:

(pes

vesj

+ f),

5:

wp,(1),

f&(s)

=

e-os)

(p:(1) implies that pos) is real. Sirce 0 is coz!continpos nd yu, (1)is po tinuotzs, the secod equation implis s a simple argument (seeExercise 4.3) shows that for each t, /p (s) ekpt:s), The last two equations imply that c, for some sc1 >nd c-# = c: so c: above apply equally Sipc tlie arguments s, so Ca has a Cauchy distributin. well to Bt1 cBta ), whre c is an# onstant, we cannot determine lc with this argument.

-C,

-+

,

-

zi-z )

.

Efrtzt-v

-

z--z)lJl--,) .

E .

n-1

1-1hzt

(Ca

.5,

.-+

n-1

E

Under

The scaling relation

The fart that Cs

=1

FI'fizti

.

=

-

=E

D

::i

,

.

conclusion.

=

< t ,, and 1et h, 1 K i K n be bounded Let t () < t 1 and using the hypothesis we have FtvvConditionin: on .

jj

independent increments.

Then Zt hms stationary Proof

.

be useful later.

will

which

trans-

:

f2k(Zf;-t;-z)

=

(1.2)intplies

The scaling relation

Exerclye 4.2. Prove

0r. 1.T.) .E/(T-c)

The desired result now follows from the next lemma (4.5) Lemma. Suppose Zt is adapted surable function 1

Z;-,)

=1

which inplies the desired

is the usual gamma function. = probabilists should recognize this as a Cauchy = 2 0, z When d the glance, fact that Bv hms a Cauchy disyribution might flrst At distribution. be surprising, but a moment's thought reveals that this must be true. To explain this, we begin by looking at the beh>vior of the hitting times (Tc a k 0) s the level a varies. where

-

=1

y #jz

-

rl

/(Ztk

E

have

(4.3)

-

Zf-,)

29

E1Zt=-t=-j

=

-xlpl

i.=:

..:

'

E

,

'

=

'

.

.' 30

Section 1.4 First Formulas

Chapter 1 frowzlan Motion

Exerclse 4.3. Suppose w(s)is real valued contirmous and satis:es w(s)w(t)= ln(w(s)) to get th additive equation: #(s) + ps + t), w(0) 1. Let #(s) for equation = #(t) to conclude that #(m2-n) m2-n#(1) #s + ). Use the @4(1).. continuity extend and this then use 0, to #() to all integers m, n Exercise

since

=

=

hy)

=

0 for y

k 0. Combining this

E=JBt)

=

)

7% >

=

Adapt the

4.4.

for po (s)

argument

.lclexpt-r,zll

=

( . . :

..

=

yo shpv

E (exptC4l)

=

The representation

The lmst formula generalizes

exptc-cxi)

Exerclse

4.5. lf u <

Ekercise

4.6. If u < r then fils

a

1)

then

.&tc -+

-->

'p))

Tc is discontinuous i

('tz,

Cs is discontinuous in

(u,r))

:i=

1.

W here

=

1.

rxerclse

Eirt

'tz

-

,

. . .. . ' . .

.

.E..'

.

.

..E ' ' '

.

.

.

'

'

..

The discussion above has focused on how Bt leaves H. The rest of the sedion is devoted to studying where Bt goes before it leaves H. We begin with the case d 1. =

(4.9) Theorem. where

If

z, y >

0, then P=Bt #)

#f (z,

(2A) -1/2

=

t)

y, Tn >

=

pjtz, y4

=

-

-p)

pftz,

-(p-.)2/2f

d

The pqoof is a simple extension of the argument we used in Secti Proof wlth .J(z) when 1 3 to prove that Pa(L K ) = 2#c(#, k c). Let J k 0 Clearly 0. zK (

f7.(T(.&f);Tn

>

t)

=

ErlBt)

-

.

.E%(.f(lt);%

S f) nd

If we let J-(z)= T(=z), then it follows from the strong Mrkov property symmetiy of Brownian motion that

>Q(#(.R);T'tlK f)

l

E=LEz1Bt-n)3 T'n S fl ' .Eb(.E/'(.t-v.c);T0 ; t) = .E%(/(#t);Tn E=1-Bt44 K g.

=

=

(

=

: . .

.

.

. . .

'

' .

'

' .

.

.

=

#rl.R

-+

-+

Exercises 4.5 nd 4.6 Hint. By independent increments the probabilitie scaling implies that they do not depend on the but then only depend on size of the interval. 'p

Let r

that

of the Cauchy process given above allows us to see L and s Cs are very bad.

paths

)

-

the frst

equality

E=hBt)

jptz,-pt(z,-p))J(p)#p p)

'' . .' .. E j. . .. .

. ..

(4.10)Theorem.

its slmple

E=1Bt

=

with

y- =

(p1 ,

4.7.

.

inft =

.

,

pd-l,

Prove

:

(4.10).

to d k 2.

Btd = 0). If z, y E H,

ll, r > tj

-pa)-

.

easily

=

pttz, #)

-

ptlz,

#)

shows

r

.

' ...

.

:'

..

.

' . .

.

.

. .

2 Stpchastic

Integation

J!

.Jfadx, To motikate Ip this chapter we will deflne our stochastic integral h o at time s and ffa price of stock a the developments, think of X, ms being the short). The which negative number of shazes we hold, may be (selling msthe relative proflts at rfresents time t, the et to our wealth at itegral h then of the inttal of chae check this note that the inqitesimal rate iime 0. To of shares we stock of = number of chge the the rate times the dIt Ih d-'' d ( hol jyjtjoduce the integrand H aj the p rdttable ( It th flrst setion yj# e w jj) version of the nti PrOCeSSeS, tht ttke number of stia#es a nxa ihematical stock and not on the future of past behavior the held must be bmsd on the will nd sectiori oi intidu intgiatis X the we perfoymance.I t h sec Irtinales COIl fair Ihtuitikely local iiiitingales.'' are tinus games while t tlocal'' Tequkqments reduce refers integrability that fact the the we to the to admit a wider clmss of examples. We restrict oui attention td the cae of simpler continuous paths with have Xg pprtingples thep'y. a to t =

.

=

q

.

E

N

,,

,

1

,

-+

.

2.i. .

Integrands:

'' . '

.

' E' ' . '

Pred 1c table Processes EE E .

:

'

.

.

.

.

. .

To motivate the clmss of integrands we consider, we will discuss integration Here, we will assume that the reader is familiar w.r.t. discrete time nartihgales. With th basics of martingale theory, as taught for example in Chapter 4 of However, we will occasionally presept results whose proofs can Durrett (1995). be fOMn d there Let Xg n k 0, be a martingale wa.t. Fn If Il, n k 1, is any proess, we define tan .

.

,

n

.

.l

'

X)a

Hmxm

=

-

Ak=1)

m=1

q

th lst formula and the restrictioz we ar abot to place on the will with example. Let &,fa, consider be indpendnt cocrete H mi we =t) k-l-a. = = = symmetrlc = 1/2, = and the 1) P let Xa Xn is P &+. , &). slmple rnndom wnllr and is a martingale with respect to Fn = o'(1,

T intivate

.

.

.

.

.

.

.

34

Capter

Section 2.1 Integrands: Predictable Processes

2 Stochastic Integration

If we consider a person iipping a fair coin and betiing $1 on heads each time then Xu gives their net winnings at time n. Suppose now that the person bets an amount Hm on heads at time m (withHm < 0 interpreted ms a bet of -Hm on tails). I claim that CH X)a gives her net winnings at time n. To check this note that if Hm > 0 our gambler wins her bet at time m and increses ltr 1. fortune by Hm if and only if Xm XmThe gambling interpretation of the stochastic integral suggests that it is natural to let the amount bet at time n depend on the outcomes of the flrst n 1 flips but not on the flip we are betting on, or on later flips. A process Hn that has Hn E Jk-1 for all n k 1 (here (0,t'1), the trivial c'-fleld) is said since its value at time n can be predicted (withcertainty) to be predictable at time n 1. The next result shows that we cannot make money by gamblint on a fair game.

Elnmple ar random

Npdop

1.2.

(

F, (f), with

P) be a probability space on which there is defined T PT < t) t for 0 K t K 1 and an independent = with P 1/2. Let 1) = P4 q= =1)

Let

variable variable

35

=

=

.

X

-

.-f'a

=

Let Xn be a martingale. .X')n is a martingale.

is predictable

If Il

.

..

.!tl

aj

u..

l

yt

) . . .

'

.

Proof lt is easy to check that CH Xla 6 fn The boundedness of the implies E 1(S' X)a l < (x7 for each n. With this establiqhed, we can compute conditional expectations to copclude .

.

=

k

'

-

z.=

=

-

since .Jfrj+l E Su and .E'(Xrz+z Xn 1.Tk)

=

'

0.

,

t

The last theorem can be interpreted ms: you can't make money by gmbliik on a fair game. This conclusion does not hold if we only assume that Hn is optional, that is, ll 6 Fn, since then we can bmse our bet on the outcome of the coin we are betting on. Exnmple 1.1. lf Xn is the symmetric and Hn = a then

ll sincefm2 1 =

.

.x)s

=

(2

=

.

j'

1

=

.

.

To check this note that $he memsure dX, corresponds to a m>s of slze >t T and hepce the iniegral is times the v>lue there. Npting nw tltat 'Fn = 0 while $ 1 we see that h is not a martingale. o

simple random

walk considered

above

n

)-7

m

.m

=

'

The problem wtt te lmst example is th same as tlte problem with the our bet can depend on the outcome of the event We one in discrete time are betting on: Again there is > gambling inteypretation th>t illustratrs what fter thewheel is spun and the ba is wiimz. Consider th game of roulette. rolled, people can bet at any time before (<) tiie ball corries to rest but not is yfteq(k). One way of guaranteeing that our bet be made strictly befpre T, condition, is to require thp,t the amoupt of money pe have bet i.e., a sucient contiuuous. lett This iplies, fbr iustauce, that we cappot ieact a t time t is instaptaneously to tpe advantage of a jpmp. in the prpcess we are betting on. The simplest left continuous integrand we can imagine is made by picking < b C E Sa, and setting a -

CH .X')a + f(Jfa+1(.Xk+1 .Xk)I.'&) CHX4. + ffn+1f(an+1 .Xkl.&l H aY)n -

xk

=

0

,

.

.

.X)n+1l.&)

1*

Xs d-''a

=

:.::.z

.

EH

tz

=

.

and each Hn is

,

1et Ft (rlXa : s S t). In words we wait until fime T and then flip a coin. with respect to Ft : However, if Xt i. a martingal we deflne the stocastic itegral h X,#.Xa to be the ordinary Lebesgue-stieltjq integral then Apd

-

(1.1) Theorem. bounded, then

0 t < T :ti

=

=

-

n

m=1

(:I

ln continuous time, we still want the metatheorem ttyou can't make money gambling on a fair game'' to hold, i.e., we want our integrals to be martingales. However, since the present (t) and past (< t) are not sqparated, the deqnition of the clmss of allowable integrands is more subtle. We will bgin by considering a simple example that indicates one problem that must be dealt with.

1

,

(+) E' '

Hs-w) .

tn words, i.s., C e

. .

=

c()i(c,5!(t)

. . ..

. .

'

:

'

. .

.

.

l

'

buy c(=) sares of stock at time a bmsed op our knowltke ten, F. W hold thm to time b and thrz setl thern all. Clearly, the eyamplesin (+) shoulb be allowable integrands; one should be able to add tro possiltities in the previons more o these, and take limits. To encompass jthe .... . . . J-fleld smallest all containmg sentence, we let 11be the sets of the form (c, ) x calld c-fleld. where .A e Sa. The 11 we have just deflned is the predictable We Fill dmand that our integrands .I are measurable F.r.t. ll. In the previous paragraph, we took the bottom-up approac jo t e e nition of lI. That is, we started with some simple examples and then extended the clmss of integrands by taking yums and limits. For the rest of the section, we approach. We will start with some natural xequirements will take a tttop-dowq''

or

we

.

:

'...

.

.

36

Capter

2 Stochutic

Section 2.2 Integrators:

Integration

and then add more until we arrive at 1I. The descending path is more confusing since it involves four defnitions that only difer in subtle ways. However, the reader need not study this material in detail. We will only use the predictabl J-qeld. The other de:nitions are included only to allow the reader to make with other treatments and to indirectly make the point that the connection the questions mssociated with continuous time prpcesses can be theoretic memsure diult. quite Let & be a right continuous flltrytion. The frst and most intuitive cntfdepending cept of on the pst behavior of the stock and not on the futpre followilg: is th formance'' per '

,

) is said

Hs,

to be adapted

if for

each

we

have Ht E

&.

.

measurable if for eah t the mapping H is said to be progressively memsurable. (s, td) Hs, tJ)' from (0,t)to R is x&

is to use '

new only

=

-+

-+

e

The distinction between the optional and predictable . ta nt for Brownian motion since in that cmse the two fact is that, in general, 11 C X. -fields

c-felds is not imporcoincide. Our lmst

Exercise 1.1. Show that if Hs, tJ) 1(c,,)(s)1#@) where .A 6 Sa, then .I is the limit of a sequence of optional processes; therefoxe, H is optional and C

.

2.2. Integrators:

Contlnuous

Local

Martlngales

In Section 2.1 we described the clss of integrands that we will consider: the predictable processes. In this section, we will describ our integrators: continuous local martingales. Continuous, of course, means that for almost every Y, the sample path s Xa @)is continuous. To define local martingale we need notation. If T is a nonnegative random variable and is any process we some define YtI'At 0: (T > 0 X1.' 0) () on (z =

(2.1) Defnltion. X3 is said to be a local martlngale (w.r.t.(.T, t k 0)) if there are stopping times Ts 1 (x) so that : is a martingale (w.r.t.(--#-ag'. t k 0)). The stopping times Ta are said to reduce X.

XI=

'

As we will now show, the ' one. We have added the

Proof Since all the processs used to deflne 11 are left continuous, we have 11:. 11 (:: To argue the other inclusion, let Hs, u7) be adapted and left continuops and 1et Snts,l .Jf(m2-n,) for m2-n < s K (m+ 1)2-n. Clearly Il'n 6 H/. Further, since H is left continuous, (:l Sts, tJ) as n Hns, (J) x.

=

c-feld Accordipg to Dellacherie and Meyer (1978) page 122, the optional J-fleld f and in the case of the natural liltiation contained in the prgressive Brownian motion the inclusion is strict. The subtle distinction between the lmst two c-fields is not important for us. The only purppse here for the lmst tW de:nitions is to prepare for the next one.

Let IP be the continuous adapted processes. if Jfts, tJ) e 11/.

11= 11/.

''i

of of subsets Let A be the J-fleld (0,cx)) x fl that is generated by the continuous and hav left limits, i.e., right that adapted processes are (r-fleld all of which makes smallest these the processes memsurable. A (.J) memsurable w.r.t. A. optional is said if Hs, be is to process H

of subsets

37

-+

ef?-

--,

tr-field

Local Martingales

=

We encountered this notiorl in our discussion of the Markov property in Chapter words, it says that the value at time t an be determined frin the 1. In The definition above, while intuitik; is iit information we have at time prodttct sf ae with enough. We are dealing a function deflned on strong of the two mesurability about function need co) X t-1,so we (0, to worry as a variables.

tsmodern'' apfroach This is a reuonable de:nition. However, the J-flld. smaller a slightly diserent deflnition that gives us a slightly

(1.2) Theorem.

Contuous

by the left (0jx' )xo that is genezated predlctable said A process H is to be

of

'

deflnition of predictable is the same as th old and proof possible. to make the next statement

0) to deal with the fact that A'a need not e We need to set XT> 0 on (T integrable. In most of our concrete examples is a constant and we can take T1 > 0 a.s. However, the more general deqnition is convenient in a number of situations: for example (i) below and the deqnition of the variance process in =

a'

(3.1).

In the same way we can defne local submartingale, locally bounded, locally of bounded variation, etc. In general, we say that a process '.F is locnlly A if there is a sequence of stopping time T'a 1 (x7 so that the stopped process XT hms property .A. (Of course, strictly speaking this me>ns we should say locally a submartingale but we will continue to use the other term.)

Why local

Section 2.2 Integrators:

2 Stochastic Integration

Capter

38

(2.2) Theorem.

martingales? working

for

are several reasons martingales. with

There

with

local

k 0). is

'

'

a martingalek

show: (

.

(2.3) The Optional Stopping Theorem. gale. lf S K T are stopping times and Xht 1. t en

Proof (7.4)ih Chapter 4 of Durrett (1995)shows that if L and times kaa is a uniformly integrable martingale w-r.t.

'

ftyzTlfzl

,

is meaningless, since for large i X' ( i pot However, it is trivial to deine a local mart gal y:

then the concept 'r deined on the whole space. there are stopping times Ts

1r

so that

.

.

.

'

.

.( .

. .

'

.

=

.

i .'

(

' .' . .

.

E'

%M n

are stopping

then

'Fz

To extend the result from discrete to continuouq time 1et Sn ((2n51 + 1)/2n. Applying thp discrete time result to the uniformly integrabl marting>le Fm = (x$ Xzamz-a With L = 2 Sa and M w s that =

=

times a (iii) Since most of our theorems will be proved by introducing the proofs ar no to reduce the problem to a question about nice fnartigales, harder for local martingales deflned on a random time interval than foi ordinary martingales. stopping

'

..

Lei X b a continuous local mrnis a uniformly iptegrable martingale

ot mrtihgale

< x,

39

.

(i) This frees us from worrying about integrability. For exanple, let X$ be a martingale with contirmous paths, and let p be a convex function. Then pXt) is always a local submartingale (seeExercise 2.3). Howver, w can cnclude that ph) is a submartingale only if we knoW .E'lw(.X)l < (x7 for each 1'$'a fact that may be either dicult to check or false in some cases.

(

,

First we need to

th

rather

martingales'

(uYv(f),.T-v(f)t

Con tinuous Local Martingale

Remson (iii)is more than just a feeling. There is a construction that makes r) it a theorem. Let Xz be a local martingale deflned on (0, and let Ta 1 r be a sequence of stopping times that reduces X. Let To 0, suppose Tz > 0 a.s., and for k k 1 let

(

'

'

.

.

.

.

.

'

.

'

.

'

.

X'xr llsa) = Xsa Letting n

(x7

..--.y

expectans

.

r

;

the dominated convergence theorem for conditicmal

and using

((5.9)in

.

.

chaptert

te (1905))

of Durrtt

foltows.

resutt

(:I ,

-almost

=

n'(f) To understand

=

t

t

-

n

(

1) Tk-l + (k 1) K t + k 1) t -

-

n

-

s s

this deflnition it is useful to write it

1

t

2

-/(@) z2 -

-

out

.

for 1-

=

'

f(-Ygkar'I.'G(,))-'G(,)

=

=

. .. .

.

,.

'

. .

P roving the desired result.

1, 2, 3:

The next result is an ing local martingles are

/,,n

(:l example of continuous.

the simplcations

that

come

j

rotn assum'

(2.4)Theorem. lf X is a continuous local martingale, we can always take the 1> n) or any other sequence sequencewhich reduces X to be Ts = inftt : 1-'V T < Ta th at h as T 1 cxnas n 1 x

o

IS71+ 2, Ta + 31

.

. g.

. ,

. .

.

:

.

In words, the time change expands g0, onto (0,x) by waiting one unit of time each time a Ts is encountered. Of course, strictly speaking y compresses (0,x) to be deflned for all t k 0. The reason onto (0,m) and this is what allows .;G(f) change with the time for our fscination can be explained by: 'r)

=

f(Xv(t)I.Tk,))

+ 1) (T1 + 1, Ta + 1) go + 1, + 2j (Tz + 2, Ta + 2)

T'a

=

1)

zk+ k

(az',l

f n t -

=

K Tk + (

Proof of (2.2) Let n + 1. Since v(t) S Trz A n, using $he optional (@) gives Exnhn lFv(t)). Taking conditional stopping theorem, (2.2), Xv(f) expectation with respect to ?v(a) we get

Proof

optional Ex

that reduces X. If s < 8, then applylng theorem to Xy at times r s A Tm?and t A Ti gives

sn be stopping Let

a sequence

=

A Tm' A Sr:)1(sa>p)l.T'(s A Q A &))

=

Xs A Q A &)1(sa>0)

the

40

2 Stochastic Integration

Chaptr

Multiplying by 1(tm>n) E Fo

Fs h

c

Section 2.2

TmJ

Continuous Local Martingales

Integrators:

41

2)p + (d 1) > 2). i.e., p < #/(d if To see that Xt is not a martingale we begi by noting that Bt z) t12(Bi z) under Pr so as 0 i probability. To x, 1#tI (:<) and Xt 0 we observe that for any R < x showthat E.XL

A %j we get

if and

-4d

only

-1,

-

-

-

=a

-

E (A-(f A Tm?A Sa)1(sa>0,z,k>c)l-'/1.s A TCA Sn )) )'

As n

=

Xs h

TmJ

A &)1(sa>0,m>:)

..--y

'->

-+

-

-.

h Tm'A Su ) 1 Ss

1 cxl Fs ,

h Tm? ), and

E ir X f

-Y(r A Tm' A &)1(sa>c,tm>0) .X'( A Tm')1(n>t) -+

for all r k 0, and 1aY(rATm'A,$k)I1(s.>(),rm>n) %rr, so it follows from the dominatpd convergence theorem for conditional expectations (see(5.9)in Chapter that 4 of Durrett (1995))

K (2<)-d/2

I!/IS.a

dy + a-(d-2) lgl-(d-2)

A-(d-2) Since R is arbitrary Erxt The integra l is convergent so msup-.. ECXL 0. sinceE=XZ > 0 it follows that E,X, is not constant and hence Xt E1 is not a martingale.

s

-+

'

.

E

''''''''

''''

E (-Y(t A Tm?)1(v',k>n) l.T1s A Tm'))

.

Xs A Tm')1(:m>f))

=

.. .

proving the desired result.

(:l y...

.

ln our deqitio n (2.1), as m inga le (wr t jkmhf. t k 0)). Wedid thi with the proof of local mrtingle

;.

..

.

'

f

(2.4)inTamitl exercise shows The next that we get the same deflnition if we assume Xj js a martingale (w.r.t.(J5, t k 0)). mart

.

Exercise

Fghs, t

.

.

a ,

6f

.

Let S be a stopping time. Then XgS is a matingale 0, if and oly if it is a martingale w.r.t. Ft t k 0.

w.i.t.

2.1.

k

,

In the flrst version of this book I

.

.

': .

...

.

fr'ny

'

:

..

2.2. .

.

'

.

(2<)-d/2e-I#-?I2/2fj

2) since (i) there is if and only if p < dld (2'mI)-d/2e-lp-rIQ/2t and large ignoring for y (ii) changingto polar coordinates we have -

r g

-(d-2)p

.

'

.

.

'''

'''

. .. .

'''

.

.

''

...

:

i

.

.

'''''''''

'''

.'

.

1''

'

. .

The last example may leave the reader wit jj t jj j mpress kon tjjt uo ot integrability will guarantee that a local martingale is aq honest martingale, tt ks th next kery usful tt hW, thi i a fls irprlssioh. E

'amount

,

lf Xt is a lpcal spbmartizg>l E

for each t then X2 is a

y j-(d-2)pdy < x ..

1

.

q

artd

r

d-1

no

dv

trouble

'

with

the

'

.

convergepce

which is bounded near 0 and

< x

lA%I

< cxl

Cleazly our mssumption implies .71A'tI < cxl for each t so all we have to ECXL 17,)k X.. To do this, note that if Tn i? a sequenc pf stopping ti.mes that reduces X Proof

checkis that

;we

txpIaa,ul

.7 then we let n tional expectations. -.+

(

sup

osasf

submartingale.

=

=

'

.

='

.

2.1. Let Bt be a Brownian motion in dimension d k 3. In Section 1/l#f Id-2is a lcal martingale. Now that X

,,

.....

by

,

.

As many people have pointed out to me, there is a famous example which shows that this is WRONG.

E l.X'f jr

...,

is pvided

3.t * will lt2 that it Bt is twtjdimnsional Show E 1.X':IF < o z.z: lok IAI is a lcal mrtiklet LS'Xg i:xnso X i ot a mrtiltgle. .

.

(2r5) Theorem. . .. .

t'You should think of a local martingale that would be a martingale if it had E l.X I < x.:'

3.1 we w11l see

Irt jtion fnotion the Xt > < x but limt-x'

lxikcls .

copnterexampl

worse

said

ms something

Exnmple

An eve

ivi

g.

jryijjg..!,l

('

x

k x-

the dominated convergence theorem for condi(:I See (5.9)in Chapter 4 of Durrett (1995). and apply

shows that the last result holds for supermartinAs usual, multiplying by gales, and once it is true for supem and sub- it is true for martingales (by pplying the two other results). The following special case will be important: -1

(2.6) Corollry.

A bounded local martingale

is a martingale.

Captez

42

2 Stochutic

Integration

Secton 2.3

Here and in what follows when we say that a process Xt is btmnded we mean there is a con>tant M so that with probability, 1, lAhl K M for all t k 0 theorm fpr lpcrl l!!qpp irumstanes F Fill ned a pnvergp marsimple result is sucient tingales. The ibllowing for our needs.

(2.7) Theorem.

Let Xt be

a

local martingale

E then Xp

=

Proof

Lt dc

=

0

limu,r Xt exists a.s. and EXz

ajFa-l)

-

S(Ak2j.'&-l)

=

-

(since

.a

yla-l

-

.

that

ro

.s.

-

-

da

Brt

-

Zirclse 2.3. Let X b a continupus local iii>qtlnal u function. Then tX. ) is a local subinartinale.

,

1rt

and

.

Exerclse 2.4. Let X be a continuous lotal maftingal stopping time. Then K = .Xk+, is a local martingale

k

b

gkk (

) (E

:

'

Xk

lt

ad with

ckl

to

respect

b

G

=

.

S(.n

=

and it follows by induction tht

=

.

y

FR+.

.y

has the desired properties. so To see tha d is unique, observe that if B is another process with desired properties, then pl.n Bn is a martingale aud da Bn 6 .Tk-1. Therefore

.

'i

=

-

.

= 72n-1

=

::

43

k 1

Xai y + E (.L2j.'Fn l ) A1

=

Exj

Proof Let y be the time scple of (2.2)and let X .Xv(). X is q.pprtingale. upcrossing The inequality and the froof of th rik r ti/ale conkigit c therem given in Sectipn 4.2 of Durrett (1995)generalize in a Ayrqightforward way to cptinus xitj a,.s. and time >nd allow t!s to Eclude th>t y limjx oncluion *e not X. limf.r Xg exists prove the lst XZ and cvefkence therm f'l then use the the drinatd t convergenc conclllde EXz = fF: E Fx = EXv E) y .q.

and Covarlance Processes

From the de:nition, it is immediate that W,z .Tk-1, 6 d,z is incremsing Xj is a submartingale), and

on (0,r). If

EX.

deflrte for n

.s

sup 1A% l < cxl 0KJ < r =

and

Vrance

= Bn 1.7k-1)

-

-

.s

Bn

=

Jlc

.s-l

-

Bz

-

fa-l 0.

=

ID

The key to the uniqueness may be summarized t'ay predictable disaq crete time mrtingale is constant.'' Since Brownian motion is a predictable maringale, the lmst statement fails in continuous time and we need another

Msumption.

'

(3.3) Theorem. Any continuous local martingale Xt that is predictable loally of bonded variation is constant

and

tintifne).

roof By subtracting v' we may suppose that Arou'z 0 nd prove that X is identically 0. Let F( be the variation of X on (0,t1. We leave it to the rader to check ,

Exercise 2.5. and EXz < x

Show that if X is

then Xt is

a continuous

local

a supermartingale.

martingalewith

X3

k 0

yl

2 3 Varlance *

@

and Covariance

Processe

E .

The next deqnition may look mysterious

: :

r

' .

.

' .

.

. .

tj

3t1r lf X is o tirmous and locally Exe*cl is continupk

of

bounded

variation

thn

r-y

14

( . . '. .

but it is very important.

s ciu inqs : Mtk zf). sincet / s irplies I.X I < Jf (2.4)iplies . Defln that the stopped rocess Mt = Xt A 5') is a bounded martingale. Now if < s t usitk well known properti of conditional expectation e.g., (1.3)and'other (see resultsin Chapter 4 of Durrett we hve '

)

(3.1) Theorem. If Xt is a continuous locat martingale, then we defne the predictable increasing (.X)f to be the unique continpous variance process martingale. and makes local = that hms 0 process A XI A a .a

=

(3.1)

To warm up for the proof of

we will prove a result

in disrte

tim,

,

-+

(3.4)

S((A1

=

Mal 2l.'Fa) y

.E'(M2j.F ,

) L 2M a F(A1'tl.'>-,)+ a ostvj ata) stsy.z y =

t

-

an equation ments since

that we will refer to the orthogopality y is the fact that the key to its validlty

(and hence EM,(Mt

-

G)) = 0).

Mi

E

,

..xyajs,l

...us

.

(3.2) Theorem. Suppose Xa is a martingale with EXn2 < (xn for all ?!. Then = - Wa is a there is a unique predictable process Wa with 0 so that Xxn martingale. Purthermore, n da is increasing. .c

(1995))

ECMS

,

of martingale

(Mt

-

.6)1F,)

incre=

0

Section 2.3

Chapter 2 Stochastic Integration

44

Letting (j

.

.

=

.

(0,t), the orthogonality K (Ef lcf1)sup.f lcjl, and the

subdivision c?&

< ta t be a tn < l martingale inequality increments, the of fact that Mas %ff imply =

E

of

E Mj2

S

E

=

M2

M2jm-z

-

.

m=1 rz

M,--,)2

-

m=1

<E

sup lMzm

v's

Mt--z

-

m

< ff E sup lMm

E ((.X

I

m

'

=

=

.

since

..s.

-

n

'

-+

=

-+

.'.'

.. ' ..

..

..

,

.'.

ECXL

Writing

Xs 1.S) 0. g Qj Qsa + (Qja =

.

.i

A

Qt

A

Q,

-

=

=1

-

Witi aie tw ppiessef ln (3.1) Suppose A >nd Proof of uniqeness z4' martihgate tht is is a contlnuous local the desired properties. Then Jk Sine Jlc = X p 0 it lpc>lly of boundd variation and hence must be constant. (:I q follows that d( for all ./4(

s

,

,

) (

.

-

tmpprtynt, ;

,

.

.

.

,

.

tet

=

.

Qr (X)

=

puxk

k=1

E

-

Xtg- ) +

(.;L

-

As(,)

)

,

y

k

. . .

,

..

k(,)

Xtk-

-

k=1

)a +

t-X

.;V.(.))2 (X,

-

-

hu

-

Xtkfst)a

-

-

2

Xtpta))

.Xp-,)2 + (.X

Xtkfk))2

-

The next step is to take conditional expectation with respect to Fs and use the flrst formula with r = k(,) apd t tk(a)+1 To neyten up the result, deflne 1 sequence a t for (s) K n ; (t) + 1 by letting uk(a)-1 = s, i(s) K i < (), aud tik(f)+1.= =

'tlrj

.

'tzf

-

,

=

.

Exz

t

4A(x)Ia)

-

.1

=

EXlLFsl

-

Qlxj

-

E

(t)+1

)C (v'u

-

=k(,)-f1

Xui-

)2 & '

j

k(z)+1

'W

= E (.V1.S) Q) (X) -

(f )

the difterence

out

.

-

,

.)

.1

jkorkjng .. ..

=k(,)+2

'

E

-

x fk-z )2+ txj xj )2

p

IA).

.

.z4

(x

(f)

,

The ln (3.1) when XL is a bounded martlngale Proof of existence uniqunrq. complicated ipjpiqe onsiderably To its th:rl xistenc pf is more reader ?orttte battle to come, we would like to point out thyt (i) Fs are the ey oo about to prove the special case that we need of the celebrated will construction lpfql prvide infprmation, and som: decopposition (ii)the Tltough and (3.8)are quike the dethlls of thk (3.1) (7.6). e.g., ah>d skip reader boldfard TEp squeamish the ts tp can pr9o are not, so the ahead. Pzoof'' flve q of Existence pages r wtth ltms = = partition < < h x, we t,, Given a ta ( to () = supt : tk < ) be the indk of the lat point before time @. (To avoid confusionlatr, we emphuize now that k.ltj is a number not a random variablt) quadratic variation by We next deqne for a protess X, an approximate ,

(3.4)one

'

-

+

.&

.

.

= (-X.(.)+z

-

.

of

is a

-

a an (j Qa )

-

k(a)

we can do the

.

-

(z)

E

=

(3.3)established

Q/(X)

=

-

E

With

-

A-a)2l.T'a)

-

-

-+

-

E ((A-,

=

the remson for

in the proof

as

45

.X'r)2 - 2(Xa .X'r)J2(.X'j A'alfa) + (uL = E ((.m X,)2IF,) + (.X', aYr)2

Art = (0 g < h1 ;'. k @'i(rz) tJ in which take a sequence of partitins the mesh Ilal = supm Ifnmtnm-ll 0 theu continuity implies supm I.&lk ; Mfz-z j 0. Since the supm S 2ff the bounded convergence therem iiplls 0 a.sk In th last Mzz-, I 0. This shows EMI= 0 so Mt fsupfw lMtz arbitrary, probability with have Mt 0 foi all rational cohclpsion t is one we so El continuity. result t. r'I'he desired now follows from

lf we

-')2l?,) -

then Xf2

martingale

-

l

Mm-,

-

(see(3.1))then

/oof To prove this, we flrst note that remsoning can conclude that if r < s < t then

= jgtwzvE

If Xf is a bounded continuous

Covarlance Procsses

and

If the reader recalls what we are trying to deflne the dehnition is explained by (a) Lemma. mrtingale.

n

Vrfance

= x2.% Qhx) -

4

-

E

J-) f=(a)+1

.L2y -

Ak2y., y-a

46

Section 2.3

Chapter 2 Stochastic Integration

(3.4).

where the second equality follows from

E1 . .

.

.

.

sj. .. .. .

'

.

g . .

.

.

j

Looking at (a) and noticing that Q)(.X) is incremsing except for the last term, the reader can probably leap to $he conclusion that we will conssxuct the desired in (3.1)by taking a sequence of partitions with mesh going process and proving that the limit exists. Lemma (c)is the heart (ofdarkness) to 0 prpof. the of To prepare for its proof we note that using (a),taking expected value and using (3.4)gives

Varlanc

Summing the squares gives h Qr ' (Q ) S Qraa' (.X')suptlapx.z

wherejkj

=

+ Xsg

k

suptj

Covarance Procepps

and

2.X,(.))

-

47

2

tj %skl. By the Cauchy-schwarz inequality

:

.

Eqtxt

(b)

Q)(-Y)IF,)

EX3

=

=

ztxt

To ispire you for the (c),Fe propise routine. of rest of the proof (3.1)is proof of

-

-

that

.

hh'

(e)lf IX 1K M

$

.

the

3

Th ot

valu

A? are two partitions, we calt Lf the partition obtained by Proof If L taking al1 the points in A and in L'. If we apply () twice and take diferepces M2-QAA QhX)-QL'Xjt is a martingyle. By definition t (y) t we see t ha t yi = is a martingale, so A' a' (A-))2 = f'Fr 2 = ('F) Eor h (-Y)

Proof of

(e)

If

r

=

in a moment, we will drop the argument drop the r when we want to refer to the

P rocess t

-+

Qgh.Since

(c+ )2K 2c2 + 22 (since2c2 + 22 (c+ )2 (c

for any real

(3.5)

=

-

Qr

.

J

(F)

-

)2k

'

A

)+

K 2Qr (Q

Q we

.

A'

L'

6Qr (Q )

(d) if IAI+

IA'l

0 then

.

EQ7L'Q*4

. .'

.

.

. .

(e.1)

0

sa < sl

=

.

Recalling

-

Q)k

=

(xaa+, -

.X,)2

-

(.X',p

-

.X,)2

-'Lp)2+ 2(a'Lp+, -L.)(A%p = (-Yak+z 2.X,) (X,s+z .X',p)(-Y,k+z+ X'u -

=

-

-

n

.-

-

.X,)

< sn

r then

=

1

J7l-Ya

2m=1

+

-

.

X,m-,)2(Qrr(A-)

QB a.

-

(X)) .

To bound the frst term on the right-hand t then some arithmetic and (3.4)imply

side we note that if IX

IK M

for all

n

-

X,m-z)*

Slx,m Xam-z)2

S (2M)2f

-

m=1

m=1

(e.2)

.-.o

. ON = 441

-

x-x

.x.2

&,

A ,.

-

.aa..,

va

m=l

K 4M 2EX2r

%4M4

-Y,m-z)a E Fam then use (b) ' botmd the second term we note that (Xa. and IIr Xanxl K 2M to get (X) #((aLm Xawk-z)2(Q1; QCa.(X)J 1.fa#a) . lzstt:rg(.x)z :r,.(x)) jau ) (xa. (e.3) #,m-,) a Exr v'Lm)a IF,m) = t.X'a.,v -

-

=

-

.

m=1

-ux,..z

Qhk+

.

-

-

< sk-v K tj.j., we have

.

A-,--z)2

-

-

To do this let sk C AA? and j 6 A so that % K sk the deinition of Q,a (X) and doing a little arithmetic

does

5-2(x,- x,--,)4

.

0.

-+

J'-!(A-,-.

.

.

that the bond

clear

2

that it is enough to prove that

see

,.

-+

. '.

and

.

it

n

.

X is bounded

n

E V(X,m

E ' .

4

'

LL' is the partition

=

=

. .

since

2A)J(,))

-

m=1

b

0) we have

LL'

Combining the lst two results about

c apd

numbers

but it does mak

a

Qr(-Y)2 r

EQr

For remsons that will become clear from Q) when it is X and we will

S 12M4

r

12 is not ifnprtant on r, A, or L'.

Edepend

and

'

EQhL'Xj2

for a11 thn

-+

Qr

E suplAlk..z + X'k

-+

-

-

(X) 2 J

xalzlsa) once (c)is established

a

'

l/2

(Qa )) K LEQ.

.Y.71.'&)

(c) Lemma. Let Xj be a bounded continuous martingale. Fik r 9 0 and lt r) Svitlk mesh of partitins 0 z g k ty < tnk czi r of L n be a equce 0. Then Q1=(.X') converges to a limit in L1. llal = supk Itnk nk-ll .

'

Whenever IAI+ I'l 0 the second factor gos to 0, enough ti it is to prove nuous, so con

,

-

Eov a *

-

-

K (2.u')2(x, m

-

2,.-,)2

48

(e.4)

-

)2(Qr(-Y)

-x

,m

am-z

Qram (A-))

-

r

m=1

s s

4M ZEXZ r

am

-

m=1

%4M4 the piof I:j

Let An be the partition wtth mqrtirtgay (by(a)), 4 of Durrett (1995))

E

(f) From this and

(c)it

StlP

sr

Am

IQ

-

:Aaj2

s

t

4sjt.)l,a r

-

:Aa r

E If (Z1' 1F,)

j2

Z,w(&)

so that

Q

a limit

-+

uniformly

..

on

g0,r).

Since Q) a convezges in L2, it is a Cauchy sequence in L2 and we can pick an increasing sequence n(1-) so that for m k n()

Using

(f) and

-

Qr

n(p)

l % 2-k

.P sup

t Kr

lQ,

Aa(p)

-

Q,

l

1/j:2

>

sstjgy

it follows th>t ELZ, 1.F,)

E

(3.7) Lemma.

Then

(xYT)

=

jpja;

Zs

=

az

and

Let X be a bounded martingale,

Let Ta be a equence of

< k42-

'i'n

side is summable, the Borel-cantelli lemma implies the only fails finitely many tims and the desird result inequality on the right ' ' ' ' '''''' . . . .. . . El follows.

Since the right-hand

E .

Zt 11,)1'

sjzy

' :

.

.

the proof is

=

XT=

.

(XT)2

-

(vY)T

is

,

...y

complete.

a

stopping

times incremsing to

1(z a >o)

is a bounded martingale

cxl so

time.

that

By the previous xesult there is a unique continuous predictable increasing prois a martingale. By (3.7)Wy = .A7+1 for K Ta so cess Wn so that (Xn)2 we can unambiguously deflne (aY): = z4y for t %Ts Clearly, (X) is continuous, predictable, and increasing. The deflnition implies Xy t.X'lzaa: is aj 1(z.>a) (.X') is a local martingale. a martingale so Xl E1 .y

-

.

-

Tt

-

End of Existence

Proof

-+

-+

(:I

local martingale

--+

-->

()

(:)

.

we By taking subsequences of subsequences and using a diagonal arzment each Q? is continuous, all Since N. Nj for uniform convergence on (0, can get a QLh is not increqing. A is, as well. Because of the lmst term (.;% .X()) dkz-ar is increasing. However, if m k n then k 'n is incremsing, so k Qka-ar continuous that it f6llows and is t arbitrary A is incremsing. Since zz is t A -

y

.gg

and T be a stopping

(X)T.

Proof By the optional stopping therem so the result follops from uniqueness.

2

Chebyshev's inequality it follows that Aa(p+z)

LP

-

.yj

Proof of exlstence in (3.1) when X ls a loal martingale Our flrst step in extending the result to local martingales is to prove a result about the quadratic variation of a stopped martingale. Once one remembers the deflnition of XT given at the begirining of Section 2.2, the next result should be obvios: the quadratic variation does not incremse after the process stops.

Proof

f lQrim

E (.Z 1F,) T = ELE Z1

-

g

Taking limits in

follows that

(g) There is a subsequence

&,

Proof The martingale property implies that if s < then .E'(g1tIFa) = Z1. Th right-hand side converges to Zs in LP. To check that the left-hand side cont ELZL we note that linearity properties of and Jensen's inequality verges to 1.,#-a) for copditional expectation (seee.g., (1.1a)and (1.1d)in Chapter 4 of Durrett (1995)) imply

.

.:2

2).

,

. i? a z-nr with () s k s 2n. sinceQtnx inQ).' poipts Chapter maximal inequality (seee.g., (4.3) usingthe gives -

(withp

check =

-.+

,m-,

.

It only remains to put the pieces tgether.

is to

49

(3.6) Lemma. Suppose that for each n, Z.J'is a martingale with respect to and that for each Z; Z$ in LP where p k 1. Then Zt is a martingale.

-X1

completes

Varance and Covarance Processes

The last detail for the cmse in which X is a bounded martingale yl. is that Ml a martingale. This follows from the next result

we

.

4M2E EX2

(e.2)into (e.1)proves (e)which

Plugging this bound and of (d)and hence of (c).

(3.4)ms before, n

l

g(x

f

over m, and using

(e.3),summing

Taking expected value in get n

Section 2.3

2 Stochastic Integtation

Capter

Finally, we have the following

extension

of

(c)that

will be useful later.

Chapter 2 Stochutic

50

In tegration

Section 2.3

(3.8) Theorem. Let Xt be a continuous local martingale. For every t and of with of subdivisions mesh IAn Aa se4uence (0,) l 0 we have

Proo

Vrfance

and

Covaziance Processes

51

Since

-->

sup ast

IQ)n(X)

(vY)al

-

Q)n(X, F)

0 in probability

-+

Proof Let J, : > 0. We can flrtd a stopping time S so that X( is a bounded coincide martingale and PS K t) % lt is clear that Q*(.X') and QLLXS) n (0, From the deinition and (3.7)it follows that (A') and (.XS) are equal on i (0, 51. So we have

tis

follows izmediately

1

=

a a(X (Qa

+ F)

an y Qa (

-

from the deflnition of (X,

.s))

#;) and (3.8).

E1

.

,%.

Our next

is useful for

result

(X, F)k

computing

.

E

sup IQ)(X) a
P

(X),I>

-

K + P

:

-+

lf X

and

'

(-Y,1')

as

(c)implks

0.

(3.9) Defnltion.

two

are

'

that the lst tern joe to 0 as (:1

locl martingals,

continuous

1 k((X + F)z

=

> e (-YS),I

-

X -

are continuous local martingales. (X, F)f process A that is locally of bounded variz A a local martingale.

.

Since X-ne. is a bounded martingale

lAI

sup

lQ)(-YS)

(3.11) Theorem. Suppose Xt apd is the unique continuous predictable ation, hmsJlc = 0, and makes a'V

ArZ'i

Welet

(.X' 'F)t)

-

Proof

From the de:nition, it is -

(X, 5')z

=

1((,Yz+ K)2

sy

tb se th,t .

(aY+ F)

-

=

.

((

-

K)2 = (X

-

F)))

is a local martingale. To prove uniqueness, observe that if Jk and /( re two processes with the desired properties, then Aj .A( (vY! (A'i$ yk) locat martingale that is lotally f bottdd is a cntinuous varition ad hence H 0 by (:1 (3.3).

-

.()

-

-emark. If X

and F are random

variables

with

mean zero,

,

=

-

-

-

,

covtx, F)

=

EXY

1 k(f'(.X + F)2

=

1 = ztvartX + F) so it is natvral,

I think, to call (vY,F)

vartx

-

the

EX

-

-

Fjzjz

From follows, X,

F))

-

covariapce

of

X

and

=

.

k()

(A'k

=

-

Xp-zlt'Ft.

-

.X,(,)

)($

-

Exerclse

3.2.

(-'Y+ F, Zjg

Ekercise

3.3.

(uY

Exercise c = b, X

Kk-,)

=1

+ (X

')t

=

1

(m Zjt

=

(F -Y)t )

-'''

--

. .

+ (F, Z).

=

.

-

get several useful properties of th covariance. I what and Z are continuous local martingales.First since V'VX= Xx, (.X

=

Q)(-Y,F)

'

Y.

Given the defnitions of (.X)t and (vY,'Flz the reader should not be surprised that if for p, given partitipn L (0 p < l < t2 .J with lims tn x, we let (t) supt : tk < t) and deflne =

(7.11)we

5$k(,))

Exercise

-

a'V,Z)

=

(X, Z).

lf c, b are real umbers then (c.X', F)t abLxjF). Taking F, and noting (yY,yYlf (.X')f it follows that lc.Xlt = c2(A')t

3.4. =

=

=

.

3.5.

(.XT,

FT)

=

(A-,F)T

then we have (2.10) Theorem. Let X and F e continuous local martingales. and sequence of partitions Aa of (0,t) with mesh llal 0 we have

r every

t

(x,'.FT)

=

(A-,5')T

-+

sup

lQ)n(Y,

') -

(-Y,F),

l

-+

0 in probability

Since we a<e lazy we will wait until Section 2.5 to prove this. We invite the reader to derive this directly from the deqnitions. The next two exercises prepare for the third which will be important in what follows:

52

section 2.4 Integration

Chapter 2 Stochastic Integration

then

Exercise 3.6. lf Xt is a bounded martingale integrable martingale.

XI

-

(c) (ff X, J'f 5')t

a uniformly

(.X') is

-

In this step pake thir flrst

and S is a stopping time then Exerclse 3.7. If X is a bounded mrtingale = with (X)s+t-(X)s. aud (F) = respect to Fs-vj Xs-v -Xs is a martingale 'I

3.8. If S constauton (S, T1.

Exercise

KT

are stopping

times and (.X')s

(X)z, then X is

=

=

-

.Jfaff Jnf

a

Bounded

w.r.t.

S

T are stopping

times and X is

constant

n E

=

2.4. Integratlon

E

we will only prove a version of The other in Step 2. Before plunging into appearance

(a).

two results will the details recall th t a t the end of Section 2.2 we deflned Ht to be bounded if there was an M so that with probability one Ht K M for all t k 0. (4.1.a) Theorem. If X is a continuous martingale and H # is bounded), then (S u'Yltis a continuous martingale.

(ff r

ln this section we will explain how to integrate predictable processes w.r.t. inbounded continuous martingales. Once we establish the Kunita-Watanbe continuous able general local will be equality in Section 2.5, we to treat a martingale in Secion 2.6. As in the case of the Lebesgue integral (i) w will begin witi te simplest fupctions and then tatt limits to get th general case, and (ii)the seqvpce of qteps used in defining the integral will also be useful in proving results later on. Integrnnds

tJ) We say Ills, (J) is a baslc predictdble process if Hs, of predictable = basic processes. where C E Sa. Let Tlo the set aud X is continuous, then it is clear that we should deflne

=

1((,,514s)C@) If .I 1(.,)C =

-

aYlz H

Hs dX,

know

we

..

.

E..

.

.X')lF,)

.A-)a

..

EH

=

= = where the last two equalities

-

CECXL

/ (c, ) and

for t

-

=

-

follow from C E F.

Exercise 4.1. lf 5$ is constant is a martingale. t K then

CH .X),l.'&) u'L)1J7,) .X',)IF,) 0

.-Y)t

ECCIXL

and

Euj

1.f-,) X, =

.

= .E'(5$1..#'a) F, whep c

We say Sts, tJ) is a slmple predlctable 8an be written ms the sum of a llnit nuzber is easy to see that if Il' e II1 then H can be '

.

''

''

.

'

.

Hs, of

)

.Ar)t,

written

as

i =1

the next thrre '

.

.

.

.

...

.

.

.

.

where

tl <

tl

<

.

.

.

<

tm aud Ci e Fti-

.

ln this cmse we let

m

E

.l .(X +F))f

.l

=

and write

bmsic predictabte

1(.t-z,t;)(s)6k(td)

=

local martingale

) ..X')t (# ..X')t +(ff

process

of

.

m

,

(b) ((#+.Jf

(x,

From this it is clear that (H ..X')t is contilwous, .-Y) G Ft and .E'I(ff.X)zl < check constant .-Y) Since is for thr martingale property we t / (c, ) can x. by considering only a < s < t % (See Exercise 4.1 beloW.) ln this case,

S41(n,t)(d)d.X,

=

To extend our integral we will need to keep proving versions results. Under suitable mssumptions on Il', ff X and F

=

st<

.l

0

.

E llc

Ks

<

Step 2: Siraple Integrands

t

(a') (H .X')t is a continuous

Ak)

-

f.l

0 St % a StSb

-n) -

cxs

=

-n(g))

c()(.x@)

=

(:r), since once Here we restrict our attention to integrating over (0, t) by intgral over (0, how to do that then we can deqne the

.

=

llc

E'

jmdxa

(#

0 CA

.-)t

e

.

.

Step 1: Bajlc

Proof

Martlngales

Bounded

w-r.t.

53

d(.X, F),

.

Exerclse 3.9. Conversely, if S (-Y)g.. T1 then (Y)s (,%

Martingales

..X') +

(#.#)f

Hsdx.

=

i= 1

ci(m

-

xj ) y-z

.Jf E 1I1 if H lt processs. .

:

Section 2.4

Chapter 2 Stochastic Stegration

54

The representation in the deflnition of H is not unique, since ohe may yubdivide the intervals into two or more pieces, but it is emsy to see that the right-hand side does not depend on the representation of H chpsel. ( (? q

,

4)

Supf

2 ,.b) Theorenl. then

(

.

.X and. Y

0se

j. C 1T1. llfnj..r IL

r

Jnarttngales.

are coptinuous

CH

.

(X +y))t

=

LH .X')t +

=

(# .A)t +

(ff

.

.

.l

Let H

.

Proof

To prove these results it Zt

uY)

H aud H1 = ff By subdividing the intervaks in the defln. (s)6U for j = 1 2. ln this case it is Hi = E'n We Can SIIPPOSe t i =J. 1(ff-y j i a equal to of each side identit# is frst the that see to easy '

=

.

t ODS

55

'

i

,

.Y)t

Bounded Afarlngafes

mn.

variation it defines a J-finite signed mesure, so we fix an a) and then integrate with respect to that memspre. In the caqe under consideration, the integrand is piecew se cons t an t s the integral is trivial suffices

(.Jf Xlztff

=

.

F)t

.

show

to -

that

t

j

.

((ff + ff ) ..X)t Proof

Integation

Jfyffa (f(7,

Fls

is a martingale, for then the Erst result follows from (3.11), the second by taking expected values, and the third formula follows from the second by taking H = ff and X = Y. To prove Zt is a martingale we begin by noting that

jf

((.21+ V2) aYlt.Jf .F)t

(#l

=

.

.X)t(ff

F) + (S2

.

The lmst t*o obervationd imply that if the result holds foi (.J.JI+#Z, Jf ). Similarly, if the result holds for it holds for it holds for (S, ff 1 + ff 2 ). In view of the results in the lmst paragraph, we can by establishing itfi th cii..l' 1(c,)C, ff 1(c,ajD, that b c c mssume (i) K or (ii) c, b d. '

.

.X')t(.Jf

.y)j

(S1 ff ) and (S2 ff ), (.Jf, ff 1) and H, ff 2), 1

1

'

m

C)tt.X;

.

-

+ lYi

A-z)

't-zll -

=1 of continuous

martingale, it follws from (4.1.a)and

(4.2.b)tat

martingales we hake

If X is a continuous martingale and H .X')f is a continuous martingale. .I is boundedl, then

(4.2.a)Theorem.

e

is a l1l -

=

continuous

f.Jf 6 l11 : -

.

our third result

(4.2.c) Theorei.

martingales

If X and 'F are bouhded continpous

lli tlien

.

t

(ff X, ff 'F)f .

.

.

Xltff

.

F)t) .

ff, ff E

.

E

=

JJ

vY)2j E =

Hsl'kl,d(X, Y),

jo

respect to (.;tF), are Remark. Here aud in what follows, integrals since Lebesgue-stieltjes integrals. That is, s (m F), is locally of bounded -+

%

() CD ((.Xa CD ((A

=

=

Zt

-

%

=

-

.;L)(K

Xc)(%

f7D(.;%'Fi

Taking expected value Ezt

-

Fcl Fc)

-

-

to check the martingale

sues

and

/62#4.1, with

.

so it note

0

t

EH

and

Y),

#,ff,dlx,

=

.

,/

-

.

'

E

')4

.

,

l

Consequently, EH

JJ

H 0, so we need to show that tff.A'ltff Case 1. In this cmse S,ffadtx, 'F) is a martingale. To proke this we bserve that if C(% Lk)D1(c,aJ then CH .X')t(.Jf .'F) = (J '.F)t is a martingale by (4.1.a)

Case 11. In this case

..

rlrning to

furtheifnore

cn

=

=

Since the sum of a finite number

row pyoye the result

diW

=

=

-

-

-

X,K -

Xa

F,)

-

a s

Kt K

.

'ctxt

-

-


k b

To do this

we

Xs)

(-Y,'F),))

e &, .E'(X -

% 11,)= CDECXtYL (Ar, -

This completes the proof of Cmseii

-

for a K s

Xutl

-

s

(aY,F))J (X, F)c)l

-

((X, F)

-

property

((.X',')t

and noting

((X, F),

-

and

')t

-

F, 1F,)

=

0 we have

(yY,K (m ')a)IF,) -

finishes the proof of

=

(4.2.c),

0 C1

Seclon 2.5

2 Stochastic Integration

Capter

58

Hn aY) is a Cauchy sequence in M 0 ms m, n x, Since ll.J:fn s'nllx aud by (4.6)must converge to a limit in M2 which we define to be the integral CH.X). To see that the limit is independent of the sequence of approximations y artd 0 with Hn x Sllx 0 and ll/n Jfllx chosen suppose llSn = H% if setjing approximating by Xn third n is Form .X ya sequence Wn a Zi Loking 0 so we have Hn .X odd, Xn Hn if n is even. llffr' ffkx at the even and odd subsequences of Hn it follows that F = Z C so the limit is unique. ln words, the lmst argument shows that if the limit exists for ANY pnique. Note that ms a corollary of the sequence of approximations, it must be construction we get -+

-+

.

-

The proofs of the remaining

,

-+

-+

.

-

-

(ff

.

equalities

two

(.X'+ F))

.

(#

=

(H .X Jf

.5')t

=

'

.X'lf + CH y)f

.

t

.-,

-+

Tfunfta-Watanabe Tnequalty

Te

0

.

,,

,

#,zf,d(x, #),

.->

--+

=

-

=

(4.3.a) Theorem. lf X is a bounded then H X e M2 and is conthmous.

and

martingale

continuous

H 6 H24.X)

.

t#n

-+

.

-->

.

-

;

w get that

p

,)

(sup Itsn

t

u .x)I

.x)

-

.

>

s

c-assup

<

E'-2

llsn .

and it follows that t

.

.-+

. .

.X'lf is

-

CH uYltI

-+

.

0

(H .A'llla

..u4

.

-

uniformly

.Ar)t converges Since this holds for any .7 > 0 we say subsequence passing probability.'' By to a we can hve in

1(#n .X')

u r xllz

-

4jI(.Jfn A')

.

E

.x)

tst#n

sup

4.4.

Exerclse

4.5. lf X is a bouned

II II

to

0

(# ..X)t

2.5.

rlnlte

ll#

X112

=

.

Ineqallty

local martigales

are

1/2

co

Isffa IdI(.X',F)1,

where dl(vY,F) j on (0,).

lastands

and

H and ff

surely

0

two

1/2

x

H, d(-Y),

K

are

ff, d(F),

0

for dk:t where

16 is the total

variation

of r

(X, Flr

-+

'''''''

and

(ff

.

H, ff

e

1I2(X)

.X')f

that S'+ff 6 1L(Ar). The triangle inequality for the norm 11 Proof .IIximplies n and n ffllx with 0 Let Hn ff e llz llff K Ilx 0. The triangl llffnr&) implies ll(JJrz CH Jf + )llx 0k (4.2.b) + ff inequalityimpes -+

-+

-

-

,

-+

-

'''''''''''''' r

.

Remarks. d1(.X',F) 1, s and (i) If .Y y and d(vY)a s then d('), (5.1) reduces to the Cauhy-schwaiz ineiuality. ()i)Notice that H ad ff r not mssumed to be predictable. We assupe only that Hfs, (z2) and ff (s,tJ) are mmsurable with rvspect to .R. x F wheie 7 is the Biel subsets of R. ht tlat tt notiop of nrtingqle ttyin this leyel of generality we ca doespot ntr int the fiof arfter the first stef. =

(Gn vY)t .

-+

(G vYlt in 442 ivhen .

Cl

=

=

ts

(

.

Proof Steb : Obserk tht if s S @,(#+F, lf Welet (., #)f, z (M, Nj3 (M, #),, then .

(

'

'''

'

'

.

.

and use the fact that

=

rraszt

:

.

and H 6 lla(aY) then

martingale

Knita-watanabe

(5.1) Theore>. If X and F mesurable processs, then almt

0 El

.

,

-+

Ip this section we will prove arl inequality due to Kunita >nd Watanabe and afply this result to extend the formula given in Section 2.4 for the covariance of two stochastic integrals.

a.s.

continuous.

((# + ff ) .aY)t = CH .X)f +

=

llznll llzIl.

0 then

-.+

-

.

+

Now 1et n x G H, ff .I + ff

llzn zII

and

-

(x)

(4.3.b) Theorem. lf X is a bounded continuous martingale, then H + ff e H24A') and

-+

1111is a norm

lf

--

The fact that H.X E M 2 is automatic from the definitiim. lf H e TIl Sine Xlt is contipus. 0 then (4.2.a)implies have Ilffn Jfjjx maxiral L2 inqulity nd the using lll#n -Y) CH -Y)II2 0, chebyshev .

Exerclse .f

Proof

=

will be given in the next section. To develop the rsults there we will need to prppexty extend the ismetg from 5111io H24.X'4. To do this it is useful to analysis. rall some of your undergraduate

.

'

''

'

X+yl

'''

.

.

E

.

''''''''''

(.X+F,X+'/),. '''

.

.

-

0

K (aY+ F, X

= (.X',x)t,

-

+

ylz

-

(.X'+

2(A', F), +

2

F, X +

(F, y)

'F),

.

60

Capter

Secton 2.5

2 Stochutic In tegration

umber of lf we llx s and t and throw away a countable for al1 s, t, and will with hold for all probability inequality last the of 0 then one sets measure rational Now a quadratic cz2 + z + c that is nonnegative at al1 the rationals and not identically 0 hms at most one real root (i.e.,52 4cc K 0), so we have that ((-Y,5'),)2 K (aY,-Y),(F, 'Fla .

.

-

'

< ta be an incrmsilg sequence of tims,' let Step 2: Let 0 tn < tl < hi, ki 1 ; K n, be random variables, and define simple mauiable proc =

.

.

.

Te Tfunlta-Watanabe Tnequallty

61

To see that this exists, note that J, is the Radon-Nikodym derivative of d(X, F), with respect to d1(.;LF), 1.Now apply (5.2)to .Jf, = Iffl and ff, J,lff, l.. El =

With (5.1)established, from Section 4.

(5.3) Theorem.

we are now ready

If H E 1I2(X)

n 1la(F) then .

...

.l (A- + y))t

,

=

.

business

H E l1z(X + F) and

'. ... .

.A-)t + CH

..

of unflnished

to tpe care

.5')z

n

.Jfts, tJ)

f()1(t-z,.j)(s)

=

= n

ff (s ,

(J)

Proof

First note that Exqrcije 3.2 implies

1 . -ft$d

)1(t i-

=

i=1

..

.

z,

f)

(s)

.

g . .. .. . .

;)

q

and the Kunita-Watanabe

the integral, the resplt of Step 1, >nd the Cauchy-schwarz From the deinition that inequality it follows of

n

S'aff ,d(X,

'),

Ifl

<-

0

=1

n

1(.X,

.

I

')1:-,

,

((A-, -Y)1j-z(F, '.F)1'-,

1.,:I

< f=1

1/2

n

<

-

=1

pzovin g that for simple measurable

. ,z,zr,d(x,z),1

1

s

-

)

=1

processes: jg

Lj

#owlet

jg

zr.7d(y-),)

Lj

tep :.1: tet

M b

a

large numbr

and let T

=

inftf i (uY)t t- (F) restkited

%

-+

Hn Z)t

x and use

(S

By

ur atttion for s k T and 1//,1,lff,l f M for s K T. llaving conveizce bouded the and usin: (F) flnite measures; so are to (0T) (X) meyurable To improve bounded holds for processes. theprem,w see that (5.2) rftemsurnbl proof) b 1et J the , fr8ess taiin a (5.2) to (5.1)(andcomp le t e such 1) that valuesin (-1, 1

t'1')t)

Proof

Hn X)f + Hn ylj

=

.

.

. .

..

n

By

.

are

zlt

.l -+

.

($.4) Theorem. If X, F ff e 112(y) then

vhich is (5.1)with the absolute value. ogtside the iptegral.

.).

(.

=

.

,cd(x),)

s

have

2((x), +

-+

-

z

t.Jfn (.X'+ F))t .

. .

+ ('.F))/j

(which should remind the reader of (z + yj2 S 2(z2 + p2)) and it follows that H 6 112(.X + F). To prove the frmula now we note that by (4.5)we can fnd Sn in I1l so that llffn #llz implies that 0 for Z X, F, X + F. (4.2.b)

'l -

.

(

i(x,')t

s ((A-)

)2. So we

-

(-'r

Ii'l 1/2

t-?(F,

l

((A-)('))1/2

+ 1')

a

a'lti

.

j/,

(5.2)

:,2i (x

1/2

(F), +

(5.1),implies

inequality,

since 2c K c2 + bz, i.e., 0 K (c

.

(x)t +

=

I(x '.F)1, s

1

%?

'i')

(x +

.

.

bounded

..'

torg

continuous

'i

'

. -,

=

.

'

..

#,k

martingales,

+

t. H E I12(X),

t

X, ff F)t

(3.11),it su/ces

.

to

Saff,dt.X, F),

=

show

n that

1

(t)

,

/-' l

dtx

,

z),1

-

y

,,d(x,y),

-

is a martingale. Let Hn and ff rt be sequences of elements of l1l that converge to # and ff in H2(.X') and 112(F), respectively, and 1et Z1' be the quantity that results when Hn and K n replace H and ff in (t). By (4.2.c), Z'J'is a

62

Section 2.6

Chapter 2 Stochastic Integration

martingale. By (3.6),we can complete the proof by showing Zsn The triangle inequality and (4.3.b) imply that Sst)p 1(.J.f' .A')t(ff

n

.'F)t

KE

l((#n

sup t

E

+

.

-

#)

-

.

.

t

g

-

To bound the flrst terml we use (i) the sup of the product yro duc t of the sup's (ii)the Cauchy-schwarz inequality, znequality and (iv)the isometry property in Exerclse 4.5: ,

; ftsup

,1

I((ffn

(ftsup 1((.Jfr K 4II(Sn x, since as n estimate shows -+

1Il A-l lsup

-

Ilsn X

''

't

q

S'IIA-

0

-+

tst)p

1(.Jf A')t((ff

is

-

than the

smaller

lk-) F)t .

Ij

0 .

A similar

.

.

F), in f,1.

.Jf,ff,d(.X, F),

-

0

lHsnff an

-

S',ff

I#,n -

0

E

f

JJ

-

Jf anff nd(X,5'), an

j

-+

S,ffadtx,

this

F),

,

za tnL

--,

0

-.+

Ilff'l11y llfflly< x. Combining ff, Id1(J, 1') lait follows that

,

shown

0

lfn I2d(F),

-.t

-+

and the desired result

fbllsvs

hom

!7I

e

ci tht

treatments (seee.g., Revuz and Yor (1991) p. 130), if M G 11a(M), H.M is DEFINED to be the unique L 6 M2 with Lz 0 =

(y, yj

a:j

s

.

ts, x)

E M2. 5.1. Prove the uniqueness

in the last claim.

(5.4) allows us to do easily soze things that we would have hatl to work to be continuous local martingales, and suppose prove in Section 2.3. Let X and .it'c yc 0. The lmst assumptipn entails no lqs of generqlity. jee Ekercise '

=

=

gn

S'al Iff,n ld1(-Y,F)

Using the Kunita-Watanabe .t

s Lj

Izc

inequality,

't

-

1#,1Iffan Jf, Id1(-'LF) 1, -

0

(5.1),the l/2

,z,I2d(x),)

(

) ( =

,

-

,

Is+

5.2:

xvclse 5.3. VY,TI$ yJI

ld l(.X', )I '

,

t

S

apd H

Exerclse

0

0

on

0 and 1S,1 lfr

1/2

j

t

:

Hnnlkandqx,Fla %

ffllx

-

,

hemark. In some

/42

Exerclse -+

.

-

.1

j

for all N

-+

F), (S .X),(ff so we have shown Hn A'latff To estimate the second term in ZJ' Zt, we again begin by putting the absolute valpes ipside, replpcing the memsure by its variation, and then using t e triangle inquality n

.

0

estimate

z,l. we have now in (3.6). -+

<

-+

simila

1/2

lffan Jf, I2d(A-)a -

llsn

x, since

(iii)the L2 maximkl

ffllxllffnllz

-

as n with a -->

-

n

n

.

;yj

.

IIff l1'i- IlffII&'

and

< E

'F)t IJ

411ffn

=

;

63

1,

.Jf,II.Jf,nldI(-Y,F) j

llffn F)tI2))1/2

Hj #)tl2).E'(sup .'112

-

.

.

-

.XII2IIffn

H)

-

ltff

.

I#,n

I

F)

.

q -

..y;

lt.Jf .X)t((f ( a

sup

'1

0 n

fartfngales

Taking expected values and using the Cauchy-schwarz inequality E

Iltff

Iocal

w.z'.l.

,

.1')1

A'lttff

.l

Z, in f,1.

-+

Integation

t

frst term is

is

ln this section we will extend local martingales antt

tinuous

a

local martingale.

Local Martingale

w.r.t.

our integral so that the iptegy>tors cap so that the integrands re in

1/a

lzcI2d(z),)

Lj

2.6. Integratlon

.

,

)

I1a(.X')

=

H

: g

Jf,2d(.X')a < x a.s. for

al1

tk0

be

con-

Captez

64

Stochutic

2

Section 2.6

In tegration

Integration

We will argue in Section 3.4 that this is the largest possible class of integrands. To extend the integral from bounded martingales to local martingales we begin by proving an obvious fact.

ln words if we set the integrand = 0 after time T or stop the mavtingale T or do both, then this just stops the integral at T. Our next t%k is to generalize our abc's to integrands iri llat

(6.1) Theorem. Suppose X is a bounded contirmous mattingale, H, ff IIa(X) and ff, Jfa for s % T where T is a stopping time. Then (S .Xla (ff X), fOr s S T.

(6.3) Theorem. If X is a continuous local martingale CH .X')t is a continuous local martingale.

=

.

e

=

6:

Iocal hkartingale.s

w.r.t.

and

at

til

H : lIa(vY), then

.

'

Proof

Write Jfa

=

Sal + Ih2 and ff

=

,

Proof

SJd(X), > n or IXtI> iz1, it suces By stopping at Tn inf ( : that if X is a bounded martingale and H E I1a(#), then CH .X')f is a cpzinlou. martingale bt this follows from (4.3.a). (:3

ff ) + ffa2 where

.rf'l

H, Clearly, H i mplies

.

X)

=

1

(ff

=

'.F)j for

.

(Jf2 .X'),

al1

(.Jf 2

=

.

and it follows from tr Combining this

.H'4 ltasz)

=

a

Jga 1(asz)

=

t. Sipce Sa2 .xla

3.8 that (S2 .X')f the qrst result and using

xercise with

(Jf

=

.

conclusion.

K T, (5.4)

((.Jf + K ) .X')t

2

.

.X'l

.

=

0 for t

k-

(4.3.b)gives the desired

14Jf 6 IIa(X)

f3 1L(F)

-(ff

extend

the integral Sa

%Tp

inf

=

let Sn be

now,

(t

IA'tl >

:

>

times

of stopping

a sequence

with

.Jf,2(f(1), > nj

n or

0 '.

k

and ' Sa 1 n. tt. Sl' z ffaltassa), and observe that if r? n, (6.1)Jmplis that Hm Xlf V: Hn X)3 for t K Sm so we can defln H X by setting CH X)t (Vn Xl fOr t S Sn To- complete the definition we have to show that if Rn and Sn are two sequences of stopping times K Ta with Rn 1 G) and Sa 1 (x7 then we nd p with the same (# .X')t Let SJ be the stopped version of the process deflned a t the beginning of Section 2.2, let Qn Rn A S,z and note that (6.1)inpli E

.

.

=

.

..

xg

.

,

CH .X')f +

=

(s Proof norm

.(x

+z)),

(s

=

.

aYlt

.A-),+

u

Tp provy the flrst formula we note Shat th trt>ngle inequality for the 1 11Ix implles H + ff E l1a(.X). Stopping at .

Ta

=

inf

t

't

IA) I

:

,

t

H,2 dt.X'la or

,2

Jf d(F)a

,

0

>

nj

0

reduces the result to (4.3.b) For the secohd formula we note that #he argument in .I e Hat.X' + 5') and by stopping we can rectuce the result to After seeing the last two proofs the reader

can undoubtedly

.

(6.5) Theorem. K e 1Ia(F) ther

lqcal martingales,

.

lf X, 5r are

continuous

(5.3)shows (5.3).

imyrove

that E1

(5.4)to

(IIR,

.X'),

.

=

(SW..

.X'), for

s

% Qn

#,ff,d(X,

=

.

.

chosen. The uniqeness

If X is a H$

.

X

result and

continuous .l

=

.

X)V

(6.1)imply

local martingale =

of stopping

H

.

XV

=

HV

and .

XV

H E IIa(aY) then

e

l1a(vY) >nd

F),

0

Using Exercise 3.2, we

Since Qu1 x, it follows that (.Jf.J)t is independent of the sequenc

H

j

LII X, ff F)z

=

(6.2) Theorem.

(ff

.

'

.

tires

q

,

then H E 11a(X + F) and

=

-

locq) in>ytirlgales. If .J ff

.rj,

El

=

'

Let X and F be cpntinuouq (6 4) Thorem. l1a(-Y) then H + zf 6 IIa(x) and .

0 for s

=

6.1. Extend the proof of (6.1)to show that if X i a bonded Execise continuous martingale, S, ff E l12(-Y) and H. A-, for S K s K T where X,T stopping (S .X), (S.X)s thn (.Jf .X), .Xjs for S % s i T. times are

To

.

for s < T

0

=

.

ffa2

=

=

to show

can

generalize

(6.5)to sums

of stochmstic integrals.

E/n=I JfJ Hi X and y Yi where the X (6.6) Theorem. Let X = E'C=1 Yi continuous local martingales. If Hi e l1a(.Xi) and ffJ 6 lIa(5Y), and arq then Hjlf'is dxi FJ), (.;Lyl = t) ij =

.

.

j

,

l

Chapter 2 Stochastic Integration

66

Sectbon 2.6 Proof

We leave it to the reader to make the qnal extension of the isometry property: '

.

Exerclse 6.2. If X is a continuous .I X e 442 and llS Xlla = IlSllx. .

local martingale

.

.

.'.

.

:

'

We will conclude this section by compttting some stochutic integrals. The first formula should be obvious, but for completeness must be derived from the deflnitions.

Ill#n=

H4

.7111

=

Ilsn

J.f11i K Mlsup

=

as n

-+

tx

by the bounded

For the rest of this

Jfa dXs

section,

(7(.Xy

=

zka zz (0

az

-

=

Exercise

local marting>le

6.4. If X is a continuous

2A%dX' .. .

.

.

< fnkta)

...

. =

..

.. .

tJ ls

0. Our a sequence of partitions of (0,f) With mesh Ir,I = supf ltJ' t7-1I theme here is that if the integrand Ih is continuous then the integral is a limit of Riemann sums, provided we evaluate te integvand c f8 lefi entlpcin of e intervals. -+

-

'

(6.7) Theorem. If X is a continuous local martingale then as n cxl

and

t

t -

-Y(f7))

#, dX'

-+

()

i

Exercise 6.5. Sltow that if X is a at the right end point then 1-12A12 zlX(f2+1) +

-

.X(f7)l

X

-

(.')f

local martingale

continuous

2X4 dXa + 2(X)t

-+

0

i

-

=

A)2

and we evaluate

-

XJ+

.

-+

.&'t7l-Y(Pz+z)

XI

=

then

(.X')z

Comparing the lmst two exercises you mightjump to the onclusion k if we evalute at the midpoint we get an integral that performs llke the calculus. However, we will not ask you to prove this in general.

Ht is continuous

--,

0

.

=

be C for

< cxl

0

<

.-.-,

=

The rest of the section is devoted to investigating what happens when we ekaluate at the right end point or the center of each interval. For simpliity we will only consider the special case in which Hs 2X,.

Xs) tg < ty

I#an .JzaI2

J

theoremt

convergence

f

j

67

implies

.

Exerclse 6.3. Let X be a continuous local martindale. Let S K T stopping times, let C@) be bounded with C@) E Fs, and deqne Jfa S d s K T and 0 othrwise. The .I E Ha(.X) and (

fartngales

..

S E 112(X) then

and

(or4.4)

Exercise 6.2

Tntegraton w.r.t. Locl

in probability

Exerclse 6.6. Show that if Bt is at 0 then

Proof First note that H 6 l1:$(.X')so the integral exists; By stopping at T = inf (s : s, (.X), or 1#,1> M) and making H constant after time T to we can suppose (A')t K M and IS,l K M for all s. Let preserve continuity, San = Iht for s e (t7 t)z+z),Iljn = Ih for s > t. Since s Jf, is continuous o (0, t) we have fox each ta that

2a

=

'

dimensional

motion zrownian

starting

1

J'-)2#(( k=0

,

a one

that in

one

+ 1)2-ntj

+ 1/2)2-nt)(f((

-

#(l:2-nt))

-+

,

sup l#,n

-

H, l

-+

0

The desired result now follows from the following lemma latr.

which

will

be usefpl

' .

(6.8) Lemma. Let X3 be a contirmous local martingale with (.X')t % M for all lf .Jfn is a sequence of predic#able processes with lSn I K M for alt sLw and . with supa lS,' .Jf, 1 0 in probability then Hn .X') CH .X) in Mz. -

-+

.

-+

.

converges in probability

Jc2f4

to

dBs + t

=

B t2

.

integral, is more conThis approach to integration, called the Stratoovich venient than It's approach in certain circumstances (e.g.,diflksion processes on manifolds) but we will not discuss it fvrther here. Changing topics we have of (6.7)that we will need in Chapter 5. one more consequence Exercise

6.7. Suppose /t

:

(0,x)

-+

,

is continuous. Show that /12ds.

a norma 1distribution with mean 0 and variance

Jpf ,

J

,

dB' has

68

Section 2.7

Chapter 2 Stochastic Integration

2.7. Change

Varlables,

of

It's

Formula

Comparing

This section is devoted to a proof of our flrst version of It's frofrmla, (7.1).In with a slicker. Section 2.10, we will prove a bigger and better version, (10.2), proot but the mundane proof given here has the advantage of explaining where the term with T' comes from. (7.1) Theorem. Suppose X is a continuous continuous derivatives. Then with probability

y(.m)y(.x)

and

+

:

.J(.t)

(7.2)

f has two

/*(dn)

(

1 -

a) 2

f''Xs4dLXLs

t)

=

variation,

=

.;Vy+, ) when

,

s 6

(f)'t;%1), ,

(.m+, xlz -

X

$

i

so that

t

-X7)2

-

G1d.'',

=

and what we want to show is

to

ot

(7.7)

'prv

any a and

(Izj

.-+

g

1 (c(c,j;;

j

Jnx, )g(x),

To do this we begin by observing that th uniform continuity of T' implies that uniformly in s, while (3.8)implies that .An, as n x we hav Gnr y&t.X'aatl probabillty to (X)a Now by taking subsequences in we can suppose converges that with probability 1, Wn,at converges weakly to laYlaaz. In other words, if (.X').at as distribution functions, then Wna and s td and regard s we fixssociated memsures converge weakly. Having done this, we can flx LJ and the deduce(7.7)from the following simple result: -->

-+

t

't

(?J,

.

=+

-+

'j'(

+

show

-+

.n

0

ajfas

-

to

69

0

=

=

t

in probability as n x. (7.5)follows from (6.8). To prove (7.6),we let Gns gl @) fncxj: Gn JN(.;V)for s k t and' 1et

:,P(*)(A7:.+,

By stopping at Ta; inf (t : 1AR Proof or (X)z k MJ, it sces lcalculus and result when the we know that for l.;Vl (aY)t < M. From there is a ctc, ) in between c and b such that ,

laj

0

j

i

want

Formula

fxsjdx.

-+

1

-.+

-

,

,

-

Vaiables, It's

=

As the reader will see in the proof of (7.1),the second term coms frolit the fact that local martingale paths have quadratic variation (.X')t while the 1/2 in frpnt of it comes from expanding f in a Taylor seriesk

J()

.w(*)(-'L;'+z A77)2

fAsjdAs

=

-

1 -2

=

Remarx. If Jk is a continuous functiop that is locally of bounded and f hms a continuous derivative then ('ee Exercise 7.2 below)

-'Ly)

-

(7.6)

of

becomes clear that we

T'(.XI-)(-Yz:-+z

j,' rtxaldxa!./' r'(x,)d(x),

=

-

local martingale 1, for a11 t k 0

(7.4)with (7.1),it

Change

'

Let t

tg < ty

.

.

.

<

t2a

=

1Xt4

.sequence

-+

/(A%)

-

J(.XLr+,) fhh:j

=

,

.;Vy+, )).

(7.8) Lemmak If (i) memsures pn on (0,t) converge weakly to px j a flnite measui; ad (ii)grt is a sequence of functions with IpalK M that have the gs), then ms n cxz property that whenever sn G (0,f) s we have gusri)

i

:,P ()(-'V7+z -

i

Xt:

2

)

-+

-+

jg Jpd/za

.f'(.;%;-)(-'L7+zXt: )

-

=

=

dpco

'-+

-

1 +2 (al) fnch:

0

-+

-

=

(7.4)

where gl

An of partitions follows that (7.2)it

number. Consider a with mesh lAnl 0. From t

be a flxed positive

t)), we can assume that al1 the yza Proof Bj letting p'n(.) = ynAllpup, construction 2 of standard probabllity A memsures. are (see(2.1)in Chapter with shows that there is a sequence of random variable: Xa Durrett (1995)) distribution pa so that Xn

-+

Xx

a.s. ms n

-+

(x)

The

convergence

of g,z

to g

Chapter 2 Stochastic Integration

70

implies ga(.Xk) theorem.

), so

gx

-->

Section 2.8 Integration

the result follows from the bounded

convergence 1:)

(7.8) is the last piece in the proof of (7.1).Tracing back through the pif, that (7.8)implies (7.7),which in turn completes the proof of (7.6).So we adding (7.5)and using (7.4)gives that for each @, see

J(.X)

'1

1Xz4

-

= g

t

J.

J'tlald-'a

+ ;; o

0

lmxstdxss

the formula is a continuous function of t, iy fpllows that with sinceeach side of equality made in (7.1). C1 holds for all t k 0, the statemeni probability1 the .

In Chapter 3 we will need to use It's formula for complex (7.9) Remaz-k. extension 1 d f The is trivial. Write f + r apply It's formula to multiply and add and the formula by i the two.

Va lle

'tz

=

.

't,

Semimartngales

Exerclse 8.1. Show that if Xt = Mt + d and X3' = Mg' + a aTe contirmMt + Mt') + (.At + A) is a ontinuous then Xt + JV ous semimartingales semimartingale. =

In this section we will extend the tegral from continuous locl martingales are three remsons for doing this.

class of integrators for inour stochastic to continuous semimartingales. There

(i) If X is a continuous local martingale and f is (72 then It's formula shows but it is not a local martingale unless us that fxt) is always a semimartingale /&(z) = 0 for all z. In the next section we will prove a generalization of It's formula which implies that if X and f is C2 a continuous semimartingale then JXt4 is again a semimartingale. ,is

(ii) lt can be argued that any explain this, we begin by defining

fsreasonable

'tz

,

'p

%v.r.j.

an easy

integrator'' is a semimartingale. To integrnnd to be a process of the

Orn1

E xerc 1se 7.1. Let for all t. If ff n is a

a11s,

ul.t

lff,n

o and sup,

process of memsurable processes

sequence -

with

be a continuous Jf,

I

-+

0 then

Ilffn

sup

Exercise 7.2. Use the fact that fb) faj and Exercise 7.1 to prove 7 between c and ( ,

.'

2.8. Integration

w.r.t.

with

Iffn (s, 0.

-+

.

=

)

.

uJ)

ff ) W)tl

-

-

total variation

Tca,

E

H

bljb

c)

for sone

c

.

where 0 Q S T1 K S Tnmz are stopping times, and the Hi 6 Fn have lsl < (x, a.s. Let lle, be the collection of bounded emsy predictable processes that vanish on (t,x) equipped with the uniform norm .

.

.

IlSllu

Semimartingales '

.

( .

.

.t

'

.h

lf th (ontippops) X 0, then the dcompotio!

semimartingal. .o

=

=

sup

1.J.f,(/)1

4)

a continuous

(8.1) Theorem. Let Xt be a continuous and M P rocesses A aTe chosen so that unique. is Mt + A

Hi 1(n,n+,)

=

=n =

-

semlmartingale if Xg can be writtn as X is +A, to be and contirtuous martingale local is a continupus Adapted prowhere T is a variation. A nice feature of continqops semicess that is locally of bounded of unheard their i. for more general cpunterparts martingalesthat is said

n

l.lt S M I K M for

=

E E

Proof lf Mg' + A is another decomposition, then A continuous local martingale and locally of bounded variation, is constant and hence H 0. -

A

so

=

,

AT? Mt is a by (3.3)A -A

easy

integrands

deflne the integral as

we

n

.l

.x)

Hixn..

=

.

xvkl

-

i=1

Finally let L0 be the collection of a1l random variables topologizrd by convergence m probability, which comes from the metric 11.X110 f(1-Y1/(1 + 1.A-1)). See Exercise 6.4 in Chapter 1 of Durrett (1995). A reult proved independently by Bichtler and Dellacherie states =

-

(ZI

X3 = Mt + A be a Remark. In what follows we will sometimes write of Xg consists of specify that the decomposition continuous semimartingale'' to the local martingale Mg and the process A that is locally of bounded variation. tslzet

For

(8.2) Theorem. If H X is a semimartingale.

-+

(.Jf .X') is .

continuous

from lle, .

0

-.+

.

for all .

.

'

then '

r'

..

.

Since any extension of the integral from flejf will involve taking limits, we need our integral to be a continuus function on the silple integrands (insome sense). We have placed a very strong topology on the domain and a weak

72

Sectlon 2.8 Integation

In tegration

Chapter 2 Stochutic

Protter (1990) topology on the range, so this is a fairly weak requirement. In his approach one gets some takes (8.2)as the dennition of semimartingale. without much efibrt but then one must sweat to prove that a very deep results semimartingale semimartingale (defined (as define ms a good integrator) is a above).

Seminiaztingales 73

mnt.

Proof

This follows easily from the deqnition of the integral with repect to the result for local martingales given in (6.4),and the fact a that the results are true when X and F are locally of bounded variatio. ln provingthe second result we also need Exercise 8.1. E1 semimartingale,

'

,

.

the integral from local martingales (iii) Last but not least, it is emsy to I1a(X) by a slightly smaller clmss of integrands to semimartingales if we replace that dos not depend on X. extend

The rest of this section is devoted to deqning the gales and proving the third of our abc's.

(8.5)Deflnition. lf X M+A and X' M' +yP galeswe deflne the covariance taY,X/lf (M, M/). =

covariance

for semimartin-

are continuous

=

semimartin-

=

be a cohtinuous semimartingale. Getting back to business, let Xt = M+d if predlctable i::: of bounded locally processes, the set We say 1l' e l11 Tn. (x, for stopping that of times Ta 1 so % there is a sequence 1#(a,u?)I% n If Il' E ll1 ve can deflne

To exptain thk we recall the in Sectlon 2.3, and note

quadratic,

approximate

QtLX) defined

and X' tf'' + A' (8.6) Theorezn. Suppose X = M + are continuous semimartingales. If n is a sequence of partitions of (0,) with mesh IAr,I 0 then l / : Q n(X,-Y ) (M, M )t in probability =

.

,1

variation

-+

-+

exists for as a Lebesgue-stieltjes integral (which martiugale that note Mt we local the respect to

zz2d(AT), 3

x.

To integrate

Proof

s n2(AT)z- <

Q. (x x')

co

So l11 (: l1a(M), we can define CH Mlt .

LH -Y)t = LH .

.flt

+

.

(#

,

(8.3)

is a continuous

If x is a continuoussemimaztingale

and 1et

note that

tionshipsbeween addition

also emsy. We name and multiplication.

it in honor

e lrl, of

then

(8.4) Dlstributlve j

-+

((ff + ff ) aY) LH (.X + 'F))t .

= =

LH A')t + (Jf .X') LH X) + LH ')

.

.

M'4 +

'

-

surely,

.

,

t

(-X','.F)t

.,)

,

-+

while

=

the flrst one hms I7It<

tx7

EI

ff, Ki E

semimartingales

-''f

=

qhxx .t.

and prpcess and J4 is continuous j urely. T do this w 0 almt

Xi and Yi are continuous (8.7)Theorem. suppose Hi = lH, X 'F rn Enj= ffJ Yi then Ef=1 1

HsiJfj dtv'

,

FJ),

0

ij .

Let Mi and Ni be the local martingale parts of X and Yi let M Hi E'rlzz:t M, aud N E.fn=I Ki Ni It follows from our definition that (.X',F)f = (M, #)f, so using (6.6)and then (Mf Nijt = (.X', F)f gives the Proof

,

.

.

,

S I7ltsup lFki-+, 'FkyI i

1

.

.

Qan (F, 7)

-+

the similar rela-

Laws. Suppose X and F are continuous semimartingales,

.

7)

continpops

As n 0 atmt cxl the second term apd the desird result follows.

(zz.x),

.

' .

an

H

semimartingale.

The secpnd of our abc's is

X

.

.t

,

Q, -(F,

.)t

and

M') + Qa(. Qhnu .t

=

it suces tt? show that if X is a locally of bounded variation then

since by the uniqueness of the decomposition this is an unambiguous deinition. From the deflnition above, it follows immediately that we have Theorem.

Since '

0 -+

a.e.

with

Aa

Ta

and Ta

g).

=

.

=

.

,

desiredresult.

(:1

Chaptez 2 Stochmtic Integcation

74

2.9. Associative

Section 2.9

Law it is useful, to

In some computations,

Let Xt be a continuous semimartingale. T Hte the integral relationsip

To exteqd to more general integrands, we will need a technical result. we k are bounded,

and

Ikdxs

t

'1

difl-eyeptials.'' where the dK and dXt are ictitious objects known ms A god xample is the derivation of th following forfrmla, which (fofobvius law: reasons) we call the associatlve '

.

.

(ff

Ar)

.

l#ff )

=

dH dlserentlals = gives #K f dXt

Proof uslng stochastic (7f aYl and observing dH

.

(.Jf aY)) .

=

Ht #t.Jf .Xlt

t

a

vdsj

ltnks yds)

=

0

Proof of (9.2) The RadozwNikodym derivative dvjdy identity is just the well known fact that

t

,.J

X

.

.

Ihh

=

.

,

F)t

=

dXt

=

Htdj?

.

)

#((#ff

.

X)t

is useful because it The above proof is not rigorous, but the computation routine tells us what the answer should be. Once we know the answer, it isxrtd then prdictable processes t verify by checking that it holds for basic onclude folloWing th extension process we used for defining the ihtegral to that it holds in general.

vdsj

/d

=

dv

-dg ,

See for

example

Exercise 8.7 in the Appendix

To for #.

simplify

the

extension

(9.3) Lemma. Let X be a Jf E lIa(X), then @)holds.

to l12(.X),

continuous

of

we will

=

ka

the desired

so

yds)

(1995).

Durret

tke th limit for ff

local martingale.

If H

Let X be any process. lf S, ff E 111 then

.

,

'

.

@)holds. .

.

. .

Since the formula is linear in Ii' and in ff separatelyj we can witht Proof loss of generality suppose H = 1(c,)C and ff = ltcjdjD and further that either are H 0 and se (i),both sides of the equation (i) b ; c or (ii) c, b d. In equal. hence In cmse (ii),

-+

'')

' .

. ' . .

'

.

(ff .X')t =

n

-

-

=

.

Sa a Sf S b b %t < cxn

-

0 LH and it follows that

.

(ff .A'))

(ff (ff .

.

CDLXL

=

X))t

CDIX: =

ttSff

.Xc)

-

)

an) -

.

Xjt

.

0 Kt Kc a Kt K b % t < cxl

-

= E

0S t

-

.

.

.

:

.

.

' .

..

l1.5(ff vY) H (ff .X) IIa2 IIH ((ff ffn) .X) Ila2 -

' '

0 Dxt JL) D1A% Ak)

llz ard

Proof Let ff n E 5111such that ff n Tf in lla(X). Since H is bouuded, n Sff S.Jf in 112(Ar), and it follows from Exercise 6.2 that CHK X c-zy tS.Jf ) X in Af2. To deal with the left side of observe using that @),we (6.4) twice, Exercise 6.2, (6.5)with (9.2),and the fact that IL is bounded we have

=

=

then

.

.

'c

and

e

-.+

(9.1) Lemma.

h and

. ttqtochmstic

H

.

p is a signed memsure with flnite total variation, t -,Jz(ds). Then let )) =

0

@)

take limits. First, however,

p((0,

eqtation t

aw

(9.2)Lemma. Suppose

0

as the formal

Associative

=

0

-

Hs

n

-+

x, .H

.

(ffn

.

.X')

-+

H

.

(ff

(9.4)Lemma. Let X be a continuous ff e l1a(aY), then @)holds.

.

.

d((.!r zn) x), -

.

X

/f2(f

E

-

4

B

s c IIff so as

.

ffanlz (f4.;y;),

,42

-

Jf n II

.X) in 442 qrtd tle result follows. local martingale.

If H E llatff

.

.X') and

.

Let Jfn Proof 6 2 imply

e

l1l

Setion 2.10

Integration

76 Chapter 2 Stochutic

that Hn

such

E

jjffn (ff -Y) H (ff .X')11z lllSn .

.

=

.

.

-

Hj

-

=

.

:

so Hn (ff observe that .

.

E

..

. To deal with the riglkt side of H (ff X) in .X') using the definition of the norm then (6.5)with (9.2) .42

-->

.

.

llffnff

.

11a2-E

Sff

-

-

0

1lHn

H llz2

-

.q

E

-.+

.

.

.. . .

;

-

X i M2

(),

we

.

.

Proof tet :r',, inftt : Jaff a2d(Ar), or #,zdtff = .Jf,1(,sa-a) Jf,1(,<,zk) and 1%T (9.4)implies that

.

. .

. .

E

.. . . . E ..

-

-

.

Xt 2t'F 2+,

-

IASI

y) .

(-'V2+, X3ylF y +

+

'f

mesh

yi )w )

-

lf ff E l1a('F)

-L;)(k+z

-

K)

-

-+

(-'',')

. .

.

.

.

.

.

.

ti

. .

.

in probability, while application of (6.7)to each of the last two that they converge to X,dXs and to X, d .

.

implies

sums

(:l

,

When is locally of bounded variation then (X, 1/)t H 0 and this reduces to the ordinary integration by parts formula of Lebesgue-stieltjes integration.

:

::*::

a

JJ

J

and

nJ. tt SJ

A )a >

.

Xt y)(K yyz

r+,

(0,t)with

of

77

(8.6)implies that

**

'j

(Ah

=

(.X+,

locl martingale.

:i=

A-P'FO

i

.

t

-

0

-+

.x

. . . HK aPP ly ing Exercise 6.2 we have Hnlz X

(9.5) Lemma. Let X be a cotinuous H Q Tlatff X) then @)holds.

.'

Xt

Salzff a2d(.X')a

(#,n

=

= SO

ffllaix

llffrt

(.Jf -X')ll2=

.

Several Semimartingales

of

Proof Let n = (tJ') 'be a sequence of subdivions goingto 0. We can write

Exercise

(6.4)and

H in l'Ialff .'X).

-+

Functions

.

- Tn

S

.

g--T=

(t

'

gn

x ) Lk

Tn

=

UsingCH (ff .X')) (6.2)uow it follows that result. G) gives the desired letting n .

.

=

)

F,dx, .

(Jfff

.

.X'lt for t

'.

.

and ILI

K Ta,

E

If S, ff E

(9.6) Associative Law. Suppose X is a continuous semimartingale. lII then H (ff Ar) = HK X. .

. .

' . .

' :

.

L'

'

Semimartingales 's

u'Vx

lf X and 'F are continuous

by Parrts.

t

t -

A'OF: =

0

JI#X, +

0

semimartingales

A-adls + (vY,F)t

sense.

(7.1):

(10.2) It's R, and 0 % c % d. If Xt, formula. Let f : Rd Xc+1 semiiaxtingales Xdj are locally of bounded and tinuous g t en .

.

..

of several ln this section, we will prove a version of It. formula for functions semimartingales. The key to the proof is the following

(10.1) Integration then

X,dF,

We are now ready to prove the following jentalization of

J

. J(Xz)

.

of Several

-

Note that the right hand side makes sense ip a path-by-path

C.OI1

In view of (9.5)and the deqnition of the integral it suces to pve the Proof adapted and locally of bounded variation. result when Xj = Wt is continuous on (0,s) lS,l, or the variation of However, by stopping the irst time Iff,lcovered 1:) something by reduces to (9.2). exceeds n the statement

2.10. Functlons

-

--,

.

.

.x'tK Ab'n

X

-+

.

=

-

1Xz4

d =

=1

0

Dilhjdx,

provided the derivatives Dj f, 1 continuous. Remark.

At this point

saving

.

.

i

.

'

)

+

K J' % d

.

1

.

-0

.'

lsflsc

and Dijf,

lhtfvk-stdq.x' 0

1 % i, J', <

a derivative or two may seem

c exist

.

X( are y vaxiation

J

X j )a and

are

like pinching

penniesbut Whenwe get to Chapter t and apply this result to utt, Bt) we will be very happy to not have to check that the partial derivatives lf and ut.w exist. . Proof By stopping, it suces to prove th result when lax'tf 1,(.X)t % M for satisfying a11 t. Since any function / the lzypotheses can be approximated by ,

78

Capter

Chapter 2 Stochastic Integration

polynomials gn in such a way that ga Dign 1 % K d and Dijgvt, 1 % j % c M)d, it suces to prove the converge to 1, Df.f, and Dij.f uniformly on (-M, allow and 7.2 Exercise polynomial. us to pmss For then (6.8) result when f is a from the result for grg to that fpr J. by linearity to prove the To prove the result for a polynomial? it suces ku zk= where /:1, 1-2, dj.. kn e (1, rponomial result when I is a z ,z superscripts, e.g., not should our krt powers, 1-1, that are note (The reader monomial might be zlzzlzlzlzz = says that then (10.2) If n = 1 and ,

To evaluate the third term (F, g) we observe that by for the covariance of stochastic integrals, (8.7), ,

,

,

.

.

.

.

.

.

.

.

.

,

.

.

n

('.F z) 1

'

f
0

,n=

m#

)

x(n+1))

l

,

Adding the last three

equalities

xkn

-

gives

ldx

=

n+1

F f Zt

,

0

-

FPZ:

=

0

i %a+1

which is trivially true. To prove the result for a general monomial, we use holds, and n Xt (m) be a monomial for Fhich (10.2) induction. Let X = l'-lmazl X3k(n+1) Applying

(10.1)gives

.

,

n+1

X,

-

2

tfxt)

m#:

t

+

Xkmt ,

m=!

.

1et

the formula

-,

Xtk

ZL =

iltxtfl

Akmt

=

,

.)

*l

(10.3)and

,j'sn-l:z m#

(,

i.J

d(A

%-'

-h

',

A

%4 '),

nzmz n 9%'.j

(Notice tht for each i in ttte sum for (F, Zjt there are two terms i and = n + 1, i in the lmst sum.) The proof is omplete. ,

.i

=

f, j

(10.2)to

F gives

Chapter n

j

s

-

i

,

Kn 0

(10.3)

so using the

C1

i#i

()

n

'

x'(J))

Z, dFa

m mt/i.

')f

,

=

fsa

+

-

1 Q

,

m=!

t .Jsa

0

y, dga

r+1

k(m)

X,

0

#(xk(f) x(J)) ,

contius mesh llnl

--k

tcal martin0 th

>

msl

t =

.t

'are

m

.1

-

(t.0) Quadrtlc Varlation. Sppose X ad 'F gales. If rj is a sequence f pititions of (0,t)Svith

my:

.yd

0

dxkit

X,klrnl

=

0

Summary

,

,

z,z

n+1

n

By deflnition,

(jjxl)

k(m)

X, ,f

1,

,

i

1aw (9.6),

associative :

.'

j

1

+2 i -

m=t

m#

n-l-

of the proof of the multivariate With the completion fprm of Is's formula, we calculus enough stochastic have to carry us through most of the applicationsin the bookj so *e pause for a moment to recall,what we've just dtje. In Section 2.1, Weintroduced the integrands ff, the predictable processes nd in Setion, 2.2 the (flrst)integrators Xt the contipuous local maitingals. with p, local In Section 2.3 we defned the variahce process assiated map covariance of and (.X, generally the two local marting>leg. more tingale, Theorem (3.11)tells us tha t t h e covar iance cpuld hye been deflned as the uniqu continucms predictable process dt that i loca ly of boundd vari>tion, The next result gives a has dtl is a martingaler 0, and makes u'V'i pathwe interpretation of (.X,F)t :

X' ktrnl dxkij

=

=

(7I

=

Applying

79

.'

'

.

Summar.y

y: , j .

Tl

Xs (m) J.X'k(rl+1)

I'I

m=1

,

Taking X time t.

:::F:'.F we see

that (aY)zis the

tquadratic

variation''

of

the path up to

82

Section 2.11

Chapter 2 Stochastic Integation

used only in The developments described iu the previous paragraph are material in Sectioh (9.3) of Chapter 4 and in Example 2.1 of Chapter 5. The would like used until 5.5. We Section 2.12 concerning Girsanov's formula is not of Chapter 3. beginning to suggest that the reader proceed now to the

To dal

extension of It's formula dtte to Veyer (1976) ln this section we will prove an Loosely speaking that hms its roots in work of Tanaka (1063).See (11.4). functions. of conkex two it says that It's formula is valid if f is a diflrence and cuntable hke a t Since such functions are diFerentiable except at a generali/>tim. modest signed memsure.. this is a second derivative that is a nice Brwnian general: if Bt is a staudard in do can best the this is one HoFqver, di the seminpytingale be mpst # t en f renc of motio and Xj ::p fBtj is a and Jacod, protter, (1se). conyex iunctiops. see (j this section follow those in Sections 4 and 5 of hapter IV o

slkazpe

cinlar,

two(1990)

developments in but things are Protter is step

since

here

simpler

we

.-,

X be

a contiuous

fvt

Ixzj

Here f') = limhtntftz) convexity and f is a z by -

(z continuous -

'

.

.

'..

.

By be the decomposition of the semimartingale. Proof Let Xt = Mt + and N N, that l.Xtl S l.l 'stoppingwe can suppose without loss of generality witlt compact upport in (-zx,0) function be all C* Let g a (M)z % N for and having j'gsj ds = 1. Let .lf

.

Js(z)

(a) Then

(b)

and

Jn is convex 1uXj4

-

j J (z +

-

Jn(A)

,(p) dp

0

Jn'(X,)

.

sets. Since

on compact

in t

difl-erentiate

(a) to

get

.X.)gy) dy yztz) n .y

:

'

tu

.

J

integrals (Eyrcise 6.2), imply E

(sup(z;'

.n)2)

zhu zlz s xsup t

-

-

'1

Kln'

4s

by the bounded subsequence we

/,

(-Y,)

-

T'(A%))2 d(M),

theorem. (Recall (M)x K #.) improve the last conclusion to

convergence can

sup

(Q* -

ft)2

almost

0

-.+

=

lJ:

(f)

sufp

-

Jt I

s

.

-

T o take the limit of the third note that (b)implies

and

0

surely

t

j,- 1Ja/(x,)

-+

By passing to a

/ 't ?' Let J1 /' (JL) dAs ln this L (-Y,) dAs and h from the bounded convergence theorem that for qlmt every o

g)

.

=

(z +

we

83

J'(x,)l dA'

final term ffy

-+

() 2 Jf

1

=

case

it follows

0 .J??IX, ) d(X), r,

we

')

C* Using It's formula we have t

the right,

=

is the left derivative which exists at a1l adapted increasing proess.

.

y?

=

uniformly

uniformly

-.+

dA', + Ik

0 .))/

on

Fomula, Local Tfiaae

' t X( co Now if 1: = s n x 4 ) 4 and z = () J'x , ) d.u , then tli; Z2 maximal inequality for martingales and the isometry property of stochastic

(e) 1'X't

=

y,a (z)

'rke

t -

(d) '.

.f(z) -+

Iht

'->

the irst term

with

hav no jupps. Our flrst

R be convex and let Let f : R (11.1) Theorem. and semimartingale f(X) is a semimartingale. Then

A(z)

fnv)

Local Time

Formula,

2.11. The Meyer-Tanalta

interval, it is easy to see that l.Xtl f N, it follows that

The hfeyer-lnaka

dxs +

'j

t

(j. 0

fn''xsl d(X),

co in (b)to get the desired Our strategy for the proof will be to 1et n continuous on each bouuded formula. Since a convex function is Lipschitz

,

ff

''k

=

fnu(.Xt) 1nu(.X0) -

-

()

IL (Xa) dX, g

Combining (c),(e),and (f) we see that almost surely the right hand sie of (g) converges to a lint uniformly in t, so th left hand side dots also. Since Ikn* is continuous adapted and incremsing the limit is also. E) If

-+

ated

we with

fix X then for a given f we call ff the increasing process associf. The increasing process associated with Iz cl is called the local =

Sectfon 2.1 1 The Afeyer-Tanaka Formula, Local Triae

Stochutic In tegation

2

Capter

84

that justifles this

tlme at c, and is denoted by LI. To prove the flrst property we need the following preliminary name, (11.3),

stopping we can suppose I.X I % N and (-'Y)t < N for all t. Having done this we can 1et N J(z)

1et fff be the increasing tl with begin hh-h = Iz-cj,so ff/+ff = processmssociatedclaim Jf Wesince is that h Ja = z c we have ffl fff2 0. To .! Our second 1#X, o ' let g) = < c which has g?(z) = 1 and note Xt A'c = see this ( the associated fft must be H 0. Let

A(z)

(z

=

(z)+,

-

c)observing by

A(z)

.

(z

=

-

,

and

(11.3) Theorem. Ljt' on l y i ncremses when X t ft' be the memsure wlth distribution function t ( : Xt = cJ.

= -+

E

. '

E

'

il . ' .

: . :

*' li c r t be' preis a sqpported by LI, then is .

'

. . ..

'

'

tf ' .

,

.

=

=

c)+

Xs

-

c)+

-

=

r

1(x.>.)

dXs +

1

'j'(f.1.

-

.q

#'(z) Starting

with

the deflnition

j

j.1

=

S

'((f

cl

-

-

and then integrating 1/2 p(vY)

lA%

cornplete.

N

Lx Jztdtll

.(A%)

-

-

Since LI for g and

We are nop rvady for our main result. Let (11.4) Meyer-Tnnalca formula. Let X be a continuous semimartingale. aud of derivative left be the functions, f of f' two convex f be the diference distribution functin = y in the sense of distribution (i.e.,# has suppose T) Then t 1 I'X'j #-Y, + /z(#c).U Ixzt

r?

0

cl c'=

-

=

p(A%) p(A%)

:

=

c)

-

signtxa

c)

-

dxs + Lta

'j

j'

Proof Since f is the digerence of two couvex functions and the formula is linear in J, we can suppose without loss of generality that J is convex. By

1

1

gives ' . . .' :

/.N

j.x

ptdc)

N

y

rt

j

signlx,

-

( . .

. .

c)

dXs

..

. .

' .

.

ydajL;

-N

(11.5) and (11.6) below will justify interchanging the two integtals in the flrt side to give right-hand the term on

=

-

-

signtz

the local time

of

-

/(m)

yda)

-N

+ =

.

N

't

1A)

1)

Zj. side vanish so L The left-hand side and the integral ou the right-hand : Xt k = and > S rational inf when T with H S holds any (t q Since this q a((t argument : Xg < c)) = 0. A similar c 1/n) for any n, it follows that c)) and a)+ : Xt > shows the proof is using (z a)- instead of (z -

cl

-

.q,

=

-

pztdcllz -x

,

a so LI Intuitivly, I-'V cl is a local martingale except when Xt Proof emsier is it to wlzen however, use the constant To Xt a. prove (11.3), is # alternative deqnitions in (11.2).Let S < T be stopping times so that (S, T1 C pplying (11.1)to /'(z) = (z c)+ we have (t : Xt < c).

xr

1 2

-

J(1

-

-

-

=

=

-

-

-

.

=

gz) Since (J g)H z a + bz for 0 on (-#, Nj it follows that fz) IzI/ N. Since the result is trivial for linear functions, it suces now to prove the result tor which is almt trivial. Difl-erentiating the dqnition of we have -

Proof

85

=

0 for

=

j

t

g'xsj

dXs +

> N (recall l-'Ll K N and 1a1 the proof of (11.4).

completes

use

1

CN

y j.x

Jtldclf4

(11.3))this

proves the result E1

For the next two results let S be a set, let c be a c-fleld of sgbsets of S, artd c$-). let ;t be a nnitemeasure on (,% For our purposes it would e enoug totake S R, but it is just as emsy to trat a general set. Tlle first step in justifying the interchange of the order of integration is to deal with a measurability issuq: if we have a family .Jf/@) o? integrands for a E S tlien we an deqne thq integrals .Jf: #X, to be a measurable function of (c,t, td). =

,

,

semimartingale with Xo (11.5) Lemma. 0 and Let X be a continuous let S/&) = Stc, =) be bounded and S x 11 measurable. Then there is a =

,

Capter 2 Stochatic Tntegrafpn

86

Z(c, t,)

e 't

x 11 such that for p-almost HI dxa

ts

J;

versionof

. @,tJ)

a Za,

every

is

a continuous .

.

Mt + Jlf be the decomposition of the Proof Let Xt = stopping we can suppose that l.l % N, lMtl S N and (.X')t a11t colusi Let 7 be the collection of bounded 11 E S x 11for which the Theorm of (2.t) the Monotone Clss hOlds Wewill check the assumptions check (i) of the MCT suppose in Chapter 1. Clearly 1'f is a vector space. To (t,tJ) where ff (t,td) e l'l and fa) e bS. In this case #tc,t, ) Jtclff =

,

.

=

t

'1 .f(c)

=

0

0

=

i.:zz

:/

.

Js

Z /z(da)

ff s dxs

TI ?t. Taking /' and zt-to be ipdicator functions of sets in H). = (X with B E i S, B of checked MCT x e Tespectively, we have (i) te with Hn 1 H and H bounded. To check (ii)of the MCT let 0 K Hn E of the The L2 maximal inequality for martingales and the isometry property stochmstic integral imply u%

so Jzf e

.4

.

sup It.Jfn'c

.f')

-

.

CH'

.

't

M)tI2 g 4sup

EkHn' (Jfn,c

= 4E By passing to a subsequence

we

sup lt.Jfnk't' variation

To deal with the bounded ..),

ltzznz'z ssup f

.

almost

have for p .'l

-

.

(ff

'

-

.

Mjt

-

ll''

Hall d4M),

-

.

M)l2 O

-+

'z..)I s

Af)t

l

Js

Iat

/2tdc)

Tf

l

A'lt

'

CH -Yl

=

.

so f ff e 7t. Taking f and ff to be indicator functions of sets in S and 1-1 respectively, we have checked (i)of the MCT with = (.A x B : 6 <S,B G 11/. To check (ii)of the MCT let 0 S Hn E 'S with Hn 1 H and H bounded. Letting 11,11 be the total mmssof pj applying Jensn's inequality to the probability measure Jz(.)/IlJzl), then mltlplying each kide by we have ,4

,1.

11,11

1 s(/

sup

s 11/211

lzlz'a z/l

ydallz

-

s E

sup t

S

Izfn'? z/ 12gda) -

Using Fubini's theorem, the L2 maximal inequality, and the isometry property the right-hand ide

0

-+

=

Izz;''c H:6 dl.lf -

0

'

every c

part we note that

Ej*

1L .X')t

=

(l/s T(') pdas =

and

'//

E

Proof Let a'V= M3 + A be the decomposition of the semimartingale. stopping we can suppose that l.Alz< N, lMl K N and (.X'): = (M)f S a1l As noted in the proof of (11.5), the result in the special cmse X .A just Fubini's theorem from memsure theory, so we will suppose the proof that X M. Let 7 be the collection of bounded H E S x li for which the conclusion holds. Again, we will check the assumptions of the Monotone Class Theorem, (2.3) in Chapter 1. Clerly is a vector space. To check (i) of the MCT, (t,Y) = where ff (t,kJ) suppose fftc,t,l Jlalff e ll and .J(a) e bS. In this cmse ZI = J(c)(ff .X')t so .

By (Af)z K N for

semimartingale.

Hs' d.X,

'

'

Section 2.1 1 The Afeyer-Tanal'a Formulat Local Time

-+

0

s 4

S

Izn'

fsup

1

7/12 gda)

-

.i'latn'a

s

sup

-

j

gda) Z.112

Cr

= 4

E

t.Jfan'l

) s by using the bounded convergence values inside the integral we have

zlz,c,(dc) stsypl/s

/-/-u:

dx- ,(dc)

-

/'/ utyda) -

-

Hslll #(;),

-

plda)

theorem three times. 2

-

j zC,(dc)I)

s / sup lzJ',.)(

dx,

0

.->

-

J

S

'

zplptdalhz ,-,

J

()

88

Secton 2.11

Chapter 2 Stochutic Integration

Taking H1

js tff'''

=

nk we have that with Jlldcl and pmssing to a subsequence Zlk'' /.lldtzl converges uniformly to ZI lztdcl. ..X') =

Js

Js for then it The lmst detail is to check that #n X converges to .I X in proprty, the isometry we cap followsthat Zf llda) = CH Xlt ln view of ffllx 0 but this is routine. showing that llffn by detail lst the complete

probabilityone LHn

.&d?k

.

.

h

1jj E 11#

j

x

0

.

.

-+

-

js

#,nztlgdaj

-

2

J Iha gldaj

ys KE

(14.Ar),

We can now

-x

(.J6

-

S:)

#(.X)a

ydaj

-z+

(:1

of L3 ms the local time at c. with

-

.

1

-M 2

(.;% c)+

=

-

(.X

-

Ald (vY): i N '

'

c)+

-

1(x.>c) dXs

-

-

-

(A% c)+ is continuous, the stochastic integral

so we only have

-

of

0

and 1et f G C2 Suppose flrst that g is continuous Proof compaTing with It's formula proves the identity in quesion. this cmse (11.4) continuous function, using the Monotone Class Since the identity holds for any (:1 memsurable Theorem proves the result for a bounded fH

=

g. In

.q.

When X is a Brownian motion the identity in t

da

=

(11.7)becomej

R

0

Ia as a mapping from R to C4(0, T!, R), the Fix a time T < x and regard a real valued functions continuous on (0,T1, which we equij with the sup norm view of ln Kolmogorov's contlnulty teorem (41.6) in IIJII supcysz IJ(s)I. Chapter 1) lt suces to show that for some a, p > 0 we have -.+

=

E 11.f?f.511/f Clc -

11+G -

.

< CE

-

There is a version

2

yz

.0114

.E'1lf.J

J

j

(11.8) Tlzeorem. Let X be a continuous local martingale. L3 is continuous. of the LLin which (c,t)

dX,

1(x.>u)

=

Suppose without loss of generality that c < One of the Burkholder Davis Gndy inequalities (45.1) in Chapter 3) implies that

#(X,) ds

0

spent at a up to time From this, we see that ? can be thought of as th time occupation time measure function for the density @,or to be precise it is a lA(.) = lx (Xa ) ds. The There are many ways of defining local time fpr Brownian motion. above is famous for making it easy to prove that there is a one we have given of c and t. It is somewhat version in which L: is a jointly continuous function martingale and it remarkable that this property is true for a continuous local is not any more dicult to prove the result in this generality.

-+

lm1K N

c

c)+ (X It is clear that (c,t) continuity to investigate the joint

'

that

we can suppose

.

-.+

.(X,)d(JY)a

=

-+

Proof As usual, by stopping for a 11t By (112)

local time LI.

with

-x

=

-

s

0

f0:(c)da

fC#(J)

=

-->

-+

j

x

=

'i$

'J

Let X be a continuous semimartingale (11.7) Theorem. function then If g is a bounded Borel memsurable co

=

,

X

the identi:cation

complete

89

Exnmple 11.1. (11.8) is not true for Xt lftl, which is a semimaztingale by (11.1). lntroducing subscripts to indicate the process we are referring to, we clearly have f,1 (t) 1,1 (t)+ f,s-'(f) for a > 0 and fg1(t) 0 for a < 0. Since 0s() / 0 it follows that a j-(t) is discontinuous at a 0. One can complain that this exampl is trivial since 0 is an nd point of the set of possible values. A related example with 0 irt the middle is provied motion. by skew Browninn Start with l# l and flip coins with probability of 1/2 and > 1 + to assign signs to excursions of Bt i.e., maximal p o p time intervals (a, ) on which lft 1> 0. lf we call the resulting process then f,e (t) 2ps0 (t) as c t 0 and f,e (t) 2(1 plso (t) as c 1 0. For details see Walsh (1978). -

bounded convergence theorem three times.

by using'the

Tlle Meyer-Tanaka Forznula, Local Time

/0

To bound the right-hapd side we note that using Schwarz inequality and Fubini's theorem:

=

E

=

(

j -

c)

b

2

L dz

j

b

d(.X'),

1(u<x.s)

(11.7)and

&

<

(

-

.E'4f4)2 dz g

ajE

(

-

/.(.14)2 d

c)2

sutjjj; rE

S(j)2

then the Cauchy-

.' 90

Capter

Secton 2.12

2 Stochastic Integration

T

fgj,

laYrr

=

zl

-

lan

-

-

zl

:

S

-

,

jg-zj-jzrzj

and

.Xb)2 + 2E

signtaL

z)

-

'.

''.' :

.%

T

st

%ZELXr

'

z) dXs. ..

-

0

whichwith tlze trivialinequalities (c+)2 :g 2/2+252 impliesthat .E'(f4)2

'

sgntx,

-

dXs

jp-zl

=

'=

Since 1 is

K 8Nl + N4

2(2X)2 + 2S((X)z)

There is a

=

J'f

f

cet

converse

of

dotldpt

=

have

we

is

(12.2)which

a martingale/p. will

be

usful

in

(12.8) Lmma. partingale/p Given m a nonnegative etwivalent pfobability memsure Q so that dQt(dP3 'E

'

constrttcting

there is '

' .

.

examples. a unique

locally

&t,

=

-+

Proof

-'->

.'&,

a

in the lemma defines the restriction of Q to .F for < frz, Wf oh C, lt t1 < ta n). Deflne the i Jlf i for be Borel subsets of Rd and let B (1 : % K e fnite dimenpional distTibutions of a measure Q by setting

is a

if and only if as

(local)martingale/o

(local)

Proof The parentheses are meant to indicate that the statemept is trt!e if ltd X q Fs. X/W, b a martingale/o, s d t the two locals aie removed. Let Z dQf. So ir Z dfl, and (ii) Zat dfl = if Z e .'Ft, then (i) Z dp and martingale/o, have we is a (i), (ii),the fact that (ii), (i)

QBt

.

.

ttfl

=

1

semimartingales and th: In this section, we will show that the collection of equivalent affected by a locally dennition of the stochwstic integral are not canonical probability change of memsure. For concreteness, we will work on the coordinate generated by the maps Xj @) space C, C), with h the qltration said to be locally Two memsures Q and P deflned on a qltration & are . equivalent, and i.e., Pt are equivalent if for each t their restrictions to Qt = dot/dfl. The Temsons In th cmse we Yt a mutually absolutely continuous. for our interest in this quantity will becom clear as the story unfoldsy

X is

The last equation

any t. To see that this deqnes a unique meure

ormula

2.12. Girsanov's

(12.1) Lemma. martingale/'.

a martingale/o

(12.2) Corollary.

and (.X')t by by the isometry property and the bounds we have imposed pn lxl heed to clzeck th iqulity i kolmogrk' stopp ing. This gives us what we proof. and completes the continuity theorem

M + A. X) The proof above almost works for semimartingals Remrk. t argument the = dAs 1(x >a) and then = J/ dMs 1f we deqne f/ 0 1fx >.) t t) 11 is continuous, so we wl11 get the desired conclusio above shows that ((z, JI is continuous. if we adopt mssumptions that imply (c,t)

91

This shows af is a martingale/P. lf F is a local martingale/o then there is a sequence of stopping times T,l 1 x so that Xv' is a martihgale/ Q and p thorem implies hence aykagk is a martingale/P. The optional stopplng that and magksarr'a is a martingale/' it follows that at is a local martingale/'. To prove the converse, observe that (a) interchanging the roles of P and and applying the lmst result shows that if pt dpt/do, and Z3 is a (lca1) Q martingale/P, then ptzj is a (local)martingale/o and (b)#t = th-', so letting result. Zt mK we have the desired C1

f1,.

of

To bound the sup, we recall the deinition

Gfrsanov's Formula

=

t

.

'

dP

B

whenever t k fn. The martingale property of tyt implies that the qnite dimensional distributions are on?istent so we have desned a unique memsurq on (C, C). Ea We

are ready

to prove the main

result

of

the

section. .

r'

(12.4) Girsanov s formula. If X is a local martingale/p f aa (a, Y) then Xt Wt is a local martingale/o. j

Jo

and we let dt

=

-1d

.z

, ,

-

'i

J

=

'

afykd.p

.

f

tnrjx

=

dPt

'Fido

=

A

W

W

=

Y,dQ,

a,F, dP,

=

.4

f

J

,4

a,Y,d#

= yt

.

.

..

'

..

..

Proof Although the formula for z1. looks a little strange, it is easy to see that it must be the right answer. If we suppose that .A is locally of b.v. and hms .zl.n 0, then integrating by parts, i.e., using (10.1),and noting (a,.A)t = 0 since W hms bounded variation gives =

(i2.5) tvj(aL-z4t)-pX:

't

t

(X,-u4,)d(,+

=

0

asdxs 0

t -

0

a,d.,+((,

A'lf

92

Chapter2

Section 2.12

Stochutic Integration

(12.6) Theorem. The quadratic variation ?')f under (aY, is the same P and Q.

made in Section 2.2 that our Eltration At this point, we need the assumption conyinuous only >dmitp martingales,so there is a continuous version of t:e to apply formula. iptegrationzby-parts which we can our martingales/p, and tlze fifst two terms op the right local (:4 Xs are Since view of (12.1), if we want Xt Wj to be m (12.5)are local martingales/P. Irt Jl. the choose that need martiugale/o, sum of the third and to so we a locll fourth terms H 0, that is, .

.

=

(Ax,

= a, d.Aa (a,-Y)

.

n=1J

(t

xla

(12.7) Theorem.

is to prove that the integral that deflnes If t K Trj then

The last detail remaining : at Izet T,z inf < =

tu-ldta,

=

E

E

yk exisss.

E

(3.9). To

< f2n

.

=

t) is

prove the

a sequence

-

x,7)2

(A')t

-+

lf H E fll

then

(#

.

.X')t is the

under

same

P

and

Q.

Proof The value is clearly the same for simple integrands. Let M and N be and B be the locally bounded variation parts the local martingale parts and of X under P and Q respectively. Note that (12.6) implies (M) (#)t Let

.

.4

,

.

and

.

in probability for any semimartingale/P. Since convergence in probability is not afected by locally equivalent change of measure, the desired conclusion (:I follows.

that it is neces-

(9.6),it is clear

.

-+

t

F'rom th last equation and the asspciative law that sary and sucient t

and hence the covariance

P-oof The second conclusion follows from the first frst we recall (8.6)shows that if Aa (0 tg < tkt of partitions of (0,@)with mesh llrjl 0 then =

-

0

(;)t

=

0

' dI(a,-Y)I, n s tx

.5

.

'

djta,xlj, 0

< x

Tn

T,z then A is well by the Kunita-Watanabe inequality. So if T = limn-ztx, = Et, defined for t % T. The optional stopping theorem implies that Safwza have > Ts noting t we at on ataza so =

.E'(; Ts K t) Since at

k

0 is

continuous

and

=

; Trj i t)

Ean

lf H G f11

and ffz

Letting n

-.+

x,

K .E'(&t;Ta S f)

we see that at

=

=

.E'tag'aiTn

0 a.s. on (T

-

=

f(a;T

But PL is equivalent to Qt so 0 = f$(T %.t) #(T < x) = 0, and the proof is complete. ,

S t) K 1/n

(12.4) shows that the collection of memsure. Our next goal is to locally equivalent to P, and H : under P and Q. The flrst step is

% t). Since t is

arbitrary (:I

is not aflcted by change Q is that if X is a semimartingale/', l'l, then the integral LH .X'l is the same

of semimartingales show

.

0 and

lll k

or

(4.5)we

j I#r

n)

can

flnd

simple

Jf,1 d(I.I+

-

z4))

=

Hm CN +

Hm

lfI)

so

that 0

-+

.p))t

.

(the left-hand side being computed under P and the right that for any H e fH the integrals under P and Q agree (x7 the proof is complete. n is arbitrary and Tn

S t4 = 0 PIT

l-l

(M),

.

allow us to pass to the lirnit in the equality

Hm (M +

so S @J,

=

-+

-

-.+

Q(T K t)

:

0 for t k Ta then by

S'IIN 11.2= SIIAT, 111.,',1

These conditions

T's K T, it follows that

=

inftt

=

.

f'tcet;T S t)

93

E

'

t,

Girsanov's Formula

Q) to conclude to time Tn Since

under up

.

E1

.

3

'

..

.

Motion,

Brownian

'

'

11

,

In this chapter we will use It js formul to deeperi ou j un tj ers ( a utjsg nian motion or, more generally, continuous local martingales.

oj

syow.

and Translence

3.1, Recurrence

f lf Bt is a d-dimensional Brownian motion and B1 is the th component tll B3 If i # j Exercise 2k2 in Chapter 1 tells us that is martingale with (#)f = Bi.1Bi't is a martingale, so (3.11) in Chapter 2 implies (# Bijt = 0. Using this R is C2 then infoimation in It's forzula we se that if f : Rd l

.

1

-+

i . :

:

..

' .

E

. ..E

f(Bt)

E

.

.

JBn)

-

'

the Laplacin

:::'z

Dl, of

.

J, we

1Bt4

(1.1)

.

.

,

0

=

quation

'd VT(#a) dm

:

j.

. i .

%(

E

.E

(

DiifBsjds

g

of

rceht

'

0

(

'

the lst

write

cn

.

E

DLILB, ldsj +

by) fbr tlw

1Bz4

-

.j .

=

i

Writik vy

'

()

i

J, ftd l i:z8 Ef.l natlf ws mre i

1

+ ;;

Lbiil for

Lfmtds

0

of two vects hd tv fo# the inr prdt i ti qrst ieim td equation. given previous mening precise of is in the that the nmctions with AJ = 0 ax c>lld harmonl. (1,1)shops that if we pmltion iult BioWian with i a lcl irtartihgale. hrmonic the futi pos a The next reult judicious choices of hl4monic functions) is the key to deriving pxoperties of Browian motin frop It's formula.

tlot

H,

(nd

inft : B j G). If Let G be a bounded open set aqd r closure continuous of G, G, then for tlke and o AJ 0 in J is k=fB.). hake J(z)

(1.2) Theprem.

J e :2 z

E

and

G we

=

=

,

=

suptlz 1. Let ff < *) Ottr qrst step is to prove that '> zl B / and then of ff G. lf z E G 1#1 z, y e G) be the diameter r < 1. Thus .&(r

Proof

=

=

=

.&(m< 1) k .P.(1f1

-

zj > ff ) =

.b()#1j

>

ff )

=

> 0

#1: & and -

Capter

96

rzowzlanMbfon,

3

Recutsence and Aansezice

Section 3.1

.II

Exl holds when k 1. 'To prove th.e lmst This shows supy k ) K (1 y) ptz, is the transition probability result by inductlon on k we observe that if for Brownian motion, the Markov pzoperty implies .&(m

=

-

(1.5) Theorem.

For

all z and y,

< x)

/.IT:

97

1.

=

/.-:,(:r < x), it suces < x) Since to prove the result shows reqection intended) A little 0. we can also suppose when y (pun (M 1)/M, and the right-hand side > 0. Now using (1.4)/.41L < Twz.) z D approaches 1 as f oo. .p.(4

Proof

=

=

E

(

pzlz, 2/).P!,('rk

.l%('rk ) S

1)

-

ty

'

:

E

=

l

-

where the lmst two inequalities follow from the induction fact that 1#1 zl > ff implies B / G. The lmst result implies that x) = 1 for all z Q G and moreover that '.

'

.

.

.

'.

.

.

'.

'

.

(1.3)

.

.

Errp

sup

<

.

(1.6) Theorem.

<

.

for

(:x7

all

0

T

For

Prm G)

j'

E rr

1tp-p

=

=

tr

0

@)dt

(r >

conclude that

. E

.. .

:

' .

' .. .. E

. . .

(

PzBt

any s < x,

=

y for some t

k

s4

1.

=

By the Markov property,

Proof

'

(xi

< p <

(1,5)to

.&('r

-

.

It is trivial to improve

the

and

mssumption

'

-

=+

)k-1#.(#1e G) S (1

K (1-

'

=

y for some t

k

s)

=

f.(Ps(',)

(Tp <

(:x)))

=

The conclusion of (1.6)implies (rgue b# contradiction) that for an# y will of probability times is 1 there a sequence t,z 1 x (which depend on with 2) conclusioh outcome that Btn we will hereafter abbreviate as so y, a the often'' i.o.'' the terminoloa of the theory inflnitely CCBt In Bt y or y onodimenonal what shown Broppian that have is Markov we processes, of motion is recurrent. '=

on (0, When Lf = 0 in G, (1.1)implies that fBt4 is We have assumed that f is continuous on d and G is bounded, so f is bounded aud if we apply 9he tiqze h,>nge n' deflned in Section 2.2, Xt =F J(#v(:)) is a bounded martingale (Wlth respect to G .G(:)). eing a ounded martingale surely to a limit Xx which hms Xt = S.(Ak lft) and Xj copyergep almt Sinc r < (k$ an1 f is continuous on d, X. Fzi fBpj. hnc EZXg = EnX. ECXO Ezfmj. E1 z EmX. Taking = 0 it flloWs tliat fkj

=

=

E Xerc lse 1 2 .

Use

.

In the rest ' of t is sec tion we willuse (1.2)to prove some rsults concernin: the range of Browni> motion fBt : t k 0J. We start with the pnerdirpension>l

(1.6)tp conclude limsup Bt

::

=

cxn

=

t-+ oo

liminf Bt I-+

-x

=

CO

In order to study Brownian motion in d k 2, we need to flrtd some apof Brownian propriate hazmonic functions. In vieW f the spherical synmtry niotion, an obvious way to do this is to let w(z) J(lzI2)and try to pick za2 z'= z2z thar lzlsince it 0. We use 1212 R so tht L + f :R is euier to diferentiate:

Case.

=

Lt

b and T

c < z <

J%l.

=

c)

=

b

=

=

tnftt: Bt 4 (a, )J. '

... .

(

z

fblfz

-

Proof hms fa)

exercise usinx the

J(z) 1,

=

=

b

J(b)

z)/(

-

-

0, so

=

c) hs

fn

=

0 in

(1.2)izplies J(z)

=

=

)

(

=

j

=

.&tfz

:::

'

. .

.

.

.

'

'

.

l .!

inftt

:

Bt

=

z).

Fom

(1.4),it

=

-+

y

.

.

.

DJ(IzI2) #'(IzI2)2z# + 2F(14l2) 2T(IzI2) .f''(lzI2)4z3 =

on =

=

(c, ), and a).

EI

Therefore. for A .

Deduce the last result by noting that Bt is a marting>le stopping theorem at time tr'. tional op =

Eyther

.

a

(c, ), is cohtinuou EcfBrj

.

'

jj

....

:

1.1

Let Tz

El

=

'r).

local rrartingale

(1.4) Theorem.

1

follows immediately tat

and

. ..

i'

= 1'

0 we need .

.

.

.. .

0= =

.

.'

.

.' '

'.

. '

.

j + zy/tjzj)) )-)(yzz(lzj2)4z/ 4lz12/'N(1zl2)

.

..

+ 2d J?(lzI2)

Section 3.1

98 Chapter 3 Brovnian Motion, .1T .!5(.h

Letting y

=

lzI2,

we can write

the above ms 4yjmy4 -d

+ 2d/'(:)

=

0 or, if y

>

for all 6 > 0, so So inf ( > 0 : Bt

0,

=

,

1 2p (#)

T''(p)

=

0 for some t 0J we have =

=

C

coosing

=

< x)

Prsz

Thus, in d there.

appro-

=

0, and thank: to

99

deqnition of

our

,

(1:10)

0 for z # 0, so by Taing J'yj = cp-d/2guazantees Lp riately we can let d 2 1og lzl p(z) = d a IzIa-d

0)

>

'Ikansince

and

Recurrence

0 for

=

all z

motion will not hit 0 at a positive time even if it starts

k 2 Brownian

=

z

We are now ready to use (1.2)in d k 2. Let Sr Since (p has L 0 in G (z : r < lzl< J0, r < (1.2) implies .&.

=

=

Exerclse 1.3 Use the continuity of the Brownian path and P=Sz to conclude that if z # 0 then P.IS, 1 (x7 as r 1.0) 1.

inft : I#fl r) and and is continuous on =

=

,

=

Enpmj

=

whrr w(r)is short for the

epwsc

5)

<

+ w(A)(1

-

PrS

<

w(z)on (z : Izl rJ. Solving

v>lue of

x)

1

=

(1.7)says

For d k 3, formula '

/(z)

=

::1:

=

now gives

.jp-# jzjz-d a a s- a uu s-a -

Ss))

f

(,S

Ss)

<

=

'

' E'.

.

.

.

.

.

.

E ..

:

.

.

.

.

here. The fact that two There is no point in fxing R and letting r that three dimensional dimensional Brownian motion does not hit Brownian motion does not hit points and indeed will ot hit the line (z : zl cxl in (1.11) we get zz 0J. If we flx r and 1et R .-+

,

pointsEimplie,

(1.7)

Pr'v

5k)

<

P (.a) P( s)

=

w(z)

-

.(p(g)

=

-+

=

'

Ih d = 2, th lmst forqmla says

. .

(1.12)

1ogR log Izl logR 1ogr

< co)

Prsr

=

'

'.. '.E .' . .

d-2

(r/lzI)

.

.

if

< 1

.i .

.

lzl

> r

-

Prsr

(1.8) If we fzx r and 1et R

(:x)

-+

in

<

5k)

=

-

(1.8),the

right-hand

1 for

any z and any r >

msv

<

(x))

proof

of

(1.6) shows that

=

side goes to 1. So

From the last result it follows easily that for d k 3, Brownian motion is trnnsient, i.e. it does not return inflnitely ofteh to any bounded set. (

.

(1.17) Theorem.

0

As t

-+

x,

l#tl

-+

co a.s.

Proof Let Wa (I#tI> n1/2 for a1l t The strong Markov property implies =

a' d repeatlng

the

q

in the sense

Twdimensional Bropnian motion is recurvent (1.9) Theorem. PrBt G i.o.) > 1. that if G ls any open set, then

lf

z#0

we llx R, let r

-+

0 in

Pzuk

<

(1.8),and 1et So = izftt > 0 Sa)

lim P=% K r-w0

< S.a)

=

:

Bt

=

0), then for

=

=

0 for some t

k

e)

=

.E'c(#s.(To

<

as n

-+

x.

X

(a4c) n =

Erpms-lsnz/u

Now limsup Wa

=

Ptlimsup

of Brownian paths implie Since this holds for all R and since the continuity x) >ll 0 for z # 0. To extend the lmst Sa 1 cxl as R 1 x, we have Pcsz < result to z = 0 we note that the Markov property lmplies

.Fl.8t

P

0

(:xl))

k Snj and

note that Sa < co by

(1.3).

.

=

0

ho inqnitely time Sn and

.this

nxcKkz

.4n)

< co)) UaO=N

=

(nl/2/n)d-2 --,

0

Xa has

k limsup #(.a)

often the Brownian path never returns implies the desired result.

=

to

1

(z : jzj K n1/2)

after EI

hav proved the following result about how Dvoretsky aad Erds (1951) fast Brownian motion goes to cx) in d k 3.

Caper

100

i.o. as t 1 x) J%(I.IK #(t)V@

accordingas

section we will investigate the any z

Suppose g(t) is positive and decreasing. Then

(1.14) Theorem.

#(t)d-2p

)*

dt

1 or

=

concerned Here the absence of the lower limit implies that we are only ttnear shows calculus that x.'' little A behavior of the iutegral X

(2.1) .Pa,

x or < x.

=

t-1 log-c @dt

:::

the

with

(ii) -P.(.

k 0) H 1 if and

0 for some

=

>

0)

H

we

have fouud is that

j

if d

only

'

/(z)

(

. . . .

..

(i)

w(z)

cxl as

-+

(ii)lw(z)l -.+

(z( -+

co as z

3.2. Occupatlon

-.+

.

.'.

.'

.

Tk

=

inf (t >

are

x if and only if d

<2

2r). For' k k 1 1et j

Tk-l : Bt E Dj Sk : Bt C G)

>

we get f6i k

jg

t)

dt

k

Ssk

s

(4.5)in Chapter

y.r

jo 1r (ft ) dt k s

Pmsk)

=

=

k1 Hst

1 it follows that lo Bt ) dt

i.i.d.

are

. ..

Since these random variables have positive mean it follows frm the strong 1aw of large numbers that lo Btj dt 0

Proof of

(2.2) x

Erj

If

f

mo

tion. As

yj

t

= -.+

5C S

pco

=

j

X

(.h)dt

=

(:x)

a.s.

p,(z, y4

function, then Fubini's theoyem implies

dt .E%T(.R)

j j*

ptz,

-

px

=

y) dt

0

(2<)-d/2e-Ir-pI x,

lo

O

is a nonnegative

= here plz,

S''*O

result.

/(.B,) dt

Times W

k 1im

=1

provig the desired

0 in d k 2.

r) = (p : 1p1 < rl the ball of radius r centered at 0. Iu Section Let D = ln this 3.1, we learned that Bt will return to D i.o. in d ; 2 but not in d k 3. .8(0,

inft

n

d= 1 log IzI d = 2 Izjz-d i k :

(i)and (ii)above

and the features relevant for

=

j

d using the strong Markov property,

1D(f

show that for

G2

.8(0

=

Sk

Sk

.

G

and

0

=

101

in d k 3

< x

K2 :

.

in d

1

=

Tk

P=

lzl

=

x)

=

0 iu d k 2. .

.

dt

Let Tn

for T1

From this and

:

properties of what The reader should observe that these facts can be traced to spherically transformations) symup to linear we have called p, the (unique metric function that hmsAw(z) = 0 for al1 z # 0, that is : .

(#f)dt

.JC1o(@) (2.1)

Writing

the basic facts about the recurr

At this point, we have transience of Btownian motion. What some

lo

x or < &

rderived

(i) .P.(#.&fI < 1 for

-b'

Jc* 1D (B t ) dt and

time

occupation

'

.

rence and

(2.2) Es P rOO f of

Zf/(1ogt)EVd-2 for > 1, so Bt goes to x faster than 81/2.81 according ms a K 1 or a view of the Bmwniarl scaling relationship Bt =d that in any a > 1 Note The lmst rsutt thau V@. sensibly exct p e at a fmster rt e we could not shows that the escape rate is not much slower. Revlew.

Occupation Times

Section 3.2

Drownfan Motfon, Tf

3

p

j j pttz,

E

p)T(p)dydt

y(p) dy 5

V

tor

/ 2 t is the transition density Brownian (2xt)-d/2so if d 2 then Jpttz,yjdt x. J

s

=

Sectlon 3.2

Chapter 3 Pzownan Moiion, 11

102

When d k 31 changing

variables

lz

=

(X

(x7

pflzl p) dt

0

t

=

1

-Ip-zl d/2 e

0 (2.mt) () s

=

Flz

(2.3)

l =

z

# l2-

s

0

r(#2 1) = 27+/2 lz

whererta) 'i'

.

z'r

=

4.1.

' '

1k.1r

y4

(?(z ,

:

g .

-

'

''

''''

''

.

.

* pttz, v)dt

j,

=

1, we let

=

ds

c

tu

then in d k

#

'

z. /-

(2.4)

2.mc =

..

.

'

..

'

G4, p)

t)

1 ,/-2

=

J 0

,

1p

To complete the proof of (2.2)now we /(0, r) lxld changing tp polar coordiuates G(0,p) D

(fp

r

=

=-

(2.5)

that taking f

observe

s

0

cd

s2-d

c

ds

=

extend

.r

j

is

(#,) ds

=

.E

Ip

zl

=

since

W

co

ds :() .

,

s-

1/ 2

(y

N

1/2

'u

z)2/23/2

-(g

ew

-

-

as

c

.sj

x

x

c=

2 z 'u

du

,,-a/ztu

-.ds

z

u

.s

ds e

z) a

-(!/

z )a

-

e

c

W

-

X

j

z

-

-

zl

2

j.)

-u

,/a

ds e-'s-

ju

2

a

-jp

=

zl

-

0 co =

dv :-

r2/ 2

a

strong Markov motion starting at 0

j

lp

) ds

m

K Eo

j

the

7-2 =

-

1 1-2 2z. 2 .

y-

=

The computation is almost the same for d 2. The only thing that changes pt(0, 0) again, then for z is the choice of at If we try c y the integrand and the integral diverges, so we let at pt(0, d1) where 0 as t ew el (1,0). With this choice of cf we get

'

'

''

.

-f-1

'

-+

=

=

(

lo

=

.

C(z yj

(fa) ds

kez-nel, because G(., yj is tht electrostatic We call G(z, yj the potential for more (1978) potentialof a unit charge at p. See Chapter 3 of Port and Stone approach another take to In d K 2, j') pf (z,#) dt H (X) so we have to on this. useful G: dennea

,

t:o

1 F=

zg )

(e-lT-#I a /2f jzt

x

j

p)

=

0

(pt(z,

-

ct)

dt

(2.6)

.

,

= 21 c -

Gz' p)

1)t-1/2 dt

-

=

applying

property at

lo with D

x

2W

y

-

< x

2

-

=

-y.r2

d-1

j =

.

the last conclusion to z # 0 observe that the exit time from B, lzI)for a Bzownian rotational symmetry we have using the antl G) (r T

Tp

=

a

(e-(p-''r)/2t

-

(e

';r

/c(z,,).f(,)o

-

*

1

=

and the integral converges, since the integrand is K 0 and /2zI gives y z) a -+ Y. Changihg variables t

y, and

.f(s,)a

p,(0, 0). With this choicek

=

-

:3, G(z, y4 < co for z

..:E .

.

When d

gamma functipn. If we deqne

'

'

(d/2)-2

0

pl 2-a

usual

is the

s-e-'ds

2s2 -3

/c(z,,)y(p)oj- ,'..f(a)a

cu

G(z, g)

-

n*

-

&

*

2,0/2

pl 2

-.lz

-,

d

-

J

dt

d/2

#l2

-

a /at

103

where the cf are constants we will choose to make the integral converge (at least when z # yj. To see why this modifled deqnition might be useful note that if f(#) dy = 0 then (mssuming we can use Ftzbini's theorem)

:12/2,: gives

-

Occupaton Tmes

1 = 21 -

=

-zg

1

jz-pja/af

-

e-1/2t) j-lj f-ldj

e-' ds

l/aa

x

ds e-3

()

jx-yja/z,

1 t- dt

=

ds c-4

(- logtlz

-

pj2))

-

=

1

logtlz

-

p1)

''

Captez

104

Section 3.3 Exit Tlnies

3 frownlan Motion, 11

To sum up, the potential

are given

kernels (r(d/2

by

1)/2xd/2)

jz

p12-d

dk 3 d'H (2.7) Gx, #) = 2 (-1/ 1Og(I i/l) = pl d 1 -1 Iz p1) where Ctpllz The reader should note that in each cse G(z, p) v is the harmonic function we used in Section 3.1. This is, of course, no accident. Gz, 0) is obviously spherically symmetric and, as we will seei satisfles z G(z, 0) 0 for z # 0, so the results above imply G, 0) + Sw(Izl). The formulas above correspond to W 0, hyhich is nice and simple. But what about the weird looking B s? What is special about tm? The answer is simple:They are chosen to make laLGz, 0) (a point mass at 0) in the al sense. It is emsy to see that this happens in d distribution' 1. In that cmse = then P (z) IzI 1' X 7* 0 (P?(z) 1 z < () Bpnzj yeaders n check that -2Jn. Mre sophisticatd if B then so pageq 96-97.) this is also true in d k 2. (See F. John (1982), To explain why we want to deflne the poten tial lternels, we return to Brownian motion in the half space, frst considered in Section 1.4. Let .I (g : yd > 0), v inftt : Bt / H, and let -

-

the formulas for G imyly that IG(z, #)T(p)ldg are qnite, so the last two equalltles are

compact

J

of

support

1 and

-

=

-

-+

=

=

.

.rztz,

,

=

(

=

=

# be the

.

,

.

-+

so Gxz,

yj

=

If z

e

H,

; k

0 has

Gxz,

pd-l,

(g 6

of p through the plane

zqection

(2.8)Theorem.

then

.

Rd

:

yd

=

and

support,

compact

0J (z

.f(z)

:

>

0J

11*

/c(z,,).f(,)o s-(/-y(s,)a)/c(z,,)y(,)o-

J

0

Proof Using Pubini's theorem which is justiied since f k 0, then > 0) C H we have fiofn Chapter 1 and the fact that (z : .f(z)

E. J

J T(f)

dt

=

J

0

/

> f)

EslBttr

pl +

-

p)

-

,,

j- j (pt(z,,) j- j (pt(z,#) -

=

(z + (z +

y)

=

y4

=

2p 2z

when

0

when

z < y

p)J(p) dy

-

< y < z

so we can write (2.9)

y)

Gn,

=

2(z A y)

for

al1 z, p >

0

Gn, yj is constant 2z for y lt is somewhat surprising that y a11points expected occupation time! is, y > z have the same =

-+

k

z,

that

3.3. Exlt Tlmes

(3.1)Theorem.

H

J

p) + z) +

-

-

atttytdydt atttytdydt

-

Gx,

p1. Separating things into cases, we see that

ntz, f))J(p) dydt

H

0

-(z -(2/

=

#1

-

=

=

(pt(z, j /0

(4.9)

p)

lz+

-Iz

=

In this section we investigate the moments of the exit times r = inftt : Bt / G) forvarious open sets. we begin with G (z : Izl< r) in which case r Sc in thenotation of Section 3.1.

0

*

=

dt

using

introduced, we invite the reader to pas Gx, yj looks like in one dimension. If you will probably Iot guess the behavior The corhptation is easier than guessing

-+

-lz

7/).f(p) dp

Gxz,

=

Witi this interpretation for GH for a minute and imagine what y you don't already know the answer, for small tallt. x. So mh ms y the answer: t7(z:#)

-p4)

(p1 ,

J(ft)df

g

=

=

,

T

E=

=

=

E1

S'ekpected

-

-&

=

E

The proof given above simplifles considerably in the case d k 3; however, part of the point of the proof above is that, with the deflnition we haye chpsen for G in th #eurrent case, the formlas and proofs an be th yame toi al1 d. p) = G (z,p) f?(z #). We think of Guk yj s the Let time at ocqpation (density) y for a Brownian motiori staztin: f z and killd leaves The rationale for this interpretation is that (forquitlble Jj it #.'' when

=

-1

lG(z, yjlyjkdy Jvalid.

'

=

'

.

.

The and

105

J

G(z, #)/.(! dp

Proof

If

1zIs r

then Ecsr

The key is the observation

l.512 -

(r2

=

-

that d

dt

lzl2)/(I

=

i =1

((lj)2

-

)

106

Capter

3 Prowpan

foton, ff

Section 3.3 Exit Times

being the sum of d maztingals is a martingale. Using the optional theorem at the bounded stopping time Sv A t we have

.

lzI2 =

Em

stopping

llfsratlz Sr A t)dl

(3.3) Theorem.

-

(1.3) tells us that Er,h < cx), and we have lfsra 12S r2 so lrtting t Srd) and using the dpminated convergence theorem gives IzI2 Er implis the desired result. ,

(r2

=

-

,

Exercise

3..

tet

b > 0 and T

c,

These local martingales are useful for computing sional Browniau motion.

infft

=

:

whici

I::l

q

)). slzowEzT

/ (-c,

Bt

x

-+

=

ab.

To get more fomulas like (3.1)we need more martingales. Applying lt's and formula, (10.2) in Chapter 2, with X3 X3 a continuous local martingile, X,2 = (.X)j we obtain =

,

i

lXt, (-Y))

J-%, 0)

-

. ..

jvblh + jo + jo =

,

From (3..2) we see that if () D11 + Da).f tingale. Examples of such functions are

J(z, y4

z,

=

z2

z3

y,

-

t

1

z4

3zg,

-

-

fh

(X)t) is

,

6z22/ + 3:2

local mar-

a

1Z

n

,

#)

Cn

rn2*-2FN

#

,

-

-

so that D=/2 of the mth term is the (n/zjthterm is 0.)

=

EzfBp-hl.n

-

(a

6./aaza

.

Ta) + 3('q

..

.

(/'n)

J

.

' . :

(1.3)and 2 the dominated2 cppyergence Usij (i)and c 4 ttzEora +3.Eq

=

-

.

-c#4t+#2j2

-cj3

js a

j

.

. .

theorem

a

.

.

we

ieranging (:I

(local)martingale

Our next result is a special cmse of the Burkholder bavis Gundy inetualities, (5.1),but is needed to prove (4.4)which is a key step ih qur pot of (5.1). :

.

.

.

.

If Xg is

a continuous

E

FFI

K n/2,

cs,o

=

1

and

for 0

%m

sp

local martingale

44 <

with A

=

0 then

3811,(.;')2.

-

1)

cancelled

3.;%(X)f

,

Xt4

-

<

(n/2)

Proo f First suppose that IXtl and (.1):are S $J' fr all t this case (2.5) in Chaptr 2 implis X4 6.42 (.X') + 3(X)2 is a rartingale, so its expectation is 0. Rearranging and using the Cauchy-schwarz irtequality .

-

-(m

=

+ 1)c,z,m+1

by Dv of the

(m+

llth.

The frst two of our functions give us nothing new (a'L and XI local martingales), but after that we get some new local martingales: -

=

0smK(n/2)

where (n/2)denotes the largest integer we pick 1 2rrI)(n cn,mtn 2m j

X3

0

(3.4) Theorem.

=

scaling

E

-

j

or to expose the pattern f

c). Then

Proof (i) follows from (3.1),so we will only prove (ii). To do this 1et Xt Bg4 6BIL + 3t2 and Ta K n be stopping times so that Ts 1 cxn and v'Vaguis martingale.Since T,, < n the optional stopping theorem implies

>)Xerc ise 3.2 Find a t b c so that B6j and use this to compute Ezrt.

(x),)(ftx)a

Dtxs,

I.&Ik

n.

Now lfaarr'al i c, so using cxl to conclde 0 can 1et n gives (ii).

(I)a)d(A-),

0, then

=

Let a = inft : (ii)Ezraz = 5a4(3.

for one dimen-

The dependence of the moments on a is easy to explain: the Browniau c1/2#j implies that va/az =d relationshipBct =d

' ' . . . .. ..

(-Y),)d'a

c2,

-+

Dallla,

.j

=

.

t

(3.2)

(i) Ezra

expectations

107

6.42 (.X') + 3(.X)2, j

(,D.. of

>lJU/ qF3/l(Jf)/

zz

6>l(JUt2(Jf)t) % 6(1lJF/)1/2(/;(7f)/)1/2 '

.

-

(.X)f re

Using the L* mxximal inequality fact that (4/.3)4 K 3.1605 < 19/6 E satl X1

(/.3) in Chapter 4 of Durrett we have

S (4/3)4fX1 X 19

LE

X1(

syuy

1/2

:

. .

.

ard (1095))

'72

E (.1)7 )

the

Section 3.3

108 Capter 3 rrownanMotion, 11 Since 1.X,1< M for all s to get

divide each

we can

X14ll

by ftsup,
-

So if Ts is a sequence applied to Xan gives

then square

S 381(f (.X)t2)1/2

sup .X4

E

side

4%t

The last inequality holds for a general martingale if we replace t by Ta = infti : (:x) and. using the t, l.;V1, r (.X): k n). Using that conclusion, letting n E1 monotone convergence theorem, *e have the desired result.

'

'

Letting n

-+

1 (%

2

as

'. ..

and using

the monotone

exptz

=

-

D11 + D2)J (1a

p/2) satisfles

a continuous

Local Martingale. If X is (3.5) The Exponential 1(.X')j) is a lcal martingale. tingle, then Cxlt = exptz) 2

4e k

0, then

=

by Jensen's inequality. martingale.

local mar-

=

.'If w let 5$

a'

.

.

xbtx

:

.''''',

. .

.1..

lo(X)), ..9;49.

....

then

...

.

.

(3.6)

Fn

-

(3.2)says

..

Kd.X,

=

... .

stochastic

T'(f) T(f)c()

J(0)

=

.

.

=

gives of

the

local martingale

a continuous

Proof Now,

Let Zt

=

inftt

:

Bt

=

a).

fb expl-n) Remat'k. to prove

with

J(.X)t) is a martingale.

(-Y)f % Mt

la(2.X')), which is a local martingale

continuous

exptX

=

=

exptzx:

-

(X)j)

is a

local maris a

lo(.X'))

-

Y'

'

,

(3.5).

e

=

K t)

t = 0

a >

=

Zg exp((W)z)

0

and

k0

-av

1/2

a (21rs )

'

transforms, you

(3.8)

can use

.-aulzs

ds

ce

=

-

..'

.

.

.

.' ..

.

E .:

'

' .

.

.

=

Ez expt#fnasa '

.

-

#2(L A Sa)/2) .

..

. .

.

'

.

.

.

. .

l

If 0 k t h q 4i ght-hand side is K expt#c) so letting n -A (x) and using the 02Taj2). Taking 0 = bounded convergence theorem we have 1 = Eo exptpc result. 2 now gives the desired (:) -

12

'i$

#2/2) and Srt % n be expt#f #nIT. < x) = 1 by (1.5).Let XL Proof stpping times so that Sn 1 (x) and Xthsn is a martingale. Since Sn % n, the optional stopping theorem implies 1

by

Then for

If you are good at inverting Laplac (4.1)in Chapter 1: J5(n

( -

Let Tc

1

PIIrPOSeS.

exptzxf

that

we see

Ia

=

(3.8) Theorem.

Jc

=

Chapter 2 now,

'

=

=

Kt

Eetting 0 E R and setting Xt 0Bt 'in (3.6),whre Bt is a one dimensional 01tj2). These Browrian motion, gives us a family of martingales exptpf mrtingales are useful for computing the distribution of hitting times associated with Brownian motion.

exptyk), which for a given continuous function a(t) has unique solution J() assumptions) suitable Jk = c(s) ds. It is possible to prove that where (upder details. for solution of Dolans-Dade Z is the only (1970) (3.6).See The exponential local martingale will play an important role in Section 5.3. Thn (audnow) it will b useful to know whvn the exponeriql loc>l martipg>lr and (1.15) is > martiug>le. The next result is not yery sophisticated (see(1.14) enougi results) for is and but of better for Yor (1991) in Chapter Vl1I Revuz

-

(2.5)in

's

=

d.;L Tis property of X As in the cse

=

(3.7) Theorem. Suppose X is and A = 0. Then X explA

.'

-

diferential notation, that dls X the right to b called the martingale exponential ordinary diferential equation

Ol1r

.

we have

.

0

i

or, in

that

. '

1F,1

Remark. It follows from the lat pr o f that if X t ls tingale with 0 and (X) % M for a11 t then 5$ mar ti/ale ih M2

..' ''''

-

Using

lsup

k

,

'>Kt

.

therem 2

sup y2

E

4e An

=

.

convergence

109

inequality

,

S 4e ut Ezthn)

a K 4f5$avk

Mt

If we notice that T(z,yj another useful result. we get

Zg the L2 maximal

of times that reduces

Supxagu

E

Exit Times

110

Section 3.4

11

rrowmlanfoton,

Captez 3

Change

-2/.1/12

then is a martingale. If 0 = is a local martingale. Repeating the proof

Let

a

inftt

=

:

Ez expt-a)

Then for

2e -cWI/(1

=

c >

k 0,

and

0

+ e-acW.)

Exerclse

3.3. Let T-c

Proof Let () Ez expt-L). time a (anddropping the subscript

inftt

=

Applying the strong Markov property at a to make the formula emsier to typeset)

=

la

c).

j#f I k

JtT-c

Zt

:

of

-a).

111

0 and exp(-(2p/?)Zt) (3.8)one gets

-

=

If c, ;t > 0 then

=

< co)

Time, Lvy's Theorem

020.2/2

-0y

(3.9) Theorem.

of

exp(-2(w/J2)

=

(

g ves

>otexpt-r); Bp + fntexpt-'rl/zctl;

Ez expt-T'c)

=

-#z).

Us tng te

xpression lbr

1

,/.()

given in

(3.8)is

of

Another consequence in Section 7.2.

a

-c)

Br t::z

3.4. Change

it follows that

e (-c, c)

Llvy's

Theorem

'

re.ult.

the dsird (3.8)now zived bit of

thatwill

calculus

l

in handy

come

.

,

lf 0 > 0 then

(3.10)Theorem.

of Tlme,

ln this section we will prove Lvy's characterization of Brownian motion (4.1) artd use it to qhow that every continuou local martingale is a time change pf Browhian motionk

expl-r)

(1+ #2.()).E

=

c)

Since Bp

Symmetry dictates that ('r,Bpj =d ('r, and we have r and Bp' are independent,

#.()

=

'

,

(4.1) Theorem. If X is a continuous local martingale (.X') % t, then Xg is a one dimensional Brownian motion. Proof

(4.5)in

By

Chapter 1 it suces

with

(4.2) Lemma. For any s and Xa.k.t X. is independent normal distribution with mean 0 and variance t.

of

-

/,

J

.za

/ zj e

va'

1

.#j

gj

a

w,

Changing

variables

=

=

t

* 0 1

20

1/2s, dt

=

2.: 2x X

()

z a4 C

B-

the integral

=

#/ 24

-

1

y-ie27.s3

ds 2s2 p/2a

above

8

.

.

.

#X1

e-a

a J.

:

1 = io

'-

e#x'. dl?

0

.

ps

t

2

'

,

cfpxv' du

)

g

.

.t

,

.

,

,

'

and integrate over

. ..

..

.E'tKavki.Al since Fo = 0. So

we

=

E CE lKaau

( .

fle

fGlavu

lh'4 ;.A)

.

.. . .

The deqnition of conditional

expectatin

=

.E'(y;.A)

.

.

.

implles

0

=

have

=

exptplzz

'z

o,

=

'.?

=

02o.2tj2j

-

'rhe Erst termon the riat, which we tet Fp' a+r and let e F. = W ill ar11 martingale with respect to .F* i a tocal To get rid of tat termlet .j Ta 1 fx, be a sequence of stopping times that reduces K, replace i by t h T,,, : ::.:.z

The exponential martingale can also be used to study a one dimensional c's + pt where c' > 0 and p is real. Brownian motion Bt plus drift. Let Zt c2 with martingale (-Y)t = so (3.7)implie that Xt Z% pt is a pt)

(7.9)in Chap-

..$.0'.

.

-

hms a

,

= z2 and consulting and the remark afe: (3.8) Using (3.8)now with c = expression is equal to the lmst of the T. we see that for the density function D right-hand side of (3.10).

-

and

F,

Proof of (4.2) Applying the complex version 9f It's formula, ter 2, to X./ = Xs-vv Xa and fzj cii e ie= we get

'i

-

0 and

-

-dsl2s2

Proof

z jyyj

oxp ( .j

=

show

to

,

t:o

a'

.)

;

.p(.,4)

-

=

()

-

tan

02 -E -

Z

a

e

ioxt

v

tfzj..x ,

112 Caper Since

gives

Ie#'l

=

1, letting n

.p(z)

.g.

t

:2 2 :2

--E

=

-

=

(ef #X

E

, v

;

''

it

PI.A)

--.

0

are

J('z)dn E

..

.

:

. .

.

.

:2/2, Since we know that IJ(s)l S 1, it follows that I.(t) y('tz)1% If conclttde and equation continuous diferentiate last j is to the is can we so j diferentiable with -p2 T Oge ther with

=

.P(.A), this

itj

2

shows

.

that

#(.A)e-#Qt/2 =

:-#2z/2

#x:.z)

j

Exerclse 4.1. A 0 and

F(p(X()l.'S)

that

'

EgvM

i

P(A1 q B)

=

expectation,

.

.

and

we

*1 .

.

.

:

have proved the desired EI

Suppose Xgi, 1

with

local martingales

are continuous

S Kd

=

(-Y then Xt

or

dp

=

,

we see

dp .E'(1s(A1)1.'S)

=

by the deflnition of conditional independence.

=

xt,

.

.

,

(0 f

XiLt =

if

=

i

otherwise

Xjdj ij a d-dimensional Brownian motion.

j

.

An immediate Ee

e B)

'

.j()

=

integrating

'tzl

=

.j(0)

and

113

A monotone clmss argument now shows thqt the last conclusion is ttue for any bounded measu.a ble g. Taking g = ls and integratig thJ lst eqality ovyr E SLwe have J7(.)#(aX1

..z-

=

(4.3)by e(-#)

Theorem

Lvy's

of Time,

.4

.

.. .

J'(f)

and hence

t

02 =

.A) du

says

=

each side of

Change

.E'(:(X1)1.T-1)

the two measures

and

../()

-

Multiplying is a constant

dp; .A

()

-z

-

epX.'

by Fubini's theorem. (The irtegrand is bounded zex, ;z), the lmst equality writing

luite.)

theorem

using the bounded convergence

cxn and

.-,

i8x1; )

Ee

Section 3.4

Afolon, 11

3 frownan

(4.1)is:

of

consequence

Every continuous local martiugal Tvith A% s 0 and i>ving (A'). > (xa is a time change of Brownian motion. To be precise if we 1et y(u) inftt : (X)f > ul then Bu = Xvtul is a Brownian motion and Xt #(x),

(4.4) Thepreim

W

=

Since this holds for

all

.A E F it

f8Xl

Ee

(4.)

ibllowsthat

=

characteristic conditional or in words, the variance and with tribution 0 mean dis

'. ' '

'

E

.

.

. (

:

Proof Since ;((X)j) t the second equality is >n ippedtate consquence of Bu To the first. prove that we note Exercise 3.8 o f C h ap ter 2 implies tt Bu2 and maytingales. yhow lpcal continuous, B%q that it is to are sp suces .

e-#2t/2 lyl()) az

.

=

'u

of VY(is

function

that of the normal

@.

Tp get from this to (4.2)we flrst take expected values of both sides to The .concludethat X; hms a normal distribution with mean 0 and variance condttional X't constant charcteristic suggests is tat function th tht a fact is idependent of FL.To turn this intuition into a proof let g be C fhctlon nd let with compact support,

-

0 is

(4..5)Lemma. Bu .Fv(u), u ,

a

.-+

'tz

lpcal martingale.

.

E .

)

'

.

j.

Proof of (4.5) Let Ts implies that if u < then

poj

#(2)

-

1 2*

1Ah1> n.

The optional

=

wherewe have used Exercise 2.1 in Chapter to replce F-tulhn by .'G(t).To let ?, (x) we obsvrve that the 12 maximal inequality? the fact that X2v(v)A v,a , imply is a zpartinggle, an the definltipn of tXlvtvlaza .2

dz

-

more than enough to conclude that

E sup X 2(vlaza n C

W(-

theorem

l.'G(u)) Xvlulac-,k ,(z))

ior

stopping

'

-+

mssumed

be its Fourier transform. We have p is integrable and hence =

j

e#'plzl

:

Exvjhn

' =

inftt

=

'p

o)

tyz

v

k 4sup EX2vtvlav'a '

E

-

Fl

= 4sup .E'tA-lvtvlwza s n

41

Section 4.3

Chapter 3 Brownian Motion, 11

114

theorem imply that as n The last result and the dominated convergence expectation L2 for t conditional is a Since in Xyt) Ivttlagk x, lt follows that f'txvtvlagu in Ehv) contraction in l.'G(u)) I.'G(u)) E1 and the proof is complete. -+

-+

'tz,

=

'N complete .

'

''''

'''

(4.6)Lemma.

(4.5),the optiozal

Proof of (4.6) As in the proof of that if u < then

-+

deqnitionof

x

'y($

(X)v(v)an lFv(u))

-

that using

we observe

then

theorem implies

stopping

2

Xvtulavk 2

(c + )

-

2

(X)v(vjAn1

-

A

c

,

(3.4)j nd the

+ 2.E'(X)v2(v)

a'vvlagk

Let Tk

inftt

(X)t

:

(X)tAn

=

'

.'

.A

=

((X)x

B f+

<

(4.7).

of

= =

(sup:1,41< huptXg <

x)

xl

Clearly, C c: B C B+ ln Sectiop 3.1 we showed thp Bpwnian potion Proof has limsupz-..x Bt = x. This and (4.) implies tliat n- C r+, pr r+, 7 x. Finally (4.7)shows that C C. EI

t7k

''''

'''

'y(u)

=

=

(x.+

Bu

(x).)

.

:

'

.

.

(' .

.

E

=

XJ D (T .

< x)

X): '

=

;

j

Jcdtxla

we have cxl

liminftff 41+

.X)f

=

-x

E

ah d there is no remsonable way to czttihue

'

.

-

.X)z

'

''

-Yvtu)

on ((.Jf

=

limsuptS zrz'

.

.

.

(

:

inftt : (X)t % it) To prove the promised extension of (4.4)now, 1et limf-... Xt n ((.1). < (x7), 1et Bt be a Browian when u < (X)x let Xx motion which is independent of (a'V,t k 0J,and 1et =

.

to deflne CH #):' fo'i t k T .

E .

'

.

,

(zz

on ((.X')x < tx7).

S n

.

'

,4

'

so

-+

'

JJ

.

E

' .

The result in (4.8)can be used to justify the assertion we mae at tie begizming of Sections 2.6 that IIa4uY)iq the lgrgest possible class of integrands. ffldtxla < co). CH X)f can be defined Suppose H 6 11 and let T = supft : for t < T and. hms

< x) > 0. In this of time (-Y)x The llrst

=

''''

:

.x),

lmt-.x 6 M i so lk Using this with Exercise 4.8 in Chapter 2 we get a'Vatr'a x zagk 2. shows and lim-.x Xt exists almost statement surely in The last exists (>a now gives x) D ((X)x < n). Letting n almost surely on (4.7)yCI

Yu

(lim-.x Xg e'xists ) *1

=

#((X)x

amount

k nJ. (3.7)in Chapter 2 implies that

(XW)i

. .

. ..

The following sets are equal almost surely:

'

.

with

surely

exists almost

limxx =

g'

'

q

it leads to a converse

.

.

.

.

Our next goal is to extend (4.4)to Xt cmse Arvtul is a Brownian motion run for an stp in making this precise is to prove

Proof

(4.8)is that

E

The proof can now be completed ms in (4.5)by using the dominated convergence (:l expectation is a contraction in L2. theoremj >nd the fact that conditihal

(4.7) Le-ma;

C1

'

.4

K 2E su-p

s cf(x)2-?(v)s c12

'

of X.

(4.9) Theorem.

lA-lvtula(rk

2 2 K 2c + 25

H

*

:

'''

idpendent

.

E SllP (Xv2(v)agk n

=

-

' The reason for our interest in

'p

. a Exnjhn

suces

:.

0 is a local martingale.

.'G(u), k 'u

(4.1)it

By

.

mtion

'''

.

u,

-

to show that Fu and FJ u are local mrtingales with xespect to the fltration J(5$ : t K u) This holds on (0,(.X)1) for kemsons indicated in the proof of (4.4).It holds on g(uY)x x) because B is a Brownian Proof

,

the proof of (4.4)now it remains to show

.

.5u2

To let n

F is a Brownian motion.

.1,2

-+

E'.

115

't?.

.:2

..

(4.8) Theorem.

of Tlme, L4vy's Therm

Cange

(X)x

'z

<

u

k (x).

Convergence is not the only property of local martipgales that caq be siug (4.9).Almost any almost-sure prprty ciitini tlt iwni rzkiuk ca. be trauslated isto a corzesponding result for lol tates.Tlkis gives us a nurnher ol- theorems about the behavior of patks o local martingales. We will state only:

stuied

path immediately

(4.10)L.k f the Iterated Then on ((.X')x = x),

Logrithm.

limsupXglLfxjgj -+ oo

Let fz(t)

pz

1 a.s.

=

2t lozlogt for t

k

c.

Chapter 3 Prownfan

118 set

of

(b)we

interchange the deflnitions and let

To prove

=

Sa

=

T

=

inftt inftt

Reversing the roles of Bs and

s 1/2

.p(,r1/2 #,

Bv*< J)

<

roles of Bt

% (#

'r)

and

t A r)1/2 in the flrst

-

1)-

1

2

.-'

Z, taking expectations

random vriable

.'

zvz)

A 5'2 A T)

1)2)

k (#2 -

('rA

-

l

.. .

!

'

..

X %2)

P

.

.

s: PX

pz

aud the above

time result mentioned

is

tn

the proof of

(a)it follows that

2

>

K J) # >(X

' ,

2

>

2#(x

)+

>

#(y

(5.5)vith Z -. K &zu't EvXI24

arld using

=

>

:

Jj

?'Jr

1)-1h-iS(#(r h S1 A T)2j A Sa A 1a)2 BLv = (#2 < x) s (#2 1)c1-2(J)2.p(S1 1)-12.p(g.1/2 > ) % p2

so

that ff :2

<

tollws

>

:-1.

Fom the

K KNEVCY)

Evk-lnq

-

..

Ev(YI8)

,

-

J)

>

X/2, X, Yl6

1 and then pick N k 0 so that 2N ihat growth condition and the monotonicity of it Pick J

-

and using

) #w()

>

. . .. s

Integrating dpj

A T)J

j,

=

-

using the stopping

119

lqumption it follows that

From or

(r A & A /l

f((r

Replacing by a nonnegative Rzbini's theorem gives

(5.5)

in the proof, it is easy to check t it -

Adapted to Brownian Filtrations

Mavtingales

'X'

:

.P((r A 5'2A T) 2

h

('rA t) 1/2> ) : (r A t)1/2 > #) : l#(r A t)l > J)

inftf

S

>

Section 3.6

Motion, 11

-

since 1f('rA T)IK (J)2 proving(b). ;

this ombining

E1

'

.

.

=

=

=

again gives

E (X)

'..

It wlll take ohe more lemma to extract (5.2)from (5.3).First, we need moderatly lncrasing if tp i? a > deinition. A function p is sajd tc! be nondecreasing fnction with w(0) 0 and if there ts a constant ff so that w(2) K ff w(). It is emsy to see that w(z) = ZP is moderately increing 2F) but (ff w(z) ea= 1 is pot for any z > 0. To complete the proof of (5.2)now it uces to show (takep 2 in (5.3)and ote 1/3 / 1)

Solving for EpX4

Evxt

=

w

EvX4

S CEeCY)

It is enpugh to prove the result for bounded p for if the result holds for v A n for a11 iz k 1, it also hols for p. #ow p is the distribution function of a measure on (0,x) that hms Proof

e()

=

j

h

dwt)

Jcxl

=

j

1(?,>) dwt)

381 Revlslted. 16. Taking 8 i:

=

%ff 82EpX) + ffN+1Xw(F)

; fffw(X/2)

now gives

-

(5.4) Lemma. lf X, F k 0 satisfy P(X > 2, F K 4) S &zPX zk2) for l1 8 k 0 and p; is a moderately incremsing function, thrn there is a constant C that only depens on the growth rate ff jo that

inequality and usipg the growth cpnditipn

with the previous

ff N+1 jaf w(F) ff

K1

E1

-

To compare with (3.4),suppose p = 4. ln this case ff 1/8 to make 1- ff J a = 3/4 ad X :z 3, we get a constant

164 4/3 .

=

87, 381.333

=

24

=

which

''

.

.

.

Of course we really don't care about the value, just that positive flnite constants 0 < c, C < (x7 exist in (5.1).

3.6. Martingales

Adapted

Flltrations

to Brownlan

Let (f%, t k 0) be the fltration generated by a d-dimensional Brownian motion with Bz 0, deqned on some proba bility space (f),f, >). ln this section Bt will show (i)all local martingales adapted to t k 0J are continuous and we =

t1% ,

120

Capter

Section 3.6 Matingales

3 Brownian Motion, 11 variable

(ii) every random integral.

2(f1,

X E

Bco #) can be

ms a stochmstic

written

,

121

1). Repeating the aud it follows from step 1 that 5$ is continuous aron (0, above and using inductiim, it follows that th result holds if F gpment hBtt lkBtuj where tl < t2 < < tk % n and h, fk are bopnded =

-

All local martingales

(6.1) Theorem.

Adapted to Prownan Filtrations

to fBt, t k 0)

adapted

.

.

.

.

.

.

.

continuoustunctions.

are coninuotls.

Proof Let X3 be a local martingale adapted to fl'h t k 0) and 1et T,z % n to show that for be a sequence of stopping times that reduces X. It suces enough wprds, other continuous, it is to show that Ta) in is each n, X(t A or whee = of .E'(Fl1%) martingales form result the 5$ holds for the e Bn We steps. of three generality in build up to this level ,

'

.

,

3. Let '.F e Bn with E 1y1 < x. 14follows from a stand>rd application class theorem that for any E > 0, there is a random vartable the zpozlotoze considered Fl < 6. Now in step 2 that hws E l.XT of the form Step

of

Xf

-

.

Step 1. . Let 'F = fBn then X /(#n), so t Markov property implies .->

=

is a bounded contitmous function. lf t k n, is trivially continuops for t > n. If < n, the .E'('jl%) = (n t, Bt ) where

) where f l

K

.ls, z)

=

1f(A' , 11'lt)f.( jz ls t ij -

s stjx

yjjsj)

-

and if we let Zt = .E'(lXf s F111%),it follows from Doob s lnequality(seee.g., that (4.2)in Uhapter4 of Durrett (1905)) j

-

-jN-.ja/a,y(yy; 1 e a/a (2vs)

=

'

dy

sup t Kn

Z)

K EZn

>

=

Slxe

'Fl

-

Now .X'f(t) = ECX' l1) is continuous, so letting e ot continuous functions, uniform limit is X(u?) so a td,

ts,

&(s,z)/ -

so the dominated convergence lBns. Step 2. Let F contlnuous.If

=

=

'

Remark.

=

dz

I

like to thank

would

above.

a

Kn

and

1 n, Itn

-

t, Bt4

f Ja are bounded ,

-.>

and

VichaelSharpe for telling

.f1(lt,).E'(.f2(fta)ll%z)

on

(t1,co).

.f1(ft,):(ft,)

=

On th

.

.

.

(6.2)Theorem.

For any X

with

e

fa2(f),lx,

P) theye

,

d

=

Ex +

J'l =1

x

are unique

a.e. I:I

the proof

me about

let t1't,

x continuous

s

We now turn to our second goal. Let Bt (.R1, Btd) with Bo t k 0) be the qltrations generated by the coordinates.

11BtlEhBtzl3Btt

=

0

and

Hi E 1la(ff)

H1 tllj

0

We follow Section V.3 of Revuz and Yor (1991).The uniquenesq is immediatesince,the isometiy property, Exercise 6.2 of Chapter 2, ifnflis thr is ozly on way to write the zero randoz variable, i.e., using Hi (s,1) > 0. To prove the existence of I%, the flrst step is to reduce to one dimension by proving : Proof

'

where :12)

z./.;)

theorem implies that as t

step 1 implies that K is so the argument from then if Yi = .E'tlz 11) aud haud, other %t1,

'tz

+

ivez

=

where tl <

=

'i

(zgjd/ac-I-I2/2.f(z

fBtlhBgu) k 1, then

zi

is function on (0,x) x R, so at t n, observe that changing

E

0 we see that for is continuous.

-+

z) is a continuous It is emsy to see that continuity continuousfor t < n. To check sd/2 zv-s, dz gives dy = variablesy c: z +

<

1 =

(2x(tz -

is a bounded continuous function,

K

=

.jg-zjayatja

t,))

d/a

-,)

y Lyjtu

d

so

f(./(ft,):(ft,)I1t)

Then X

%fl

=

,

.

.

with

EX

=

0,

and

1et Xi

=

S'IXIF.I.

.

Geometrically,see (1.4)in Chapter 4 of Durrett of jection X onto f,?, the subspace of mean zero random Proof

for t

Let X e f,2(t'1, Bx #) .X + + Xd.

(6.3) Lemma.

Xi (1995), variables

is the promemsurable

)' ;'(' y' 1'tj' ;y' q' k' $' J' l')' (' 111E1151;* ;'))' jjz-'yjjjd tyf y)' t'lqjd tjd tg-' t'jjd t.)' r' (grljjd ((::,p' ti i.(.(k1:))'. ..j( ;yjyjj, ( . ... . iE . ry(),y) '(..!(;:..gyjj.jjy.j!yjjyjjjj y......jgi . . -.'...(.'E.,'..'.,;.r.,')y(j(rkItp!('E(j..... ...

jjrygyjjjd tjj,,:,;,d ,'114:(4* r,,-,;,j,d (lqjljjjf y' jd .y' yjd ;'yd 41::4)* (' jj::yyd Eiii'd r' t't' 'jytd 'j' y' j;yd j'@' jjjyy'jjd t'yjd jtdyytydt.yyjd jyd .j' q' 11:::::::::11* ' 'E (i (i .E!.'(''. )' #' ))' j)d y' jyj-jdy.y'jd jrd jj'j' j'jjd ;'('.11:;;' yjyjtd )' y' j-jd (' q' 1* :. ..('.' .!'.ljE!(((jj)). jy (E.yj . @; E ; .:lqj. .. :)(E r.yy.. y(.EE .i qE(. .r(yji.jE!)j'q ' ' 1. j y j. . y.. ..j',''E;'E.)E'.Ei).,!'i;(i))l?..1,!1E'j)..!,..'..iqi.,..E.,y.. jj yj y j1*. tjr .. j.......j.... jj1*. ...' jjj

-;1jji;,' (' L' 5* )' 5* tjljd ('J'yjd j's')d !j' i'j'(' dll:::::l,kd ijyd (j' :,,,:,,,,:,;,' !j' ::::::::::::::' (j!::;i!!'' !' .1iii,,' ygd ..j.' -t'-,,' ,,jjj!yjjj;,' jrljy.:l'''i,-':.,,d .,.;i!!q:;j!r,''' ',.,;j!rqjji;,,' ,;,,,y' .-,' .,.-;;2:!1,;2::r..* .,,.,..,..,...' .-.!:111::* --,' .-.'

.' '

'' ' '

:

i:

.

:

. -

-E : .:

.. . . . E y .

.

.

y

:

,

,

3 Pzownlan Afotlon, 11

122 Caper

.

=

.

.:2:

.

.

,

,

.

'

'

'

.

which can be writte real pd umbes

dlmenslon Let I b the set of integrands where the j and sj are 1 (oniandoml) as Eni= J (az,agl < sn For any integrand 0 sc < st < e I c 1EI2(.8), we can dfi' the Jfa dmj and use the isometry property, Exercise 6.2 stochmsticintegral X = iude Usig (3.7)and the remark afer its proof, C ptz. Chapter 2, M to conc of f(F)f martingale exponential of thr integral, Zt de:ne the further we can that iiriyli Z'q i Zt an d concl de that Zt e M2. tt's forrla (3.6), soclative law (j.6) t sink the an ' za IF ' Recalling yk have Chapter 2, we

Proof in

one

./f

=

.

JJ

=

-

,

Jn

.s)

.l

=

.

.

ZL

-

Zo

'Eiili.i.ilpjrlj'tiik.yiy/yljyljyijl

. E .. . . g : , .j gg . ..

g. .

lf F E f,?, and Z E L) then y and Z are independent so 0. The last result shows tht the spaces f,,? and L e EYZ = EY EZ l with a1l elelents f,l, in inean the together /2 Since span orthogonal. E 1Z1 (6 7) follows.

with zespect to lx

,

E

-

(

so

a

=

..

a lj real z , so complex variable par iytE.t ucularv(i) = (), when = o I,z dp uuder the map qt ' ii('i:iE.''qt.!i!jq,'j!r!g!j!(Ij'gy'j..j(j'j.) . jj.)j) .t j y

442

we

have EZ3

1

=

=

and

Zc

it

,

iiitlliytjjj jy..j ;

Zt

.j

.kE.'.i!.k:!!;(!jt(i.,iyj,;, E

y ij.lriiiyykjy.l,jy.,tqjlyj j ( Ey, :jj! tI E E . iy . .. I i . . jk E .( . y ; .E ... j y . y y y q Eyiq y yy,.,'.gy..- jy-. .. .ty(,

EZ +

To complete the proof of

g,.Jf, 1y,j) (s)##,

()

.

.

,

;'l..,''.'lli;!'ti)l:)';t:j is the zero kj jy k ...

measure since its Fourier transform vanishes. vauishes Btv , Bt a-, ) = o'Bg, , tjjjljjjjrj , Btvs on o'Bt ), r.!jj jjj!jj j@ :.E!E!Ik ill:':lll:':llk,. '1kCi5,': (:11.. 'dIt:;. ; .. (II;::,,. .j(. jjjy. '... yj)rjj;yyj.jpjjjjjjtrIt'; . q.....(...!..)...;.,,:)),:.;.;.,.tt.t. ...... E. . . : E;qg ! . . g. '

j( . E,')y('. ';'(E,'':;y',!!jl!!'1y'g:',l'j'j'!''t'j')E''1 . .. .. E , EE ( ),, : : E. i ;! : i ' q . ! . . i E. k!;i EI y-y .y(- ..-... q-y..E.yy-. . - j'I - -. -.. - .y . E! . y . Ej q (.q . y -. y y .. yjyy.ifq. j,.E. .

.

:

.

:

.

.

-

-

.

.

.

.

.

.

,

:

,..

- .

.1:::::. dlt::::),-

.

.

This shows that 1,'gdp Btn ) and the proof of

-.

.

-

-

-

..y . . 5'p . . yyjyyj (;@ ! .; ...y...yyj....:,r).j)..),.jjjj(jj.)jjj,j,j....y.,y.j,.y.q.(y....,y ! ; p (:111::-,. ( r y) j ) ! r q . . . r g . (6 5) .g ' . ...'..' ljrkj;.yj )'.yjj,y.. j jy.j.y ;E yy .tj.j(. .. ;))).j ji- ;i rj ) gr;iy:; j)$1*41:gjj5,, 41(g4j5,, ' '.;jjr.jy:j.rjj!j!.:.j.iyq )..;. j(.... r,t)'. . (y.jj(y j ; jy -

, ., E.l'i.itlEiiljjllitEE'iyEjEzjEy

E

.

i! -

i-

.

.

!

:

.

EEEi gy

.

,11:::::1,1 .11::::,1,,

E

-

q-

----

jt js cuar that

!

iE

.

. .

:

jy.

: y y

yy.

.

. . E E. i ... .. . - . EE. . . - y.. .. . E .. E E. . . . . 'j'.. . yjj ; q y..... - .. - ..y.y... . j . . . . E . .E,:EyEi:,:iltitrjliltjjjjjj ' .

-

.

is

.

a subspace. x

.

-

-

-

.

-

.

-. -

..,k;;:!jij;:::...

...;j!!Ejjr;;rI

Let X

6

:,,,:,,;.

f with

jjr::;jj!r.

..,,fjj),f'...,,,,:','''.;'k.

,

:

,

. .

,

:

.

.

:j,,:,,;.

j .. .

.

. :::,y

.

E yjg . E EE ,). y . . q y !E.. q- y . -: . -y:i. E E g ( y... -. . . E ' y ii ! ! : E E -! . .)EE..E')'Ei,@Ej'@ii,j:j;r)gryt,:ty/i : : i E: q ,ilytjltktyyEj qy. . .

j y( .y .''.' (jqjj(jjky jj'i7j!yj;j;.(j.. . ( .j)jyj(;(jg.jjj;jjr....j-i...g(....(....jykg..y.....y.r ' .!.'..!'ry . y. . :

(j

.

.

. .

-.

L2 the isometry property then (6.6), cjjapter2, we have and

yyt),q(yyyr,..y

j

i E!) !

E : . .. . ' :.E : yy. . . . '. . :. . . E i . . E :. y. : E . i . : q.;ii2(rE;#..ip;i:j!,'jl;)?t)t)l@yE y qyljEjttiti,El)

:

-

.

.

.

elementary

-+

'jjg.yjElEjjjrrijtr. ...'rE(yj.k !ji ' (y..j. q.E j'l (1. . ''.. (.('(ji.l(!lj!y1 ( .jj y'. ;jj ;)..!ryy.... .... y.!.. ...yj.. .. ! yjqjyttk a ,';iq!'.7l'i;i'!.))r!)jiy!y.j. jjj ; j. . k(jqr;j jytijyj.g, . . ji( .ryj . j..,E.i.. yy l(..);j ! : q -..E.)Eq. ;. y.jjjkjj'.jg't... yj...j(yq... y '. (.qyj. ;y yyjgjy.. y. . j j.r.;j 'tEE'))t't.E ..... . j.q jy.jy!j:rq(jrt. .q....y.... j . . ...

.

the

about the variance, integral, Exercise 6.2 in

.

(6.5) Lemma. Let be the ff, d%. written as EX +

.

-

each of

the

x m )2 (sx. -

variabl

random

t

the

set

.

-

.. (

Z E

a

.

iEE

E!.E . .t-l'j 'f'j; yE .

i

- .

.

:

-

'

!!E!

: E E E:E:q !iii .

E.

:

E. .

Bta.

E

E

. . -.

.

-

.

q.

-

-

E

-

.

i - .E: . .E. .. E E. . .. y , .g i i- . - y -. . y . q. y . -- - - . E- gy j. . y y y. -E-. . -- -.. . ! jqy . : E . - : ..i. .yy .

. - - -

.

-

i

.

-

-

. -.

--

.

-

.

.

E; ; Ei E ' g . .. j . i!j 'y y( - . . -. E . . ;.yy .

. .

-

.

.

-

.

-. -

.

:

-

-

-

-

E

j ii

;

.

.

EE ; q E.. :. . q; EE . E' .

-

-

-

-

.

. .

. . -

-

-

.

. - . . . -

i

. -

.

-

.

.

,

.

. -

.

.

.

. -

-

.

-

-

-

.

-.

-

.

y y ..-

.

-

-

.-E -- - -

yy . . EEi i . EE q -..;E.- g q.( y - . : ( E .i Eqy E y -i y . . (. : g : ( . - .. . - : iy,....j.i)j?,!.))r)t').4t. ..;y,!..j,g,jj(..q..,.,E,.((y.,....,i,g.i y . g j . '.i.,.@l;y.(.r,yy.(.,;r,!',!k!j,)j')(jIj)).j.j ..j.j..;..q;y;y..,,y@....;j),y;j.:;;,,..y.,E,..y..y......yy..j. i ' .: . . : . .

-

.

.

-

.

-.

-

-

.

.

-

'.....,t), jjI '.

-

.

y

.

i

.

-

-

- . . . .

y -y y.j .. g y.- y E . yE- E-g.y.y. y y y - y.yy - : y. .. . -. y g y- .-. . Eyy ( . - - y. yq y y .. . y ... E . .iE (y E y yE- . y.q yg y-. yy. . . .- yy -j - . . y. y yyy -. -. .- y yy.. y y E.y g.. ! yy .y . -E -

y

.

.

.

.

.. .

E - -. . .. .r.: :. y. : . :. -

-

.

-

-

.

.

i

.

.

-

.

-

.

.

E. --. . .. E ;

-

. .

-

-

i . EE -. -. .. - -. - EE. . . -i . . - .' -.. . i.i . . 'EE j'. ' : : . i iE i !I qtj -

-

y

E-

-

.

-

-y . E .y y. - -. .. . y (! : . . g . y E EEj.- -y-y : i.E.y . . . g .. . . . y . -. y. y yy

g

I

-

-

q y. g

. -

-

.. . . . E. E - q

.

-

- .

-

-

. i E . . EE g , - Eii E E E E i E EE g ; :

E

EE

:

g

-

-

=

.

-

.

. E E q .. . ,;- -. i. E. E. .

-

is emsily seen to be analytic in C, i.e., w(z)is represented by an absolutely 0 for we have pz) convergent power series. By the assumed orthogonality,

.->

-

.

. . . .

-

E Ei. . ' i . E . -E .-

.

z)

,

.y

. .

.

-' . . E.

(.R,

.

. .

yilyE

-

-+

,,...,........,g,,,,

,,.,,,..,,.,,,

.:. : q. ij,.,,,,''j'y,y,'qE,q'y.r

. .

-

n

-+

-

,,-;jjj:(jjjjg,,,.

.

qjjrlyiyt jtyE

-

,

.:2

-+

,,,;jjjqjj;jjr,...

. -

. . - .

-

j

,jjj;;-

E

-

j-)

E

.yyryq, jy . j k;Ey'E;!E;jlEl)))j!l$!1jljl!i

:

.

-

-.

yE -

. . - .

.

-

.

E

-

.

.

-

-

'.,t,yy,.r

Proof of (6.4) We will show that the meuure I/Fdp (i.e.,the memsure p to show th>t To do this, it suces with dpldp = *F) is the measure. rero Bt,. ) for any nnitejequepc: 14:dp is the zero measure on the c-fleld c(.&, < a. Let j, 1 % j ; n be real numbers and z be a 0 tc < tl < 02 < complex number. The function

(y;))I

-)t..t.(..,.y...(.,.

-. E i.

-

.

be

a

,,rjjr':l,'r'.k.

.

-i q . .- . .E E... . y .E'Eiitp' . y. .r.E...'.'.,.',.r@;)Ejij.I@,pl .,.y.(yj..(....,).yqy.,..;.y.rq

can

.,.;j!,:i;jiq'''.

Ak)2 . ;i jj jjjji.. i. .... 0 and EXn EX, it follows sucexn x ju jmplies Exn .. k.'4)!..';r:..!:.(t': yj'yj j(jjj yy. :;!E!Ilt-. jjjjy,rjj., '..'..j(.jj;jjyygii:)j!(j;jjj'.j..( j....yy..jjyyq.yy.j.y. jjs jyj jj. ... j.jp'. 1i111'1)!t ...jjj.jlj..., i. The completeness of 112(.5) see the remark at the beginning of Step 3 in ..EE)i,.(.i!.j,l;:1)l.,)..II (( pj:;Iy 0. . . . r. .... iilkrEi',lydjilEllijjl'j'j'jliyljilk,jljjtjl section2.4, implies that there is a predictable H so that Il#n #lls . j of ; jy; y isometry implies that property Hm H B B y converges the to ; use .. (ililt')i)jljttt . jg.j.j.yygy..j.....j wuother . . .. in M a va ksg jjmjts irj (6.6)gives the representation for X. This completes r;r(i(E;iy).yy(!)j(!)t.j!j!Tj)())jE( k . tljy i('jyj yy; jy'l IZI tjje proof of (6.5)and thus of (6.2) ' i )'iiT'l(jkk7jt1 ! .!. :@ . k.;!!.;:j.j.(.j)jlijqrj-jjjj.jjyy.y y.y...(yy.(.;y;.j).j!(.q;.y..y.(..y.j .. y(.(jy(yrjyj:j,jkjj . . ; j! j. yj. yyy... j .yy..y j. . yji .!;r!ry).j;t'yj !.gl)jj.y jj'. yp-i -. .qj ry.)jjy..y..j.. . ;y().;.!yjjji.jy.jjty ( )); .j ( y g j . . j ; y ...q'i..i.ijjy..r)'.. y j . )l;y; (jqyy. . jj.jj.;jjjj.y.;rjj.y.y.j.jjjyy y!gj;j;'. .jy. ' ..' .r.j..lq ));. .y....y(.g(y.jq;yl(jjp:y'. i. (;.yjyg;y. ..g.yy.( l . y j .. .;.(.)l;gj.y!;y)y..yj, rjyyi :jj.!q..j...jy (y .. y y.. '... q;(j)jjj!g;.tk'rrj...y.jr...j((j.yj;y.yj)ryy.yj.yyjy(y..jy...yy .; j .j . ktr . y..q.E.gy;.)kijyjjjyjjyjjjjryj'. jy;.j.;j(ygj.).yyjjjj.y.y..... jj. )) (jy(y j.. y . ; q y . . )'. . ; ... ! j!;;ijjy;jjjy lj; j(y.y.jr.yy;.;yjyjyyjy.yqyjy.g.y..y...gy j!jjyj'. y ; j .'. l'(j.q(qj.yjq.!.lyjj.pjjrjjyjj..jjj...j.!.y(j;.ypjjjjjy.g! . . j. .j!.yjyy.y.(y.....g ..i!!. yj;! .j'('!:. .g. jy ;'(4!tj. j'j. t'j. r... yj .... .. .-...'y)..E..'.y!(.j(.iy...jyy(y;y(j'!. yjjyyjjjyjjj.y.yyj.y . rr'. j yj.. . r . . y jyyy..j.rj.j..yjyj;(y.y.y.y ...yj . ......!.l)(.;ry)yi:j,yy,jy'. .-.:.j .y(.i.kiy(yjjj.;y.j(y.jjjIjg.j)yjgjyj.yj.jyq.yjqyyjgyy.!.;y.;(:j.j..q..(j..j(..yr .. y. (ry.jyjj.yj). .....;;'ry:..j.jjr.jyjjj'. jj.,$'jj'. y .jy j yy. .. yjy .j.y.y j..,j . E

that

-

()

y

E' ' EE i .. ..-'

.

-

......,..,....

.

--

then Wf H 0.

.'

zxg sau))2

.xm)

-

.

.,-;1!:111(j;,..

.

y - y : y. y . E.i.i E. EE. . Ey y . yE.. -.! y. y . . . . '(E.'''Ek';l:Ei'i)))'.k'j'jtyi('tj';jk'y .

' : .

,

sttxs

.

!- E q

.

-

to show

a1l

0 for

=

-

-

=

sx.)2

.

.-

. .

r.......i,i;..)E....(;.i;.,);.k.,

it sues

-

.

. : . E .Ejy,lij,t,jjtjjjyjyjyjyyj . E. . . ,,jy,,.jyyyj, y -

.

...

f

z

-

-

collection random variables X of Then is closed subspace of L p

E lF' exp

s(x

E i.EE,,E,jE(!gj'!.!jl)rl)ll!i y y E,. y

.

.

=

-

.

..

:

.1,2

w(z)

-

-

:

(6.2)now,

If #F e L2 hms .E'(g#F)

.

.

(sj, (w) sj, (w), sj.(w) sja-,(w))

. .

'.) jjj;ryjyj'j ))jj yq y'y .. y . .... k ;j. .

E

.

w

-..

.

.

=

,

r ; jijy.j,ypiljijyylj.tiqitjyjl ; .. .. )y r)(;jij .jy'jjj . yy .

'liilil,lll'i.tiijyltyqyj

i.e., the-desired representation holds for .' = (f(ff B)t : .Jf e 1, t k 0).

x

,

...y..(.)....(.:(

..(

.

.

(6.4) Lem-a.

123

-1.

EE,jijpjlyrlyjlltltjrt.yrjj

-

(r =

Filtrations

theory tells us that p must vanish identically. In The lmst equality implies that the image

lEiE .illiltj,jtll)l ; jj.qE.

'.l..'jj:);)t!),jy))(yj,

1. Since Z3 E

Aaztlrgales Adapted to lzownan

kllkyyrtyj.yjylytjjlq.j,Etiy 'y

'

E

0,

=

section3.6

y..

,,,y.i,jjttily

0

'Fn = 0 and (F)n follows that

. . - . .

,t'.....).'..'

'iqliljjljjt.jljEri.EEE

X in Using an fact suppose that Xn ..........,,g..y,.y.r('(.pj!jgyi.yy.j);j;jjj.;..jj.yj.jjt,.r.j..y...y..,..y,..,...y..y.y j illiy.Ei'rliljllrkl.jtl.ljtttpjyyl and of stpchastic

Zs S, dm

=

..'

.

.

. ; -

-

y,yy.j,.j,yyyyjgy,r,j.yjyg...y,j (!. .rjy.yj )jyjy jy . .E.! .;rr.jtrq'lt:!)ri k. ..y!. . .y .. .y...(.(y..(jy;yy rjj):iy'j y-tj jj.. gyyyyyg.yjj.y;jyy . j ('-. y j!y.yj.rjj.jy...yygjyyjy.jy..y.yyjyg.j.jj..j... .j. Iyyjj.gy jyj. ....;... ( !jt,jjjt.jrt); . y jj ! . k j jjj.)j..j(.yr((.y qj..ryy.r(..(.g (ti yjj. yyj.y. jj (yjjy . .ri (!i!!j.jrj! j yjjyjyy .yy..... ... tjjr y ( j lj j j j I j . g . k yjj-))y'. j..y. ;jjyyjj)yjyyy..j. .....ii..yj.(j.!..lkl .jjyj. j.yyy..j yyy; y. jyyyy-p jy .(.r.(g;).).I);rytjjjt'. y yg jyj. ;.jy;, y (jyyjy .

.

(: I ; . i p'ji. . ;jyt)': .

-

E

. .

-

q

.

q .

-

.

-

q

.

.

-

-y

. !-y

E . E

,

-

jy -

y .. y !-

-

-

.

- . . , -

-

-

,

,

-

-

q

yE -y... -q ! E y j . gEi- E.q(-- y-y. . .. ! yyy j-; . y . y.-.(yyy- y yyy.y - . y , - - j y-y -(gy y. y.- yy - - jg y -. y y- -yy yy yg.--. --(.; .

.

.

i y i .(.E i. : ,;,;$. ( ( IIi

-.

. -

-

-

--. . . . E. .

!.. .

-

.-

(. i :. q: g y . ..y. gy. yy... . y ! E g yy yk , y y. -

ttE

EE . i E; . E.q ( q E . .: E!(E E.Ei.-EEE ! E - -E r E E

..;

-

.

,

-

.

.

.

.

.-

.

.

.

.

4

Equqtions

P ar tial Digerertial

Paraolic

Equations

In the flrst third of this chapter, we will show how Brownian of the followihg equations: solutions usd to contruct (clmssical)

uf

motion can be '

1 -.AiI

=

2 1

-l

uf

=

+ g

2

y

g

u in of

+ cu

-Au

:

2

x Rd subject to the boundary condition: (0,x) Rd (0) x and u(0, z) J(z) for z 6 Rd. Here,

u is continuous

at

ch point

=

Au

(.)21Il =

a +

oz

.

.

+

.

:2

'tzd

t'lzda

and by a classical solution, we mean one thyt hms enough derivatives for the equation to mke snse. , That is, u e C y a th functions that have oe cont j neach of the zi. The uous derivative with respect to t and two with respect to continuity in in the boundary conditipn is needed to establish a connection IR.J . and u(0, z) c.z: J(z) which between the equation which holds in (0, ) x holds on (0) x Rd. Note that the boundary cpndition canpot possibly hold R is continuous. unless J : Rd W will :ee that the solutions to these equations are (undersuitable assumptions) given by '

,

'tz

-+

s.y(s,) + /' gt s-(y(st) J

-

s,

0

(/,' z-(z(s,) exp

ct

-

s,)

ds)

s,s-)

ds))

Capter

126

Partial Diential

d

In words, the solutions may be described (i) To solve the heat equation, run

(ii) To introduce a term gzj

add

a

s

= .X'0 >

follows:

Brownian motion and 1et u(,

the intgtal

of g along

z)

path.

the

=

N

(X i xJ) )

'tz

continupus

Proof

lf

'tz

in

(0,x)

at each point

satis:es

Applying lt's

.

.

of

(1.1a),then

Ma

:

E '. . .

. .

.

Rd and u(0, z)

utt-s,

A'tIt

dBr

.

'

Br) dr

r,

-

.

t

=

otherwise

0

'

.

'i

..

:

If there is a soltioh

..

.

.

.

' '

.

of '

.

.

=

sihce

JLu

+

-.'tzz

B,) is a local martingale

0 and

=

E1

theorem.

(1.1)tht ..

.

oudd

i ..

'E

.

.

the it must be .. .

'

.. .

.

EcfBt)

>

use n for a

'

'

.

.

: .

.

. .

.

..

('

.

. .. .

.

..

is bounded, then M,, 0 K s < t, is Proof If we now assume that martingale. The martingale convergence theorem implies that

/'(z).

the

while

j Sd

equation

.

(t.3)Theprem.

j. . . . . .. .. . .

=

r, #r

-

,

1%

r

Our next step is to prove a uniqueness

''

x Rd.

(0J x

Tnt

(1.2) follows easily from the It's formula the second term on the right-hand side is ' local martingale. '

yzy

#r) d

r,

-

Here H means that the last equation deqnes r. We will always generic soltztion of the equation and w for our special solution.

from the fact tht if the This equation derives its name, the heat equation, units of measurement are chosen suitably then the solution ult, z) gives the temperature at the point z E R d at time t when the tempratpre proflle at E time 0 is given by J(z). rhe frst step in solving (1.1),ms it will be six times below, is to flnd a local martingale.

(1.2)Theorem. on (0,8).

E

'

.

.

r

y q uajou

and .Xr0has bounded variation, Brownian motions, so

-dr

=

.

r(@,z) equation:

t

-w

=

()

To check this note that duho X%vwith 1 K % d are indpedent

Equation

ln this section, we will consider the following

(1.1b) is

Bz)

u(0,

-

ea j

d gives

<

.-

,

('

lal'tz

B.j

<

...-

-

In the qrst three sections ot thischapter, we will say more about why the expressions we lkave written above solve the indicated equations. In order to bring out the similarities and difl-erences between these equations and their elliptic counterparts discussed in Sectins 4.4 to 4.6, we have adopted a rather robotic style. Formulas (m.2)through (m.6)and their proofs have been develand at the end of most sections we oped in parallel in the flrst six sctiohs, discuss what happens when something becomes unbounded.

=

s,

-

.

0 1 s + 2 c

timer.

'uf

3

-

,

'?z

B jbr 1

+

-

(1.1a) e (71,2 and

(t

=

'

-c(r,

4.1. The Heat

Xi>

-

=

z at

s and

-

).

EcfBt

=

exptf c(t (iii) To introduce cu, multiply fBt ) by mt s, f,)ds) before taking expected values. Here, we think of the Brownian particle as having mass c(t and when we 1 at time 0 and changing mass according to mj s, #,)m, expected values, we take the particle's mmss into account. take An alternative f cl t interpretation when c K 0 is that expt s B , ) dsj is the probability the z) giye> the killig rat Jvhn th Articl particle survives until time t, or

is at

;g

Sectim Y.1 Te

Equations

'u

Mt :'

H

1imM, ,1

a

bounded

exists a.s.

this limtt must be JBt ). Since M, ip pntformly integrable, lf v s>tisfles (1.1b), it follows that 1?(t, z) ETMZ utt, z) I::l EZMt =

. .. . '.

=

=

.

Now that (1.3)has told us what the solution must be, the next logical step is to flnd conditions under which v is a solution. It is (andalways will be) easy to show that if is smooth enough then it is a clmssical solution. 'v

formula,

(10.2)in

Chapter 2, to

ulzo,

.

.

.

,

zdl

with

(14) Theorem. .

supposeJ is bounded.

If

%'

e c1'2then

it satisqes

(1.1a).

128 Proof

Chapter 4 Partial DiRrential

The Markov property implies, >Q(T(ft)l.T-,)

=

Section d.1

Equations see

Exercise 2.1 in Chapter 1, that

fb.(T(#t-,))

'J(t

=

s,

-

1?(t s, Bs) -

'p(f,

,

=

-

0

1

(-'pt+ jAr)(t

-

=

#,) Dfpttz, y4

The left-hand side is a martingale, so the right-hand side is also. If then repeating te calculation in the proof of (1.2)shows that .80)

(2,rt)-d/2c-I.-,Ia/2t

pttz, y) here calculus gives little 5V

r,

1)

e

=

-(zf

C1,2, Dipttz, p)

=

(z

p)

=

(z

Dpttz, p)

=

Diilhz,

Brldr

Writing D

.

yi)

-

t !n)2

-

0l0zi and Dt

=

129

0l0t,

=

pttz, p)

t ,t(z, p)

-

a tytzj a

-

ffeat Equation

Te

yj ) pftz, p)

-

t -(f/2 + Iz p12 2t a t -

+ a local martingale The left-hand side is a local martingale, so the integral on the right-hand side and locally of bounded variation, is also. However, the integral is continuous and L'v are surely. Since H 0 almost must it be by in Chapter 2 so (3.3) l.aL'v H 0. For if it were at 0 ontinuousy it follows that + some point # (, z), then it would be # 0 on an open neighborhood of that point, and, hencej (:1 with positive probability the integral would not be H 0, a contradiction.

lf

1 is bounded, then it is

j

'pf

to

easy

see

ID.pt(z,p)

pttz, p)

that for a

=

i,

,

or

t

J(g)Idy k cxl

-kn

lt is emsy to give conditions that imply that keep t e exp sition simple' we nrstcosider the ,

(1.5) Theorem.

If

f is bounded

Y-@1/2# Bt Proof 0 and variance 1, so if tn implies that .&)

-

-+

and

continuous,

where N and zn -+

satisfles (1.1b).In order to situation when f i bounded. 'p

then

'p

satisfles

(1.1b).

hms a normal distribution with rfiean 0 theorem z, the bounded convergence

and is continuous in Rd, so (1.6)follows from the next restlt on diferentiating Jmder the integral sign. This result is lenjthy to state but short to prove since work. Nonethless w will :ee we mssume everything we need for the proof to that this result is useful. (1.7) Lemma. Let (S, 3, m) b a J-qnite memsure space, meuurable. Suppose that for z e G an open subset of Rd have:

(a)u(z)

(b) Jf (z* + ltei 'zf(z)

is a solution is to flnd conditions that The flnal step in showing that smooth. is ln this it that case, the computations guarantee are ot very difficttlt'. 't/

Proof

If

f is bounded, then

'v

E

6'.1,2and hence satisfies

By deqnition,

1'(t,z)

(1.1a).

an

tj

=

p)

ErlBtj

=

Jpttz,

y41y4 dy

Gx S

ff (z*,y4

-

-.,0

jark (z. og

nnloziexists

-+

=

mdyt

R are measurable

is continous

yyjgtyjj

ogj

go mtty;

at z* and equals

w(z*).

.y.

,z(z+)

with

y G

at z.

%.

<

Using the desnition of lz in (a),then is justiqed for I#I%ltz by (d),we have -

functions

K hz and Jl jA-z (z*+ oei,p) do for 1/z1

Proof

lzei) ,u(z*.+ =

:

kAJs (z,p):(p)

(;; j s jyo

Then

,

-.+

xtkjgmdyj

where ff and (Hkvoz

(c')

(1.6) Theorem.

Js ff

=

R be g : S and some > o 0 we and

(b) and

Fubini's theorem, which

zf(z*,p))p(p)m((lp) /s(zf(z* h = j Js (z* =

+ ltei,y)

(?.rt'

:z

-

+ 0e' #)#(#) mdDt do

Capter

130

Partial

d

Dividing by h and letting

Section 4.2

Equations

Dilrential

0 the desired result follows from

-.-?

(c).

IZI

Later we will need a result about diserentiating sums. Taking S = Z with all subsets of S, and p is cotmting lnmsur S in (1.7),then setting g > 1 and A(z) = K (z,n) gives the following. Note that in (a) att (c)of (1.7)it is implicit that the integrals exist, so here in (a)and (c)we mssume that the sums converge absolutely. =

(1.8) Lemma. we have:

Suppose that for

z 6

G

an open subset

of Rd and some

0

>

'u(z)

=

and

-+

*)

-

=

'tzf(z)

j

and(d)E,zJ-p! o

nulozi

kT:

&yith

n :

'tl

-

is continuous.

/'(z).

J is continuous. e C1,2unless

Here gt, z) '

,

=

=

a is a solution of the equation with 1 = A and pj so we can restrict our attention to the cmse J H 0. Having made this simpliflcation, we will now study the equation above by following' the procedure used in the lmst section. The frst step is to llnd an asociated local martingale. =

.q

If

u satisfles

=

(2.1a),

Ethen

4

(*

j

M, is a local martingale

u(z*). .. .

E

' ' : .

:

(

Proof

'tIt

=

s, Bs) +

-

.s'.ly(.st)I -

/

Izl-2log-b I.f(z)I -+

0

as z

Replacing the bounded convergence theorem in convergence theorem, it is not hard to show: (1.9) Theorem.

If

J

is

continuous

4.2. The Inhomogeneous In this section, to the equation

and satis:es

1ttt

-

s, Bs)

'tztt,

Bz)

-

!

=

0

'tzt

=

lal'tz

' .

co

(+),then

+ g in

'p

satisfles

(t.1).

happens when we add a function lmst section. That is, we will study

(0,x)

x Rd

.

.

(1.5)and (1.6)by the dopinated

gt,

(2.2),since

1

(-ut +

jAtI)(t

Tut

r, #r)

'

r, Bc)

-

dr

a local martingale.

Azkin the

hext

0

-

.

dBr

-1.:

+

-eu

''''''.

Ly

-g

=

.

.

apd the second .

.

the .

g

(

right-

'

result.

.

>

E=

.

t

a solutin

gt

'tJ,

.$1t

gt

q:z ()

(2.1)'that is

:

s,

-

,

) ds

M,, 0 % s

<

t

limM,

of

'

jo

Proof Under the assumptions on g and bounded matingale and u(0, z) M so H

on

. .

Stippose g is bounded. lf there is (j 3) Therez. boundedon (0,T! x Rd for any T < x, it must be

Mt

trm

cl

step is a uniqueness

.p(t,z)

z)

(1.2)gives

3

which proves '''hand side i:

Equation

we will consider what we considered in the

(2.1a) e C'1,2 and

-+

0 '

+

f is

r, Bv) dv

.=

Applying It .'s formula ms in the proof of

ta.)la/ae-I--,l-'2tly'(p)Id,

that guarantees this for locally bounded

gt

(0,8).

on

.

'u

'u

=

Our flrst step in treating the new equation is to observe that if ul is a solution of the equation with 1 A and g 0 which we studied in the lmst sektion, >nd ua is a solutioa of the eqttation with f = 0 and gn then ul + uz

Unbounded f. For ome applications, the assumption that f is bounded is too restrictive; To see what type of unbounded f we ca allow, we observe Since that, at the bare minimum, we need .GIJ(.lh)l < tx7 for all

a

z)

'u(0,

apd

We observed in Section 4.1 that (2.1b)cannot hold unless 1al'u so the equation in g= (2.1a)cannot hold with

(2.2) Theoren.

z

+ oei) do < %.

exists at z* and equals

condition

(0) x Rd

at each point of

.

=

Then

continuous

131

'

ErzA(z) where fn ofnloziq G R, n E Z are memsurable functions 01n (b) A (z* + e) Jn(z Jc ari (z*+ oeijdo fbr ll K hz and a; (z) is continuous at z Er dy (c) (a)

hz

(2.1b)u is

The Inhomogeneous Equation

-

s,

Bs) ds

t, defined in

(2.2)is a

Partial Dillential

Chapter 4

132

Section 4.2

Equations

=

't?

Erlh

Ezlh

=

=

Again, it is easy to show that if v is smooth (2.4) Theorem. Suppose is bounded satisfles (2.1a)in (0,c=) x Rd.

1)(t,

z)

enough

it is a solutio.

continuous.

and

.

Proof

If

Take home message ist it is not enough to mssume is continuous to havr a locally in That n : C1'2. We must assume g to be HBlder continuous (k7), which may dpend otk is for any N < x there are constants C t% (0, e whenever t such that 1p(,z) gt, p)1 K Clz N.

C1,2, then it

e

'v

.

'#

1

j =j ' j -

J(f

Br ) dv

r,

-

,

r,

-

gt

-

:

jv

dr +

-

'p(t

-

s, #,)

#c) +

gt

c

=

z)

=

:

-

(-r

r,

use

s, p)

-

dy '

+

1 2

-'n

=

rrc

+ p)(t

-

To translate between notations,

r, Bvj dr

'''''''' '''

'

'

. .

.

.

.

Th en ext step is to give a condition that guarantees that satis:e (2 1b) As in the lmst section, we will begin by considering what happens when everything is bouuded. 'p

.

lf g is bounded, then

then ms @ -+

1)

satisfles

for write

1.p(@,j ' 1,(.. -

.

;, r)g((,

Zz,

set Ta

=

0, D

=

p-rtz,

rldt dr

Rd,

=

)

Because of their importance for the and change variables s = y = method, parametrix the dilerentiability properties of volume potentials are results well known. The we will state are just Theorems 2 to 5 in Chapter 1 ofpriedman so the reader who is interested in knowing the whol story (1964), '' .. ' '' .( inissing detail thre. 8 firtd the

(2.6a) Theorem. If g is uous on (0,x) x Rd. Proof

'r,

.

E

.

.

. .

'

.

'

Use the bounded convergence theorem. 't?

C so that if ovloi have lD17l

a constant =

s

.u't

-

0

D1)

Dfp4tz, yjgt

=

(j

-

. .

r(t, z)

is

.

contin-

E1

jpl% M is memsurable, then % CMt1/2, are continuous,

t

s,s,)lds

.

bounded memsurable function, then

a

(2.6b) Theorem. There is the partial derivatives Di and are giveil by

(2.1b).

0

.s

D

Zz, @;(, r)

'

z)Is

p,(z, ylgt

again

'and

.

7(z, t)

Br4 dr

:

lglK M,

ds

g is bounded

mssume

-Ip-rIQ/2,

side The left-hand side is a local mattingale, so the integral on the right-hnd varitin, and boundd locally of is also. Since the integral is continuous (3.3) in Chapter 2 implies it must be H 0 alpost slzrely. Again this implies that assumptio impl# that thi quantity is cntinuous (-w # 1aAr+p) H 0, for ourpoint . :nd z) ID if it were # 0 qt some (f, we would have a cpntradititp.

If

C1,2 is to

=

-

Proof

e

.l)

(2.6c)and

of

where 1?atz, yj The expressipn we have just writtn (2xs)-d/2e would call a volume potentlal antl would above what is Priedman 'p (1964)

6 C1.?,

'??

+ a local martingale

(2.5) Theorem.

be found in th proof

#

. 'p(t, -

can

0

The left-hand side is a martingale, so the right-hand side is also. If thn repeating the calculation in the proof of (2.2)shows that -

s

j

r(,

s, l,)

,p(t

1

-

The remson for this in (2.6d). The first step in showing Fubini's theorem to conclude

.,#-,

sr ) dr + ss.

r,sc)

-

assumption

.--

gt

1

plE'

Using the Markov property, see Exercise 2.6 in Chapter 1, givqs E=

133

The final step in showing that is a solution is to check that v E C1,2. Since the calculations these properties are quite tedious, necessary to establish we will content ourselves to state what the results are and do enough of the proofs to indicate why they are true. If you get dazed and confused and bail out before the end, the

Since Ms is uniformly integrable, 'tztt, z)

The Inhomogeneous Equption

s, y) dy ds

Dirential

Partial

134 Chapter 4

(1.6),the

Using the formula for Dips from

Proof

.

1

-

(z,

.,o

(2vs)-Nz

*

Section 4.2

Equations

v;

-

-

--'

e

.

t

al-.k-la 'm

./1

'

;n.

Wtthoughthe last formula looks

0,

fxltzf so

=?

haye

all right.

is really

everything

B'tg

-

t

i Bsjat

l.I d

because M, then

)1K

,

ffzlz

s,

-

''

.

'

'' .

.

.

1' .

E

'

'

.

'

.. .

.

.

(z

Ez

.

.

-

.

l . .. '

. .

s

.(

j

.

..

. '.

(

E

.

-

f#(t

s, #,)

-

Sa

l

'

<)J 1

'

:.

.

'

(:1

E

'

.

so

(2.6c)

=

CMsll

=.:

E

E

.

.

fj

#(f.r,- s,

-

pf g one can show the above is S Cs--alz Using the Hlder continuity its integral from s 0 to t convergs absolutely, and with a little work (See follows. Friedman (1964), pages 10-12, for more details.)

,

-

,

s,)

f,f )2 s

=

at z, because the fact

E

' :' i '''' '''

.s,

!

(z

E=

= near

. .

#)2 a

-

Sa

sa)) s-l

continuous

=

-

.

E

integr>ting

we >re

' :

'

:

,

...z

w

.

135

.

The lmst detail now is: ,

: ds -E g

Sj)#(

l(zf -

S

' i

.

. .

.8,)1

%

s,

-

are continuous,

< x

have

and

.

. .

(2.6d)Theorem.

2CMt3I2

under the integral sign,

Using our result on dilerentiating the partial derivatives Din exist, :

-

s

spspicipus

B

x

s, ylay as

-

'

ds

a

s,

-

E .

'

'

'

-

If

.

'egtf

-l'.g(zf

=

E

We can overcome this problem if g is Hlder #,)2 that Erzi s allows us t write

is

side

right-hand

The Itthnmogeneous Equatiim

0.v z) ot- (t,

(1.7),it follows that

.

Let g be

as

in

(2.6c).Then 0.H0texists, f

cci

g(t, z) +

0

p)g(r, yj dy

-aptmrtz,

dv

.

()

,

.( ,

the idicated form.

To take the derivative

Proof

and

w.r.t.

,

we rewrite

'p

j

j. .

.

as

j

Things get

even worse

whn

1?(t,z)

derivatives.

seond

take

we

:

pt-rtz,

=

(2.6c) Theorem. Suppose that g is bounded and Hlder continuops locally in are continuous, and t. Then the partial derivatives Dijn oanloziozj =

DiiT

DfJ#,(z,

=

#)J(f

()

,,

.

t at

yi )2 s s2

(2'ms)-d/a (zf =

(t consulting e

(2f -

E=

()

6) for the

.

tj

g

f)2

-

'

8

gt

s2

-

This time, however, .E'.l(z

,

(ys

.E

= +

r

s,

#,)

ds

o

ytgr, j wpt-rtz,

f

.

j f j (2v)d/2(j

1l

w.r.t. z).

y4 dy dr

2T(

f

= z o

dr j

vkgr,

-

. ..

pj2j

t-jz -

exp

r)(d+2)/2

r))-d/2

-

.

. .

.

1Z

2(t ft-r)

#12

-

-

2(j

=

r)

(

-Llz

r)2 exP

2(t

dv

+ ()

:(j

j)2

.

sl

sltll

(zf fj)2 2 -

ds Ec

=

)2 -

-

s

=

(x7

11so

' .

.

'

.

.

k; .

.

.

-

#12

r)

)

tsts '

gr, !/) dudr

Xa(Iz ft-rI2.(r,#-r)) -

r)a

-

''-

,(y., yj

,

In the second integral, we can use the fact that E+ lz Bt-r I2) C(t like the frst, but cancel one of the t r s antt make the second exprssion if we do this, t dv -

-

. . . .

the t gives two terms. Diserentiatmg Diflkentiating the integrand and using

p) dy dr

-:/2

..2.

-d

'

.

-

yjg

.s,

'

:

prmula fpr

-jz-yjaya,

-

-

f

.i.

=

.

Diferentiating the right-hand side upper limit of tlle integral givs g, a Ormu la from (1.6)givs f

jot

.

'

stmplicity Iarof supposelbr right-hand side is Dffp, , the

0

s, #) dV ds

-

yjgr,

0

'

.

..

-

0

f-r

=

X

=

-

r)

to

even

Caper

136

4

Parlal

DiRren tial Equations

Sectlon 4.3

and the remedy This is the diculty that we experienced in $he proof of (2.6c), is the same: We can save the day if g is lllder continuous locally in @. For further details, see pages 12-13 pf Friedman (1964). E1

source can be consulted for a wealth content ourselves here with an Eynmple 2.1. Let #(z) We will now show that

Unbounded

g. To see what type of unbounded g% can be allowed, we will restrict our attention to the temporally homogeneous cmse g, z) = (z). At the bare minimum, we need

(2.7) Theorem.

z-j,' l,(s-)Ids

of information about

(lzI)where (r)

=

for

r-F

=.

Formula

137

these spaces. We will

r

K 1 and 0 for

r > 1.

: ffa if p < 2.

The special case p

< x

The Felman-ffac

1 is the Coulomb potential

=

which is importgnt in physics.

Proof If t:v < 1 then changing to polar coordinates is largest when z 0 we have

and noticing

the integral

=

E z.a

fa

jo

tf, ) ds

0

-+

su '

If we put absolute varllzes ipside the integral and strengthen #he last result to uniform convergence foi z E a then we gt a deqnition that is essentially due A function is said to be in ffd if to Kato (1973).

/' I,(s-)I#s)

sup t0 lim

kJo

r

By Fubini's theorem,

the

we can write

liy tt

itz,

sup a,

where

above

-

ctt

r

when p

=

0

jw-pju.

2. In d

s ,

-r,

0 and

Iz -

g12/0

-+

-(d-a)

c,

= p1) I(p)1dp 0

d a r 2 log d r :(r) d 1 r of The equivalence (+)and (++)is Theorem 4.5 of Aizenman and Simon (1982) or Theorem :$.6 of Chung and Zhao (1995).Section 3.1 of the latter =

-

z

=

=

which

holds in d

# 2, then

the

r1-F

2 Fhen p

<

2

dr

0

-+

p we

get

?rl-p

logr dr

(j

-+

0

El

function

4.3. The Feynman-Kac

Formula

In this section, w will consider what happens when we add cn to the right-hnd side of the heat equation. That is, we will study

(3.1a) e C1'2 and (3.1b)u is contluous 'u

where

=

1

Remark. We will have more t say about these spaces at the end of the next section. For the developments there, we will also need the space ffaloc, which is defned ip th pbvious way: f E ff alocif for every R < x, the spatially trunated ./i.(1zI<s) ltd.

t

=

-

<

r-(d-2)

w(r) by

= cd

p)I(p)ldp

wtlz

and

fr

as

-ftz,

lirj sup

r-F

= Cd

()

(2xs)-d/2 e -Iz'-pI/,# t) y) a t By considering the msymptotic behavior of condition analytical in a more form as we can cast this kt (z, y)

=

Ir-#IfG

If we replace /J(r) by above

RE',

.s(

wtlz pI)I(p)I y / cdjkvrtkrtd-dv 0

/

u

=

laLn + cu

at each point

(.

(0,x) x Rd. of (0) x Rd and

in

u(0, z)

=

J(z).

If ctt, z) % 0, ten this equation describes hat flow with coolig. That is, u(t, z) gives the temperature at the point z e Rd at time t, when the heat at z). z at time t dissipates at the rate -c(t,

Chapter 4 Paztial Dirential

138

;' )'

Equations

Section 4.3

The flrst step, ms usual, is to find a local martingale.

(3.2) Theorem.

Let

ca

JJ

c(t

=

M.

is

a

local martingale

Proof

xd-i-l =

8

=

.tI(f

Using dcrt =

c(t

with

x

t

=

r, Br) exptctr) dv +

r,fr)

(.X

s,

-

=

' ()

Arj

sj

=

exptcr)t dctr +

-ut

xpttrlvult

0

-

hut

j'

()

.. !

. .

.

='

=

-

J'

otherwise

0

1 + cu + -A'u

f

aztd

(3.1a).

2

c

Br4 dBr .

rtt

explc

).p(t

s,

-

B,) exptc,) ' 1 (-p + c'v + j'??)( = () + a local martingale .p(f,

-

s,

redlt.

Bs)

z) - s.(y(s) .p(t,

a solution

of

.8c)

-

=

ETMO

=

c

is bounded and

r,

-

Bv) exptctr) dr

f is bounded

(

that

(3.1b).

satisqes everything is bounded. an

j

'p

continous,

then

'p

'

.'

:

E'

..' j .' .

.

'

('

=

r(,

z)

:

eM$

,

so exptcl)

1 as t

-+

=+

0.

=

-

K s < @,is a bounded mam Since M is uniformly integr>ble

ErM4

c-Aft K exptcl) < lcI% M, then result this inaplies 11.f11xsup.IJ(z)I

If

I.E'aexp(c1)J(#f) ErfBtlk

on c and u, Ms, 0

J(# ) exptc).

Proof Letting

(3.1)that

exptcll) '

=

C C1,2,

*

.

tingaleand Aft H lim,-. Ms it follows that 'tz(f,z)

1)

.

,

Proof

' .' .

-

,be

Under our assumptions

.

.

The left-hand side is a local martingale, so the integral on the right-hand side and also. Since the integral is continuous is locally of bounded variation, (3.3) rely. Azai this inplies that in Chapter 2 implies that it must be H 0 almost (-1n + cr + laA1g)( r, Bcj exptcr) H 0 for ou4 mssumptions imply this quantity is continuous and if it were # 0 at some point we would hay a contradiction.

is

(3.5) Theorem. If satisfles (3.1b).

(3.3) Theorem. Suppose that c is bounded. lf there is is bounded on (0,T) x Rd for any T < cx), it must y'

.

exp(c1--,,))

exp(c,).Eb.(y(lt-,)

=

The next step is to give a condition that guarantees As befpre, we begin by considering what happens when

The next step, again, is a uniqueness

. .

and continuous.

The left-hand side is a martingale, so the iight-had side is also. If then repeating the calculation in the proof of (3.2)shows that

Br) exptcr) dr

r,

-

that

is bounded

implis, see Exercise 2.7 in Chapter 1 >nd take

explclllla)

ErfBtt

Kd side

and

-

y

,

the ridht-hand

(t

r?

-

.

.

is bounded

The Markov property c(@ r, z), that

=

139

is smooth enough it is a solution.

.

(3.4) Theorem. Suppose If 6 C1,2, then it satis:es

for 1 < i / d,

v

'

q

lr) exptcr) dv

r,

-

y

t if 1 <

XiLt '

r, Br) d.r and rearranging,

-

. . .

.

#(,)

-

-

since we have

:

.

Proof tr, z)

'ult,

0

jo

. :

(3.1a),then

u satisfies

4

-

.

=

-pj(f

utt

.

Bs) exptc,)

s,

-

J

+

.

As before, it is emsy to show that if

. ...

'p

lbrmula

s, #,) exptc,)

=

(' '

(0,).

on

Applying It's c J gives that

tz(.-

Br) dr. lf

r,

-

..

The FeynmanulLac Formula

(1.5) implies that (f,z) ErfBtj the desired result follows. ..-,

S IlJllx .E:l exptcl)

is cpntinuous

This brings us to the problem of determining

be a solution.

11

at each point of

when

'p

0

-->

-

is

(0Jx Rd

smooth

and E1

enough to

Chapter 4

140

If c is (3.6) Theorem. Suppose that J is bounded and Hlder continuous. bounded and Hlder continuous locally in t, then G C1,2 audy hence, satisfles 'p

(3.

j

ja;

(

y

g

g '

.

Proof To solve the problem in this cmse, we use a previous the cms. We begin by observing that to c(f

:

0

Taking #(z)

l

i

: .. ' . ' E

' '

trick to

'

l

'

'

.

'

.

.

'

1?(f z)

.

'

''

reduce our result

H r1(,

'

z)

formula,

So It's

(10.2)in

M1ultiplyil g b y

-

1

.

'

.

f

s, Bslvt

-

-u

,

Sallds

z)

+ r2(t,

z) l'?2(f,

S NI:= for t?1 1et ht be a

.

0

p)1

'pzt, -

.

. chtious

expt-calctt

o

s, Bs

-

,

t/

#

Byownian motion starting

..

.

whenevr

..

.

:

0

at

. . .

+ A) 1y+ 't))l I.E1T(z .E'IJ(z+ t ) 1y + :,)1 s clz

=

and

-

;

-

p1

#)1

't'2(, -

To get a similr estimate observe that since f is H3lder

Y(t,) dc,

#(cfc) =

=

Erfct

plugging into the previous

and

1

i

Irz(t,z)

-

f

equation

'. The first term on the right, t?1(t, z), is C1,2 by (1.6).The second term, ctr, z)'p(r, z). rztt, z), is of the form considered in the last section with gv, z) observation with and that if c is If we start the trivial f are bounded, thn bounded on (0,Nj x Rd and apply (2.6b), we see that

j

expt-cl)

j

)+

ErfBt

=

,

'

the last

of

'

'

e-r we have

=

value

141

=

variation.

(' . .

-

.

Taking the expected one, we havq

Formla

'v

and locally of bounded is continuous Chapter 2, implies that if lt E C1 then

tcl )

.E

Bc) dr

r,

-

(

.

.

cf3 =

The Feynman-ffac

Section 4.3

Equations

Partial Diential

-

.

pl''

-

z) is Hlder continpous locally Copbinipg the lmst two estimates we sre that suppojed and w h>v tha j ctr, z) js jsjaer in t. Since c and are boundd z) is Hlder continuous continuous locally in t the triangle inequality implies gr, r2(0, z) 6 C1'?k and E) hk locally in t. Using (2.6c) w n ow, (2.6d)

) ds

'p(?',

'v

exp (c1)givs

E

E

'

E

'

E

j

e kptc') t

-

1

c(t

=

-

s,

Bsj

explct

c. As in the last section, we can generalize the rsults above but for simplicity we restrict our attention to the temporally ?l(z). Given the formula (+)above, which expresses homogeneous cmse ctt, z) volume potential, it is perhaps nt to surprising that the appropriate 1) s a mssumption is c 6 ff a. The key to working in this generality is what Simon

Unbounded

ts

c,)

-

0

to unbounded

c's,

=

Pllzgging in the definitions of ct. and ct we have 't

c(@

exp

-

0

:

r, #r) dr

c(t

1+

=

0

-

(t

s, B,) exp

r, #r) dr

-

ds

(1982)

calls

E

:

and using Fubini's values, Multiplying by j'Bt ), taking expected which is justiqed since everything is bounded, gives

theorem,

(:.7)xhasminskil'slemma. a

.i

'?(f,z)

=

ErfBt)

+

t

r

j tctf E=

-

Conditioning on Ss, noticing c(t s in Exercise 2.7 of Chapter 1,

./.

(c( -

s,

l,) exp

(/,' c(

-

r,

-

s, O,) exP

s,

g

Lj

c(

Bs4 G .T-, and ,

-

#r)

r,

drl

J(.%)) ds

Then

dr)

J(lt)j

Fs(

suzp Ez

be a function on R,d with

k

jv

gB,

) ds

< 1

(/',(s,)ds) s-exp sup 0

s (1

ahd

the nonnegativity

-

using

r

the Markov property Proof

Br)

>

Let g

=

c(t

-

s, l,)'p(t

-

s, Bs4

The Markov property sup f. .'r

.

.

ds

.

0 < z< ,s

.< .-

a a<

.

.

.

dsu p(#,,)

a)-t

of g .

.

-

imply that

glfaa

)

-<

a%

.

.

dsl

.

0
=

= ('

Summing on

(slj

in

Since the integrand is symmetric

n now

.<

Tl-

-

.

s,z) we

,

.

dsn

g(#.N

z

)

.

.

.

.

.

ds

.

.

dsn gtfaz

9B'$ ,

=()

From

this context.

'h

.

glBsn )

.

4.4. The Dlrichlet ln this

,

'

last result, it should This condition

be

clear

that

wy

.

assuming

r

' .

.

.

.

in

atural

/z E ffa is

.

.

sup E.

exp

Jr

.

. ..

.

.

Equations

B. Elliptic

0

l(#,)I ds

-+

1

with developing

the theory q

'

E

'

.

continuous

and

let ht,

we

can

(4.2) Theorem. inftt > 0 : Bt Let r Mt 1ltftl is a local martingale on (0,r). =

Proof

Applyitl: lt's tI(#t)

=

at each point of

0G,

'tz

equations

is

are

suitable

(under

t =

TuB,)

(4.2),since

1 fOr all z (t) P ('r X) which (4.7b)below will show =

..

c(s,)ds))

u(#o)

The second step is a uniqueness ms-

.

,'-z(s-) ,(s-)ds) + jo' .s (z(s-) exp s-(z(s-) Lj,-

-

for t < r. This proves martingale on (0,r).

J.

to these

z)

f.

=

=

then

u(z),

satisdes

the

<: .

j GJ. If

(4.1a),then

u satissej

formula gives dB, +

.

0

-1z

We will see that the solutions sumptions) given by

'tz

Thus, is an equilibrium temperature distribution when OG is held at a flxed temperature pro:le 1. As in the flrst three sections, the irst qtep in solvirtg (4.1)is to find a local martingale.

1

-Au + g 2 1 = 0 + cu 2 set G subject to the boundary condition:

we will study

A

=

=

0

That is,

'tl

1

=

'u

at

0 = -Au 2

'fz

(4.1a) e (72 and hu = 0 in G. (4.1b)At each point of 0G, is

:

In the next three sections i of this chapter, we show how Brownian motion k clmsslcal solutlons of following equations: construct the used be to

in an open continuous and

the Dirichlet problem.

.

j

With these two results in hand, we can proceed mch ms we did ih the cas of bounded coecients. E

..'

To see what this means, note that if heart equ>tion

0,

0

and, hence, that

Proble'n

'u

( .'

zuarantees

s.(/'I,(,)lA) +

.'

-

,

we will cosider

section, .

gives the desired formula.

SuP

.

)

.

143

=

=

n

d'S

. .the

'r

glBsn )

,.::r0

,t=0

.

.

.

The Dirichlet Problem

where inftt > 0 : Bt / GJ. Tp see the similarities to the solutions given for the equations in Part A of this chapter think of those solutions in terms of spac-time Brownian motions B, t s, Bs) run until time = infts : Bs / (0, x) x Rd). Of course, f' *

have

.

1 n

.

.

t

1

u

.

.

w
.,

--

Section 4.4

Equations

Chapter 4 Partial Dirential

142

L'tt

0 in G

=

t

1

w

LuBslds

0

the first term is

and

a

local E:l

theorem. For this

we

introduce

=

.

.

..

.

.' '

. .

is necessary for uniqueness. Chap ter 3 implies bounded sets G satisfy (t).

,

Note that

(4.3) Theorem. Suppose G satisses (t). If there is a solution bounded, it must be r(z) H EcfBpj

of

(1.3)in

(4.1)that

is

lf is bounded Proof local martingale. Using 'tz

.

bounded

(4.1)then Ms 0 K s < r, is a (2.7)in Chapter 2, (t), and (4.1b),we have satis:es

and

,

.

.

.

Section d.#

Equations

Dirential

4 Patial

Capter

144

.

'

.. .

.

.

Mr u(z)

H

lim Ma a1r ETMZ

=

=

. .

.

. .

.

..

...

( .

.

. .

fBr)

The frst step is to show

(4.5:) Lemma. If t

..

if

r(z)

=

zrz

(z)

H

Er

(.J(#r)1(r<x))

.

..

As in the first three sections, it is (4.4) Theorem. then it satisfles

Let G be

(4.1a).

emsy

into

(x)

.

Let

account.

l

.

.

.

J

'.

'

.

.

to

PcB.

e G'

.Pptr K t

-+

for f)

-

t))

(E,

s E

some

6 C2,

If

prtz, pl#,tr

=

is bounded and neaurable

.

#,(z, #)

f is bounded.

That is,

/ f)

liminf#.a(r s t) z J'atr

to phow:

any open set and suppose

.P.(,r K f) is lower semicontinuous.

--.

By the Markov property

Since y '.

z

n''ex

Proof =

0, then

>

<, then

-.-,

C1

blt to drop (t) we need

(t) is not needed for the existepe of solutions, modify the deinition to take the possibility of r

145

E j

..

.

.

E

.%Mp

=

... .

Proof

Te Dirichlet Problem

f)

-

dy

and

-d/2 c -la-pIa/2e

(2TE)

=

Kt

it follows from the dominated convergence theorem that

z

-->

C

Pcm

6 G for some s E

i

(:,@))

is continuous for each .7 > 0. Letting e t 0 shows that z K ) is an incremqing limit of continuous functions and, hence, lower semicontinuous. (:) .!%tr

Proof

> s)

(-r

implies that on

The Markov property

Ez (T(f,.)1(m<x)1.F,)

(#)

=

-+

'

'''''''' '''

The lefhand side is a local martingale on (0, so (#,) is also. lf then repeating the calculation in the proof of (4.2)shows that, for s E r),

i7(#a) -

1

'p(fo)

=

j'

'pt

6 C2,

If y is regular for G that if zn v, then

and

t

j, then fjt'r K i)

>

1, so

=

-+

from (4.5a) tt tollows

'19.9'.

.

(0,r),

liminf n-x

.

.

.

.

. ..

''

Lb.B

r)

dr +

it is easy to complete the proof. Sini f is boundd With this rstablished, contiuotts, it suces to show that if D(p, J) (z : lz pl < J), then

local martingale

a

=

The left-hand sid.e is a local martingale on (0,r), so the integral on the rightand locally of bounded hand side is alo. However, the integral ls continuotls variation, so by (7.2)in Chapter 2 it must be H 0. Since A is continuous in ' G it follows tht A'J H 0 in G. For if A # 0 at some point we Would have a El contradiction.

(4.5b) Lemma.

If y is regular for G and zrt Prn ('r <

(X),

and

-

y, then for all &> 0

-->

Bv E DD, f))

1

--+

1

Up to this point, everything hs been the same ms in Section 4.1. Diferences condition longer (4.1b),sin it is nosatisfy appear when we consider the bouudary continuous. must a G set and The open sucient for f to be bounded regularity condition. Deflnition.

.

1 for all t > 0'

('rK t)

.&n

'

.

.

A point y 6 DG is said to be

polnt

a regular

if #vtr

=

0)

=

1.

(4.5) Theorem. Let G be any open set. Suppose j. is bounded and cohtinuous fy). y, then :(zn) and p is a regular point of 0G. If zn E G and zn -+

-+

Proof

Let

6

so small that

0 and pick

>

Jb Since

/ tr (r S ) r&

-+

1 ms za

-+

sup

0s,st

-

J 2

<

:

y, it follows from the choices

liminf Pru (r < x, Bp E .D(p, -IXI n

liminf Pcv ('r K t)

n-..x

1#,l >

)) k liminf n ''*

-

15

(X)

sup csas,

fa

v

I#fl >

that

above

%t, pgsup j 1#, -

y

-

6 2

>

<

1

-

f

zn

I%

-:

146 Chapter Since

E

Paztial Dirential

4

this proves

was arbitrary,

Equations

(4.5b)and,

Section 4.4

hence,

(4.5).

Te Diichlet Problem

immediately that if we let Ta = inftf : Bt 6 (2-n,2=n+1) x rj, csld-lj and pick small enough, J%(Ta then x) < 1/2, c,, K 3-n Now Eactt3-n = 3-1(3/2) = if 1et inftf j : Bt > r so we / Gj and = inftt > 0 : Bt j (-1, 1)d), then Fe hlve =

.

(4.5) shows that if every point of OG is regular, then :(z) will satisfy the boundary condition (4.1b)for any bounded continuous f. The next two exercises develop a converse to this result. The qrst identi:es a trivial cmse. Exercise

4.1. If G C R is open then each point of OG is regular.

2 and 1et y E OG have Let G be an opn set in Rd with d with continuous function J(p) 1 and fzj < 1 .!5tr 0) < 1. Let J be a of points zu y such that for a 11z # y. Show that there is a sequence Exerclse

4.2.

=

=

-+

liminfa-.a (za) <

1.

s(r

''

r',Thus Polv > 0) 9 Pzlv

rtlzli

conditions

will

-

=

=

L.

=

.

.

.

..

.

.'

.5

.

.

.' :.

.

.

Tllorn.

c)

(z : z

=

Let d k 3 and let

=

y+

Now that we have defined a is spherically symmetric,

ov +

e 7(p,

Clnlm. lf

cn

(x7

term is the

t 0 suciently

single

(0J.

point

PnBt na K liminf tt0 *

=

In

=

(z : zl e (2=',2-'+1) .

)

za

=

za

.1.

z

and

'?p,

Izl<

c)

c))

'',

=

0

ea >

s

j

1#, -

gl > r

0

=

Combining the lmst two results with a trivial inequality we have

fmst, then 0 is not a regular point of 0G.

Pzlhl, .R3) (0,0) for some t > 0) Proof Brownian motion Bt starting at 0 will not hit

(0,cr),

y, polnting

vertex

where ea is a constant that depends only on the opening c. Let r > 0 be such that F4p, c) n D(p, r) (: G'. The continuity of Brownian paths implies

,

=

0G

0

an r >

the rest is easy. Since the normal distribution

cone,

PyBt

z) where

lim Py sup -.0 where the n

D

show that if Gc is too small near y, then y may be shows that if Gc is not too small near y, theu y is

. .

'p,

'

=

Lebesgue's

4.2.

irregular pint.

.

'??,

Exnmple

t is an

1/2 and

.

.

-

'

.

''

7'4g;v

Disc. Let d k 2 and 1et G H:z D Exnmple 4.1. Punctmed (0), wheie = x) f%(Tn 1 then D = (z : 1zl< 1). If we let Tp = inf (t > 0 : Bt 0J, regular of point 0D. One can get an by (1.10)in Chapter 3, so 0 is not a D ff phere example with a bigger boundary in d k 3 by looking at G = ff (z : zl z2 0J. In d 3 this is a ball minus a line throgh the prigin. .

n=1

Pxoof The frst thing to do is to deflne a cone with ln direction with openlng c as follows:

give two examples.

=

.j

,.

.

' .. . .

. ..

y.

fo a point to be regular?

ln order to answer the flrst question we

...

< xl s y'latrz'r, s1

,)

(4 5c) Cone Cpndition. If there is a cone V' havin: vertex y nd such that F' n Dy, r) C G', then y is a regular point.

Do irregular points exist? What are sucient

'

<

tr) >

=

The lst tWo exmples irregtlar. The next result ... .

it is sucient From the discussion above, we see that for to satisfy (4.1b) situation necessary) regular point. each point of almost OG This is a that (and questions: raises two

=

Co

' . . ..

't7

147

probability

so with

0,

=

1,

'a

e

GC)

K lim/tr t0

% ) K .J$(,r = 0)

and it follows from Blumenthal's zero-one law that =

.

.

.

=

a

=

0)

Since Bt is transient in d k 3 it follows that for a.e. u2 the distance between (#, : 0 K s < x) and % is positive. From the lmst observation, it follows

.' ..

'

. ' :. ..: ..' . .

: . .. .

.

' . .

.

(.

' (4.5c)is ,

Exerclse

for

=

0)

'

.

=

1.

.

for most

ucient

4.3. Let G

each p E

./5(,r

(z : #(z)

cases.

< 0) where g is a C1 function 0G. Show that each point of OG is regular. =

with

Tgyj

0

Chapter 4

148

Partial Dirential

Section 4.4

Equations

However, in a pinch the following generalization

be useful.

c>n

'p,

gzt

.

. .

.

' ('. .

..

E

.

-.s

.

.

k '

.

.

E .

:

. .

:

.. .

..

Let G

e any

(4.1a).

open

i

.

then

ounted,

set. If J is

. 'p

i'

E

.

=

.

.

and,

So

is a

.

. .

.

.. . .

; ' . .

jo jo

times g

cpnstant

conclude

.ts(,,r;

(z +

dv rd-1#(r2)

hy)

=

/z(z)

and

(z)+

tnf

.

+

yi

'tavexaging

E

i

.

(4.6a) Lemma. Let G be an open set property the averaging

(z) when z

e

G and J

<

r

where

=

arnd

/ be a bpunded function which hms

.Js(z,,) '(p)

% > 0. Then

''rtdpl

E C* and Lh

=

0 in G.

Let # be a nonnegative inqnitely diferentiable function that vanishes Proof (J2, x) but is not H 0. By repeatedly applying (1.7)it is routine to show

on that

:(z)

1/7(1#

=

D(r,J)

-

zl2)(p)

dy

e C*

of

i

yi

zf)(pJ

-

j

-

z.flDltzl

subtiacting (z) froin

that /t(z)

=

.

=

z)

Ey,

r) w.r.t.

dp/1D(0,

r)1 and using

hz) + 0 + C2A#(z)r2 + O(r3)

each side, dividin

0.

+

k

z)1 Czrt. Integrating over. D(z, where Ic(p, the averaging property we have

(z) memsure. where is surface meuure on D(z, J) normalized property'' of harmonic functions. Now a simple The lst result is the takes over to show 5 6 C* analytical argumnt

version

zlDftzl

-

E%

to be a probability

,r(#a)

that A

now

,

';r

a)

-

( . . .

(@: Let z E G and pick J > 0 so that D(z, J) C G. If we let o' = Proof details Markov' that implies property more J)) strpng D(z, the then Bt 4 (for See E X ample 3.2 in Chapfei 1)

+ z) dz

hence C 0 we note that the multivariate Taylor's theorem implies that if Ig zl K r then

ro

.

.

.

C

-

dv rd-lktra) J

.'

.. ..

. . condltion,. . we noF jurp Having our discussion of the boundary to determining when ip is smooth. As in Section 4.1, this is tte our attention under minimal assumptions on J.

(4.6) Theorem. hence, satisfes

=c

=

-

.

completed

=

J

When d = 2 this says that if we can find a line segment ending at z that lies ifi!z each potitt t he complement then z is regulr. For example this ifnplies that the boundary of the slit disc G = D (z : z: k 0) za 0J is regular. .

-

#(IzI2)( Js(,,,)

-

.

149

Note that we use l!/ zl2 = E(!n z)2 rather than j!/ zl which is not mooth at 0. Making a simple change of varibles, then looking at things in polar coordinates, and using the averaging property we have -

tz) of 74#, b, a) to b' th intfection 4.4. Deflne a flat cone y, : 0 E contains 0'v the line (# + 1 dimensional hyperplane that RJ. ullat ucone'' cone.'' replaced by is Show that (4.5c)remains true if

exerclse with a d

The Dirichlet Problem

.

.

bj r2 and letting ,

. .

.

-+

r

.. .

0 it follows .

.

.

c)

Unbpunded G. As in the previous three sectins, 4ur last topic i to wliat hpbens when somethinx becozes nbounddk Thi time tve $v1ll d ignjre J. combining d tocu p (4.s),(4.5), (4,6)we haki

diss

(4.7a) Therem. Suppose that P oint of OG is regular. If for al1 lttn of bndd

(4.1).

.f

z

is boupded and contiuous 1, ttten e G, .J$tr < x) =

'p

d that each is the uique

To see that there might be other unbounded solutions consider G = J(0) = 0, and note that u(z) cz is a solution. Conversely, we have =

(4.7b) Theorem. Suppose that f is bounded and point of OG is regular. If for some z 6 G, P.('r < x) (4.1) is not unique.

continuous <

and

1, then the

(0,co),

that each solution

of

Section 4.5

DiEerential Equations

Chaptez 4 Partal

150

given in (4.6a), x) has the averagin: proprty = Proof Since @)= regular, each point y a trivial 0 in G. Since it is C* and hms Llt e OG is comparison and (4.5a)implies .&t'r

)=

.

limsup Pztr

x)

=

!/

> 1)

K limsup Pztr

K Ppt'r

q

1)

>

0

=

show that /z is a solution

The last two observations completes the proof.

.

By working a little harder, we can is the only way to produce new boudel

that

show

(4.1)with f

of

&.&('r

adding

=

H

0, which I::I

x)

to

solutions.

(z)

-

< x)

Erlmtkr

=

+ tJ%('r

We will warm up for this by proving the following special

(4.7d)Theorem.

'

:'

....

..

:

..

'

.

.

. .

.

cmse

If u is bounded and harmonic in Rd then

in

.

=

Rd.

o(z)

ppxr

=

boundary function

.f

(4.1)with

=

x),

e 0. boundary

follows.

k 0. theorem

the

on

martingale

hBtj

at

=

/z(z) -+

=

.2.((.B);m

(z) k

cxl proves

t)

>

-MPrr

=

1 -MJ(r

> @)

x).

''

..

'

.

'''''

:

'

. ..

r .

.

.

..

.

.

,

'r)

'tz

is

constant.

imply tlzat uBtj is a bpunded Proof (4.2)above and (2.6)in Chapter 2theorem &; implies thay s t martingale. So the martingale convergence u(#t) Ufn Since U is memqurable with respect to the tail c'-fleld, it follows from (2.12)in Chapter 1 that P.( < Ux < bj is eithi H 0 or H 1 for any The lmst result implies that there is a onstapt c independnt of z so that a< values it flloWs that ,i(z) c E=U. k c. E expected Ku,z c c) 1. -+

-+

.

.

,

'akig

1z(#f) is > bounded localrs y (4.7c) From the= proof of (4.7d)we see that x4, hk U, iii J( r) so i-) On (i. < lipnyr w n (0, s U. frprtingay of tlt st i a cstat Wa ce ifdepekt h t we needthatto sh + is that ot th co). this i Intuitikely, qen #r a o (r pointBo so msymptotic c'-qeld, but the fact that 0 < Pztr = cy?) < 1 makes of the triviality extract tltl. in Chapter 1. from (2.12) tt difftc ult to around identiqed in the previos p'ra raph, Fe 4i1l the diculty To get ' in doing this aie t steps whole space. The two extendu to the EcfBpj (a) Let #(z) = utzl r < x). (4.6)and (4.5)imply that # i with boundary function ). H 0. bounded and satisfles (4.1)

Proof of

(4.1)with

satis:es

To do this we use the optional stopping time r h and note (#z) 0 to get

.

G

1t?

Equation

Because of the Maxkov property it is enopgh to show that for all z and t we hake tptzl S .E'.o(#t). Sinc w k 0 this is trivial if tr / G. Tp prov this for z E G we no te that (i) implies J/l = wBt) is a oundect local martingale on (0, and I'Ip'g 0 on (r < x), so using the optional stopping theorem

'. '

which

to G,

's

(iii) 'tp(St) is a submartingale.

x)

=

(ii) w

Letting t

'tz

'z(z)

(i) When restricted

.'

Suppose that f is bounded and coninuops and that each (4.7c) Theorem. nd tisfl (4 1)''i G , th thie i PO '!1t 9f OG i? regular. If is bounded stht 4 such that

i

To complete the proof now, we will show in four steps that from which the desired result follows immediately. The proof of (4.7b)implies that Mprr < x) satisfies function f > 0. Combining this with (a) the desired relt

E

.

Poisson

=

o(z)

=

.Q(I#,.Af)

(iv) There is a constant

p

=

.E%('tp(ft); r

so that o(z)

=

t4 K .#Q('tz?(ff))

>

ppzr

tx).

=

Since w is a bounded submartingale it follovs that as t s+ x, wBtj to a lin'lit W'x The argument in (4.7d)implies there is a constant ftW'x p) 1 for all z. Letting t x iy .

=

=

converges

#

so that

-.+

'tp(z)

=

ErwBt

);r

t)

'>

'tz(#t).

'titlk

'thet

arid using the bounded

convergence

gives

(iv).

(:I

=

'

'

'

-

(b) Let f

=

llllx

and 1D(z)

look =

at

(z) + Mpzr

(c

=

x)

z z

eG s cc

4.5. Poisson's

Equatlon

In this section, we will see what happens when we equationconsidered in the lmst section. That is, we

(5.1a) u E 6'2 and lzLu (5.1b)At each point of 0G, =

-g

add a function of z will study:

to the

in G. u is, continuous

and

'u

=

0.

As in Section 4.2, we can add a solution of (4.1)to replace 0 in (5.1b)by u = f. As always, the first step in solving (5.1)is to find a local martingale. 'tz

=

4 Partial Dilerential

Capter

152

Poisson

Section 4.5

Equations

's

Equation

153

side is also. The left-hand side is a local martingale on (0, so the right-hand If ?? 6 C2, then repeating the calculation in the proof of (5.2)shows that for s C (0, 'r),

j

tfk nBt ) + =

is a local martingale Proof

(,

or

v(s,)

E

yj 'u(#c)

/

=

-

%nB,)

< r.' hand side

The next step is to prove

(5 3) Theorezn. .

(5 1) that is

.

result.

,

re bottnded

. '

?

If tht'e is a soluti

f

w(z)

H

. .

.

.

.

.

.

gBtldt

Ez

'

'tl

I-fl K Ilullx+

rllpllx

(2.7)in

'

Chapter 2 and

.

H

tztzl As usual, it is (5.4) Theorem. then it satisfles Proof

emsy

=

lim Mt

,(#)

o

=

)

EzM;

=

(J* J 0

implies that on dt

l

j

J

J-

('r > sJ,

gntt

J 0

If

'p

E

di E

E

-

the conditiops

corfl

rtded

ms no suipris.

to

.

-+

We begin by observing: (i) lt foliows from (4.5a)that if e7> 0, then Pcn (r > :) s+ 0. (ii) If G is bounded, then (1.3)in Chapter 3 implies C supz E=r <; hence, 11'p11x < cllpllx x.

,

Let 6 > 0. Beginning with some elementary Markoy property we have

0

n

-+

x,

.

'

+

'?(s-)

and,

T .(#,)

+ E==

ds ; r

> e

e '

.

and using

(x)

inequalities then using the

!

ds Ip(#a)I

e') S ellpllx +

II't?Ilcx,JD.-(r

> t')

'

(i)and (ii)proves (5.5)since

e: is arbitrary.

E1

Last, but not lemst, we come to th question t smoothness. Foi thee developments we will assume that g is deflned on Rd not just in G and hms Recall we are supposing G i bounded and notice that the compact support. values of g on G are irrelevant for (5.1),so there will be no loss of generality if we later want to suppose that dz 0. We begin with th case d k 3, because in ihis case (2.2)in Chapter 3 implies

J#(z)

=

Co

(ll

<

=

Letting .

a)

(lr) dr

(5 5) Theorem. Suppose thxt and are bounded. Let y be a regulaz point of 0G. If za E G and z,z 0. y, then r(zn)

K EzI'p(zn)I

r(z)

(5.1a). sz

discuion lii the lt setiop, onditins that the boundary hold should guarantee

K e'll#llcz + .E'a,-(l'??(.e)l;'r' >

Suppose that G is bounked and g is continuous.

gnt,

.q

rpartingale

tter tlteektensive

.

to show

The Markov property

+

The left-hand side is a local martingale on (0,r), so the integral on the righthand side is also. However, the integral is continuous and locally of boundd vaiation, so by (3.3)in Chapter 2 it must be > 0. Since + g is continuous in G, it follows that it is H 0 i G, for if it Were # 0 at some point then we Fpuld have a contyadiction. I::I

dt

=

t1r

ErMZ

a local

Tht'

(5.1b)imply

T

Mv

,

Proof

'tz

is integrable,

(.j1A.p

/'

=

f

Proof If satisfles (5.1a)then Mt defined in (5.2)iq a lpcal martingale on (x) r). tl G. < then in Chapter implies for If G is bounded, 3 Err : z (0, a (1.8) E nd g are bounded then for t < r If

side

/,.

4v

-+

.

0

Since the right-hand

' gBr,

.

'

:

.

..

T

-

+

Jlr

I:I

a uniqueness

Suppose that G and must be

/

Lumjds

the flikt term on the right-

and

-g

=

pg

J 0

o

(0,r).

bounded,it

.

.

laLu

This proves (5.2),since is a local martingale on

j

:;-

dB, +

.

J 0 )'

for t

,,?(s,)

-

r).

formula ms we did in the last section gives

Applying lt's 'u(#t)

'r),

g Bs ) ds

1(z)

=

Ez 0

:

Ip(f)I

<

(x,

Chapter

154

Pattil

z.l

and, moreover, is

a

Dirential

Section 4.5

Equations

bounded function of

This means

z.

we can

=

gBt) dt

E,

Dij

() use the

strong

Alarkov property o(z)

and change

notation

=

conclude

to

T

Ej

.

#(#)

ln this cse,

lz pI2-d (2 =

-

tpt-(plz

-

the estimate used

)

dt + EwBp

above

IDllz

to get

's

Equmtion

155

As in Section 4.2, txopble starts when we consider second derivatives. If # j, then

deflne

X

1t)(z)

Poisson

p1-d-2(zf

-

yjj

-

leads to

pI2-dIS c'lz

-

yilzj

-

p1-/

-

which is (justbarely) not locally integrable. As in Section 4.2, if g is Hlder continuous of order a, we can. get an extra lz g1t' to save the day. The details are tedious, so we will content ourselves to state the result.

,

-

(*)

'p(z)

o(z)

=

E=wB.)

-

,

(4.6) tells us that the second term is C in G, o nepd pnly prpve that w is, a task that is made sipple by the fact thart (2.3)lrt Chpter tl iv) te followingexpllcit ttfnl

(2.4)and

,,(z) cd lz pI2-d,(p)dy =

If g is bounded and has compact support, C Svhich only depends on d so t at

.

S CII#IIx lfh'u'(z)I

f,#0J

jz

.; .

..

dy

<

pjd-:

-

,

then y

&'''

Di Iz pl2-d -

=

d

2

(z.j

yj )

-

J

2(zy

'

s jet reader

(zf Iz -

yij pld

.

.

jz

-

pj d

side is conyergent

-

yi)

by differto apply

o(z) where G is the potential

dy

=

Gz,ylgy)

j

=

logtlz -.1. ' -lz pI

G was defined as

(,g())

lz

y yy..z pl

-

pI)

d d

at were chosen

the

lim E=

T-x

2 1

fla

to make the integrarl =

::.z: =

So if

converge.

r

jg dz

=

1, it is sucient

0, we

p

Using this iuterpretation of w, we can emily show that holds, fiblem is reduced to provinj that w is C2, which is a pioblem Once all the computations holds in d are done, we fln.d that in d

=

.(#,)dt

(+)

< x

then

continuous,

dy

-f,t(z,f) ./r-

G(z,p):(p)dp

since

-

.

kernel deflned izl (2.7)of Chapter 3, that is,

G(z,:)

where the see that

y

p(p) (j y s jj,jj.

is C a

. g

-

-

-

.

,

'.. . .. '

-

w

The last result settles the question of smoothness in d k 3. To exted result t d < 2, we need to flnd a substitute for (+).To do this, we 1et

-

.'So diFerentiating undr the integyal sign .' . ''''.Diwj . . . Cd(2 d) . (zy 'yi). gy) , Thr integral on the right-hand

then

contizmous,

flzld a proof either in Port and Stone (1978), pages 116-117, or 'Pr4ldinger (1977), gives pyges 53-55. Combining (+)with (5.6b)

.yy

2

=

is C1 anct

w

g j . . y y ( .. .. . k . . .

.

Proof M?ewill content ourselves to show that the ekprsion entiating under the integral sign converges and leave it to the (1.7) to make the argument rigprous. Now -

can and

lf g is Hlder

(5.6) Theorem. Suppose that G is bounded. If g is Hlder and satisfes (5.1a). C2 hence 6 'p

j'

2

The reader in Gilbazg

-

The flrst derivative is easy. (5.6a) Theorem. there is a constant

(5.6b) Theorem.

(5.6)

to assume that g is ontinuous.

po again our in catctl. %2 and that The reader can flnd

details for d cited above.

=

Equations

Parla/ Diflrential

Chapter d

156

1 in

(4.5)of

Chapter 6,

for d

and

either

k 2 ih

of the sources

we will consider

1

what

when

we add c'tz

to the left-hand

we will study

in Section 4.4. That is,

considyed

=

cos

.

1=

ltappens

Te Schrdinger

Section 4.6

The general solution is Wcos z + B sin bzj where b = boundary condition to be satisfled we must have

Equatlon

4.6. The Schrdinger In this section, side of the equation

E

c

cost-c)

.

+ B sin ba + fsint-a)

then subtracting

and

Adding the two equations

2 = 2.4 cos ba From this solve for

we see

that B

=

0 always

0

the

we want

B sin ba

-

them it follows that ZB sin

=

and

works

2v. So if

Wcos ba

=

Equation

c

be

we may or may not

able

to

..

.

'

. . .

If os ba

If cos ba # 0 thrp

Proof 'tI(.h)

Let ct -

explct)

on t

=

-

exp(c,)V1t(l,)

=

However this cannot possibly hold for hapd side is % 0 torsome valus ot z.

formula gives expto)

'tz(1,)

dBo +

.

A'tz(#,) expto) 0

',

c6nidering

is

1 -u& 2

+

'y'tz

=

.

.

H

=

0,

and

the

E1

>

(-c, c), =

0

c

H

n',an

'a(c)

=

'

'

'

'

'

'

'

. .

,

E.

''

''

'

while ,

(1

.

.

.

. ..

.

.

.

'

.

,

il '

. .

.

.. .

.

1It-r-cl

X

=

1

.

1

example

ThO

=

.

i:l

.

'

.. . .

'

'

the'imst

.

c(#,)#s

frexp

:

5ye

.. :

.

let

''''''. ...''.

then w H co in (-c, c). The rest of this sctipn is devoted to showing that if is flne.'' The development will require several stages. % / x, then The flrst step is to show (6.3a) Lemma. Let > 0. There is a p > 0 so that if H is an oprk Lebsgue measure lffl'' S it' and rn p inf (t > 0 : Bt / ff) then .

f

the right-

.

deverything

exptcrl)

EzfBr)

E

l

k: 0

.... .. wtt . .. Exampte j.i)) tht the trouble x/2 then c H y is too large, or to be pmise, if

(agatnin o(z)

We will not do this, howeyer, because the folloFing simple this result is false. 1, G

.

see betow

;

,

''

r(w)

sine

'llrst

At this point, the reader mlght eypect that the next step, wsitr hps been fve times before) is to mssume that everything is bounded and conclude that if there is a solution of (6.1)that is bounded, it must be

=

.

ab > x/2

ds

E

6.1. Let d

.

:

for < r. This proves (6.2),since d% = c(#,) #s, laLn + cu term on the right-hand side is a local m>rtingaleon (0,r).

'p(z)

''

exlmpl is that if c

j

1

' ' . ..

'

Wewiil

dca

0

0

Exnmple

then

cos z/ cos ba

=

)'

Applying It's

+

.'

ab < x/2

9.1) that if

;'

cB'lds.

.j-

cos z/ cos ba is a solution.

=

r(z)

(0,r).

1t(#())

u(z)

(inExample

We will see later is a local martingale

'

.

0 then there is no solution.

=

shows that

eollztiox

We

'f

i . ..

(

.

'

'''

g

:

.

sup E. (exptpcwl) s2 '

g)

laroofpiic

.t,

>

zrn

() so that ee,

>

-/)

<

jn

s 4/3. clearly,

1 e (2za)d/2

-Ir-pIQ/2-?

dy

1S1

s (2za)d/as

1

set with

Chapter d

158

Paztial Dirential

Section 4.6

Equations

pl2A.7)dl2 if we pick Jl so that < 1/4. Using the Markov property proof of (1.2)in Chapter 3 we can conclude that

Prrx

>

n')

1 s : sup 9 So it follows by induction that for

''

( > (

Espnxrn

=

>

Pvrn

integers k

all

d V 1z'(p)S : Fhere

k 0 we have

K

-

=1

(x)

4

s J7 k=1

1 4:-:

=

we

cxl

1

J7 ak-,

. .: : .

% n') .

=

4

k=,

Careful readers fnay have noticed that we left r.t However, by the Blument 'h a l 0-1 1aw either Prg computation is valid, or Prrx = 0) 1 in which

(

..

1 .

1

-

za

r

=

tb(z)

Lt c* : s> Ic(z)l.By (6.3a), we can pick rc o mall tltat itrr I.& Bz I > r) and r a, then l'xexplc*tr'r) < 2 for all z.

Since 1/z is cnvex,

:+

r

(6.3b) Lemma.

inft

:

Let 25 K a.

11-D(z, 2J) C G and y E D(z,

Th e remson for Olr interest in this is that it for y e D(z, J). If Dy, r) C G, and r 1D(:)

=

K ro then th

Splexplcg'rl'tz'l./ (Tr)))

= .;'plexptc*Trll

o(z)

:'

'

xdwyt

*'

'E

Co

k ckl imptis

2 (:x)

and using

(++)

jopfy,vj

s2

.

c.d

a simple covering

,

w

k 0

we

ttltzl

1/2 and

integrating from 0

dz

have dz

(++)and (+)it follows that

j

(6.3b) and

rd-l

s(.,aJ)

Co k 2-1 (2J) a oy.n)

C

k

1t)tzl

--

D(z, 2J) D Dy, J), tp(z)

by

1

-1

.(g)

dz

0(,,,)

o k 2-1 (2J) d 2

memsure,

l/.zalexptc*garl)

tp(z) k 2

tp(z).k :-z co (2s)a

jopLu,vj 1stzlrtdzl

to be a probability indepehdehtk

are

ztptzlvtdzl

k

where again Co = d/cd. Combining ijlies

the f>c4

Jehsen's inequality implies

d

wij

z and using

=

aDztr)

Combining the last two displays, multiplying to 26 we get the lower bound

K .G(exp(*Tr)to(#(Tr)))

where z' is surface memsure on oDy, r) normalized since the exit time, Tr, and exit location, #(Tr),

dy

'

), then

Varkovproferty

strog

o(a)r(dz)

and

k f.(ex#(-c*X)+(#(Tr)))

j

f'alexpl-c*Trlz

Rearranging

shows

dz k 2-1

.E'.(ekp(c(s)u)(#(Tr)j

(2J)d

wybK 2d+2o(z)

Proof

:

s

-

:::i

= fxlexpt-c *Tr))

2

=

=

rd-l

1). ltearranging we have

=

D(p,O

z)

s(,,,)

(z : 1zI 'tz'(z)

.

0 out of the expected value. 0) 0 ln which cmse our = 1. E1 cmse Erexporu)

=

of

cd

,.,(

where Co Vld is > constant that depends only pp d. Repeatirk the flrst argument in the prof with y gives that cr. k -c*Tr

have

. . '..

4

-a

is the surface area

the last inequality by

j

-1

.

Equatiim

=

Eexptpalftl 1)-/ 4 rn .

(72

(+)

DO

Ez expldlw)

and

%a

1)-/)

-

1 sup Patrn > n') < w 4 = eo K 4/3 by mssumption,

Since e' is incremsing, and

(

D(g, J) c: G, multiplying integrating from 0 to J gives If J

> y)

1)y); rn

-

ms i!l the

The Schrdinger

.

-1

8d wyj Co

argument

:ic

:-(d+2) wy)

lead to

D

Chapter 4

160

(6.3c) Theorem.

Partial Dirential Let G be

Equatiops

a connected

1o(z) < x Proof w < (x) an open a subset with zrp so wy) onc

Sectfon 4.6

open set. If w

for

all z 6

/

(:x)

then

(A3)

G

(6.3b),we

E1

< x.

WC

MYOCCCCI to tMc SRiQRCRYS

I'CSl;It 5Ve ,

5V1t

to

lf there is a solution 'p(z)

%

of

(6.1)that

ErfBpj

explcrl)

toof (6.2)implies that Ma Since f, c, and u are bounded, letting s gence theorem gives ,tz(/,)

=

'u(z) =

th: lt

Strengthen

161

x.

see

--,

CeSX l s ion.

/

(6.3) Theorem.

that if 1t)(z) < x, 23 K a, and D(z, 2J) C G, then tlJ(z) < XJ is on D(z? J). From this result, it follows that Gz = (z : considered subset of G. To argue now tht Gz is also closed (when mq of G) we observe that if 25 < a, Dy, 8) (EEG, and we have z,z E Go large, y 6 Dtza, J) and Dtzrj, 2J) C G, y e G then for n suciently From

w

Te Schrldfnger Equaton

ErlBvt

exptcr);

r

is bounded, it must be

exptc,)

1r

is a local martingale on (0,r). A t and using the bounded conver-

< t) + EsuBtj

explc);

r > t)

!

Since J is bounded and u?tzl Sxexptcr) < x, the dominated convergence thepremimplies that as t x, the first term converges to EofBpj exptzl). =

-+

(6.3d) Theorem. Let G be 1G1< x. If / i2$ then

a connected

inite Lebsgue measure,

opep set with

To show that the second term 0 we begin with the observation that since conditional expectation and the Markov property (rt,> t) E Ft the dennition of -+

,

't

1D(z)

qup

,

< txl

imply

'

'

E

'.

fpr Proof Let ff c G be compact so th>t 1f? ff l < y the const>nt in (6.3a) 0 c For each z e K we can pick a ta so that 25. K rc and D(z, 2&) C G. The open sets D(z, &) coker ff so there is a qnite subcover Di, &), 1 K % f Clearly,

ErltBt)

-

=

exptcr); r > t)

.

=

.

'tt7(zf)

suo ls-sz (6.3b)implies

wyj

K 2d+2o(z) for M

If y 6 H = G Markov property

-

=

< cx:h

p 6 Dzi,

sup wyj

vK

ff then Ey texplc*rJzll ,

.;),

.

K 2 by

r) + Ev

a1l y E

'r(p)k

(6.2a)

.

so using

rn (exptcolzpllra); 1t?(p) Ey (exptcrx); f'f'p Bpn ff 2 + K (explcrx); e ) i 2 + 2M =

=

Now we claim that for

so

< cxl

BvH

.

the

.

. .(

.

'

.

strong

l'

.

=

(A1) G is a bounded (A2)

f

and c are

connected

bounded and

open set. continuous.

t)

G expt-c*l.!%tr

S 1) k e > 0

r > fz.tl'ztftllexptctl;

t)

:-l.E'.(lu(#t)l exptcr); s e'-'lleullxfwtexptcrl;

S

x, by the dominated convezgence theorem since as and < x) 1. Going back to the first equation and the proof is complete. shown 1ttzl -+

u)(z)

.l%tr

=

'p(z)

(6.3d)established,

r > > f)

The flrst inequality is trivial. The last two follow easily frpm (A1). See the first display in the proof f (1.2)in Clzapter 3. Replacing wBtj by 6,

: Ik-4

we are now mpre than ready to prove our uniqueof the results that follow we will now simplify result. the statements To ness list the mssumptions that w will malte for the rest of the section.

With

.

EzkztBt) explcrllF); EruBt ) exptclzstlt) ;r

=

r > t) v > t) =

-+

0

Es exptcr)

< tx,

in the proof, we have Cl

This completes our consideration of uniqueness. The next stage in our always furtunately, is it has been. Recall that here and in program, ms emsy as what follows we are assuming (A1)-(A2). (4.4) Theorezn.

lf v

e C2,

then it satisfies

(6.1a)in G.

Capter

162 Proof

d Patial

DiRrential

The Markov property

('r >

implies that on

f.(exp(cr)J(lr)l.'&)

=

=

The Schtdinger

Section 4.6

Equations

to the previous Emse.We begin with the identity established for a11 t and Brownian paths LJ and, hence, holds when t

s),

explollbxlexplcrlytlzl)

r

exp(c,).??(.s,)

cBsjds

exp

The left-hand side is a local martingale on (0, so the right-hand side is also. If 6 C2 then repeating the calculation in the proof of (6.2)shows that for

=

0

T

1+

0

Equation ther,

163

which hold

r(tJ)

=

T

ctrldr

cB,) exp

ds

,N

'r),

Mgltiplying by JBp)

'tl

and taking expected

values gives

,

s e ( r) ,

,

'?( s-lexpto)

/' ()as

.p(.o)

..

.

+

-

-

r(z)

cs)

0 + a local martingale

(srlexptcrlfr

Conditionzg ..

.

.

:

' '

L

on

'?()

.

The left-hand side is a local martingale on (0,r), so the integral on the righthand sid is also. Howeker, the integral is continuous and locally of bounded variation, so by (3.3)in Chapter 2 it must be H 0. Since E ()2 and c is 1Ar continuous, it follows that + cv H 0, for if it were # 0 at some point then ? EI we would have a contradictlon.

(.nr) +

Erl

=

j (c(l,)

ctsrlfrj

=

a

EnlBvt

> 1?1(z),

J(lz)1(,<,.)) ds

exp

.

,

ve can write

and using the Markov property,

'p

The 'Ilrst term,

LjT

* Er

+

fb(c(l,)'p(f,);

r > s)

the

abov

ms

ds

0

r1(z) hF r2(z)

is C* by

(4.6).The

secopd term is

T

Havzg proved (6.4),the next step is to the boundary As in the last two sections, we need the boundary to be regular. consider

'p2(z) =

condition.

c(f,)1?(f,)

Em

ds

0 ' .

.

.

c(z)v(z) then we can apply results from the last section. If c so if we 1et gz) and f are bounded and w / x, then is bounded by (6.3d),so it follows from results in the last section tht ra is C2 and has a bounded derivative. is C ad has a bounded Since E C* and G is bounded, it follows that c(z)'p(z) is I'Ilder continuous, derivative. If c is Hlder continuous, then gzj from the last section to conclude and we can use (5.6b) E (72 and hence =

(6.5) Theorezn.

'tl

satisfles

(6.1b)

point of 0G.

at each regular

'p

Proof Let p be a regular point of 0G. We showed in (4.5a)and (4.5b)that if zn 1 and 1 for al1 Jr > 0. 6 D(p, )) y, then J%a(r K 4 Since c is bounded and / is bounded and continuous, the bounded convergence theorem implies that .&a(#z

-+

-.+

-+

f'ralexplczlllfr);r <

1)

-+

E

C=

(exp(cr)J(#r); r

> 1)

if lcIK M in (6.3d)

we observe that of w established

< eTII.fIlx.E'.-(o(1); S eMIlTIlxlIu'IIxP.-(r

=

'pa

If in addition to (A1)-(A3), (6.6) Theorem. C2 and, hence, satis:es (6.1a). E 'p

Jy)

from the rest of the space, To control the contribution then using the Markov property and the boundedness w have

'p

'p1

r >

1)

> 1)

0

-+

E1

This brings us finally to the problem of determining when is smooth enough to be a solution. We use the same trick used in Section 4.3 to reduce 'v

C. Applications

to Brownian

c is Hlder

continuous,

then

Motio

In the nxt three sections we will use the p.d.e. results proved in the last three to derive some formulas for Brownian motion. The first two sections are closely related but th third can be read independently.

Caper

164

Patial

d

Dirential

.

In this section, 4.4 to find the

we will use results for exit distributions for D

(7.1) Theorem.

If

Summing the lmst expression on

for the Ball

4.7. Exit Distributions

.

...

.

.

..

.

' .

.

.

.

:

.

.

.

.'

l1u(z)

:

.

-

r(z)

.

memsure.

If we replace the 1 by

then

n solves

(7.2a)In D,

wE

C2 and Lv

(7.2b) At each point of 0D, The llrst, somewhat

(7.3) Lemnla.

-

e op

z

1)

is

=

painful, step in doing this is to show

If y 6 OD then Lky

'' :

rhj,

0 in D.

=

-

-

Iz

-

'

'

kd

:

E

(z

-

the 1g12,

.

.

(z

y4

-

-

collapses

expression

(-Iz

pl d+a y4

-

.

-

pl2 + 21z12

-

2z

-

.

lpl2 lzl2)

y +

-

.

2z y #

IzI2

=

.

-

=

0

.

.

.

.

' .

' .

lpl2. ' '

.

opep

set D,

'

' '

.

'tl

.

fz).

'p(z)

and

continuous

(12

-

of tlt t- So bringingthe is a linear combtpatlon (tnsl e the v ) iliflkrrptiationupder the integral (andleaving it to the reader to justify tltts uqing (1.7))gives .

0.

=

=

p12 =

since Iz

D

z 6

(4.1):

the Dirichlet problem

lt-#tz)

-

J(z)

+ 2d

-

to be a probability

Jdo ,(z)T(#)r(d#)

y

.

,f+z

-

-

clmss theorem shows that if of the monotne Proof An application (+)holds for J e C* it is valid for bounded mesurable f. In view of (4.3),we can prove (+) for 1 E C* by showing that if V(z) = (1 Iz12)/lz pIf@nd =

.

-

l:!) 12 dT(p)zr(dg) an Iz pI

is surface measure on OD normalized

+ 4:

-

-

.:

';r

Iz

1zI2 z 14 !/1 -

pIa

-

dz + (1 IzI2) Iz p1a+a jz gjdx.a -2dIz d+2p12 + 4d(Izl2 z y4 + (1 lzl2) 2d = Iz pI Iz pIdA2 Iz p1d+2

then

-

where

-2d

=

165

gives

.

=

1

=

.

the Dirichlet problem proved in Section (z : lzI< 1). Our main result is

and measurable,

f is bounded E=1B.4

'

for the Ball

Exit Distributions

Section 4.7

Equations

y sqtisfies

(7.2a). To

(7.2b),the

check

flrst step is to

1 1z 12 ag't#pl

show

,

-

Proof

f(z)

To warm up for this, we observe Dlz

-

pIF

=

Di

(

yp/2

zj

p.f)2y

-

p1z

=

p1P-2(zf

-

yi)

-

=

an Iz -

pl

H

1

apprpah but we prefer to use a soft ncmpzpputational This is just 1, I is invariant under rotations, instead. We begin by observing that f(0) with and LI 0 in D. (For the lmst conclusion apply the result for f H 1.) = : inftt If I > rJ. To conclude f > 1 now, let z E D with l1 r < 1, and 1et r Applying (1.2)of Chapter 3 with G D(0, r) now shows fcalculusy''

:'::z

i

'p

=

so taking p

'-.d

=

Fe have

Di tx (z) Diferentiating

=

=

(-2z) Iz

1

.

-

:1 a

-d(zf

(1 IzI) Iz

+

-(d

+ 2) gives

=

yi4

(-2) Iz 1pl d + 2 (-2zf) Iz pId+a :w)2 dtd + 2)(z d a lzl) (1 di4 + pjd-l-z pl jz Iz -

=

.

.

=

:1 a+a

-d(zf

Diikvz)

yi)

'

-

and using our fact with p

again

-

a

-

.

f(0)

=

EoIB.)

=

f(z)

where the second equality follows from invariance under rotations. To show that r(z) ly) as z p E 0D, we observe that if -+

-+

-

-

-

-

-

-

-

1 Iz12

0

-

kz z)

=

--,

Iz -

zl d

Ip -

zl d

as z

-->

y

z

# y,

Chapter # Partlal Diential

166

4.8. Occupatlon

From the lmsi calculation it is clear that if J > 0, then the convergence is niform for z e Bz6j = OD Dy, J). Thus, if we 1et B3(J) = OD #n(J), then -

since .J(z)

=

Ir(z) T(p)I

jy

=

r(z)

=

-

+

sup

'

'

.

whre .

lJ(z) J(p)I

pl

-

Axlog

-

. .

j t= r(#2 .

'

E

.

'

.

.

.

..

yI

p la.(j

-

d d

1 2

= =

cj

k 3

1)/2gW/2. .

.

y

.

l

z

-

-

'

.

.

. .

.

.

.'

. .

.

-

zE.8t(J)

-+

small

it &is. This where

Exercise 7.1. The point of this exercise is to apply the remspning of this sectipp 1et to r inftf : B3 / HI where # ((z,p) : z e Rd-l y > 0). Ff 0 E

.

.. . atE

.

.

1 lzl2Gz, jz zja -

.

y.

tert

-

zepsinjE

.,

meurabl

r z-/-,(s,)a /cs(z,,),(,)o

Cl

.

tuzded and

lt j i

(.1) Therein.

The derivation given abpve is a little unsatisfying, since it starts Fith the ad thn verifls it bt it is simpler thafl aid' With lflvir's ! ( . 100-103 in Port ad . . trsfoimations It Stone pages (seewhy (1978)). ls rh'. . th merit of explaining 1-p(z) is the probability density of exiti: g: ki is a nonnegative harmonic that has :(0) 1 and kyzj 0 when z e OD and z # y. z ansWr

lz

-

:z7

Cd lz

J(p) tzlzlxldzlj j

ztzl'mtdzl

is,

Gtz , yj

The Erst term 0 as z y, while the sup in the scond is shows that (7.2b) holds and completes the proof of (7.1). -+

'E

.

In the lmst section we considered how Bt leaves D (z : lzl< 1). In this section will investigte how it spends its time before it leaves. Let r inf (t : Bt / we of Chapter 3. That y) potential and be the kernel deined ih let Gz, (2.7) DJ .('

now we observe

ztzlltzjxtda)

f 2lIJIlxfo(J)

.

167

=

fy)

-+

Ball

q

for the Ball

.LL:i ..

.

0 and

-+

1. To prove that -

Times

-

/so(,) a(2)''r(d2)

Occupation, Times for

Secton 4.8

Equations

Gn4, p)

.

Gn, p)

=

=

p)x(dz)

-

.. .

.function

=

-+

roof

-+

=

Rd-1E

=

ptz,p) where Cd is chosen so that

1

zz

,u(z,,) J

(Iz

j' do :,:(0, 1) =

where

Cdy :1a 'F y ald/a

-

/

=

(b) Show flz, p)

=

(c) Show that if zu

=

J -->

do

k, pn

(d) Conclude that Ez,yjfBpj

H -+

0 =

We thin'k of Gnlz. ''''''''''

.

.

.

0)

1.

thenutza,

pn)

J do

ylfo,

ptz,

-+

'k

J, 0).

''

.

'G

.''#

''*

0).

=

.

.

.

.

=

m v. uosa;,

G(z, y)

z

.j

az

-

-

-

z + 1 G(1, yj

-

pl +

z+

z

pl + 1

k y and

(1 (j

z)(

-

.,;(j

Geometrically, if we 'Ilx p then z that it is linear on (0,p) and on GD(#, #) 1 #2. =

-

.

Jx

cases

u)

.

.

0 in H. (.2)

..

.

.

.

-Iz :z.c = -Iz

(1.7)to conclude Lu

(7.1).

with

at z spends at y expected amount of time

Considering tlle two

0 in H and use plz,

. .

&otz, yj

is bounded and contzuous.

(a) Show that Llta

Section 4.5

as the exoected amount of time a Brownian motion exiting G. To be precise, lf before c GD tn the starting spends exiting G is in X before Bt (z,yj dy. Our task in this section is to compute Go(w, p). In d = 1, lvhere D s (s1, 1), (8.1) tells us that '

1 and 1et

deltkz,yt-f,

(+)in

Combine

1

z

z

-

: 1

t7(-1, yj <

-

:

y + 1)

zy

K y leads to

1 + yj + z)

-->

!/) +

-

: -

(1

1

-

= -1

1

g

u

y z

gz <1
p) is determined by the conditions with G0(0, yj = G0(1, p) = 0, and 1) (p,

Gnz,

(1z1

Partial Dirential

Chapter d

168

In d k 2, (8.1)works 1) and the expression

well

in

=

(8.3)

Gotz, 0)

=

Gz, 0)

Equations

When y = 0 for then Gz, 0) is constant (8.1)reduces to

G(e1, 0)

-

Cd(IzI2-d1) =

=

Laplace Transfom,

Section 4.9

(- js/g; jtjg

jzj

dk 3 d 2

on

Proof To check (A) and (B), we observe that if we let Lzt dnote the Laplacian acting in the z variable for flxed y then y)

hrGz,

i

(a) Since To get an explicit formula for Gnz, y) in d 2 for y # 0 we will cheat and of making look up the answer. This weakness of character hms the advantage yj nothing point is the that Go, more than the Grern's funtion for D Fith (jjqyju conditions. Folland (1076), Dirichlet boundary kage j(), function for. D to be the fpnction ff (z, yj on D x D dtermined by th following properties: kE

0 for

=

p/1p12

(

4,.

(+)implies

D,

(b) Let z e 0D. Clerl#, the right-hand vanishes, we note

sid

oe rjeekjs

(B) For

each p

D, ff (.,y4 = G(., yj is C2 and harmonic in D.

e D,

if

zn

z E 0D, fftza,

-+

yj

-+

G(z, p)

The last

0.

i

cotiu6us

-

t

.

..

Cd

Ipl2-dc(z, p/Ip12) =

Iz

.

.

.

z

lzlpl -

e D.

z

To ee th>t it .

.

y

=

j'

.

...(ak.a)

jgjz

t

-

pld-2

z.

for

. .

jgjauua

Cd

Iz

.

C'd

-,

p1a-2

-

-

(A) For

#y

z

that the second term is harmoni

=

each y E

169

.

:r

,

Arcsine Tzw

p1p1=114=2

V0 E

follows from a fact useful for the next proof:

equality

' :

..

Follahd notation to c torm Remark. For convenience) we have changed and of and roles interchanged the z to ours y. This interchange rrtalies no symmetric, since that is, Ik-, y) Green's function is ff y, z). dilerence, the (See Fol lan d (1976)page 110 ) 3

=

(8.6) Leznzna. If Proof

Using

lzI

1 tllen

=

lzl2

lzlpl plpI-1I Iz pI. lzI2

z z and then

=

=

-

1 we have

=

.

-

'

.

,

Proof

Using

(8.1)and

(7.1)

$ve

t?otz.zl

G,

z)

This is harmonic in D by

=

(8.7) Theorenz. z)zr(d#)

=

-

(4.6).(4.5)implies if Ez=GB.

,

p)

-+

G,

z,z

-+

z

e

#)

D . . ..

'

.

. ..

E

..

. . ..

(8.5) Theorem.

ln d k 3, if 0 <

lpl

<

cstz,!,lc(z,,) =

1 then -

I,I2-dc(z,p/Ipl2)

Proof

Again

we need

-

1p1<

- 1

(lnlz

=

.

to

check

22 # + '

+ 1

lz12 Iz =

(1976)%ve

of Folland

2 if 0 <

0D, then

. ..

made

=

Goz, p)

-.L

'gj2

ttguess''

1 then -

p1

-

ln

Izlpl

p1p1

y

+ ln

-

-

1

1)

(A) ard (B).

.

the connection in (8.4),we cn now flnd G.o by guessing with properties (A) and (B). To tguess,'' verifying'' functions we turn to and and fnd 123 in Folland (1976) page Having

ln d

.y

-

= 1p1

p)

-ExGBv,

IzI2lpl22z 2

Again turning to page 123

1 lzI2 aGy, p1 Iz ao

-

-

have -

-

p1p1-112 =

Izlpl

(a)

Gox, p)

-

Gz,

=

-

1 g'

ln

z

-

a 1p1

1p1

Again the frst term is harmonic by (+)since p/1p12 / D. The second term does not depend on z and hence is trivially harmonic. (b)Let z e 0D. Clearly, the right-hand sde is continuous at z. The fact that it vanishes follows immediately D from (8.6).

170

Chapter 4 Partial Dimerential Equations

4.9.

Laplace

In this section

Trqnsforms,

Arcsine

we will apply results

Section 4.9 Laplace Transforms, Arcsine Tzaw

Law

computations in Example 6.1 imply that in this solution.

from Section 4.6 to do three things:

Exerclse

(a) Complete the discussion of Example 6.1. (b) Prove a remarkable observation of Ciesielski

and Taylor (1962) that for Brownian mtion starting at 0, the distribution of the exit time from D ( : lzl < 11 irt dimension d is the same ms that of the totql occupartion tipz of D in dimensiond + 2.

9.1. Show that if,.;

E.

=

Since costz)

=

.'

(c) Prove Lvy's 7

law for the occupation

arcsine

..

time

of

:

(0,x).

0 then

expt-/r)

e-f7)/2

t=c,

(9.1) Theorem.

$ (-c, c)).

Let r

inf (t : Bt

=

E If y

k

Proof

72/8a2 '

a),

=

=

this is what we get if we set

.

.

.

e'T

=

0,

y >

H

If 0

< y <

and

12/8c2,

J

1 in the

H

then

coslzv?JY)

cosjgyj

.H

'

.

'

.

?,

-u 2 . To check tllat w / local martingale on reduces i'J't then

G),

n

(x)

=

costz

2y)/ costc

+

n,'u 0

'tz(-c)

=

we observe If we

Several properties

2y)

(0, 'r).

=

=

u(a)

=

(-p/0

.

.

of n are immdiately

and

Exercise for a1l z. dimension d k 3):

1

(a)

E=

=

(1t(fnAn)e'Y(YkZ''))

(iii) The of

J(r)

=

r2-dJ(1) + (1

strong Markov property

1Lv

i'

1*r, it follows from Fatou's lemma that

-

r(z)

=

.

0

=

.52/8a21

The lwst equation implies w / x, so (6.3)implies the result for y < H x when y k x2/8c2, suppose not. ln this case To see that exptczl) implies '?J(z) ErfBp) is the unique solution of our equation

(6.3) but

=

1 + r2-d

for z

J41)

To express the lmst equation in terms of

Ezenr

1

(6.3)imply

and

pv

(inany

at the hitting time of D and

r2-d)

-

obvious

.f(tzl).

Chapter 3 gives for r > 1

1ttltle-t that (6.2)implies that Mt is a 1et Ta be a sequence of stopping times that

u(z)k =

.

1s(a)a)

= (i) Spherical symmetry implies (ii) Using the strong Markov property

and noting Ta A n

'tz?(z)

.

'p(z)

1ltzl Letting

'

':

tit couise,we on ly have to compute t?(0) but as in Example 9.1, 9.1, we will do this by using a diFerential equation to compute

,

-+

.

. .

x.

solution of

1

(9.1).

1

'p(z)

a nonnegative

in

=

.

s(z)s-exp

.

u(z) is

-p

-t'

:

' . .

-

then Eze'tr

By Example 6.1 !

Jr

c(z)

coshtz 2#) coshta 2p)

=

)

1, G Exnmple 9.1. Tlke d problem cosidered in Section 4.6.

D

Exnmple 9.2. Our second topic is the observation of Ciesielski and Taylor (191). Th proof f the yeneralresult, see Gtor and jhaiye (1g70),setiops 5 ad. 8 r Kight (t981)1 pak 88-89 requires fnore thi a littl fpiliaiity with Bessel functions, so we will only show that the ditrtbuti of th ekit time from (=1,1) in one dimension starting from 0 is the same as that of the total occupation time of D = (z : Izl< 1) in three dimensions starting from the origin. Exercise 9.1 gives the Laplace transform of the exit time from (-1, 1) so we need only compute ''

The third topic is indepertdent of the flrst two.

(ef?+

there is no nonnegative

case

that

.

'p

(y(1) -

=

ofvtz)

=

eD

for z 6 OD

J

we note that

if p(z)

y?(Izl) jzj J , (lzl)

1

jzj

1)

is the unique solution

2

Divz)

(1.12)from

z: -

jzja

+

,,(jzj;

T

z;

jzja

=

/(1zl)then

172

Capter

Patial

d

Section 4.9

Equations

Dirential

which shows that (b)holds. To complete the solution, we have t pick C(c) to make f E C1. Using formulas for Tr) for r < 1 from the previous display, for r > 1 from (a)iand recalling # 3, we lzave

Summing over i gives 1

'q'f

'j''p Multiplying by 2

we see

J''(r)

(b)

=

+

tl

gt

1

--

,

(lzI)+ zjzl T (171)

=

sltis:es

f

that

1

/(1-)

.

g-1

J'(r)

r

'

'

2p.f(r)

-

() for

=

.f'(1+)

r < 1

.

'

'

'

'

'' ' '' '

'

'

'

'

(ik) n is C

.

and hence

-(/(1)

'

'

'

to

F(r) is continuous

at r

1.

=

Exnmple

=

mlPtzz + m + 1)

m=0

arcslne

=

=

-

co (ar)2m c(c) I'---2 + 1)! (2r?z m=0 check

Pzlh

f'rj

=

Ca) -

Jr

r

J ?? (' .)

=

C(c)a r

2 ? .f''(?.)+ -J (r) r

s i,,l,tc,.) =

27 this matches

=

=

Clclc r

a

(1951)derivation

ra

sinhlcr)

Remark. The not depend on ahead does not value for which

dr

-

7-(1

its

'7r

-+

the otlzr

see

arcsine

#1y,x)(z)

-a

=

-

1 -Ar + 2

can now forget about we have

2 arcsintu)

-

-

(9.3) Lemma. Let c(z) C3, bolmded, aud satisqes

# 0.

coshtcr) =

ca

.f(r)

+

2C(c) Jr :,

sinhtar)

(4.2)in

law,

with a,#

Chapter 1. To

k 0. Suppos

'p

is

-1

cv

=

Then

origins

'p(z)

dt

=

e-Gff r

e-pll't

0

Poof Our ssumptions about imply that + c(z)'p(z)) dz in the sense of distribution. So two results from Chapter 2, the Meyer-rfanaka formula (11.4)and (11.7) imply 't?''(z)

sinhtcr)

of

reader should note that by scaling the distribution of S/t does t, so the fraction of time i (0,) that a Brownin gmbler is converge to 1/2 in probability as f x. Indeed r = 1/2 is the r) is smallest. the probability density 1/x r(1

For the last equality prove the irst we start with

for a11 z

=

r)

-

-2(1

'v

2C(a) -

27 and recallipg 1/2 thr solutior

cr

'

the result in

l(s 6 (0,t) : B k 0) Iklf 0 K 0 K 1 then

=

1 a g' 0

% 0t ) =

c(a) sinhlcr)

the solution we want, we that it works. Diferentiating

tRc)coshtcr)

=

sinhtc)

I::l

law. Let Ih

Proof

is one of the happy families pf Bessel flmctions. Letting c 1), we see that when d 3 and hene v r(z) = (z 1)r(z hms a simple form

and simply

c

9.3. Our third and llnal topic is Kac's

(9.2) Llvy's

Stwell

tfound''

Since

sinhtc)

C(c)c-1

-

.

=

Having

1

=

=

Facts (ii)-(iv)give us the information we need to solve for /' at lemst i p'rinciple: (b)is a second order ordinary differential equaticm, so if we specify J(0) = C and J/(0) 0, the latter from (i),there(x7).is a unique solution fc on (0, 1). Given Jc(1), (ii)gives us the solution on (1, We then pick C so that (iv) holds. known'' fact that the only soT0 Carry OU.t O1lr Plan, We begin With the ltltions Wlich tO2#) stay bounded near 0 have the form crl-td/zlf (ayalmy(r (b) where x (a/g)v+2m S(Z)

-

C'(c)c-1

-

-,1)

Solving gives C(c) = 1/ coshtc). Exercise 9.1 when z 0.

-

,

C(.(z)coshtc)

=

=

and the reaij we note that by applyink (b) together, 'p tie equatipns (a) with ad uking of (t.6) (Which rfers prf > the 0(0, 1 t q (iii) ) iu (5 Ca)) * tn cntlde '

Laplace Transforms, Arcsine Tzaw 173

=

1

$

j

.j

r(#t)

jo 'p/(.8,)

.?/(.1)

-

=

dBs

-

.

j (1

+

c(#,)r(#a))

ds

Section 4.9

Chapter 4 Partz'al Differen tial Equations

174

in c(f,) ds and using the integration by parts formula, (10.1) Letting q = exptct), of bounded which = locally and is = with Xt Chajter 2, X q variation, w have

JJ

It is a little tedious to

these

solve

'

,

W-#-#-W

W=

Jo

explcalz,/ll,l

'p(.s(,)

z?(f)exptct)

=

-

t ./r-

explc-lfl

(&

+

dB'

gives a solution.

c(s,)s(s,)Jds

What

+

jv '??(/,)

(9.4)

'D(0)

exptct) + fn'exptca) ds is a local martingale. is bounded. As f 0, lh x, explq) K e-' K marting>le and boppded cpnvergence theorems gives

boupded sincey isusing the 0, so

a

p(z)

EZMZ

=

=

ECM.

(9.3) tells us that

'pt)(z)

=

=

Cezs/*

1l(z) .

J

'tl*

-/'p

=

To flhd

1)

.

+ 1 we write

y'pl

=

'/S0

=

1

,?

-r1

2 1 -K0 u 2

'p

=

1?0

+

2(.+p)

Be= zjw+

-z

+

xl-t

where

rl(z)

This identity

O

.. 1/7

and

+

1

.+p

z >

'v

buded, ()

x < ()

&

C

1 1 = B+ a a + # B/(= 2(cf + p) -

so

up by

we have

-z.2/a

1

1

= a + p W'a = From

2/y

e

=

n

(9.3),(9.4),and

the

-

2

.

-

1 2

.

tfz

z. z--2

g

=

:-(E'+#)'

*

1 =

yy

(x,

j

e-tzt

t

,

t

-

Z'

0

s

dt ds

-#a

-

0

s(t

-

s4

ds dt

of Laplace transforms it follows that

uniqueness

again

CK' e-G(-')

W

o

Ez expt-p.Jftl

.t

p)

+

Fzbini's theorem imply

and invoking the uniqueness

and B we note that we want v to b Jl

171

+ 1

Picking the signs to keep

de -w

(x)

dt

= 'y

j.

with

z>0

To solve the equat ion and note that

so

to fid a bounded C1 function

we wnt

.-@f

d

-

.

.

(X)

) ds

exptc,

Er

=

ata

To go backwards from here to the answer written in changing variables z = 2y, dz = (1/2) 2.tjt dt to get

0

0 using and Fubini's theorem leads to the c(z) nosv Plugging in the dehnition of El formula given above.

we have

warm

1

=

p

+

-.+

-+

'

(9.2),we

asl-lTn

1

W+

=

explc,ldc,

vtz)

:. '

that

interested in is

we are

t

.'

check

to

easy

s=W-W-+-#

+ 7):/J

j

=

but it is

eqations

'p(#t)

'

Mt so and q

Laplace Transfoms, Arcsine fgaw 175

t =

t

-

'm n we

-7'

e s(t

-

s)

ds dt

have the desired result.

C1

jjd

' . .

5 Stochqstic '

' ..

.

'' :

-

.

.

.

.

DiFereptial

'

. ..

.

.

.

.

E .' .

.

(

..

' .

.. .. .. .

..

..

..' .

.E

.

.

.

. ..

Equations ''''''*.

'

'

.

.

.

.

-.

.. . .

.

..

'

.

..

(

..j( '

. .

.

. .

5.1. Exampls ln

Ethis

chapter

(i SDE's) that

we will be interested in splking stochastic E informlly ks are writtn

ilations

diseretial

'

dXs

(X,) ds + c(.X',) dBs

=

or more formally, as an integral

equation:

..'

j

j

(*)

Xt E

'.

: .

.

.. .

' .

' .

' .

unt.dm

1/(.X'a)dq +

Ab +

=

.

.

.

E .'

E

.: .

..

' ' .

l

:

'

;. ..

i

'' '

. . .

.

' ' . . . ..

... beginnipg of, Sectiop 5:2. We will give a prcise forpulatiop of @)at . the In @),Bt is an m dimensional Brownian motion, c is a x z?z matrix, and to make the matrix operation wcyk- out zight we will think of X, and B as column vectors. That is, they are matrices of size d x 1, d x 1, and r?z x 1 respectively. 'Writing out the coordinates @)says that for 1 K i % d .

. .

.

.

. .

.

.

.

.

.

.

.

m

Xgi = ,

xc +

biA) 0

ds + q

'E

J ajy

0

njxj)

dB a

the equation Note that the number m of Brownian motions used to cdrive'' generatibtto additial d. This ds not less than or more be may dilculty, is useful in some sttuations (seeExample 1.5), and will help us distinguishbetween cT which is d x d aud T which is m x m. Here C'T denotes the transpose. of the matrix c. To explain what we have in mind when we write down @)and why we want to solve it, we will now consider some examples. ln some of the later examples Tltls is cstomarily deflned to be process. we will use the term dxusion ttstrong with continuous paths.'' larkov this section, llowever, in process a 'any

'ase

'

Captez 5 Stochatic Difirential Equations

178

Section 5.1

the term will be used to mean tta family of solutions of the SDE, one for each starting point.'' (4.6)will show that under suitable conditions the farrly of solutions desnes a difusion process. rownlnn

Expnentlal Exnmpl where .X is a real number motion. Using lt's formula .1.

Xg

=

a'Y +

j

pX, ds +

a standard

j

'f

cXs dS, +

1

j

j

/2*,

. . .

:

.

.

zi

c() .

.

'

integrals,

't

(8.7)i

Chapter

j

Jkk(x,)ok(A',)

ds

0

k

aitA)

=

i j

z z

+

Substituting dXk J

=

Xia dXja +

,

0

k(x,)

ds +

0

J7nj ,.

i

.

..

zibiz)

we

get

+ zd/tz)

gives

j

t (.E'(.;v't.zq, i ) -dt

.

.

.j

Ex,

-

iexjj

j

,

.

odyytz)

yst)

justifying the pame lnflniteylmal covarlnnce. vaz-lanc and c(z) c(z) When d i= 1, we call c(z) the lnfnlteslmal stnndard the infniteslmal dviation. In Example 1.1, l(z) is o'z proportional to the stock price z. The ininitesimal drift (z) = (/I + (>2/2))z is also proportional to Note, however, that in addition to the drift p built in to the exponential there is a drift c4l2 that comes from the Brownian motion. More precisely it comes from the fact that e= is convex and hence exptcft) is subinartingale. a

)

Di lzl

j

Xj, dXi +

+ aijz)

f(-Y,) ds and digerentiating again =

.

.

.

+ zbjz)

f=0

Subtracting the lmst two equations

:

j =

zibiz)

=

.

....

.

(atla)jjotl

yt

'.'.p,

X3iXj

.

1.2k Th Bessel Ptoces.es. Eknmpl Let W'i be a d-dizensional Brownian motion with d :$. 1, and Xg = I#J$1.Diferentiatint gives

ds

0

by parts formula

Usipg the intgration

(x,) ds

.

.

then the formula for the covariance of two stochmstic 2, implies =

aij

0

'

J'0

.

't

=

'

.

(aY, Xjt

,

=

J(z)#T(z)

=

(x ) da

gives

g

.

i

.

+ E

txa)

drift. is alled th infniteslmal Because of the lmst equation, To giv the meanihg 6t the the cocient w hote that if

.X',J

.

d EXixil j

=

.

0

diferentiating

.

-+

=

0

179

values

Xsibj (.A-a ) ds + E

.

=

i d f(z) W.,;'.;L1=()

E

''-

Using Exji

t

i

+ (.

expectd

J z z + E

=

'f

&(z)

=

)

If aij is also continuous,

ds

= Exponential cz. so Xt solvej @)with (z) (p + (c2/2))zand Brownian motion is often used ms a simple model for stock prices because it stays nonnegative and Quctuations in the price are proportion>l to the price of the stock. Tp eyplpil the lwst phrme, p retprn to the geperal equation (A.) apd qupIf .X$q z then stopping at t ATa for slzitable P ose thatb and tr are contmuous. (x) we stopping times Ta 1 x, taking expected values in @)and letting n have t zi Exji ds + E () So if is continuous =

i

ftlja'i

Lt Xg Xt ekjtpt-l-ls) one dimensional Brownian

Mtlo.

Bt is

and

using the mssociative law, and taking

Examples

t.xiXjj t

(X,)dBi,

,

j. .

!

gzi

Dlzl

jzj

k.

=

jzj

-

z,)

jzj?

g ,

Zlzl . . .'

E

.

.'

'

..

y uu

=

. .

'

.

'

' ..

So using It's formula Xg

-

A'o

=

S

t I/vf '

#F, (, I I

dea +

t

1 -

2

()

d ) ds 1#F,l -

.

j.

jzj

Section 5.1

DiRrential Equations

Chapter 5 Stochutic

180

side, since Bt is a We will use Bt to denote the flrst term on the Tight-hand i.e., Bt is a Browniau motion by Lvy's theorem, local martingale and (#) (4.1) in Chapter 3. Changing notation now we have =

,

d 1 ds 2.Xa -

(1.2)

Xt

@)holds

so

X0

-

Bt +

=

(z) (d 1)/2lzIand

with

=

-

0

o'(z) .

Exnmple 1.3. The Ornstein Uhlenbeck sionalBrownian motion and consider

(1.3)

dX3

.

.

.

E

'

. .

:

'

Eetht be

.

:

.

.

.

.

. ..

' ' .

.

difnen-

a one

.

..

Xt

=

dBt

#

ea'

.X%+

dXg

(1.7)

d To

.

'tl)

=

e-q'''p

U1

=

i'

Xt dt

j

dt

1

-

.

n2

take -.y

Jtz,

'

-p/2

g)

=

z

Note that c is but we need the drift a unit vector tangent to the bz, y4, which is perpendicular to the circle and points inward, to keep ( X) from iying ofl'. always

circle

,

:

t

=

1

-

Xt df

=

-z/2

(z,p)

,

-.-XdBt

=

@)we

this. in the form

write

#fa

0 check To this informally, we note that diferentiating the flrst term in the proddt and diferentiating the second gives # dBt To check this uct gives -aXL formally, we use lt's formula with fu,

. ..

.

Exnmple 1.5; Brownlnn Motion on the Circle. Let Bt be a one dimen= sional Brownian motion, 1et Xg = cos Bt and X lin 1, Bt Since XI+ (.;% X) always stays on the unit circle. It's formuela implies that

-a

(1.4)

.

j

.

.

of the velocity of a Brownian particle which is which describes one component experiences slowed by friction (i.e., an acceleration timesits velocity aLI. where solution explicitly. This is one cse we can flnd the c-Gf

j

,

-ce.iLit +

=

Here V( is an Ornstein-uhlenbeck velocity process, and 5$ is an observation of the position process subject to observation error described by a d#Ft, where W' is a Brownian motion independent of B. The fundamental problem here is: how does one best estimate the present velocity from the observation (K : s K )? This is a problem of important practical signifcance but not One that we will treat here. See for exmple Roger and Williams (1987) 327-329, Wksendal Chapter VI. or p. (1992), .

Process.

181

,

1. E

Examples

Exnmple

...'

'

.

1.6.

Branding

Feller's

'

.

.

Disuslon.

This Xj

k

0

:

and satisfes

:

(1.8)

.

conclude

to

dXt

pXt dt + c

=

Xf dBt

To explain where this equation comes from, we recall that in a byanhing d idetically process, each individual in generation m hms a idepedent distibuted number of ofspring in generation m + 1. Consider a sequence of branching processes (Z;j r?z k t) in which th prbability of k children is p2 ,

(1.5) Theorpm. -Gf

ze

When A

and variance

0.2

=

Jf()

e-2Gr

the distribution dv. z,

of

X is norm>l

with

pan

Proof To se that the integral has a normal distribution with mean 0 and the e-atz indicated vari>nce we use Exercise 6.7 in Chapter 2. Adding natuk El n ow we have the desired result. =

Exnmple

(1.6)

1.4.

The Knlman-Bucy dl d/i

=

=

Filter.

Consider

(A1) the mean of pz is 1 + (A2) the

variance

of ,2 is

(A3) for any 6 > 0,

d#f

(#a/n)with pn

d

Ek>,s

J2a

with 2p2

Since the individuals in generation tributed number of children E (Z1nlZl

tdl'li

Pkdt + -cI4 dt +

,

and suppose

,2

-.+

p

-+

>

0

0.

-+

0 have an independent and identically disnz

=

var(z1nlg;

=

)

nz

nz

=

)

=

nz

1+

.

.

%2

p n

-

182

Section 5.2 It's

Chapter 5 Stochastic DiFerential Equations

So if we let X1

=

of

l

( j XlN/s

var

''=

X''( 0)

2)

-

zj

zJa2

=

lI, pages 176-191 for E.

.

'

5.2. It's

n

-1

=

183

'

z7a

'

E

(0)

XH

z)

others.

some the

-1'

(X1N/a

E

(1981),vol.

arise from genetics. See Karlin and Taylor

then hngtjln

Approach

Approach

.

n

(' :....*.

o?

anb

variance the The lmst two equations say that the inflnitesimal mean rescaled process X;t are z#s and zok2 These quantities obviously converge to those of th proces Xt in (1.8).In Exgmple 8.2 of Chapter 8, we. will shw that under (A1)-(A3) the processes (X1' t k 0J converge weakly to (.X t '0J.,

by giving some esqential defnitions and iptroducing a will kylin th eed fr sme of the fifnaliti. Shat W will coterekmpl uikiks rutt mai prctd exitence bsint the d to ou. nit' tor then entaikld SDE, which wws proved by K. It long before the subjt i brile of the general theory of processes. the complications To flnlly beome precisv, a solution to olr SDE @)iq a triple (.X, Bt where Xf and t are continuous processes adapted t a flltratin so that 7

.

setion ..

.

''

. .

'

. .. .

,

,

We begin this

.....

''''''

. .

.

.

.

.

'

. ' . .

. .

.

.

.

.,&')

,

.-&'

Exnmple

Dlffusion.

Wright-Fisher

1.7.

This Xt E

(0,lj

and satisfies

(i) Bt is a Bqowpiyn mption with respect to &, i.e., for e>ch s amd ff-j.a Bt of and is normally dtstribvted with men 0 nd variarp: s. is indpent -

dXg = (-a.X

+

7(1

,

u'V(1 aX) dBt

.;V))dt +

-

.&'

-

The motivation for this model comes from genetics, but to avoid the detail of that subject we will suppose instead we have an urn with n balks iu it that with may be labelled .A or c. To build up the urn at time m + 1 we sample aln with probability replacement from the urn at time r?z. However, we ignore the draw and place an c ilt, and with probability #/n we ignore the draw and tb mut>tions. ink 1I genetics terrns the lmst two events correspond plac ap pf when the the number of /'s in the urn at time m. Now Let Z;b on a given trial is .A's in the urn at time is z, the probability of drawing an ..

'

-

,fraction

.

-be

.

.

.4

. ... ( .

ps Sine

we are

sampling

z

=

.

( ) 1

& + (1

z)

-

.

g. . . . . .

('

j'

)'

@) ..

.

E

E

,

,

.

'' ' E .

' .. .

We will integral

(

.

A

,

''

' .

:

.

'

..

E'.'

..

#

' ' '' .

' . ..

' ' ' . .

:

' '' .. . . .

' ' . (.. .. . . . .

.

l

. .'

.

tr

0 ..

.

.

'

'

'' :

.

.

=

=

nz ) = npn nz) = nyatl

' ' 'i .

:

'i' . .

.

and

(.X'I'/s z var

j

=

(Xy/aj Xn(0)

=

z)

=

=

Exnmple 1et

2.1.

and

t-(>z+ #41 -

pa

'. . . . ..

i

.

,

' '.

(

.

.

'

.

.

.

,..

, ' . . .

'

.

.

(..

' ' 'i ... . .(' . : .

' '' . .

'

' ..

.

. .

.

'

g

' ' ' : . . .' ' . ': : . . .: . .. . . .. :

'

i .

' .

.' .

r ' .

and ldcall# bouzded so the

'

.

g

E'

.

'

'

(

,

. ,

.

XtB

.

,is

..=

p,,(1

j ' .' .

'

.

it will hold for h It is emsy to see that if the SDE holds for some adapted #f). When Xg is the flltration generatd by Xg to FF the filtration = generated by the Brownian motion, then we can take h an d X g is called solqtlon. could satisfy (..) witholtt strong dicult imagine how It Xj to a B eem' qdapted pl b th is famws dtle H tp h tp , T>np> showing ut re a being that this can happen. i y E

y.

-y;

z))

.

vyj

1

E..

.

y yj

,

y

q

,

0

=

:

:

sgntNa

=

0

'

) d
f

where sjnt) . :i 1 if z k 0, sgntz) ::'z: if z < 0. Since Bi is a local martingale with (.8)z= t, (4.2)in Chapter 3 implies that Bt is Brownian notion with re1, the mssociative the flltration generated by 141. Since sgnlzlz spect to F law implies that -1

'

.

=

,

j

't

.

Now pa x so again the infnitesimal mean and variance of A'l' z as n of irt (1.9)but the rigorous connection has to wait until those X3 to converge Example 8.3 of Chapter 8. This is just one of many of the diflksions that can

..

E

E

E' .

1

.

,

Let 144 be a one dimensional Brownian motion with I/'zn ,

'

Xn(0)

-

.

:

.

Bt z)

''

.; ..

.

k

, : '

are measurable '

dh

&(xa)

0

'

.&'

-s

pa)

-

!

,

(-'L) ts #,

'

'

.&'

then hnvtjln

=

E

-+

+

that

lways assm ezstsk

.

t

a'tq

=?

,

(glnjz; var(z lA

E

'

.

,

with replacement,

E

If we let .X1'

-s

-

:

'.

(ii) Xg satisqes

-+

0

So 14:is

a solution

of

sgntea ) d%

@)with

b > 0,

J(z)

d
=

=

I/li

0 =

sgntz),

and

X

=

h

B'

=

FXB g

.

,

5 Stochastic Dilrential

Chapte

184

To se that JF is a not a strong in Chapter 2, with Xt mula, (11.4)

Equations

solution

#Fk and

=

Section 5.2 T's fom

IzIto get

(2.2) Theorem. If for a11 j, z, and we have l(%lzl c%(p)l K ff ustochmstic pl and l(z) then the dilerential equation (p)lS ff Iz

Bt

(*)

we apply the Meyer-Tanaka

flzj

=

Approach

=

,

-

185

lz

-

pl

-

,

j

t

L't

1m1 -

''

sgn(<,)

=

d
=

0

has

''' 111'* ..jjL'

densityformula in wliqrqLt is the local time of INl at 0. The ocupatiqn time establisheb in locat of continuity and the te tlme (1t.7) pf haptey), (1.8) of Cltpter t, imyly tht '''

'

..

.

.

.

.

.

.

.

.

. .

.

.

''''''

''

'

.

.

.

'

.

.

'

''''

'' '' .

'

. .

.'.

.E

:

:

.

,

t

1

.

:

w) *6

(j

' ' .

gs

l.(ll
xu

(

E

6

1

=

;;; ou

pk

6

...

c

..

Lg* td hhye

t utu

E

r

z

as

-

witlk

ippt

nt

E

j

E L

)

g

:

I1z1 f,o is melurabl so 0. b# 1*41,k Howevei, I'F ls clirly gnrtd by generated 1m1. ..

po

..y

i

:

((

...

.

g .

'h

:

'' .

qttrtin .

.

j

=

,

.

.

k.'. .

uniqueness

holds.

the solution by a successive

approyimation

scheme

that

t

Z

Xz +

=

t

l(-Ya) dB. +

(Xa) ds

to describe oze iteration. In words, we usr. the pwcess (Xo t k 0) ms input for and the output th process (v'% tlte coecient c >nd t k 0). To solve @) we begin wiih th initial duessX: H z and dehne the successive guess by '1s

t:;

=

0

and pathwise

We construct

(I,) ds

0

,

ltltit jo

j

t

Pathwise uniquness

Proof

c(A%) dm +

z +

reducesto the clmssical Picarb iteration fnetotl for splving ordinary diferential equationsin the special cmse c p 0. We begin by irroducing the zptation

i

: E

As usual when pe have an equation, we are intvrested in having a unique solution. For SDE there are several notions 6f uniqenessk Our flrst, pathwis unlqueness holds if whenever Xt and X3 are solutions of @)with Xo v'L= z driven by the ame Brownia'n motion Bt then with probability one Xg X; for all t k- 0. Here, the fltrations St and F; are allowed to be diferent.

(2.1) TheoreG.

solution

a strong

(E

t daptd toth it

.

Xt

=

does not hold in Example 2.1 .

show that P/f W observed in Example 2.1 that 1/7 was a solutionk =i = 0, ih sgnl-z) ekept when nt olutin, tht We -W is alo a and side right So is side is 1 which cmse the left the ro

,

iteration:

xa4.l

...,z

sf

ft k c The rest of this section is dvoted to prving that this procedure constructs a stiong solution and that pathwise uniqueness holds. To help the reader digest the argument we have divided it into sectis. Oh of the claims is emsy. lt is clear by induction that A'1' is adapted to LFF >

3

.

The baslc

estlmate

which

(2.3) Lemma. Let r be a stopping B (4td + 16d2)ff 2 Then =

is the key to the proof is: time

with

respect to

&,

let T

<

(>:)

and

1et

.

-sgnl)

EL

-1.

'

( j 't sgn(-N,)d#, .

=

-

j

z

sgntI#,l dB, + 2

!

j

jo

1fyv.=o) d,

E

0

1(e.=p) dBs

Turning to positive results:

0

jo

IF,

-

h l2ds

*

.

2s

(c

So the left-hnd

side of

s

(2.3)is

.

t

E

A

7012 + BE

-

The inei uality is trivial if the right-hand side is infihite each term on the right is fnite. To begin we recall that that suppose 2a2 + 22 Using this twice 2a2 + 2( + c)2 2c2 + 42 + 4ci + + c)

'

=

Z 12 S 2E IFn

E

=

2

-

Proof

To prove that the second integral is (andhence the rightzhand side is -W)) recall that #Fk is a Brownian motiou. Example 3.1 in Chaptrr 2 impls that (s : W'a= 0J hms measure 0, and hence f

If

rcssystpv'az

1(11:.::z:)ds

=

E 1(sK t I'F, = :

0JI1:z

K 2.E'1F:

1:1

-

(a)

A

j

12+ 4E nssuyspz

j

+ 4E cssuyspz

:Ar

tK )

-

tA r

bzs

''

2

c(X,) 2

) ds

-

J(Z,) dB'

so we

n

(c+ )2%

5 Stochastic Dirential

Chapte

186

To bound the third term in k lmplies that (

(a),we

(K)

E

:

:

biz't

ds

S

((K)

0

()Sj KT

ontinuity

with the Lischitz ' 4E .

.

.

'

l 'r. . ..... .

(c)

': .

:

.

the

0

)'

' E

(K)

J7

.

fr 1

y To bound the

second

(y(F,)

.

rrpr

,,

j

s

(a),let

term in

'

..

i

Let

.l .. '' . .

=

X

=

70n

IAC

SIIP

-

0 K, K 11

''-

12

XH-

,!

(2.3)implies

that for n k 1

T

-

(

X j tjs

Aa(s) ds

%'B

0

.

1

1

,

,

E

i

1.X-,1 xa0j2 <2

y

-

12+ 1(z)sl2) tjgtzlla

Using the fat that

bi(Z,)) ds

=

As+1(T)

0

!

4 g

ggjk

f(

,)

n

(

.

j

To get started we have to estimate l1(s)I. Ciu erlir inequality f?r squares implies that for norms lc+ la K 21cl2 + 2112 So we have

:

*,

V

bi

ntttrfl

f =1

n

(2.4)

$

su>

.

. . ..

:

bzs) ds

.-t

YO

artd observe that since

142(t)

sup 0 K ST

A i'

)-l

S 4T E (

E! .:

of approxlmatlons.

sequence

Zr&()

2

4E

s

..

.

of

187

lds

.

give

assmption

tar

sup

y

m

jj '

Convergence

ar

f(Z,))2 ds

-

d

l'??()I2 <

sup

i

(

A,r

Approach

(a),(c),and (d)proves (2.3)k

.

and using

Taking supnssz

(b)

:

2

0

Combining inequalities

that the Cauchy-schwarz inequality

observe

. .

fAr -

Section 5.2 It's

Equations

.

jy gg

..y

s

kthiw

b the

j

q

L t'/2 l,(z)s,l su> 0KaK

.q

t

,

1t

and that the rightllkand

side

sup

0K,K1

Io.lzlsl

hms flnite xpected value

we

have

tAT

Mt

(c.(#-,) Jf(,))

=

0

an

d let

for the

i maxtmal ra = inttt : lMtl k n) A r. variance of a stochmstic integral imply that

X

y

z

(11)

SIIP

OKKTAJ',X

z

i ; 45 (W/ara

y

)

-

k

1(f) S Ct +

inequality

'he

E

dBs

.

-

>nd

hy

jgtu)

4s 0

ttk formula

Combining this with

)j2ds

(2.5)

y

yk (a

-

(2.4)and Ar,(t)

f2)

using inductior we have 2@N+1

tn

< Bn-c

+q

V

(n+

1)!

:'

Letting

n

then uqipg

x,

-+

.

'

E

l

LE sup

j

.

nszsc-

Ar

(b) >pd E

E

(d)

d

nssc-

s

16s

<

16d2.Jf2E

.

tBs

.

'

()

-

j 0

Thr

sup

osfstr

(n(Z,))

.

dBs

IXJ'

(2.5) implies the right-hapd

-

7/-11

>

'

..

'

'

4f a litnit. Cieb#sev's .

. .

emsily get the existehc

'

'

%22ns(T)

2rn

side is summable,

ipquality

so

the Borel-cautelli gives

.&n-11

c.lzallz

l(n('i&)

0

(2.5)we

2

Tar

J

:

.

P

(tn(K) -

=1

'

Flom S lk OWS that

have

f Ar

J'l sup =1

w

2

c(g,))

-

d

K 4f

onthpity

( . .' . . .

.

(c()

0

Lipschitr

IK Zs 12ds -

P

ds

Since

Ea 2-n

uniformly on

<

(:x)

(0,71.

sup

lar

-

>

2-n i o.

it follows that with probability

=

1,

0

A'C

-->

a limit

XT

DiFerential Equations

Chaptez 5 Stochatic

188

We begin by showing that

The llmlt is a solutln. occurs uniformly in L2

The last contradiction implies that the desired result follows.

also

convergence

.

al1 0

(2.6)Lemma. For E

Km

< n

-

It's

Approach

E)cP,

E

> 0 is arbitrary,

vt) K (X +

2

k=m+1

,

,

,

=

.

a k (r)1/a

J7

/

so IZI

,

a

but

189

To prove pathwise uniqueness now, let (X Bt Si) and Zt Bt FI) be two solutions of @) Replacing the hi by Ft h v h2 we can suppose h .,V. To prepre for developments in the next section, we will use a more general formulation than we need now.

x,

X,n l2

1x,m

sup

OKaST

S

Section 5.2

=

g'

Ilg112 EZ24l2. If n < x, then it follows from monotonicity expectedvalue and the tripngle ipequality fpr the qorm 11llathat

Proof

Let

=

azld Zt be continuos (2 8) Lemma. Let to Ft with processes adapted and : inftt r(J4 y = Zz 1et > Suppose J. that Bt is z IMlor lZI and Zt satify @)on (0,r(JMj.Then Brownian motion w.r.t. Sg and that X Zg on (0, 'i'i

of

=

.

.

n

p
=m+1

=

s x,,- l '.

,

''<

s Letting

n

To

-+

see

E

.

z

I.&n+1XrI2 < BE

sup

IA'n

lima'v n+1

=

X3co

=

(t)

.

..

If we let

Proof

,

#()

sup

inf (t : Let r tp(r) then < cx) r

=

.

=

Xr

E

.

Let

1aY,n

+

E

.

B ...

#?(t) .B#(t)

+ B

=

j

,

+

s

' y(s) ds >

jo

(2.3)implies

and Z3 are solutions,

zA,r(a)

w() S

BE

j j, llr

XT 12

s

0

-+

so (2.7)holds

./

1F,

E

E

.Er

JJe

..

ds for .

E. .

(

.

..

. .

with .A = 0 and it follows that p '.. . :

.

''''Temporal t''''''''''''

coeciets

InhomoMenelty. .

.. . .

i .

Ip. some t dpend n tin, :.

.

d

.

!

.

#(s) d

+

s

j' e(s) ,

'

'

.

.

.

j vs,

ds

0.

H '

.

t

E

tdf

applic>tions

it is important to

. ''''''' . ' .

allow

' .

the

that is, t6 copsider 't

(p

.,4

s' p

,

y

t

.

d

-4

.

sp

E

The last result implirs th>t twp solutipns must agree until the flrst time thi implies that the two t lt'e ba 11o f ittius R us one them leaves exit at must the same time.) Since solutions are by deflnition continsolutions X,d we in us t hav' '(J4 1 x as R 1 cxj and the uous functions from (0,x) desired result follows.

yl1( k 0

,L

.

% 12ds

-

zrli

-

of

. .

.

X

j

are continuous, w(t)k #(t)J Since # = sihc impossible #('r), but this is =

:

. .

.

(2.3)to get

in

'

and

.

4(z) z +

E

12ds -

vt) % .A + Xcpl. () % ..

4(t) =

.' . . '

Since

.

(z4+ eleft then

=

X*J

0s,sg'

Spppqs

.

.

-

J

The key to the proqf is

Grppwall'j Inequallty. Tlin is continus.

Zt

and

A'J

:::q

-

(lststr

.

.

Prof

g

'

'

.

.

:

,

that XT is a splution, we let $

Uniqueness. ....

:

. .

+: &.' cxn and using Fatou's lemma proves the result for n

-co by (2.6),so Xt

d

.

.

, .

=m+1

v=-=-

=m+1

K BT E

7) (2 . .r

:

g

,

'

'i

=

-

.

'

?

n

T1 ''''!

'

'r4.)

11 57 tu, x-krllllaa,(z')1'2 Il--su-p--lxrk

11 cssu,spylxam xsnI s z -

=

ds

it follows that if

w(t)

(*8

Xz

=

an

+

0

bs, .X,) ds +

In many cmses it is straightforward example, by repeating the argument can show

9

s, X,) dB.

to generalize proofs from abpve

with

@)to (*.). For

some minor modi:cations

one

190

Section 5.3

Chapter 5 Stochutic Df/l-erenfal Equations

(2.9) Theorem. If for a1l fort : r, l(qltf,0)IK Jf, Ic'J(f, z)

-

z, g, and

=

< cxn there

T Ikt

If(f'0)I<

S ffzlz

p)1

att,

j,

,

pI

It,

and

z)

-

bit, p)1 ; ffg'lz

(*z.)hms a strong

then the stohmstic diferential equation uniqueness holds.

ffv' so that

is a constant

then

E

pl

-

(ii)in (3.1)holds

We begin

witi

Proof of

(3.2)

./4

with

.

.

.

zd)

,

.

e-frnwtzl

=

and

=

,

.

.

,

.

zd)

:(

we have

Comparing with (2,2)one sees that the new result mply msstme, Lipshitz continuity locally uniformly in t but neds pew cozditions; lcutt,0)IK Tfg in the tmpprally homogeneouj cas. and jft, 0)j S fz that am automatic

t

-A

w(X t )

e

The dependence of the coecients on t usually does not introduc any new diculties but it does make the statements and proofs uglter. (Here it would and Stroock and Varadhyn be impolite to point to Karatzms and Shreve (1991) result where (1979).) So with the exception of (5.1)below, we need to prove the we will restrict our attention to the temporally for time-dependent coecients, homogeneous case.

5.3. Extensions

X:0 t

formula with

Using It's tftzp,

..

IzI2.

1+

=

w(.X)

-

d

:

+

-B

=

C

.-.Ba

(p

(A- ) ds ,

d

j

gax'fdyi a a +

0

f=1

-Ba

0

j .s

a

g

..

f=1

: g (y

'

.

.

that are only locally In this section we will (a)extend lt's reslt to coecients dimensionat result due SDE for one to Yamada and Lipschitz, (b) prove a results shoW exathples in (2.2)and and. the that to Watanabe, (c)introdce sharp. (3.3) are fairly

d

.1

e-B'

+

-#w(.X,)

tzX,klxal

+

0

+

f=1

implie, onrvsumptiop t ilg; 1

Sllpermar

:y

y yj

.

;

f (3.i)( iti UICYJ:? .

.

,

that the lwst term is

r-.Awl.ml

0 s?

s

cff(.X,)Jds Z

y

.k

k

,

Cxij

(

,

,

nd i.t:rodillil

G'R

b'R

and

Withy

(

(a),' (z) i':= #( j) j bs (z) u j (z) sy jaj c p JR(z) a(z) pi O and 0 for IzIk 2/ (b) 1#,,(p)1 : = K ff'l (c)ly4,,() s Jf/lz p1, I#,(z)

Coecients

. ... .

E ' . .

.

.

'' ' :' E .. . .. . .

'

g . ..

. .

.

,

:

.

.

.

. ..

.

E

'

.. . .

.

'

:

.

..

j

.

=

'

,#(#)1

-

y'

'.

''''

.

In this

*e will xted

subsection . . . .

(3 1) Theorem.

.

Suppose

*

lcu(z) -

(%(p)1

. ..

(i)for f ffnlz

(2.2)in thq followig .

any n < -

. . '..-

*

way.

pI

.

.

ltz)

-

?4(p)1

-k

j

fr'l

.

.

.. .

.

. . . -

:

i/l .

. .

.

.

Let a

=

o''I'

and suppose

do

Eor an explicit constructih

(

xe:lse

k

3.1.

I?i(z) :(p)1 < C'lai

'suppose

f =1

af(z)).

(z) i:z

2R

-

IzI (Sz/Izl) .

R

Izl2)

7/1when

f:r R

IzI%

.&.

Let

K l
.

.

' .

.

(z) costant c an

d

=

0 for lzl k 2S. Show that = 2c1 + l(0)1.

(*n)

is Lipschitz continuous

.d=l

Fix a Brownian motion and let

K f(1+

-.1

-

g

(2zf(z) +

pI

..

.

.

.A < cxl rid a functioh w() o when 1z1, IpI%n d (ii)thre is constant Tltn -xf j a j o k;a j s p ermltjngal of i then that if Xt is a solution e p ) @) @) hms a strong solution and pathwise uniqueness holds. (3.2) Theorem.

-

..

.

m we have

is a local

s

'' ..

a. Locnlly Llpschitz

ja

= local martingale

';'

.

191

the emsier proof.

solution and pathwise t

w(z)

B and

=

Extensions

x

.x

=

+

jo'

u''y

n(x,)

.

' .

,'

be tlte unique strong solution of ds +

/' ,

,r'(x,)

dB.

.

on R? with

192

Capter 5 Stochastic Dilrential

Section 5.3

Equations

Let Ta = inftt : lXf l k n). (2.8)implies that if m < n then A'r t K Tm. From this it follows that we can deflne a process Xr so that

=

XT

ar

=

for t

X1 for

-+

.

=

=

-

,

k e-#t/lTs

s t)

:ji,njfaa #

e

Exnmple

s t) s

eAtg'tzl

=r

(:x) Since Wehave supposed v wsy (x) which proves the desired result. n -+

cxn it

-+

explosion

w(p)

inf p:1p1

follows that #(Ta K t)

-+

0

-+

as C1

p

1) to have p + 1 1/(5 s. 1).

:.=

,j;j

b. Ynmada

and ..

j' .

explosiop

Watanabe .

' ..

.

in d

x)tI2 s

' Note that since our condition concerns compensate for a larger variance. Exnmple 3.1. Consider d c(z) S B (1+ z2) + 2z4.

=

1

(xf)z(A-.f)z

.

. .

c(7/)IK ptl

-

C

with

p1)

p-2(u) du

p(0)

=

0,

a drift topqrd the origin

(z) czi

and suppose

(z) (1+ :.:=

..

.

.. . .

=

in

.

incremstng continuous func0 and for all 6 > 0

x with

1(z)

-

(p)I%

=

6

a sqm,

-z3.

Then separately

3.2. Consider d = 1 and let =

.

(x)

(3.2)holds the

x-1('tz)du

can

conditions

if

Then pathwise uniqueness

=

(x)

holds.

Remark. Not that there is no mention here of existence taken care of by a result in Section 8.4.

Proof The lrst strp is to deflne a sequence lzl. Let c,z t 0 be de:ned by ac 1 and

pn of smooth

=

c(z)

r

less smoothness

0 (ii) there is a strictly ircremsing and concave function p1) where x(0) xtlz 0 and for ll e' > 0 -

.

.

show that when considered The next two exaples and a are cloye to the best possible. b on

Exnmple

.

1

is a strictly (i)therewhere

Suppose that -

.

1

>

:

In one dimension one can get by (3.3) Theorem. tion p with 1J(z)

for J

an# J < txl

'

?

$

=

..

.

.

.

ccus .

E.

(

Jp we get a slutin

(1+ 1zl)-f (2.t)implies there is no (seeExample 6.2 and (6.3))that

xplionrfof

n

=

-

3.3. Suppose b = 0j and o. = when 8 K 1. Later wt will show

in d k 3 ih d K 2

cf)r+1

(1 -

# = 1/45 =

.

I(-Y

=

.

.

th In Words, what we have shwri is that for locally Lipschitb' coecints h.. If Ts 1 txl solution is unique up to T,z the exit time from the ball of radl we get a-solution deqned for a1l time. lf not then the solution reaches inqnity ! in fnite time and * say that an kpllsih tts ccried. explpsiqn Intuitively (3.2)says there is no if, Fhen Itanabe inequality implies aj diagonal

ap

tfyi/d Setting c = 1/p and tn that explodes at time 1/c

Rearranging gives .P(T,z

=

=

-

n

%Ts

=

w(z)k E

193

XL Suppose a' = 0k > 0 so t is strictly increasing. Let Ta inftt : Xg n). While Xt 6 (n 1, n) its velocity bxj ) k nl so T,z Ta-z S n-J. Letting n-J Tx = limTa and summing we hav Tx % EQI < x if 6 > 1. proof since orfly We like the last depends on the msymptotic behavior it xplicitly, which hms the of the oecints. One, can alsp solve the equatipn ot explosion giving exact the time. Let 5$ 1 + kt. tfi'k = %?dt so advnte ctl-r = then yk if we guess (1 -

limnx Ta To complete the proof and X will be a solution of @)for t < Tx + (x7 a.. If Artl z then using the optional to show that T1 now it suces stopping theorem at time t h Ta which is valid since the local supermartingale is bounded before that time, we have (e-A(AT-)w(x,azk))

Extennions

Ja-l

lzl)J

with J > 1

Ja

p-2(u)du

,

=

n

of solutions.

That is

approximations

of

Equations

5 Stochastic Dfsrezlta.l

Capter

194

Let 0 % /rz4u)K 2p-2(u)/n be an d

function

a continuous

#n(u) d

For I in have

(a) we

1

=

#n integrates

of

=

yu(-z),

to 2 ove:

c-l)

(cs,

(d)

JJ#a(u) du, for z

=

= 0 for IzIK an ; 1 for ca < 1zI< ca-l 1 for cn-l s lzI

>

0,

t

,(xa2))

-

.

' . .

yo

dBs +

-(at) j ' w'-(a,)da, -

+

j

jo

l'la I s

martingale

A(t) is a local

so

)j w:(ak)

pn

lzl E (ar,,

-

(a) we

lf2(f )I S

j

.j, :

.

'' .

e lc

-1

.

(z)dz

=

whrr tx7

..

'

. .

for

.

f

is

any

'

.'

6

>

.'

.

.

'

. E'

<(z) 0. k hen ftj

ctptipupo,

0 is H 0.

>

x(#(f)) with g(0) Proof Let g) be the solution of p?(t) e. It follows gt) proof of show from the that J(i) S that g is small when 6 is small To (2.7) note that q is strictly ipcrewsipg and 1et lt be its inverse. Since g(()) z, we ' '(()) H'1/x(p((z)) 1/x() nd hence if c < (z) 1/ g h =

=

.(xa2))a

ds

'c

??

:::':

:p-2(ja,j) ,?z

A

g:

'

()

-

#(c)

gi

1/x(z) dz

=

r

u

ln words the right-hand side gives the ainouht time it takes g to climb from Taking c = t' and b = 6 > 0, tlte assumed divergence of the integral c to ipplies th>t the amoupt of tipe to climb to & approaches (x, as .7 0 and the E::l desird result follows. .

there are stopping times Tm 1 tx

t

a slight improvement of Gronwall's inequality.

.

+ f2(t)

=

with

-.+

0

note that #?(z) = n and is 0 otherwise. so usmg (i)

in

ds

,.t:

=

,

''hsjfcxs)

E Alf A Tm) .12

d(a),

1 lzI)we have

/, (.sIa,l)

j'olx(J(s)) ds

(3.4)tepma. Suppose Jt) increasing for z > 0, an has .

tvfll.ks -w--(a-)t(-'t.l)

(b)

To inish the proof now we prve

'

and using the

(xn

-

> A(t) + .I()

with cn-l)

(-Y,i)) ds

-

-

0

To deal

ttat)

ds

-

-+

= .X2 z driven t)

=

dB' /a(A,)f&(-V)&(.X'.7)J

+/+ 1/'

ds (-Y.7)I

-

-+

:

=

j

.E'I(-Y))

(x) and using Fatpu's lemma, then letting Letting m n monotone convergence theorem (recall pn (z) k 0 and yu(z)

,

folrnula gives

.

of x we

(a)-(d)we have

-

so It's

195

so

'

.

j

concavity

the

' xx(la,l) s x(sIa,I) /' /s

Combining

and it follows that patz) 1 Izlas p 1 x. SuPp ose now that 71t and X2 Ay tskospttitions wlth X1() by the same Brownia inotion, ad 1et At = X1 Xt2.

a,

(ii)and

so using

1,

' g

dz *uzt

w's(z)

and

Iy$(z)Iis

(c(x))

I#a(z)IS

ds

dp

wa(z)

=

observe

.E'I.n()I s

E

xi

=

.

Extensions

Ja-l

Since the upper bound in the defnition uch a function e sts. jae (

t/u(z)

(cw,ca-1) .

Ga

By deqnition

in

with support

Section 5.3

4a(lzl) K 2p-2(Izl)/nwhe .E

..j.

P z (Iy

s

j; vgs

?z

c. Exnmples ..

' .(

' ' .

.

.

.

.

the comWe will now give examples to show that for bounded coecients bination of (2.2)and (3.3)is fairly sharp. In the llrst cmse the counterexamples will have to wait until Section 5.6.

Dilrential

Chapter 5 Stochatic

196

Exnmple that

(z)

Consider

3.4.

Equations

(z) lzl;A 1. (2.2)and (3.3)imply

0 and

=

=

k 1/2

holds for

pathwise uniqueness

Secton 5.4

in d = 1 in d > 1

Jk 1

Example 6.1 will show pathwise

Weak Solutions

197

the Browniau motion the solution at the same time. In signal processing applications one must deal with the noisy signal that one is given, so strong solutions are required. However, if the. aim is to construt diflksion processes then there is nothing wrong with a weak solution. Now if we do not insist on staying on the original space (f),F, P) then we the map LJ (X @),Bt (u?)) use to move from the original space (Q, F) can to (C x C, C x C) where we can use the fltration generated by the cordiaates to1(s), 2(s), s K t. Thus a weak solution is completely speciqed by giving the joint distribution (.X Bt ). Reoecting our interest in constructing the process Xj and sneaking in some notation we will need in a minute, we say that there is uniqueness in distrlbution if whenever (X, B) and X', are solutions of SDE(, tr) with Ab i.e., PX e a1.) PLX' E x1.) XL z then X and X' have the same distribution, tr) for all E C. Here a solution of SbE(, is something that satises @)in the sense deqned in Section 5.2. .and

-+

faits if

uniquenes

(s

1/2

<

in d ju

< j.

Lj

=

o

1 j.

'

,

= 0. If k l then Exnmple 3.5. Consider J(z) = 0, (z) lzl;A 1, and solution: Xs p 0. However, if J < 1 there (2.2) implies that there is a uniq CtP inlt are ot h ers. To flnd one, we guess Xf -'

=

::

.

.8/)

=

=

=

./1.

dX,

F

mke

soto

dX,

Cpsp-k ds

=

(x,) ds we

=

taYl

Cnspn

=

set

(p 1) = -

Cp

=

p

i.., p

(X

i e., C .

J) p-1/(1-O

1/(1

=

=

,

-

lf J < 1 then p > 0 and Jv have created a secoud solption starting at 0. Once construt inflnitely many. Let % 0 d let we hve two olutions we can .' .

'

'.

..

'

' '

..

..

Xa tk a weakening of LlpsThe main reson for interest in (t ) ls th ..- . .. ... .. . . . act als condition continuity att inyrovemnt, is chitz .of However, . . . . ..the . . . . on counterexamples. Show and further sharpens the line between Georems (a) holds when C > 0 and in t.hat

xerciso

.2.

,r

-

..

.

.

.,

.

.

.

(ii) (3.3)

s(z)

=

(b)cflsider

4,(p) where

t

C'1OZt1/')

fO'

c(e-2+ z) oi

'

z

with p > 0 atd g f) = expt-l/p) pp(log(1/p))(p+1)/p. =

S k

d-2

show

''

'' .

.

.

'

.

(4.1) Theorem. tribution.

that g'tt)

i:

(#(tj)

.(,

,

.

If pathwise uniqueness

holds then there is uniqueness

suppose(x1,s1) aud (xi, sc) are two solutions wit x x z. proof remarked above the solutions are spciqed by giving the joint distributiqns (A =

,

#)

.

) Since CC c) is a complete

seprabl

,

conditionaldistributions Qi, in Durrett (1995)) so that

.)

Pxi

deflned for .,1.1#i)

e

=

fhetri space, we can flrid regular : C and e c (seeSection 4.1 .4

Qf(f,.)

Let Po be the memsure on C, C) for the standard a measure k on (C3, C3) by g'(.c x

.1

x .Aa)

=

jgo

ciii

(.d

Brownian motion

.

Intuitively, a strong solution corresponds to solking the SDE fo: a given Brow. weak solution we are allowed to constrttct producing nian motion, while in a

in dis-

Remark. Pathwzse uniqueness qlso implies that every solutiop i.s a stropg lution. and the conlpsion However ij pot that proof is more complicated so important for our purposes so we refer the reader to Revuz and Yor (199t), p. 340-341.

(Xi

c-2

5.4. Weak Solutlons .

Exnmple 4.1. Since sgntzlz 1, any solution to the quation in xafn/le 2.1 is a local martingale with (X) H f. So Lvy's theorem, (4.1)in Chapter ( 3, implies that Xt is a Brownian rriotion. The last result asserts that uniqueness in distribution holds, but (2.1)sllows that pathwse uniqueness fails. The shows that the pther next result, also due to Yamada and Watanabe (1971), implication is trne. =

'''''''''.

d.&@c)Q1(tJc,

.1)Qa(tJc,.A2)

and

define

If we let y/@n,

2)

1,

and a diagonal matrix A with diagonal entries c UVKU. a is invertible if and only if a > 0.

then it is clear that

tJf(t)

=

.!.

(yl

-

,

,

y0)

(x2 Blj

aud

(y7jy0)

-

,

kE

a

.

.

a

.

k 0

so

.

=

:.::z

cvT(z)

jz 1

2

.y

#=

x

=

=

= Take C or .A1 = C respectively in the deqnition. Thus, we have two solutions of the equation driven by the same Brownian motion. Pathwise uniquenerss implies F 1 '.F2 w ith p robability one. (i)follows since

X-1

2

and recall from Section 5.1 thaj

(:1#12 and (4.3) Theorem. If oTazjo k llc(z)al1c(p)ll S Clz c1/2(y)jj (C72al/2)jz pj for z, y. then Ilc1/2(z) s

cf(A-,) ds

=

We say that X is a solution to the solves MP(, c), if for each f

-

ar tcal martingales. :

bixs,

Xgixji

and

ds

:

pk'

. :

t -

0

0

(

.. .

bl

(

'

j

.

b

m foi

..

nd

or

,

(xa)ds .

.

f

.

.

..

.'

'

0

' y'

'

To

' . .

.

.

To be precise in the desnition above we need to specify a flltyation .J for the nlf iptersted lpcal rrtaxtlnkales. ln posing the martiugale ptoblem we a the undetlyig geherality of without that loss the proces Xt so we n assume = and generted by Xt the with flltratitm is the Xt@) tlf one spie is C, C) o'1$ and foi above bmsic data the b are In t he formulation a t fflarobtaiping with of problem tr fropz c. the problem. This presents us tingale pypiti of intoduing solve by thi problem we begi Tt? sqme !: jizlce i j j y aij i: at tht symmtric, if l, is, is z a;i (# X )t (aY X )f, a '

tn

'

'

/

.

see why we need a

lzl. In this

.

.

..

.

:

E

.. .

.

.

.

.

:.

.

'

.

4: gt ar ipsc j ( r

case

J(z)

e c2 to

=

'

2

clpl

pj for all z, y.

-

-

. . . .

.afctzlpl

then llc1/2(z) c1/2(p)j1 < d(2C')1/2jz

. . . . . .

.

c./(A,) ds

=

g'

Ip

mp.x lKtKd

j

. .

.

to

.

suppose more smoothness

Suppose aij E (72 ad

!

equiklent E

When a can degenerate we need Contilmtms Sqlare yoot. (4.4) Theorem.

(

. . ... . ..

.

.

:

E

it

The second conditioh i, of coure, (.X', Xiit

-

-

martlngal and j,

simplyX

Xt

'

p1 for all z,p

-

-

(.X'..;'Jf

that

E

Uwv-h, UVhU, so if we want c = co'r we can take tr (4.2 tells us a entries matrix with where A is the diagonal (4.2)tells us how to flnd usual algorithms of The for fnding U are such that root the square a given c. J(z) c(z) will measurable resulting apply them then the if we square xoot to a c(z) sophistiation smooth with measurable. start takes also. be lt to more and construct a smooth l(z). As in the cmse of (4.2),we will content ourselves to just state the result. For detailed proofs of (4.3)and (4.4)see Stroock and Varadhan (1979), Section 5.2.

.42

Let c(z)

k

t

199

=

CX1 l1)

.,

Weak Solutions

Section 5.4

198 Capter 5 Stochastic DiEerential Equations

alz

upschitz

get a

1

lzl/2 is Lipschitz continuous

d = 1 and a(z) = only if k 2. and >t 0 if

consider

'.

,

Returning to probability theory, our flrst step is to make the connection between solutions to the martingale problem and solutions of the SDE.

.

=

E

E

,

o

zia . z g .

i

J' ;

So c is nonneqative useful information.

z i aj k c.J, k z J

-

.

-

-

zi tzk k

--

,

k

i l 2-l

defnite

and linear algebra

(4.2) Lemma. For any symmetric orthogonal matrix U (i.e.,its an

,

square

enlarge-

.

.

,

,

(4.5.)Theorem. Let X be a solution to MP(, c) and c a measurable root of c. Then there is a Brownian motion Bt possibly defned on anJ). ment of the original probability space, so that (X, B4 solves SDE(,

E

> 0

E

Proof lf and let

o.

is invertible at

X this is easy. Let 'Ftf

each

-

i

provides t!q with the following

(a)

Bf

=

)'-) 0 J

From the last two deflnitions and the nonnegative deflnite matzix c, we can llnd vectors of length 1) rows are perpendicular

t

oj-

t

1

0

xi:

mssociative

=

J$

'n

-

=

jt() i (x ) ds

law it is immediate that '1

,) c(X dB ,

-

(x,) dni

'f

(b)

=

X,

-

A-a

bvn4ds

-

0

,

.5j

are local martingales with (.&, Bijt = in Chapter 3 implies that B is a Brownian motion. To deal with the genral case in one dimension, we let

so @)holds. The

z'

jt so

=

1

(todeflne thi.

may

-

f =

'

we

,

o o.x

,

)

J dpl/

dy + *

'

of qo(4.5) shows that there is a 1-1 correspondence between distribttipns ..,,.. . . problmj Whiclk b# definitipn lutions of the SDE and solutions of the martigale ar distributiohs on (C, C). Thus there is uniqueness i distribution for th SD2 solutton. if and tmly it the maa-tlngau problezfz hs a unique our fal topic is an important remson to be interested in uniqueness. . .

..

'

.r

'

Our result for one dimension implies that the Z,i may be realized s stochstic integrals with respect tp independent Brownian motiops. Tracing b>ck through the dennitions gives thedesire representation. Fo) mre details se Ikvcta and W>tanabe (1981), p. 89-91, Revuz and Yof (1991), p. 190-191, or Karatzl ind (ZI Shreve (1991), p. 315-317. ;'

=

B

201

.;,'

I., let I/'/a be an independept Brownian motion need fo eplarge the probability space) and let

let Ja

Exercise 4.1

#0

1 if J(X,) 0 otherwise

(

=

Wak Solutions

Section 5.4

200 Chapter 5 stochaticDiEerential Equations

'

() t

. Bt The two integrals are well deqned since thelr . vrinc. processe ae . K martingale with is Chapter implies that Bt 8 in 3 loal so a is a (4.1) (#) moti. Biwhin To extract the SDE now, we observe that the associative 1W ad th fact H 0 imply that Jsxjj .

. .

.

''' ..

:

''

. .

,,k.

.

:

.

'

.

.

. .

.

' ''

. .

are locally bounded functions and MP(, c) Markov property holds.

(4.6) Theorem. Suppose a and has a unique solution. Then the

strong

=

E

't

.1

(r(.X',) dB, 0

f, dF,

=

= In view of (b)the proof tltis Fe note that

>;

be

will

t

X

jo

tfyk

=

E

r.t

L JF,

-

when

complte

2

Js

'o

-

,z,2

jo

=

zt

(zf ,

zJ)

't

kt

H

0. To do

jo

ax,

) ds

=

Urxs)

=

1 if

=

orthogonal

matrices

dF,

=

o

01. 1.&)

con-

Exl'jv

=

.

() To begin to turn the lmqt paragraph into a proof, fix a starting poit will follow thq stand>4d complicated simplify rather argument, this To we za. practice (seeStroock and Varadhan (1979), p. 145-146, Karatzs ad Shrk and (1991), p. 321-322, or Rogers Williams (1987), p. 162-163) of giving the details only for the case in which the coecients are bounded and hence

.#'tii

'. '' '

'''''

=

x

't

and

0

0

xi-x.i

aij

-

0

we t

l

ds

=

0

have

htnixs) ds

. .

.

..

(a'c) da

are martingales. We have added an extra index in the frst cmse so we can refer to al1 the martingales at once by saying ttthe Mi be a yegular condiConsider a bounded stopping time T and let Q(Y, FT. given 0) distribption (.X+f want fo# We to claifit that t k tional ''

otherwise,

.r

(x,) ds

-

.)

j

.

t =

.

0 =

E (F

Mfo

'

0

.7)

.

(I(y)a

,z,2

Ui,kxsjak,txsthljx'a,

=

0

hj

JJJa dF,

we show

To h>ndle higher dimensions we 1et J(z) be the structed in (4.2)and 1et

lntroducing Kronecker's

(t

t

E

a'

=+

0

.

Propf For each z E Rd let Pr be the probabiltty peasure on (C, C) that gives 4h distribution of th uniquewsolytion. starting from z. If you talk fmst distributioh of the proof is easy. If T is a stoppmg time then the conditional (.X+,, s k 0) given Fr is a solution of the martinjale problem starting from Xr and so by uniquness it mudt be Pxv). Thus, lf we 1et 01, be the rndom R we shift deined ln Section 1.3, then for any bounded neasurable F : C property the Markov strong ave

,

(4.8) Lemma. For .P.0 a.e.

f.J,

the

all

i.i M.l.

-

o/.l r

=uij

t

o

ov

Capter

202

5 Stochastic DiRrential

Equations

Section 5.5

Change

of

Meuure

203

.).

are martingales under Q@, To prove thi we consider rational s < t and B stbppig theorem implie that if .A 6 Fr thh .

(((.6 i

Ezz

-

Mii 1s) ) ,

'

e

Since (F): < .Mt, (3.7)in Chapter 3 implies that invoking(12.3)in Chapter 2 we have deqned Q.

'

Ai

01' 1x)

j

o

''

that tlie ptitmal

and note

.'&

=

i' l

0

=

J

tef

1

-

'

'

.

.

(

'.

'

'

.

.

'

..

. .

:

.

'

...

.'

. .

'

E

.

..

: . .

.

.

.q .

.

(1

o

g. .

So by the fovmula fp: the

ch ange

o f Measure

)t

since

'

cls, z)

j .zl.t

Jcp(#,)ds, let

-

cts,

)

uj

5$ .which exists

(z) (s,z)

and

let

o

x,)

.

d

t -

aa

-

t

cfls,

=

0 t

=

uner q.

ij t

csacs,

X4) ds

=

0

0

Zc-lts,

z) by our assumption that I c l % M. Letting tions of Q nd P to & we deine our Itadon-Nikodym

ts,

.

d% dPt

=

a

=

exp

yk

-

1 .j(J')t

d (a, i ),

A'jl ds

% ML

Pt be th restrir deiivttv t be

Qt ad

.

'.

'

'

'

'

'

.

.

=

0

'

ce, jtsj Xa) ds

f

.x-

j ) (s , X a ) d s

=

()

0

t

E

=

X1

-

0

1

xJ),

pj (x,) +

bjs,

.;Ll ds

but the other half is

easy.

We note

j

aijxs)

=

and (12.6)in Chapter 2 implies that the

.;L) d(X, X),

Xslcjs,

1

J (s,X,) ds

(.x

since

(F)t

,.#)a

(

T

is a local martingale/o. This is half of the desired conlusion that under P

.

cts,

=

a

:

d(-Y'

(c a)j (sk.Xal ds

t

Tt

E

vn

integrals

Using Gixsanov's foxmula now, it follows that

,

y

0

0

=

ts,

s

=

.;@)

Let

of stochastic

'

z). Ij. follows that

c-1(z)(s,

=

'heorem.

Piof

0

=

(5.1) Suppose Xt is a solption of MPI#,), which, for ioncfeteness withut of and loss generality, we suppose is deflned on the ianonical prbability td. Suppose c-1(z) space (C, C,P) with & the fltration generated by lTa-lts, z)l z) and memsurable and exists for-all z, that is has % M. We probability e4uivalent to P, so that under Q, measure Q locall# can defne a Xt is a solution to MP(# + c). u'kj

*

()

cea cts,xa)

=

t

E

i cea ct s , Xs ) dX-,

=

't

(a,.X'

Our next step is to use Girsanov's formula from Section 2.12 to solve martingale problems or zore precisely to change the drift in a' existin: soltiok T accordate Example 5.2 below we must consider the cmse in which tv a'ddd t: t drift b depends on time ms well ms space.

.

covyxiauce

'' .

'

5.5.

j

ce, dFa

=

.q.,

.

aud

)

.#,

PTOOf.

is a martingale,

To apply Girsanov's formula ((12.4) from Chapter 2), we have to compute a-1 and dte X),. It's formula th asspiative law imply that 0 a '

Letting B ruh through a counta le collection that i closed under intersectiop it fotlows that .J'., the ekyected vale of tM!fl r.- .#.f,U)t..s zd genertds whlt inflies 0, ul4enss Vsing now gives (4.t) and tofnflt.kl.. tr (4.8). i

tn

ds

cov>riance of Xi and Xi is the same '

,

n

Exerclsp 5.1. To check yom understanding of the formulms, consider the cape in Fhich b dpeq nqt depend on s, start with the Q cozstruted abpye, change to a measure R to remove the added drift and shop that dRtldot = l/m so R = P. ,

' .

'

.

Exnmple 5.1. Suppose Xt is a solution of MP(0, f ), i.e., a standard Brownian motion arzd consider the special situation in which (z) = VJ(z). In this cmse

204 Capter 5 Stochastic Dirential recallingthe deflnition UX

) we

have

,1

.

K

j

=

(5.1)and

of

X in the proof

of

'

. TUCXS)

X'

.

=

Section 5.5

Equations

V(Xz)

l' ..

. '.

U'(X)

-

dPt

exp

l .

jl

k

.

de ldp, 2

.

AV(X,) ds

(g(m) )/' -

hux-,

+

n

In the special case of the Ornstein-uhlenbeck is a positive real number, we have J(z) expressionabpve pimpliqes to

proess,

-a1l2

=

d% dfl = In the trivial and Luzj

=

t 2

I

-al-'

exp

dPt =

It is comforting sa tisfy

to note that

Qt(.;% =

p)

should

=

be since under

(5.1) deals with

da t

(z)

-

p

=

1

.

p Xz

exp

.

p A .

-

.

when

A

=

exp

Jz y .

-

t:e

the

and

-2da,

=

-

-

p z .

Q, Xt has the

-

same

:r

=

d

we

.

have Uz)

p

=

+

between solutions to MP(,

c) and

MP(0,

c).

.z

Remark. Combining (5.3)with (3.1)and (4.1)w see that if in addition to the cT where o' is locally Lipschitz, then MP(, c) conditions in (5.3)we have c upique solution. using hms a By (3.3)instead of (3.1)in the prvious sentehce result in one dimension. we can gt a better

=

=

y4

:::i

jjz/2j)

is

Proof Let Z and a be as in the proof of (5.1)and 1et Ts inft : IARl > n). From the proof of (5.1)it follows that cttt A Ta) is a mrtingale. So if we 1et cvt A T,z) then under hn be the restriction of P to hhn and let dolldln c) solution of MP(, a up to til'ne Ta. Qn, X3 is(txdftic Th growth odition implies that solutiops of MP(, c) do not ekp 1o d We will po* show that ih the abqence f ekplsios aj is martingle. First, to show at is integrble, we nte that Fatu's lemza imlies

;yy

ms with

1-1

Bt

-#

'

#@.

.

uniqueness. colresp'

Eat

%liminf f'tatt

A Tr,))

n'-x

1

=

, ,

where E denotes expected value with respect to P. Next we note that

t

.

'E

1

1. Suppose P and 15 are not locally equivalent measures. equivalent to 1l, Q1and Q2are not locally equivalent locally is Qi be equal.

,

(a)

ndep

,

CASE

Izl2)

.z4(1

-

.

Let P and P2 be solutions of MP(7 c). By chage of measue * c producesolutions Q yd Qz of MP(# + c). We claim that if .F # Pk theh Q # Q2.To prove thls we consider tw cmses: Proof

K

=

distribution

nder the hypotheses of (5.1)ther (5 2) Theorem. between solutions of MP(#, (I) and MP(# + b, tz).

cf(z)).

z''L.

Then there is a 1-1 correpondence

at

our next result

of solutions,

+

i =1

2 1p1

//

'eorem.

jzda

:azjxs

I#tI2 Ptvj

-:

-

(5.1)

Suppose k is a solution of MP(, c) constructed pn (C, ). c-1(z) exists, Tc-1(z) Suppose that is locally bounded and memsurable, and the quadratic groFth copdition frorp (3.2)hplds, That is,

z, the one dimensional density functions

(2JrI) -a/2 exptr-lp

existence

where

-2az,

=

(5.1).

(zztzl

VV(z)

=

o

lt will

(5.3)

p

j

= as it

I+

drift, i.e.,

cmse of constant 0 so

dQf

2

+ al-'

ds)

AJ(z)

q # q,.

The nekt can considerably relax the boupdedness condition in cover many examples. lf it does nt cver yotlrs y te tat tlte min of ingredients of the proof are that the conditions hold locally, and we c) of solutions MP( 1 do ot explod. knw that l'estl

Ivv(x,)la (z)

and

Ohe

,

u(xa)

-

205

2. Suppose .P1 and PL are locally equivalent. Izl the definition of tn, (F)f and and X are independent of Jl by (12.6) (12.7)in Chapter 2, so doldp

Nikodym derivative: -

Measure

CAss

to

Thus? we can get rid of the stochmstic integral in the deqnition of the Radon-

dQt

of

=

.' -

formula

applying lt's

Change

Then since

!

(b)

f'lm

-

aavkl

=

ftla

-

a(Ta)l;Ta

S E (a(Ta); Ta S

The absence of explosions implies that

%t)

@)+ Eo

as n

-+

fltyl; Ta

-

(:x)

and cannot

E (a(Tn); Tn K

)

=

Qn(Tn K @)

-+

0

>

t)

This and the fact that fta;

(d) sn e.g.,

x.

-+

a(t A Ta) '

.

..

Tk

t)

>

'

1

(b),(c),>nd (d)imply

(5.6)in Chapter 4 of Duirett

and from this that a', 0 dameas that f (5.1)ad

t:vtt

.

)

Ta K

A Tn)

(5.2).

1

-+

Let Bt be a Brownian motion startig = 0, and let :7 > 0. Then

.

El

(

'

...

.

Our lmst chore in this sectitm is t complete the proof 1 by showing

(5 4) Theorem.

It folloFs (see

.

The rest of the propf is the

a martingale.

:

'

'

(2.10)i

f

at

Chapter contin-

0, 1et g be a

In the last section we lqarned that we could altr the b in the SDE by changinj the measure. However, we could not alter o' with this technique because the quadratic variation is not alected by absolutely pntinuous chapge of measure. In this section we will introduce a new technique, change of time, that will allow To prepare for devrloprprnts in Section 6.1 us tp altr the difiksipp coeient. will allow torthe possibility thatour processes are deqned on a iandom tize we interval.

#(t)1 < e

-

and

i,

.'

. .'

.

. .

'

. . . . . '

.

. . .

.'

''

.

' .'

. '. .

are local martingales

E/2

=

d11 =

exp

dBs

,

j.

() :

1,l '

.

=

'

'

5

'

.

.

'

-#(t)

'

-

,

.

.

e(c) / -

ddop: .

lc dn

c

j

2

( j dQt dy

=

-

exp

()

gxs

= g

) ds <

cxa for all

t

ds)

and suppose

(

<

-

s

j

eu-t

that

.

=

=

.X'(v,) for

,

respect

to Pt

f

j

..u

0

(.'L) da = yj f

= Changing variables t

l,I a

ds(

'i'

u

0 qt

f(yr)/g(Fr) dr

-

f

.

yu gives

=

-

'/v

0

bi (yr )/#(yr)dr

=':

Xi

f(#,)

-

v

0 .

since the are stopping times, the rkht-hand and we have checked one of the two conditions 'yu

change

h (y..) ds

-u

'it

and

step

=

11

Lj (/'l,1a J0

wit

t

eah

(Xa ) gs

aij

-

Let g be akposttive function

dzl

expectation

2, dBs

E exp

j

q

=

.p(c)1'2

t

dPt

(. That is, for

F((n), then in the second We begin by noting.that Xt A'a = A'v(c) 5$., to get variablesr = n dr gxjjds, Xi

/ Lddopttt

To complete the proof now we 1et E denote observe that if l,1 S M for s K t then

j XLXL

and

Proof

1/2

2

s

<

.

ls a stanard Brownian potion. then it follows from (5.1)that pnder Q,Bt <; in Chapter t imptles that Le t G (1./ #(s)l : for s < J! Exercise 2.11 inequalhy that G(G) > 0. It followq from the Cauchy-schwarz .

fpr t

= inftt : tn > s r t k (J nd 1et F Deqn the invers of a. by < = ct Then K is solution of MP(/#, c/#) for s < 3.

a dB'

o '.

c)

'ya

j

1 -

Ml,

t

=

j

(0,().

on

Proo? We can qnd a dl'function lt with (0) 0 and l/t(s)= y(#)l < for s % @,so we ca suppyse without loss of generalijy that g E C3 and let b. = #(s). If we let X ji b, df, and deflne Qt by

dQt

bi(A%) ds

-

()

.

o?

be a qolutiop (

'' j

0

>

tt x

.

('

sup I.Dt 0KtK1

:.

.

of Time

5.6. Change

(61) Theorem.

uous function with g(0)

P

and hence has exthat a = g'sj is a (a

'

'

(a Iaas )

JJ

-

.

cu in L

-+

that (1995))

% s K t is

..

.

that

A zk) c:h E

at

.

f(a(Ta);

-

(JJ

Change of Tlrzie 20;

dsj 24 dB. since exp (1/2) I2,I? is a martingale pected value 1. For this stqp it is important to remember non-random vector.

implies

is ar martingale . =

Sction 5.6

Equations

5 Stochastic DiRrential

Capter

206

' . ..

E

ds l

. .

is a local martingale on (0,) to be a solution of MP(/#, algj. side

Chapter 5 Stochastic Dierential

210

('

Then MP(0, c) is well posed, is no explosion.

Equations

i.e., uniqueness

in distribution holds and there

Proof We start with Xt = Bt a Brownian motion, which take g(z) = (z)-1. Silce gBtj % Gl for K Ts inftt gBj ds < (:x) for any t. On the other hand 0 ,

=

:

Jf

' ..

''

* :(f,)

./-

ds k

fR-1l(t: lftl S 1Jl

lfl

(64) The-emk

Conider

*...

.

.

.

.

(i) 0 < Ea %ay) % CR (ii) (z) is locally (iii) z (z)+ c(z) <

E

izflesion d .

<

(x7

when

Ebounded

.4(1

.

Then MP(,

c)

is

well posed.

can construct

+ z2)

One Dimpnsiona

solutions

1 and suppose . .

IplK R

.

.

:

k

blguslons

of E1

6.1. Constructloh

of MP(, c) in one

ln this section We will take ai =

.

.

cxn

=

since Brownian motion is recprrent, so applying (6.1)we have a solution MP(0, c). Uniquene follows from (6.2)so the pof is complete. Combining (6.3)with (5.2)we dimension for very general b and c.

6

. MP(0, .J) and > J?Jwe have

solvs

E

j

''

..

.

approach

to constructing

solptions

pf

'. .. '

.

''''

j.

dXt

(Xz) dt

F.=

+'J(.m)

.'

dBt

that is special to one dimnsion, bt that will allow us to obtain a very detailed understanding of this cse. To obtain results with a minimum of fuss and in a generality that encompmsses most applications we will assume: E

(1D) b and c'

are continuus

>pd

c(z)

J2(z)

=

S 0 fpr >11z

The purpose of this section is to answr solutio exist whn holds? Is it' niqe? inay b utlihd as folW:

the follpwing questions:

(1D)E

.

. .

'

g. .

.

. .

.

y

y. y .

.

.

,

'

:

.

.

lsa polqtion

(i) We deflne a functioz p so thqt if X) is @.local martingal X + w(aV) on (0,). h (#)z+ ds s (ii)oi by time changing a Brownian motion.

Jc tkal

'.4

.

(iii) We deqne Xt

w-1(K) and

=

. check

.

j.. .

. ..

of the

(

. .

j. . .

DE

. ,

.

. .

on

. .

.

. .

.

(0,) then

(

y

ktiuct t t '

Does a Or approach

Does it kpbde?

'

tg

'

joloh

that Xt is > soltitio .

of

P(0, ltj

of the SDE.

E

To begin to carry out our plan, suppose that Xt is a solution of MP(, for t < f. If J e C2 It's fimula implies thatfor t < ( 1

(1.1) ' .

. .

(

J(Xi) .zziJ(.X)

+

j9

J'(.X',) dka

= local

# pj

mart. +

j

.1

2

ja y''(.;L)d(.X')a

z,J(x,)

ts

(I)

Captez

212

Section 6.1

One Dimensional Df/fu:ons

6

where

T(z)

(1.2) .

1

,,

: c(z)T (z)+

=

,

bzjl

nd c(z) = (8 - (z). From thi we see that fXt4 is a local martinlale on 0 and noticing (1D) implles c(z) if and only if J(z) > 0. Setting L1 gives a Erst order dilerential equation for f'

(0,() >

=

fj' Solving this equation

I

J

sexp

tht

J(z)

.,4.

+

j'a

S

dz4

(./* o (/'

.

dz)

c(z)

n

Any of thes functios call be called the natural 0 and B = 1 to get .

(1.4)

-

scale.

(6.1)in

<

X

yz/

Using the first equality (1.3),we have

:

i

y

:

=

ct.xal

ds

(F,) ds

=

0

0

that

we see

w

=

14J'4w)

t k ()

'k

=

is a solution of MP(0,

,(y,(z))

V

--L

(z)

zz

z (#(z))#(z)

'(4(z)))a/

(

and p

j

/(p)

<

#(K)

/('i1)

-

.

For the second term =

-

a(#(z))

w-1)

S

that

we note

1

,,

combintng

'j'# (p)(p)

(1.1),and (1.5)imply

formula,

#'(F,) d'

j

=

.(/(w-'(p)))2c(w-1(p))

tl So if

an

we

1et n

amount

=

>

=

infts

:

(F),

of time (F)4.

>

t) then g

=

0 is continpolw

5'r, is a Brownian motion run for

follows frrri

2(#(z))

1

,(#(:)))aq

j

E

(recall#

which

(z) (w #?? show # e C2 so It's

These calculations

f9r f

-(2(g)/c(p))p'(p),

=

=

j.

:

+ '

'

.

i

l .

that

i

j

the

#''(F,)(K) ds .

. .. . '

ormulasfor

.(?.

an

gives

(#(p))

To deal with the flrst term observe that I is a solution of MP(0, It) up to time /z(F,) dBs. so by (4.5)in Chapter 5 there is a Brownian motion B' with dya J(4(g)), Since 4/(p) &(p) letting Xt ./(K) we have for t < =

where

1/,

1

=

-

,

p'X,)2

and

(

<

=

gq

g

(1.5)

tr > s or

=

'

We will usually take

d,

/(x,)2d(A4a

=

inf (t :

.q

.

ToEftnd

(F),

Chapter 5,

#t (z)

it wijl be conyepient to replqce the 0's at the lower Note tat our mssumptipp (1D) lmpliesp is /2 so (1.1)is justifed. martingale, results in Section 3.4 imply that it is local Since /(.;% a '.K ) ! motion. of the time chang function we note change Brownian time is a that (1.1)and #h4fo4mula for th varianc of a stochastic integral imply

pituatlons point. b y oter timits some of formula in lt's our use

Howevr, 11.4 som

/ (', r)J,

:

K x.

r

J'I''

inftt

=

'

a(z)

n

#(W'a) ds for t

=

dy

ab) y-exp Lju-lbzb n

wtzj

(

=

E

=

m

a Brownian motion,

,

-25(Z)

exp

vbe

#(z)

-x

'#

.

c(z)

.a

(a,#)

) for q To deine Xg now, we 1et be the inyerse of p nd 1et Xt /(X). To check that X solves MP(j c) until it exits from (&,#) at time we diferentiate w(#(z)) = z and rearrange, then diferentiate again to get

Using

r(,) =

Letting n

-2''(z)

andit fbllws

(a,

:

we flnd, -

w-1(1$).

't

-2

=

0

213

above: To construct solutions of MP(, d) we will reverse the calculations will use a time change of Brownin motion to construct a solvs which X we and then 1et X%= MP(0, With Exampls 1.6 and 1.7 of Chapter 5 in mind we generalize our set-up to allow the coecients b and tr to be defned with interval < < > Since t 0 0 for a11 z, the image #. on an open r) unde with of is < interval f, ; 0< an open #) K

)

(z)

(

Construction

=

=

x'V X c -

=

j ,(x,) /' (x,) ds, +

,

ds

214

Section 6.2 Feller's Tet

Capter 6 One Dimensional Disions

The last computation shows that if 5$ is a solution of MP(0, ) then Xt #(X) is a solution of MP(, c), while (1.1)shWs that if X3 is a solution of MP@, c) then X = pxjj i a solutiim of MP(0, #). This etablishes there is 1-1 correspondence between olutions of MP(0, #) and MP(, c)z Using (6.2)of Chapter 5 now, we see that =

E

.

(1.6) Theorem. Conside: C, Cj inf(.X'() / (a,#)J. Under (1D), on (0,q.,p)). with

result

Using the uniqueness

let Xg () vt Let (: < # and ktpj uniqueness in distribution holds for MP(,

and

=

(4.6)of Chapter

.

each z E

j

=

'q.,#)

Tn)

<

a dz mlzl

' (a2)f'i (.3) Pzl

.

.

0 for som

>

.

.

>

e (0,#)

z

.

w(z))< x 0 for all z e (0,#)

qep)

Tc)

<

-

condition The key to the pioof of otzi suient foi no kplqsions givp sucientl# of in (7.1) Chapter th fact that 5 if .A is large was (3.2) and

': '

.

..-/t

'

St

,

.

-x.

=

c)

,

(a,7).

1 foy all z E

=

=

r oof

(a,#) let

x)

=

tx then the frst integrand is H co and the integral is Remarks. lf ppj x. Similarly the second integral is x if w(a) The statement in (a)means that the following are equivalent

(a1) Prl

5 we get

.&('r(.,p)

are infnite then

=

Pc be the 1aw of Xj t < rom where L is the cemetery state of Section for k r,p) pxoess hms the strong Markov property:

(1.7) Theorem. For (1.6). lf we set Xg 1.3, then the resulting

(c) If both integrb

215

'

.

'

' '

2

..

(1+ l.X'fI )e

=

( . .

'

'

'

js a su/ermartigte

.

.

,

To get the optimal result on explosions we have to replce 1+ lzl7by a function that is tailor-made foi the procs. A naturat choie that ives p nothing is a flmd i On SO (j that # k 'g'

6.2. Feller's

Test

E

( .. '

......

I n the previous to MP(, c) on gives necessary Let

we showed that under (1D) there were unique solutions The next result qa,#)) where q.,#) = inf : X = conditions (x7 a.s. and sucient for no explosions, i.e.,

/ (&,#)J.

(

(0,

'r(o,m

for p E (a,#) n lim Tp

=

=

G

E .

=

c-.J()

(2.1)we

benouement. Once we have f the conclusipn of the >rgpment is motivations we begi witlt the end. Lt expla in our .

J(')

-y.

Ewe

(2.1) Feller's test. Let mtzl = 1/(w'(z)c(z)). '

e(z)

T0) is positive for

(b) /,:I:rL < :1) is

be the natural

scale

-

J.

:

. .

positive for some 0

-

(a1l)z E (0,#) if and

j /' dz mtzl epb w(z))< .

deqned by (1.4) and 1et '

some

dzmtzl

(all)z

. E

cx)

and (2.6)

ttk

' '

.

clulatious

on y if

the

the

nd of

.

.

.. .

:r;

.

.

inft

=

f'r

=

:

yt

lim f;

' ..

=

yj

and

' . : .

(

'

.

that

only if

<

only

< x.

f if f)

m.

(

(2.1)follow emsily. g

.

..

.

.

.

..

..

.

for y 6 (f r) f'z limf; ,

=

p$r (z,r) llz/$ l?r =

'

easy

ltmJ(p) p'yr

=

the conclusions of

stablished

are

.

:i

ftijofwittsiktk

(a)is finite if apd in (b)is flnite if and

(2.2b) The integral

:

.

at

J(r)

..

.

Uce these :

(a, 0) if and

7

(

1imJ(g) ptz

=

'' ..

.. .

(2.2a) The integral in

if

only

cxn

.

(e(z) e(a)) < -

is a local martingale

p#

pptxd 0 e (, #). have With6t loss of generality If you are confronted by an (a,#) that does not have this property pick yotzi Changing variables favorite y in the interval and tr>nslate the system by when a1l of the 0's are replaced by y's. holds that in the integrals see (2.1)

.'

'

.

To flnd such a g it is convenient to look at things on the natural scale, i.e., let K w(.;V). Let f = limxt. w(z) and 1et r limwjp w(z). We will ;nd a C2 function Jx) so that is decreasing on (', 0), f is mcremsing on (0,r) apd

=

=

<

. .

.f

=

(a) #.(Tp

jy ' . .

.

is a loc>l pprtingale

=

p) Tp inf (t : XL lim Tw and T. plG

ln stating

e-fg(.X)

section

ptz

Let .

so to

One Dimensional Dihsions

Chapter 6

216

Section 6.2 Feller's Test

f) = J(r) = x. Let ca < 0 < bv be chosen so that = and let Qa A Z,t. Since e-'J(F,) is bounded before J(cs) Jbuj n optional stopplng apply theorem at that tirre to conclude the A time we can n), that if y e (cn, then

(c)

Proof of =

Suppose

so we want

'ra

=

'ra

etytp)

Pp (Tn < f ) f Letting Proof If

we conclude

(x)

-,

frj

Ppttjla

false then

(1)

s

esfyj fbn)

=

.G(c-ray('i)) 1 + Jlrl.Eiy

' < '''

Rear#atljing gives

k

X Q< =

(:xn

j-

Tba

; ft

w

is impossible,

<

E # (c-f'ba ; fa < :&) a contradiction,

Proof of

(a2)implirp

(b)

t Ey

Jtf'r implis (a3).(a8) uhless 0

<

is identical to tht of

TCV)

1

ytg;

letting n

and

we

This show, .

,

a (vhih i.

=

(a).

-

=

jo

l-c-'J('F,)

+ local mart.

g

.

.

=

+

1

c-'J''(',)(',))ds

a=?

.. . .y.

..

Proof

in

liz)

,

fnz) E*r,=o

.f(z)

=

=

=

:

' . .

for any R

(x,

and

r'(z) a

*

2ya-,(z)

=

n=0

.

.

..

. .

2A-u1(z)/(z) for

=

J7ln (z) I''''I ,,

< x

,

.

.

.

'

=

(z)

n=1

1 and

7

2y(z)

(z)

.

with

some simple

properties

:

of the

fn

.

.

fu is convex with J,$(0) 0 fn is incresing on (0,r) and =

since (z) >

(2.4).Using

decresing

on

(', 0)

0 (2.6a)folloWs emsily by induction from the de:nition in (2.6a) the deflnition, (2.6b)follows. (2.6c)is an immediate ' .. ''

consequenceof (2.6b). ' Or next stp

=

e-'y(yk)/.('F)

'

y.

lA(z)l <

ISJKR

'' T Jn thn 1/?(z) :Fp Esud (z). Using A'(z)= 0 it folloWs that

(2.6c)

The search for f. To find a function f so that e-IJ(') is a local martingale, we apply It's formula to fz, z2) c-rlJ(z2) with Xjk = f, XI X, and recall X is a local martingale with variance procesq given by (1.5),so 1

:

(2.6b)

'

; (r'r < :&)

ft)

J'''!sup

(2.4a) ln k 0

#w-,(p)(T# < T0). This iq triyial so the proof of (a) is (>1) <

g

2Tn-1(2)

(la

we >et

k1

n

x

(2.3). move(2.5),we begin

E !E

dyj' ' j,

(1.8) in Chapter 4 implies that if

i-e-j J' satisfes To

0

cxl we have

-+

(e-Tr

>

.'

1 and deflne for

H

=

1''%t

(e-fa

fL) k

h

.'

.

iqta < (&)

-

let

lnz)

(2.5)

z).

we

0

=

-

(2.4)

cxn

=

.

shows that comp ete.

ms n

-.+

xt applying the optional stopping theorem at time e -%f X) % fbuj for t S a) we conclude that

1 Jy)

whichis

0

-+

.J$t7-t. < :&) #.-!(p)(o < ksfalse, i.e., (a1)implies (a2),

<

Noting

< cp).

Pw-z(,)('q.,#)

=

1 u j (z)J (z) J(z)

(2.3) To solve this equation

*

=

rL A t)

<

co we concllzde 0

that if (a2)is If ;(r) < justifled since

< x) wFtAz,rl

'->

'ra

=

-+

n

T!

< y < r,

tat conclude

tetting(

0=

0 as

'-+

It A Za 1et bn 1 r with l > y, and 1et gk Let 0 optional stopping applying A t, time at theorem the cx)j

(a)

of

217

E1

E

is to use induction to show

(2.7) Lemma. A(z) ; (A(z))n/n!,(2.5)holds,

1+ l

and

S f K explll)

antt Proof Using (2.6c) (2.6a)the second and third conclusions are immediate of frst The inequality A (z) i (J1(z))n/n! is obvious if the ozw. consquences

218

Chapter 6

Sectlon 6.3

One Dimensional Diffusions

n = 1. Using (i) the deflnition f and h, (iv)the result for n

Jn(z)

of

Jn in (2.4),(ii)(2.6c),(iii)the

1, and

-

(v)doing

little

a

deqnition we have

calculus,

of

To evaluate the first integral v'yj

,,dy fv dz :ys-,(a)

j jo S * *A-,(p) jv# /' = j d,Ja-,(,).fI(p) 'r (J,(p))n-1J1(p) s jo (,,

=

/j(r)

dz

dy

z)

-

1)!

*

z) ,

-(1

+

dz

+

exp

=

1?)1+J

(y,(z))n

ts

p)1+J

j

exp

*)1-88

-1-

dv

n!

d-

d'tf exp

s

1+

1l)1+d

+

-

+ 1

+

so

)

1+

-

yggkq

y.qo

To the proof now we only have to prove (2.2a)and (2.2b).In of view it sces t proye these rrsults with f repl>cd by hi Changing (2.7) = = variable z w(r),dz p z('p)dv then y = tp(u), dy = p z('u)j u *e have

--p

1/7(p)

=

'u)1+:

-(1

1+ ,

X

+ 1

..4.

1 so my)

=

*

1

-

219

0

> -(1

- /- (41+ )/-tl ( - 1' j* (t1 s ) -

=

#

that if y

we note

so w(x) < x. To use (2.1)now, we note that c(p) thre is no explosion if and only if

2

,

j

exp

=

Recurrence and Aanslence

z1+4

-

complete

n(z) j-dy /' dz p (#(z))l4(,/( (?dy 4(p) dv 2 =

= cx) if

c

j, jo w,(s).(s) j - j,

=.

Letting z

1r

=

))

,

(,

where in the lmst step we hav changed variables z n + 1, y of the l's. The next lemma estimtes the last expression and

q'rj

u

d,?,yl,,l

(2.8) Lemma.

2

(s

X(r)

=

E

=

j :

p d'D

w?(s)u(s)jv

I'p

j

y.p

2

(x)

y

ts

-E

A similar argument shows that the second and the proof is complete. factor of 2) f) To see what

(2.1)says

concrete

expmssion

case

=

1,

=

=

-x

in

i

)

y

y

x

g

t?

y

(2.1)is (exceptfor

a C)

1zl)J/2

where 6 > no explosion.

-

z

:

0. When We will now use when J > 1. Whn occurs integral in (2k1) is x.

+ 1 to get rid that it is

0 < &K 1

variables

z

#

l+

6

1+;

-

z

1+J -+

(1+ J) =

zy-: y+

1+4

1+ 8

w' dw

z

-

=

y-

ms y

x

-+

have

we

v-vzy-n

x

J

K zy-ny

1

dz exp

a

w

-

,

dw

g

+ zy-nj' z ms y p the dominated convergence theorem. followsfrom

Since

6.3. Recurrence K 1 the

exp

y

dz exp a. JVY'V

dz

#

Changing

(

'

jj

We considr

Exnmple Let (z) (1+ coecients are Lipschitz continuous, so there is (2.1) to reprove that result and show that explsion and the secnd y < 0, p'yj k 1, so e(-x) 2.1.

c(z)

in a

s

(p(g) /(s))

mtsl

Proof

d'?z/(1z)

'p

shows

If &> 0,

.,(s).(v)

and using Fubini's theorem

=

.-.+

-+

x,

the desired result El

and Translence

The naturalvscale, which was used in Section 6.1 to construct one dimensional diflksions, can also be used to study their recurrence and transience. Let Xt be a solution of MP(5, a) and suppose

/(z) Let Tp = inftt

scale

and 1et p be the natural

&2(z),

=

0 : Xj

>

j

=

z

j

exp

!,) and 1et

=

(3.4) Corollary.

with c'(z) > 0 for a1l z

continuous

(1D) b and c' are Let c(z)

Section 6.3

One Dimensional Disions

Chapter 6

220

.r

v

a

vy

(z) v

z

show

T. A T. The flrst detail is to

=

''' '''

(3.1) Lemm.

If

z < b then

c <

< x)

.&tr

1.

::ti

if and only if w(R)

X is recurrent

Dansience

and

221

R.

=

This conclusion should not be surprising. Xr b=:pxj ) is a local martingale so it is a time change of a Brownian motion ru2' for a random amount of time and if w(R) # R that random amount of time must be fnite. To understand the dividing line between recurrence anb transience we consider some examples. We begin by noting that the natural scale only dpends through the ratio 2/c. on b and

defined by

zjtz;

-

Recurence

f MP(0, &) where 61 that $ w(-'t) i a slution Proof we saw in section Chaf in continuous. and Thus positive tr 5 implies (6.1) &given by (1.5)is motio, change of Bt ; Brownia constructed ms a tim that '.F can be =

Exercise 3.1. (i) Show that if byj al1 z > 0. (ii)Show that if blyjlay) for all > 0. ' .

'

.

.

.

P.ITP

. .

< x) % 0 for y k yo then : > 0 for y P.IT: < then yz k k .

3tik The exercise identzes Exnml blayj 0 as y x. Supse

With

,.4

.

the interetinj ce that for z k 0

.'=+

=

x) ' .

.

as beig

E

byj

1 fr < 1 ( '. . '.. E

...

k 0

1

where

.8(-/a)

y,

=

w

inf (t :

o

> s) and tr

2(z)/c(z)

dslltmj

=

0

If

with probability one, X will exit exit (c,bj with probability one. and will X one,

will exit

Since Brownian motion

with probability

(w(c),tpbjj

=

(w(c),p())

r >

we can study the recurrence and trausipr is a uniformly bounded marngale,

fp(z)= Erv-j

(p(a).&(Tc

<

T5) +

w()(1

.&(Tc

-

<

(3.2)

T)

<

e-K

Pb

P(z)

-

e )

-

p

paj

:r

(n

<

z.)

P(z) =

0

dz

+ zl-r .

=

>

.

w(x) F:z x.

and

-Jf lf

-x

'

. ..

1

r <

and it follows that

Th)J

fP((1)

-

al1 y >

for

w() =

0

tlie

so

and solving we have .PZIT.

>

aq we

fjhhhp) lmplies =

..

k

C, r

where

X

-G1 so p'yj

With (3.1)established did for Brownian motion. optionalstopping theorem

1 then

C(1 + zl-r

=

* 0

ln the borderline cmse r

pLb) w(a)

exp

=

1-r

1 the

=

C

-

((1+p)1-r

-

depends

outcome

on

lj

the

dy

< x

value

of C.

-

''' ..'#''

Lettinx (J)(x) = lims-.x w() and w(-x) since e is strictty incremsing) we haye ''''.''''

'F

.

'

''i.

(3.3) Theorem. PaIT. .&(T

. .. . . .

.

< <

''

'''.

''

''

...

.

(thelimits exist

.

.

.

.

SO

Suppose x) 1 if and only if w(x) = xi x) 1 if and only if w(-x) c < z <

limc-e-x paj

= ''

=

=

=

ln one dimension we that (3.3)it tollows

-x.

say

=

g

.g

(1+

p)

dy

< x =

(X7

jj g > j 1 fc

EI

s

To check the last calculation note that Example 1.2 in Chapter 5 shows that the radial part of d dimensional Brownian motion has (r)/a(r) (d 1)/2r, while Section 3.1 shows that Brownian motion is recurrent in d K t and transient in d > 2, which corresponds to C % 1 and to C > 1 respectively. Sticklers for =

..

.

(x)

w(x)

.

that X is recurrent if PrTv

. '.

. .

< x)

=

.

.

. .. .' '

.

. . ..

1 for all y. Frrit

.

-

.'

Capter

222

One Dimensional Disions

6

Section 6.4

detail may complain that ve do not hve any Brownian motions deined for dimensions 2 < d < 3. However, this analysis suggests that if we did then they would be transient. '

,

'

.

.

.

Exerclse 3.2. Suppose a and i.e., c(p+ 1) ay) and (g+1) condition for Xt to be rqcurrent. =

!'

. .

.

'

.

.'.

'

1,

.'

'

suppose vz is a solution of VP(, c)) where b and c satisfy (1D). tet jz < :8:z let D be real n mbers, which you should not confuse with the coecients, (c, ), apd 1# r = inftt : X) / (c, )). 0,ur goals in th sectiop, whiqh >re accor lishedin (4.t)and @4) elow, are to show that (i)'if g i bounded apd .

.

.

memsurable then

mj r

gx-)

.

ds

.

'

..

.

.

.

-

..

.

..

...

.

/ cotz,

.

.

..

.

ytgyt

The frst step is to generalize (4.1) Lemma.

supzctcj)

(1.3)from

..

' ' . .

;

: .

. ..

Since 0 K Jz)

n

x

-+

.

. .

. .

.

and use

.

.

.

J(z) k

< :2/: for

z 6

the monotone

'p

Proof

Let Mt

=

,p(-'L)+

y(z)

=

(z2 (z -

-

zjjp,

=

have P/p theorem.

convergenc

,??(Ab)=

mitingale

on

-

so Mt is a locat

t')

..

zrjv ''gxst

C gX,4

v(.m)

h :)k Nqw, 1et E)

1 Errn 1)

with

f

(

ds

(i) and (1.1)imply that

ds.

loal mart.

(0,r).

.

If

'p

-

j

and g are

' gx't

for t

< r

ds

bounded then for t

< r

'rllpllx

+ l-tflK 111711x

(4.1) implies that th 4ight-hand i Chapter 2 and (ii)imfly

side

is

integrable

an

random

variable

so

(2.7)

E

T

M.

-

p

=

A n)

-g

=

Ezr < x.

(e(c),e()))

'

+ Ern

we (w(c), w@)J

=

Chapter 3.

v = (w(c)+ w())/2 be the inidpoint of the interkl, f = (w() w(c))/2 be half the length of the interval, y 6

g

.

Ezln-hnt

#(#t) dt

lim M3 =

>

t1r

()

=

:

jz

(4.2) Theorem. Suppos a is bounded. If there is a function (i) e C2, Ln in (c,bj r() 0 then (ii) tl is continuous at c and b with r(a)

T

r(z)

=

inftlp)

pztza) ds

-

-

Proof Let p be the natural scale deqned in (1.4)and consider $ w(-;V). t #(F,) ds where By (1.5)yt is a local martingale witli (F)t (p)is positive = continuous. : estimate inftt To r and / (w(c),w())) we 1et

=

formula implies

It's

=

i

r(z)

+ (z)y?(z) !a(z)y??(z) :

-2/p.

=

223

Our choice 9 p implie? ltyj k p, so J(X) + t is a local supermartingale on inf (i : (01 r). Usink the otional stppping theorem at time a A n where a . .. 5 / (e(c)+ l/n, pb) 1/n)) we have

o

,

=

''

loc>l mart.

=

-

.

and (ii)to give a formula for the Green's function GD (z,:v). As in the discussion of (8.1)of Chapter 4, we think of Gstz, g) ms the expected amount of time spent at y before the process exits D when it starts at z. The rijorous ztieaning of the lasf sentence is given by (i). The analysis here is similar to that in Section 4.5 but instead of the Laplacian, we are concerned wit

zJ(z)

T(X) T(Fn)

'

satisfy (1D) and are periodic with period (p)for al1 y. Find a necessary and sucient

Funtions

6.4. Green's . .. .

=

E

and note that 1 k 0 in (w(c),w()) with J&(z)

Functions

's

Green

E

>

=

EnMZ

=

E.M.

#(Sf) dt

Ec

=

0

deriving the solution (4.2).The next two pages will be devoted toand verify the answer who equat.ion content Readers in to are guess (4.2). to the skip to (4.4)now. To solve the equation it is convenient to introduce Solvlng

can

,

m(z)

=

1

,(z)a(z)

224

Chapter 6

where

w(z)is the

(4.3)

1 2m(z)

natural

(

.d..

1

p'z)

d

scale

and note

zd;

.c(z)

tfz

+

(fz2

2

Breaking the first

that

dz.f

( ) -w'z(z)

c(z)

ld oytz) dz

p'z)

2

-2(z)/c(z).

tet

(8)

-g

-

' :

.

an d integrate to conclude that for some constant

1 sy)

.

:

(a)

In order to have

#(w(z) w(c)) =

/'

-

j .... .. . ,,(z)/E E !

.

:

.

.

2ja

-

' ..

Plugging the formula for

r

we have

dysyt

j

dgsv

dzmtzlplzl

Fe have E

.

(Pt

.

/(c)

-

. . ..

w(z)) (1w() w(c) (w()

=

.

=

-

dyqv

.

E

m

have

we

b

2,u(z) /.

q

'

dysyq

j

tr

dzmtzl,tzl

p

2(1

-

-

j

u(z))

z.

c

j

to

z, and recalling

that

r(z)

=

2(1

u(z))

-

(d)

2u(z)

+ dzmlzlptz)

dzmtilft)

writing

w(z) w(c) < /( ) w(c) s r (zj zq;

j

y.b

/

'

j

dysv

and

r

dzmtzl,tzl cancel with the

=

tfzmtalptal

dy s(p)

-

; dzmtzlplz)

/.

dysyj

Using Fubini's theorem now gives

s(z) 2(1

(e) we

j 'r

dz?.,,(z),(z)

2u(z)

denne G(z, z) to be

jz j

dysv

.

''fs

ilp

+ If

u(z))

-

/.

dzmtzlglz)

e(z) w(tz) 2 :() /(c) (w() w(a))m(z) vb, w(z) (P(Z) P(t')) rr(Z) 2 ) vaj e

15,

dysy)

-

=

-

.

-

when

z

k

he.

z

sz

-

(4.4)

-

'

-

(f)

z

-

then we hake

(b)

/

,,(z))

-

r

-

=

(1

r!/

Multiplying the lst identity by 2 ja dz m(z)#(z),

.'

/- /-

# into (a)>nd

u(z)

a

b

'

. . ..

=

dus''b

va)

j' dy s(p) j' dz,n(z),(z)

Using this in (c)we can break ofl- part of the second term flrst one to end up vith

j.!/

dysy)

b

2

vbt

f/yj

=

0 we must have

-

have

/.

a

-

'p()

jy

zyp (smtal,tzl

=#-

r =

,

p

Multiplying by sy) on eah side, integrating y from 17(c)= (1and s = p' we have 'p(z)

q

syj

Ralling

r

J'= %-. LJ :

d.v

(z, ) we

z) and one over

,

.-xam/zov,

=

-

dz

,

+ 2u(z)

'a1

,

1 dv s(z) dz

225

an integral over

=

=

,

ja6dy into

s(z) 2(u(z) =1)

=

sine the deqnition of the natural scale implies w&(z)/#(<)= is called the (density function of the) speed measure, thogh as we willse in Example 4.1 the rate of movement of the proce har is inkerel# proctinal comparison with our source for m(z)! simplify notatiop ald To facilitate to to s(z) this material, Karlin and Taylo (1981), Vol. II, w p (z) be the drrivative of the natural scale. ro solve equation Lf c.::: no&y, we use (4.3)to wrlt y d .-.

section 6.4 Green Functions 's

One Dimensional DiEusions

r(z)

=

jj

G(z,z)#(z) dz

W

Section 6.4

226 Captez 6 One Dimensional Dihsiops Combining the

calculations

(4.5) Theorem.

above with

E=

jo

,(-Ya) ds

Gx' a)p(z)

Gz,

dz

Taking g

By the monotone class theorem it suces Proof to prove the xesult when is show continuous. To do this it suces 1) deined in that the to (f)satisfles the hypotheses of (4.2).To check (ii),we note that s z b, G(z, z) 0, c or z r(z) and G is bounded so the bounded convergence implies theorem 0. To check that v (22 we note that

1, c

=

.-#

-+

-+

2m(Z)(

Z)

=

=

z

Letting

'qc,,)

,

=

inftf

:

e

eb

w(c)

=

r(z) (w()

2

=

ja(w(z) w(c))

(p(z))

-

+ (w(z) w(c))

b

pbt

-

Diferentiating we have

and noting

w()

-

2

e(c)

dz :

.'

,

,

-w

=

ja(w(z)

(z)

-

b

J.(w()

w(c))m(z):(z)

(z))m(z),(z)

-

,p (z)

,,(z)

-v

=

-

:

.

.

.

. .

/z e'zteb) -

dz

2

y(z)(w() w(a))p -

=

This shows that

'??

satis:es

(i)in (4.2)and

(z +

lt

(z

z

(

mtzl

r- h

z

(4.5)with

g

=

z) mtzl

-

-%

z z

becomes

k

sz 1 we have

dz

'

.

:

+ h) m(z) dz

.

.

m(z)#2 xit a small interval

,

'

'

.

ew

interpretation of rlz note that (11.7) in Chapter 2 implies that time at z up to time t then for a bounded memsurable .(

.(-Y,) d(X)t

=

=

g-lax't

1 so mtzl

1/c(z)

=

:

.

ds

0

and taking #(z)

J(z)m(z)

=

$

T(z) dx

.

.

(

:

''

=

0

mtzlf4 .

-

gives us the time that X3 takes t

=

T(;,) ds

.

'r(c,,)

(4.7) Lemma.

,(z1).(z)

gz)

Let 0 <

=

<

p. Then

0

/. (w(z) w(a))m(z) -

so Lv

-g.

using

$

z

z a

Exnmple 4.2. Second Proof of Feller's test. Let T(x= limrta T. and Tc AT:. We cont. ent ourselve of (b)in (2.1).The flrst of two steps is

c(z)/2,

-a(2)

.

'

when z

'lk us mu 1tip ly 1ngthe local times by mlzl cortverts thm irto ocupation times. This and the previous interpretation suggest that it would be natural to all m(z) dz the occupation However,, it is too late to try to change measure. the original name, speed measure. El

=

=

,.

a)

to be precise, the ratio of the time fpr Xt to that for Brownian

When Xt hms no drift p') we have

w(c))m(z)J(z)

-

0. Thus as centered at z, or P otion. For another if LLis the local

.

When

.

dz

n

w() w(a)os(z) 2

=

fzrgtzldz

-

-

)), and

.+?,

=

'

( ' . . ' .

dz

the lmst two equations by (z) and e C2. Multiplying and usipg Lp the deflnition of m(z) We have 0 adding,then

This shows

'

.

j=Lpzj w(u))m(a),(z)dz c

+ w''(z) (w() w(z))m(z).(z)

Xg $ (c,

J)

when wheu

-

B=r(z-s,=+hb

.=>

have ,,

m(z)(z + /z z) m(z)(z z + j

=

+

a'

+ #(z) we

wlzlim(z)#(z) .. .

vbt w(c) (z) z DiFexentiating again

-

-

the lmst expression

+

'

dz

-

aljb

-

-

z)

(4 6)

that the terms from the limits of the integrals cancel

-

.v

m(z):(z)

-

d)/(

-

z)(z

-

and b =

-

-->

Z)(2

-

2m(z)(

G(z,

,

-

227

=

b

j

=

Rmctfons

Exnmple 4.1. Speed mnsure? If X has no drift (e.g,,Bmwnian motion or any diflksion on its natural scale) then w(z) z and (4.4)becomes

(4.2)shows

lf g is bounded and memsurable then r

's

Geen

the result follows.

(:l

dz

< x

inf (t Vto establish =

:

Xt

=

z),

variant

Capter

228 if

and only

Proof

One Dimensional DiRsions

6

if inftx
T:)

<

0 nd

>

Section 6.5 .FQIT. A T:) < x.

supo,zu<

Letting r

15(L so info
(4.8)

T

<

vbs ) w(#) w(c) =

w(a) >

b

w(c), (s() ebj w(c) w(z) 2 ebt .() w(c) (e(z)

j

-

-

-

Taking g

'

.

'.

.

E' '

..

=

''

(4.5)

1 in

.

Erpxu'r

S

(1

:)

-

> sup a <# <

sup #(r

>

;r'E(a',)

Using (5.7)in Chapter 1 of Durrett the moments of the exit time r.

p(a)) mtajts

=

.

=

(1.2)in Chapter

+

-

'

.

k 1)M); r > AT) .Pptr > k 1)M) -

-

c) k

K (1

kM4

-

(1995)nqw,

we get an upper

/-

-

t'tlrrttzel'z

6.5.

tE

a. So

&'qc,&)stays bounded

(?n

a11 EI

undry

Behavior

In Section 6.1 when w constzycted a diflzsion on p. half lipe pr > bmpded intervalj and we said that we gave up when the process reached the budary I!I it had exploded. in Fhich Fe cap this section, we will identify situations extend tlke life of the process. To explain how w might contizpe F begip Fith

as

-x

0

(w(z) w(a))m(z)dz < -

E

a

El

cxl

The. two conditions in (4.7)are equivalent to f%(To Th) < x. To complete the proof npw we will show

<

T:)

a t r Wial example.

at 0. Suppose (z) H 0, and c(z) is boundary co). dBt (0, To deqne a solution to dX3ig(-z) tr(Xt)c(z), teflecting d' to R by boundary at 0,:' we exted tting with a let 5$ be a solution of dyt c($) dBt on R and then 1et X3 1l$1. 5.1. Refectlng positiveand contkmous on

Exnmple

0 and .E(Ta A

>

=

::

=

=

Jf ,!5(T.

T)

<

0 then Ezl

>

A T) < x.

j'

jt

PaIT. < :J) > 0 implies .&(Ta < cx)) > 0. Pick M large enoug so Proof that #OIT. K 1') k c > 0 and that .J%(T S M) k E > 0. By considering the flrst time the process starting from 0 hits z and using the strong Markov ( prfrt#, it follws that if 0 d z < then '''''

-

.

.

.

P 0 (1b S

)

=

f'o(#.(T

K t = Tz' );T S

).l3n(:r',, s t) s

.z'wtn

s A similar argument

bound

.

-.+

.

3 now, we have

Z'

-

The frst integral always stays bounded ms a apd a if arpd only if w(a)> c

(4.9) Lemma.

229

lt follows by induction that for a1l integers k k 1 ;

-x.

(

2 e(z)

kM)

>

the proof of

=

0 if and only if

>

E

.

AT: and repeating

g'(0)

-

.

n

.&('r

(3.2)implies

Ezra,sj

=

Boundary Behavior

/

.

.

f)

.

.

.

.

.

''' .

To use the lmst recipe in general we start with Xg a solution of MP(, solves MP(0, pxtl and let p be its natural scale. ) where y j

h(#)

.

'

K t)

=

s i)

k

Combining the last two results

< z <

e =

Changing variables y

=

w(z),dy

>

0

for all ce <

z <

w'(z)dz,

=

-

/ (.(,) 0

(e -

() J

E

.

.

,

0

that

shows

Pz (:Q A T K M) k

for

. ..

=

1.E,q-,,,) J5(Ta K t)

.....

t

g

.

To see if we can start the process K at 0 we extend lt to the negative half-line by setting (-g) (p)and let Z3 be > solution of jMP(0, lt) on R. If we let ?h4g) = 1/(lpl) be the speed measure for Zt which s on its natural scale, we r-tpl to conclude rW(-p) can use (4.6)and the symmetry

shows

#ZIT.

(/(P-A(#))l2c(/-1(g))

=

'a.tt/

c)

=

#)F-?(#)

,

e: =

.(z)).,(zj.(z) -

dy

w(J) the above

4z

Section 6.5

Capter 6 One Dimensional Dilsions

230

1/#(z)c(z), Now, (i) recalling that the speed measure for Xg is m(z) changing notation p'z) = s(z), and (ii)using Pubini's theorem, the above J

=

n

s(z)

Introducing M ms the condition in Feller's test (5-1)

i . E

2

=

j

Proof

propity

< x)

If J%(4-,,,) :'

.

!

dz

s(z)

(M(z)

-

.. .

. .

. .

.. .

cannot

with the

.

.

.&(n-,,,)

supE'.t'1

.

5' ' '

.

(j

.

.

..

.

( E.E. . ..

(.

:. . . .

' ..

.

;

' ..

'

. .

. .

. .

.

Varkov

'

-

5.2.

' .

:

'

l

P(z)

= onto

I

JV

,

where r z

z.!

:

K x, 1et yyy

q

-

E

v u.a. mbz)

(0,r),

and let

I

=

iT(0))

s(z)

dz

we can

get IN to the

boundary point:

The analysis above shows that when J

< x

px, dt +

.

Le$

xtdt

0.

r

y

a

(0, =

y

exp

<

=

jlol rn(z)(/(z)

-

we can get OUT from

dz

C aZ ......-t7'

-2pp/cQ

dy=

e

d!l

2

1 27 (

e-zpmlcl

-

)

The speed memsure is

1

=

w(z)c(z) ,

we note

P(0)) dz

=

2p./,a

e

az)

/(J

that

1 jlo 1

(e2#'/G

a

1) dz

-

2pz

< oo

since the integrand converges to 1//2 as z 0. To calculate J we note that a 1/(J and J z) ms z m(z) 0 so M(0) x. The comb jnation I < x and J = (:x) says that the process can get into 0 but not get out, so -+

ew

=

-x

=

0 is an absorbing To investigate the boundary at x, l

the boundary point.

>

-xa

0

2j2p).

0 sinc the 0. See Exercise

#

-+

(T(z)

Feller's test implies that < x

natural

entrance

j

0

when I

x

=

To investigate the boundary at 0,

4

ttpk? vbvl)

=

r

rn(z)

from which the desired result follows.

(

.

=

0

whichmaps (0,x)

(0,r)

absorbing

=

and then conclude that for each integer 1-k 1

Consider a diffusion on

cxn < cxl

=

Of course we want to suppose o' > 0 but we will als sppse calculations are somewhat different in the cases # 0 apd p 5.1 below. Using the formula in (1.4),the natural scale is

&

-

name

regular

n

Fellqr's brmlc-hlng dlffuslon.

dxg '

sthg :

1

Exnmple

(

i .

.

were named by Feller as follows

which

J <

231

The second cmse is called absorbing because we tan get in to 0 but caot get out. The third is called an entrance boundary because we cannot get to 0 but we ither can sta'rt the process there. Finally, in the fourth ase th procss c get to nor start at 0, so it is reasonable to exclude 0 from the state space. We will now give examples of the various possibilities.

'E

Vsingtht

0.

that

note .

!

M)

>

E

< x. .

>

=

.

from 0, we .

.

' .

.

.

dz

escape

.

) Pick so that >ct'q.u, < .:J') proof of ih the s (4.9)we frst get tr 6 (

s(z)

0 then .E'r(-e,

lk

E

M(O))

I < cxl < (x) = x = x

dz

of m to bring out the analoa

To see that this means that the process repeating the proof of (4.9)shows .

m(z)

(2.1),we have

.E'nq-e,e)

(5.2)Lemma.

z

jvjo

,

antiderivative

;'

/.'

m(z) dz

dz

This leads to four possible combinations,

and

=

Boundary Behavior

boundary.

m(z)(e(=)

-

that

we note

O

w(z))dz

=

j* 1

1

2px

dz

=

cxh

Section 6.5 Boundary Behavir

232 Chaptet 6 One Dimensional Dihsions To calculate J we note that J x. The combination I

into or

out of x,

and J

x

1/(&2z) so M(x) = (:x) and says that the process cannot get

m(z)

p k 0,

when =

=

(x7

=

k

so

x is

a natural

boundary.

Exercise 5.1. Show that 0 is an absorbing boundary when (a)p 0, (b)p < 0.

boundary and

t:xn

is a natural

Exnmp e 5.3.

The flrst conclusion is reascmable since in d k 2 we can start Brownian motion at 0 but then it des not hit 0 at positive times. The second conclusioh ca be thought of as saying that in dimensions d < 2 Brownian motion will hit 0. Exerclse 5.2. Show that 0 is n absorbing bounda if y K =1. We leave it to the reader to ponder the meaning of Brownian motion in dimension d K 0. Exnmple

=

Consider dXt

on is the lndex of the Bessel process. To explain the restriction and to prepare for the results we will derive, note that Example 1.2 in y on Chapter 5 shows that the radial part of d diiensional Brownian motion is a Bes:l prpcess with 7 p (d 1). The natural scale is Here 7

Power noise.

5.4.

process.

espe

-1

233

(0,c=).

1/(v

The natural

'(z)c(z))

=

z-2J

scale is

AJ

=

(p(z)

dBt the

z and

=

speed

mrmsure is m(z)

: '

so

>

1

I

=

0

z 1-2: d z

< o:) = cxn

=

if < 1 if 6 > 1 -.

-

(z)

exp

=

=

1 *'

-

1

-%

d#

#

1

v/z dz

dy

ln z

=

(2 1-v

'

=

y)

-

if y

jj.

=

vp

From the last computation we see that if k 1 then w(0) < < 1 we observe that the speed memsure To handle y .f

=

-x

1 1

1

Combining the lpst two and I

=

m(z)

=

So taking q = 1 in the defnition of f 1

f

=

.C1-'y

o 1

-

To compute J we observe that for any y 1

J Combining the lmst two

=

z'Y

7

Z v+1

dz

natural absorbing

-v

we see

entrance boundary regular boundary

regular

dz

=

z=+1/(y + 1) and

< cxl

that 0 is an if if

2' 1

j

.-.

2:

dz

g

.z

(x)

th.t the boundary poiht 0 is

we see

if if if

<

8C 8C J E

h'

(1,x)

g1/2,t)

(0,1/2)

,

(x7

M(z)

>

(

conclusions

x.

Exerclse -1,

z 1 a 7+

conclusions

<

=

Wften&< t/2

x.

=

The fact that we can start at 0 if and only if < 1/2 is sggeste by (3.3) and Example 6.1 in Chapter 5. The new inforrpation her: is that ths solution resutt, recall the can reach 0 if and only if J < 1. For another proof of tltat dolutionsi f time change recipe in Example 6.1 of Chapter 5 for constructiht the equation above and do

z%

=

w(z)c(z) t

hence J

E'

-1

1

and

-x

J

))/(j.

-

When J k 1/2, M(0)

y C 7 E

(1,co) (-1, 1)

5.3.

.J.I':=

Consider

rr'o -2J1 a

B,

'z'c

(s.s1) ds

Show that (a)for any J > 0, Pllj (c) Plh p.m) = 1 when J k.1.

H'J

and < x)

Exercise 5.4. Show that the boundary at and entrance if 6 > 1.

=

x

1.

.5-2:1

=

()

(b)Ellj

(s.>1) ds

a

<

(x7

when

<

1.

in Example 5.4 is natural if 6 ; 1

Chapter 6

234

One Dimensional Disions

Section 6.6 Applications to Higher Dimensions

Remark. The fact that we cannot reach x is (5.3)in Chapter 5 implies that these processes do not explode for any J. The second conclsion is surprising at flrst glance since the process starting at cxl is a time change of a Brownian motion starting from x. However the function we use in (5.1)of Chapter 5 is gzj = z-2J so (2.9)in Chapter 3 implies expected.

for

all z with

1zIK r.

(Ah)

-

d cos h (z) E=l (azf). Using It's

Let

=

=

a

0

ds

z-2:

=

0

1

which stays bounded as M Exercise

5.5.

.

Wright-Fisher

dlfihslpn.

2z dz

6.6. Applicatlons

'

Noting that coshtp)

to Higher

.

coshtp)k max

7 0 p e (0,1/2) p k 1/2 .

'

the cmse a .. .. .

0

= '

and

then

rE

:

' ..

' .. .

.. ' .

and

(ii) c i.

=

i,j

i

sinhtaaLlf (.;L)

(E'a(x,)

y

C

.'

.

'

jince ltxsvht)

K d coshtar)

it follw

p

each

i.e.,

z,

(i)

and

t

=+

-

'''

k

a

'

.

Using the optional stopping

a#(Sr

f))

that

A t)

K coshtarl/a

cxl gives the desired result.

(:I

*

cfJ(z)pJ

(C.1) Theorem. Let Sr = inftt so that E=S. K C for all 4 with Assumptions

Ltting

'

g0,Sr).

on

S (z) K Erfhlvvhg)

Ersn for

If(A-a)I)k' acoshtaxj)

-

tad is a local submartingale at time Sr A t we have .-l

.

when

0

>

y

#

The frst detail is to show that X3 will eventually

la:I:S'-

so hlaj thorem

t:y

''

and nondegenerate

k jsinhtgll

:

..

locally bounded

i

Proof

+

k a coshtaxa )

'

optinuous

ee*# ,

2

a i ..j- coshtax,lcfftxa)

arguing

'

.

Dlmenslons

:::1

(i) b is measurable

ey -j-

our choice of a implies

ln this section we will use the tehnique of comparing a multidimensional difconitions fusion Kith a one dimensional one to obtain sucient for (a) no explosion, (b)for recurrence and transience, and (c)for diflksions to hit points limr-..x Sr Throughout this inf (t : I-'LI rl and vb or not. Let Sr will section we suppose that Xt is a solution to MP(, a) on (0,Sx), and that s>tisfy the coecients =

k 1 and

=

..

.

ds

=1

Recall the deqnition given in Ex-

by flrst considering Hint: simplify calculations that the value of a is not ifnportant. '

d

1 . 2 i tr coshtax,lcfftx,) + I'''-I (?

mple 1.7 in Chapter 5. Show that the boundzy point 0 i absorbing if if regular entrance if

)(A%) ds + local mart.

-:

cxnif and only if J > 1.

-->

sinhtaIj

=1

-

J(#,)1(s.sM)

EM

formula we have

d

t

ltvl

235

:

l-YtI

=

r).

1zlK r.

Yalz)

hemark? actng cosn so tlkt

exit from any ball.

C

Then there is a constant

(ii)imply that &

0

we can pik

an tr

lrje

s

<

(x,

that i?

back

througtheproof k -as(z) 2

To see that the last estimate Exnmple 6.1. The natmal 'pcyle

Suppose d has

shows C

l(z)l

*(z)

where

is

Iktzll

1+

is fairly sharp consider a special case

=

t, c(z)

=

1, and

Cr

k 1+

ositorrl/a

=

e2I*

= 0

'*-

dy

=

1

(z)

(d2#r1) -

2:

-#sgn(z)

=

with

#

>

0.

Chapter 6 Onq Dimensional DiEusions

236 for z

-w(z),

k 0 and w(-z) =

Using

(4.8)now

b = r,

with

the spqed measure

so

rrl(z) c

1

=

e (z) z

w(r) w(0)

d%

c

=

0

=

w(=r) w(r) w(-r)

=

So

that

noting

an

2w(0)

-

2e(r) w(-r)

Our plan is to compare St

-apjal

,

-r,

=

Section 6.6 Applications to mgherDimensions

introduce the natural

we

.1

((),x)

on

nd speed measure

mtzl for Zt:

=zgy) oj j (-

exp

=

bLztjdNt

w(z)

scale

s(z)

-

-

-

1.412with

=

/z(Zt) dt +

=

237

,a(,)

w(z) ji=s(,) dp m(&)1/(s(z)p2(z)) =

we have E 0 (-r,r) 'r

r 1 ezpz e-zpz dz gp o 0 1 (-n2#IzI e2#r) :-2#lzI dz + gp

(d2#r

=

)

-

+

=

In view of Eellez''s test, (2.1),the next Tesult says that if Z3 finite time then X3 des not explde. .'

-r

equal

The two integrals are

r

j.

E 0'r(-r,r)

=

so

.

.

.m

-

.

u.

(6.2) Theorem.

,

lj dz

2V(r-Z)

-fj

'

.

t a2J(r-z) - 2#2

,

.

-

-

z

p

r

1

=

2p2

'

(j

(elpr *

-

,

-

-

* dzm(z)(w(x)

./r,

r

p

1 and (z) c: =#Qrt(z), th To compare with (6.1)now, note that when c(z) remark after the proof says we can pick a = 27 + 2 so the inequality in (6.1)is ::

E 0q-r,r)

S

coshttz; + 2)r) gp + :

,

,

With the little detail of the fniteness f Ersr difl-usion and a one dimension>l duce a compyrison between a multidimensional which is d We begi by ihtrodciik thi yiis t Khsminskii (1960). on, in of deqnitions that may look strange but are tailor made for coxrijutatitm and of proof (d), (e), (6.2). (a)of the ' . .

..

'

a(z) v(r)

=

sup

=

zzTctzlz

t

a(z)

/z(r) =

'

'.

tetting

.(z))

-

x

-

:

.

=

limryx %, then

.

t-j

r( gn a )p y n <

t

cxl and

s t y j;

n

-+

-+

c-f.J(')E

.

-+

..

.

.

.

.

.

.

.

.

..

.'

..

.

#(.%)

. .

=

-+

.

-

j

e-'.(z,)

ds

f

+

=

and

p(r)

=

2p(r)

''

:'= < x we have Pss. sj 0. To deine the function g, we follow the proof of (2.1).Lt X z.= pzgj, is a local use the construction there to produce a function f so that martingyle, thn take g = J o p where p is the natural scale of Zt It follows frin ui ssmptio? nd (2.2a)that grj (:xn as n. Since g 6 C2 it follows from It's formula that ( .

-

rj

p(r)s'(r)

' .. .

in

-+

e-%gztt

and

#(z) : jzjz

'

(:x7

=

-+

#(lzI) 2. e out of th way We now intro-

reach

Proof To prove the result we will let Rt IaLl2 and flnd a function g with (x) as g(r) (y) so that e-'g(Jj) while Rt is a supermqrtingale r 1. To d use the optional stopping see that this is enoug, let Sr = inf (t : Rt pu; zr) theorem at time S1 A Sa A 8 to conclude that for IzI> 1 j

a. Exploslons

can't

..

There is no explosion if

:

lj'

.. .

.

+

e-'g'zslyzs)

0 1

j

ds + local mart.

t -,

e

g

u

(g ;#a(z; d,s ,

,

238

1 2 (z)J (z)+ =g0 ,,

p(z)p''(z)

?.z(z):

,

(z)

#(z)p(z),'(z)

+

b. Recurrence

this as gzj

=

Before plungiug into the proof we need one more property which follows from (2.6c) and the fact that our natural scale is an iucremsing function with w(1)= 0: #J(r)

(b) A little

k

0 for

r

k

=

.i

=

2

Using It's

formula now

c-#(af)

-

glj

Translence

conditions for transiene and recurrerce The next two results give sucient Let > 1. ln this part we use the formulation of Meyers and Serrin(1960).

a(z)

=

z

IzI

.

c(z)

z

Izl

Fj7

ds

+

0

i

1

..s

+2 -

i

()

.

J

e

,

p

-,

()

c

g

,

ds + local mart.

,

2z

(z)+

.

tr (c(z))

z(z)

we can write

'

E

we

a

< cxl

t

' .

.

. . .

..

=

exp

-

t :(s)

the last equa-

ds

.%

1

l

whkt the conditions say note that &tj To see but if p % 1 and this corresponds to not > 1 p

ds ( a ;:(jyy(x,) above

.

6t)

.y,

Uqing the functions a(r) and 7(r) introduced tion compactlf ms e-Xc(.!) cflin) = local mart.

:

.

' ... ...

1 + 2 -

=

In the next two results we will >pl# the deqriition to

Lxg)d a (p izjxxjayj ,

x

(j) dt

1.

0

e -'.'(.R,,)2Arji(A%)

,

,

tekz?

.):..' (yjgyr'jd

-e-'#(A,)

=

d

and

.-,

have

we

E1

2 ms lzI T formulate a resllt that allows de(z) to Dll function' if and oly if deflnition: J(t) is s'id to b

l

?/

=

while

ivhr tr (c(z)) i:z E cf#(z) is the trace of c. We call detz) the effpctive dlm' sih at k beuse (6.3) nd (6k4)will sltw tht there will be rettrrence 2 in a neighborhood of infnity. r transiezce if the dimension is < 2 or >approach (x) need

p'(IzI2)2z .lJ#(Izl2) J''(IzI2)4zjz.j # .Dfptlzl ) g (IzI2 )4z7+ g (Izl2 )i .Dp(lzl2)

in

1

gives

cal.culus

is a local supermartingale

239

:(z)

=

we can rewrite

Using the deflnition 9 0 and p npw

(a)

Combining (d) and (e) shows that e-fg(.&) Rt k 1 and completes the proof.

it follows that

Since e-fg(Zf) is a local martingale

to Higher Dimensions

Secton 6.6 Applications

One Dimensional DiRsions

Chapter 6

=

E(s)

(1gt)-F =

is

a Dini

function if

p/log s.

Suppose detz) k 2(1 + e(lzI))for lzIk R and J(t) is a Dini thn is transient. To be precise, if IzI$ R then .&(Ss < x) < 1. Xt fukction,

(6.3) Theorem.

-

(d)

I

+

e0

f-#(l,)

'

+ #(-%),'(A,)

# a(*a)J''(,)l u

To bound the integral when Rz k 1 we use (i)the deqnttio of v and (ii) the equation in (a),the fact that g k 0, and the deqnition of p -#(.Ra) +

(e)

t

pt-Ya)

-:(./,)

<

-p(

s

,

zxjg

(a ) + ,

y,(sa)) a(x,)

+ ('z(l,)#'(Aa) t#(X') + p(A,)

.a,)

+ g''llsjk

Here Sr

inftt

=

explosion.

:

l.Xl

=

rJ, and

being transient includes the possibility

of

V

(b),then

for 1zIk R and 5(t) is not a Dini Supjose detz) % 2(1+c(1zI)) (s.4) Theorm. precise, if recurrent. then be is T Xt function, lzl> R then Pz < x) 1. =

Proof of a(x,)

jpt.aal s0

ds

(6.3)

The flrst step is to let exp ',(r) j* =

r

(-/' 1

e(') dq

t

-ds

s

240

which hms

r

P g(r)

exp

=

r

P z/(r) =

-

eXP

r2

and hence satisies

r(r;;jy(r;

vpuLvj (y. y = IAhl? and using It's formula Letting Rt four times below: use (6.5)Leppa. t-7(IzIz)) - S 1.X 6 I< '?, p '

'

((

'

. . /oi

If

ranrj

(

. . forppta Itt's

.

.

gRzt

=

i 1 +2 ij 1 +2 =

which

' .

(' .

)

(

(..

l

..

ds + local

-

-,-

Wl1q jle

suprrmltinyaly '

i

.

.

.

.

+ g,R,)

exp

1

(sj' 1

dt(

Ctf)

t

.

?-f

.E

s

we again have

=

:(r))'/'(r)

zk) r ''( + (1.+ '

0

=

..

but this time #J(r) k 0. Since we have mssumed detz) K 2(1+6(lzl)) for Izlk R while Rt k R2i If it fllows from (6.5)that #Rt ) is a local supermartingale stopping using optionpl Sc h S, ((6.1) thorep time r R Kr d s < x the 1) we see that < x) implies that

r.t

.&('r

=

g. . . ....

j thi frpft o? (6rty .

(de(z)

.

(6.3)follops.

s and

..

.

. . .. .E ..

' :

.

y

,

' '' g .. .

. '' .

.

ak #(IzI)

(

mart.

Ez:*Rvt

=

#(r z tprs

Rearranging gives

j u g (.R.,)4.X',.X',adj'l x ,

Prsc

) (js Letting

g

#(S,)2cf(X,)

1

jo

.

.

we get

de(A%)

t

=

for

de(X,) :

?

?, Rag (Jj) + g (S,)

2J(.X,)

%t

y(.a,)

s

givesthe

ds

'

.

.

.

<

g

,)

+

#(s a )(j.

.pts

-

r <

s ))

,k 4(s2)a 4(lzl2) a q ) #(r )

,

.

,

-

z

#(s2)

that

-+

x

since J(t) is nt

v(a,)

(6.6)Theorem.

s ()

If detz)

x k R5. If =

Sr A Sa

k 2(1

/c exp

E1

while R' (6.5) implies that pRtj is a local supermartingale optional stopping using at time r theorem < the .R.K r K s cxn then x) 1) implies that < we see that ((6.1)

=

then #a(Tc

< oo)

=

f(lzI))for 1zIK z? ad

-

(-/,

1 ) dz) --YJcxl

C Z

=

a

y

0.

.&(r

=

w(IzI2) k Emvlvl k

w(r2)#z(Sr

<

(4 7) Theorem. .

S,) g

q .

< 1 P=% < Ss4 % w(IzI2) a

vr )

a Dini

function El

=

n

#e(X,)

aud

.=

The proofs of (6.3)and (6.4)generalize emsily to investigate tidimensional difusions hit 0. Let Ta inft > 0 : X3 0J.

.

la'LI <

<

noting desired result. (x,

-.+

5.,)

<

c. Hittlng Points

ds

:

the result.

Rearranging gives

=

-

two mssumptions on g, ?,(s,)

Eprove.

' . ' . .

#/(jzl2)

(u2,?2) apd

i

local mart. +

.'asg

-y/(r)

following which we will

14> lockl

.

:'(A,)2At(.X,)

-

our

.

(c)fri

##(r)

Since

t)

-

Using

o

'

:

.

.

j'

#(r)

()

241

This time we let

. .(l

then gat)

with

(6.4)

=

get the

we

Proof of

dt

j

zz

0 for r : u-

=

0

,

t -

n'(r)g'(r)

+

lzl q ('u,r),

Fhen

tising

glkt

.y

..y

r f(f)

jj

-

/

dt

j

1 + e(r)

,

The upper bound is independent of

r(j)

Jy

-j

q

Section 6.6 Applications to Iligher Dimensions

One Dimensional Dihsions

Chapter 6

(

q

then #.(Tc

lf #e@) 1

j,

.

< oo) >

2(1

0.

exp

for Izff and E(IzI)) n

-

( =

1

:

(z ) da z

)

.yd

y

kyu ,

whether

mul-

Caper

242

6

One Dimensional DiEusions

Sectjon 6.6 Applications to Higher Dimensions

We have changed the sign of 6 to make the proofs more closely parallel the previous pair. Another explanation of the change is that Brownian motion

so the boundary is now

(6.6)

Proof of

a

little below 2 dimensions rather than

a

(3.2)and

the Markov property

(6.7)

Proof of

hits potnts in t k 2 doesn't hit points in d k 2

is recurrent in d K 2 is transient in d > 2

Using

This time

#(r )

above.

little

Since

Let

's'rj

-w'(r)

we

=

lzlk

the conclusion to

p. C1

let

we

j

=

now extends

243

1 :(j)

r ex p

Ja )

-

dt

t

-s

ds

#'(r) k 0 and

have

r#''(r)

e(r))#'(r)

(1

+

-

0

=

1.X12and using (6.5)shows #Rtl is a local supermartingale while Letting Rt Rt e (0,p2). If 0 < r < s K p ujing the optional stopping theorem at time v = Sv h Sa we see that =

which has

i)(tiiI2) k tz,.iRrb y

=

#(r2)1%(,$'r < &) + #(s2)(1= Prsr

Rearranginz gives

-

and hence satisqes

r/'(r)

1Ah1 Letting Rt 2 and Rt e (0,p2). If 0 < r = Sr A Sa we see that =

0 (1 e'(r))#(r) us tzk g (6 5) Aws pR,4 is a local supermartingale +

using the opsional

Kp

w(IzI2) k ErvRr)

while

ssopping

k vrzlprsr

<

thqorem

at

<

,S,)

Letting r

-+

0 and noting

w(r2)

(:x)

-+

Letting r

-+

(3.2)and

#.IT:

K w(lzIz)/w(r>)

0 yives

-z+

< x)

0 for

all z.

now extends

(ii)d

or

lzI<

0 for

>

the Markov property

(6.8) corotlary. I? (i) d k 3 =

-

-

Pz (Tn < c) Using

S,)

&)J

,)(Izl2) ss) k #(s a 4(r a ) )

#(r2)

0 and noting

time

Rearrapging gives Pcsr

pksv

<

=

-

.

< s

,)(s2)

'

<

=

p

the

2 and

c

conclupion

is Hlder

to

IzIk n. El

continuous

then

UThU where A Proof Referring to (4.2)in Chapter 5, we can write c(0) matrix with entries is a diagonal matrix with erries f Let r be the diagonal UVT. of and 1/ covariance stochmstic 1et The formula for the F' integrals f implies that Xt = I/a'z hms =

gives

.

#.ITO

<

Sa )

=

0 for 0 <

tu-l

ru

: :

lAhl

lu'L1

=

=

=

l

0 and for

To replace Sa by txnin the lmst equation let cb

zk = inftt > tu = inf (f >

IzI <

n

k

1 1et

(0)

PJ -+

-->

'rh+1)

< x)

=

0 for 0

<

1zI

< p

I

So we can without loss of generality suppose that c(0) f. In d k 3, the continuity of c implies detz) d as z 0 so we can take :(r) H 0 and the desired result follows from (6.6). If d 2 and c is Hlder continuous with 6(r) Cr5 J To the desired result from (6.6)we take extract exponent w can note that -+

-

=

=

p/j)

The strong Markov property implies that the n a are i.i.d. so o'n cxl as tral and hitting 0 in lkit the in is impossible fo 0 It to procss lcu, zt x. hms probability t)so we have k,,, #.IT:

77'c(0)F

=

-+

=

=

'exp /' cz,-z dz -

.

ad

-

/' (-vc'(1 exp

,,))

-

ad

-

x

244

Capter

One Dimensional Dihsins

6

in the integrand converges to a positive limit ms y

since the exponential

(6.7)that

It follows from

0.

1:l

a two dimensional diflksion can hit 0.

Exnmple 6.1. Suppose (z) H 0, the matrix with eigenvectors

c(0)

1,

=

1

,t?1 IzI(zl,za) =

-+

1)2

and

for 0

lzI<

<

1/e let

c(z)

These deqnitiops

take e(r)

j

1

1

=

Izj

2p/log

-

,/-

-

/,

=

atgdz

(f

''Ay -''

z

j

Processes

zl) Izl(-z2,

2 2p/log(lzl) so th>t detz) Plugging into the integral

are chosen

p/logtr).

j z/eexp

=

as Markov

1

=

and mssociated eigenvalues t

DiFusions

be

and

-

,/e

exp

(plog Ilog ,1)

l-pl lolplp

=lz

hezc we can

d

' .'.. ...

' ' .

.

ay

(f

Q

=

if P

ln this chapter unique solution says, we will be their symptotic

we will assume that the martingale problem MP(, c) has a and hence gives rise to a strong Markov process. As the title interested in difusions ms Markov processes, in particular, in behavior as x. -+

,

7.1. Semlgrups

hnd

The resultq in this mewsuryblr f, pd

Generators

'

hold for any M>rkov processj Xg Foy any bppnded 0, 1et

section

tk

.

. .

> 1

The problem of whether or not difusions can hit points hs ben Remvk. investigated in a dilerent guise by Gilbarg and Serrin (1956), see page 315.

'l'tle Markov .'

So taking

implies that

proprty

.

expecte

'

.

'

EE' .

za+tJ(z) for s,

.

values gives the semigrouy

k 0. Introducing the

we see that Tl i a contractlon

=

property

Ta(=y)(z)

IITII sup lT(z)I and

norm

L=

=

semigroup

on L=? i.e.,

=

< cc,1 (/' : 11J11

,

S 11T11 IIZTII

(1.2)

Followin Dynkin (1965), but using ;ne the domam of T to be

di/erent notation,

slightly

we will de-

,y,

p(T)

=

fl e L* : IIZJ Jll

As the next result shows, this is

-

a reasonable

-+

0 ms t

-+

0)

choice.

(1.3) Theorem. p(T) is a vector space, i.e., if h, A E O(T) and cz, ca E R, ch + (:2/2 e O(T). D(T) is closed. If f 6 O(T) then Ta; : Dl'j and s TaJ is continuous. -+

7 Dihsions

Capter

246

as Marov

Section 7.1 Semigroups and Generators

Pzocesses

Remark. Here and in what follows, limits are always taken in, and hence continuity is dellned for the uniform norm 11II. .

247

Kxnmple 1.1. The property in (1.4)looks innocent but it eliminates a number of operators. For example, when k k 3 is an integer we cannot have Af = J the th derivative. To prove t his we begin by noting that gkz) = 1 z2 + z ha,s a local muimum at 0. Prom this it follows that we can pick a small J > 0 and deflne a C* function fk k 0 that agrees with gk on (-J, 6) and hms a strict global maximum ther. Since A1k (0) EI > 0, this contradicts (1.4). -

,

follows Since 7ltcllt + c2J2)= cptllll + czt/lla the llrst conclsion with let t)(tT') closed, T(T) is inequalit To fn prove from the tiiangle e/3. O(T) pick .7 < large lf > Let 0. so llA Jll n we lu e llla ;II 0. < E/3 for t < (). Using the triangle inequality and the contraction Illrlia ln 11 property (1.2)now, it follows that for < tp Proof

,

--,

-

-

.

-!

=

-

+ IIQA All + Illa TII IIZJ TIIS 117l/ 71/nII E/3 E/3 + < e K llJ AII+ -

-

-

-

=

and cqp-

1

-

!

Prof Let f e 'p(T) and ga last deflnition, we note that

=

JJT,.Jds.

-zll

1111s

fl

c

T&f h

J

-

AJ Its domain O(.A) is the set of which there is a g so that

Before delving into properties that cannot be generators.

T.Af ds

=

Remark. Here we are diFerentiating and intejiting functions that take values in a Banach space, i.e., that map s 6 (0,x) into L*. Most of the proofs from calculus apply if you replace the absolpte value by the norm. Readers who want help justifying ur cofnputations can cnsult Dynkin (1965), Volme 1, pages 19-22, or Ethier and Kurtz (1986), pages 8=10.

is deflned by

of a semigroup

generator

%f

1im

=

At0

f for

as

the limit exists, that is, the

which

c

-+

0 since

f fpr

1

-

-

g

0

-.

as pz -+

J e p(T).

nga

.

Proof

1 e /(.)

Since zp is a maximum,

J(zc) k fyj 1Jl/'(za)

)

Now c+

c

%+s1 ds

=

0

=

TaTds

ga + tl

:'

i'

E

ln,-s -

'' .

,-

-

v-g

-

111 ds

-

0

/-+jjz'-y nyjj

y)j yt

s

.

.

for all y then

.J(zn) %0.

;(zo) K 0 so d.f(zc)

K 0.

ds

-

-

+

.

1

j

ds IITJT T11 -

-+

0

This shows that ga e 'p(.A) and Aga Taf f. Since the first calculation gala and DT) shows 0 the flrst claim follows. converges to f as c e =

-

for the

()

. (( .

(

of generators, we pause to identify some operators

and

the motivation

explain

J-llz)z- ttlds-

'

If

To

so we have

0

.

(1.4) Theorem.

Xxf

=

-

Rzirk. While T(T) is a reasonable choice it is certainly not the only ohe. A common alternative is deflne the domain to be Cft, the continous flmctions which converge to 0 at (y) equipped with the sup norm. When the semi-group We will not follow :Il maps Cz into itself, it is called a Feller semi-group. this approach since this assumption does not hold, fpr example, in d = 1 whrn (x) is an entrance boundary. The lnllniteslmal

f e 'p(.A) then Af e p(T),

t

Ttf

ll7l; T,JII llT,(TlsaT T)Ilf II:l-,r,T T11 =

If

.

.

-

of actual generators, we have

(1.5) Theorem. #(.) is dense in p(T). hd C D(X) 1)1 . d . WXf XXf :,

=

To prove th last claim in the theorem we note ttlat the semigroup traction properties, (1.1)and (1.2),imply that if s < t

Returning to properties

-

-.+

Section 7.1

Chapter 7 Dl/lbsloms as Afarov Pzocesses

248

141h6

The fact that Af e 'D(T) follows from (1.3)which implies Tsf p(T) and O(T) is closed. To prove that If e we note that the tion property and the fact that f 6 'p(.A) imply -

'p4d)

jjn71t-11

Ev (;(Xt-a)

jjm Ln-xyjjs jj s jjny -

J

()

-+

-

-

mxyjj s

jj jjm-, LrslyIIT'T 1 T xyj

Our flnal topic is an ipport>pt play a minoi role here. For each bounded measurable 1 by

.

p(z)

s

() fince J e O btatned by integrating the one

.-y

To make the conlwction

+

-

awd

(

jjm-stxy zxyljj

llxy nzyll

0

.->

-

e

p(r).

The

it.

abve

snal equatin

.

'

. .

Clearly

in

(

.

'

.

'

.

.

.

probleps

(1.6) Theorem.

J

.u'/e is a Pp martingale

with

gxg)

respect

fxz )

-

j

-

we

ad

To by Xt

memsurable

or each

is integrable

are bounded,

..r.

g)ytzl dt

e-ty(.m)

Ec

=

.

6

for

dt

1-1 onto ,1)(.A). tts inverse is

(

.A).

-

of U and changing

we have

*

e-tm+sy

..

that Uf

conclude

will

fact is that

subtle

nlldt

-

(:o

1

-

ds

to the flltration Fj generated

for some developments but operator 0 dehne the resolvent

concept >

J)//l. Plugging in the definition =

1 =

memsure la

Alxs)

Tr.J(p) dr

j

0

0 h 1 c = h

.u1

and prool-sincey A >

nf-s

=

Juzl

If

e-t

U maps /(T)

Proof Let Alf vriables s = t + li

will now prove 6 ,27(.A)then for any probability

p-, -

*

J

:z:

ll&'lllS lIJII/. A more

(1.7) Theorem.

'

(t.5j is

EI

between generators and martingale

249

drl

.J(Xr)

J0

'

..

A.f

ygyvj

-

avjj

-

S

+

-

Genertors

wkichis 0 by (1.5)

'

'n-sl

t-a

-

=

jjr;

j

-

= rJl-aT(!8 T(p)

SIl.4; and that the rkht drivativ of If The last equality shows that A%! is %Af. To see that the left derivative exists and has the sarri value w note t at jy

1Xz4

-

1

=

and

But for any g, it follows from the defnition of Tr and Fubini's theorem that

contra-

./)

mxyjj

-

Semigrpups

'2)(.)

-4

Taf ds

e and

Auf

0

f;

=

-a

f

-

T, J ds

e

-

we note that

.

sinceusl is '

r'

and compare the last two displays to

conclude

that S 11/11) (using1171J11

'

( .E'

( .. '

zpul 1.,--) M1 + s. -

By the Markov propezty the = Ex.

(.f(xt) j'

second

(J(.;V-,

lx-t

-

-

j

y)IIs c,y j

.z-,)

-

Ilxsgy (uy -

-

#

side

term on the right-hand

) fv'%) -

dr1

./(.;)

-

f-,

yyfyvj

ts;

+

is

J

()

1

(1 e-')1IJll ds + -t

.j

jn

c-,jj/jj

j

I

j

()

-

..

IIT'T Jll ds

0

-+

-

(

n-,jjyjjas

,

h

0. The formula for Auxf impltes th>t as and its inverse is -+

/

ds+

,

-

Ajlf

=

J.

Thus Ux is 1-1

Chapter 7 Dihsions

250

To prove that U is onto, 1et f

z (J

-

A14

=

*

d

j*

e-' w

0

e

= e-f

-/-

Integrating by parts

Ildt

Section 7.2 Examples

Afarkov Pzocesses

as

'p(.)

yjty

e-f

-

j*

d

e-z

7lJ(z)

pg dt

w

-

l

,

AJj

=

f

;

q

Markov processes >pd

-lte-

compute

-

1+

. .

(c-z

t)

-t

-+;0j

jj7lltz) J(z)

-+

1)

.

.

jpz',

.

dvgv

.p(z

,

jpz,

dyj is a

-'.

dvgy)

-

t as

-

ykj)

..-

.

1 , an d

=+

c

-

Let (S, Sj te a memurale txnmple 2.1. Pure Juinp Varkov proesses. x) be a transitlon iernel. That is, lnd let Q(z,.A) : S x c (0, space t$', each .A e (i) for x Q(z,.A) is memsurable (ii) for each z e S, .A Q(z,.) is a inite measure

o)y(p)+ o(t2)

pz,

jlytz)

1+

-

+ -f

of

jpz,

dylgly)

.

=

Sitce t -1 (c it follows that

j

e-

it follows that

and the proof is El

7.2. Examples ln this section we will consider two f>milies their generators.

c-AJ(z) +

=

-

0

-

E=1Xtj

7lT(z) lzj t

vje-tlldt

+ c

=

Doing a little arithmetic

G)

x

Wt;

Combining the last two displays shows complete.

=

Agj dt

pgdt-

(e-'mJ)

=

X%defnes our pute jump Mazkov process. To compute its generator we note that Ng suptn : Ts K ) has a Poksson distribution with mean t so if f is bounded and measurable

(1.5)implies

and note

251

o(t)

+

uausition mobability

ytzlljj

-+

c

0. Recalling the deqnition of Pat, dy) it follows that

-->

-+

(2.1)Theozem.

'p(.A)

'p(T)

=

=

L*

and

-+

(iii)Q(z,(z))

=

0

Al)

Intuitively Q(z,.A) gives the rate at Fhich our Markov charins makes juppp frpm since invksitle. z into and we exclude jumps from z to z with they are To be able to construct a process Xt a minimum of fuss, we will = supgE Q(z,S) < x. Deflne a transition probability P by supposethat

=

JQ(z,

dylly)

-

J(z))

..

.P(z,.) .P(z,

1 yQ(z,.)

=

1 lzJ) y( =

-

if z

(

.

Q(z,5'))

.

.

,

.

.

.

u

.

,

event.

Exnmple 2.2. Dlfrusion processes. Suppose we have a family of memsures Px on (C, C) so that under Pc the coordinate maps Xt () = f.J() give the unique solution to the MP(, c) starting from z, and St is the flltration generated by the XL Lt Cxa be the C j: functions with compact support. .

be i-i-d. eyponential with parameter that is, Pti > t) = e-'. Let tl t2, = .jza j; e a ar kov chain and let Tn Let Let Ta = tl + for 0. 1 + s n k with transition probability P(z, dyj (seee.g., Section 5.1 of burrett (1995) for defntion), and let a Xt = ya for t G gTa Ta +1 ) ,

We a:e now xeady for the main

(2.2) Theozrm. Suppose a and b are continuous Then Cx2 c @(.) and for a1l f Cx2 e Alz)

=

'j'

1

aitztlhjft

+

and MP(, c) is well ppged.

biztDilzt

253 Proof

Chapte 7 Diffusions as Afarov Processes If

J e C2 then It's T(-Y)

J(a')

-

formula implies

Let #(p)

Proof

t

J7 i

=

ds + locat zilart.

Dilunjbix''t

(.;%)

Ej(!& an)2.

=

1

+ ij7

Diilxstaitxst

(.X)

-

ds

+

' .

.' ..

.

the local martingale,

of times that reduce expcted yalue gives

:E

.

J(z)

-

stopping

the coecients and the bounded copvergence

zslxtt -

.

y(z)

't

=

4

y

+ J7 i,J

Ez

=E

step

are

bounded

-

is to bound the movement

t

j,

-+

For

-

zfltxa)

+

-

B T

+

W

g

ck(X,)Jds K ztrB

.4)

+

r? it follows that

=

Pzsc

.4)

K t) K ztvB +

result.

E1 -

thE

,

z

/

ff

,

-

J(z)

-

uytzlj

,

jtjxytx,) s tsa

fzj

0

=

-

and

Af

lz)

=

0. So

xytz)

t

.

using

suj.

TQ

%%

(2:3)again

E=1Xt4 t

ds

i C

sup Prs

gives that for

sup

DEOK

comes from' using the strong

wlierethe second equation time Tx. (2.4)shows that <

t)

sup Pzsj %:DEff

)

<

< @) -+

0

as

z

/

ff

< /)

s cpcvn

-

r8K

mxytzlj

-

K 6 + 2C

E=fXt4

the diflksion process.

K t) G t 2 .

i

-

I

()

-

sup Prsr

E

r which is the desired

Erf-)

=

=zK

J-l(2(X,

..

0

-

(2.4)Lemma. tet r > 0, Jf c: Rd be compact,cft) and ffr (p : lz pl S z e ff ). Suppose 1(z)l% B and Ef % .A for all z ffr. If r frletsome r) : we Sr inf (t l.X A Ik then =

v

=

.E%.f(-Y,)aJ(A%) ds

xyjj

-

of

2cf(X,) ds 0

Proof of (2.2) Pick R so that Jf (z : lzI S R 1J contains the support : coeciets of f and 1et ff Since S are J e cx2 and (z IzI AJ. and hms compct continuous, Af is continuous This implies Af is support. uniformly continuous, that is, given : > 0 we can pick E (0,1) so that if lz pl < &then I.4J(z) .A/*(p)1< e7. If we let C = supz l..J(z)Iand use (2.3) it follows that for z E ff

ds

tl

J(z)

J-l i

=

Loilx-tbix-,

jo'zyt-v-lds

=

'f

E=

=

Since /1 k 0 and ltXs.)

now we have to show that

conclusion

jjEzfxt) The first

'

5-.7./,, J .j.

To prove the

ds

then the integrands theorem implies

1

(2.3)

fat-xasr)

.

are continuous

E .

j

..-+

ds

Dit.f-nlaijvn)

;

f e G%2and

xilbi (-Y,) ds + local mart.

Letting Ta be a sequence of tims that reduce the local martingale, stopping at t h Sr A Ta taking expected value, then letting n x and invoking the bounded convergence theorem we have

tAT,X

.

If

-

:

0

+ 1J7.&. 0 i

2(.X'>

0

,

a/(.X,)(-Y,)

Er

=

1

have

we

.

tATa

Ezlunsn

-z

formula

253

'1

J7

=

0

'

Using It's

-

0

'

lf we 1et Tn be a sequence at time t A Tn and taking

Section 7.2 Examples

Pys

<

t)

Markov property t

-+

0

at

Chapter 7 DiEusions

254 Since

0 is

>

f

Afarkov Pzocesses

as

Section 7.3 Diferentiating

the desired result follows.

arbitrary,

For general b and a it is a hopelessly dicult prpblem to compute ,27(.A). To make this point we will now use (1.7)to compute the exact domains of the semi-group and of the generator for one dirpensional Brownian motion. Let C be the functions f on R that are boupded >nd uniformly continuous. Let Cu be the functions 1 on R so that 1, #, .fM 6 Cu .

(2.5) Theorem. Afzj

For Brownian motion in d for al1 J e '2:)(.A).

1, D(T)

=

Cu,

=

Cu2 and

'p(.)

=

,

r(z)/2

=

=

sense

X

l:1T(z) TtJ(p)1K 11T11Ipttz,z)

-

J

my,

The right-hand side only depends on lz pl and converges to 0 as lz !/1 0 so lf E Cu If J E O(T) then ll1JlJ J11 0. We claim that this imylies < E/3 then pzck f e Cu-. To check this, pick E > 0, let t > 0 so that IIZJ pl < then l7lJ(z) Tl.f(p)l < EI3.Using the trangle inequality so that if Iz now it follows that if lz pl < J then

e-x-evy)

'Zi'l /

dy

2J(z)

-

J-x

lt is emsy to use the dominated convergence theorez to justify diferentiating the integrals. From the formulas it is easy to see that g, g/, f e Cu so # '6 /2u and gtgqyj .:y(z)

.:,(z)

#/

gtz)

=

-

(z)/2.

J(z) by the

(1.7),it

proof of

follows that .A#(z)

=

.

To complete the proof now deine a function f E Cu by

we need

/(z) lt

=

-

Uf

to show 1

tz)

=

=

u

j

D.

,1)4.)

%2.Let

E

Q

and

(y;. .

satisfles

1 'i'y (z) ,?

-.+

-

-

dy +

Jl

The function y

z)l dz

255

!

pT

e-u-rsvi

pflz, z) for the Brownian Proof If t > 0 and f is bounded then writin transition probability and using the triangle inequality we have -

Probabilities

again gives

Vi-'

Since .A:(z) is larger than Cu2and consists of distribution and lies in Cu

For Brownian motion in d > 1, T(-)

Remark. of the functiops so that Lf exists in the (seeRevuz and Yor (1991), p. 226). .

g''j

Aanslon

ptz)

-

0

=

-+

-

.

wyyy

.f11

-

-

-

A11solutions of this diflrential equation Since / Ux1 is boundett it follows that

.ry

have the form + Be Uvf and the proof is complete. -e

=

-

.

(:1

-

,

IJ(z) J(p)l K IT(z) TtT(z)l+ I7lT(z) TlJ(p)I+ ITlT(p) -

-

To prove that -p(.A)

implies

Ufj

=

ct =

me

Our frst tmsk is to show that if we have

1 E Cu then

jn

gzt (2) -,/a =

e-tp-xlzi-.ytpl

dy +

(2)-,/2

(r = r

e-u-rtvby)

jr

=

e-@-p)ziIy(g)

Uxlkj

dy

r -X

e-(=-OYiQ'J(!4dy

a and

that if

b are

continuous

and

MP(,

c)

is well

d

forJ e cx2or ,

z)

'p(t,

changing notation

=

T1J(z), we have

d'v -dt z.p(j,z)

(2.1)

=

-X

'

-

we saw

-jX.f(z) = ZXT(z)

and noting that the terms from diferen-

dy

In the last section posed then

Probabilities

dy

Uvjf E C,. Letting g(z)

T

Differentiating once with respect to tiating the limits cancel, we have

'(z)

j

c-Ir-#IVWy(p)

7.3. Transition

< e

(3.10)in chapter3

begin by observing that

(2)-1/2

.f(p)1

-

-

where .A acts in the z variable. In the previouj discussion we have gone from the process to the p.d.. We will now turn around and go the other way. Consider

(a) e C'1.2 and (b) u is continuous 'u

(3.2)

'uf

=

on

Lu in

(0,x)

(0,x)

x Rd

x Rd and

,u(0,

z)

=

J(z)

Captez

256 Here

we

7 Dihsions

have returned L

=

as

Afarov Processes

Section 7.3

to our old notation 1

so (a) tells us that that u(t s, Xsj implies -+

J7cfJ(z)ZkJ(z) + J7(z)D.f(z)

'*'

ij

-

j7

To explain the dual notation note that (3.1)says that r(t, 6 ,i7(.A)and satisqes the equality, while (a)of (3.2)asks for u E C1l2. As for the boundary condition (b) hote that J e z)4.) c 'p(T) implies that 11Tt1 JlIc.7.. 0 ms $ 0.

,utt -

fxjj

s, .Xa) is a bounded martingale ms s t, so the martingale -+

utj,

.)

-+

-

.

To prove the existence of solutions

(3.2)we

to

Section 4.4.

(3.3)Theorem. Suppose a and b are bounded is, there are constqnts 0 < J, C < (x7 so that

n'ct Suppose also that

IcfJ(z)

fOr

a1l z 1 y

-

cj.(,)Is cIz

a is uniformly

p1,

-

Itz)

elliptic,

-

tplls'elz

that is, there is

p1,

lpI2

k)

Then there is a function ntz, p) > 0 jointly continuous in t > 0, z, y, and C2 which atisfles acting of dpldt function Lp L z on the z variable) (with as a so-that continuous bounded >nd if is aud J

theorem

o

Izl

=

Js

pf (z,p) dy

n(z, y4 is the trnnsition

density for our disusion process. Lt D J0. To obtain result for unbounded coecients, we will consider

<

(3.5)

i,i

(0,). (b)implies

on cnvergence

swytx)

=

.P.l.X g B )

(a) E 1'2 apd duldt Lu in (0,x) x D (b) is continuolp on (0,x) x b with and u(t, y4 0 when t > 0, y E OD 'u

--lptl.tzlpl

CUE,

257

=

i.e., so that

a constant

z)

Probabilities

solutlon of the parabolic equation tzt pj (z,yj is called a fundnmental solutions L since it can be used to produce for any bounded continuous initial data J. Its importance for us is that (3.4)implies that

That

continuous.

=

.

(1j75),

turn to Friedman

and H3lder

.

'Iansition

=

(z :

=

'u(0,

'?z

z)

=

J(z),

=

=

'tl(, z) satisqes(3.2).The

maxnum

To m>ke the connection

ptlz,

=

the SDE

with

It's ut

-

Erfhh

=

z)l K 11J11. l'tzlt,

solution

s, X,)

-

A)

'

=

du dt

--(t

to

(3.5)we

turn to Dynkin

Voly II, (1065),

and suppose that (HC) and (UE) (3.6) Theorem. Let D = (z : 1z1K ptRt, y) > 0 jointly continuous in t > 0, hold in D. Then there is a function C2 as a function of z which satis:es dpRldt Lp (withL cting on D, y z, e the z variable) and so that if )' is bounded and continuous =

of MP(,

c).

If u is a

utt,z)

)

formula implies 1z40,

of solutions

.0

we prove

a nonexplosive

'utt, z) Proof

implies

principle

(3.4) Theorem. Suppose XL is bounded solution of (3.2)then

:)T(:) dy

To prove exiqtepce pages 230-231.

/p1'(z,

p)T(p)dy

=

satisqes(3.5). -

r,.Xr) dr

the' connection

To mak

with the SDE we pzove

4

+J-1 i

0

ofzlt-vrlftxr)dr + Ditnxlaitxr)

-u

0

dv

of MP(, c) and let r

(3.7) Theorem. Suppose Xt is any solution D). If is any solution of (3.5)then /

XL

.1

1

+ )-l ij

local mart.

'tz

stj,

z)

=

>;.(y(a'

);

'?-

>

t)

=

inftt

:

258

Chapber 7 Dihsions

X implies

Proqf The fact that u is ontinuous in (0,t) x It's formula implies that for 0 % s < t A r u(t

-

s, aYal

-

u(0,

A%)

Section 7.4 Brris

Marov Processes

'as

'

=

du

--a

0

+

)C 0 1

.&)

-

dv

r,

.;

+ y''l -2

(t

it is bounded there.

DutxrlftA%l

dr + local mart.

-+

-+

l'.(J(.A$); r

=

>

.,4

E S, z

-+

for fzxed z 6 S, .A

-

u(i,z)

for fxed

dv

-

-

the next, we will mssume that the reader is familiar with the bmsic theory of Markoy chains on a countable state space as explained for example in the irst five sections of Chapter 5 of Durrett (1995). This section is a close relative of Section 5.6 there. We will formulate the results here for a transltlon probablllty deqned s). Intuitively, .P(Ak+1 e dl-'k = z) = ptz, d). on a memsurable space S, Formally, it is a function p : S x J' (0,1) that satisfles '

so (a)tells us that u(@ s, Xs) is a bounded local martingale on implies that 1It 0 as s 1 r and u(t ).X,) s, A%) s, .X'al ('r > t). Using (2.7)from Chapter 2 now, we have

(0,tA r). as s

t)

259

-+

#

Dituxrlaqxr)

Chains

-.+

(b)

In our

applications

-->

ptz, .A) is memsurable ptz, .A) is a proba j; ility memsure

to diflksions

t on

n

we will

ptz,.A)

=

take S

Jx

pztz,

=

Rd,

and

1et

f) dy

where pttz, y4 is the transition probability introduced in the previous section. By taking t = 1 Fe will be able to investigate the asymptotic behavior of Xn (x7 through the integers but this and the Markov property will allow us n as real numbers. to get zesults for t thmugh the x We say that a Markov chain Ak with transition probability p is a Hrrls chnln if we can flnd sets .A,B E &, a function q with (z,p) k 6 > 0 for z G y C B an d a p robabiltty measure p concentrated on B so that: (i) If inftn k 0 : Xn 6 .4J,then Pz < x) > 0 for a1l z a eS =+

Combining

(3.6)and (3.7)we

-+

see that

.4,

Emlxj Lettig R 1 (x) and ptt, y) ;NR(z1 #) R is increasing, -+

r > f) =

we

lima-xpRt

=

Jo

pllz,

p)#(p) dy

(z,y), which

have

,

(3.7)implies

(3.8) Theorez. Suppose that the ma4tingale problem for MP(, c) is well and satisfy (HC) and (UE) hold locally. That is, they posed that c and hold in (z : 1z1 K JO for any R < x. Thn for each t > 0 there a lower semicontinuous function p(t, z, y) > 0 so that if f is bounded and continuos then pflz, yj dy Exfxtj .

'is

=

Remark. The energetic reader can probably show that pt (z,y) is continuous. However, lower semicontinuity implies that pttz, p) is bounded away from 0 on compact sets and this will be enough for our results in Sectiop 7.5.

7.4. Harrls

Chalns

trx Jc

=

exists since

.

ln this section we will give a quick treatment of the theory of Harris chains to prepare for applications to difusions in the next section. In this section and

(ii) If z e and C B then ptz, Cj k qlz, #)p(dp) c In the difl-sis we consider we can take X = B to be a ball With radius r, p to be a constant c times Lebesgue measure restricted to B, and q, y) mtz, #)/c. See (5.2)below. It is interesting to note that the new thory still contains most of the o1d one as a special cmse. .4

=f

Eknmple 4.1. Countable State Sace. Supppse Xu is a Mrkok chal on state space S. In order for Xrt to be ountabl a > Hryis hai!z it is eqessary with Pzt-''a and sucient that there be a state u fi some rl >- 0) % 0 for a1l z 6 S. 'tz

=

pick r so that pty, r) > 0. If we let Proof To prove suciency, and B then hold. To prove ncessity, 1et ('u) be ('p) (i) (ii) /7((u)) > 0 and note that for a11 z =

.

Pkn

=

u

for

some

n

k 0) k .E'.((vYzz)p((u)))

The developments in this section make the theory of Markov chains on a

>

,4

=

(zIJand

a point

with

0

l::l

are bmsed on two simple idems. To countable state space work, all we need

(i)

E(' 115,:* .1:::::,,* 1111::* .1:::;1!!-* 11:::::.* .1::1111:.* j'.1::1111)* :,,' 11::::211::.* .1:!5115;;2.'* 11r::!:;' .1:::111!'* 11(:::211,:.* dl;:!!iii;l?'d 11!:::21.::7* ilts:'d :,,' 11:::::1,* .1211,::* :jj' ,,,' :,,' ;::;i!:::,' 41:::2)11::.* ::::::::::::::' klp!i,lk--d ::;:::::::::::' ;::;1!:::-* ;:::i!:::!' :,i' ;:;;1!:2:y* ;:::i!:::?' ;:;;i!:::,' !,,' dl!:::qlll;;?.'d .11::::((::;7.,* :,,' ;:::1!:::,* ::::::::::::::' ;::;i!:::t' !y' 41::1111(.* ;::;1!:::,* ils)lk.,d :bii!p;d 1I!::::Ii:!'' ..jI!::;,'' 'E11:':;IL.' .,,,' ,;;i!!::;Ii' ilt::!qllli;?''d (::;1.:::-* j'y':111:':111.* :llEsllk.k:,i!,;d (J1rq')I;q'r!1i.' (iii,d il::l)d ..-ii!!EEEE::)I'' 'llkzird ;I!EiI1;-;:Ei!;;' 112E52::* 'lIi:E::1,' (:11::--* ,111:;.* :i1!F::I,:' 11I:Eii!!.' ;1pE5i1k,,' 11::4)* jyd '1EEiE:'' 411:)* ,,.;i!!E(((:;;I'' .,.ii!!55((E;;I'' :1I(r7'' ,,-;i!!E((E:;1I'' (:11::-.* dllsE'd 111:4* (11(:7-* (iii.d ()1(::-* j't')'k';:!!5!1t..' jyyd .j;jjiijii;' ((.yjj' -',,' .j' 4512:.r:11:.::1k.* ,-,' -11:;,* .-,' ,;i!jiiii;7' .:1!:,;,.* .4,* .,j,' .,,,' .-.!:111::* ,--' -.41!:::;1.* .-.' y.' ,jj,,,' -,r::,,-,::j,,' ,;jj,,,,-' -j4!!::;p'' ..-1144)* .-,' .;j!ji!ii;7' ,-,.jj!'''''' ....'.........' ,...t-.111!1...* ---' ,....-..:1!1.,.* .''..'.'.'....'

:,11::-* (::11...* .:i!:,;j'' )'.1Ik!!!!IIj@;'' $' jiii,d .j' j'y' rjll::-d -;!!i!!I;7' liid (:!IIIEEE)-.,' '1IE::::,h' 11!::::((E:27'-* .1!::::E2E:;71'* (()1(..* -1,;:115E;12-* iii.d (411:.2,1;.* (' y' .14:::E2(2:;?1,* .-;i!iiiii;7' ::1IEi7'' qlllEd )' (lli::llk.d ':IIii''' 11!:::E(:(:271,* tltd .;i!jiiii;7' ((::1i' t':lllq':ll:r':llk.d ((:11i' (2::11* ((:),' ((()1h..' ((:)1i' (4:)1h* ((::)p' ((::)i' ..'IlE()' .'j'(tf r'-1!::::1h:!'' .,,,' -112!1,11-.* jjy:::,,d -j,::,k-' ,'.i!!!::;1''

.' Chapter 7 DiEusions

260

.112:;.

!11::k.

dl;::::p,l

-jj1!::;;,.

' dlEEEEEEllii

.;jjiiiii;7'

;:;;i!:::t. :II

11::(. :,,. -;:;!!::r'

l!llEq' ':1IiE7''

11::::)1.::. :.I

.;1!!E11--'' 11!::::: 11:::::,,1 !1E::::E

4t

:

:II

also follows easily 'rhesecondthenequality then this is the same n d r:ll:r':ll:r'll-.'1It:k. ()1iy'r1h,. .111:k,

'112:k-

:jiisl. 11EEjj:,. )ll:!-. .!II,j:1Ik. (II:':1Ik. IIIEEE:::

',!i.;''i.;p-'

41:::::.

':jlii'''lll::'.

-

.

. EE ( .. . y.( j jg:: yy y j yy E ErqEil,,jli,rjjiijljrllylyyk,jjyjyji.,y,,g :

:.

.

g

-

qj k g;. ( j j .y jj y r

:

. .

:

: :

.

.

.j . .. jj gjjj g Ei EE. Ei. i l g q E(r q !.. !y .!.jjyjjyyqyyyiy.yy!y(yyyyyyyjyyyg.jyy. E E j;.iEji. j 'y @ !j j .j .. . .. q:. q y . yjqy y g j ..y y j y ( . .q. yyjE-yyjjyjj. jj,jj;yy y g j.yjy y j.y...g.j.....yyy..y.y..y 'E. .'jEjy ; j ' E y yyj .Ejy E.q . E.! i . ! i . ( q . .E..j...(.ri....;).q.qiy..)yj yyjjj.,(jyj.....-jyjy.yqj ..Er y.. j . ;yy!j(!yqjjyj..;;yjyy. y.jj.,yj..,,.j, y.. (yy ( .. y j.jj.jjy.y..jy.yyyg..y ' .. ..yy..q. gjj;yyjgjy y j . .'. . Ej..iyjj;jyjy j j. j y.jgy.j. . . jjjyjjjyg . . ... y. j.q!qjjqgyyyyyjjjy yyijgyjjjyyjj..j'jj.jy.;j.ygy.y.jy.y. ( ; ;yrr.iqi,rit,ykj,yyjygyl.yk.,y.jyj,,.,gy,yjj . . yg .(r..jyjyyj)yyyqyjjkjy jjjg j jy..j-jyyyj.jy. y . j y y y j y y j yyy j yyyjj y ,l,E,E,iyq.ljqgyjyyyy,jjyj,jjjjyyjgj,jy,yyy,yj . . yjj yjjj jy yjyj ygyyy. y.. yyy.jyyjEy jgjyjy . .,'.'.:.,,(,..;y.).i(,.,p.,i..,.,jpii,,.,y;,,k.,..ykyy,,,.

y . . y y, . )' j)j y. jy ... ))ikyy)(.-(y .j . -

-

-. . .- -

-

E-

. . -

-

-

-

. -- -

.

:

.

.

.

.

y g. - -.

.

-

--

-

. -

- -

-.

.

.

.

E -(-'.

-

.

.

.. E. -

j(j. j.

:. : y . :y . - . - -g --y -; - . . y---. - E .

-

-. - . .. . E

.

,

,

'

E (;E .. . . !

!

-y..

y

.

.-.

E..

.

:

E. y yy

. y

-

Pv'z

making

6

'p.

-

-y

y yy y

jg

..

-

E!

.

. y .. -

.

.

.

-

-y

(

-

.

y

.y

.)

-

- -

-

,

-

.

-

E .. - .

-- -

-.E

. . -.E . . E-.E j E..EE. .. yy. yy . -: y: . ... E E.E E -yy. . : y -. ..g . gk gy y g j . g g g.j.j j .g g . ig g. -y y y y j. g -i -

.

E

.

-

.

- . .

'g . .g

-

.: , -. g . .. . .-

.

q

-

.

,

.: - q ! .y. . E . . . . ..y .. -

.

.

-

-

-

E.

.

.

.

-

y

y

-.

'

'

.

y...

-

EE

-

y. y

. ,

.

.

12

,

..

.

g

j j .

-.

-

.

-

(0: l

-

-

-

-

'Ey,Erlqy,yyjyp.l.qijyjijjpjjjj,yyiyjyyqjj,yr,yjj,jjyy,yjyjy,qy.,ly,qjy

. . ..

j jjy,

y. ;-.

-

.

.

-

--

-

.

.

-

- - --

-

-

- - -

. .

- .

.

.

y j: ( ) rjy j jj j j rr . y l (!y.(y . .. rjyjy . j .jj. .j.j)j.y jyjy . . j.jyyjyjj.;jjyj'y jj j j yy.y . .jj ;g.jyjyjjjyjjyy .. yjjij:jyjqi(jj(jyy j ) j yy jj yy y.jy;yjy.yj . j.j

j

.

p' is the transition probability Lebesgue meaure on B4 but

.j.jg y.;yyj,yyyj. y.yj gyj j. yyyyyjyjyjgyjy y..E.jjlyjyjjjjijlyjjygljjjjyyjjgk,,ijyjyyEyjyq,jjjyjyjyyjyjjyjyyyjEyyyyj . . y q( (y.,Eiyylty,ylljyiyyy,jjjij.jq.yj/,ji,..,yyjjyy,yjjy,ljjggyglyyyjlklgl,jyyj,,,y,, ....jj.j)(y ...

. . . . - ...

,jj,.,,,:.,,,-*.

.

(4.2),and

in

. E

y - . .

j ,.

y

. y y. ' i. i . E

-

.

.

-

as

vk

-

-y -. - . .

i

y. .

'

.

::.z

.

-

..,jj:::,:,

=

.

. .

.

Xn and R are constructed roof bserve tltaf if and S) 1 then a'Y Xa is obtained from Xo according to

'y

-

.. E.

E

.

.,j,,,

(,%S4 then .G/(#s) on

E. : . EE E i. EE : E E.(yr,.)E;ty)jEij,.j(,;gy.;)'i)y'(!jrj,gyj : yy

. .

-

E

.

.,,jr,,,,

eS

by . ... yj..) jyi(. a transition j j '.'rqy!(lik;yjjryjt(yj;y)y.jyy:((y.r(y.j..;.(.jj. ;r...,,,.Ir.r(,;y,.,,r!;j.y,j)I,qjrq(,yyyyyyyyj . (. ry.. . (y j y c1 , r .. '...E'.((gry..j).( (yjj:jgijqg.j',yjgE..y.y.j...y..j . yjy j.. . . Before deyloping the theory we give one example to explain why some of E) (yj gq;j j! ly yy;jj (yjyy .jq.jjyyyy ,...,..j,.y,.,rj.(..,,,,,.j,.,y,.,,....,.y.y.,,,.,.,..,,,.y.. . )y;;.y(ry .;yy.jyg j..j.j.. h t t a temnts to come will be messy. e s i . i !g)jjyjjjrjyjjj jy jg y.. ...yyj.j.y.jjyjgjjgj;yjjjjgy j.jy .yy.yjjgj.jgjjjjgygg j y..jy y j g g j . ..y.E. .(yj.yyjyjyjyy iy'Eyjr.;;yy(.jjjIg)jlj!.yyy .. i). Exnmple 4.2. Perverted Brownian motion. y. '.:jyj. i)j(j;y For z that is not an integer gj.jj .;;;.(j..j.jyqyj.yyy.jj.j E. E ( k :jy q ; ;.:g ) . ..:l j . j j q(,.y.j.y. . k 2, let ptz, be the transition probability of a one dipensional Brownian .....;. y.qyyy.;yy...;yyj,yjjyj)yjj.gjjy.jyyyj...jyjyy.y.yyj.yy.yy y;.q. ;!ipyq,y motion. When z k 2 is an integer 1et yyjyrjj. ,y.jy.yyy(.y. i yy .r.. ,..yj.j.. '..'j'r.;)(.i.!(;;g)::)yyr)y!;;.rj;yry yjj..j r yjgjy j .y yjy.yy j . y y . j . j . y .;'('.;.y;y!jy!yj;yy(yjj;j(jqj.j.y.g!y.j.y.j.y... ' ' . j( F(21, + 1l) = 1 z-2 . . ... (.j( ; j, y y . . ... .jj .jy(jjjj.jjj .j.j. j. jj gy.-y. . i .jy (. j j y y j.jjy j.g ;y jjjjyjyjyjr;jyjy.yjjjjyj.y jyjjjy .y ..,y,,..,.y.y,y..jj.,yyj.j.y.y,y,j.j;,yjyyyyyjyyy.jyg.ygyy.y,.,,..jy ptz, Cj = z-2IC r) 1) if z + 1 C yg j.jjgyy.gy ........',.!.l..;q.,.,.,j,!jr.yI,yj(!yjqy.y.,,,;.y,j,..yjj;yr,y....(........, yy ( ( j(;j.;jjy j,y j gyj E . Ey y i iy y

'

..'.. j .g

.

=

. - ..

.

.

-

.

-

.

Eplxnt

:

-

:

-

.

'

mobabilitymemsure

. -.

z

1 dp

.

- -.- -

.

for

-

.

E-

. .

=

g

--

-

-ty

.

.

:

..

-

If p is a

/(a)

J(z)

:

,

.

.

(4.3) Lemma.

=

,

,

.

'

-

J(z)

:

-

. .

'

-- .. -- .

.

-

.

.

--. !

yj .. j ' j. ... .gyjyy . .. j . .. yyyy. y.y y

il:EE!!EEE!i rillq'''ll:::::,,-

.

-

,j.jj.yy,j.

. -

.

.. .

-

-

).

.

.

' : . .

.

-

y

-

-

:

.

.

.

.

:

.

.

(4.2) shows that there is an intimate relationship between the asymptpttc behavior of Xn ad of Xn To quantify this we need a deflnition. lf f is a bounded measprable function on S, 1et j' = ?g/', i.e.,

qq ,rqyqylyqiyiljjiEjiyjyljyjljl'yjyjy,yjylj.jEyyyrjy.yyyijy : ( jy y. y

(Illi'''lll::::l,11::EE!EEE!E

.

.

... r..).!jj y.j y yjj. j.jjj jj.jjg

i !EE E (EikgjyqrrylEgrqjiryjjjgjjyqyy,yjyy ;. . y - . y y.. . E.' ( ( ( . g. . q : y,jEyyqjlj.jjEpjilyjj,yljy.yryyyy,, .( y: yy yyy

- ' . ..

.

j

:

tnstead

#at.Xk

of a Harris chain

jjrjjgygjjyjyggjjjrjjjyjjyjjj,jyjyygjjgyjyjggyrjjjjjjyjjyyyyjyjyyyjyjg,jj g g = .. . g. jyyyy gy . : . : .EE.EE.i(.: Ei-ig! : y E!r E . g qjjj gy . !((.i.(E.!(y g ;y ..yrq -y y.y- gy .j k.g . . ' (. ' j'g(Eg!....j.jggE;.gg.!.j.;Ej.j:jy.(jjq.y.gy.rjjq y . y y j . .. . y gi.j ;y;... jg.y E .. - .(q .E y. yr( y-yy - .y .gE.E.g ;.y jjy.j((..q.- il.jlj..r.l j ( j. j y-y gyq;yjgjgyj :..tyt jjjy j . yjg .ygtsin -. E i. Ei E -. y(. (,;jy- -!y j y y y y- y.yg. .(....,,q...ygyyjy,jgjyyjj,,j,jjtyjjjjjyy.y.yygj.j g . y j y yaaaayygyyayyyyyyyy.yyyayygyyyyyyyy.yyyyyyyyyyyyayayyyyyyyjyyyyyyyyyayyyyayyyyyayy g

.

.

.

.

..

.

n

+ 2 for

/

(takeW

all n) >

=

B

=

(0,1),

p

z

0

-

:

:

-

.

.j

..

.

. .

,

:

.

.

.

(2:1,1 ,. dennition.

'

.

.

.

.

.

.

.

.

.t

-y

-

. .

-

t'y jj.j.jjgyj..jj

-

from the ms one trnsition

.1121,.

.11@:;.

j.

y

.

11::f1112.

(:11(::111.. ilii!ll il!ii!lk,. Tllr:':ll.. (:11::7.

y

-

y

.

=

-

:

. y

-

-

,

=

!::::::::::::;.

.

. .

.... gy.. jj(. j.j jgg j . . . (tg'(!

Before giving the proof we would like to remind the reader that ma- . .,,y. .:y, on the zezt, t.e., . zn . .oue nrqv cuww sures , multiply the transition probability . accorclh: . want t show pr/ to anu tneh yp. lf we frst make a transition S O C one according to #j this amollllts to OI1e tllaflsititm accordill Eiil. ;11!1!1k-. :12!511--. IIE!E!;. :1!E5,1k-.:1EiE5;. '1It:k, (::!,p?,''' 'lliF:'llir''.liis!r. ...ki!!EE(:(;:1''

:

-

.

-

E.

.

.;j!r::;p,.

11::(( 11::4) ::;iIr:2:'

- .

y

,;jjqiiji;'

il::iill;.

!

.

.

.

. :

.EIIii''' 11::E!j4E44 jl:))i (jii). ..-1Iq(' tll)ji.. 14::() 11::4) -:1!:-;,-((:1,1 (q:1p.

Proof

-

-

:

.

.

:j,

.:10:,;,1

:. :

.

11::4* :::i1!2::-

11::4( ':11:':11..

,;jijiji;:

':i!:-;pI

.

.

.

''

--,-

-,-

:

-

-

,1:::::)1-

:211r:::111::::11k.. :111:2::11r::2111.. 414:(). (q:),1

.

-

-

fen

2111,. .

:.E :

-

Here and in what follows, we will reserve J1 and B for the special sets that elements . will of S. We ocur in the deflnition and use C and D for generic simplify notatiou by writing #(z, c,) instead o >(z, f.)), ?z(a) O o 11::(( 41::::)1.::. -ij1!L,-iI:. 'dI@:;. ((:1,1 Our first step is to prove three technical lemmas that will help us cafry out the proofs below. Deflne a transition probability )) by

.

.

:

-

-

,

.

=

tj::ll. ((:111

11::(

..

:

.

,.

-,41t:::;,-

:::::::::::::::

.

.

.1(::::(2E:;?.-

,:jjji@2i;7'

:::::::::::::::

:

- :.

.

-

-

:II

::::::2:::::::.

-

.. . -.. : : y .. : . .. : : .

.

:

;:;i1!::)'

.;;@4r:;;,-

::::::::::::::E

.

.

':lliir''dl:::::),, 11::().1(:::EE((::71il:EEE!i!!i (2:),1 .. ((11i.' 11::44 !1(:q(

.;;i!ii2ii;'

-.

-

.

-. .. : . . :y

=

!II

.. -.

:

. : .

,'

..!!iiEi;r'' 11::(( '2111((( 11::EEEEEEE1

-

:

--i-i-hi-hit. -I,-i- -.---t-h------p-i-ti(ii),,, -u---iu---t-z--p--ttbeing which corresponds such point below) in B manufacture to a a (called rlll;;. occmsionally :1!111;.'1Ii::::,,. out needed plan notation this iiii. is '1IL,(:I1,. The to carry ''1i';.'ik;,'' obnoxious but we hope these words of wisdom will help the reader stay focused . . on the simple ideas that underlie these developments. with chain cha will construct Markov Xa Given a Harris a on susj, we (S, B U (a) : transition probability # on (,V,Sj where ,V S U (aJ and J' B e d). Thinking of a ms corresponding to being on B with distribution p, we .. . deflne the modifed transition probability as follows:

i-

y( . lj$jj4jy14j)j)jjy4j ..k... y .. Section 7.4 Harris Chains 261 l'( y (. ( y . .y. . (4.2) Lemma. Let 'F,zbe an inhomogeneous Markov chain with p2k z? and l nk-j.l zp #. Then Xa = 'Far,is a Markov chain with tr>nsition probability # and Xn = F2s+z is a Markov chain with transition probability p. (, .

-

Afarkov Processes

us

' :.E' : .-

! (' 'EE.EE(E@ ..'...(gj':;jjij). (' .; E' ;yqyry . yy ry y,g .jy y ( ' .. Eg.Ey j j j(jj.jg y jjyg.yyy. j.jq..jy..j. . .((E. .EEj;.(jjjy . y .; yjjyj.j;jg !y.qy y ; jy j....y..yg.jjjy.....y.j.yy. (y.;'..g. yy;ijj)y qiiEljlyyl'.yil,'lyElliyyjyjjjj,E,yjjgyljyq.Ejyryq y j ... .. .... yy(jjjjj.qy...jy.. j..jy.. ..y. i (E(Ej) :. ....(.(E-.y;..E. jyiE:;;ErIyyj.;i.(.y..jyqy;yy..Egy.jy..y... yy . j y y ) ',......,..,,..,...,,,.r,.,,,,,,;,,,,,,,,,,, jj . .j r j . ..j jjjyrjjt. .y j.y.j.:.y . i y.. E; E E j E ( ( ( .,'E'E'Fi:El(:',k.iE,i;jij)krijyjjjyy,,y(,,! ...( .yj.yy(jjyjy;jj)yyjjy. jjjj..(y.(.g ( ( ;( .r,!','iii,;;':;l)l'qi,k,(y.jj.:,j',!.(.,((. . ... j ).( jg(yjijjyj jyj.y.y.y.j .. yyj(.jy yy yEEEy('Ej;l((y,yr.rE).y),yjjEyjq,jy yjjgyy y y.yjyjjyy.jjy j.jyy.j....y.... yy .y.y.y.. jj'j . g y q j . y .. y y EE,'llqijlylEyrirElkyjyjy,,jyjyrylyyj.,,ljyyjlr,yyyyjy jyjy j gy . .....gjr.gy;gyjyyjj.. gy.y..yyg

.

-

-

.

In words, if # acts to p since n

according

.

-

I can sympathize with the reader who thinks that such crazy chains will not arise xpplicationsj'' but it seems easier (andbetter) to adapt the theory to include them than to modify the ssumptions to exclude them.

jyyyyy y;.;.. gyy yyi.y..jl,..irjr'jiiyljyjy,jpyjjjy,,y,ryriyylyjjyyyry,jjj,,yyyj,jy,y,g.,yyyylyjj,yjy,,,y .jyjjjkjjjjyj yyyj y.y .

..... t'-.

-

.

-

E

.

g --; . .

-

-

-

'','.r')'i''.::)''.!.;.'i..ii;,,;y(yjyqjg)yj..,yj

jjj jj j ! jg j(jjjykjjjy.jy r..r)j!yyjjjj('. 11:::::::::11. jj.yy..yjyj.jyyyjj.. .. . j)y j .. ; .jj g . .. ..;.j)jjjjjy j j . y j g ..yjy y. .jj .yyj.jy..j.y j y:jy..yjyy jjryssyy. j..(;j.;.j;kjj(yjk . j jj.jy .yj j...j . j jjj.jj kjj jj j y . y y)y yy . .. .i.i ;..jjy j . j y.. y E .- j. q. ;-j - g ; ;q y E.! (;- ,- (;y - jyy y j gy ( y - gj y . y y yy yjyjy; j yjy y ;yy .yy.g. y (ii . y y g.y.yy. ...( :: kgiy.(i((gi jj: y.j ; y y y kyk; j(yyyy . . y y y y.j y. . :

::

:

:

.

.

,

,

,

-

.

.

a. Recurrence

and

translepc

g

-

-

,

, .

.

-

-

.

.

-

.....,qy....,jq;.,.-p,.,t..ji,i..(;(j.)..jj.y,j.!(gr,,,.,,,).yyjgj,..j,..j...;).g.,y.

R

=

We begin with the dichotomy between recurrence and transience. Let a). inftn k 1 : Xn lf .&(S < x) = 1 then we call the chain =

. .... rk.)qy(rj'y (y) '. . rjj yt(-. ( j yy ...r....y.....;r2.q,j!..j.j;.j.y;.!jyjj!j,.jyyy.j.jj.,yjjy.j;.jj.jj,.jjjyj,.,.j,y,...yjyyq( rtyyjjjyjjjjyyj.jyjj.rylyyyyj,,lryyj.yl.j....y.

' ' (. y(Ey y jyjjEy E ..E.j' jjjE y; l .jgjrjjg::!.j7.jj. lp:gyEy ... Ej' ;jg j. yg.y..yj.( (,;,rqjE.!7r)yty)j!I)j,;.;yjgkj.rjjjj;,;(jyyqq.y..j.y,j,jkjjj.y.(j.qjq:. j y!gk);))'. .......;.(.q..E...:,j(j.y,,.j.yj.j.jjjjjy.j.yjy.j.yj,.j,j.yyjrjyjyy,j.yj..y,.y;,.yj.y.y. !

:

:

:

-

:

:

:

,

-

:

.

y

.

.

. : -- -

y y,

g

-

.ti;.E

rjjk (. yyy .: ....t..tE..j.jt) ..,.yq.jj...,,.Ek.j.j,);..y.jj.,!tj!j.jy.q.jrjjyy(yy,,.,.jg.j.jy,y,y.,.y,.,r.,j..,.j.-,y,....

.( ;; il1l!li l E : . :i ..

.

.

i @ E iE: .

. ....

262

Captez 7 Dihsions

as Afarov Processes

Secton 7.4 Hazjs Cllains

otherwise we call it trnndent. Let Sz R and for k k 2, let recurrent, Th strong Rk = inftn > Sk-l : Xa GJbe the time of the th return to = #tz45 < xl < x) Markov property implies Pllk 1 so #.lxk a i.o.) in the recurrent case and is 0 in the transient case. From this the next result follows emsily. =

=

.

=

=

(4.4) Theorem. Let (C) Ea 2-n#n(a, C). Inx)the recuzrent cmse if (C) #.IXS 1. < then 6 C i.o.) = 1. For a.e. z, =

263

0 then the deqnition of implies .&(Xa E C4 = 0, so 0 for a11 n, and pC) = 0. We next check that # is Izet Gk,&= (z : #(z, a) k J. Let Tc 0 and 1et Ta = inftm k Tn-1 + k : Xm e Gktsj. The deqnition of Gk,&implies Proof

If

(C)

#.IXk e C, R J-finite.

=

> n)

=

=

.plTs

nlT'a-l

<

n)

<

>

/

(1 ) -

.&(.2

so' if

=

Remark. Here and in what follows F'(z, C) #.IX. E C) is the nth iterate of the transition probability defned inductively by 751= # and for n k 1 =

6') #n+'(z,

=

/#(z,

6')

inf (n : Tr,

k Ta) then EN

< 1/. Since we can only have T. + k for some 0 K n < N it follows that (C%,4) % VJ. Part (i) of the deflnitioh of a Harris chal implies S C Uk,mlGk,l/m and c-flniteness follows. .

dpny,

let N

we

Xm E E

. .

t?k,J

=

with R > m when

:

.

.

'

Ts S

m <

E '

'

'

Next we show that ##

#.

=

.

:

.

CAss

1. Izet C b a set that does not cohtain a. Using the deflnition Fubini's theorem.

and

,

The frst copclusion follows from the followinj fact Proof Cha P ter 5 of Durrett (1995)). Let Xn be a Markov chaln ad P (um'xkamlt-Yme fmll -Ys) k 6

on

txa

(seee.g., (2.3)ip E

supose

e .An)

E

-

'

H (aJ and P((.Xk E Bn i.o.)) Then #((Ak e da i.o.)) 0. Taking Bu H C gives the desired result. To prove the second conclusion, let D (z : .F(J@ < x) < 1) and observe that if pnt&, D) > 0 for some n then =

-

jpfpy,

)

=

a i.o.)

K

x)

#n(a, dz).Pa)(S <

<

since

/

t

C and #.(.X

> n)#(p,

dy, R

c)

n + 1)

#(c)

=

=

a)

1.

=

2. To complete the proof now it suffices to consider C

#(d#)#(g,a) =

1 Remark. Example 4.2 shows that we cannot expect to have .&(S < x) for all z. To see that this can occur even when the state space is countable, consider a branching process in which the ofspring distribution has pc 0 and E kpk > 1. If we take .A B (0) this is a Hxrri chain by Example 4.1. Since p''40, 0) 1 it is recurrent. =

)(2 Pa(Xn6 dy, A

=

=

(a).

> n)#(p, a)

n=0

Ea(a = n=0

=

=

jp-k e

.

n=0

CAss

1

u azg

> = E a(-1k+,6 c, R

.,z

=

.&(Xm

S

=

=

n + 1)

1

=

=

#(a)

=

b. Stationary

where in the last two

C

measures

(4.5) Theorem.

Let R

=

inftn

k 1 : Xa

R-1

#(c) s =

deines a (4.4). p

J-flnite =

pr is

stationary a

J-flnite

a).

In the recurrent

cmse,

=

5-7

1(x-Ec)

n=0

=

=

e y'.7at-1k

c,R

> n)

n=0

measure for #, with # << stationary mesure for p.

,

the measure defined in

(a)

=

only

the n

equalities

0 term

=

we have used recurrence

contribptes

and

the fact that when

in the deqnition.

Turning to the properties of y, we note that it pv and #(a) 1 so p is using To check pp note that the deflnition p p, we p'v and (b) (i) and using = using using in the desnition (3.1),(ii) (a) (4.1),(iii) ## #, (iv) =

J-flnite.

in p /t?: =

=

=

=

#P

=

To iuvestigate uniqueness

(/'D)(/$

=

ji35'D

of the stationary

=

FT

=

CI

p

measure

we

begin

with:

Capter

264

as Afarl-ov

7 Dihsions

Processes

Section 7.4 Harris Chains

stationary measure for p, then zz(.A)< (4.6) Lemma. lf v is a J-flnite stationary measure for # with p(a) < x. p vl is a J-llnite

(x)

and

The last term is nonnegative,

letting n

so

=

Proof We will flrst show that zz(W)< x. If lz(.) cxl then part (ii)of the deqnition of a Harris chain implies vC) x for all sets C with pC) > 0. and with vBj < (x7 then pBi) If B = tlf 0 by the lmst observation pBj So zz(.A) < cxl ad 0 by countable subadditivity, a contradiction. lz#(a) p(a) E#(.A) < x. Using the fact that vp v, we find =

=

C) To turn the

=

C)

=

p#(C') iz vC)

-

6

vAjpB

nyC'l

equatity the lmst subtraction being well deqned since ?z4.) < x. rom the lat 6p(.), it is cle'ar that v is J-fnite. Since i7(a) it also follows that pr v. To check p# p, Fe observe that (a)of (4.1),the last resglt, and the deflnition of p imply p# = vp = p# = :ic

=

=

(4.7) Theorem. then v = p(a)/z Proof

for #

p(a)

=

=

Suppose p is y is the

where

If v is a constructed

recurrent. memsure

J-flnite

tatiimary memsure in the proof of (4.5).

stationary By (4.6)it suces memsure to prove that if p is a J-finite with p(a) < (:x) then p = p(a)#. Our qrst step is to observe that

x we have

-+

alpcj

k

.

k into

=

=

265

an

=

we observe that

/(

p(dz)#n(z,a)

=

k

p(a)

pdzlgnz,a)

j

p(a)#(a)

=

p(a)

=

since (a) = 1. Let Sa = (z : #n(z, a) > 0). By assumption Uask = S. If :(D) > p(a)p(D) for some D, then LD n&) > p(a)#(D n&) for some n and it follows that p(&) > #(a), a contradiction. (:l (4.5) and (4.7)show that a recurrent Harris chain has p, J-finite stationary measure that is unique up to constant multiples. The next result, which goes in the other direction, will be usful in the proof of the convergence theorem and in checking recurrence of concrete examples.

(t

.

8) Theorem.

If there is

is recrrent. Proof

Let #

which

'm#,

=

(i)of

a stationary

k

probability

a stationary

distribution then the

distributipn by

(4.6),and

chain

note that

the deflnition of a Harris chain implies #(al $ 0. Suppose that the part chainis transient, i.e., Pll < x) q < 1. If Rk is the time of the 1th return =

C)

=

p(&)#(&, C4 +

s-(0)

dylpy, C)

then PRk

< x)

=

qk so if z

#

Cr

Using the last identity to replace ptdpl

on

the right-hand

side we

Js-(.) X''d#)X#' + /s-f.)

p(&)

'7(&)#(a'*)#(#'

C) +

p(a)#(a,

=

have

Cr

E=

1(# a

n=0

C)

t:e (0

P.lRk

=

=.)

<

=)

=1

The last upper bound is also

valid

1

v-l

<

-

=

j-

=1

when z

=

q

a since

./s-fal

Vtd''

= Continuing in the

p(a).p.(#,

+

obvious

#,(-'%

Ct

e c) + p(a)Jk(-'e, # # a,-'e, # a,-'% e c)

way we

a,-'ea

Co

X

Ea

e c)

1(xu=a) a=0

=

p(&)

S

Pakk

m=1

+ Pvkk

#c

for 0 i k

for 1 % k < n

< m,

< x)

XmE C)

+ 1, Xrj4.lC C)

1

qk

-<

0

Integrating with respect to # and using the fact that tion we have a contradiction that proves the result

have

#a

Co

P.(&

1+ =1

x

1

n

p(C')

=

-

q

>-E e

=

S

a=0

#(t)

1

1

-

is a stationary

x

1(#a=.) n=0

''

=

=

cxl

q

distribu-

.1::11111.* 41225:1:* :11:;;* .15515,:* !:::::::::::::' :,,' ytd ::;:::::::;:::' ::;:1.:::,* !,,' qjd ::::::::::::::' :,i' :::;1!:::,* :,,' :.,' i::;.!::,?d yyjjyjsy.d i:::'',' dll::!:::d :11:;;* 11Eir2' ((IIirr11L.' 111::)* ;::;1!::)-* ::1,..:;1,.':4,:.* 114:)* 114::4* .j' j'p' ::11:.:,1k.* ;11!E,11-,* :41r::-* '1Ik.j:lI.' :lljii!lliiisii,d '!It,(:11.' ::11..* ::::4!:* (:11..* g' (' .1Eii5!,' CIIES!Ik,-:E!!!I;X '1Ii.:11k.' (iii,d E:ll:::llt,d ;1E!5!1;' ;ll:r'd jl'jtd ::...:1111:2.'* :1Ir!!'' 11EEE!,t' .1EEii!':' :Il(:!'-' :)11r:::111:::111..* 11IEE,:,' pj.y' yyd E:I(,.' j';11!E511--* 1j' y' $' )' j'(('.::;1.:::,* (' t';::;1!:::)* jjd j'y'(jIl;::,,' jjf

y' )' t't' (211:.:)1k.* dl!!::E:El:;?.'d .1!::::E2222?.-* dll!:E::EE:;?.,d dl!!:E:llE:;?.rd (111:::11..* ,;;jiijii;r'''':'l'' (:11(2q* ill:iilll.d 11(E2E11,* IIE!E!,;X iii.d 11(::E:1.* fiit.d y' .;;idriii;;' jjd t' ((:)1h* ((:)1p' ill:E)d

.(' .-.' ,;;!iiiii7' .-'' ..-;i!!::iii!!:' '-'' '!:;':.:'''d .d ':iiiiii;k'' '''''''''' .' -(' ;' 111::::2* 411555:2* llrrr:rlir!'d :,,' :::::::::::' g' :,,' ('y' ,'y' 111:::::* kll::i!!,d

.' '.!r::t!!l

.1::1111:.

(4.9) Lemma.

There is

.

.

(5.4)of

For a proof see

(4.10) Tieorem. :iii.

''lIi:::),I

(iii,. '1It:;,

an mp < x

''''

E

ty) >

0 for m

f

111::::)14! ill::::ll,l

In view

of

k

.

(4.8)

) ... y. ..

.

-'.

a

#

.

.

. .

11:::::1,.

''!!',;'''

.. .

g'#

=

% (-)11

')

:5ii,.:jiil;-

Let 52

-

s

.

=

m=1

-...

=

114::::),111(i2::

dl::ql 11::(( -:;:i!:::,'

i:iiili (:11::-.

j. j. )' jyj j i.

y

q!

j .. .jj...jjjjjyyyy . .. i ! .i .' ' E.E'i'.i.E(1.y'yj.).;y..;grIryjjjjj!jqr!jy.;yy;i .j.g.yg; i.. ' ; ,r r; ..'y..'.j......;yyg...ryyyyy(y(yy.y(jqjgry;jyy;yjj.jgy.gj;y. ..(....jqjryq.y-y..y.q.jg. .. ....(.'..i...y.y.'yyE'y.jy;..y'.yyyy!-yyjy.yyqy.yyy;.y.yy.yy. yyjyyyyyy.y.yy '' qjryj. j)j jy)j (.;(gy ..'.@'i.E(.)i('gi.r,;r!))),j)j)',',,.!,.., ..-.-. --y-.y.j.--jjy. .yg.tyjt, -.;- , ;; -.), ; ! y yyk(. yjy.r-.-..r .y;;j y .

E

j

-.

-

-. E - Ey E .: . . !..y - .y! y --.EE.i;-- . - y . . -. - -- . E ! . . . y- y . y y Eyy.. .. yy-. . y -y yy -j y . EEE . Eyy y .

-

.

.

.

.

(C1 x C%)

'j

-. -y

- -

';

.

-

-

*(C);'i(C)

=

.

'

-

-

:

:

.

.

.

-

-

.

j.j -

-

-

. -

-

.

--

-

deflnes a stationary probability distribution for #, and the desired conclusion tollows from (4.8). (Xs '.Q)and ' To ptov the convergence theorem now we will write Zconsider Zn with initial distribution & x #. That is Xu t:v with probability 1 and hs the stationary distribution #. Let T inftn : Z-rt (a,a)J R(4.8), (4.7),and (4.5)imply i x # << so (4.4)implies '

-y -.Eiiiiy.'lEyiiqrikEiyiyijj.llyyijkyyyEyyjyyy.yyjyyjygyyEyyyy ---

.

.

.

-

y -

,

yj; yy

yj;jj j y yjyyjjjtittjyyyjyy . ...........j(@jj.jjy.j(y.jj-jjjyjyjjrj yyjjy --

--

-

yy,;.yy.y.- jyyyy, y-y.y--j-j jyy.gyy.y. . y. (q.q. , yyggy Ej ' .'': ' j(.y y j .(. j y ki j jj.yj y . . E. -j g [email protected]'i'iqjy'jiyjjlryy'lji ( qjEgEy;.y;q(y(y. q(g. .. yyyy . y :y. jq::E ... . . .q'jj(:y jjy;yy. yy g i..g . .y:jr. .y..jyy. y . qy :.j ..y jjy.( y.:..yyjy g gjj . jjj yj.gj.g..g.gj. j j g E . . y-;g- y . .y y . . . .. y . ggj y j -.i jy y - y. jy;jEj y j.y. y -

-.

-

-

, E

.

... ..y(.rjyjr.yyj.r ..l)! ). .y -

: .

-

n

tj. .

.y

:

:

,

-

. . - -

.

-

. -

=

: .. .

then

. .

- -

.

.

. . . . . -y . -

E- E : -

.

:.

.

. ,

:

=

=

-

f.x(T

-. - . -

.

-

.

,

'fk

jjj j -

-

-

.

---

-

: . .. .

-

,

-

-

=

=

yy. . yjjj.j.jj. yj yjyj j.y j g ' ..'(.'..j.'..-q-...;j;i(jq!j;ir.,i.;j;iy:k:..yj;jlj(jjrqq--.yj. .jyp.y.y...-;..y.y jygE ; yjyyy (i;j;(jk .. E...Ey(.'. ..(y .. .(-'(..y;..;j(.g;!j.jj..j.:y.j ..!(......E..,.;y.y(j.j..y..y.E.jy.. .. ..EyyE'y,'E.Ey,lyg!(g)y!kj;ygjy,.E,y'igijjEyy'y(l.,,yE.q gg jttj. . r .. yjq. y ; ; E'iy.Ej;j(Eiy (jj;;.yi:yggEjj.E(j .jq.(yyyjj.j Dropping E i r. . .'E..(.'E.'.....,;y,,i(((;.yrjyjjy.yg,,j,j@!(.),,;r.jy;yj r.j.y.j.jj y(;yj((.qjir).gjyE.j .j.(Eyy(q.jy . ,

.

. y..

<

)

=

1

.

.

- :

-

.

-

j y.j.yy,lyy, . :. jj;jyjj. .j.j y...yy r..j.j.jy j y . y ( j.-j.. g..yj...y ....j;(i.'.j..k'g.i.j)..)yjji.kq.y(jgj.jy ( (j( ... j q .y :-

-:

y.

-

-

E

-

:

:

the subscript

x

'#.

frofn P and considering the time T,

y

,y.i,;,i,;.(.,,;)q,jyyp,;,yyy(;,,

a-m 11::4( (r:1p((2))1 .;jijiiii;;

41::::)1.,7.

.y,.yr.,

E.E.: iy

- ; -. - ( q qy !y.yEj - Ey,jj.. jg-. yy . .y ; y j -(jy y'.. g yy. j . . .j'...... Ejj k!(Ej.yy.. . E :. ( ( E : .. . yj i.y . ..E-. ; q g E. E.. ..E 'E::E!:. :. i E:. .... E . .. . k ! . .E. (: jq:-.E-: ;. .. . -g ;. q-jj-.. -

-.j --

-

.

.

.

-

.

...

. .

,

.

:

.

:

:

.

-

11::()

-

. :E y. r

:

.)11

-

-

i

.

i i qE jtttlEE y g

y.-

-

.k

-.

.

s

-

-

-

.

.

. - . -

-

-

.

.y

-y

.

y

-

- -g

-

-

-

-

i

Ei -

E

-- ( .. i : -. - . - .. E . . - E- . . - y j , . y .- - y gy -. ( gy -- -. . -y . . Eg EE--. ;.. . -y -. -. - . -. . .. - .- . - -. . - -

-

-

. . .

ig y . Ey. y.-.

-

(2:1,1 (::),. 'tproduct

-

.

-

-.

)j.jj jj .j

y -

.

y y . -. -y E,i,('EE'E'i.,'Egi.l;lj(j!ky.jj!;,k;.,y,jy; i q - .

-

.

:

.p(7a

.

'' .: :.

.

this the

'

.

y . ..

-

E: : : E. E E . ..E. EE . .y g :: .. y.. yy Ej y .yy y Ej ' : :y.q ! E. :: . . . i.E . y y :.y E yj y . . . y y r E E y . y g r y. y y E jg y (j y y . (.E.ElEqy'yrjrijjiyi'.ilqE,ijiylljiqijiElp.yyyyjjj y yyyy

:::4!: :::!!: :,-

.plJ@a c,T e

.

,

x S by

c)

=

-

yy

.

call

E PT

m)#n-'n(&,

-

. . .

-

:

moves independently. we will and denote it by a measure) Chain'' of product (think

:

=

-

.

-

.' .: :

'

That is, each coordinate

=

-

..

,

E5It..

s

n)

.. . . . . .- - - -- -. E . . . - y. . .'.'.''.'ii''.)i...',..iyj.i..j);r,(;lyg::i:i.'pj',:,i,'y,;. .

-

# on S

e

c,T

. g-. . y y . E E- q y.. : E : g : y y r .y yy y . y E E EiE . E y..

E

.

: : . :. . . .

-)

-

-

'

-

.

-

-

E i. E y. E- E.E

- :

'

r'z)II#n-'''(ce1e(

rt

Pvk

mpl ) .. j .. j . ..i.r.i(q.!.((qiy,;(jyyjrr.qj;;jy:)(r.jjiy.jj.jjyjj.-g.yrg(.jy..gjyg.i.:..j.. j ( g . q ; y ; . . j ( . . (r j. ( g .(j(jjrjyj)jrjyjy;.yy;y.yyjyy. since on (T j.(!)(jE;.j;j.yy(:2y;;yjyjyjy.yj(y.... y;j.jq... (..y mJ, Xm = ik = a. Now .. ....yjy,jyyyg,yyj,jyyyy,gygygyyyy,gjj,y,yy,yyg, y . jki(i ! yyyyjyjy .!I;k.,yiy ,y .yjyyj. ;j ' :. . r'. y.r(jyjj.yy;jj jj ,ryi,,,jy,y.,y!,..y .j y E #(Xs E C) = >(Xs 6 C, T K n) + PukL G C, T > ?i) '(. (y..y(. rij(j y q ( g y y ...j(.jjytjlrq ...(i ),!(jj;jyyjj.jj)y(!jjqj iy...jjy . . j I(.yjj((jq n) + Pun .. . i.'. i.!.j .j(j;.j.jyy((jjj...i j;jyjyygy(.y. : c, T e T > n) j.y( r = (...:'..i(Eg..q;yqjy(y.jj.yj.jjyjyjyj.j)yjj.y.j yjyyyyk.y..jj jjj .. g y .(.;y. y.(y,.j.j...;jg.jgyy y y r.;..jj.gjj.jjyjjy.j;yj..gjy.jy.gyj.j.y...jyggjjyyyyjy.yj..jyy.....j %#(fk e C) + #IT > n) y y. . jtyjjyyqj yj j.. ; ! ; . j;rj( jl .....r(y(jyi...!j.j!yy)jyj(y;jjy.y);yj,.jjyy.g.yjy.jyyyy;r;(yyy;yj..qj lnterchanglg the roles of X and F we have j y .y . ....'..j.q..yEqj.)(!-j(Ej.r.jyqjy;;i.yjljj)y'. yyjjy ..yyj.jy . .jljjjyjy ..jyj.gj.y .. ;. j'i y p !...'... j ( ( ( j , j q. ; .y ) . y j k y ) j. . j . j .. j.jyj. -.yjj .'...qrqjg...y.r.jjyyijy(jq;i..jyjjyjy!jjjyjygy)yryE.yyq,jjy!(yj.jjy. Pku .ryyjj t'y y!rjj . J3tt e j .. . . j)j yjj,jjy jy e c) + PCT > n) . . .jjy)y j y y . .(i''';;,','E,i.!'7'i:i.'i!)II;!jjjj(Ejg.j .. y. yjj.y. q ))! . q jyjyjjyj y q yj(r;,jy y y y y .j . jryjjyyyj.j.y jjyjyjytgj . yjyjjjji,ltjijjjyyyjj,jyjjjjj,jjr,yj .. and it follows that ;(y j ) j r g j j j. . j .jig y . yyjy j . yjj.j.jj y y j y l ( . ; y ( . . ..'..q'.. y(j.j.y(j(jy.j!!jjjyjy!rjy,y.. ;jyyyyj.j .y y y);jjyj)j((y.j,yyj; jj.jy.yjj.jy.yg .yj .. .'...('.EE.y.yy;.y..y.ji,pjjyg;ij;y'j ;k yy. yy jjjyy;yyjy-jjy (j.y jyykjiyjj.jjrjyjyyj. .. j j!g y j y j'y y y j gjj y ,y .........j.;.y;y(;.jjyjyEy'; y. .. yjy yyj;y , yy g y. jy yjyg yjyjy ;j . y p g .!.q.y)(jqj!;E-kl'y y yyjjy.y yy y g y . y ,j;( jij1. y j j , g;gjyyyk. . . .....(...!)Ej)iij'('('5 .iiyi j!j..y... .yj!yy.j;;).yy , ;rk .y!)jj-gryi j;.j!!j.,yjy.y.;j .

-

(2:11.

::)),'.:::: (:11:,. :::2!,:

is recurrrnt. To do this, ditributio fr #, ahd th two

=

.

,

'-1i,;'Ii';,''

.;;@iiiii;'

chal

'#

y

. .

..

0

:. .

.

..

..

0

..jy,.y.tjy.y

E .y @. . .... - -.gi q.q EE.y ... y y . ..s,gyss.yjyjsjsyjjgyjyyyysygyyyjg.jysgygjjysjy,sgyy

-

>

a) >

:

.

..

(111.. (211:':11-. ,t (41144::),! S x S. Deflne a transition probability

:jisl;. .1:::::)1-

..

(a,a)

:

:

,

:

.11.1k.

'!!i',p,'.

a)#N+M(a, k #f'(z1,

y g g g gjgy yyg ,E'.lp.Ey,,,l,ijlEyiiyr,jylig,qjjjjkyig,jqggyyylyy,j.y.y,,yqj,yy,, . yg yy.,.g y

: - .

.,,;i!!::E7;1I'',:;,I::,.

-!%(.?

a)

#A-(z1, a)#f'+3T

y..,.jy..yy.j,y.ygy.jy,jj.jyyjyjjj.y..yy

.

t:v

k

,

, ..,.

then

dlisir,.

II#''(z,

.

y

y

... .

.

'#

a)

>A-+T'+M(za,

'Cl!E;')..;'yj;jiyj(j!jEiyEyy!!qjE.y;((,.yyr.y

.

''''''''''''''i!'-'.

:11'':1/':)1,.

''!i.F''ik'-' 11iEiE!t: ljiii!r.

y

:

.

11::4)

((:1)1

.

: . : g

za), (a, a)) > 0. and hence jN((zl ..( .).q'.,q(.!:(j.',j.';.',l;jy(rjyj,gyygjy,.,,.j.,,j.. j((.y...,(.(....,.. The next step is shpw th,t the product .gjj jgy jss,jysjyysjssj y....... yjy zr# is a stationary hot that (4.6)iiriplies .. ..jyy.yyjyjjjy.y jj... ; !.j) COOr independently m ove so .. .qj.y!rj-(.k.;j.jyjy;.yyj.yrjj k)(j..g.jy. dinates y yj y .

(T:)1' 11. 11.111:E( 11::E 11. 111 (::11. z

.

:i y:

-

,

-''

is to note that if

:

....y..,y.,.,y.y.y..;yjjjj.jygjg.;(yyyy.j.y.y,,

.

to qhow that if

#A-+f'+3T(z1,

:

g q r j; E .... j.g.j.y...j.jjyj;jjjjjjjjjjj.j.jj j.jy.....j

-

.

-

. .'..i..i.,.!..rq.:(j....q,,.jyr..rj,q.j(.y..yy. . -

.

.

.;i!ii!ii;7t''''''.'

Our second reduction

y .

:

..

.

. .

'''''''''''''.

: .j

-

..

. .... with stationary

-

(4.6)it suces '''

:

.

'.

..

. .

j;.j.,).y,.(.j .

q(q;.yy;;.y,yjqr(j;;y,;.Ejy(,kyjyyyj

E(

'.j.jyj.j..jjig.i-Egjgg.yy.. E. !qj . . g...E(jqi.y.jjqjjyj.y.g(.-jj.j.yy.y.j.y.j (......,..;...g.(yy(yj;.jjqj)..(j.;.!.y,r.j . (' ..'E';yj.g.y.EyE. yygjjq j.y,,.yy.. .. . ;... j.j. .. yg.y. jy

'''':'''''''''i'k''.

'-'

.

.

.

i!E'.r.'.).;jjj(iyEyE..j(.qjyij.4;.yyj,.

...

mp.

14::::::,11::::::),-

.'''''.'''',.'i!,--.

:

.g j jy;,,, .j ..Ey,'.j'.y.j;y;,j,gjiiy,,r,j,jy,,,y,,)y,;(,jyj,yyyyyqyjy(;yy;,.yyyyy

.

.

,. apd

(

.

.

E: . . g.y . .

. Remark. Here 1111 denotes the total variation distance between the measures. (4.4) guarantees th.t a.e. z satis:es thehyythesis, while (4.5) (4.7), d (4.8) imply g' is absolutely continuous with respect to P

=

.

'iiiiy,lijqllryq.jjlqqiliyjjrriqi,jl

-

.

-

.

noW ready to pove .

(1995).We are

. 11. 11. (::11. 11. 11::E( (::),' 111 . ''''''''''''''

:

:

.

411@:k-

:'''':'.'

-

-

(y . 'E;y.'qj',r'.'E,g:,.'jiI',ijE!.i, j. .g . j.gj j yy r..y 'jt, .

.

. :

.1::::::,..::::::),.::::::::::::::: '

,;i!r::;;''

-

'

,EE'(@i)(qE!Eiy!@,i(;l'1''i),(El.(E(E,@y,.iyq(r'g

''''

Let Xn be an aperiodic recurrent Harris chain r:lliir'' Elll:::li,. !;p,'';I!;. (:111,.. (111:':11.. ,.-;i!!:E:::'I'';::.!:,,.

.--.

.

y

.

.--1:::::::::q

.::;,-117-..

.1:::::),1

:

:g r.. yg. .y.. . .. E. E !. - . .. ;q. yq . y : : q (' y. y .-' y -' i. - . - - ! . i. -(qqyyjjjjyjtjjjj ! ...j.y(,;,; .. E- yyyg y, . .

q.;:ij.k .. ....,.,..E.y,y;E..E(.irj.,.!jy(.E4j..E,r. y

Chapter 5 in Durrett

.,.;i!!::iii!!:.

.111:;-

. -

.

-

that #*'ta,

so

. -

. . . .

.y.

-

=

.

-

-

-

Section 7.4 Harris Chains 267 lg.liIj j jjj .' . ;''.lyyi'kyjyj:jjj1yiy..y. yj...y...y j... ' . ')ik )(.y.lj((. &)) and We claim that the product chain is a Harris chain with .j....:)..) . .yjyyjyj.. ((a, j...y:jy.jy..,.j.s;jssssj. y.gy h .. .t yjj = check B x B. Clearly (ii)is satisqed. To . );.y,jt.j( (i) we need to show that for all (z1., z2) there is an N so that N((z1, za), (a,&)) > 0. To prove this lt ff and (j . j jk!j.jjjj(j(yj jj ( g .. L be such that #X(zl, &) > 0 and #f'(z2, (z) > 0, let M k ma, the constant in ti ,.,r..,.,...,:,,,,.i.,.,,,,,.,.,,k;.,,. E (4.9), and take N = ff + + M. From the deflnitions it follows that ..

.

11(::::).1

.

y

(. . ir...r(j. yg. E.

.

.

-

-

-

'

:

-

.

:

-

.

If I is a set of positive integers, we let g.c.d.(.0 be the greatest common divisor of the elements of f We say that a recurrent Harris chain Xn is aperlodlc if g.c.d.((n k 1 : p (a,a) > 0)) 1. This occurs, for example, if we can = of take B in the deqnition for then pta, a) > 0. A well known consequence :5ii.. (iii.. (::!i,?'''' ;1!!5111-- '1Ij2;.4

i. EEE.

.

.

. --

'11It;:.

'll.r'':l.:.

-

i

:

.: .. . -

-

(lllizErll... .11::::),.-

Onverg .111::::::::::.

c.

W

E: .E!' E: . .

' kk;y! !y . y .'..-.. l.)y)j.

j;jq'.;y.rj.j j.y.y'.).yy;iy ..j..y( EE(ij';jj@ . . .@(yij(!;jj;yy.jjgjy .!. jrE.j ..jEj.yjyjjjy. ;)((rj ;!(il;qy j .yj.j . .kjj..g.j..y.(..

-

,

.

-.

-

-

-

- .

E y

-

E

-

.

.

-

-, y

y

.

i i -. E -E i

.

. .tj)),y,,,) ,

.

-

. y jy y -g y Ei - . . y . y E EE- y y . -. --ty gy y- . . -. . . - -. .y y. - ; - yy- -j..y. --j . . - - y-. -j-. E . E.. E , E. q . - . ( yj'ij g: ; g. -- -. - -i -. g. . -- .

-

!. - . E - y .

..

-

y

.

-.

.

.-

. -

'

.

-

-

.

y .

.

-.

.

y- -.. . y;. y .E

jy

. . .

. .

-

-y

(

-

-

-

-

.

-

-

.

'

.

-

-

.

.

.

c) s

c,

s

n)

.

Capter

268

7 Disions

as Marov

ConvergepceTheorms

Section 7.5

Pzocesses Letting n foltows.

-+

x and using the monotone

269

theorem the deskqd result

convergence

In '

Taking utzl

&S

n

'-#

Though

Theorenas

In this section Fe Fill apply the theory of Harris chains dkeloped i'l th previsection to difpsion processes. be able to use thattlyoiy we wilt szppse 'tb

us throughout tlii

sec iton

'

t af

q

MP(, c) i? (5.1) Assumption. (HC) and (UE) locally.

norma

=

Proof

:

.

.

By

.

l . .

.

... '

(3.8)there

is

gives the following

.k'. .

i

..

.

F= 6

'

.

. .

.

.

function pftz, p)

>

0

: .'

. do

. . .

i '. .

th>t

=

for

-e

ff c

z 6

=

.

(z : IzI>

r)

sharp. -c/z

=

=

=

=

-1

=

-

Erxv

=

Rearranging, then letti g t gepce theorems, we have

s.(qz,)

-

and using

(:x7

-+

(z, s) aj

.Q'r(1,)

pftz, y) dy

.)

e

'

(z) K

.

.

Exnmple 5.1. Consider # 1. Suppose c(z) 1 and (z) for lzl 1 where c k 0 and set ff 1). 1/2 :7 = 2c- 1. To with > then If holds c (-1, (5.4) compute ENTK, suppose c # 1/2, let : = 2c 1, >nd 1et utzl = z2/:. Ln in (1, ) so u(a'V)+ t is a local martingal util tife (t,l p inft : X3 / (1, )). Using the optional stopping theorerp at time q1,,) A t w have

-.,

rilmsie on Itarris chain.

2z

to prove! it is remarkably

emsy

21

' .

lower semicprtpnuop:

.F'.(a'

(5.4)was

2

.

From this we see thqt .P.(S continuity implies #:(z, p) k #(a, a) > 0.

a and b satisfy

coecients

(z : lzIK and p is Lebeskl pewsure ther Xn is p.n aperiodic

=

a

th

well ppsed ard

r)

TheoreG. If X B lized to be a probability .

lzl2/E

(5.4) Corollary. Suppose tr c(z) + then ErTK ; IzI2/f.

X.

7.5. Convergence

(5.2)

=

.

2: =

-

2

A t)

the monotone /.x2

-

bounded

and

conver-

r(,,s)

X

1) > 0 for each z so > 0 wtzen z, p E d so

(i) holds.

(it)holds

Lower

yemi-

and we have (:I

theorem (4.10),we To flnish checking the hypotleses of the convergence distribution zr exists. Our approach will be to have to show that a stationary ielii ili (4.5)ptduces a statihary show that ER < x, so the constructio sligllt check (x7 generalization ER < with fnite total mass. To we will use a of Lemma 5.1' of Khasminskii (196)k '

E

for all Suppose k 0 is C and hws Ln i (5.3)Theorem. = inf : Xi A-). Then > Tx compact set. 0 Let a k is (t e -1

'u

'tz(z)

z

To evaluate the right-hand side, we need the natural convenient to let start froz 1 rather than 0:

/

ff

where

/(z)

z

=

(3.2)in

=

ff

EmTK.

z

2

-

dy

=

CJ(p2c+1 -

1)

(

=

1 it follow ttiat

j2c+1

z2c+1

-

-

t:

zcdz Z

1 y2C

this it is

1

Chapter 6 with c

>Qq1,,)

-

1

=C Using

p

exp

and doing

scale

ac-j.z

-

1

z2c+1

1 '

-

-

52

1

-

-

'

2.+1

r

-

1

:

The second term on the right always converges to ms b x. The third term on the. right converges to 0 when c > 1/2 and to cxl wlien c < 1/2 (recall that e < 0 in this case). -1/6

-+

supermartingale

Proof It's formttla implies that 14 u(.;V)+ t is a local Tx). Letting Ts 1 Tx be a sequence of times that reduce 14, we have (0, =

'u(z) k Ezun

h Tn) + Ecn h Tr,)

k Ern

h Ts)

on

Exercise

5.1.

Use

(4.8)in

Chapter 6 to show that

ErTK

=

cxl when c

% 1/2.

Chapter 7 Dillsions

270

Pzocesses

Marov

t'zs

1 and (z) K -Vz: for z k 1 Exerclse 5.2. Consider d = 1. Suppose c(z) where : > 0 and 0 < J < 1. Let T1 inft : Xt IJ and show that for z k 1, 1)/6. Ezl' K (z1+J =

=

=

-

5.2. When d > 1 and c(z)

Exnmple

I the condition in

=

(z) % IXt 1,and use

To see this is hms

1et K

sharp,

=

+

.

8 SVeakConvergence

6)/2

-(d

z

(5.4)becomes

lt's

fz)

with

formula

Izlwhich

=

,

Dif

zf/lzl

=

Diil

z?z/lzl3

1/IzI

=

-

to get Xa

'.%

y

5$ =

, We have

+

c

.

1.x,I

(xa)ds +

t

be a one dimensinal Brownian and write -4d

.(z)

the

=

condition . .. ..

-

z

.

.

.

motion B

1

(z) <

-(d

supxEs

EZTK

is

+

<

8.1. In Metrlc

tht we

d

with

(#)t F= t)

dt + dBt

2

example

to the pievious

and shows

that

sharp.

(z : lzI K

rl

'

and H

< cxn to

ER

(z : lzl K

=

< x.

r

+

1J.

If

=

=

To estimate

(7m

Jm-,

llg--z) K Cz

=

sup E=Tx

inf

xZK

*

> 1) -

k IffI=,y*K '

inf

pI+1(z,y)

=

6:

>

(3.6)implies

0

wherep71(z, p) is the transition probability for the process killed when it leaves the ball of radius r+ 1. From this it follows that PM > m) < (1- 0l=. Since (:I Um F'm % 1 we have EUM < (1+ Czll&z and the proof is complete. -

J

=

E

(1.1) TheoreG. (i)

r6W

M, we observe that if rn is the exit time from H then Pa,trx

=

-.->

=

-

=

=:$

< x.

Proof Let Uz 0, and for r?z k 1 1et V'm = inftt > fm-l : Xt E ff ), inftm k 1 : Um E Z). Since Um inftt > Fm : lml / H or t e ZJ, and M Xumj E .A, R S Uu. To estimate EJR, we note that .X'(Jm-1) e .A so E

We begin with trextment 9 we>k pzyergene on a general space S with a wittk and i.e., me tri p a tunction (i) plz, z) 0, (ii)plz, y) py, r)z),witlk (iii) z). yj-py, z) ptz-, ptz, yj : balls < Open deflned by (g r > 0 are ptz, k and the Bcgelr sts S ar the g-fleld genr>ted by the ppep balls. Throughopt this sectton F Fill supppse S ij a mtric spae arld S, is the cpllectiop of Brel sets, even though several results are true in much greater generaltty. For tater if there i a countable dense set, and use, recall that S is said to be separbl complete if every Cauchy squence converges. A sequence of probabilit# msures #n on (S, Sj is said to converge each cpntinupuf if fpr bounded ftpction JJ(z)pn(dz) r..-, J(z)>(dz). weakly situattons it will be coveniept I n this cmse We write pn to deal p. I many di reC tl y with randont variablrs X a r>ther thn with $he Msociated distribuE .A). We say that Xu onverges weakly to X and tions JIn(.A) Px ale if EfX). X X for any bounded continuous function f, .E'J(.Xk) write called the Portpanteau result, sometirrws gives Our flrst Theorem, fve ' t ( equivalentdeqnitions of wak' onvergene. qa

1

-

Spaces

n j

.

Let ff then ER

#s

Ix a j

.$

. .

=

(x)

1

-

intiim,

The lmst detail is to make the transition from Erlk

(5.5) Theorem.

()

(itis a local martipgale

(.Xz) +

.

reduces tltis6)/2

1+ c)/2

j

d

t x dptz I.X,I ()

=

. Xt 1.Xh l

dyk =

t

1

+

,

'

l

z

()

tjj,s

.

1/7 here for a d dimensional Brwnian

written

let

If

Xs 1-X-, I

EJCXA)

The fllowing staterents EJX)

-+

sets G, ltminfa-x

(iii) oi

all

(iv) For

all sets .A with P(X E t'9.)

pn '

(v) Let D)

PX

e

D))

= =

;

'

=

Pxtt e t?) Px .

.

' '

'

'

.

'

-+

E1X4.

''

.

0, limsups-xPtAk

the set of distontinuities of J. For

0, E1Xu)

functibn

E ff ) K PX

Px

,

.

quivalet.

are

for any bounded continuous

(ii) For a1l clqsed sets ff limstlps-.x

'

a11

J.

E ff ).

e G). 6 W) = PX

.).

e

bounded functions with

Section 8.1 In Metric Spaces

Con vergence

Chapter 8 Wa

2T2

Remark. To help remember (ii)and (iii)think about what can happen when Pxn = za) = 1, zs z, aud z lies on the boundary of the set. lf ff is closed fpr we can have za $ ff for al1 n but z e ff lf G is open we can have zs E G all n but z # G.

The deflnition of the cu implies

Proof We will go azound the loop in roughly the order indicated. The last step, (v)implies (i),is trivial so we have four things to show.

and this inequality holds with X in place of Xn Combining haye limsup I.E'J(.Xk) .E'J(X)l K

273

-+

0 s J''laf/l-'%k e =1

Efxn,

.)

.

-

sr our conclusions

.

we

'26

-

(1) lmplles (i1) Let ptz, ff ) = inftptz, y4 : y e ff ), z-b= maxtz, 0) is the positive part, and let h (z) = continuous,and k lx so

(1 Jr)+

'thrj

=

-

where

neX

ff )). h is boundedj

'thpz,

but '''Ii'' ..'iII1''.

n'ex

Now

h t lx

as

lim Eh (Ak)

e ff ) K

limsup Pxn

j 1 x, so letting j

ELxj

=

R**X

(ii).

cxl gives

-+

'only

.

interior of

tiiilnd

that OA = ff

nte

.,

.

.

.

..

.

...

.

.

E

..

.

:

.

of the . .

(1.1 is e.

''v

.

.

'

.

'

.

.

(1.j)cotlnuous mapptng theorem. juppose pn on (, ) opvrrge Fpily gbp) 0. to /I, au d let v : S1 S' have a discontinuity set Dp wit

Jlp

P sl

O

':. .

.

:E'

.

ren

=

-

# oW tF:A

'

.?...

.

..

Prpof t t follows from

J

i

:

(v)in

imyly

so our assttmptions

G,

::

=

=

-

.: .

E1

-+

(ii) ls equivalent to (iii) This follows immediately from (a)#(.4)+#(u4c) z if ylc is closed. 1 and (b) is open if ad f Aj & Ae' the the closure txibly(iv) tet zf (11)and =

so the proof is complete.

An important consequence

'

' .. .

.

is arbitrary

6

..

.

.

.

c.y R be bounded and coptinuous,

sice

(1.1)tltat

.

.

DJOV?

c

Dp, it

.

Jpa(dz)T(w(z)) Jp(dz)T(w()) -'+

E ff

PX Using

)

(iii)nov with the

(ii)and

Pxn limsup N*eX

E

e W) = PX

PX

=

last

liminf

6 G)

PCX,Z

PeX

n-

v

ff ) E G)

%PX

y

=

=

.P(

6 .A)

PX

6 .A)

At this point w have shown .

..

'

'

nrex

n

.)

e .A) k Px

which with the triviality liminf

.' .

E

yg

.. ' .

.

'

.. '

.

o

s-ltgajytzj

sincethis holls

for any bounded

k

e

. . .

i

.

lipqup Pun amx

%limsup gives the desired

.

'

:

6 .A)q

In checking useful.

convergence

result.

' '..

(t.2) convergingtogether

i probability

then 'F,z

z:z).

.k

s

.y

o p

y

jjjgjyj

. . . .' .

continuous

.

. ..

lim )n f Px

.

Changing variables gives

e ff ) K PX

.'

liminf #(Ak E X)

(

we have

equality

Pu %limsup N=+x

a4)

G4

6

J, the desiretl

result follows.

in distribution the following lemma iq sometimes

lemma.

Suppose Xn

:::).

X and p(.Ya,t)

: pty, ff Proof Let ff be a closed set and ff: (g .JfJ ) K ), where jtp, ff ) lf and only z) if there ls a y > : inftptg, z E ff ). It is emsy to see that z / ffJ closed. and = y) With this little is so that Bz, n ff 0, so (ff,)cis open of detail out the way the rest is emsy.

f

aipuh =1

E y1f)

aipx

-+

i= 1

0

-+

X. =

=

;

l

.4)

6

g

.

.

#tys 6 K ) S Pxn

y. .

E ffd) + #(p(Xn

. .. .. . 's) ,

.

>

)

274 Letting

Capter n

-+

aud using

8

Weak Convergence

Section 8.1

the second term (ii)in (1.1)we have cr, noticing

limsup .P('Fs n

oo

-+

e ff ) K limsup n

the right goes to 0 by assumption,

PCX. E ff4)

e f)

.

E ff,)

K PX

,

.

t PX

ta'

j

'i'a

Let Fa(z)

rmsy.

Fs-1(g)

=

inftz

:

=

Fs(z)

.

.

z)

yzatsx,

k

'

...

.. . ..

.

'

'

(fJ)

x

t

So

(1.1)implies

that Xk

:zty

K p(.f1/k)

.

ff it follows that

txkE ff ) %ptff )

limsup

y and it follows that X has distribution y.

D

Proof of (1.4) Construct a decomposition Xk of the space S ms in the preceding proof but this time require that each Ak hws Ilnrluhk, ) 0. For ttnluik ?J. each n construct decompositions (I'II of g0,11 so that tf2z ) arrange the intervats in an appropriate way we introduce an ordering in wich tlj if b, < (c, % c and demand that IL' < 4 if and only if fr. < ITj. Let , )E Aktj and Xknwj deflne limk-.x Xkn k,j for td e Iknj As bfore, Xn zk t/ and hms distribution pn. exlsts EJ pn(.AJ) IlnAkj 1 1 0, we have Since

yj

ro

=

, tJ

-

'*'

''

'

N

.*

)J

' .

@

*

'-''

'4

'-''

'#'

.

.12

)

l'

We will warm up for the real proof by treating the special case pn

.

.

. .

H

. .

.

.

'*.-1

'v

=

-

I(.J:l)

-

i

(.2g)l

'

=

-

lpw(.k,J)

=

p,z(.Ak,./)I

=

i

=

l

,=

.

.

p.

/4

'

,

'

'*

=

=

=

*'

..

,

Proof

J'-! IlAk,jj

=

JE.S(k,.&')

nd noting ffl/k

-4

'

'i'x

Gls

(-n(=) e .,,J)

=

be the distribution

-+

=

Letting

'k'

td

=

E ff ) K (.X@) E Lhsk,xjAk,jj

izskbhej

e'

for tl e (0,1) be a random variable that is uniform on (0,1), almost jurely. (For a proof, see (2.1)of F,71(J). Then Fa W invite the reader to cohtemplate the question: Chapter 2 of Durrett (1995).) there an emsy way to do this when S R.2 (oreven S (0,1)2):?'' while we givethe klneialprof. The details are somewhat messy and hot important fr the developments that follow, so if the reader fnds herself msking ttWho cares?'' she can skip to the beginning of the next section.

let IJ() and let 'Fs

e

s J'-! = J''!

'

..->

1

EEL'7

(1.1).

oGplete septheorem. Suppose l (1.4) Skorokhod ' s representatlon ::::> random variables JIs metric Fa on px then we can deine space: If arablr with the Borel se (0,1) (equipped ts and Lebesgue zeure ) with distributi surely. # Fx almt pn so that

R tis is

,

jzskbKj

'

Remark. When S funC tion let

.

proof of (1.3).

) as 8 t 0 so we have shown

E ff

The next result, due to Skbrokhod, is useful for proving results about weakly, because it converts weak convergence into that converges qequence '''' a ' . . . . . . tinst ure conveigce.

=

.

=

#(Frz e ff ) K PX e ff ) limsup n-G) The desired conclusion now follows from

275

Let zk be some point in Ak and 1et Xktjwj = xk,j for (J E Ik,t Since Xk A%+1. is contained in some element of Ak which by dehnition hs diameter K 1/, it is a Cauchy sequence. So X() = limk A%@) exists and satisfles p(Xk, .X') K 1/. To show that X has distribution y we let ff be a closed set, 1et Sk, ff ) = fj : Aktj n ff # 0J, and 1et ff: (z : ptz, ff ) % JJ ms in the ,

OQ

-+

Memsure theory tells us that PX

on

Tzlhketric Spaces

2 7(#x(W,J)

pn(W,J))+

-

J

E

'

where #+ = maxtp, 0) is the positive part of y. Since the px oAktjj c 0, (1.1) implies pa (.4k,/) pnAk,j) and the dominated convergence thepre!n implies -.+

(1.5)Theorem. Eor each defned on (0,1) with .P(F

memsure p there pro babtlity = p(.4). E

ts

.)

variable y

radom

J7 Itfr

(E 1 6) .

of S iht k Proof For each k construct a decomposition Ak (.Ak,1, i.e., eac disjoint sets of diameter less than 1/ and so that Zk reflnes set .Ak-1,f is a union of sets in Ak For each k construct a crresponding = itAkkjj and arrane decomposition Ik of the unit interval so that a4k-1, and only if Ik-,i D Ik,j D Ak,j if things so that .)

=

.k,2,

.

.4k-1.

,

.

tfk,j) .

,2f

)

(Q )1

0

-.+

-

,;f

as n

(k7

-+

.

t4'

Fix k and lt aa be the left endpoint of IL% and let Sjzj :r:z j : 11 4 < Iljo which by our construction does not depend on n. (1.s)implies that .,

'*' t4

ax

=

JE Sioj

lfkx 24 l

=

-

u

,

1im

n-cxl

'

lzn? I 1,

JES(Jn)

=

v

lim cen

n--.cK7

'''r

''<

l

)

Section 8.2

Wea Convergence

Chapter 8

276

Similarly if we let pu be the right endpoint of Rjo then prt ' of Hence if td is in the interior of IT4 it lies in the interior it follows from the deflnition of the Xn that

7x

x. for arge n, an d

-.->

1LIj

limsuppx

,

Aul

n-cr

s

t

E

ms n

-+

E'

277

Combining this With the f>ct that is closed under intersection it is esy to see /z(tJ,=1.f Given an open set G and an e: > 0, choose sets that patu,xzAl ). al.k t'. Using the convergence just proved it so that /Itus=zAl > p(G) -1 .4

-+

,

.

.

-

.

followsthat

2/

;&G4

set of points that is x we see that if (J is not one of the countabl wich I::1 of Xx , proves (1.k). some IT,.i7we have Xn the end point

Letting k

Prokhorov's Theoreins

-

e < ptuf-l.f)

lim /,stukzzfl

s

s liminf

/z,,(c)

-+

-+

Sine

is

6

arbitrary,

(iii)in (1.1)follows and

the proof ist/omplete.

C1

( .

roof for Rd This is a fairly straightforward generalization of tlte argument k of d (See 1. e.g., (2.5)in Chapter 2 Durrett (1995).) for In jjt ls cae we can rely on the distribution functions F(z) yfy < z) where for two vectors y K IR.J wit ratinal coordinates. z means yi K zi foi each i. Let Qd be the points in By enumerating the points in Qd >nd yhen using a diagonal prgument, we an flnd a subsequence Jk. so that Fak () converges for eac h q e Qd Calt the limit G and deqne the pzkoposed limiting distribution functiop by .

8.2. Prokhorov's

Theorems

.. .

.

.

.

::.::

Let 11be a family of probabilit# measures on a metric space (S, p). We call 11 relatlvely compact if every sequence prt of elqments of 11 contains a subs!r that converges weakly to a limit p, whic may be / I1. W call 11 quence pnk 6 for all each for tight if e: > 0 there is a compact set. ff so tht Jltff ) k 1 proofs of section devoted is the This to y E I1. -

If 11is tight then it is relatively

(2.1) Thorem.

(2.2) rheorem. Suppope S is pact, it is tight.

coinplete

compact.

and separable.

lf 11is relatively com'

but th secohd okl is The frst result is the one we waut for appliations, comforti: since it says that, in nice spaces, the two notion are equivalent. .

the result successivelyfor Rd rtx a couwtable finally, a general S. This approych is somewhat union of compact sets, lengthy but to inspire the reader for the flbrt, we note that the llyst two special cmsesare important examples, and then the last two steps and the converse are rem arkably emsy. Before we begin we will prove a lemma that will be useful for the first two exqmples.

(2.1) weshallandprove prool'ol-

)

1

.

!E!)

.

F(z)

=

inftGtl

:

qG

d

Q

and q > zl

where > z means qi > zf fpr each f. To heck that F function we note that it is an imrediate a distribtion cpnsequence of the deqnitiop that (i) if z K y then F(z) % F(p)

ts

(ii)iipuw#(p),

=

F(z)

.

where y t z mns yi t zi for each i. The third function is emsier to prove than it is to state:

conditim

d) (iii) let = (c1, 1) x x (cd, be a rectangle d)., >nd sgntr) 1et (-1)n(c,v)where . atx (tld, .

t

.

distribution

a

with vertices I/' = (a1, x = ntc, is the number of a's appear ir b. The inclusion-exclpsion formula implies that the piobability .

.

.

j

for being

of 2. is

Evuvsgn(1?)F(r)

,l)

.

k 0.

.

.

'p)

'

'. .

.. . ..

.

y

. .

The inclusion-exclusion formula says m

P(UP1,-1.i)

.P(.f)

=

f=1

#(.

-

i<j

n+)

+

.

.

.

+ (-1)m+1.P(nt,-l.f)

. . .

j ' .. .

y

The fact that (iii)holds for the Fn implie.s that it holds for G and by taking limits we see that it holds for F. imply that F is the distribution function of a measure Conditions (i)-(iii) on Rd, which in our case will always have total mmss K 1. Define the ith where the ltmit is m arginal distribution function of F by Fi@) =.: lim Fyj = coordinates takn through vectors y with yi z and the other yj x. F is a one dimensional distribution function and hence its discontinuities X are a cotmtable set. Call z a good point if i / D for each i. We claim that F is continuous at good points. To see this, let w < z < y, 1et zi be the vector which uses the flrst 's

-+

Proof

. :

y

.)

-

.

=

g

g

F(p)

Fwj

J7F(z)

F(z-1)

-

S

f=1

J-lF(w)

ptz,p)

i= 1

and

of F now follows from the monotonicity

The continuity the Fi

F('tz4)

-

continuity

the

9

Our next

p,

9d with

r 6

,

p < q

F(z) Sihce . .

<

'tz

.. . .

.. .

'v

is that for good z, Fr,.(z)

-

e

.

.

.

F()

S 6

-

G'

.

'

S

F(1?) it follWs tht ..

:

.

.

.

..

e

.

.

S Gqt K F(z) S G(r) %F() +

e

,

-.+

'It* the product (2.5) Lemma. The Borel sets S : dimensional firiite sets (z zi e A for 1 % K J.

'

.

.

w)

=1

=

'

. .

.

-+

'.

.

S F(p) f F(:) K F(z) S F(r) K F() +

implie #(u)

.

.

z < r and

.

'<

To prove this pick

F(z).

-+

2-po(7,

=

It i easy to see that the induced topology is that of coordinatewise convergence, zf) for each and hence this metric makes (i.e., zn z if nd only if z7 and complete separable. A countable dense set is the collection of points R= ' with initely many nonzro coordinates, a1l of which are rational numbers.

.

claim

.

'

' '

'

.

-

-

=

279

of real Proof for R= Let 1R,* be the collection of a11 sequences (z1,z2, numbers. To deine a metric on R= we introduce the bounded metric pot, yj Iz pI/(1+Iz p1) on R and let

from w and note that the monotonicity components from y and the last d properties that come from the interpretation of F as a mesure imply -

Prokhorov's Theorems

Section 8.2

Weak Con vergence

Chapter 8

278

'

Gqj for q E Qd Fnk (j) Using Fa() % Fn(z) % Fa(r) and the convertnce the claim follows emsily. lt enters into Up to this point we have not used the tightness assuzption. tlke f a piobability distribution fpction only to guaiantee that F is t he p :7 for 1l i, memsure. Given .7 choose a compact set ff so tht Fk (ff) k 1 associ>ted where- Fatff ) is short for the probability of ff under the mewsure witi ire., one with the distribution function Fn. Now pick a good rectangle the ibabilitk of ai bi $ D for all i, so that ff (:: A. The formula in (iii)foi about pointj good at imply convergenc .A and the previou, claim

J-lleld

,

gnerated

by the

'

(z :

P#of To argue that d D 1t* observe that if Gi are open then Gi for 1 K K J is open. For the other inclusion note that

zi

e

-.+

n

l!/

K 51

p(z,p)

:

=

nrzzz y

.,

,

F(d)

ruli

Finally. we

t

wltich

=

Eav to

of interet

lim Fas(.A) -yx

prove

k

thatJks

Fntff timinf '

n''ex

. : .

.. .

)k i

-

'

F. For thi w)

=>

in its oWn right.

6

.

.'

' '

' .

E

:

te

.

.

. .. .

=

EI

-

Rd be the projection zutzl Let g'd : R* (zl, zd). Our first claim is that if 11 is a tight family on (R= then fy o g-i t ;t e ll) is a tk ht family on tRd, 7zd). This follows from the followhlg general result. -+

=

.

.

.z

'

-2*)

,

a -1

tight family :

y

e

l1) is

Proof Given E, choose in S a compact ' c'= ? If ff ) then ff is compact and all p E Il.

tff

theE

:::.

/ n' 1/n) e 'J=.

: pz,y) n') us==z.(p

p(z, y) <

(2.6) Lemma. If 11 is from S to S' then (p o

g

assoiated In Rd probability measure pn y if and onl# if (2.4)Therem. a11 good F. for F(z) #atzl for that are z functions have distribution

:

.

fllwirig

t

solp

e 'Jx

m=1

'oi)f

-

K E z-'npotzm,pml

:

if on (S, S) ad a tight family on set ff so -ltff

?)

is a continuous (S?, S').

that ptff

)k

1-E

map

for al1 ;t E l1. k 1 :7 fr

?) D K so p o /z-1(.Jf

-

El

-+

'

. ..

E'E. . . '

E

.

'

'

'

.

:

i

.

.

'

.

' .

'

= 0, o if If z is goo for F, then Proof (g : y % z) has p0A) We the F(z). Fs(z) of iz;y p then To implies cnverse prove (iv) (1.1) pn will apply (2.3)to the collection v1 of finite disjoint unions of good rectangles. (i.e.i rectangles, a1l of whose vertices are good) We have already observqd is /l(.A) for a1l .A E To complete the proof we note that that ps(.) closed under intersection, and any open set is a countable disjoint unio of good rectangles. =

.

E

-+

.

-->

.4.

.4

Using (2.6),the result for Rd, and a diagonal argument we can, for a given converges to a sequence ps pick a subsequence Jlnj so that for a11 d, Jlst o limit vd. Since the memsures va obvlously satisfy the conslstency conditions of Kolmogorov's existence theorem, there is a probability measuve v on (R* j T*) g'- 1 vd. so that v o We now want to check that ynk v. To do this we will apply (2.3) lz(:./1.) with the finite dimensional sets W with 0. Convergence of fnite ?z(.) and dimensional distributions for al1 .A e #. (1.1)imply that ynk (.) 'md

,

=

:z:)*

.4

=

=

-+

280

Captez

8

Weak Convergence

Sectfon 8.2 Pzoorov's

The intersection of two fnite dimensional sets is a fnite dimensional set and t'9(.An Bj c: (').Atl OB so is closed under intersection. To prove that any open set G is a countable union of sets in we observe ptz, yj : > < there 8 that By, if G is 0 (: E a so that p ) G. Pick J ) (z CG 2J) large that Gc R= By, 6 can 0. so we suppose G' # 0.) If n # so pick N so that 2-N < J/2 then it follows from the deinition of the metric we that for r < /2 the flnite dimensional set =

.4

.(p,r,#)

(p : Iz

=

!&1< r for 1

-

.N'Jc

S

By, J)

c

G

.

.'

.

.

.

.(g,

.

.

.

yi,

=

=

-y)

-

,

.

=

completesthe proof

of pug

z::)-

(7I

v.

Before passing to the next case we would like to observe that shown

have

we

-+

=

,

,

-+

The boundaries of these sets are disjoir!t for diferent values of r so we can pit r, N) hms v measure 0. r > J/4 so that the boundary of of th be an enumeyatipn Let y pa, As noted above R= is separ>ble. dense set that 1ie in G and let Jl members of the countable rf, #) G suppose be the sets chosen in the last paragraph. To prove that Ufyk and pick G, that B > G 0 let C a point yi sp yhat so y tlf-f z g (z, Bz, yi) < yl9. W claim that z E Wf To see this,note that te trilpgl 719, which inequalityimplies Byi, 8v/9) c: G, so & k 4v/9 and r > J/4 ,

.

tzw)

; i

tzn)

=

,

.4,

281

lf the points zn p(z, qi) and hence z in S then limptzn qi) (z). Suppose z # z/ and let 6 = ptz, z/). If ptz, qi) < V2, which must be true for some i, then p(z?, qi) > E/2 or the triangle inequality would lead to the contradiction :7 < p(z, z9. This shows that # is 1-1. Our fnal task is to show that /z-1 is continuous. lf za does not converge :7 > 0. lf p(z, qi4 < 6/2 Fhich must be twe for to z then limsup ptza z) f) ptza again then limsup the triangl ineqality would jiv spme qi k V2,oz contradicition, and hence does not converge to (z). This shows that a. (zn) /z-1 if continuous. then is zu z so C1 (z) -->

.4

=

Teorems

-+

lt follows from (2.6)that if 11 is tight on S then (p o : p E 11) is a tight family of measures on R& Using the result for R= we see that if yn is a sequence of memsures in 11 and we 1et vrt pn o ?-1 then there is a convergent subsequence vug Applying the continuous mapping theorerp, (1.2),now tt? the = function weakly. it follows that puk o vng converges v -1

.

=

.

=

The g eneral plii) k 1

Whatever S is if 11 is tight and we 1et Ki be so that case 1/ for a1l y C 11 then a1l the measures are supported on th Uff and this caze reduces to the previous one. set Sz (a ,

'

-

J-compact

Proof

=

of

(2.2)

We begin by introducing the intermediate statement:

(H) For each t', J > 0 there is a flnite collection W1, so that ptUksa.Afl k 1 :7 for al1 y 6 H.

.

.

.

.,z

,

of

balls

of radius

J

-

(2.7) Theorem. ln R* weak convergence is equivalent dimensional distributions.

to

Proof for the c-compact We will prove the case reducing it to the previous one. We start by observing

result

,

flnite

of

convergence

in this

cmse

by

We will show that (i)if (H) holds then 11is tight and (ii)if (H) fails then 11is not relativley compact. Combining (ii)and (i) gives (2.2). Proof of (i) Fix e and choose for each k flnitely many balls d 10 da k of radius 1/ so that p(Usapd,) closure all 1 for l1. Let K be the k ;t e V2 of n2'.1Uuns .Av Clearly ptff ) 1 1 &. ff is totally bounded since if Bb is a J ball with the same center as A) and radius 2lk then G?,1 K j %nk covers ff Being a closed and totally bounded subset of a complete space, ff is compat and th proof is complete. I::I .

lf S is c-compact

.

,

-

.

(2.8) Lemma.

.

-

then S is separable.

.

.

.

'

. '. .

.

Proof Since a countable union of countab le sets is countable, it stzces tp of radius t/n, let znZQ, pr ove this if S is compact. To do this cover S by balls and check subcover, of tha t the m of balls fnite the the centers be a m K ms E1 countable set. dense are a .

i

Proof of (ii) Suppose (H) fails for some t: @nd J. Enumerate the memers of the countable dense set :1, q2, let A Ulkjdf Bqi, J), and Gn For each there is a measure yu E 11 so that Jza(Gs) < 1 6. We claim that yu hs no convergent subsequence. To prove this, suppose pnu :::t. v. (iii)of (1.1) implies that for each m .

.

.,

=

=

'rt

-

(2.9) Lemma. If S is a separable metric space then it can be omorphically into R*

embedded

homc

.

Proof Let 1, 2, from S into R= by .

.

.

be a sequence of points dense in S and deqne

(z) (p(z, =

z), p(z' :a),

.)

-

.

a mapping

'

zzttpml %limsup pnk (Gm) -.cxl

s

limsup paptG,,sl k-+ oo

s

1

-

e

.

Capter

282

Weak Convergence

8

However, Gm 1 S as

8.3. Th

so we have vsj

1x

m

Section 8.3

<1

:,

-

jjwjj

.suj

0%

l <

(3.1) Lemma.

sets.

.

' (

.

'

'

.

.

.

(, i) to

functions from

82 <

.

.

jw(j)j

(3.2)

B is the same as the

Xd

E

orel subsets

Rd

of

E

ttd: II

-

IlK e 1/nJ n =

-

((J

Ia'()

:

-

()l

The lat

-k

:

1/nJ

-

u

-+

-

-

.

Let 0

K

l <

ta

<

.

.

<

.

trj

S

1 and

.

-aa(tJ)

=

m

:

C

(Rd)n be deqned by

.-+

.

(*(1), .

'

.

y.' r .

g .

.)

-

'

.

.

.

.

.

,.

. . .

:

t

.

...

. .... distqibtp

memsure p on (c,c), the memsures p o m-l:' which give te . . uder g, are called the flpite dlmepjtozwl . of the vectors (-'%z , , .Xa) tion dlstributions or f.d.d.'s for short. On the bmsis of (3.1)one might hope tliat of the f.dkd.'s zriight be enough for weak convergence of the yn. convergence However, a simple example shows this is false.

diven a

.

.

.

.

,

.

.-'L.)/zak dw)

.

Jt.A-t,

-+

tpdusion

..x'k)v(da,)

,

.

.

implis that p.k o z' v o zrj so from the assumed conkrkehi f the f.d.'d.'s we tbat the f.d.d's f v are deteriined and hence on: subkequential limit. there i To e tht thi ifnplies tht te whole sequence converges to v, we use the following result, which ms the proof shows, i kalid weak convergence to< on any space. To prepare for the proof we msk the reader to do -1

-1

;:z)*

,

Bxetcl 3.1. Let iu b a sequnce of real numbers. If each subsequence hms a fmther subsequence that converges to then r ra r.

of rr,

-+

'

,(tn)) .

lxtz

.

.

'onlj

the intersection is over al1 rationaks q 6 (0,1). Letting n x spws (tJ : II II< e) 6 C and B c c. To prove the reverse inclusion oserve tlkat if the are open the llnite dimepsional set (u?: w(f ) e A) is open so th g' theorem implies C C B. (:1

wher

('

Pl*oof lf pn is tight thn by (2.1)it is relatively compact and hence each subsequence p,,. hms yurther subsequnce pak a ( converges ( o a jkm yj v If J p R R is boppded apd continuous then fu .X%k) is bounded and continuus frm C to R and hence ,

fuction

a given continu6s

.c.

uosa

, Observe that if # is

Proof

(Y)px(d+)

=

-+

'

E

.

C genrrated by the fipite dimension>l

J-feld

,

dimensional

K )

the

,

1 /* 0

Theo4p. Let yn 1 K n k be probbility memsures on C. lf the flnite r:xn distribut (ons o y pu converge (o oy yy aud jj tjw yg as (jgu then y'k z::> #x.

(

.

for 1 %

K 1 and A

tk

<

.

.k

=

'

of Borel subsets C. Itroduce the toordinate ratlor and (8) deflne the flnite dimensional sets by

i Xti @)

()sp(d)

Rd

of

(tJ

where

ts 1

collection =

->

'

=

Let B be the vayiarbles.X((J)

-..+

..

fR(0,1), Rd) be the space of continuous with the norm eiuipped Let C

283

As n x, A() fn H 0 but not uniformly. To see that pn does not converge weakly to yzx, note that h(a)) = supcsfsl (t) is a continuous function but

a contradiction.

Space C

The Sppce C

..

.

(3k3)Lemzk If each verge.sto v then gn =y.

subsqence

of itn

v.

has a further subsequence

that con-

Proof Note that if J is a bounded continuous functionk the sequence of real numbers J()ps(#w) have the property that every subsequence hms a further subsequence that converges to ftvdw). Exercise 3.1 implies that the whole sequence of real numbers converges to the indicated limit. Since this holds for any bounded continuous the desired result follows. (:I .f'

J

.f

Exnmple

3.1. Let cn

=

1/2

-

1/2n,

a :::F1/2

mmsson the function

-

1/4n, and let pa be the point '

.

0 zlntz

cn)

-

A()

F:z

4s(1/2

0

-

z)

z z

z z

cnl

e (, n) e a, e ps, 1/2j e (1/2,1)

As Example 3.1 suggests, pn will not be tight if it concentrates t at oscillate too much. To find conditions that guarantee tightness, duce the moduls of contlnulty ta@)

=

supt

ju7ts) -

(t)l : ls l K J) -

on paths we intro-

.

Capter 8 Weak Convergence

284

(3.4) Theorem. The sequence are n(), M and J so that

(i),a(1(0)1 (ii)pnw6

> > c)

M)

yn is tight if and only if for each e > 0 there

< .: for

all n

k

Proof

From

no

'

'

. .

(3.5) The Arzeia-Ascoll

and

only if supwux

(see .

.

.

:

.E

page (1988), :

'

.E

..

.

and lim,-.c supwix

:

...

AL ubset .A of

'heo/em.

< x 1kv,(0)1

Royden

.g.j

.

p)41 +

169) .

p,

a.s

.ts(u)

c

'

'

.

:

.

Piqk N so thay Jf2-N/(1 tt foltows that >(,t 7k e)

8.4k

'

.

'

.

.

To prove the necessity of (i) and (ii),we note that if p i tijht and E > 0 we can choose a compact set ff so that pntff ) k 1 6 for al1 n. By (3.5), ff C (X(0) %M) for large M apd if e > 0 thp ff c: (ug:%E) for smarll J. of (i)arnd (ii),chopse M sp that pn (IX(0)I To prove the suciency lt M) S e7/2t1 for all n, It we let f / 2 for all n and choose %.so that ynlnk > 1/:) % ff be the closure of (X(0) % M, K l/- for al1 1-) then (3.5)implies ff is all n. CMITIPaC t- an d yn 1 EI foy e k (ff) -

-

.l

-

E '

.

K 2-N/(1

-

l.Xk(U Xn (r)lS

.

..

tz

.

.N

.

L

..

Vkorokhd's

*.

q

E

!

-

.

E

'

: is triv ja j w jjeyj For example zt Condition (i) is usually emsy to check. result, .Xk(0) cp za and 4a The next useful z as n -..A x. (3.6),will b J'andpm condition result of formulate in checking Here, in ih we terms (ii). variables Xn taking values in C, that is, to say random processes (Xa (f),0 K t K 1). However, one can ealy rewrite the result in terms of thei: distributions ps(.) Px 6 .A). Here, d E C. -+

=

(3.6) Theorem.

enough so

rl-t

for

2-)

s

ksc.

:

Exlstence

p) > 0

.

then

s

2-(1-n)N

Theorea

2-*/) then with

-

t, r c Qa f3 (0,1) with

-

285

that

3 2(1-n)-/(1

=

we have

-

(1+

-

./1.

2-)

2-(1-9)X

so jhat A%

%6

and

o

for SDE

In this section, we will describe Skorokhod's approach to constructing solutions of stochmstic dilerential equations. We will consider the special case (*) dXt = 'uk

)dft

.

'tzl4k

Existence Theorem for SDE

py)

-

Chapter 1 it follows that if

probabilityk 1

'

.

(1.7)in

's

and pick p > 0 small -

'

.

Skorokhod

= (1

=

We begin by recalling

<

no

Remark. Of course by incremsing M and decremsing we can always check the 1 but the foizfmlati i (3t4)eliminates th need fr tha condition with no Also by taking e: = p A ( it follws that if p is tight thn qnal adjustment. there is a and an nc so that yrkon > p) S ( for n k nn. Proof

Let y

&/#,

E

.

e for all n

s

Section 8.4

where o' is bounded and continuous, since we can introduce the term (.X) dt by change of measure. The new feature here is that c' is only assumed to be continqous, not Lippchitz pr even Hlder continuous. Examples at the end of Section 5.3 show that we carmot hope to have uniqueness in this generality. Skorokhod's idea for solving stochastic diferential equatipns wms to discretize time to get an equation that is trivial to solve, and then pass to the limit and extract subseqential limits to solve the original equation. For each n, defne .Xk(t) by setting Ak(0) = z and, for mz-n < t K + 1)2-n,

tm

AQ(f) Xa(m2-n) + c(Xn(m2-n))(#t =

-

#(m2-''))

.

E

Suppose for some a,

#> 0

Since Xn is a stochmstic integr>l Fith respect to Bzownian motion: the formula for covariance ot stottkmstic intgrals implies (.A-n

1

xijt rt

0

k

then (ii)in (3.4)holds. Remark. The condition should remind the reader of Kolmogorov's ontinuity criterion, (1.6)in Chapter 1. The key to our proof i the observation that the proof of (1.6)gives quantitative estimates on the moduls of iontinuity. For a much diferent approach to a slightly more general result, see Theorem 12.3 in Billingsley (1968).

t =

tc.rc.jkltxatgzrzsj/zallgs

t

= wherems usual' c = O'O'V. If followsthat if s < then

cfj(A%.((2nsJ/2n))ds

0

that cjtz)

we suppose

< M for

,

I(A'n' -Yjlt ,

-

(.Xk' Xni ,

lal ;

Mtf

-

s)

all

,

j

and

z, it

Chapter 8

286 so

(5.1)in E sup uE(:,)

Taking p

=

Weak Con vezgence

Chapter 3 implies

Section 8.5

Skorkhod's representation theorem, (1.4),implies ! that we can onstruct processes 'F with the same distxibutions s the Xgk) on some probability space in such a way that with probability 1 as k converges to F(t) x, Fk (@) uniformly on (0,T) for any T < n. lf s < >nd g : C R is a bounded continuous function that is msurabl with rspect t ,F,, then

E

llk (u) xni(s)IF s cpl'l(xj ) -

-

'

(x,$),1F/2 s

cp(.&J'(t

-

s))#/2

.-+

-+

4, we see that E sup lAk (u) Xni(s)l4K CM, 2(t j1 ' uE(.s qf: =

'.'

-

s)2

E :(F).

E '

T(K) T(K) -

J'

-

@

,

Uing (3.6)to check (ii)in (3.4)and then noting that .Xktl so (i)iik (3.4) is trivial, it follows that the sequence Xrt i tight. Ivkink Pfohorov's Theorem (2.1)now, we can conclude that there is a subsequence Xnk) that converges weakly to a limit X. We claim that X sqtis:es MP(0,c). To pypve this, we will show (4.1) Lemma.

If

16

fAt

C 2 and

IAj

-

k,

J(z) cp

J

j

-

Since this hols

theoremshows

is

a

local martingale

'

;

izj. '.

J(z)

lxnj

=

J''! i 1

lL'ilutd-Mj t

--

+ i' )! -

ij

So it follows from the definition of Xn

.

,

xplr

.

'

l.'

.

0

,

-

1

xjlr

that

-

.

.

s

sm/.zriifif t linear

=

t

a.f(Ak(E2nz(2-'z))du

Thorem.

rovptanmption:

As

n

m/n

=

e (m/n,(m+ Bn

x,

-r,

singt

' .

'

..

.

'lz

1

n ta

-

n

'l.l

,

.

*

*

.

n

B Fhere B is a ptapdard

:::>

,

'

'tis

(32) .

Convergenc f flnlte dlmnslonal vatiables, sum of (nt)independent randrr theorem that if 0 < 01 < K 1 then .

1)/p1

y

,

We w ill p rove thi iesult .

rj r12

.

.1'

B

Prof

))s it ytwa ; z-sztr) )7cf3(Ak(E2 it Lnfr) dr is a local martingale. J(.Xk(s)) =

.(.

0

=

.

indexed s s

( hn

so if we let

then J(Ak(t))

=

/,'

.''z

Let &,&, be i.i.d. with E = 0 and 1. Let Sn = (1 + + a be the partial nth sum. The most n>tural way to turn Sm 0 S m % n into a process = sLntjly-.n where (-z) by 0 t 1 let is the largest integer z. continuoud To have a trajectory we will instead 1et

(5.1) Dopsker's

r

(Ak

.

te

.

.

ihigs

are t*o

dlstributlons. lt follows frin

.n

n

-

m-z

,.a

)

Z:'

(B tl

3

B t2

-

Xt

1. ,

*

*

-

B tm

To extend the lmst conclusion from hn to Bn, we begin by rn nt (nt)then Bn hnt + ra knjj-vl-n =

to do:

Since Syj is the the centrl limit

'

-

B tm-z

.a

-

lmss '

Theorem

(r)

,

f th motpne

z-ytyrlkrj z'al

-

c

,

'

Dijlxrjjdqv'l

7,(y,)

-

.

-

=

'z?

=

Proof lt suces to show that if f, Dif, and Dijf are bounded, then the martingale. above It's formula implies that is a process .f(.Xk(t))

zsytyr) dv

la

8.5. Donsker's follws by applying the result to

/'

-

'

(4.1).

which proves fz.f(A%)ds

gnk)

-

g, an appliction

foi any cotinuous

.s

t

3

.

(J()

then a Efyait o fy g

#r))

T(Fr)

) y(x)

,(y

= lims

J.

=

conclusion

Once this is done the deslred z and

.

Donsker's Theorem

-

=

pbserying

) that if

!jl'..

t

! Capter

288

8

t!

Weak Convergence

Section 8.5

Since 0 % rs < 1, we have rnyq-j-l/v-'n 0 in probability and the converging together lemma, (1.3),implies that the individual random variables B7 u:). Bt To treat the vector we observe that

whereC

-+

.

CB

f''

B

-

i

in probability,

''.f=

z

) (t''. =

1'.f-) (ft''.

-

B: 1

-

,

Btnm

.8y.

-

,

.

.

.

,

tn.

-

)

,')=

CB

-

.,'',-

-

z

z

)

0

-+

..-

z

) c:),

B3

-

,

, ,

.

.

.

,

Bt

'

Pwof

Bt

-

.

m-

(iii)needs

-+

. '

.'

..

'.

.

'

'.

..

.

'

.

.

.

'

.

,

.

..

. .

.

.

.

.

.

.

.

,

'.

'

Tightness. The L2 rpaximal inequality for mprtingales Chapter 4 of Durrett (10j5))implies 2

E Taking '

P or

writing

oslsz

.. .

.E

:

'

xpax oKlKa/m

(seee.g., (4.3)in

things in terms of P

,

.4-

(i) P

# i.

(ii) 1, (iii) lf M k -

,

'

f)

From this,

(4ponly

(ii),apd

.

.

= .

.

Iltn

B,n I >

be i.i-d. with

<

:

-

mran y.

f)

.

'

.

aM)lM2

=

.

lI

1 and a(M)

>

M)

.

K aM)lM

t,jz fpuj4 -

%

CJMZ

=

:+

'

'

.

+ (2:/2

and Ca only to make it clear. that

Since ELo pu)

z::':

-

0 aud the

are

4

care about p zrtJtk

s

-

'

.

()

-

-

#M)

M

2M2

2z.P(l6

I>

E

L -

+ pwzl .

.

z) tfz

2.:1*2p%

=

puj* K 16(#>4 + 242pwy) '''

Since we have supposed M k 1, this gives the term CJM2 .

.

.

complete.

We will apply

(5.3)

(5.4)

artd

the proof is El

(5.2)with

M

Part x-n.

=

(

npfi convergence

# -f) < n

(i) implies

asy-n) -+

.

sz,,

theorem. Since y

nljuzl

n

.

at.zrl &/T

=

t 1i$ lt

t

xt',l?l . .

z) d

t

!t

:?

the fourth moment we notice that

(16I >

4z3P

.

,

(

4 Efj 2

+

#wz)2K 41v2

j

.

.

i k

2, 4) then

=

11

' .. .

So using our inequality for the pth power again, we have

by the dominated

< 1 then E

Ii

Let 1(It.1s.&.r),-let .E'(4?). Efi, and pv z

=i

=

s

4/m:2

.E'(l(I2; Ifl % ?)? let pu

II > s M-2.E'(lfI2;

K ftlflj

M

-#'?)

N

'. .

-

This gives tlze term Cz'2. To estimate '

1y E

...

it follows that wssumptions

our

(

E

-

c1

3- pu t

putp

-

Ef- J

4/m:2

-

:

pwzl

integer

y

max >XtX>+l/m

(.2) Lemma. Let l (2 S T Fl: ' let a(.T) .t. #. Ten w have

t

an even

E

Since we need m intervals of length l/m to cover (0,1) the lat stimte is not good enough to prove tightness. To improve this, we will truncat d cozrtpute . fourth moments. To isolate these details we will formulate a general result. .

lf p is

('

15'I> EWn

E

.' .....

.

-

S 4E%2

-k

z

(V

j1)

289

jlwzj.

+

=1 4 z = XEfi- gwz)+ 2

'

.

z

=

.

and using Chebyshev ) s inequality it follows tltat

n/m

=

max Sj

.

.

' ...

to be proved.

.&t-jz tpvl4

za) Using the tct that (zl,z2, + zm) is a zl + (z1,zl + za, continpop? i.papping and invoking (1.2)it follows that the finite dipensionql t dlstristln . . of Nn converge to those of B. .

' .

. . .

-

.

6% = 24

.

Only

)

,

and

. ' '' ' .

independent Bt z Btu

(#4ay

i7zT)

+

Remark. We give the explicit values of t hey only depend on pu and pv.

'

(1.3)that

so it follows from

(fn l BjnQ

=

,

'

.

32

(.

.

-

,

=

Donsker's Theorem

=

0(./)

0 0 part

(ii)implies

Capter

290

8

Weak Convergence

Section 8.5

x. Using the L4 maximal inequality for mattingales as n (againsee e.:., (4.3) in Chapter 4 of Durrett (1995))and part (iii),it follows that if ' = n/m and n is larte then (hereand in what follows C will change from line to line) -+

E sup

1,9k wzI4 < C'flst -

oKkKf

'#wzl4

-

sup

0KK;

lxi

l k EWn/9

kpu

-

B = Jsj/v-n and Ik,m probabilitiesin (5.5),it follows that Let

limsuoP -x

a

....

max

0< < m

I.n max E Iy '

,

.

@

-

m

To turn this into an estimate

J

H*

-

r

+

1 =

rqo

<

'

-

c6r4

of continuity,

2

s

+

-

-

*45.6)

and

J so that 6:-442 limsup nY

+ 1/m)

m

#(1z?1/m(Wn) >

6/3)

msn

-+

cx) s

sup

ostsl

I'fn htnI> -

e/3

limsup n-X

> e)

f

of

Let

Donsker's theorem.

maxttt)

=

max osmsn

k

c)

: -+

0 K t % 1J. It is easy to see that R. is continuous, and (5.8)inplies

SmlWn M =>

J5(Tc

=

that it is continuous E

-

15(M1

-

e7

R has the property

-.+

S

max Bt cstsl

Y ,

To complete the pictuye, we observe that by of the right-hand side is 1)

(3.8)in Chapter 2.P0(11

=

k

1 the distribution

c)

txnmple 5.2. Let #(tJ) = suptt < 1 : (f) = 0. This time # is not continuous,fpr if oz has tJE(0) = 0, E(1/3) = 1, :(2/3) = :, w(1) = 2, and linear on each interval (,/,i + 1)/3) then #@c) = 2/3 but #@,) = 0 for E > 0. lt is ewy to see that if #(w) < 1 and (f) hms positiv and negative values in each mterval (#@) &,#@)) then # is continuous at tJ. By arguments in Example 3.1 of Chapter 1, the lwst set hms memsure 1. (If the zero at #(*) isolated left, would

J(,)I

-

.&

K e7 contluity

-+

the triangle irtequality implies .p(o,/m(n)

'*oj

we note that

g : it follows that

To convert this into an estimate the modulus of by observing that (5.3)ard (5.4)imply P

5.1.

-

-

-

If we pick m

> e)

the proot

1'/(*) #()I E 11 #11so 4 : c(0,1)

i (k+ 1)/m K t i (k+ 2)/m

I/(i) /(s)1K I.f(t) lk + 1)/m)l + IT(( + 1)/m) T(/m)I + IJ(/m)

than the maximum

so

ws-v,inhn)

s

completes

(5.8) Theorem. lf # : C(0, 1) Ptra.s. then lBnj @ 4(.8). Exnmple

1

-

i

291

,

'tt'1/m(J)K 3 ctrjaxwtyyj,rrj: a.x..IJ(s) J(/m)I since for example if klm

(ii)in (3.4)and

smaller

Theorem

The mairt motivation foi proving Donsker's theorem is that it gives as corollarie? > number of interesting facts about randpm walks. The key to the vault is te continuous mapping theorem, (1.2),which with (5.1)implies:

1)/rn1. Adding up rrl of te

el9

'

of the modulus

:2

C%-*.

hunt-I >

-

uqnnj

j is

E

l/rlz,(k +

=

(5.7) This verifes

Usiug Chebyshev's inequality now it follows that P

,

.Ptu)j/amtfnl limsup n-O

n2

-

.

of Bn over

oscillationof hn over Ilnsl/n,(gnf)+ 1)/n) and it follows that

2 1 s c(zM2 + z2)< c mo+

(5.5)

Now the maxhnum oscillation

Donker's

sc

wms of

hn

we begin

on

sptm

the

it

not be isolated on

%n Sm-l Sm K Qjlt o L :

.

The distribution of L, given in 0

(4.2)in

=

the right.) Using

suptt

; 1 : Bt

=

=

c)

I

0J

Chapter 1, is an arcsine law.

Bxnmple 5.3. Let /@) = j(t C (0,1) : tJ(f) > tz)l. The point that # is not continuous but it is emsy to see that # is continuous with l( e g0, f) : () = clj = 0. Fubini's theorem implies that

.EI( e (0,1J: Bt

(5.8)now

1

=

jo .!aa(.st

=

c)

dt

=

()

H a shows at paths tu

292

Capter

8

# is continuous

so

Weak Con vergence Pn-a.s. With a little work

llmf Before doing that that I(t 6 (0,1) : Bt

n

work

Proof

:

Sm > c/fl/n

(5.8)gives

m

ca/fi)l

I(fe (0,1) : Bt

=>.

l(mK n

Chapter 4 shows

:

Sm

>

i'orsxedu5.1, Example

0

>

E

EW)

K nptllml

>

Kn-

Evh'j

>

'

wtmaxmss Ixmls (1.1)it follows that

we have apd

The right-hand

(Bllv-ntk

' .

:-2.E'(xm2 ;

Ixml> :W)

.

.

and on

convergence,

.

IAk l K EVRJ, w (maxmsa

.. .

.

.

.

5 ' .' . .'

hv

.

l .

'.

.

.

.5 .

max

() Kj< 1

dt

'

.

in probability

(5.s)so

using

M

n

and it follows from the converging

n

-.+

0

together lemma

n

j'''lsk m

-l-(/2)

m

m=

=

0 by

m=1

.

c.zr;)1 I(t e (0,1) : B;%> (c+ e)4:J)l / ll(m s ?z : sm > n 1) : B1 > (c e)./fi)l K I'(t6 E0,

l# l <

j''qSkm

n-1-(/2)

-

0

.

0

-+

.-,

side is close to 1 if M is large so

i .

+1))0y

293

(k

m=1

n-w

if

n

y'lsm,s

limiyf .P(Ga(M)) k P

note tht

Te Space D

.,r-t,+2)z'z-)

> c).l

cW)Iwe

-

1

s

by dominated

/-

,,-1-(,/a)

dt (stnl/.z,;lk

that for any c,

a result about

Ptm!.x IAkl

.

1

c)I

(9.2)in

we would like to observe that 0Jlhas an arcsine 1aw under

I( e E0,1) : B.1 >

(5 9)

I( e

we have

(0,1) : Bt > .&.

>

Application of

To convert this into then

zc.)-

(5.8)implies

Secton 8.6

1

1 0

(1.3)that

skt dt

It is remarkable that the lmst'result holds under the assumption and EX,? 1, i.e., we do not ned to assume that .E'l.X'/l< x.

that EXi

=

0

=

Combing

this with the flrst conclusion of the proof and using the fact that a with probability one, we arrive -+ l( e (0,1) : Bt > JI is contluous at conclusion. emsily at the desired =

Exnmple cotinuous

5.4. Let #@) = J(a,ljtJttld so applying (5.8)gives

where

rl

1

j (f1'/W)

dt

=>

j

k

>

0 is

an

integer.

# is

Bt dt

To convert this into a result about the original sequence, we begin by observing that if z < # With lz pl K 6 and Izl IpIK M then -

,

1z From this it follows that c,,(M)

=

-

p

:.#.r+1 p lzl+l dz S k+ 1 k+ 1

ls

'

'

8.6. The Space D In the previoussection, forming the piecewist linear approzmation' was n >L noylng bookk-eeping detail. In the next sectlon vWhen w consider proces that jump at random times, making them piecewise linear will be a genuine nuisance. To deal with that problem and to educate the reader, we will introduce the space D((0, 1), Rd) of functions from (0,1) into Rd that are right continuous and have left limits. Since we only need two simple results, (6.4)and (6.5)bc'Chapter loW, d e an flnd this material in Chapter 3 of Billingsley (1968), and of Chapter Jacod Shiryaev 3 of Ethier and Kurtz (1986), VI or (1j87),we will content ourselves to simply state the results. W begin by deflning the Skorokhod topologyon D. To motivate this consider u

Kxnmple 6-1k For 1 %n K x let

on

max lxml M-k+ztv-n, mKn

s

max mKn

< MWn I,$'m

A(f)

=

t

0

t 1 t

e (0,(n+

s

1)/2n) 1)/2n, lj + g(n

,

294

Capter

8

Wak Convergence

where (n+1)/2n= 1/2 for n 1 for all n.

=

Section 8.6 The Space D

We certainly want

x.

f,t

Jx

-+

llA-/k 11

but

y. .

.

g

Let A be the class of strictly incremsing continuous mappings of (0,1) onto itself. Such functions necessarily hate (0) = 0 and (1) = 1. For J, g E D defne df, #) to be the infmum of those positive : for which there is a E A

so that

sup

l()

-

l< e

and

fu and

=

= ,

tm

d(Jn,Jm)

r?z+ 1

-

=

2m

2n

+ 1)/2m so

;

-

1 2n

fm in Exampl '

'

Exnmple

6.2. For 1

Kn gutj

ln

order

to have

and ((n+ . .'

.

=

,

:

1)/2n)

,'

< cx) 1et

=

=

:

gml

dgn, T e po in twise limit of gn is g. to show

H

=

(6.1)Tliikl. sie tfliky 'E i ' . .. .

(1/2,(n+ 1)/2n) ((n+ 1)/2n, 1)

1

s

llogtn/mll) under

the metric #n.

' ..

we must

:

' .

:

.

=

:

'

1

'

.

..

.

'

(

..

dfj are

equivalent,

d we rtoti

i.e., they give

(.

.'

:

r

(0,1). For

.'

jgj

gjjg

.

:cx

0<

Tise

Where

t the

gup

(! (j)

tsjj

.

: Sj

...

i C

Sj

'

.

.

''

tight:

(. '

< 1 put

'thf

1/2)

(1/2)

and

discussed the simpler metri

.

! .

.

j .'.

have

The metrics d D.

.

'. . '

why we

)

inf m>x (t) o
=

,,+z)(#)

'ttlp

f

the infmum exteuds ove: the qnite sets

(tl

with

-

ayyj

0 but dgn, g.)

6.1. If h C D then liminfa-.x

mintl,

=

The space D is complete

o

for each S C .

=

1.

0

Weleake it to the iarder

The analoge (.2)

Exercise

<e

. O fifst tep i studying weak . cohvergence in. D is to charaterize Gfiialiing ns. a deqition from Section 7.3 we let . .

.

in the deflnition of dgn, gm) (m+ 1)/2m so

g((t))l

-

so they no longer form a Cauchy sequence. ln fact, there are no mor problems.

(0,1/2)

< 1

dclla, Jm)

For the reader who is curious

2m

-.+

0 te 1 t e 0 t6

lJ()

sup

e for which there

-

-

When rrl (x) we have dfn Jx) 1/2n so fn fn in the metric d. We will see in (6.2)that d deqnes the correct topology on D. Howevex, in view of (2.2),it is unfortunate that the metric d is not complete. =

.

..

(6.1) Theorem.

1

and

e.

295

It is easy to see that dn is metric. The functions a gr in Example 6.2 ave

-

It is emsy to see that d is a metric. lf we consider 1 6.1 then for .7 < 1 we must take ((n+ 1)/2n) = =

llll <

IJ(t) p((t))I < .7

sup t

n+ 1

Eor J',g E D define #()(z, y) to be the inqmum of those potive is a e A so that

=

dgtz, #)

>

0.

.

(q

.

To fix the problem with completeness be close to the we require that identity in a more stringent sense: the slopes of a11 of its chords aTe loqe to 1. lf C A 1et (@) (s) Illl sup log f #t -

=

-

.9

are

tc

=

of

au

tl <

(3.4)in

'horei.

a0j

<

.

.

.

<

r

and

ti

-

>

i-l

5 for 1 %

Kr

the current setting is

The sequenc

yn is

tight if and only if for each

6

>

therr

y

> M) < e for all n (i)pn(I=(0)l k no pzrltut a1l 6 for n k nc (ii) 4< 'te thktthe only kigefencefrom (s.4)is the .>

get the main result

we ned.

'

in

(ii). If we remove

'

the we

Caper

296

(6.4) Theorem.

If for each

6

> 0

there

> M) K : for all n k (i)pr,(I*(0)I E) (ii)pst'trj > < .7 for all n k no

then pg is tight and

are nc

M,

,

=

where Bz, c) due to Stroock

:

.

'

.

.

'

.

' . .

..

.

E

.continuouq

we

time time

=

:-;r1

I

(g : lp

=

yi

Ip-rIS1 :f 1

(p

f).

clfftz,

-

dp)

zflzfstz,

-

dy)

:)c)

.s(z,

zl <

-

zltp.f

-

Suppose

coeciets (A) cJ and bi ar cotinuus for which the martingale problem is well posed, i.e., for each z there is a unique measure #. on (C, C) so that the oordlnate maps .X@) = tJ(t) satisfy #.(.X z) 1 and

and

.

.

in discrte in

Qsz, dy)

=

)(z)

that

V>radhan, about usion procede. ere i ur t': mez ttie will * and of discuss Varadhan but Stroock Chapter that 11 follows in (1979) both discret and continuops time in parallel. Our duplicity lepgthepp thv proof but is pzeferable we betieve to te usual yomewhat kangvous)copsomewhat, jt e other case. i) with changes argument. minor hanlles t the out: same of transition probabilities ln discrete time, the bmsi data is a sequen jz values in Ss (: Rd j. chain takug 0, dy4 M>rkov r?z for a 11?j(z, m?j e 'A.d z) Tl,(z,.A) for z ss,A .P('F(m+,), .

ct(z)

to DiFuslons

we will prove a result, convergence of Markov chains to

' . .

=

1) (z) = Ij,

section

297

and deflne

(1968). (3.3).

of yn has a further subsequence

(6.5) Theorem. lf each subsequence then p r, cz;> y. convergej to y

In this

(z,dy)

1.

(6.1)-(6.4) are Theorems 14.2, 14.1, 15.2, and 15.5 in Billingsley We will also need the following, which is a consequence of the proof of

8.7. Convergence

l1?&(z,dyjlh

'E

Jf limit p hs yC)

to DiFusih

in the limit then we hav weak convergence. To make a precise statement need some notation. To incorporate both cases we introduce a kernel

8 so that

and

nc

subsequential

every

Con vergence

Section 8.7

Wak Con vergence

8

=

Xf

:

t

f

bixjj

-

ds

Xtivji

and

=

aij

-

(X,) ds

((

,

,

.

.

,

.,

.lym,

e

=

local martingales.

are

,:

=

,

e

=

g'

(7.1) Theprem. Suppose in that (A) hol tls an d for each

(i)linutp

In this cse we deflne Xgh = Flpyy i.e., we make Xth constant on intervals (''n+ 1)). E'zz., dp) In continuous time, the bmsicdata is a sequence of transition rates Q?&(z, Rd, values Xth i.e., in Sh C for a Markov chain k 0, taking ,

' . .

W

e

s Al-'L

=

z)

=

Qsx,

.)

for

z

e &,

.,1.

e

7

d

w kth z

,

y

.x

(iii) limta If Xt with

i

E

.

.

(B) For any compact set

z'f supwox Qs(z,Rd) ,

< x.

. . ..

''

,

< x,

IctJtzl cj(z)l

=

-

and

in

either

case

0

.

A)

suplxjss z

-.+

j R

then

we

lzll

=

(z)

=

have X)

=

0

.

0 =>

X3 the solution of the martingale

problem

z.

::.q .'

is excluded from the interpretation, we will allow Note that z e hich ccur at a ar invilble) case jumps from z to z (which Q(z,(zJ)> 0 n w chain ttpse will well Markov behaved we positive rate. To have a although

zts

= a'

..

.

.,1.

.' .

,

(ii) limhtn suplzlss I)(z)

,

d PCX,?,

supjxlsa

. . . ..

time that (B) holds, and 6 > 0

continuous

..

.

.

tlere

denotes convergence in D((0, 1)tRd) but the result can be triviallygnetalized to give convergencr D((0, T), R d) for all T < x. In (i),(ii),and (iii)the sup is taken only over points z e Sh. Condition (iii) annot hold unless the points in Ss are getting closer together. Howevei, there is no implicit mssumption that Ss is becoming dense in al1 of Rd. Renaz-sk

=>

'

Proof The rest of the section is devoted to the proof of this result. We will and 6 > 0 frst prove the result under the stronger mssumptions thas for all we have .i

,

Chapter 8 Weak Convergence

298

sup. la,ttzl (ii') lizl?jl.tlsupwlltz)

c2J(z)I 0 (iv)sups.. 1at(z)I< cxl f(z)l 0 (v)sup.. I)()I k & zs',(z) 0 (vi)sups,xl) (z) < x

(i') linutc

and in

continuous

=

=

Q(z, Rd) k

time that (F) supr

'

.

x.

Lhlzj

E

s vJ(z) +

a. Tlghtness If

J is bounded

and memsurable,

Lhfz) As the notation

may suggest, we expect this to converge

ZT(z)

=

1 x...-x aiiq:blhjfq 7. i,i

+

;;

(7.2b).1n continuous

Llxb fxh kh ) E-l j=() Fz

fxj ?1)

f

-

JcL

:

7.a(z)DT(z)

r .

f is bondd

(

and

?t

T(Xa

) is

ls

) is a

martingale.

'*

what

The first step in the tightness proof is to see

Lh to

A,y(z) k',lz =

-

g) where

#(z)

=

A(z)

(1 -

()

=

z2)(1

g(lzl2/e2),

-

z)3

0 z

hppens

E

wh

dyj

i I'?zl 11 =

1

j

yi

uini i

js 1.u11.p1

llruj'11gives xijyj

-

zjlfhj-flzlj

-

<

,J

z/J(z)

,

we apply

s

y

-,

Ip-rI :f 1

+ B6

z)

s

1p zI2IIDfjJ(z)II -

zf,tz,dy4

lp zI2Ksz,

dy) +

-

lp-rIK1

Bz, 2II.fIIxff?(z,

1)9

Now clf

(7.7)

(z)

f recalling the dennitions of so

follows.

To estimate

=

Ip-rIS1

) aud

c 0

(iv)-(vi)the

A, then using ,

of the qample paths, we will let

inf (

>

=

inftn

k0

=

mintra

=

maxtlftt)

=

lg zI2Ik (z,dy) -

the oscillations a N

(7.4)

uimijut

'

j

ad

K ; 1 k1

ztlihjfzr,yllLz,

-

it follows that

e

y

zijyj

jj

sup

(7.5)

,

(7.2a)follows emsily from the deqnition of Lh and inductipn. Since we ?. E1 (B have ), (7.2b)follows from (2.1)and (1.6)in Chaptr 7. Proof

zfktz,dy)

Noticing that the Cauchy-schwarz inequality implies

(7.6)

martindale.

z)

i

.

'

and the deinition of

-

time

=

x-x

that if

J'ltp j -

Ip-=lKl

Ilmlll (( '

we note

ly-rIS1

-

over the set

=

to

'

To explain the reason for this definitin measurable then (7.2a) In discrete t ime

Let Ae = supz

J(z))

-

y

.

(z,dyj

299

B, 1)c) + 2llJIlxff?&(z, where lVJ(z)l, and Bz supa llDuT(z)Il

1et

ffstz, dyjfy)

=

ffN

-

-

=

-

supa

wherezr,v z + crbvy z). Integrating with respect to lp zl K 1 and using the triangle inequality we have

=

-

(iii')limstc

Section 8.7 Convergence to Disions

a -

-

:

t : Ju

ra-l =

la' > IJ -

:

X.h

1; n

x(t-)1

jk

R-1

c/4)

K Nj :

0

<

t

s 1)

desired result

o a

=

0,

Chapter 8

300

To relate these deqnitions to tightnessj

'

. ''

Proof To cmses to copsider 1. zk

<

. ..

. ' .

.

. .

.

.

:

s <

a-' 1 ;

(7.11)

.

' ' .

t

s <

-

6 there are two

-.IrlT(,u)

lu

St

<

J(,s)l5

-

-

s <

.

Pyo

k

6/4)

1 f j, sup 1I?,(z, Bzt

c

6/4)

=

f#' :/4(.Xk?,) + tl/4.1:/1,k

Pvc

=

Evfly,qnuhnt and using r h

Q/4J

Our next tmsk is to

0, 1, 2, time

at

) %supzl

L Lej4z)

(7.8)ig

) K k sup

;

.

.

.

is

r A

J

+ Cel4r h 5))

=

LlltjhK

e-J

%J) + e-&P # (q e-J(1 Ce/4) s b146+

Ev (e-n.)i

.!5l'q

-

> H

Pvo

we begin by

J) <

PyN

>

t) s cx

+ e

-

(

k

:/4)

K1

-

:

expt-

.

.

sup

by (ii/). The flrst step in estimating do this we begin by observing that

.

.

#(z,

Qs, Pvc

> ahd

%sup A)/4(z)

6/4)f))

(7.2b)

&)is to estiniate (7.3)imply

1

h) + C t t k 0 is a ,/4

,

-+

0

>,(n K J).

To

at

submartingle

time r h J it follows that

+ Cel4r A J))'

Eylvtsln-ss) The repainder

of the proof is identical to the

one

k 1

in discrete time.

observing

1:1

onvergnce t

-

). To do this,

)+

,

' .

b.

>

<

apd

k fjtrt K 8) k .G(1 Jp,:/4(XzaJ)J PyN

.&(r

Using the optional stopping theorem

6K6

estimate

)< e

eke-rk

have

r

ly,el4X

1

k

n

'

a submartingale where

)S

0

-+

.Jj('61 S J). To

J) is to estimate (7.3)imply

>

(7.2a)and

Using the optional stopping theorem Boticig Xpbt = a'Gal it follows that

(7.10)

Pvc

it follows that Eve-rv

Tt g tness proof, continuous tlme. One of the probabilities in (7.8) is trivial to estimate. Since jumps of size > 6/4 occur at a rate smaller than supxllhtz, B, E/4)c) and k 1 z (7.9b)

by (iii/).The first step in estimyting do this we begin by observing that

or rearranging

in

One of thq prob>bilities

tlme.

discrete

(:I

-

-

< 1) <

(7.10)and (7.11)we

=

-

-

.!5ta

=

301

If we pick k so that e S e'/3 and then pick 8 so that Csk& K E'/3 then (7.12), (7.9a)j and (7.8)imply that for small lt, sup, Py(t@4A') > 6) S t'. This veriies (ii) in (6.4).Sine Xt is complete z, (i)is trivial and the proof of tightes in discrete time. E1

IJ(t) T(s)l S IJ() /(zk)I+ IT(,u) /.(rp-)1 + IT(ra-) T(ra-1)l+ IT(a-1) J(s)l S .7

(7.9a)

Combirting

kj

>

26/4

'ra.f.l

-

Tlghtness proof, trivial to estimate

PyN

(7.12)

l.f(t) J(s)1 f lT(t) J(a)l 2.

%6.

'

< raml -

cwsB

' '

this we note that if o. > & and 0 <

check

CAss

'

'

wxhj

0 % c/4 then

and

>

Iterating and using the strong Markov property and

we will now prove

.

(7.8) Lemma. If

Convergenceto; Dihsiotjs

Section 8.7

Weak Con s'ergence

'

Since we have assumed that the martingale problem hms a umque solution, it suces by (6.5)to prove that the limit of any convergent subsequence solves

the frst

martigle

ptoblem

is to

show

(7.13) Lemma.

If

step

fe

to cnclude

Ci then

that the whole

IIEA.J Lf 11x 0. -

-+

seqence

converges.

The

Prof 1: -

zl

Using

Section 8.7

Weak Convergence

Chaptek 8

302

(7.4)then

to ffhtz, dy4

integrating with respect

the set

over

r(e)

(z)ofy(z) zytz) J'lt', i

+

Ip-rl

(' '.

..

(J(p)

>1

T(z))Ikx,

-

.. .

.

zt4

-

-

+ . . .. .

.-.

zityt )7(!n j

Ip-rI S 1

.

.

o.J Jz.

,)zf,(z,dp) ,

. .. .

. .

j .

E

. (

r

biztihlzt

-

I)(z)

K sup

i'.

. ..

.

J

r

;i

(7.15) + (7.5)the

gr

z

.

uniformly in z. Now the diference between the second term in smaller than is

By

.f(z)I I2J(z)l

-

aijztlhil) . .

Ip-rl %1

.. . .

E

yi

'

fJ zf)(pJ

-

' .

j

'

(

-+

0

r

'

(714) .

nd

jbitfz)

E ij

lcf.f(z) sup J

d#)

c,blzll

hnL gj

n

-

y

Ixjlj,

To pmss to the lizit, property implies

# (xn)

=

0, 1, 2,

.

.

.

is a martingale

property by introduciug a we rewrite the martingale R which is memqurable with respet g : D = + 1 and n = ; Lilhnj + 1 the rparting>l Lslhnj

J(.X'ha zaa

-+

y

:

) Jxkhn ) -

na

ln -

-

1

)-')Fzaznytxk j

)

a

j=ks

=

()

theorem (1.4)implies that we can Let x 0. Skorokhod's reprqsentation P with cxi the same distribution as the Xh= 1 construct processes %n % 'O almost surely. Combining this with (7.13)and using the that Fn so bounded convergence theorem it follows that if f E Cx2 then =

'

,

ij

If%J(z)I

:(10)

J('0)

.

(p

. -

s r(e')

zityi

.

. -

ztt

Ip -

-r1

1:

X

E

.

tfhJT(zr,p) lhilx. -

zlzfflzlz,

dp)

-

,1

.f(A-,0) -

Llxr'j

'';z:

dr

0

4

Since this holds fpr all continuous followsthat if f E Cx2 then

fft,

'

g

:

C

-+

R meuurable

respect to Fs it

with

t

fxjnj

LfX,0) ds

-

0

(7.16)

k

J=0

bopnded continuous function to F qnd obkerve that if kn

E

which goes to 0 uniformly in z by (iv).To stimate the secod teiz iil (7.1) zl unifprmly we note that for auy e the integral over (p : e < 1g- K 1) goes to 0 in z by (vi)and the fact that the integrand is uniformly bounddk Usin: (7.6), the last piece

i.i

.

conklfgs

.

th>n

smaller

-

Ip-rlse

0 so that Xlt=

-%

s

-+

Iirst term is

.

' .

j

k-1

s ) Ttxkla

E

-

ij

.X'O.

,;

,

flhilzz,vt fhJT(z)l ffhtzl

zJ)

-

-+

f discrete time. Let (7.2a)implies

PrOO

..

ct(z)DJT(z)

-

-

.

C Onvergence weakly to a limit

dp) .

llfh.fJ(z.,,) DJJ(z)lI

..

Recalling the dflnition of A1y ahd using (vi)it follows that the third terrft goes to 0 uniformly in z. To treat the flrst term we note that (7.5)apd (v)imply )(z)D/'(z)

sup Ip-rlKe

=

Now zz,v is on the line segment connecting z apd y, Dijf is continuous and hms 0 as T 0. Using (7.7)and (iv)now the quantity compact support, so r(:) in (7.16) is complete. (:l converges to 0 and the proof of (7.13)

=

(7.14)

303

where

1 we have

S

Convergence to Disions

is a martingale

'' .

.

dpl

Applying the lmst result to smoothly truncated versions of fzj z and J(z) .X0 is a solution of the martingale problem. Since the follows that it zizj martingaleproblem has a unique solution, the desirqd conclusion now follows =

from (6.5).

=

1Zl

Capter

304

.XO.

to a limit

verges

Jtxa )

* Le t tlme.

continuous

proof,

Convergence weakly

Secton #.8

Weak Convergence

8

(7.2b)implies

Lh%% Jtlaa

-

0

) ds t k

0

.

is

.

.

that Xhn

so

con-

a martingale

yetj j s a so

uuou o

(e

marC1

c. Locnlization To extend the result to the original msjumptions, we 1et pk be a C* function that i 1 o (, ) and 0 ort X( j k + 1) and dejj ne l

g

,

.#' *' %'

dy)

Ik,k,

yutzlffptz,

=

dy4 +

(1 -

pklj&mdy)

,

& it

a

fiht N

''''

irs

'

.'

It is

at z. 'N

asy

'flrst

#'

vth

='wk(z)c(z)

ctz)

=

We have supposed that the martingale problem for c and has a unique to show that nonexplosive solution A'q In view of (3.3),to prove (6.1)it suces X#'a 10 To subsequence W that is there 0 iten any sequence a n a so each sequence Xhn is tight, argument diagoyal since that this note a we pove 10,#., ?n so that for each k, Xa' shows tllat we can select a subsequehce tf.J 1). : set Gk is 0 open Gk Let so if an ), K t K (f) e 0, ' a' implies th>t then d X X an If c C s open (2.1) .

:::).

-+

.

'k

::4.

.8(0,

=

:::ty

'n'

liminf PX n''-x

e Gk n #) k

PX

0jk

e Gk n

ff)

( .

=

PX

0

e

.8(0,

.

'

.8(0,

a

aVe

Ptxi nf lim n'etr i

: #)

sincethere is no explosion audthe proof is complete.

k

liminf #(.X$ n'ex

the right-hand

: Gk side

z + n

.k ya

When e' > n -1/2 Ll% r an d (ii)We note that =

r

=

n S) k .P70

incremses to

PCXD

G Gk

n Hj

e #)

as

t- 1 x

with th

Efollowing

/n c1

() z

-n

zg 2n z

-

111/a (z

z

,

=

+ zW :n

check

conditions

n

n-f/2)

-

n

1

-

n

.

n

z

-

Wn -

(iii)holds.

0 for all z so

=

1 =

'fkl-n

'

(z)

,

l/n(z)

)

. '*

Wn

u-n

uu e' .

N

L

+

1

n

.

z

-

Wn

1

'''

2p.

n

=

-

2n

.

u

y

.

'-

.

2n -

To

Wn

ug V

-z

E

-

=

27:

n

=

(i)

1

The limiting coecients 1 are Lipschitz iptiuous, jo (z) yln the martingale problem is well posed and (A) holds. Letting Xjt/a sjj/s and applying (7.1)now it follows tat =

and

-z

a(z)

::.;q

=

1/n (8.1) Theorem. As n x, Xg converges process Xt i.e., the solution of .->

weakly

to an Ornstein-uhlenbeck

,

Gk rl 1I)

1-), Xo' is a solution of the martingale problem since up until the exit from and and hence its distribution agrees with that of 10 Now Jlp to the for b a k a, of exit k-) X ? hms the same distribution so we from flrst wsthat X

'

'he

/k(z)(z)

tz)

(7.1)beginning

of

applications

Exnmple 8.1. Ehrenfest Chain. Here, the phycal system is > box flled with air and divided in half by a plane with a smll hole in it. W iitodet this mathematically by two urns that contain a total of 27: ball, which we think of as the air molecules. At each time we pick a ball from the 2n in the two urns at random and move it to the other urn, which we think of ws a polecule going t h roug h the hole. We expect the number of balls in the left urn to be about n + Cv-n, so we let Z be the number of balls at time r?z in the left urn minus 'm/a i stite t space S1/s = < k < n) n, and let 1/a i Zlv-n. while the transition probability is lIl/n(z,

the to see tht if Ii stisfles (i)-(iii) tw oarts' yf the' prof it (i/l-ttii/) aud (ivl-tvi). Frorft the IL 1, sati:e limit solves follows that for each k, Xh,k is tight, and any subsequential ' martingale problem with coecients F hre

305

8.8. Examples In this section we will give very simple situation.

.

the discrete time proof shows tht and the desired result follows.

and repeating tingale problm

0

-->

n

Examples

dXt

-Xj dt + dB3

=

Next we state a lemma that will help us check the hypothses two examples. The point here is to reflace the trucated moments moments that are emsier to compute.

ju(z)

=

yi

lz)

=

yi

yvh (z)

=

-

-

lp -

zijyj

-

dp)

zilliz, zlrff

zllffhlz,

?,(z,

dy4

d

in thr next y

nary E

.

Clpter

306

(8.2) Lemma.

8

Weak Convergence

If p

k 2 and for

(a)limptnsuplalss ItJtzl (b)limhlo suplwjss lJ)tzl (c) limhto suplmjss .

'

'

.

5

.

.

..

ot)

'

Section 8.8

a11 R <

(x7

cf/tzll

=

p 0

(z)I

=

-

0

Proof

We frst prove the result under the usumption (A3#) for any J > 0, Ek>,s1-2,2 = for larg it

0

=

then (i),(ii),and (iii)of (7.1)hold. Proof

Taking the conditions

lJ(z)

-

bt

order, we note A)

in rverse

(z)I<

Ig $1

zlfftz,

-

jf

tl

307

References. For a classical generating function approach and the history of the reul, see Sections 3.3-3.4 of Jagers (1975).For an approach based on semigroups and generators see Section 9.1 ? Ethier and Kurtz (1986).

have

we

Enmples

(z) K

E-pplzj

Tlten we will argue that if n is large then with high probability we npver s any families of size larger than nJ. satisfy (3.3)in Chapter 5, sp using (4.1)there, we The limiting coecients see that the myrtingale problem is well posed. Calculations in Exarmple 1.6 of Chapter 5 show that (a) and (b) hold. To check (c)with p 4, we begin by noting that whez Zl = f, 72 is the sum qf f independent random variables t'Jl /2 With ren 1 + pn/: k 0 and varine Let 6 > 0. Under (A#) We have in n addition lJ' 1< n& for large n. Using (c + )4< 24c4 + 2454 we have =

dy4 S Ag(z)

.

Using the Cauchy-schwarz inequality

Ittzl

-

the triviality

with

ctJtzllK

1(:f#

zilyt

-

S

1: >1 -

lp-rl

zl

a

=

fflz,

yk

-

zJllffplz,

d

S

-tb

zk)2

K Ig -

z12

1/n 74 (z)

dy)

=

n

st?z-4tzy

.E'((Z2

-

nztl

+

and

pnlnjj4kzl

=

nz)

,

date back to '

of br>nching Consider Processesk sequence which probability children of p2 hms mean the k processes(g;)jm k 0) in 1 + pnln) and variance %2 Suppose that

Exnmple

To bound Efl

Branchlng

8.2.

Efl

-

(A1)

pu

(A2)

c.n

-+

-.+

p

:

r :

(-x,x), (0,x),

(A3) for any J > 0, Following the motivation Our goal is to show

(1+

/k/n))4

j

=

nJ

22 P(IC

-

2p2

--$

lH

3252,25 n 74 (z) S

0

in Example 1.6 of Chpter

5, we 1et Xtgjn

=

y; Zwjln.

(8.3) Theorem. If (A1), (A2), and (A3) hold then X3l/a converges wea kjy to of Feller's branching difusion Xt i.e., the solutio ,

=

pXt dt +

o.

A't dBt

nzEL.l

=

,

..j.

=

nz)

j

(1+ /k/n))4

yyjzy

note that if 1 + nl

-

ajyy

-

z)

we see that if + 960252n-1 n

dz

y yjyjjjy

.jj

Xjn S

4z3#(l(J1 (1+

(1+ #s/n)l

Combining the last three estimates

Ekx,a

dXt

S

2(n)2

#a/n))4, we

-

,

.

(1+

#a/n)J4Ig;

nz(1+

-

that

we note

The desired conclusion follow. easily from the last three observations. Our flrst two examples are the lmst two in Section 5.1, of the subject, the begiihgs ee Feller (1951)t

nzl

=

s 16n-3(z#r,)4 + 16n-3F((Z2

To bound the second term

(z)

rlxjz;

-

nl

yk/nlj

> z)

dz

%252%2n2

IzlS R then + j6(z/k/n)4

Since 6 > 0 is arbitrary we have established (c) and the desired conclusion and followsfrom (8.2) (7.1). To replace (A7/) by (A3) now, we observe that the convergence for the special cmse and use of the continuous mapping theorem as in Example 5.4 i ITIPy l ' .

n

..2

n-1

m=0

1

znm

.x

=)>

,

0

d&

Chapter 8

308

Secbion8.8 Examples

Wak Con vergence

(A3) implies that the probability of a family of size larger thau nl is o(nr?). Thus if we truncatze the original sequence by not allowing more than 6rtn children (A3J) will hold and the probability where % 0 slowly, then (A1), (A2) and will (:1 converge to 0. of a diference between the two systems

(8.5) Lemma.

all

R

<

(x)

(a) lirrtc supjxjsa

-.+

Exnmple 8.3. Wrlght Fisher Dlllhslon. Recall the setup of Example 1.7 in Chapter 5. We hav an urn with n letters in it which may be or c. To ample replacement with from the 1 the time + at build up we urn t m urn time n but with probability aln we ignore the draw and place an c in, and with probability #/n we ignore the draw and place an .A in. Let Z be the humber a Our goal is to show of W's in the urn at time n, and 1et X31/n Zuty/n.

If for

.

309

have

we

It(z)I 0 (b) limta suplwlss l)(z) f(z)I then (i),(ii),and (iii)of (7.1)hold with =

0

=

-

cutz)

0.

H

.4

Proof The new (a)aud + 2 we note that p

(b)imply

?tz)

/'' As n x, X) converges weakly t of solution diflksion X3 i.e., the 1

-+

y

.

the Wright-iser

.

,

dXt

=

-Yt(1

7(1 Xt)) dt +

t-'wX +

-

-

-Yt) dBt

and

-

satisfy (3.3)in Chapter 5 so using (4.1)there problem is well posed. Calculations in Example 1.7 and hold. To check condition (c) with p = 4

The limiting oecients

we se that the rflartingle

in Chapter 5 show that (a) (b) of Z? is' the same = now, we note that if Zl n then the distribution = where 1) 6 + l + G (0, are i.i.d. with Ptj n of Sn (1, z)7/n. a/n) + z(1 (1 .

.

.

.

.

.

k

(.s, ) (s, i) (s, )

,

ms

that

1)

=

=

(s+ (s (s,

-+

-+

1, ) 1, + 1)

-

-+

Esn

npl4

-

E/

E

=

-

=

p

=

nE

pl'n < 1 for all since Ej 2454 it follows that 24c4+ -

nEfzlln uniformly for

-

z

zJ4lZ)'

e (0,1) and

Our fnal example lemma is useful.

=

?'n

z)

-

=

p) 4 + 6

n

E

2

and

k 0. From this %nt-az

the proof is

+

#(1

-

-

p) 2

S Cn

ln(z1,za) ca2

2

the inequality

z))/n4

(c+ )4 %

+ nCn2/n4

E -+

0

.

=

J

1 '-i-?z

f&ntl

'

1 -y

n1

-y

zl

-

z2) +

-

1/a z2) l (z1,

=

fpnzlzz+ vnzlzal

.

1/a bz (z1,zal

=

antl ct(1

pnzlzzl

pnzzzz

'

n

= the following

#si/n

and all three terms converge to 0 uniformly in r z2 % 1). To check (b)in (8.5)we observe

complete.

hms a deterministic limit. In such situations

i4

-

,

=

j=k

s

-

In words, each immune individual dies at rate a and is replaced by a new susceptible. Each susceptible individual becomes infected at rate p times the fraction of the population that is infected, n. Finally, each infected individual xecovers at rate y and enters the immune class. Let XL 1/n (S:jn .Jr&/n). To check (a)in (8.5)we note that

1/n(z1,<2) t'll ln(z1,za) c12

4

n

atn

1)

-

-

-

ntz)

=

Exnmple 8.4. Epidemics. Conslder . a . populatipn With . a flkd... size. of n individual and think of a disease like measles so that recovered individuals are immune to furthr infsction. Let S; and 11 be the number of susceptible and infected individuats at time n, so that n. S1 + Q) is the number of recovered individuals. We suppose that S:, Q) makes transitions at the following rates:

Ethier than 2 alleles. Proof

(8.2).To check (c)of (8.2)with

-

Again the flrst results are due to Feller (195i).Chapter 10 of Kurtz (1986) treats this problem and generalizations allowing more

References.

in

ones

Ip zI2ff(z,dy)

Fc:

=

(8.4) Torem.

the

-

zl

Jnzlza

= pzzz

-

.

-

1

n

-

.

-

Jnzlzz

-

n

za)

-

1

za)

z1.-

-

((z1,z2)

=

pzlza

-

Jnza

vza e

2(z)

1 .

-

n

H

1(z)

:

1 .

-

n

zz, z2

k 0, zl +

Chaptet 8

310

Weak Con vergence

The limiting coecients on rk Their values outside r are Lipschitz continuous extend continuous. Lipschitz them be lt follows that the to are irrelevant so we and well posed problem have is martingale we :

.

'In (8.6) Theorem. As n x, X ordinary diferential equation -+

converges

( ':

. .

.

Solutions

to Exercises

of th

wea kl y to Xt , the splttion

'

.

..

di (a'%)

dx3

=

Remark. lf we set a = 0 thn or epidemit model reduces to one considered in In addition to (8.6)they prove Sections 11.1-11.3 of Ethier and lfurtz (1986). 1/a ' X). central lipit theorem for V-nvj a -

..

q

Chapter

1

((J : 1.1. Let X (X C Bj : B E 7Ct1,2'''')). Clea4ly, any @(t1) .A e X is in the generated by the fnite dimensional sets. To complete the proof, we only hyve tp check that is a c-feld. The frst and msir step is Bj if note t to ( ((t1), (z), e then A' = (td: ((t1)j td(t2), (tJ E : is closed under countable unions, 1et To check that ./c) E be an ordrihg of (t;l, : n, G S,)., let tlj 2, ((y), tJ(y), k 1). and (tJ (tJ ((t),tJ(t2), : : E En' so Uaal.,z note that we can write da = =

.)

,(a),

=

.

.

-feld

.4

'

g

E

.

.

E

.

.

.

E

.

E

(

(

.

E'

.

.

.

.)

.)

.4

=

.

.

.

.4

.4.

.

=

.,j

.)

.

k

.

.

'?'n

.

.)

.

tilj t

((t1),

.)

.

e tlafkl

E

.4.

E

'

=

.

(

'

E

.

= 1.2. Let (td: there is an s 6 (0,1) so that 1# Bs I % Clt sl7 when sl 1< klnj. k + 1 let Eor n K I f+ i 1 i : j k 1 0, 1, B yk.jn max B + n n Bn = ( at lest one M.,ais S (2i 1)C/n-) then Again c Bn bu4 yhistime if p 1/2 + 1/, i.e., (1/2s y) < -

-

.rj

-

-

.

-

=

=

-

-

.

.

.

-'-

-1,

-y

.n

.

. ...

'

... .

l

''

.

'

.

.

.

.

.

..

:

.

:

-

. 'rhe

..

f n#(I#(1/n)IK (1 1)C/n%) K n#(If(1)I/ (2 2'= 1)6'n1/?=-)?

PBu

1.3.

.

.

:

.

c,yjk(1/2-v)+1() -+

(1.3)implirs

qrst step i? to obqervr that th sc>ling relatipnship

l

ctl

myr;

2-n/21

1,0

while the definition of Brownian motion shows EL2 o = t, and E (Al a @)2= C < x. Using @)and the deinition of BrowniAn mption, it folloFs that if anl hve lean 0 qo i # m then. 12k,r! it-l'z and 12rrtln i2-n areindependnt .s.

l

V

2 (m,n lsmsza

''

''''

'

-

-

E

''''

-

2-n)

2

;

=

J'l

lsmszn

E (A2m,n

..-.

t2-n)2

=

2nC2-2/

312 Solutions to Exercises wherein the last equality shev's inequality imply

@)again.

have used

we

J'')

P

12

< n2

1/n

>

t

-

mjn

-

side

Integrating both sides conditional expectation,

J-l am2 mK2%

z 1/n

t

-

la

izmnitely

often

k.6.

l

tinceJca tr,#r)

2.2k When Jj = .&, Sj- is norfnal ,

'

fz)

.

iB'..F -E'z.(#f

,

Ss(,)#j-,

=

jotrj

Br)

E

'

.

)

=?

En

Bi

.

l

-

)

f-,

Sj

=

a

'

.

and .Bi :-a

2.3. Let F the Markov

z-,

=

that Ta o 0

xives

property

value

2.4. Let F = 1(x>1-t) property gives

=

1 + @1.T)

=

o

o

. Ineans :iz

:

i

y'

:

)

and

j

1(a>1+:). Using

thp.t 'F

S il.f)

'z1

o

h

=

s

pl

(z,y)

gives

Jlsi

2.7. Since

1

t)

-

y4

=

p,(0,

pt(0, p)Pp(Tn > 1

-

=

at time n gives

=

Pen (T

>

tr,Bv ) dr

Fs

let 'F@)

- >

=

hs +

ul

'u,

dn

and note

#ts +

os

=

jo*clfr)

,u,

#a+u) du

c

,

implies La

Br) dr F,

tr,

=

EB. ()

dr C Ss,

we

hs +

'a,

#u) du

have

s,) /'c(s-)(s) (

e'xptc-ls-tztstlexp -

J

4

lj k--.Ci

gives

) dy

'o

0n jFa)

,

't

./a

...

j-,

=

=

dr

ctgul d'uj and To evaluate the second term we 1et y@) J(#-,) expt/ .. = c(#,+u) fB3) dul F the prpperty implies note o % so Markpv rxptl -J

Ez =

o

s.(z(s)exp(c,)Is-)

n.' By mssmptiim it holds for 2.5. We will prov th result by induction 1(v'>1,.,Ex). 'F@) Applying the it is fo Suppse 1. n. now tet true n o

z

'

1(z
=

>s, (Tp >

and: zecalling

.15tf-S )

.E'.(y

so the Markov property

pllz, :48.35(Tc> f) dy

=

and npte

Jantf,

y)

=

we

.

(T > t)

PB

tr,#r)

.

E=

> 1+

Taking expected value now

With

'

1 so

-

now and recalling .J7.(f1

J%(A

Markov property

)v

/a

.'5

->

,

> .17.4.2

Taking expected

R

=

the second term

evaluate

a

are indep endent norfflals

1(g'o>f)'and nte

To

BiB: s

=

(

.sf

.,

have

under

sice

with i # J' 2.1 becomes

izj

=

the desnition of

and the desired result follpFs.

+ E ir

.E'.(.I.'&)

2.1 becomes

,

with mean zi. When ..

since un er .P

e Xa,we

Jvh-aj

=

.

d.

dv

f

0

=

side is smaller than #(.a+1)

'

2.1. Txke F(td)

),

.s

313

.E%I'Fo pr&;.r,) k a.P(.rz)

the Borel-cantelli lemma implies

so

= (T > n, Bn E ff recalling oyer and uslng our mssumption show

c2-n

.

..-

The right-hand

is summable

.P

Cheby-

and

The last result

1KmK2R

The right-hand

for Chapter 1

Ansuws

1, B3 G ff )

J(#f) expl .

c(#r)

tk)

Fa

=

.Eb(,)

JBt-aj

c(=y) du4

expt g

:

2.8. (2.8),(2.9),and the Markov property impl# that with probability e there are times sl > tl > s2 > t2 B (c) that converge to c so that S(sr,) s,zl there is a local maximum. and S(s) > #(c), In each interval (s,z+1 .

.

=

.

,

2.9. Z H limsup/n #(t)/J(t) 6 Ft so each c, which implle: that Z is constant

(2.7)implies surely.

almt

Pzz

> c) E

(0,1)

for

314

Solutions to Exercises

Azlswrersfor Chapter 1 for some n k CVLLrelation (1.3)imply

2.10. Let C < x, ta t 0, Ax (#(ta) k X = xAx. A trivial inequality and the scaling =

#n(.N)

k

.Fb(#(fN)

CW7)

k

.Pn(/(1)

=

#)

3.3. Since constat time. are syopping times the last three statements from the frst three. (5' A T K f) (S % t) tl (T K f) 6 Ft T (,$'v S tJ (S S tJ n (T S J q Ft fS+ T < t) = u,rEq:x-rxctlxs> < :) n (T < r) h

and

=

k C4 > 0

e

3.4. Defne

=

, so

osasl

it follows from symmetry J

,1

< e/2

ISa l < e/j,

1/,1<

sup

m<

z e (-V2,V2)

P= Iterating

and psing

sup

osfs;

.

.

.

k 0

ad

l.hI

T,

<

6/2,

m>

0

oKtKnl

I.R1k

inf sup Tm n

=

mkn

liminf Ta

and

=

n

inf Tm sup n- mka

B6

e,

j

.

e (-e/2,

6/2)

r,,

.

.

z 6

e (-/2,

k

.

:/2)

.

.

k

r

%t .

=

eh

.

since J1

1/n))

-

E Sg

'

continuous

e Fs

s A t.

(,s'< tJ n (,$'< sl (5'< i) n t < s)

:/2)

'

.

3.6. (A % ) = LS K f) n

3.7. (i)Let

0

>

(

(-:/2, E

.

Us (.A n LS

=

' .

=En,,

= =

(s' < rl e Fr c F. (5'< r) e F. c Fa

' .

n

i:

This shows (S < t) and (S we get fs k J, ( >

tions

0

%t)

),

are in Ss. Takipg complemepts fs z;z ) r sai

tkz

and

apd

inteysec-

(iM(S < TJn (5'< tJ tlx:f (S < ) n (T > ) e h by (i),so (5'< T) e Fs. (5' k T)n(T < ) U T) irz Fs n Sr. Takihg complements and intersections we get (S k T (S K TJ, and (S = TJ ate iu Fs F =

.

'

'

' .. . Uctlffn where ffn are ctsed. 3.1. . . Suppose Let .Sk . U'R cll'6. ffn 't rn In view closed is inf (t : Bt E Hgj is a stopping tlme. so (3.4)implles Ts of (3,2)We cap complete the proof now by showing that Ta Tx ms n 1 t Clearly, Tx K lirztF'n for all n. To see that < cannot hold let t be a time at which Bt e Then Bt 6 1% for some n and hence limo ; Taking the inf over @with Bt e we have limTk %Tx :

=

.

.

:

.

.

.

=

=

.

..

=

n

,

.

.

./1.

.

3.2. If m2-n

..

On the other hand if .A n (S < t) e Fg and the qltration is right 1/n)) then .An LS K ) (S < + t (. n e nn Ft-vlu = &.

4 > 0

=

.

.

sup

'.'

.jk

SitKe

tw

limsup Ta a

the bottom one for z ; 0 shows that if

Exercise 2.5 gives that if .

Pc

fl Ts t

''''''

''''

.

.

.

.4 n (5'< t)

'

.

.

0

'''

Using the top result for a)

.

''''

.

.

0 S J KJ

=

,

3.5. First if W E Fs then

nsasd

= P0

=

'

2& > 0

H

that

sup

T1 Rn Sn-1 V Tk Repeated use of Exercise 3.3 & by that 1?,ais a stopping time. As n 1 (x7 Rn 1 sups Tn so th frst conclusion follows from (3.3).Define Sa b# S = T1, Sn = Sn- A Ta Repeated use of Exercise 3.3 shows ttwt Su is a stopping time. As rt 1 x, Sn infn Tn so the t second conclusion flloWs from (3.2).The last two coclusions follow frm the .&1

' shows N

2.11. Since cooydinates are independpt it suffice. to prove the result in ne dimension. Continuity implies that if 8 i small

1

.

:p

-+

sup

follow

.

(x) and notitk Letting N Ag t .A we have .J%(/) k #p(.81 k C) > 0. Since S1 it follows from (2.7)that /:(.A) 1, that is, limslp-.c f(t)/W k C . e probability with the prof is cmplet. one. Since C is arbitrry

.Pp

315

<

i (m+ 1)2-n then fSu

<

t)

=

LS < r?z2-'.') 6 Fmn-u (: Ft

.

3.8. If

.4

e T thezt

fBSn) e

.)

n (& K )

=

Unsmszattsk

=

ml2n

n fBmlzn)

.)

e

: F3

Ansvez' foz Chapter 1

Solutions to Exercises

316

by (i)in Exercise 3.7. This shows (.B(&) E =.+ (:x) we have Bs = lima Bsj n e naFsa 3.9.

(3.5)implies Sshn

to show

it suces .

.

uat. n tTa

=

yhat X n (Ta

.

Jt n (gk > s) n (rs A s since e Fs implies implies(Tn > :) E h

n fS

.4

.

<

>

t)

X) =r

Fs

<.

.

=

y

fh

J

4.1. We begin by noting symmetry

,4

Ss

e

.

.J%(A K 1 + t)

Since

.. tTn > .

.

2

=

and

(2.5)imply j

#x:o unzo

s.

,

,V)n LS

.'v=

g

q<

wvowxvw

t)

.

.

j.

j yy

.

y .

.

. .

.

j . . y

. .. .. .

' j . . . . . .

t:r',, vj q kg >

'

ds dy

1

p1(0, g)

2

()

s)

=

15(Tn

s) dy ds

=

()

()

by Fzbini's theorem, so the integrand gives the density .J%4.2 Pv (% = t) = .J%(Tp f), (4.1)gives

y

6 Sq and the fact that Ta i,s >

.Pp(T0 g

1 + t). Since

=

'

j. .

(j

=

E

xu...x,u

wu

m(0,yj X

= ;, uo

t mI.o y

Letting

.

x

< = u
<

e Fsv

.

S4)

>

rhn

.

so B%4

e Fsv

=

the other inclusion 1qt

C Fs. To argue

.

.)

317

#c(S

.

1 + t)

=

=

'

e

0 W7 *

1 = 2%aya

time

ytopping

j

x

2

()

j

-p2/a

z

-.y

ye

2<3

a 1+t)/af ye -p (

Lt

y

/at dy 1

.

2* a/a

(

j.

t +

j)

.

4.2. Translation invariance implies that if r

3.10. Since s #(f,) ds

C Fs we have

ds1

s-(/-S

gn-t

Ez,ctfBp. /S

gn-t -

Ez,zlfB.. The rest of the argument to 5'4td)

Applying the strong Markov property T

Ez

ds

ja

=

T@) #(tJ,)

ds

f'(0,p)T(Oa-r) =

fwn)

-

20)

at

time

and

'rr,

Oa)I.'&r) .E'(0,0)T(fawr) =

-

is identical to that for

(4.4).

0, ps) 1, so for some & > 0 we h>ve (pe ($ > 0 for all s % J. 4.3. As ? . equation psjpt) the ps + @)we can ow conclud by induction that sing ?pi p1l n, w(s)> 0 for s < nJ, and se can take logarithms wltlput feellng guilty. 4(2-(n=1)) and inzcdon tells us that T lte rest is emsy: 442 ) + #( 2-a) while 1)2-rz) 2-r'4(1), + /(2-n) #(m2-n) and induction 4(2-n) #((m and does not vanish, ps) m2=''#(1). continuous Since is implies#(m2-n) esired result and continuous follows. the is #(s) w(s)

have

we

then

.-+

-+

Jc

=

t F@)

Applying the strong Markov prperty using the lmst identity we have

.'-s)

-

f0)

-

< s

=

-r&

J(O,) ds

$s

:::

=

''(O..)

,

=

=

-

=

=

1 if

l()

'Pk(* -

s <

,

2c

-

r < tdt

=

otherwise

s) <

2c

-

u

E

4.4. Let

(

o

psjl7s)

=

Eak's

=

.E(exp(-T.)).

(4.4) nd (j

.5)

imp jy

. .

,

.

=

Eah

=

f'rtY's o psla)

w(c)w()

.

Sypmetry of the normal distribution implies E% F= Ea% j so it we let S intts < t : B, = c) and apply the strong Markov property then oit (S < x) ErYs

w(c)

=

As before the flrst equation = = 1 and c Vi gives cb

=

vxa

implies =

'Zfcl

)

wlc)

and

w(c) =

=

wc2(1)

exptca)

this implies c

while =

the

with

second

-.sV-.

4.5. Let Ml = maxnsasl f,. Exercise 3.10 implies that with probability 1, Ta Tc is discontinuous at c M. This shows J5tc fl > B and hence a and result hint. 1 as n follows from the the is discontinuous in (0,n)) cxl -+

Taking expected values now gives the desired result.

+

=

-+

-->

-+

Solutions to Exercises

318

Answers for Chapter 2

inftt > 1 : Sa2 > .G). Since Let M1 = maxcsasl Bs2 and T rfl the strong Markov properiy and BI (4.3)imply that with probability #g2 at is discontinupt!s M. This shpws Ca yo s one # 1 s % (x) ahd the result follows from the hint. is discontinuous in (0,n)) 1 as n 4.6.

=

<

integrability

check

2.2. To

we note

319

that

.8v2

.!5ts

=+

=

-+

tromthe strong

4 7 If we let Iz) = T(), then it folloFs and symmetry of Brownian motion that ErlBtt

'r

E'

=

= equality

holds

::

0 when yd

=

and the desired formula

k 0,

2

1.1. If e Sa to converges .

and

a <

2.1. The martingqle

tJ)

b then Ils,

ms n t(.,)(s)1x@)

proprty

pptional

is

1(c+1/a,#+1/a)1g

=

x.

-+

E

a.

j

1p1

The integral is convergent and as we showed in Example 2.1, .&(I#f I> R) arbitrry. result follws from the fact that R iq as t cx s th desited

2.3. Let Ta = inftf : l.Xl > nJ. By (2.3)this that Xn S n. Jensen's inequality for conditionl implies in chapter4 of Durrett (1995)) Eexj

ar fltration

w.r.t.

.A)

=

. .

.. .

'

..

.

.

'

but this is 0 for r < s and true for r Fs then Fsaa it follows that if .

.4

n IS

,))

A

S'rl

k

j

s. If

=

-

'

is a martingale .

with respect . .

l.Tk+a)

2.5. Let Ta be a sequence that reduces X. The martingale property shows that if .A e J;tl then E t.Xagk 1(za>a); X) E (A 1(gk>c); .A) '

'

=

C

Fr

.E'txtamtlarrk 1(za>a)l.Tk+,) =

Letting n orem gives -+

XaS

-

= 1(n>a)f lX(amtlan 1(n>a) Yshsn1(sn>c) = .Xtam,lan

a

.

>

,

.0),

=

..

to .

x and using Fatou's lemma and the dominated convergence ECXL

; d)

%limf n-

Ezhn

1(gk>()) ; .)

=

.E'(.X;.)

f pe see that EXt K EXz < x. Since this holds for al1 Taking ECXL 1.T) To replace 0 by s apply the lat conclusion to K t t which by Exercise 2.4 is a local martingale.

E Fz we

.4

=

other hand

ECXtSCAn ('S S

.x.

.:1)

=

f(A1(s>o);.

n (S S

s))

=

ExssA

n (S.K sl)

.

$

=

X,yi

,

14 is increasing, we only have to rule out jump discontinuities. 3.1. Since Suppose there is a t and an 6 > 0 so that s < t < u then F(u)-- 7(s) > 3: for -+

Adding the lmst two equations gives the desired conclusion.

the-

.)

lim f(A1(,zk>n); S n-G)

on the

Note

(1.1.b)

.fk+,

ftasa 1(s,x>n)lG)

. .

XL. .g.,

.&)+.

.A)

.

(d n IS

1

wtxTal

=

G is that

.E'IJ',S;

will reduce expectation (see

sequence

gk)I.faa,) k e s (xnt js nx' )) =

.

.

-+

If s < t 2.4. Let Ta % n be a sequence which reduces X, and Sn (Ta the (Ta > > R Ta involved S on (& 0J then definitions (note SR (:: and imply fact stopping the 1(gk>.a) theorem, optional e

.

,

(3.5)in

l

x

dy + (log.R).P.II#f l > R4 k (2'mt)-d/2 :f l 1og 1p1

Ecx,

and

Chapter 1 implis that Fshs C Fs so if X( is Irtinale with reject t Ss it is a frtartihgle With respect t: Fshz To agr the tlii .. directin 1et E F. Weclaim that n (S > s) E Fsa,. To prk thi w hk#e t check that fr alt r ./1.

j jog jpjjpdy <

(

=

.E'(xzS;

for al1

=

-+

/4!,)

since

followsas i (4.8). Chapter

,,

.-+

.(/(#i);

The last

p E lXt 1

sizlce llpg 1p1 Ilst integrable near (x0 aud e -Ip-zI2/2 t p,kes care of the behavior for show lArge y. To that ErX) we observe that for any R < x

Markov property

E=LEz1Bt-u.r4 r K ! g. / Erkzo.%-pl ) = r S ) E=Bt44

S f)

:z.j)-1e-lt/-:r1Q/2f

-+

-+

320

Answers for Chapter 2

Solutions to Exerclses (.X+f

Xs)2

((.X)s+f

(.X)s). Using the lmst qquality

321

all n. lt is easy to see that 7(u) V'(s) is the variation of X over (s,u2. Let will contradiction derive now that the variation over (s0, e < t < un. We a is infnite. Pick 6 > 0 so that if Irvl < &then lX Xr l < E. Assuming sn and un have been deflned, pick a partition of lsn un) not containing t with mesh < 5 and variation > 2E. Let sr&+1be the largest point in the partitiop < t and shosvs that the be the smallest point in the partition > t. Our onstruction ua) always of @a+1) variatih X 6, > is over the r,+1, variation ttl so (sn, 'tIp) contradiction which is in:nite, implies 14 is contluous. t a over (sn,

Let Zt

3.2. (aY+ 'Flz

3.8. By stopping at Ts inft : IXtl > n) and using txercise3.5 it suflices to prove the result when X is a bounded martingale. Using Exercis 3.7, we can t suppose without loss of generality that S 0. Using the L max imal inequality optional stopping and throrem martingale the Xj61n, on the martingal on the (X)r it follows that X;

-

'upl

-

' ..

=

-

3.3. .-XCZL iq a.,local martingale, so lf 5$ desiredyeslt follows from Exercise 3.2.

martingale.

3.4.

ah

)(X)

abx,

F)

c tXt$

=

(X, F)t) is a local

-

((Ar)s+t

. .

. .

.

we get .

.

. .. . .

.y

:

. .

.

.

.

-

-

-

=

='i

-

stopping theorem and Ekercise

whr thelast equality follows from thq optional 3.6. This shows that Zt is a martmgale, so (F)t

/ '

'

d

'

ihe

(vY)s+f =

=c

(X)s

.

=

.

s

-

. .

=

-Xo then (F, Zj3

=

.E'(-Ys2+, A'ja.a

-

-

.-2

(a'LZLt (F, Z) is a local

-

. .

=

=

'arzml

-

-

(I)s)I.'&+a)+ l.Xma 4 1,)2 A%)2 -Y.l+, + (X)s + (.X+, f(-Ys2+t (-Y)s+zlFs+a) = -Ya)2 Zs -(X)s+, + (-Y)s + (Xs+a

f(ZtI.T+a)

,

-.+

-

'

. .

martigale.

it i'otlows from (3.t) and 3.5. If T is such that X v' is a bounded rartigale the dennition of (A-) that (-YT) = (XIT. For a geperal stopping time .-?Tlet Ta = inf (t : 1.XLl > n), note that the last result implies (.X'TATa) = and then let n x. The result for the covariance process follows immediately from the definition and the result for the variance procqss.

,

-

E

sup

t Xn

XLv

%4ELX;hnj

ELLX,ltarr 0 =

=

(

'

(XITAT,Z

.-,

3.6. Since X1

-

(.X) is a martingale E (.X)t

so letting t dominated by

(x)

-.+

3.7. The

an

optional

=

5X/

we have .E'(.X)x

integrable stopping

random

l.Xtl K M for

and

EXol

-

; M2

variable

.

all

Letting n

-+

3.9. This is

(x)

now gives the desired cpnclusion.

an

immediat

4.1. It is simply a matter of patiently 1. s < t K c. '.K= Y E Sft so

K M1

This shows that XI (X)t is and hence is uniformly integrable.

(3.8).

of

consequence

a11 the cmses.

checking ,

-

2.s
theorem implies

f(XI.'&) f(f'(Xl.&).'&) F(J1 1.z', ) =

=

=

'n

E (Xs+t

aud Xs e Fs

l.f+,)

=

=

Xs+. 3. s < if c % s

(: Fsma so

E (Xs+t

-

AI.&+,)

=

Xsma

-

Xs

-

Xs)2IIs+,)

=

=

we n6te q

A+a)2lFs+,) .E'((A+t X,)2 + tAma .E'(-Ys2+t xs2+, Aa)2 IFs+a) + (A+, -

-

<

> .

so this reduces to case 2 if s' < a or to our mssumption = @.E ($ I.T-a)= .E'('F1F,) c: K

X F 4. b S s < .

ftFc 1Fa) y,

=

.

(X)t defines a measurv. 4.2. lf we fzx w then L2 of that measure implies -+

i.e., K is a martingale. To prepare for the proof of the second result that imitating the proof of (2.4) E ((Xs+

,

=

j

#

s2

1/2

1/2

(I(.X')f

+

-

-

Now take expected

value.

j

.Jf,2(f(.X')j

%

j

The triangle inequality fol'

/ LHs JG)2(f(aY)j +

1/2

Answers for Chapter 2

322 Solutions to Exerclses 4.3, X e Af 2 impl ies E sup conditionlet Ts be a sequence and (A)n % n. The ojtiopal .12

IXn

E Letting n

iearrang

x and hence EX2() < x. To check the other of stopping times 1 x so that X1'= is bounded C stopping theorem implies t <

(.X)Ta)

-

1(Ta>c)

EX-Z

=

it fpllows that .E'(.X'x2 (aY)x the displayed equation to conclude x

-+

Fu:

-

EXZI.

so X is L2 bounded 7. .. 4.4k The triagle inequality implies

(ff!

=

LH2

'.F) for all t. Since S.2

ffa2

=

=

0 for S

To prove the converse

-Y),

2

(ff

=

.X)a

0 for S

=

Ks KT

.

conclusion.

=

,

Il.J.rTa

.

IIzIlK Ilzn1+1 IIz

.

'

.

inftt : la'v 6.2. Stop X at Ta I > n or Jct H2 d(.X) , ) to and an integrand in lIa(X). Exerise 4.5 implies partingale

so
-

% s S T, E

.

-

.xTa

.

.

/Ta

11a E =

jo

..r.z/

get a bounded

j(.;')t

j

'

Using the L2 maximal inequality it follows thut

r,11 Ilalll lirrlizllrIIal

ck

For the other inequality note that

.

.

follows from Exercise 3.8 that (Jf? .X')f and (ff2 X)t are cppstant on givts the desire (S, T). Combining this with the first reslt and using (4.3.b)

(AA1(vk>a)) s E.M+ z7(A4x

E

.

nd it

1(n>c)

'

Clearly, .4S1 .X') (5.4)implies

323

E

IlznllS Ilztl+ Ilz,z IIso

Ta

uzpalV X)1

S 4E

'

-

llzllk lirnsup Ilzall

Letting n

jo

Sf2 d(.X)t

x now we can conclude

-+

4.5. Let Hn e 11l with Ilffn Jfllx 0. Since Ilt#n .X'):=zCH X)II2 0, using Exrcise 4.4, the isometry property for l1l, and Exercise 4.4 again -+

-

.

:

Ilff Xlla -

-

.

.

lim ll#n X112= limll#n

=

.

-

lIx Ilffllx

With this in hand

=

we can start

5.1. lf L, f/ E M2 have the desired properties then taking N L f/ we have = (f,, L f/) Uj H L Using 0. Exercise 3.8 so (f, now it follows (f/, that L U is constant and hence must be Y 0. =

-

EH

.&)

-

-

-

.

with

.X')z2

=

a

E

'

-

FT

ltaszl then X = S k X and 5.2. If we 1et ff, H 1 and ffa ff F. (It is 'for this remson we have assumed v'L = Fp 0.) Using (5.4)now it follows that =

=

j

=

't

1(,sc-) #(.X',F),

=

(X,

F)J

5.3. If we let S'a = 1(,sz) and Jf, 1(,>rr) then XV .I X and 'F 'FT K F. Using (5.4)now it follows that (vYT, F H 0 'TIf so the deired rsult follows from (3.11). =

=

.

.

x

-+

to get

and ff ,

=

ff al +

.rf,2

where

d(a'):

result.

-

6.3. By stopping we can reduce to the cmse in which X is a bounded continuous martipgale alzd (.X)f K N for a1l t, which implies # 6 112(X). If we replace S and 1aby Sn and Tn which stop at the next dyadic rational and 1et Il'k c C for Sn < s K Ta then H.: e lI1 and it follows easily froz the de:nition of the integral in Step 2 in Section 2.4 that

=

j

-

6.1. Write .Jf, = ffal + #,2

thedesiied

.;:2

.

=

(Ah FT)

and let n

j

Ta

.,.7

dxa

=

cxn

-

xs-) .

the proof now we observe that if 1C@)1

'ro complete

ff then

:

H

;

=

ff 41

=

Jfaltssaso

=

Jfa ltssasz)

IlSk

-

Jfllx

S ff2(.E'((X)n

-

(-Y)g') +

.

.

ttxlsn(X)s)J -+

-

0

Solutions to Exercises

324

Answers for Chapter 3

theorem. Using Exercise 4.4 and (4.3,b) x by the bounded convergence as n it follows that 11(.Sk vY) CH vY)IIa 0 and hence H.n dxa zzadxa. Clearly, CLXLU Xsj Cxk A) and th desired iesult follows. -.+

.

-

J

-+

.

-->

-

=

XI XJ

6*4.

=

aY2(jP y) '+

=

-

.f'

-.+

.X'2(jP) t

of the Riemann approximating sums for the x by the convergence as n of integral. Convergence the means and variances for a sequence of normal random variables is sucent in distribution (consider the for them to cnverge characteristic functions) and the desired result follows. -+

t

.

7.1. supf l((ff ''

=

(-Y'(t7+z)

-

)l2+

-Y(P,

i

2Ar(t7)(A-(t7+T)

-

A-(tP,

)J

i

(3.8)implies that the flrst term

l'cAti

=

converges in pobability to (X). The prkious t shows second exercise the term converges in probability to 2X, dXa

6.5. The dtfl-erence betwen evaluating J7'(-Y(f7.it)

2i

at

the right

X(t7))?

-

l(t2-nt) and Dk = Bk ck n Bk + 1/2)2-nt) a 6.6. lzet1/2)i-af). Bk + In view of (6.t)it sumces to shw tllat ms n =

-

Su

J7Ck,nck,u

=

+s

k,a

k

.

)

.-+

+ 1)2-nt) x

As 8 0, GaJ Exercise 7.1. -+

' .. . .

('

-

t/2)2

(

.

V Ck2 )

2-T:4/2)2

:

-+

implies (6k7) (:x$

we have

fty)

-

-+

0

i

in probability. variance

The left-hand

i

side

of

-+

(0,) with

t -

tP) t

j:

' .

.

.

uniformly

(j

-.+

0

=+

132ds a

the desired

so

0

mean 0 and

conclusion

from

'

.

E

E

and

+.A( is pntinuouj

.t

vl

0.

aPrB1,

=

c)

=

.

theorem implies that '

m/z

=

=

(

=

z)/('-

-

'

(1 PrBr

u) +

c))

=

-

c).

1.2. From (1.6)it follow that for any s and n Ptsupk, Bt fz) 1. This implie Ptsupfk, Bt x) z 1 for all s and hente limsup 4-. tkj B # cxn a.s. =

'

=

'

. . .

.

EE

..'

.

'

'

.

5'r(Y)

< x

with

'

.

'

.''

:

'

E

'

: .

positike probabilit#

t

.

'

.

.

j

en

.

..

.

#t(o,)()

-+

3.2. Let J(z, )

::.z

z6

-

=

E

01z2

a2

+ bz2t2

a4t

Dtf

=

=

-(Iz

(1/2)DraJ

W .

+c -

+ 52

n

*

=

.

+c

ct3. Diserentiating

4 + 2z 2 t = 3ct a 15z4 6cz2j + = -

j2

c

gives

0

=

3.1. Using the optional stopping theorem at time T A f we have EftB7)2 sy .E'c(T A ). Letting t convergece x and using the bounded and monoton theorems with (1,4)we have E 0T

with

(0,t)

on

dA'

.

1 @(td)which is 1.3k If a'ndwe have a contradiction.

peh

joC

=

'

.

s.,d.%

hms a normal distribution

(tP :,2 't ;% z+ 1

.'

1*

.

that if )' is a sequence of partitions ht:Bt:w,

.

Solving gives PrBv

.

If we let

(' . .

z

,

2ncj2-2n + Ck2a .D2 k,a S

-

1a

T(.o)

-

.

l

,

'and then use Chebyshev's inequality. 6.7: ms n

.

.

.

2-nj/2 + Ck nDkvg

-

T:

?. f (Xa)

-+

1.1. The optional stopping

E

k

V(C2 = E

.

.

E

=

I.AIK M.

when

suces

i

. ..

Chap er 3

2

E'n

0

-+

local mrtinxale 8.1. Clearly Mt + MIis > cotinuous nd hs dp + adapted, locally of boundd variation

-

in probability. EC2 ,n = 2-n,/2 and kck sfjk a = 0 so ESg = t/2. To complete the proof now we note that the terms in the sum are independent so ,

l

ff,

-

t

tjg

...

,

Ifn

supa

to prove the result then as in (7.4)we get

it

Jk+,))

T(A)

2(A-)t

-+

.

.

the left end points is

and

ff ) .A)fl % M

-

stopping

7.2. By

Gj

,

Ja

325

=

Solutions to Exercises

326

Answers for Chapter 4

Setting c = 15, 25 = 6c, and = 3c, that is c 15, we have 15, b = 45, c (1/2)D=.f martingale. 0, fBt is local t) Using the optional so DtJ + a stopping theorem at time a A t we have =

=

=

EnfBnht

4 lsfv.oazt'&

-

-+

x

the first term on the right

Letting t cxn and using the boun with the results in (3.3)we have

15./r3 fl

=

c6

2

)+

A

.-.+

E

ed and monotone c9(1

=

.G('D(.Wa); r >

.

!

A i)

h i) ) = 15c(a

15c4.&a + 45c25r2 J

-

,

2

45#aa(a j

,

3

and it follows that for large n,

theorrs

convergnce

15 + 75)

=

#(z)

gyt

so

61/15.

=

j

=

t'

Ltting lmt

n

(x)

-+

kulely

and poting that on d hnce Zt tks a..,

IT-U

-

x)

have Ztlt = csf/ the desired ilt fllows. =

y

+p

we

-.+

p

y

-+

=

*

J

=

=

.

.

,

=q

.

-

.

.

.

,

=

.

.

.

0z

.

y4 tfp

-

.EE

:

=

.

.

JJ

=

,

=

=

7.1.

(a)lgnoring

Cd and dilerentiating

Dz.k:

=

d

-y

gives

Dxa.:

(1z

-

y (Iz :1z + y a)(a+a)/a

-d

=

0i4y #1a + p a)(a+a)/2

2(zf .

.

Dyho

=

(lz

1 #lz + yzjalz ..

-

#)2p #(#+ 2)(z 2p)(d+4)/a #j2 + (jz -

+

-

,

'tz)

follows.

-

2

# (Iz #I2+ !P)td+2)/2 d(d+ 2)p/ # + 2yi + (d+a)/a z pI2+ y ) (jz #ja + ,a)(a+4)/2 o

,

-

,-

d

.

-

-.

.

-

-.

.

-d

=

.

(1z

4

-

4.1. Let r inf (t > 0 : Bt $ Gj 1et p E 0G, and 1et 4 inftt $ 0 :'# ciii p). (2.9) in Chapter 1 implies #p(Tp 0) 1. Since r % Tp it follows that y is regular. =

=

,

=

Adding up we see that

=

d-1

X-! f=1

Dxrp

4.2. .Fjt'r = 0) < 1 and Blumenshal's 0-1 law imply .J$tr = 0) 0. Since we yj 0 nd hence 6 1 EyfBpj > are in d k 2, it follows that PvB. Let trn = inf (t > 0 : Bt / Dy, 1/n)). a'n 0 ms n 1 (x) so Pycu < 1. t =

=

e

=

EvfBp)

=

.G(J(#.r);r % tzrjl

+

-

.

+

r > .G('??(#c.);

cu)

1) -lJ

-

N I r;z u pj2

(-d) wa(d+a)/2

.y..

=

'

d(d+2)p(Iz -

--+

-

((-d)(d

=

l

:+

'r)

1

+ Dvuhe

The fact that

u(z)

=

,

-

'

Dyyhe

Chapter

..

=

,

,

#r2dr: By mpdifying Fzaafter with (.X), = time t we cn suppose withopt loss of grnrality tht (A-)s ::: c then Xvtulls a BrWnln motibn. now we see tlkat if Sipce (A'), >nd hnce th time change n'(u)are deterministic, the dsired 4q!1t ,/,4u)

p))

-

(, 4.4. Let r = inft > 0 : Bt e ?(p, 1), a)J. By translatzon and rotatzon we can ::':: suppose y = 0, r (1,td0, 0), and the #- 1 dimnsipnal hyperplane is zd 0. Let Tp inf ( > 0 : B. 0). (2.9)in Chaptr 1 implis that .J%(Tc= 0) 1. If c > 0 thenfor som'e k < (x) the hyperplane can be covered by k rotations of J-/ so f%('r = 0) k 1/i and it follows from the 0-1 law that .J%('r 0) 1. F.:.Fz

-

Xa is a local martingale

Vp(p + p(z

-

.

4:2.

.

'vj

-

,

a

-

.

Jf

-

-

E E

Rd be piecewise constant and have 4.1. I#,I 1 for a1l s. Let = 'i E i 0 dXia The formula for the covariance of stochastic itegrals shows '.Kis a local martingale with (F)z so (4.1)implies that M is a Brownian motion. Letting 0 tc < t1' < < t,z and taking 0s for s E (tl-, t./) for 1 %j S n. shows that a'Vz X3o .X'z. Xgn- are independent multivariate n-llf matrices (tl tilf j normals with zean 0 nd covarince and (tr, it follows that Xt is a d-dimensional Browniap motion. Let 0 (0,x) :

K1

-

.

l-eartl

1=

r(z)

E

-

74g, Tgy), 1). Calculus gives us

=

1

1

-+

Continuity of Vg implies that if Iz p1 < r and z e U then Tgy + p(z y) k 0 so #(z) k 0. This shows that for small y)) 0z r the truncated cone U n Dy, r) C G so the regularity of y follows frprp (4.5).

3.3. Using 'the optional stopping

theorem at time T-c A n we have j Eo expt-tz#/J

cu)

infxEdotp,l/s)

4.3. Let y E OG and consider U -

Ezvt

0, so

-+

,

.

6

As n

327

0 follows from

( jz

(1.7)Aas

-'.

J

.

31# E

pI2+ y2)

p jz + y z)(,+.)/z

in the proof of

=

()

(7.1).

:

y

Anslvers for Chapter 5

Solutions to Exercises

328

(b) Clearly J do ptz, p) is independent of z. To show that it is independent of p, let z = 0 and change variables 0i = ypi for 1 S K d 1 to get -

jdossz,v/

(gpjwjca wdyyzl,,,

(c) Changing variables 0i

zi + rf y and using

=

D(.,4e

p)

ptz,

d0

1)

-

-

1

-,

Dz,slylr

1)

rto,

dv

-

...:11#11x,,

-

=

-

-

since for To extend the last result to Lipschitz continuity on sexvingthat if z e D1, y e 1)2 and is the joint of bD betweenz and y then

0

-.+

=

-

.

dominated convergence

=

: z E D lIIIx,f sup(I1-(z)I 1,and

I#(z) (p)I: IlJIIx,2l#() .(p)I+ I/(z) J(p)lIl.lIx,a s 1 czlz 7/1+ 1 !/1 C1z pl + l#(0)l/A)lz .%(26% z e D1, I(z)l K I(0)l + CR.

jdvsro,

.yd-dv

-

Combining the lmst two results, introducing using the triangle inequality:

,

(d) Since 9.1. r(z)

c(z) =

< x)

.&('r

-#

<

H

0

(4.3).

1 for a11 z 6 .I, this follows from

=

so zptzl

Eie -#r

=

K1

Ece-pr is the unique solution of

an d it follows from

IA(z)

(6.3)tht

(p)l:i I(z) / cllz

-

Guessig

j

u(z)

=

r

B coshtz) 1 -u& 2

if b =

with b > 0 we flnd

pu

-

2p. Then we take B

=

,1Vwq

the lie

o

(z)l+ l(z) (g)l al + c2lz pI (72Iz pl Iz p1, This shows that

-

Bb2

=

coshtz)

PB

-

2

3.2. (a)Diferentiatint 0

=

-

=

h/(z)

=

is Lipschitz taking

-

argument

-czlogz

-llog

x&(z)

27) to satisfy the boundary condition.

1/ coshtc

=

have

we

x(z)

by obr segnient

-

=

-

-

egin

.

since C % C2 and lz zl + lz pl continuous with constant (72 on IzIK 2S. Repeating the last lzl % 24 and y > 2./ completes the proof. -

g

329

z

0 if

<

=

C

-

-67z

>

0

if 0 0

z >

< z < e-1

26:-2 And so x is strictly increasing and cpncave on (0,e-2). Since x(e-2) /P(e-2) = C x is stric'tly increastn: and concave o cr). When :7 K e-2 (0, =

Chapt

r 5

1

is Lipschitz 3.1. We frst prove that ..a-1l(0)Ion oa (.R lzl K 2J0. Let #(Xz/Izl).lf z, y e D( then

with

conttuous

s

=

J(z)

constant

2R

=

-

IzI)/S

Ca and

=

:(z)

1 IT(z) T(p)I IIpIJ, lzll S qilz

y

-

Since is Lipsch itz continuous of z onto D we have

op D3

(lzlK J0,

l#(z) .(y)I I(./lzI) -

=

c,

pl

-

=

j

R

-

Apj

and

Az/lzl is the projection

-

,

(r

=

y

5.1. Under

-1

-c-l

Cz logz dz

expl-l/tr)

.

g

t

(j)

Q the

=

with

p t j)-j.y

zje

0

=

x

Llyljp

1/@p ) = pg()(log(1/g(@)))

E

.

coordinate

maps

=

vL(u?)satisfy MP(# t

/, -'ez /' txa)

u't, x, g1

loglog

=

p > 0 then

expt-

=

(lp/lpI)I

Izl 1p1s cllz -

0

=

(b)lf #()

=

-

f

2C1 +

#(x,)

-

-

0

+

ds

(x,) ds

+

,

J).

Let

y

Solulons to Exercises

330

Since w

intrested

are

Anslsrezs for Chapter 6

in adding drift

let

we

-

c

c-l

=

ms

before, but let

so w(z)k z, w(x) = x and (ii) If (z)/c(z)k e ylzen

j

J

=

c(#,)

-

n

.

p

t

c(Xa)

-$

+

=

a-l(x,)

+

eXP

(a'L)ds

.

0

-yk

=

dX,

so

ds.

0

w(x) <

follows from

recurrence

''-

0

dz 2(a)/c(z)

x and transience follows from

331

(3.3).

< e-zep

(3.3).

3.2. The condition is dg = 0. Let f jl dy. //)(-c=) y/tn + y) = enlp'y) so if f > 0 and if f < 0 then > < n. w(x) lf I = 0 then pty) is periodic so 0 < < p'y) < C < x, and it follows that c w(x) = x and w(-x) =

Ja

-2(p)/c(p)

since#i

an d 9 dtfter by a process of bounded varirtion

('.f')t ('.F)t =

The

change

*e

ltve

t

c-'(x,)

=

ds

-x.

0

5.1.

of memspre ip this cwqe is given by

(a)If #

0 then

= .

1

Yj

j .

. .

'

=

Xg

a j

-

/j j) O Nw

-X

show

to

suces

that'f

n'-Y

.

1 + -(F)f 2

::

0

.

ilg

M(0) = that p(x)

=

=

=

M(x) 0,

&+1

=

(:x) =

#(z)

0

ds

k

J-22n(Sm

m=0

0

=

c2l7lr/JQ ,

.

-+

rn(z) =

g' 2 (2lpIr/c

=

-

e

-2lpIr/trQ

exp (-j' 0

da)

k 1

.

..

.'.' (

To deal

with

note

(x7

1)

/&2 z 1

=F

).

e-alpla/r

-

ajpjz -+

-2I#lp/cQ

e

dy

C2 =

21p1e

dz

< cxl

f(oo)

-

T(z)

J' -

2

-zjgjm/ga

e

Iplz

-x

so J

w(x)

=

=

(:<)

-2IpI,.r/cQ

Since most of the integral comes from z K y S z + 2(z)/c(a)

.

since the integrand converges to 1/12 ms z 0. Again M(0) = nd it follows that 0 is absorbing. To de>l with ckl w note that f = x. To evaluate J we begin by notin: that

a

3.1. (i)If (z)K 0 then

a

1/511

-

Cr

6

.

so

m(z)(w(z)w(0))dz

-

*

.

and it follows that 0 is absorbing. so I = J = x.

1

:

-

..

c'-2 dz < &

.&m)

-

=

-

1

To evaluate I for the boundary at 0, we note

where th number of terms, N+ 1, hms a geometric distribution with f4#+1) .P.m)2 2n, and the Sm .P.mare i.i.d. with Esm Sm) = 2-2n and Esm (5/3)2-4n Using the formula for the variance of a r>ndom sum or noticing that e-1 ms iz PN k 2n) x one concludes easily that limsup k C > 0 a.s.

Chapter

.

so

(xh

=

N

1f,1-1ltlsxjsa-a)

'

=

=

n

1/J2z

=

'zz(z)(w(z)e(0))d

e (z) =

=

=

>nd m(z)

z

.

0

>

0

To prov the lMt result, 1t Rz inf ( > Rrt : I#fl 2-n), 0, Sa 0) for n k 0, and N sptiz : R < TiJ. Now inf (t > Sii : Bt

-+

so J

-x

(b) When p <

I#,l-11(js,I
limsup

=

1

'

a.s. n

r

,/(z)

-

..*

= exp 6.1. It

-

-2(&)/tz(:)

=

-x

C-zit

follows easily that

x so

Solutions to Exercises

332

Since #(z)

c-2l#lr/J?

Answers for Capter

it follows that J

=

(x)

=

and cxnis a natural

boundary.

-x

=

-x

r?z(#) =

-1.

26 >

-

-a#

#

.

:44)

y)

-

ap-l

y.x w#

ew

-+

=

=

T./aAT2.

l-2 ds

0

(Ta/a A T2z) o 0n

does not depend on z. Letting Sa Markov property it follows that the distribution S=

1-2 ds

=

(x)

-+

in

(4.7)of

j*

-

using the strong

and

(5.3)to

of

Lu

under P

siuce

R*.

does not depend on z. Pick J > 0 so that Plr k J) > 0. The fa-n are inde1 pendent so the second Borel-cantelli lemma implies that Pltfa-a k 6 i.o.) with < x) > 0. but this is inconsistent =

(z) i

Chapter

=

p

-x

>

0.

Chapter 6 we have

.FQqljxj.= 2(' (z) w(1))

5.2. We apply =

-+

7

m(z)

-6/z#,

=

v(z)

=

gjkpz)

1/fpJ(z)c(z)

=

z1+J

(1 + 8j8 2

dz +

w(1))mtzl

-

l

=

The result now followsfrom m(z)

under P,,

J

exe

-

5.1. Letting b

-1.

1. To do (c) lf # > 1 then .S > H so it suflices to prove the result when J at 1, starting motion = is Brownian z-Bnr, that if Bz this, we note a z then of distributiop so the

I

*

C

=

t (#)J(#)

0. From this we see that I < cx) if p < 1/2. When # 0, M(0) = Cz2# as z M(0) x. When p > 0, Mtzl 0 so J < (x7 if Comparing the poibilities for I and J gives the desired result.

sy so .J

Chapter since 6 < 1 implies 1

-

1

-1

5.3. (a)ffj K Ta < x. (b) (2.9)in Chapter 3 implies

333

Hexe and in what follows C is a constant which will change from line to line. If (p40) and f # k 1/2 then w(0) = Cz1-2# x. If p < 1/2 then w(z) ahd =

in Example 5.3 show that (a) I < t:xn when 7 < 1 and 5.2. The computations and hence J = x. Consulting the table (b) for 7 < we have AT(0) = of deinitions we see that 0 i an absorbing boundary for y <

8

=

dz

a-2c.

1 and note that

-

zl-l

+

.

6

(1+ 5)/2 % 1, and

2J-1

z;

(z)(1+ )-.7 <

-1

% 1.

8

.171(./1

5 4 pz) henceJ = .

*

*

=

z

co.

o

w(x) #

z1-24 =

2J

1

Combining the results for I result. =

and

f

When J

x.

=

svhenJ > 1/2 ,y

5.5., JYhen a

(x)

and

J

-

1

-

1

<

exp

(x3

and

.j,+

-

dz

if 8 K 1

cxl

=

and consulting

the table

--

1/2

2p ,C

-,p

dz

=

Cy

rnk so

that

of rnv, so we

-+

6.1. If dgu, #) 0 then can only take the values 0 and 1. In Example 6.2 we showed that h cannot be H 0 so it must take the value 1 at some point c. Since is right continuous at a there must be some point b 1/2 where # () = 1 but in this case it is impossible for dgn, h) 0. -+

:

#

1

=

jf 6 > j

< x

0

y/(z) =

1/2 M(=)

3.1. Suppose rs r. Then there is an e > 0 and a subsequence lrn': rl > t for all k. However it is impossible for a subsequence have a contradiction which proves rn r.

we

ltave the indicated

-+

Referencs M. Aizenmau and B. Simon (1982) Browniaa motion and a Ilyrnack inqpality for Schrdinger operators. Comm. p'tlrc All. Xf. 35j 209-273 P. Billingsley (1j68)Convevgence ofpvobability New York

Measures. John Wiley

Mariov Pvocesqs R.M. Blmepthal and R.K. detoor(1968) Tlteqvy. Academic Press, New York

anb

Sons,

ePofenicl

and

D. Buzkholder (1j73)Distribution function inequaltties for marhgatest Probab. 1, 19-42 D Burkholder, R. Gundy, >pd M. Silverstein (1971) A m>xtrat HP. class acterizatiop of the Trcn.s A.M.S. 157, 137-153 '' .. . '

.

'

K.L. Chung and Intvoduction .J. Wiltiams (1990) edition. Birkhauser, Boston second

K L Chung and Z. Zhao (1995) From Brownian Molion ion. Springer, New York '

g

. '

.. '

function char'

ochqstc

t/ tl

Integmtion.

Schrsdingev's fpzc-

'

.

Wnn.

.

.

First pmssage times and sojlly Z. Ciesielski and S.J. Taylor (1962) tipes for Brownian motion in space and exact HausdorF memsure. Tmns. .u s .

103

1

t34-450

!!

-

.

.

'

Semimartingates and E.. Cinlar, J. Jacod, P. Protter, and M. Sharpe (1980) Markv proceses. Z. fuv. Wahr. 54, 161-220

(1983)On Brownian slow point. Z. fnr Wahr. 64, 359-367 and P.A. Meyer (1g78) pvobabilities c?,d potentials ? xqrts oellackerie uol-

B. Davis

c

*

land, Amsterdam

.

C. Dellacherie and P.A. Meyer (1982) Probabilities Mavtingales. North Holland Amsterdam .

and

Poteniials B. T/lct?o of

1

C. Dol>ns-bae (1970)Quelquesapplications de la formule de changement variables pour 1es semimartingales. Z. fnr Wor. 16, 181-194 > ro a iility:Theovy and Ezamples. Second R. Durrett (1995) Press Belmont, CA )

edition.

de

Duxbury

336

Referencs

References

M. Kac (1951) On some connections between probability theory and diserential and integral equations. Pages 189-215 in Proc. 2nd Berkeley Symposium, of California Press U.

Some problems on random walk in space. A. Dvoretsky and P. Erds (1951) Pages 353-367 in Proc. 2nd Berkeley Symp., U. of California Press

(1965)Marko'u Pvocesses. Springer, New York A criterion for invariant measures of Markov processes. P. Echeverria (1982)

E.B. Dynkin

1. Karatzms and S. Shreve (1991)Brownian Motion Second edition. Springer, New York

Z.

kkr Wahr. 61, 1-16

Markon Processes: Ccrccterzcon S. Ethier and T. Kurtz (1986) and Sons, New York John Wiley wergence.

S. Karlin nd H. Taylor (1981) W Second Conrse in ' demic Press, New York

Con-

and

T. Kato (1973) Schrdinger 135-148 13,

W. Feller (1951) biflksionprocesses in genettcs. Pages 227-2246in Proc. 2nd Bevkeley ,S'pflzp.,U. of Califorhia Ptss G.B. Folland (1976) Xn Inkvoduction Princetn NJ U Press t *

)

E

d D 9gsion. Holden-bay,

an

Pariial Dilkrcnicl A Friedman (1964) Hall, Eglewood Clifs, NJ

Equakions

'

.

'

E.

.

' .

:

F

..

.

'

. '.

of

.

.

.

D. Gilbarg and N.S. (1977)Eiliic Scond Ovdr. Springer, New York

(1969)Reql and bs .

rccf Analysis. Springer, New

(1981)Stociastic

Drereztticl North Holland, Amstrdam Lprocesses.

Equations .

(1987)fimif Theorems

J. Jacod apd A.N. Shiryaev Spr in/er, New York P. Jagers

(1975)Bvanclting Pvocesses lnit

j

..

.

and

.D;#k-

.

for Soc/zcdc

rocesses.

Biological Applicaiions. Joim Wiley

and Sons, New York

Pavtial Dkrrenql F. John (1982) Ne:? York '

Equalions. Fourth edition. Springer-verlag,

s'occsic Pvoceses.

Aca-

Isvael J. J/.

and

Dlrkdon. American Math.

(1969)Siochastic Integrals. Academic Press, New York (1976)Un ours sur 1es intgrals stochmstiques. Pages 246-400

P.A.. Meyer Lcture Notes

bessel

kqnaiionsof

Calculns.

H.P. Mclfean

ln Vath. 511, Springer, New York

'

in

N. Meyers and J. Serrin (1960) The exterior Dirichlet problem for second order elliptic partial diserential equations. & fc. Meclt. 9, 513-538 B. Oksendal (1992) Stocltastic Dlrerencl New York

order eniptic

Partial Dlrrenicl

'ludinger

S. Watanabe

second

s'occc

with singular potentials.

F. Knight (1981) Essentials of Brownian Motion Society, Providence, RI

..

( . ' ' ' ' 1(' .. .

of Brownian motion and

o.

.

E

olume

diliccfons,

and

.

type.Prentice-

Paraolic

Gilbarg and ,. serriu(1s56)on isolated singutarities o diferetial equations. Ji d'Analyse n. 4, 309-340

and

.

' .

Excursions R.K. Getor and M. Sharpe (1979) 47, Wahri 83-106 Z. proeses. fnr.

E. Hewitt and K. Strom erg York

San Fran-

.

:

Stocltastic Dlrerentfcl Eqnations A. Friedman (1975) 1. Acadmic Press, Ne* York

N. Ikeda sion

'

,

Brownian Motion D. Freedrpan (1970) cicd CA

operators

and

R.Z. Khmsminskii (1960) Ergodic properties of recurrent diflksion processes and stabilization of the solution to the Cauchy problem for parabolic equations. Tlteov. Prob. Appl. 5, 179-196

Partial Dllkrcnticl Eqnations. Prince-

o

337

Eqnations. Third

S. Port and C. Stone (1978) Bvownian Molion Academic Press, New York Stochastic Inkegration P: Protter (1990) New York

and

and

edition.

Springer,

Classical Jbenfcl

Tlteo'y.

Dlrrenicl

Equations. Springer,

D. Revuz and M. Yor (1991)Continnons Mcringcle.s and Brownian Motion. Springer, New York L.C.G. Rogers and D. Williams (1987) Dtkgsionst Markov Processes, and Maw tingales. Volnme 2: It Calculus. John Wiley and Sons, New York

(1988)Real Analysis. Third edition. MacMillan, New York B. Simon (1j82)Schrdinger semigroups. Bulletin A.M.S. T, 447-526 D.W. Stroock and S.R.S. Varadhan (1979)Mnltidimensional Dirkdo?z

H. Royden

cesses. Springer, New York

Pvo-

338

Refezences

additive functionals H. Tanaka (1963) Note on continuous Brownian motion. Z. kkr Gcr. 1, 251-257

A diflksion with a discontinuous loil J. Walsh (1978) 37=45

of on-dimensional

tiiri. Attviqne 52-53,

Index

of solutions On the uiqueness T. Yamada and S. Watanabe (1971) to stoihastic diserential equations. J. A. Klloto 11. 1, 155-167: lI, 553-563

adapted 36 arcsine law 173, 291 Arzela-Ascoli theorem 284 : 1aw 76 nxqociative averaging propertyl 148

D, the space 293 difeventiating under >n integral 129 difusion process 177 Dini function 239 Dirichlet problem 143 distributive laws 72 Dolans measure 56 Donsker's theorem 287 Dvoretsky-Erds theorez 99 q

E

basic predictable mocess52 Bessel process 179, 232 BichtelemDellacherie theoxem 71 Blumenthal's 0:1 law 14 boundary behavior of dilusions 271 Brownian motion continuityof paths 4 de:nition 1 Eldex continuity 6 Lvy's characterlzation of 111 6 nondilexentiilisy quadratic variation 7 write.s your name 15 ! Zeros f 23 Burkholder Davis Gundy theorem 116 '

,

;

esectiv dimension 239 exit distributions ball 164

Ein

.

.E

half space i66 exponentialBrownian motion 178 exponential martingale

108

Feller semigroup 246 Feller's branching diflkston 181, 231 Feller's test 214, 227 Feynman-lfac formpla 137 fltration 8 flnite dimensional ets 3, 10 Fubini's theorem 86 fundamental sollztion 251 .

C, the space 4, 282 Cauchy process 29 change of measure SDE 202 change of time in SDE 207 CiesielskivTaylor theorem 171 cone condition 147 contilmous papping theorem 273 contraction semigroup 245 converging together lemma 273 covaxiance mocess50

.

E

.

generator

of a semigroup

Girsanov's formula 91

246

Green's functions Brownian motion 104 one dimensional disusions 222 Gronwall's inequality 188

Tndex

346 Index harmonicfunction 95 averagingproperty 148 Dirichlet problem 143

zaxris chains 259 heat' equation 126

inhomogeneous 130 hitting times 19, 26

half space 31 optional process 36 in discrete time 34 optional stopping theorem 39 Ornstein-uhlenbeck proces 180

stopping time 18 strong Markov property for Browni>n motion 21 for dilusions 201 strong soltltion to SDE 183

pathwise uniqueness for SDt 184 theorem 11 'm Poksson's equation 151 potential kernel 102 predictable process 34 basic 52 in discret time 34 simp Ie 53 C ( predictable c'-feld 35' progressivley measurable process 36 Prokhorov's theorems 276 P unctured disc''146

tail c-fleld 17 tight sequence of memsures 276 transience, see recurrence transition density 257 transition kernels 250

-

5*

ininitesimal drift 178 infnitesimal generator 246 infnitesimal varianc 179 intjration by parts 76 isometry property 56 It' S fOrmll la 68 multivariate77 i !1 E

.

' E

'

:

;E

'

. .

.

' .

' . . . . . '.

.

.

'

.(

:

Kalman-Bucy fllti 180 Khmsminskii' lmma 141 Khsrnihskii's test 237) Kolmogorov's continuity theorem 5 Kunita-Watartabe inqtzality 59 '

''

.

E

.

law of the iterated lojarithr 115 Lebesgue's thorn 146 Lty' theorem 111 37; 113 local martigal local tiine 83 continuity of 88 locally 4uivlent memsures 90

qua dratic

variation

recurrence nd transience Brownian motion 17 98'. one dimensional difusions 220 Hafris chains 262 reflecting boundary 229 reflection principle 25 regular point 144 resolvent opertor 249 relatively compact 276 right continuous fltrti

E

.

.

E

E

'

E

'

'

1

'

.

Markov property 9 martingale problem 198 well psed 210 Meyer-Tanaka fothmla 84 283 modulus of cotinit# monptone class theorem 12

'

E

.

scaling relation 2 Schrdinger equati 156 semigroup pibperty 245 semimartingale 70 shift transformatios 9 simpte predictable process 53 skew Brownian motion 89 Skorokhod representation 274 Skorokhod topology on D 293 speed measure 224 E

.

natural scale 212 nonnegative definite matricts '

'E

occupatiori

.(

timsE

ball 167

i

198 . ..

Cr

E

'

weak convergence 271 Wrightisher diflksion 182, 234, 308 Yamada-Watanabe

50

E

i '

'

.

E

E

E'

'

'

.

.

variance process 42

'

'

.

.

uniqueness in distribution for SDE 197

.

E

'

i

('

'

q

theorem 193

341