Lectures on diffusion problems and partial differential equations

Lectures on Diffusion Problems and Partial Differential Equations

By

S. R. S. Varadhan

Notes by

P. Muthuramalingam Tara R. Nanda

Tata Institute of Fundamental Research, Bombay 1989

Author S. R. S. Varadhan Courant Institute of Mathematical Sciences 251, Mercer Street New York. N. Y. 10012. U.S.A.

c Tata Institute of Fundamental Research, 1989

ISBN 3-540-08773-7. Springer-Verlag, Berlin, Heidelberg. New York ISBN 0-387-08773-7. Springer-Verlag, New York. Heidelberg. Berlin

No part of this book may be reproduced in any form by print, microfilm or any other means without written permission from the Tata Institute of Fundamental Research, Bombay 400 005

Printed by N. S. Ray at the Book Centre Limited Sion East, Bombay 400 022 and published by H. Goetze Springer-Vertal, Heidelberg, West Germany PRINTED IN INDIA

Contents 1 The Heat Equation

1

2 Kolmogorov’s Theorem

11

3 The One Dimensional Random Walk

15

4 Construction of Wiener Measure

19

5 Generalised Brownian Motion

31

6 Markov Properties of Brownian Motion

33

7 Reflection Principle

39

8 Blumenthal’s Zero-One Law

53

9 Properties of Brownian Motion in One Dimension

57

10 Dirichlet Problem and Brownian Motion

63

11 Stochastic Integration

71

12 Change of Variable Formula

91

13 Extension to Vector-Valued Itˆo Processes

95

14 Brownian Motion as a Gaussian Process

101

iii

iv

Contents

15 Equivalent For of Itˆo Process

105

16 Itˆo’s Formula

117

17 Solution of Poisson’s Equations

129

18 The Feynman-Kac Formula

133

19 An Application of the Feynman-Kac Formula....

139

20 Brownian Motion with Drift

147

21 Integral Equations

155

22 Large Deviations

161

23 Stochastic Integral for a Wider Class of Functions

187

24 Explosions

195

25 Construction of a Diffusion Process

201

26 Uniqueness of Diffusion Process

211

27 On Lipschitz Square Roots

223

28 Random Time Changes

229

29 Cameron - Martin - Girsanov Formula

237

30 Behaviour of Diffusions for Large Times

245

31 Invariant Probability Distributions

253

32 Ergodic Theorem

275

33 Application of Stochastic Integral

281

Appendix

284

Contents

v

Language of Probability

284

Kolmogorovs Theorem

288

Martingales

290

Uniform Integrability

305

Up Crossings and Down Crossings

307

Bibliography

317

Contents

vi

Preface THESE ARE NOTES based on the lectures given at the T.I.F.R. Centre, Indian Institute of Science, Bangalore, during July and August of 1977. Starting from Brownian Motion, the lectures quickly got into the areas of Stochastic Differential Equations and Diffusion Theory. An attempt was made to introduce to the students diverse aspects of the theory. The last section on Martingales is based on some additional lectures given by K. Ramamurthy of the Indian Institute of Science. The author would like to express his appreciation of the efforts by Tara R. Nanda and PL. Muthuramalingam whose dedication and perseverance has made these notes possible. S.R.S. Varadhan

1. The Heat Equation LET US CONSIDER the equation (1)

1

1 ut − ∆u = 0 2

which describes (in a suitable system of units) the temperature distribution of a certain homogeneous, isotropic body in the absence of any heat sources within the body. Here u ≡ u(x1 , . . . , xd , t);

ut ≡

∂u ; ∂t

∆u =

d X ∂2 u i=1

∂x2i

,

t represents the time ranging over [0, ∞) or [0, T ] and x ≡ (x1 . . . xd ) belongs to Rd . We first consider the initial value problem. It consists in integrating equation (1) subject to the initial condition (2)

u(0, x) = f (x). The relation (2) is to be understood in the sense that Lt u(t, x) = f (x).

t→0

Physically (2) means that the distribution of temperature throughout the body is known at the initial moment of time. We assume that the solution u has continuous derivatives, in the space coordinates upto second order inclusive and first order derivative in time. 1

1. The Heat Equation

2 It is easily verified that (3) 2

u(t, x) =

! 1 |x|2 exp − ; 2t (2πt)d/2

|x|2 =

d X

x2i ,

i=1

satisfies (1) and u(0, x) = Lt u(t, x) = δ(x)

(4)

t→0

Equation (4) gives us a very nice physical interpretation. The solution (3) can be interpreted as the temperature distribution within the body due to a unit sourse of head specified at t = 0 at the space point x = 0. The linearity of the equation (1) now tells us that (by superposition) the solution of the initial value problem may be expected in the form Z (5) u(t, x) = f (y)p(t, x − y)dy, Rd

where p(t, x) =

1 |x|2 exp − . 2t (2πt)d/2

Exercise 1. Let f (x) be any bounded continuous function. Verify that p(t, x) satisfies (1) and show that R (a) p(t, x)dx = 1, ∀t > 0; R (b) Lt p(t, x) f (x)dx = f (0); t→0+

(c) using (b) justify (4). Also show that (5) solves the initial value problem. (Hints: For (a) use

R∞

−∞

2

e−x dx =

√

π. For part (b) make the substitution

x y = √ and apply Lebesgue dominated convergence theorem). π

3

3

Since equation (1) is linear with constant coefficients it is invariant under time as well as space translations. This means that translates of solutions are also solutions. Further, for s ≥ 0, t > 0 and y ∈ Rd , (6)

u(t, x) =

|x − y|2 1 exp − 2(t + s) [2π(t + s)]d/2

and for t > s, y ∈ Rd , (7)

u(t, x) =

|x − y|2 1 exp − 2(t − s) [2π(t − s)]d/2

are also solutions of the heat equation (1). The above method of solving the initial value problem is a sort of trial method, viz. we pick out a solution and verify that it satisfies (1). But one may ask, how does one obtain the solution? A partial clue to this is provided by the method of Fourier transforms. We pretend as if our solution u(t, x) is going to be very well behaved and allow all operations performed on u to be legitimate. Put v(t, x) = uˆ (t, x) where bstands for the Fourier transform in the space variables only (in this case), i.e. Z v(t, x) = u(t, y)ei x−y dy. Rd

Using equation (1), one easily verifies that 1 vt (t, x) = |x|2 v(t, x) 2

(8) with

v(0, x) = fˆ(x).

(9)

The solution of equation (8) is given by (10)

2 v(t, x) = fˆ(x)e−t|x| /2 .

We have used (9) in obtaining (10).

1. The Heat Equation

4 Exercise 2. Verify that

4

! t|x|2 p(t, ˆ x) = exp − . 2 Using Exercise 2, (10) can be written as v(t, x) = uˆ (t, x) = fˆ(x) p(t, ˆ x).

(11)

The right hand side above is the product of two Fourier transforms and we know that the Fourier transform of the convolution of two funtions is given by the product of the Fourier transforms. Hence u(t, x) is expected to be of the form (5). Observe that if f is non-negative, then u is nonnegative and if f is bounded by M then u is also bounded by M in view of part (a) of Exercise 1. The Inhomogeneous Equation. Consider the equation vt −

∆v = g, 2

with

v(0, x) = 0,

which describes the temperature within a homogeneous isotropic body in the presence of heat sources, specified as a function of time and space by g(t, x). For t > s, u(t, x) =

1 |x − y|2 exp − 2(t − s) [2π(t − s)]d/2

1 is a solution of ut (t, x) − ∆u(t, x) = 0 corresponding to a unit source at 2 t = s, x = y. Consequently, a solution of the inhomogeneous problem is obtained by superposition. Let v(t, x) =

Z Zt

Rd 0

5

i.e.

! 1 |x − y|2 exp − g(s, y) dy ds 2(t − s) [2π(t − s)]d/2

5

v(t, x) =

Zt

w(t, x, s)ds

0

where w(t, x, s) =

Z

Rd

! 1 |x − y|2 g(s, y) dy. exp − 2(t − s) [2π(t, s)]d/2

Exercise 3. Show that v(t, x) defined above solves the inhomogeneous heat equation and satisfies v(0, x) = 0. Assume that g is sufficiently 1 smooth and has compact support. vt − ∆v = Lt w(t, x, s) and now use s→t 2 part (b) of Exercise (1). Remark 1. We can assume g has compact support because in evaluating 1 vt − ∆v the contribution to the integral is mainly from a small neigh2 bourhood of the point (t, x). Outside this neighbourhood ! |x − y|2 1 exp − 2(t − s) [2π(t − s)]d/2 satisfies

1 ut − ∆u = 0. 2 2. If we put g(s, y) = 0 for s < 0, we recognize that v(t, x) = g ∗ p. Taking spatial Fourier transforms this can be written as

v(t, ξ) =

Zt

1 g(s, ξ) exp − (t − s)|ξ|2 dξ, 2

0

or

! ∂ˆv ∂v 1 1 = = g(t, ξ) + ∆v = g(t, ξ) + ∆v . ∂t ∂t 2 2 Therefore

∂v 1 − ∆v = g. ∂t 2

1. The Heat Equation

6

1 Exercise 4. Solve wt − ∆w = g on [0, ∞) × Rd with w = f on {0} × Rd 6 2 (Cauchy problem for the heat equation). Uniqueness. The solution of the Cauchy problem is unique provided the class of solutions is suitably restricted. The uniqueness of the solution is a consequence of the Maximum Principle. Maximum Principle. Let u be smooth and bounded on [0, T ] × Rd satisfying ut −

∆u ≥0 2

in (0, T ] × Rd

u(0, x) ≥ 0, ∀x ∈ Rd .

and

Then u(t, x) ≥ 0 ∀,

t ∈ [0, T ]

and

∀x ∈ Rd .

Proof. The idea is to find minima for u or for an auxillary function. Step 1. Let v be any function satisfying vt −

∆v >0 2

in (0, T ] × Rd .

Claim . v cannot attain a minimum for t0 ∈ (0, T ]. Assume (to get a contradiction) that v(t0 , x0 ) ≤ v(t, x) for some t0 > 0 and for all t ∈ [0, T ], ∀x ∈ Rd . At a minimum vt (t0 , x0 ) ≤ 0, (since t0 , 0) ∆v(t0 , x0 ) ≥ 0. Therefore ! ∆v (t0 , x0 ) ≤ 0. vt − 2 Thus, if v has any minimum it should occur at t0 = 0. Step 2. Let ǫ > 0 be arbitrary. Choose α such that h(t, x) = |x|2 + αt 7

satisfies ht −

∆h = α − d > 0 (say α = 2d). 2

7 Put vǫ = u + ǫh. Then ∂vǫ 1 − ∆vǫ > 0. ∂t 2 As u is bounded, vǫ → +∞ as |x| → +∞, vǫ must attain a minimum. This minimum occurs at t = 0 by Step 1. Therefore, vǫ (t, x) ≥ vǫ (0, x0 ) for some

x0 ∈ Rd ,

i.e. vǫ (t, x) ≥ u(0, x0 ) + ǫ|x0 |2 > 0, i.e. u(t, x) + ǫh(t, x) > 0, ∀ǫ. This gives u(t, x) ≥ 0. This completes the proof.



Exercise 5. (a) Let L be a linear differential operator satisfying Lu = g on Ω (open in Rd ) and u = f on ∂Ω. Show that u is uniquely determined by f and g if and only if Lu = 0 on Ω and u = 0 on ∂Ω imply u = 0 on Ω. 1 (b) Let u be a bounded solution of the heat equation ut − ∆u = g 2 with u(0, x) = f (x). Use the maximum principle and part (a) to show that u is unique in the class of all bounded functions. (c) Let  2   e−1/t , if t > 0, g(t) =   0, if t ≤ 0, ∞ X g(k) (t/2)x2k u(t, x) = , (2k)! k=0

on

R × R. 8

1. The Heat Equation

8 Then u(0, x) = 0,

ut =

∆u , 2

u . 0,

i.e. u satisfies ut −

1 ∂2 u = 0, 2 ∂x2

with u(0, x) = 0.

This example shows that the solution is not unique because, u is not bounded. (This example is due to Tychonoff). Lemma 1. Let p(t, x) =

1 |x|2 exp − for t > 0. Then 2t (2πt)d/2

p(t, ·) ∗ p(s, ·) = p(t + s, ·). Proof. Let f be any bounded continuous function and put Z u(t, x) = f (y)p(t, x − y)dy. Rd

Then u satisfies 1 ut − ∆u = 0, 2

u(0, x) = f (x).

Let v(t, x) = u(t + s, x). Then

1 vt − ∆v = 0, v(0, x) = u(s, x). 2 This has the unique solution Z v(t, x) = u(s, y)p(t, x − y)dy. Thus Z " f (y)p(t + s, x − y)dy = f (z)p(s, y − z)p(t, x − y)dz dy. Rd

9 9

This is true for all f bounded and continuous. We conclude, therefore, that p(t, ·) ∗ p(s, ·) = p(t + s, ·).  Exercise 6. Prove Lemma 1 directly using Fourier transforms. It will be convenient to make a small change in notation which will be useful later on. We shall write p(s, x, t, y) = p(t−s, y−x) for every x, y and t > s. p(s, x, t, y) is called the transition probability, in dealing with Brownian motion. It represents the probability density that a “Brownian particle” located at space point x at time s moves to the space point y at a later time t. Note . We use the same symbol p for the transition probability; it is function of four variables and there will not be any ambiguity in using the same symbol p. Exercise 7. Verify that Z p(s, x, t, y)p(t, y, σ, z)dy = p(s, x, σ, z), s < t < σ. Rd

(Use Exercise 6). Remark . The significance of this result is obvious. The probability that the particle goes from x at time s to z at time σ is the sum total of the probabilities, that the particle moves from x at s to y at some intermediate time t and then to z at time σ.

10

10

1. The Heat Equation

In this section we have introduced Brownian Motion corresponding 1 to the operator ∆. Later on we shall introduce a more general diffusion 2 P ∂ 1P ∂2 process which corresponds to the operator ai j + bj . 2 ∂xi ∂x j ∂x j

2. Kolmogorov’s Theorem Definition. LET (Ω, B, P) BE A probability space. A stochastic process 11 in Rd is a collection {Xt : t ∈ I} of Rd -valued random variables defined on (Ω, B). Note 1. I will always denote a subset of R+ = [0, ∞).

2. Xt is also denoted by X(t). Let {Xt : t ∈ I} be a stochastic process. For any collection t1 , t2 , . . . , tk such that ti ∈ I and 0 ≤ t1 < t2 < . . . < tk and any Borel set Λ, in Rd × Rd × · · · × Rd (k times),. define Ft1 . . . tk (Λ) = P(w ∈ Ω : (Xt1 (w), . . . , Xtk (w)) ∈ Λ). If {t1 , . . . , tk } ⊂ {s1 , . . . , sℓ } ⊂ I,

with

l≥k

such that (0) (1) (1) (k) (k) s(0) 1 < . . . < sn0 < t1 < s1 . . . < sn < t2 . . . < tk < s1 . . . < snk ,

let then π : Rd × · · · × Rd (1 times) → Rd × · · · × Rd (k times) be the canonical projection. If Eti ⊂ Rd is any Borel set in Rd , i = 1, 2, . . . , k, then π−1 (Et1 × · · · × Etk ) = Rd × · · · × Et1 × Rd × · · · × Et2 × · · · × Rd 11

2. Kolmogorov’s Theorem

12

(l times). The following condition always holds. (*)

Et1 . . . tk (Et1 × · · · × Etk ) = F s1 . . . s1 (Π−1 (Et1 × · · · × Etk )).

If (∗) holds for an arbitrary collection {Ft1 . . . tk : 0 ≤ t1 < t2 . . . < tk } 12 (k = 1, 2, 3 . . .) of distributions then it is said to satisfy the consistency condition. Exercise 1. (a) Verify that Ft1 . . . tk is a probability measure on Rd × · · · × Rd (k times). (b) Verify (∗). (If Bm denotes the Borel σ field of Rm , Bm+n = Bm × Bn ). The following theorem is a converse of Exercise 1 and is often used to identify a stochastic process with a family of distributions satisfying the consistency condition. Kolmogorov’s Theorem. Let {Ft1 ,t2 ,...tk 0 ≤ t1 < t2 < . . . < tk < ∞} be a family of probability distributions (on Rd × · · · × Rd , k times, k = 1, 2, . . .) satisfying the consistency condition. Then there exists a measurable space (Ωk , B), a unique probability measure P an (Ωk , B) and a stochastic process {Xt : 0 ≤ t < ∞} such that the family of probability measures associated with it is precisely {Ft1 ,t2 ,...tk : 0 ≤ t1 < t2 < . . . < tk < ∞},

k = 1, 2, . . . .

A proof can be found in the APPENDIX. We mention a few points about the proof which prove to be very useful and should be observed carefully. 13

1. The space Ωk is the set of all Rd -valued functions defined on [0, ∞): Y Rd ΩK = t∈[0,∞)

2. The random variable Xt is the tth -projection of ΩK onto Rdt .

13 3. B is the smallest σ-algebra with respect to which all the projections are measurable. 4. P given by P(w : Xt1 (w) ∈ A1 , . . . Xtk (w) ∈ Ak ) = Ft1 ...tk (A1 × · · · × Ak ) where Ai is a Borel set in Rd , is a measure on the algebra generated by {Xt1 , . . . Xtk }(k = 1, 2, 3 . . .) and extends uniquely to B. Remark . Although the proof of Kolmogorov’s theorem is very constructive the space ΩK is too “large” and the σ-algebra B too “small” for practical purposes. In applications one needs a “nice” collection of Rd -valued functions (for example continuous, or differentiable functions), a “large” σ-algebra on this collection and a probability measure concentrated on this family.

3. The One Dimensional Random Walk BEFORE WE TAKE up Brownian motion, we describe a one dimen- 14 sional random walk which in a certain limiting case possesses the properties of Brownian motion. Imagine a person at the position x = 0 at time t = 0. Assume that at equal intervals of time t = τ he takes a step h either along the positive x axis or the negative x axis and reaches the point x(t) = x(t − τ) + h or x(t) = x(t − τ) − h respectively. The probability that he takes a step in either direction is assumed to be 1/2. Denote by f (x, t) the probability that after the time t = nτ (n intervals of time τ) he reaches the position x. If he takes m steps to the right (positive x-axis) in reaching x then there are nCm possible ways in which he can achieve these m steps. Therefore, 1 the probability f (x, t) is nCm ( )n . 2 f (x, t) satisfies the difference equation (1)

f (x, t + τ) =

1 1 f (x − h, t) + f (x + h, t) 2 2

and (2)

x = h(m − (n − m)) = (2m − n)h.

To see this one need only observe that to reach (x, t + τ) there are two ways possible, viz. (x − h, t) → (x, t + τ) or (x + h, t) → (x, t + τ) and the probability for each one of these is 1/2. Also note that by definition 15

3. The One Dimensional Random Walk

16 of f ,

f (h, τ) =

(3) 15

1 = f (−h, τ), 2

so that f (x, t + τ) = f (h, τ) f (x − h, t) + f (−h, τ) f (x + h, t).

(4)

The reader can identify (4) as a “discrete version” of convolution. By our assumption, (5)

f (0, 0) = 1,

f (x, 0) = 0 if

x , 0.

We examine equation (1) in the limit h → 0, τ → 0. To obtain reasonable results we cannot let h and τ tend to zero arbitratily. Instead we assume that h → 1 as τ

(6)

h→0

and

τ → 0.

The physical nature of the problem suggests that (6) should hold. To see this we argue as follows. Since the person is equally likely to go in either direction the average value of x will be 0. Therefore a reasonable measure of the “progress” made by the person is either |x| or x2 . Indeed, since x is a random variable (since m is one) one gets, using (2), E(x) = 2E(m) − n = 0, E(x2 ) = h2 E((2m − n)2 ) = h2 n. !n !n n n P n P n(n + 1) 1 1 n n = , = ) (Use m Cm Cm 2 2 m=0 2 4 m=0 Thus

E

16

(

) x2 1 h2 n h2 = E(x2 ) = = , t t nτ τ

and as t becomes large we expect that the average distance covered per unit time remains constant. (This constant is chosen to be 1 for reasons that will become apparent later). This justifies (6). In fact, a simple h2 argument shows that if → 0 or +∞, x may approach +∞ in a finite τ time which is physically untenable.

17 (1) now gives f (x, t + τ) − f (x, t) =

1 { f (x − h, t) − f (x, t) + f (x, h, t) − f (x, t)}. 2

Assuming sufficient smoothness on f , we get in the limit as h, τ → 0 and in view of (6), (7)

∂f 1 ∂2 f = ∂t 2 ∂x2

h2 (to get the factor 1/2 we choose → 1). This is the equation satisfied τ by the probability density f . The particle in this limit performs what is known as Brownian motion to which we now turn our attention.

References. [1] GNEDENKO: The theory of probability, Ch. 10. [2] The Feynman Lectures on physics, Vol. 1, Ch. 6.

4. Construction of Wiener Measure ONE EXAMPLE WHERE the Kolmogorov construction yields a proba- 17 bility measure concentrated on a “nice” class Ω is the Brownian motion. Definition . A Brownian motion with starting point x is an Rd -valued stochastic process {X(t) : 0 ≤ t < ∞} where (i) X(0) = x = constant; (ii) the family of distribution is specified by Z Ft1 . . . tk (A) = p(0, x, t1 , x1 )p(t1 , x1 , t2 , x2 ) . . . A

p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk for every Borel set A in Rd × · · · Rd (k times). N.B. The stochastic process appearing in the definition above is the one given by the Kolmogorov construction. It may be useful to have the following picture of a Brownian motion. The space Ωk may be thought of as representing particles performing Brownian movement; {Xt : 0 ≤ t < ∞} then represents the trajectories of these particles in the space Rd as functions of time and B can be considered as a representation of the observations made on these particles. Exercise 2. (a) Show that Ft1 ...tk defined above is a probability measure on Rd × · · · × Rd (k times). 19

4. Construction of Wiener Measure

20

(b) {Ft1 ...tk : 0 ≤ t1 < t2 < . . . tk < ∞} satisfies the consistency condition. (Use Fubini’s theorem). 18

(c) Xt1 − x, Xt2 − Xt1 , . . . , Xtk − Xtk−1 are independent random variables and if t > s, then Xt − X s is a random variable whose distribution density is given by ! 1 1 −1 2 p(t − s, y) = exp − (t − s) |y| . 2 [2π(t − s)]d/2 (Hint: Try to show that Xt1 − x, Xt2 − Xt1 , . . . , Xtk − Xtk−1 have a joint distribution given by a product measure. For this let φ be any bounded real measurable function on Rd × · · · × Rd (k times). Then E(φ(Z1 , . . . , Zk )) Xt1 −x,Xt2 −Xt1 ,...,Xtk −Xtk−1

=

E

Xt1 ,...,Xtk

(φ(Z1 − x, . . . , Zk − Zk−1 ))

where E(φ) is the expectation of φ with respect to the joint disXt1 ...Xtk

tribution of (Xt1 . . . , Xtk ). You may also require the change of variable formula). Problem . Given a Brownian motion with starting point x our aim is to find a probability P x on the space Ω = C([0, ∞); Rd ) of all continuous funcitons from [0, ∞) → Rd which induces the Brownian motion. We will thus have a continuous realisation to Brownian motion. To achieve this goal we will work with the collection {Ft1 ,...,tk : 0 ≤ t1 < t2 < . . . < tk } where ti ∈ D, a countable dense subset of [0, ∞). 19

Step 1. The first step is to find a probability measure on a “smaller” space and lift it to C([0, ∞); Rd ). Let Ω = C([0, ∞); Rd ), D a countable dense subset of [0, ∞); Ω(D) = {F : D → Rd } where f is uniformly continuous on [0, N] ∩ D for N = 1, 2, . . .. We equip Ω with the topology of uniform convergence on compact sets and Ω(D) with the topology of uniform convergence on sets of the form D ∩ K

21 where K ⊂ [0, ∞) is compact; Ω and Ω(D) are separable metric spaces isometric to each other. Exercise 3. Let pn ( f, g) = sup | f (t) − g(t)|

for

0≤t≤n

f, g ∈ Ω

and pn,D ( f, g) = sup | f (t) − g(t)|

for

0≤t≤n t∈D

f, g ∈ Ω(D).

Define ρ( f, g) = ρD ( f, g) =

∞ X 1 pn ( f, g) , ∀ f, g ∈ Ω, 2n 1 + pn ( f, g) n=1 ∞ X 1 pn,D ( f, g) , ∀ f, g ∈ Ω(D). 2n 1 + pn,D ( f, g) n=1

Show that (i) { fn } ⊂ Ω converges to f if and only if fn → f uniformly on compact subsets of [0, ∞); (ii) { fn } ⊂ Ω(D) converges to f if and only if fn|D∩K → f|D∩K| uniformly for every compact subset K of [0, ∞); (iii) {(P1 , . . . , Pd )} where Pi is a polynomial with rational coefficients 20 is a countable dense subset of Ω; (iv) {(P1D , . . . , PdD )} is a countable dense subset of Ω(D); (v) τ : Ω → Ω(D) where τ( f ) = f|D is a (ρ, ρD )-isometry of Ω onto Ω(D); (vi) if V( f, ǫ, n) = {g ∈ Ω : pn ( f, g) < ǫ} for f ∈ Ω, ǫ > 0 and VD ( f, ǫ, n) = {g ∈ Ω(D) : pn,D ( f, g) < ǫ}

for

f ∈ Ω(D), ǫ > 0,

then {V( f, ǫ, n) : f ∈ Ω, ǫ > 0, n = 1, 2 . . .}

4. Construction of Wiener Measure

22

is a base for the topology of Ω and {VD ( f, ǫ, n) : f ∈ Ω(D), ǫ > 0, n = 1, 2, . . .} is a base for the topology of Ω(D). Remark. By Exercise 3(v) any Borel probability measure on Ω(D) can be lifted to a Borel probability measure on Ω. 2nd Step. Define the modulus of continuity ∆T,δ D ( f ) of a function f on D in the interval [0, T ] by ∆T,δ D ( f ) = sup{| f (t) − f (s)| : |t − s| < δt, s ∈ D ∩ [0, T ]} As D is countable one has N, 1

Exercise 4. (a) Show that f : ∆D j ( f ) ≤ 1k } is measurable in the σalgebra generated by the projections πt : π{Rdt : t ∈ D} → Rdt 21

Proof. The lemma is equivalent to showing that B = σ(E ). As each of the projection πt1 ...tk is continuous, σ(E ) ⊂ B. To show that B ⊂ σ(E ), it is enough to show that VD ( f, ǫ, n) ∈ E because Ω(D) is separable. (Cf. Exercise 3(iv) and 3(vi)). By definition VD ( f, ǫ, n) = {g ∈ Ω(D) : Pn,D ( f, g) < ǫ} ) ∞ ( [ 1 = g ∈ Ω(D) : pn,D ( f, g) ≤ ǫ − m m=1 =

∞ [

m=1

{g ∈ Ω(D) : |g(ti ) − f (ti )| ≤ ǫ −

1 , ∀ti ∈ D ∩ [0, n]}. m

The result follows if one observes that each πti is continuous.



Remark 1. The lemma together with Exercise 4(b) signifies that the Kolmogorov probability P x on π{Rdt : t ∈ D} is defined on the topological Borel σ-field of Ω(D). 2. The proof of the lemma goes through if Ω(D) is replaced by Ω.

23 Step 3. We want to show that P x (Ω(D)) = 1. By Exercise 4(b) this is N,1/ j equivalent to showing that Lt P(∆D ( f ) ≤ 1k ) = 1 for all N and k. j→∞

The lemmas which follow will give the desired result. Lemma (L´evy). Let X1 , . . . Xn be independent random variables, ǫ > 0 and δ > 0 arbitrary. If P(|Xr + Xr+1 + · · · + Xℓ | ≥ δ) ≤ ǫ ∀ r, ℓ such that 1 ≤ r ≤ ℓ ≤ n, then P( sup |X1 + · · · + X j | ≥ 2δ) ≤ 2ǫ. 1≤ j≤n

(see Kolmogorov’s theorem) for every j = 1, 2, . . . , for every N = 22 1, 2, . . . and for every k = 1, 2, . . .. (Hint: Use the fact that the projections are continuous). ∞ T ∞ S ∞ N, 1 T (b) Show that Ω(D) = {∆D j ( f ) ≤ 1k } and hence Ω(D) is N=1 k=1 j=1

measurable in

π{Rdt

: t ∈ D}.

Let πt1 ...tk : Ω(D) → Rd × Rd × · · · Rd (k times) be the projections and let d d Et1 ...tk = π−1 t1 ...tk (B(R )× · · · ×B(R )). k times

Put E = ∪{Et1 ...tk : 0 ≦ t1 < t2 < . . . < tk < ∞; ti ∈ D}. Then, as Et1 ...tk ∪ E s1 ...s1 ⊂ Eτ1 ...τm , where {t1 . . . tk , s1 . . . s1 } ⊂ {τ1 . . . τm }, E is an algebra. Let σ(E ) be the σ-algebra generated by E . Lemma . Let B be the (topological) Borel σ-field of Ω(D). Then B is the σ-algebra generated by all the projections {πti ...tk : 0 ≤ t1 < t2 < . . . < tk , ti ∈ D}.

4. Construction of Wiener Measure

24

Remark. By subadditivity it is clear that     P  sup |X1 + · · · + X j | ≥ 2δ ≤ nǫ.

23

1≤ j≤n

Ultimately, we shall let n → ∞ and this estimate is of no utility. The importance of the lemma is that it gives an estimate independent of n. Proof. Let S j = X1 + · · · + X j , E = { sup |S j | ≥ 2δ}. Put 1≤ j≤n

E1 = {|S 1 | ≥ 2δ},

E2 = {|S 1 | < 2δ, |S 2 | ≥ 2δ}, ...

...

...

...

...

...

...

...

En = {|S j | < 2δ, 1 ≤ j ≤ n − 1, |S n | ≥ 2δ}. Then E=

n [ j=1

E j , E j ∩ Ei = φ

if

j , i;

  n  [ P{E ∩ (|S n | ≤ δ) = P  (E j ∩ (|S n | ≤ δ)) j=1 n[ o ≤P (Ei ∩ (|S n − S j | ≥ δ)) n X ≤ P(E j )P(|S n − S j | ≥ δ) (by independence) j=1

≤ ǫP(E)

(by data).

= P{E ∩ (|S n | > δ)} ≤ P(|S n | > δ) ≤ ǫ

(by data).

Combining the two estimates above, we get P(E) ≤ ǫ + ǫP(E). 24

If ǫ >

1 1 ǫ , 2ǫ > 1. If ǫ < , ≤ 2ǫ. In either case P(E) ≤ 2ǫ. 2 2 1−ǫ



25 Lemma . Let {X(t)} be a Brownian motion, I ⊂ [0, ∞) be a finite interval, F ⊂ I ∩ D be finite. Then ! |I|2 P x Sup |X(t) − X(σ)| ≥ 4δ ≤ C(d) 4 , δ t,σ∈F where |I| is the length of the interval and C(d) a constant depending only on d. Remark. Observe that the estimate is independent of the finite set F. Proof. Let F = {ti : 0 ≤ t1 < t2 < . . . < tk < ∞}. Put X1 = X(t2 ) − X(t1 ), . . . , Xk−1 = X(tk ) − X(tk−1 ). Then X1 . . . Xk−1 are independent (Cf. Exercise 2(c)). Let ǫ=

sup 1≤r≤1≤k−1

P x (|Xr + Xr+1 + · · · + X1 | ≥ δ).

Note that P x (|Xr + · · · + X1 | ≥ δ) = P(|X(t′ ) − X(t′′ )| ≥ δ) for some t′ , t′′ in F (*)

E(|X(t′ ) − X(t′′ )|4 ) (see Tchebyshey’s inequality in Appendix) δ4 C ′ (t′′ − t′ ) ≤ (C ′′ = constant) δ4 C ′ |I|2 ≤ . δ4

≤

Therefore ǫ ≤

C ′ |I|2 . Now δ4

P x ( sup |X(t) − X(σ)| ≥ 4δ) t,σ∈P

P x ( sup |X(ti ) − X(t1 )| ≥ 2δ) 1≤i≤k

= P x ( sup |X1 + · · · + X j | ≥ 2) ≤ 2ǫ

(by previous lemma)

i≤ j≤k−1

2C ′ |I|2 C|I|2 = 4 . δ4 δ  25

4. Construction of Wiener Measure

26

Exercise 5. Verify (∗). (Hint: Use the density function obtained in Exercise 2(c) to evaluate the expectation and go over to “popular” coordinates. (The value of C ′ is d(2d + 1))). Lemma .             Px  sup |X(t) − X(s)| > ρ = P x (∆T,h    D > ρ)     |t−s|≤h   t,s∈[0,t]∩D

≤ φ(T, ρ, h) = C

 h  T  + 1 . ρ4 h

Note that φ(T, ρ, h) → 0 as h → 0 for every fixed T and ρ. Proof. Define the intervals I1 , I2 , . . . by

Ik = [(k − 1)h, (k + 1)h] ∩ (0, T ], k = 1, 2, . . . . Let I1 , I2 , . . . Ir be those intervals for which I j ∩ [0, T ] , φ ( j = 1, 2, . . . , r). 26

Clearly there are [ Th ] + 1 of them. If |t − s| ≤ h then t, s ∈ I j for some ∞ S j, 1 ≤ j ≤ r. Write D = Fn where Fn ⊂ Fn+1 and Fn is finite. Then n=1

      Px     

sup

|t−s|≤h t,s∈[0,T ]∩D

          ∞     [           sup |X(t) − X(s)| > ρ |X(t) − X(s)| > ρ = Px                |t−s|≤h  n=1

t,s∈D∩Fn

        = sup P x  sup sup (|XI j (t) − X(s)| > ρ)     n j t,s∈Fn

27

≤ sup n

r X j=1

! P x sup (|XI j (t) − X(s)| > ρ) t,s∈Fn

 T 

 C(2h)2 sup +1 h (ρ/4)4 n ≤ φ(T, ρ, h).

by the last lemma

 Theorem . P x (Ω(D)) = 1. Proof. It is enough to show that N, 1 Lt P x ∆D j ( f ) j→∞

! 1 ≤ = 1 (See Exercise 4(b)). k

But this is guaranteed by the previous lemma.



Remark.

1. It can be shown that the outer measure of Ω is 1. Q 2. Ω is not measurable in Rdt . t≥0

Let P˜ x be the measure on Ω induced by P x on Ω(D). We have al- 27 ready remarked that P x is defined on the (topological Borel σ field of Ω(D). As P x is a probability measure, P˜ x is also a probability measure. It should now be verified that P˜ x is really the probability measure consistent with the given distribution. Theorem . P˜ x is a required probability measure for a continuous realization of the Brownian motion. Proof. We have to show that Ft1 ...tk = P˜ x π−1 t1 ...tk

for all

t1 , t2 . . . tk

in [0, ∞).

Step 1. Let t1 , . . . , tk ∈ D. Then

P x (π−1 t1 ...tk (A1 × · · · × Ak )) = P x (τπt1 ...tk (A1 × · · · × Ak ))

for every Ai Borel in Rd . The right side above is P x (π−1 t1 ...tk (A1 × · · · × Ak )) = F t1 ...tk (A1 × · · · × Ak ) (by definition of P x ).

4. Construction of Wiener Measure

28

Step 2. We know that T t1 ,t2 ...tk = P˜ x πt1 ,t2 ,...,tk provided that t1 , t2 , . . . , tk ∈ D. Let us pick t1(n) , . . . , tk(n) , such that each ti(n) ∈ D and tk(n) → tk as n → ∞. For each n and for each fixed f : Rd → R which is bounded and continuous, (n)

E F1 28

,...tk(n)

˜

[ f (x1 , . . . , xk )] = E Px [ f (x(t1(n) , . . . , x(tk(n) )))].

Letting n → ∞ we see that E Ft1 ,...,tk [ f (x1 , . . . , xk )] = E Px [ f (x(t1 ), . . . , x(tk ))] for all t1 , . . . , tk . This completes the proof.



The definition of the Brownian motion given earlier has a built-in constraint that all “trajectories” start from X(0) = x. This result is given by Theorem . P˜ 0 { f : f (0) = 0} = 1. ˜

Proof. Obvious; because E Px [φ(x(0))] = φ(x).



Note. In future P˜ x will be replaced by P x and P˜ 0 = P0 will be denoted by P. Let T x : Ω → Ω be the map given by (T x f )(t) = x + f (t). T x translates every ‘trajectory’ through the vector x.

Time

Let us conceive a real Brownian motion of a system of particles. The operation T x means that the system is translated in space (along with

29

29

everything else that affects it) through a vector x. The symmetry of the physicl laws governing this motion tells us that any property exhibited by the first process should be exhibited by the second process and vice versa. Mathematically this is expressed by Theorem . P x = PT x−1 . Proof. It is enough to show that −1 P x (T x π−1 t1 ...tk (A1 × · · · × Ak )) = P(πt1 ...tk (A1 × · · · × Ak ))

for every Ai Borel in Rd . Clearly, −1 T x π−1 t1 ...tk (A1 × · · · × Ak ) = πt1 ...tk (A1 − x × · · · × Ak − x).

Thus we have only to show that Z Z Z ... p(0, x, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk A1 −x Ak −x Z Z = ... p(0, 0, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk , A1

Ak

which is obvious.



Exercise. (a) If β(t, ·) is a Brownian motion s tarting at (0, 0) then 1 √ β(ǫt) is a Brownian motion starting at (0, 0) for every ǫ > 0. ǫ (b) If X is a d-dimensional Brownian motion and Y is a d′ -dimensional Brownian motion then (X, Y) is a d + d′ dimensional Brownian motion provided that X and Y are independent. j

(c) If Xt = (Xt1 , . . . , Xtd ) is a d-dimensional Brownian motion, then Xt is a one-dimensional Brownian motion. ( j = 1, 2, . . . d). τ(w) = inf{t : |Xt (w)| ≥ +1} = inf{t : |w(t)| ≥ 1}

τ(w) is the first time the particle hits either of the horizontal lines 30

4. Construction of Wiener Measure

30 +1 or −1.

6.4

2. Let {Xt } be a d-dimensional Brownian motion, G any closed set in Rd . Define τ(w) = inf{t : w(t) ∈ G}. This is a generalization of Example 1. To see that τ is a stopping time use ∞ \ {τ ≤ s} = lim {w : w(θ) ∈ Gn }, θ∈[0,s] n=1 θ rational

where

( ) 1 Gn = x ∈ Rd : d(x, G) ≤ . n

3. Let (Xt ) be a d-dimensional Brownian motion, C and D disjoint closed sets in Rd . Define τ(w) = inf{t; w(t) ∈ C and for some s ≤ t, w(s) ∈ D}. τ(w) is the first time that w hits C after visiting D.

5. Generalised Brownian Motion LET Ω BE ANY space, F a σ-field and (Ft ) an increasing family of 31 sub σ-fields such that σ(∪Ft ) = F . Let P be a measure on (Ω, F ). X(t, w) : [0, w) × Ω → Rd is called a Brownian motion relative to (Ω, Ft , P) if (i) X(t, w) is progressively measurable with respect to Ft ; (ii) X(t, w) is a.e. continuous in t; (iii) X(t, w) − X(s, w) for t > s is independent of Fs and is distributed normally with mean 0 and variance t − s, i.e. P(X(t, ·) − X(s, ·) ∈ A|Fs ) = Note.

Z

A

|y|2 1 exp − dy. 2(t − s) [2π(t − s)]d/s

1. The Brownian motion constructed previously was concentrated on Ω = C([0, ∞); Rd ), F was the Borel field of Ω, X(t, w) = w(t) and Ft = σ{X(s) : 0 ≤ s ≤ t}. The measure P so obtained is often called the Wiener measure.

2. The above definition is more general because σ{X(s) : 0 ≤ s ≤ t} ⊂ Ft . 31

5. Generalised Brownian Motion

32

Exercise. (Brownian motion starting at time s). Let Ω = C([s, ∞); Rd ), B = Borel field of Ω. Show that for each x ∈ Rd ∃ a probability measure P sx on Ω such that (i) P sx {w : w(s) = x} = 1;

32

(ii) P sx (Xt1 ∈ A1 , . . . , Xtk ∈ Ak ) Z Z Z = ... p(s, x, t1 , x1 )p(t1 , x1 , t2 , x2 ) . . . A1

A2

Ak

. . . p(tk−1 xk−1 , tk , xk )dx1 . . . dxk , ∀ s < t1 < . . . < tk . For reasons which will become clear later, we would like to shift the measure P sx to a measure on C([0, ∞); Rd ). To do this we define T : C([s, ∞); Rd ) → C([0, ∞); Rd ) by

   w(t), (T w)(t) =   w(s),

if t ≥ s, if t ≤ s.

Clearly, T is continuous. Put

P s,x = P sx T −1 . Then (i) P s,x is a probability measure on the Borel field of C([0, ∞); Rd ); (ii) P s,x {w : w(s) = x} = 1.

6. Markov Properties of Brownian Motion Notation. 1. A random variable of a stochastic process {X(t)}t∈I shall 33 be denoted by Xt or X(t). 0 ≤ t < ∞. 2. Fs will denote the σ-algebra generated by {Xt : 0 ≤ t ≤ s}; Fs+ = {Fa : a > s}; Fs− will be the σ-algebra generated by ∪{Fa : a < s}s > 0. It is clear that {Ft } is an increasing family. 3. For the Brownian motion, B = the σ-algebra generated by ∪{Ft : t < ∞} will be denoted by F . Theorem . Let {Xt : 0 ≤ t < ∞} be a Brownian motion. Then Xt − X s is independent of Fs . Proof. Let 0 ≤ t1 < t2 < t3 < . . . < tk ≤ s. Then the σ-algebra generated by Xt1 , . . . , Xtk is the same as the σalgebra generated by Xt1 , Xt2 − Xt1 , . . . , Xtk − Xtk−1 . Since Xt − X s is independent of these increments, it is independent of σ{Xt1 , . . . , Xtk }. This is true for every finite set t1 , . . . , tk and therefore Xt − X s is independent of Fs . Let us carry out the following calculation very formally. P[Xt ∈ A | Fs ](w) = P[Xt − X s ∈ B | Fs ](w), B = A − X s (w), 33

6. Markov Properties of Brownian Motion

34

= P[Xt − X s ∈ B], 34

i.e. P[Xt ∈ A | Fs ](w) =

Z A

by independence,

1 |y − X s (w)|2 dy. exp − 2(t − s) (2πt)d/2 

This formal calculation leads us to Theorem . P[Xt ∈ A | Fs ](w) =

Z A

1 |y − X s (w)|2 dy. exp − 2(t − s) (2πt)d/2

where A is Borel in Rd , t > s. Remark. It may be useful to note that p(s, X s (w), t, y) can be thought of as a conditional probability density. Proof.

(i) We show that Z fA (w) = A

1 |y − X s (w)|2 dy exp − 2(t − s) (2πt)d/2

is FS -measurable. Assume first that A is bounded and Borel in Rd . If ωn → ω, then fA (ωn ) → fA (ω), i.e. fA is continuous and hence Fs -measurable. The general case follows if we note that any Borel set can be got as an increasing union of a countable number of bounded Borel sets. (ii) For any C ∈ Fs we show that Z X Z Z exp −|y − X s (ω)|2 −1 (*) Xt (A)dP(ω) = dy dP(ω). (2π(t − s))d/2 C C

A

It is enough to verify (∗) for C of the form  C = ω : (Xt1 (ω), . . . , Xtk (ω)) ∈ A1 × · · · × Ak ; 0 ≤ t1 < . . . < tk ≤ s ,

35 where Ai is Borel in Rd for i = 1, 2 . . . k. The left side of (∗) is then Z

p(0, 0, t1, xt1 )p(t1 , xt1 , t2 , xt2 ) . . . p(tk , xtk , t, xt )dxt1 . . . dxt .

Ai ×···×Ak ×A

35

To compute the right side define f : R(k+1)d → B by f (u1 , . . . , uk , u) = XA1 (u1 ) . . . XAk (uk )p(s, u, t, y). Clearly f is Borel measurable. An application of Fubini’s theorem to the right side of (∗) yields Z Z dy XA1 (Xt1 (ω)) . . . XAk (Xtk (ω))p(s, X s (ω), t, y)dP(ω) A

=

ZΩ

dy

A

=

Z

f (x1 . . . xk , xs )dFt1 ...tk , s

Rd ×···×Rd (k+1) times

dy

A

=

Z Z

p(0, 0, t1 , x1 ) . . . p(tk−1 , tk , xk )

A1 ×···×Ak ×Rd

Z

p(tk , xk , s, xs )p(s, xs , t, y)dx1 . . . dxk dxs p(0, 0, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )p(tk , xk , t, y)

A1 ×···×Ak ×A

dx1 . . . , dxk dy (by the convolution rule) = left side. 

Examples of Stopping Times.

6. Markov Properties of Brownian Motion

36

1. Let (Xt ) be a one-dimensional Brownian motion. Define τ by {τ ≤ s} =

∞ \

{w : w(θ1 ) ∈ Dn , w(θ2 ) ∈ Cn },

lim

θ1 ,θ2 n=1 θ1 ,θ2 rational in [0,s]

where

36

(

) ( ) 1 1 d Dn = x ∈ R : d(x, D) ≤ , Cn = x ∈ R : d(x, C) ≤ n n d

Exercise 1. Let τ be as in Example 1. (a) If A = {w : X1 (w) ≤ τ} show that A < Fτ .

(Hint: A ∩ {τ ≤ 0} < F0 ). This shows that Fτ $ F0 .

(b) P0 {w : τ(w) = ∞} = 0. (Hint: P0 {w : |w(t)| < 1} ≤

R

|y|≤t−1/2

2

e−1/2|y| dy ∀t).

Theorem . (Strong Markov Property of Brownian Motion). Let τ be any finite stopping time, i.e. τ < ∞ a.e. Let Yt = Xτ+t − Xτ . Then 1. P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A] = P(Xt1 ∈ A1 , . . . Xtk ∈ Ak ) · P(A), ∀ A ∈ Fτ and for every Ai Borel in Rd . Consequently, 2. (Yt ) is a Brownian motion. 3. (Yt ) is independent of Fτ . The assertion is that a Brownian motion starts afresh at every stopping time. Proof. Step 1. Let τ take only countably many values, say s1 , s2 , s3 . . .. Put E j = τ−1 {s j }. Then each E j is Fτ -measure and Ω=

∞ [ j=1

E j , E j ∩ Ei = ∅ j , i.

37 37

Fix A ∈ Fτ . P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A] ∞ X = P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A ∩ E j ] j=1

= =

∞ X

j=1 ∞ X j=1

P[(Xt1 +s j − X s j ) ∈ A1 , . . . , Xtk +s j − X s j ∈ Ak ) ∩ A ∩ E j ] P[(Xt1 ∈ A1 ), . . . , (Xtk ∈ Ak )]P(A ∩ E j ) (by the Markov property)

= P(Xt1 ∈ A1 , . . . , Xtk ∈ Ak ) · P(A) [nτ] + 1 . A simple calStep 2. Let τ be any stopping time; put τn = n culation shows that τn is a stopping time taking only countably many values. As τn  τ, Fτ ⊂ Fτn ∀n . Let Yt(n) = Xτn +t − Xτn . By Step 1, P[(Yt(n) < x1 , . . . , Yt(n) < xk ) ∩ A] 1 k

= P(Xt1 < x1 , . . . , Xtk < xk ) · P(A)

(where x < y means xi < yi i = 1, 2, . . . , d) for every A ∈ Fτ . As all the Brownian paths are continuous, Yt(n) → Yt a.e. Thus, if x1 , . . . , xk is a point of continuity of the joint distribution of Xt1 , . . . , Xtk , we have P[(Yt1 < x1 , . . . , Ytk < xk ) ∩ A] = P(Xt1 < x1 , . . . , Xtk < xk )P(A) ∀ A ∈ Fτ . Now assertion (1) follows easily. For (2), put A = Ω in (1), and (3) is a consequence of (1) and (2).

38



7. Reflection Principle LET (Xt ) BE A one-dimensional Brownian motion. Then P( sup X s ≥ 39 0≤s≤t

a) = 2P(Xt ≥ a) with a > 0. This gives us the probability of a Brownian particle hitting the line x = a some time less than or equal to t. The intuitive idea of the proof is as follows.

Time,

Among all the paths that hit a before time t exactly half of them end up below a at time t. This is due to the reflection symmetry. If X s = a for some s < t, reflection about the horizontal line at a gives a one one correspondence between paths with Xt > a and paths with Xt < a. Therefore     P max X s ≥ a, Xt > a = max X s ≥ a, Xt < a 0≤s≤t

0≤s≤t

39

7. Reflection Principle

40

Since P{Xt = a} = 0, we obtain ( ) ( ) ( ) P sup X s ≥ a = P sup X s ≥ a, Xt > a + P sup X s ≥ a, Xt > a 0≤s≤t

0≤s≤t

0≤s≤t

= 2P{Xt ≥ a} 40

We shall now give a precise argument. We need a few elementary results. n P Lemma 1. Let Xn = Yk where the Yk are independent random varik=1

ables such that P{Yk ∈ B} = P{−Yk ∈ B}∀ Borel set B ⊂ R (i.e. Yk are symmetric). Then for any real number a,   P max Xi > a ≤ 2P{Xn > a} 1≤i≤n

Proof. It is easy to verify that a random variable is symmetric if and only if its characteristic function is real. Define Ai = {X1 ≤ a, . . . Xi−1 ≤ a, Xi > a}, i = 1, 2, . . . , n; B = {Xn > a}

Then Ai ∩ A j = ∅ if i , j. Now, P(Ai ∩ B) ≥ P(Ai ∩ {Xn ≥ Xi }) = P(Ai )P(Xn ≥ Xi ),

by independence.

= P(Ai )P(Yi+1 + · · · + Yn ≥ 0). As Yi+1 , . . . , Yn are independent, the characteristic function of Yi+1 + · · · + Yn is the product of the characteristic functions of Yi+1 + · · · + Yn , so that Yi+1 + · · · + Yn is symmetric. Therefore P(Yi+1 + · · · + Yn ≥ 0) ≥ Thus P(Ai ∩ B) ≥ P(B) ≥

n X i=1

1 P(Ai ) and 2

1 . 2

 n  1X 1 [  P(Ai ∩ B) ≥ P(Ai ) ≥ P  Ai  , 2 2 i=1

41 41

i.e.

 n  [  2P(B) ≥ P  Ai  , i=1

or



 P max Xi > a ≤ 2P{Xn > a} 1≤i≤n

 Lemma 2. Let Yi , . . . , Yn be independent random variables. Put Xn = n P Yk and let τ = min{i : Xi > a}, a > 0 and τ = ∞ if there is no such i. k=1

Then for each ǫ > 0,

(a) P{τ ≤ n − 1, Xn − Xτ ≤ −ǫ} ≤ P{τ ≤ n − 1, Xn ≤ a} +

n−1 P

P(Y j > ǫ).

j=1

(b) P{τ ≤ n−1, Xn > a+2ǫ} ≤ P{τ ≤ n−1, Xn −Xτ > ǫ}+

n−1 P

P{Y j > ǫ}

j=1

(c) P{Xn > a + 2ǫ} ≤ P{τ ≤ n − 1, Xn > a + 2ǫ} + P{Yn > 2ǫ}. If, further, Y1 , . . . , Yn are symmetric, then

(d) P{max Xi > a, Xn ≤ a} ≥ P{Xn > a + 2ǫ} − P{Yn ≥ 2ǫ} − 2

1≤i≤n n−1 P

P{Y j > ǫ}

j=1

(e) P{max Xi > a} ≥ 2P{Xn > a + 2ǫ} − 2 1≤i≤n

Proof.

n P

P{Y j > ǫ}

j=1

(a) Suppose w ∈ {τ ≤ n − 1, Xn − Xτ ≤ −ǫ} and w ∈ {τ ≤ n − 1, Xn ≤ a}. Then Xn (w) > a and Xn (w) + ǫ ≤ Xτ(w) (w) or, Xτ(w) (w) > a + ǫ. By definition of τ(w), Xτ(w)−1 (w) ≤ a and therefore, Yτ(w) (w) = Xτ(w) (w) − Xτ(w)−1 (w) > a + ǫ − a = ǫ if τ(w) > 1; if τ(w) = 1, Yτ(w) (w) = Xτ(w) (w) > a + ǫ > ǫ. Thus Y j (w) > ǫ for some j ≤ n − 1, i.e.

42

7. Reflection Principle

42

w∈

n−1 [ {Y j > ǫ}. j=1

Therefore {τ ≤ n − 1, Xn − Xτ ≤ −ǫ} ⊂ {τ ≤ n − 1, Xn ≤ a}

n−1 [ {Y j > ǫ} j=1

and (q) follows. (b) Suppose w ∈ {τ ≤ n − 1, Xn > a + 2ǫ} but w ∈ {τ ≤ n − 1, Xn − Xτ > ǫ}. Then Xn (w) − Xτ(w) (w) ≤ ǫ, or, Xτ(w) (w) > a + ǫ so that Yτ(w) (w) > ǫ as in (a); hence Y j (w) > ǫ for some j ≤ n − 1. This proves (b). (c) If w ∈ {Xn > a + 2ǫ}, then τ(w) ≤ n; if w < {τ ≤ n − 1, Xn > a + 2ǫ}, then τ(w) = n so that Xn−1 (w) ≤ a; therefore Yn (w) = Xn (w) − Xn−1 (w) > 2ǫ. i.e. w ∈ {Yn > 2ǫ}. Therefore {Xn > a + 2ǫ} ⊂ {τ ≤ n − 1, Xn > a + 2ǫ} ∪ {Yn > 2ǫ}. This establishes (c). (d) P{max Xi > a, Xn ≤ a} = P{τ ≤ n − 1, Xn ≤ a} 1≤i≤n

≥ P{τ ≤ n − 1, Xn − Xτ ≤ −ǫ} − P = =

"

n−1 S k=1

n−1 P k=1

n−1 P k=1

#

{τ = k, Xn − Xk ≤ −ǫ} −

P{τ = k, Xn − Xk ≤ −ǫ} −

n−1 P

P(Y j > ǫ),

j=1

n−1 P

P(Y j > ǫ)

j=1

n−1 P

P(Y j > ǫ)

j=1

P{τ = k}P{Xn − Xk ≤ −ǫ} −

n−1 P

P(Y j > ǫ)

j=1

(by independence)

by (a),

43 =

n P

k=1

P{τ = k}P{Xn − Xk > ǫ} −

= P{τ ≤ n − 1, Xn − Xτ ≥ ǫ} − ≥ P{τ ≤ n − 1, Xn − Xτ > ǫ} −

n−1 P

n−1 P

(by symmetry)

P(Y j > ǫ)

j=1

n−1 P

P(Y j > ǫ)

j=1

≥ P{τ ≤ n − 1, Xn > a + 2ǫ} − 2

n−1 P

P{Y j > ǫ}

(by (b))

j=1

≥ P{Xn > a + 2ǫ} − P{Yn > 2ǫ} − 2 This proves (d).

P(Y j > ǫ)

j=1

n−1 P

P{Y j > ǫ}

(by (c))

43

j=1

(e) P{max Xi > a} = P{max Xi > a, Xn ≤ a} + P{max Xi > a, Xn > a} 1≤i≤n

1≤i≤n

1≤i≤n

= P{max Xi > a, Xn ≤ a} + P{Xn > a} 1≤i≤n

= P{Xn > a + 2ǫ} − P{Yn > 2ǫ} + P{Xn > a} −2

n−1 P

P{Y j > ǫ}

(by (d))

j=1

Since P{Xn > a + 2ǫ} ≤ P{Xn > a} and

P{Yn > 2ǫ} ≤ P{Yn > ǫ} ≤ 2P{Yn > ǫ}, we get P{max Xi > a} ≥ 2P{Xn > a + 2ǫ} − 2 1≤i≤n

n X

P(Y j > ǫ)

j=1

This completes the proof.

44

 Proof of the reflection principle. By Lemma 1 (  jt  ) p = P max X > ≤ 2P(X(t) > a). 1≤ j≤n n

7. Reflection Principle

44 By Lemma 2(e),

(   !! ) n X jt ( j − 1)t p ≥ 2P(X(t) > a + 2ǫ) − 2 P X >ǫ . −X n n j=1 ! ( j − 1)t are independent and identically distribun n t ted normal random variables with mean zero and variance (in particn ular they are symmetric), !! !  jt    t    ( j − 1)t >ǫ =P X P X −X − X(0) > ǫ n n n  t  =P X >ǫ . n Since X

 jt 

Therefore

−X

  t >ǫ . p ≥ 2P(X(t) > a + 2ǫ) − 2n P X n

P(X(t/n) > ǫ) =

Z∞ ǫ

=

√ Z∞

1 2 2t e−x / n dx (2t/n)

√ √ ǫ n/ (2t)

or

√ √ 2 e−x ǫ n/ (2t) √ dx ≤ √ π x

−1∞ Z

2

xe−x dx

√ √ ǫ n/ (2t)

√ 1 −ǫ 2 n/2t (2t) P(X(t/n) > ǫ) ≤ √ e · √ . 2 π ǫ n Therefore nP(X(t/n) > ǫ) ≤

√ n√ 2 (2t)/ (πn)e−ǫ n/2t → 0 as 2ǫ

n → +∞.

45

By continuity,     P max X( jt/n) > a → P max X(s) > a . 1≤≤n

0≤s≤t

45 We let n tend to ∞ through values 2, 22 , 23 , . . . so that we get 2P{X(t) > a + 2ǫ} − 2n P{X(t/n) > ǫ} ( ) ≤P max X(t/n) > a ≤ 2P{X(t) > a}, 1≤ j≤n

or



 2P{X(t) > a} ≤ 2P{X(t) ≥ a} ≤P max X(t) > a 0≤s≤t

2P{X(t) > a}, on letting n → +∞ first and then letting ǫ → 0. Therefore,   P max X(s) > a = 2P{X(t) > a} 0≤s≤t

=2

Z∞

√ 2 1/ (2πt)e−x /2t dx.

a

AN APPLICATION. Consider a one-dimensional Brownian motion. A particle starts at 0. What can we say about the behaviour of the particle in a small interval of time [0, ǫ)? The answer is given by the following result. P(A) ≡ P{w : ∀ǫ > 0, ∃ t, s in [0, ǫ) such that Xt (w) > 0 and X s (w) < 0} = 1.

INTERPRETATION. Near zero all the particles oscillate about their 46 starting point. Let A+ = {w : ∀ ∈> 0∃ t ∈ [0, ǫ) such that Xt (w) > 0},

A− = {w : ∀ǫ > 0∃ s ∈ [0, ǫ) such that X s (w) < 0}. We show that P(A+ ) = P(A− ) = 1 and therefore P(A) = P(A+ ∩ A− ) = 1.   \ ∞  ∞ [ ∞  \       +  sup w(t) ≥ 1/m A ⊃  =  sup w > 0 n=1

0≤t≤1/n

n=1 m=1 0≤t≤1/n

7. Reflection Principle

46 Therefore

    P(A ) ≥ Lt sup P  sup w(t) ≥ 1/m +

n→∞ m→∞

≥2

Lt

n→∞m→∞

0≤t≤1/n

sup P(w(1/n) ≥ 1/m)

(by the reflection principle)

≥ 1. Similarly P(A− ) = 1. Theorem . Let {Xt } be a one-dimensional Brownian motion, A ⊂ (−∞, a) (a > 0) and Borel subset of R. Then P0 {Xt ∈ A, X s < a ∀s such that 0 ≤ s ≤ t} Z Z √ √ 2 2 = 1/ (2πt)e−y /2t dy − 1/ (2πt)e−(2a−y) /2t dy A

A

Proof. Let τ(w) = inf{t : w(t) ≥ a}. By the strong Markov property of Brownian motion, P0 {B(X(τ + s) − X(τ) ∈ A)} = P0 (B)P0 (X(s) ∈ A) for every set B in Ft . This can be written as E(X(X(τ+s)−X(τ)∈A) |Fτ ) = P0 (X(s) ∈ A) 47

Therefore E(X(X(τ+ℓ(w))−X(τ)∈A) |Fτ ) = P0 (X(ℓ(w)) ∈ A) for every function ℓ(w) which is Fτ -measurable. Therefore, Z P0 ((τ ≤ t) ∩ ((X(τ + ℓ(w)) − X(τ)) ∈ A) = P0 (X(ℓ(w)) ∈ A)dP(w) {τ≤t}

In particular, take ℓ(w) = t − τ(w), clearly ℓ(w) is Fτ -measurable. Therefore, Z P0 ((τ ≤ t)((X(t) − X(τ)) ∈ A)) = P0 (X(ℓ(w) ∈ A)dP(w)). {τ≤t}

47 Now X(τ(w)) = a. Replace A by A − a to get (*)

P0 ((τ ≤ t) ∩ (X(t) ∈ A)) =

Z

P0 (X(ℓ(w) ∈ A − a)dP(w))

{τ≤t}

Consider now P2a (X(t) ∈ A) = P0 (X(t) ∈ A − 2a)

= P0 (X(t) ∈ 2a − A) (by symmetry of x)

= P0 ((τ ≤ t) ∩ (X(t) ∈ 2a − A)).

The last step follows from the face that A ⊂ (−∞, a) and the continuity of the Brownina paths. Therefore P2a (X(t) ∈ A) =

Z

P0 (X(ℓ(w)) ∈ a − A)dP(w),

(using ∗)

{τ≤t}

= P0 ((τ ≤ t) ∩ (X(t) ∈ A)). Now the required probability P0 {Xt ∈ A, X s < a ∀s ∈ 0 ≤ s ≤ t} = P0 {Xt ∈ A} − P0 {(τ ≤ t) ∩ (Xt ∈ A)} Z Z √ √ 2 −y2 /2t = 1/ (2πt)e dy − 1/ (2πt)e−(2a−y) /2t dy. A

A

The intuitive idea of the previous theorem is quite clear. To obtain 48 the paths that reach A at time t without hitting the horizontal line x = a, we consider all paths that reach A at time t and subtract those paths that hit the horizontal line x = a before time t and then reach A at time t. To see exactly which paths reach A at time t after hitting x = a we consider a typical path X(w).

7. Reflection Principle

48



49

The reflection principle (or the strong Markov property) allows us to replace this path by the dotted path (see Fig.). The symmetry of the Brownian motion can then be used to reflect this path about the line x = a and obtain the path shown in dark. Thus we have the following result: the probability that a Brownian particle starts from x = 0 at t = 0 and reaches A at time t after it has hit x = a at some time τ ≤ t is the same as if the particle started at time t = 0 at x = 2a and reached A at time t. (The continuity of the path ensures that at some time τ ≤ t, this particle has to hit x = a). We shall use the intuitive approach in what follows, the mathematical analysis being clear, thorugh lengthy. Theorem . Let X(t) be a one-dimensional Brownian motion, A ⊂ (−1, 1) any Borel subset of R. Then # Z " P0 sup |X(s)| < 1, X(t) ∈ A = φ(t, y)dy, 0≤s≤t

A

where φ(t, y) =

∞ X

n=−∞

√ 2 (−1)n / (2πt)e−(y−2n) /2t .

49 Proof.

Let En be the set of those trajections which (i) start at x = 0 at time t = 0 (ii) hit x = +1 at some time τ1 < t (iii) hit x = −1 at some later time τ2 < t (iv) hit x = 1 again at a later time τ3 < t . . . and finally reach A at time t. The number of τ’s should be equal to n at least, i.e. En = {w : there exists a sequence τ1 , . . . τn of stopping times such that 0 < τ1 < τ2 < . . . < tτn < t, X(τ j ) = (−1) j−1 , X(t) ∈ A}. Similarly, let Fn = {w : there exists a sequence τ1 , . . . , τn of stopping times 0 < τ1 < τ2 < . . . < τn < t, X(τ j ) = (−1) j , X(t) ∈ A}

Note that

50

E1 ⊃ E2 ⊃ E3 ⊃ . . . , F1 ⊃ F2 ⊃ F3 ⊃ . . . ,

Fn ⊃ En+1 ; En ⊃ Fn+1 ,

En ∩ Fn = En+1 ∪ Fn+1 .

Let

"

#

φ(t, A) = P sup |X(s)| < 1, X(t) ∈ A . 0≤s≤t

Therefore "

φ(t, A) = P[X(t) ∈ A] − P sup |X(s)| ≥ 1, X(t) ∈ A 0≤s≤t

#

7. Reflection Principle

50 =

Z

√ 2 1/ (2πt)e−y /2t dy − P[(E1 ∪ F1 ) ∩ A0 ],

A

where A0 = {X(t) ∈ A} =

Z

√ 2 1/ (2πt)e−y /2t dy − P[(E1 ∩ A0 ) ∪ (F1 ∩ A0 )].

A

Use the fact that P[A ∪ B] = P(A) + P(B) − P(A ∩ B) to get φ(t, A) =

Z

√ 2 1/ (2πt)e−y /2t dy−P[E1 ∩A0 ]−P[F1 ∩A0 ]+P[E1 ∩F1 ∩A0 ],

A

as E1 ∩ F1 = E2 ∪ F2 . Proceeding successively we finally get φ(t, A) =

Z A

X X √ 2 1/ (2πt)e−y /2t dy+ (−1)n P[En ∩A0 ]+ (−1)n P[Fn ∩A0 ] ∞

∞

n=1

n=1

We shall obtain the expression for P(E1 ∩ A0 ) and P[E2 ∩ A0 ], the other terms can be obtained similarly. E1 ∩ A0 consists of those trajectries that hit x = ±1 at some time τ ≤ t and then reach A at time t. Thus P[E1 ∩ A0 ] is given by the previous theorem by Z

√ 2 1/ (2πt)e−(y−2) /2t dy.

A

51

E2 ∩ A0 consists of those trajectories that hit x = ±1 at time τ1 , hit x = −1 at time τ2 and reach A at time t(τ1 < τ2 < t). According to the previous theorem we can reflect the trajectory upto τ2 about x = −1 so that P(E2 ∩ A0 ) is the same as if the particle starts at x = −2 at time t = 0, hits x = −3 at time τ1 and ends up in A at time t. We can now reflect the trajectory

51

upto time τ1 (the dotted curve should be reflected) about x = −3 to obtain the required probability as if the trajectory started at x = −4. Thus, Z 2 /2t/√(2πt)

e−(y+4)

P(E2 ∩ A0 ) =

dy.

A

Thus φ(t, A) =

∞ X

n=−∞

=

Z

n

(−1)

Z

√ 2 1/ 2πte−(y−2n) /2t dy

A

φ(t, y)dy.

A

The previous theorem leads to an interesting result: "

# Z1 P sup |X(s)| < 1 = φ(t, y)dy Therefore

"

0≤s≤t

−1

# " # P sup |X(s)| ≥ 1 = 1 − P sup |X(s)| < 1 0≤s≤t

0≤s≤t

52

7. Reflection Principle

52

= −1 − φ(t, y) =

∞ X

Z1

φ(t, y)dy,

−1

√ 2 (−1)n / (2πt)e−(y−2n) /2t

n=−∞

Case (i). t is very small. In this case it is enough to consider the terms corresponding to n = 0, ±1 (the higher order terms are very small). As y varies from −1 to 1, h 2 i √ 2 2 φ(t, y) ≃ 1/ (2πt) e−y /2t − e−(y−2) /2t − e−(y+2) /2t . Therefore

Z1

√ φ(t, y)dy ≃ 4/ (2πt)e−1/2t .

−1

Case (ii). t is large. In this case we use Poisson’s summation formula for φ(t, y): φ(t, y) =

∞ X

2 2

e−(2k+1)

π t/8

Cos{(k + 1)/2πy},

k=0

to get Z1

φ(t, y)dy ≃ 4/πe−π

2

t/8

−1

53

2

for large t. Thus, P(τ > t) = 4/πe−π t/8 . This result says that for large values of t the probability of paths which stay between −1 and +1 is very very small and the decay rate is 2 governed by the factor e−π t/8 . This is connected with the solution of a certain differential equation as shall be seen later on. 

8. Blumenthal’s Zero-One Law LET Xt BE A d-dimensional Brownian motion. If A ∈ F0+ = then P(A) = 0 or P(A) = 1.

T

Ft , 54

t>0

Interpretation. If an event is observable in every interval [0, t] of time then either it always happens or it never happens. We shall need the following two lemmas. Lemma 1. Let (Ω, B, P) be any probability space, Ca sub-algebra of B. Then (a) L2 (Ω, C , P) is a closed subspace of L2 (Ω, B, P). (b) If π : L2 (Ω, B, P) → L2 (Ω, C , P) is the projection map then π f = E( f |C ). Proof. Refer appendix.



Lemma 2. Let Ω = C([0, ∞); Rd ), P0 the probability corresponding to the Brownian motion. Then the set {φ(πt1 , . . . , tk ) ∈ φ is continuous, bounded on Rd × · · · × Rd (k times), πt1 , . . . , tk the canonical projection) is dense in L2 (Ω, B, P). Proof. Functions of the form φ(x(t1 ), . . . , x(tk ) where φ runs over continuous functions is clearly dense in L2 (Ω, Ft1 ,t2 ,...,tk , P) and [ [ L2 (Ω, Ft1 ,...,tk , P) k t1 ,...,tk

53

8. Blumenthal’s Zero-One Law

54 is clearly dense in L2 (Ω, B, P). 55



Proof of zero-one law. Let Ht = L2 (Ω, Ft , P), H = L2 (Ω, B, P), H0+ =

\

Ht .

t>0

Clearly H0+ = L2 (Ω, F0+ , P). Let πt : H → Ht be the projection. Then πt f → π0+ f ∀ f in H. To prove the law it is enough to show that H0+ contains only constants, which is equivalent to π0+ f = constant ∀ f in H. As π0+ is continuous and linear it is enough to show that π0+ φ = const ∀φ of the Lemma 2: π0+ φ = Lt πt φ = Lt E(φ|t ) by Lemma 1, t→0

t→0

= Lt E(φ(t1 , . . . , tk )|Ft ). t→0

We can assume without loss of generality that t < t1 < t2 < . . . < tk . E(φ(t1 , . . . , tk )|Ft ) =

Z

√ 2 φ(y1 , . . . , yk )1/ (2π(t1 − t))e−|y1 −Xt (w)| /2(t1 −t) . . .

−|yk −yk−1 |2 p . . . 1/ (2π(tk − tk−1 ))e 2(tk −tk−1 ) dy1 . . . dyk .

Since X0 (w) = 0 we get, as t → 0, π0+ φ = constant. This completes the proof. APPLICATION. Let α ≥ 1 A = {w : For, if 0 < s < 1, then 56

as

Rs 0

R1 s

R1 0

|w(t)|/tα < ∞}. Then A ∈ F0+ .

|w(t)|/tα < ∞. Therefore w ∈ A or not according

|w(t)|/tα dt converges or not. But this convergence can be asserted

55 by knowing the history of w upto time s. Hence A ∈ Fs . Blumenthal’s law implies that Z1

α

|w(t)|/t dt < ∞ a.e.w., or,

0

Z1

|w(t)|/tα dt = ∞ a.e.w.

0

A precise argument can be given along the following lines. If 0 < s < 1, A = {w :

Zs

|w(t)|/tα < ∞}

0

= {w : sup In,s (w) < ∞} where In,s (w) is the lower Riemannian sum of |w(t)n |/tα corresponding to the partition {0, s/n, . . . , s} and each In,s ∈ Fs .

9. Properties of Brownian Motion in One Dimension WE NOW PROVE the following.

57

Lemma . Let (Xt ) be a one-dimensional Brownian motion. Then (a) P(lim Xt = ∞) = 1; consequently P(lim Xt < ∞) = 0. (b) P(lim Xt = −∞) = 1; consequently P(lim Xt > −∞) = 0. (c) P(lim Xt = −∞); lim Xt = ∞) = 1. SIGNIFICANCE. By (c) almost every Brownian path assumes each value infinitely often. Proof. ∞ \ (lim Xt > n) {lim Xt = ∞} = n=1

∞ \ = ( lim Xθ > n) (by continuity of Brownian paths) n=1

θ rational

First, note that " # " # P0 sup X(s) ≤ n = 1 − P0 sup X(s) > n 0≤s≤t

0≤s≤t

57

9. Properties of Brownian Motion in One Dimension

58

Z∞

√ = 1 − 21/ (2πt) √ = (2/πt)

Zn

2 /2t

e−y

dy

n

2 /2t

e−y

dy.

0

Therefore, for any x0 and t,   " #   P sup X(s) ≥ n|X(t0 ) = x0 = P0  sup X(s) ≥ n − x0  t0 ≤s≤t

0≤s≤t−t0

(independent increments) which tends to 1 as t → ∞. Consequently, " # " # P0 sup X(t) ≥ n = EP sup X(t) ≥ n|X(t0 ) t≥t0

t≥t0

= E1 = 1. 58

In other words, "

# P0 lim supX(t) ≥ n = 1 t→∞

for every n. Thus P(lim Xt = ∞) = 1. (b) is clear if one notes that w → −w leaves the probability invariant. (c) P(lim Xt = ∞, lim Xt = −∞) = P(lim Xt = ∞) − P(lim Xt > −∞, lim Xt = ∞).

≥ 1 − P(lim Xt > −∞) = 1.

 Corollary . Let (Xt ) be a d-dimensional Brownian motion. Then P(lim |Xt | = ∞) = 1.

59 Remark . If d ≥ 3 we shall see later that P( Lt |Xt | = ∞) = 1. i.e. t→∞ almost every Brownian path “wanders” off to ∞. Theorem . Almost all Brownian paths are of unbounded variation in any interval. Proof. Let I be any interval [a, b] with a < b. For n = 1, 2, . . . define Vn (wQn ) =

n X i=1

|w(ti ) − w(ti−1 )| (ti = a + (b − a)i/n, i = 0, 1, 2, . . . n),

The variation corresponding to the partioin Qn dividing [a, b] into n 59 equal parts. Let Un (w, Qn ) =

n X i=1

|(w(ti ) − w(ti−1 )|2 .

If An (w, Qn ) sup |w(ti ) − w(ti−1 )|, 1≤i≤n

then An (w, Qn )Vn (w, Qn ) ≥ Un (w, Qn ). By continuity Lt An (w, Qn ) = 0. n→∞

Claim. Lt E[(Un (w, Qn ) − (b − a))2 ] = 0. n→∞

Proof. E[(Un − (b − a))2 ] 2    n      X 2 [(X − X ) − (b − a/n)] = E  t t j j−1      j=1  X E[( (Z 2j − b − a/n))2 ], Z j = Xt j − Xt j−1 , = nE[(Z12 − b − a/n)2 ]



9. Properties of Brownian Motion in One Dimension

60

(because Z j are independent and identically distributed). = n[E(Z14 ) − (b − a/n)2 ] = 2(b − a/n)2 → 0. Thus a subsequence Uni → b − a almost everywhere. Since Ani → 0 it follows that Vni (w, Qn ) → ∞ almost everywhere. This completes the proof.  60

Note . {w : w is of bounded variation on [a, b]} can be shown to be measurable if one proves Exercise . Let f be continuous on [a, b] and define Vn ( f, Qn ) as above. Show that f is of bounded variation on [a, b] iff sup Vn ( f, Qn ) < ∞. n=1,2,...

Theorem . Let t be any fixed real number in [0, ∞), Dt = {w : w is differentiable at t}. Then P(Dt ) = 0. Proof. The measurability of Dt follows from the following observation: if f is continuous then f is differentiable at t if and only if Lt

r→0 r rational

f (t + r) − f (t) . r

exists. Now Dt =

∞ [

m=1

w:|

w(t + h) − w(t) | ≤ M, for all h , 0, rational} h

and √

M h Z  Xt+h − Xt 1 2/2 P w:| | ≤ M ∀h ∈ Q, h , 0 ≤ 2 inf √ e−|y| dy = 0 h h (2π)



0

 Remark. A stronger result holds:   [   P  Dt  = 0. t≥0

61

Hint:

[

0≤t≤1

Dt

( ! ! ) k k−1 71 w:w −w |≤ n n n m=1 n−m i=1 k=i+1,i+2,i+3

[[ =1

n+2 [

and   P 

! ! ) ( k−1 71 k w:w −w |≤ n n n i=1 k=i+1,...,i+3 n+2 [

 √  const/ n

This construction is due to A. Dvoretski, P. Erdos & S. Kakutani.

10. Dirichlet Problem and Brownian Motion LET G BE ANY bounded open set in Rd . Define the exit time τG (w) as 61 follows: τG (w) = {inf t : w(t) < G}.

If w(0) ∈ G, τG (w) = 0; if w(0) ∈ G, τG (w) is the first time w escapes G or, equivalently, it is the first time that w hits the boundary ∂G of G. Clearly τG (w) is a stopping time. By definition XτG (w) ∈ ∂G, ∀w and XτG is a random variable. We can define a Borel probability measure on ∂G by πG (x, Γ) = P x (XτG ∈ Γ)

= probability that w hits Γ.

If f is a bounded, real-valued measurable funciton defined on ∂G, we define Z u(x) = E x ( f (XτG )) = f (y)πG (x, dy) ∂G

where E x = E Px . In case G is a sphere centred around x, the exact form of πG (x, Γ) is computable. Theorem . Let S = S (0; r) = {y ∈ Rd : |y| < r}. Then surface area of Γ πS (0, r) = . surface area of S 63

10. Dirichlet Problem and Brownian Motion

64 62

Proof. The distributions {Ft1 ,...,tk } defining Brownian motion are invariant under rotations. Thus πS (0, ·) is a rotationally invariant probability measure. The result follows from the fact that the only probability measure (on the surface of a sphere) that is invariant under rotations is the normalised surface area.  Theorem . Let G be any bounded region, f a bounded measurable real valued function defined on ∂G. Define u(x) = E x ( f (XτG )). Then (i) u is measurable and bounded; (ii) u has the mean value property; consequently, (iii) u is harmonic in G. Proof.

(i) To prove this, it is enough to show that the mapping x → P x (A) is measurable for every Borel set A. Let C = {A ∈ B : x → P x (A) is measurable}

d d It is clear that π−1 t1 ,...,tk (B) ∈ C , ∀ Borel set B in R × · · · × R . As C is a monotone class C = B.

(ii) Let S be any sphere with centre at x, and S ⊂ G. Let τ = τS denote the exit time through S . Clearly τ ≤ τG . By the strong Markov property, u(Xτ ) = E( f (XτG )|Fτ ). Now u(x) = E x ( f (XτG )) = E x (E( f (XτG ))|Fτ ) Z = E x (u(Xτ )) = u(y)πS (x, dy) 1 = |S | 63

Z

∂S

∂S

u(y)dS ; |S | = surface area of S .

65 (iii) is a consequence of (i) and (ii). (See exercise below).  Exercise ′ . Let u be a bounded measurable function in a region G satisfying the mean value property, i.e. Z 1 u(x) = u(y)dS |S | ∂S

for every sphere S G. Then (i) u(x) =

1 R u(y)dy. v ∈ lS S

(ii) Using (i) show that u is continuous. (iii) Using (i) and (ii) show that u is harmonic. We shall now solve the boundary value problem under suitable conditions on the region G. Theorem . Let G, f , u be as in the previous theorem. Further suppose that (i) f is continuous; (ii) G satisfies the exterior cone condition at every point of ∂G, i.e. for each y ∈ ∂G there exists a cone Ch with vertex at the point y of height h and such that Ch − {y} ⊂ exterior of G. Then lim u(x) = f (y), ∀y ∈ ∂G.

x→y,x∈G

64

Proof. Step 1. Py {w : w(0) = y, w remains in G for some positive time} = 0. Let An = {w : w(0) = y, w(s) ∈ G for 0 ≤ s ≤ 1/n}, ∞ ∞ S T Bn = Ω − An , A = An , B = Bn . n=1

n=1

10. Dirichlet Problem and Brownian Motion

66

As An ’s are increasing, Bn ’s are decreasing and Bn ∈ F1/n ; so that B ∈ F0+ . We show that P(B) > 0, so that by Bluementhal’s zero-one law, P(B) = 1, i.e. P(A) = 0. Py (B) = lim Py (Bn ) ≥ lim Py {w : w(0) = y, w( n→∞

n→∞

1 ) ∈ Ch − {y}} 2n

Thus Py (b) ≥ lim

Z

√ 1/ (2π/2n)d exp(−|z − y|2 /2/2n)dz

Ch −{y}

=

Z

√ 2 1/ (2π)e−|y| /2 dy,

C∞

where C∞ is the cone of infinite height obtained from Ch . Thus Py (B) > 0. Step 2. If C is closed then the mapping x → P x (C) is upper semicontinuous. For, denote by XC the indicator function of C. As C is closed (in a metric space) ∃ a sequence of continuous functions fn decreasing to XC such that 0 ≤ fn ≤ 1. Thus E x ( fn ) decreases to E x (XC ) = P x (C). Clearly x → E x (Fn ) is continuous. The result follows from the fact that the infimum of any collection of continuous functions is upper semicontinuous. 65

Step 3. Let δ > 0, N(y; δ) = {z ∈ ∂G : |z − y| < δ},

Bδ = {w : w(0) ∈ G, XτG (w) ∈ ∂G − N(y; δ)}, i.e. Bδ consists of trajectories which start at a point of G and escape for the first time through ∂G at a point not in N(y; δ). If Cδ = Bδ , then Cδ ∩ {w : w(0) = y} ⊂ A ∩ {w : w(0) = y} where A is as in Step 1.

67 For, suppose w ∈ Cδ ∩{w : w(0) = y}. Then there exists wn ∈ Bδ such that wn → w uniformaly on compact sets. If w < A∩{w : w(0) = y} there exists ǫ > 0 such that w(t) ∈ G ∀t in (0, ǫ]. Let δ∗ = inf d(w(t), G − 0≤t≤ǫ

N(y, δ)). Then δ∗ > 0. If tn = τG (wn ) and tn does not converge to 0, then there exists a subsequence, again denoted by tn , such that tn ≥ kǫ > 0 for some k ∈ (0, 1). Since wn (kǫ) ∈ G and wn (kǫ), w(kǫ) ∈ G, a contradiction. Thus we can assume that tn converges to 0 and also that ǫ ≥ tn ∀n, But then |wn (tn ) − w(tn )| ≥ δ∗ .

(*)

However, as wn converges to w uniformly on [0, ǫ], wn (tn ) − w(tn ) → w(0) − w(0) = 0 contradicting (*). Thus w ∈ A{w : w(0) = y}. Step 4.

lim P x (Bδ ) = 0.

x→y,x∈G

For,

66

lim P x (Bδ ) ≤ lim P x (Cδ ) ≤ Py (Cδ ) (by Step 2) x→y

x→y

= Py (Cδ ∩ {w : w(0) = y})

≤ Py (A) (by Step 3) = 0. Step 5. |u(x) − f (y)| = | ≤

Z Ω−Bδ

≤

Z

Z Ω

f (XτG (w))dP x (w) −

Z Ω

| f (XτG (w)) − f (y)|dP x (w) + |

f (y)dP x (w)| Z

( f (XτG (w)) − f (y))dP x (w)|

Bδ

| f (XτG (w)) − f (y)|dP x (w) + 2|| f ||∞ P x (Bδ )

Ω−Bδ

and the right hand side converges to 0 as x → y (by Step 4 and the fact that f is continuous). This proves the theorem.

10. Dirichlet Problem and Brownian Motion

68

 Remark. The theorem is local. Theorem . Let G = {y ∈ Rd : δ < |y| < R}, f any continuous function on ∂G = {|y| = δ} ∩ {|y| = R}. If u is any harmonic function in G with boundary values f , then u(x) = E x ( f (XτG )). Proof. Clearly G has the exterior cone property. Thus, if v(x) = E x ( f (XτG )), then v is harmonic in G and has boundary values f (by the previous theorem). The result follows from the uniqueness of the solution of the Dirichlet problem for the Laplacian operator. The function f = 0 on |y| = R and f = 1 on |y| = δ is of speR,0 cial interest. Denote by ∪δ,1 the corresponding solution of the Dirichlet problem.  67

Exercise.

(i) If d = 2 then R,0 Uδ,1 (x) =

(ii) If d ≥ 3 then

log R − log |x| , ∀x ∈ G. log R − log δ

R,0 Uδ,1 (x) =

Case (i): d = 2. Then

|x|−n+2 − R−n+2 . δ−n+2 − R−n+2

log R − log |x| R,0 = Uδ,1 (x). log R − log δ Now, E x ( f (XτG )) =

Z

πG (x, dy) = P x (|XτG | = δ),

|y|=δ

i.e. log R − log |x| = P x (|XτG | = δ) log R − log δ

69 Px

(the particle hits |y| = δ before it hits |y| = R).

Fix R and let δ → 0; then 0 = P x (the particle hits 0 before hitting |y| = R). Let R take values 1, 2, 3, . . ., then 0 = P x (the particle hits 0 before hitting any of the circles |y| = N). Recalling that P x (lim |Xt | = ∞) = 1, we get Proposition . A two-dimensional Brownian motion does not visit a point. Next, keep δ fixed and let R → ∞, then, 1 = P x (|w(t)| = δ

for some time t > 0).

Since any time t can be taken as the starting time for the Brownian 68 motion, we have Proposition . Two-dimensional Brownian motion has the recurrence property. Case (ii): d ≥ 3. In this case P x (w : w hits |y| = δ before it hits |y| = R)

= (1/|x|n−2 − 1/Rn−2 )/(1/δn−2 − 1/Rn−2 ).

Letting R → ∞ we get P x (w : w hits |y| = δ) = (δ/|x|)n−2 which lies strictly between 0 and 1. Fixing δ and letting |x| → ∞, we have Proposition . If the particle start at a point for away from 0 then it has very little chance of hitting the circle |y| = δ. If |x| ≤ δ, then P(w hits S δ ) = 1 where S δ = {y ∈ Rd : |y| = δ}.

10. Dirichlet Problem and Brownian Motion

70 Let

Vδ (x) = (δ/|x|)n−2 for |x| ≥ δ. In view of the above result it is natural to extend Vδ to all space by putting Vδ (x) = 1 for |x| ≤ δ. As Brownian motion has the Markov property

= 69

Z

P x {w : w hits S δ after time t} √ Vδ (y)1/ (2πt)d exp −|y|2 /2t dy → 0 as t → +∞.

Thus P(w hits S δ for arbitrarily large t) = 0. In other words, P(w : lim |w(t)| ≥ δ) = 1. As this is true ∀δ > 0, we get the following

t→∞

important result. Proposition . P( lim |w(t)| = ∞) = 1, t→∞

i.e. for d ≥ 3, the Brownian particle wander away to +∞.

11. Stochastic Integration LET {Xt : t ≥ 0} BE A one-dimensional Brownian motion. We want 70 R∞ first to define integrals of the type f (s)dX(s) for real functions f ∈ 0

L1 [0, ∞). If X(s, w) is of bounded variation almost everywhere then we R∞ can give a meaning to f (s)dX(s, w) = g(w). However, since X(s, w) 0

is not bounded variation almost everywhere, g(w) is not defined in the usual sense. R∞ In order to define g(w) = f (s)dX(s, w) proceed as follows. 0

Let f be a step function of the following type: f =

n X i=1

ai X[ti ,ti+1 ) , 0 ≤ t1 < t2 < . . . < tn+1 .

We naturally define g(w) =

Z∞

f (s)dX(s, w) =

n X

ai (Xti+1 (w) − Xti (w))

n X

ai (w(ti+1 ) − w(ti )).

i=1

0

=

i=1

g satisfies the following properties: (i) g is a random variable; 71

11. Stochastic Integration

72 (ii) E(g) = 0; E(g2 ) =

P

a2i (ti+1 − ti ) = || f ||2 .

This follows from the facts that (a) Xti+1 − Xti is a normal random variable with mean 0 and variance (ti+1 − ti ) and (b) Xti+1 − Xti are independent increments, i.e. we have  ∞   ∞  Z   Z      E  f dX  = 0, E | f dX|2  = || f ||22 .     0

71

0

Exercise 1. If f = g=

n X

i=1 m X i=1

ai X[ti ,ti+1 ) , 0 ≤ t1 < . . . < tn+1 , bi X[si ,si+1 ) , 0 ≤ s1 < . . . < sm+1 ,

Show that Z∞

( f + g)dX(s, w) =

0

and

Z∞

f dX(s, w) +

0

Z∞

(α f )dX(s, w) = α

0

Z∞

gdX(s, w)

0

Z∞

f dX(s, w), ∀α ∈ R.

0

Remark. The mapping f →

R∞

f dX is therefore a linear L2R -isometry of

0

the space S of all simple functions of the type n X i=1

into L2 (Ω, B, P).

ai X[ti ,ti+1 ) , (0 ≤ t1 < . . . < tn+1 )

73 Exercise 2. Show that S is a dense subspace of L2 [0, ∞). Hint: Cc [0, ∞), i.e. the set of all continuous functions with compact support, is dense in L2 [0, ∞). Show that S contains the closure of Cc [0, ∞). Remark . The mapping f →

R∞

f dX can now be uniquely extended as

0 L2 (Ω, B, P).

an isometry of L2 [0, ∞) into Next we define integrals fo the type g(w) =

Zt

72

X(s, w)dX(s, w)

0

Put t = 1 (the general case can be dealt with similarly). It seems natural to define (*)

Z1

X(s, w)dX(s) =

0

Lt

sup |t j −t j−1 |→0

n X j=1

X(ξ j)(X(t j ) − X(t j−1 ))

where 0 = t0 < t1 < . . . < tn = 1 is a partion of [0, 1] with t j−1 ≤ ξ j ≤ t j . In general the limit on the right hand side may not exist. Even if it exists it may happen that depending on the choice of ξ j , we may obtain different limits. To consider an example we choose ξ j = t j and then ξ j = t j−1 and compute the right hand side of (∗). If ξ j = t j−1 , n X j=1

Xξ j (Xt j − Xt j−1 ) =

n X j=1

Xt j−1 − (Xt j − Xt j−1 )

1X 1X = (Xt j ) − (Xt j−1 ) − (Xt − Xt j−1 ) 2 j=1 2 j=1 j n

n

1 1 2 [X (1) − X 2 (0)] − as n → ∞, and sup |t j − t j−1 | → 0, 2 2 arguing as in the proof of the result that Brownian motion is not of bounded variation. If ξ j = t j , Lt

n X

n→∞ Sup |t j −t j−1 |→0 j=1

Xt j (Xt j − Xt j−1 ) = 1/2X(1) − 1/2X(0) + 1/2.

11. Stochastic Integration

74 73

Thus we get different answers depending on the choice of ξ j and hence one has to be very careful in defining the integral. It turns out that the choice of ξ j = t j−1 is more appropriate in the definition of the integral and gives better results. Remark . The limit in (∗) should be understood in the sense of convergence probability. Exercise 3. Let 0 ≤ a < b. Show that the “left integral” (ξ j = t j−1 ) is given by Zb X 2 (b) − X 2 (a) − (b − a) L X(s)dX(s) = 2 a

and the “right integral” (ξ j = t j ) is given by R

Zb

X(s)dX(s) =

X 2 (b) − X 2 (a) + (b − a) . 2

s

We now take up the general theory of stochastic integration. To motivate the definitions which follow let us consider a d-dimensional Brownian motion {β(t) : t ≥ 0}. We have Z √ 2 E[β(t + s) − β(t) ∈ A|Ft ] = 1/ (2πs)e−|y| /2s dy. A

Thus E( f (β(t + s) − β(t))|Ft ] =

Z

√ 2/2s f (y)1/ (2πs)e−|y| dy.

Z

√ 2 eiu.y 1/ (2πs)e−|y| /2s dy

In particular, if f (x) = eix.u , iu

E[e (β(t + s) − β(t))|Ft ] =

=e 74

Thus

−s|u|2 2

.

75 2 /2

E[eiu.β(t+s) |Ft ] = eiu.β(t) e−s|u|

,

or, 2

E[eiu.β(t+s)+(t+s)|u|

/2

2

|Ft ] = eiu.β(t)+t|u|

/2

.

Replacing iu by θ we get 2 /2

E[eθ.β(s)−|sθ|

2/2

| Ft ] = eθ.β(t)−t|θ| , s > t, ∀θ.

2

It is clear that eθ.β(t)−t|θ| /2 is Ft -measurable and a simple calculation gives 2 E(eθ.β(t)−|θ| t/2| ) < ∞ ∀θ. We thus have Theorem . If {β(t) : t ≥ 0} is a d-dimensional Brownian motion then exp[θ.β(t) − |θ|2 t/2] is a Martingale relative to Ft , the σ-algebra generated by (β(s) : s ≤ t). Definition. Let (Ω, B, P) be a probability space (Ft )t≥0 and increasing S family of sub-σ-algebras of F with F = σ( Ft ). t≥0

Let

(i) a : [0, ∞)×Ω → [0, ∞) be bounded and progressively measurable; (ii) b : [0, ∞) × Ω → R be bounded and progressively measurable; (iii) X : [0, ∞)×Ω → R be progressively measurable, right continuous on [0, ∞), ∀ w ∈ Ω, and continous on [0, ∞) almost everywhere on Ω; θX(t,w)−θ

(iv) Zt (w) = e

Rt 0

2

b(s,w)ds− θ2

Rt

a(s,w)ds

0

be a Martingale relative to (Ft )t≥0 . Then X(t, w) is called an Ito process corresponding to the parameters b and a and we write Xt ∈ I[b, a]. N.B. The progressive measurability of X ⇒ Xt is Ft -measurable.

75

11. Stochastic Integration

76

Example . If {β(t) : t ≥ 0} is a Brownian motion, then X(t, w) = βt (w) is an Ito process corresponding to parameters 0 and 1. (i) and (ii) are obvious. (iii) follows by right continuity of βt and measurability of βt relative to Ft and (iv) is proved in the previous theorem. Exercise 4. Show that Zt (w) defined in (iv) is Ft -measurable and progressively measurable. [Hint: (i) Zt is right continuous. (ii) Use Fubini’s theorem to prove measurability]. Remark . If we put Y(t, w) = X(t, w) −

Rt

b(s, w)ds then Y(t, w) is pro-

0

gressively measurable and Y(t, w) is an Ito process corresponding to the parameters 0, a. Thus we need only consider integrals of the type Rt f (s, w)dY(s, w) and define 0

Zt 0

f (s, w)dX(s, w) =

Zt

f (s, w)dY(s, w) +

0

Zt

f (s, w)b(s, w)ds.

0

(Note that formally we have dY = dX − dbt). 76

Lemma . If Y(t, w) ∈ I[0, a], then Y 2 (t, w) −

Y(t, w) and

Zt

a(s, w)ds

0

are Martingales relative to (Ft ). Proof. To motivate the arguments which follow, we first give a formal proof. Let 2

θY(t,w)− θ2

Yθ (t) = e

Rt 0

a(s,w)ds

.

77 Yθ − 1 Then Yθ (t) is a martingale, ∀θ. Therefore is a Martingale, ∀θ. θ Hence (formally), Yθ − 1 = Yθ′ |θ=0 lim θ→0 θ is a Martingale. Step 1. Y(t, ·) ∈ Lk (Ω, F , P), k = 0, 1, 2, . . . and ∀t. In fact, for every real θ, Yθ (t) is a Martingale and hence E(Yθ ) < ∞. Since a is bounded this means that E(eθY(t,·) ) < ∞, ∀θ. Taking θ = 1 and −1 we conclude that E(e|Y| ) < ∞ and hence E(|Y|k ) < ∞, ∀k = 0, 1, 2, . . .. Since Y is an Ito process we also get  k  Rt  θ2  Y(t,·)− 2 ads   0   < ∞ sup E e    |θ|≤α 

∀k and for every α > 0.

Step 2. Let Xθ (t) = [Y(t, ·) − θ Define φA (θ) =

Z

Rt 0

ads]Y(t) =

d Yθ (t, ·). dθ

(Xθ (t, ·) − Xθ (s, ·))dP(w)

A

where t > s, A ∈ Fs . Then Zθ2

φA (θ)dθ =

θ1

Zθ2 Z θ1

[Xθ (t, ·) − Xθ (S , ·)]dP(w)dθ.

A

Since a is bounded, sup E([Yθ (t, ·)]k ) < ∞, and E(|Y|k ) < ∞, ∀k; we |θ|≤α

can use Fubini’s theorem to get Zθ2 θ1

φA (θ)dθ =

Z Zθ2 A θ1

[Xθ (t, ·) − Xθ (s, ·)]dθ dP(w).

77

11. Stochastic Integration

78 or Zθ2

φA (θ)dθ =

θ1

Z

Yθ2 (t, ·) − Yθ1 (t, ·)dP(w) −

A

Z

Yθ1 (s, ·) − Yθ1 (s, ·)dP(w).

A

Let A ∈ Fs and t > s; then, since Y is a Martingale, Zθ2

φA (θ)dθ = 0.

θ1

This is true ∀θ1 < θ2 and since φA (θ) is a continuous function of θ, we conclude that φA (θ) = 0, ∀θ. In particular, φA (θ) = 0 which means that Z Z Y(t, ·)dP(w) = Y(s, ·)dP(w), ∀A ∈ Fs , t > s, A

78

A

i.e., Y(t) is a Martingale relative to (Ω, Ft , P). To prove the second part we put Zθ (t, ·) = and ψA (θ) =

Z

d2 Yθ (t) dθ2

{Zθ (t, ·) − Zθ (s, ·)}dP(w).

A

Then, by Fubini, Zθ2 θ1

ψA (θ)dθ =

Z Zθ2 A θ1

Zθ (t, ·) − Zθ (s, ·)dθ dP(w).

79 or, Zθ2

ψA (θ)dθ = φA (θ2 ) − φA (θ1 )

θ1

= 0 if A ∈ Fs , t > s. Therefore ψA (θ) = 0, ∀θ. In particular, ψA (θ) = 0 implies that 2

Y (t, w) −

Zt

a(s, w)ds

0

is an (Ω, Ft , P) Martingale. This completes the proof of lemma 1.



Definition. A function θ : [0, ∞) × Ω → R is called simple if there exist reals s0 , s1 , . . . , sn , . . . 0 ≤ s0 < s1 < . . . < sn . . . < ∞, sn increasing to +∞ and θ(s, w) = θ j (w) if s ∈ [s j , s j+1 ), where θ j (w) is Fs j -measurable and bounded. Definition . Let θ : [0, ∞) × Ω → R be a simple function and Y(t, w) ∈ I[0, a]. We define the stochastic integral of θ with respect to Y, denoted Zt 0

by ξ(t, w) =

Zt 0

θ(s, w)dY(s, w)

θ(s, w)dY(s, w)),

79

11. Stochastic Integration

80

=

k X j=1

θ j−1 (w)[Y(s j , w) − Y(s j−1 , w)] + θk (w)[Y(t, w) − Y(sk , w)].

Lemma 2. Let σ : [0, ∞) × Ω → R be a simple function and Y(t, w) ∈ I[0, a]. Then ξ(t, w) =

Zt

σ(s, w)dY(s, w) ∈ I[0, aσ2 ].

0

Proof.

(i) By definition, σ is right continuous and σ(t, w) is Ft measurable; hence it is progressively measurable. Since a is progressively measurable and bounded aσ2 : [0, ∞) × Ω → [0, ∞) is progressively measurable and bounded.

80

(ii) From the definition of ξ it is clear that ξ(t, ·) is right continuous, continous almost everywhere and Ft -measurable therefore ξ is progressively measurable. 2

[θξ(t,w)− θ2

(iii) Zt (w) = e

Rt

aσ2 ds]

0

is clearly Ft -measurable ∀θ. We show that E(Zt ) < ∞, ∀t and E(Zt2 |Ft1 ) = Zt1 if t1 < t2 . We can assume without loss of generality that θ = 1 (if θ , 1 we replace σ by θσ). Therefore Rt [ξ(t,w)− aσ2 ds]

Zt (w) = e

0

.

81 Since a and σ are bounded uniformly on [0, t], it is enough to show that E(eξ(t,w) ) < ∞. By definition, ξ(t, w) =

k X j=1

θ j−1 (w)[Y(s j , w) − Y(s j−1 , w)] + θk (w)(Y(t, w) − Y(sk , w)).

The result E(eξ(t,w) ) < ∞ will follow from the generalised Holder’s inequality provided we show that E(eθ(w)[Y(t,w)−Y(s,w)] ) < ∞ for every bounded function θ which is Fs -measurable. Now E(eθ[Y(t,·)−Y(s,·)] |Fs ) = constant for every constant θ, since Y ∈ I[0, a]. Therefore E(eθ(w)[Y(t,·)−Y(s,·)] |Fs ) = constant for every θ which is bounded and Fs -measurable. Thus E(eθ(w)[Y(t,·)−Y(s,·)] ) < ∞. This proves that E(Zt (w)) ∈ ∞. Finally we show that if t1 < t2 .

E(Zt2 |Ft1 ) = Zt1 (w),

Consider first the case when t1 and t2 are in the same interval [sk , sk+1 ). Then ξ(t2 , w) = ξ(t1 , w) + σk (w)[Y(t2 , w) − Y(t1 , w)] (see definition), Zt2 Zt1 Zt2 2 2 aσ (s, w)ds = aσ (s, w)ds + aσ2 (s, w)ds. 0

0

t1

81

11. Stochastic Integration

82 Therefore

θ2 E(Zt2 (w)|Ft1 ) = Zt1 (w)E(exp[θσk (w)[Y(t2 , w) − Y(t1 , w)] − 2

Zt2

aσ2 ds)|Ft1 )

t1

as Y ∈ I[0, a]. (*)

θ2 E(exp[θ(Y(t2 , w) − T (t1 , w)) − 2

Zt2

a(s, w)ds]|Ft1 ) = 1

t1

and since σk (w) is Ft1 -measurable (∗) remains valid if θ is replaced by θσk . Thus E(Zt2 |Ft1 ) = Zt1 (w). The general case follows if we use the identity E(E(X|C1 )|C2 ) = E(X|C2 ) for

C2 ⊂ C1 .

Thus Zt is a Martingale and ξ(t, w) ∈ I[0, aσ2 ]. 82

Corollary .



(i) ξ(t, w) is a martingale; E(ξ(t, w)) = 0;

(ii) ξ 2 (t, w) −

Rt

aσ2 ds

0

is a Martingale with Zt E(ξ (t, w)) = E( aσ2 (s, w)ds. 2

0

Proof. Follows from Lemma 1.



Lemma 3. Let σ(s, w) be progressively measurable such that for each t, Zt E( σ2 (s, w)ds) < ∞. 0

83 Then there exists a sequence σn (s, w) of simple functions such that  t  Z    lim E  |σn (s, w) − σ(s, w)|2 ds = 0. n→∞   0

Proof. We may assume that σ is bounded, for if σN = σ for |σ| ≤ N and 0 if |σ| > N, then σn → σ, ∀(s, w) ∈ [0, t] × Ω. σN is progressively measurable and |σn − σ|2 ≤ 4|σ|2 . By hypothesis σ ∈ L([0, t] : Ω). Rt Therefore E( |σn −σ|ds) → 0, by dominated convergence. Further, 0

we can also assume that σ is continuous. For, if σ is bounded, define

σh (t, w) = 1/h

Zt

σ(s, w)ds.

(t−h)v0

σn is continuous in t and Ft -measurable and hence progressively measurable. Also by Lebesgue’s theorem σh (t, w) → σ(t, w),

as

h → 0, ∀t, w.

Since σ is bounded by C, σh is also bounded by C. Thus Zt E( |σh (s, w) − σ(s, w)|2 ds) → 0. 0

(by dominated convergence). If σ is continuous, bounded and progressively measurable, then ! [ns] σn (s, w) = σ ,w n is progressively measurable, bounded and simple. But Lt σn (s, w) = σ(s, w).

n→∞

83

11. Stochastic Integration

84 Thus by dominated convergence  t  Z    E  |σn − σ|2 ds → 0  

as n → ∞.

0



Theorem . Let σ(s, w) be progressively measurable, such that Zt E( σ2 (s, w)ds) < ∞ 0

for each t > 0. Let (σn ) be simple approximations to σ as in Lemma 3. Put Zt ξn (t, w) = σn (s, w)dY(s, w) 0

where Y ∈ I[0, a]. Then (i) Lt ξn (t, w) exists uniformly in probability, i.e. there exists an aln→∞ most surely continuous ξ(t, w) such that !

Lt P sup |ξn (t, w) − ξ(t, w)| ≥ ǫ = 0

n→∞

0≤t≤T

for each ǫ > 0 and for each T . Moreover, ξ is independent of the sequence (σ0 ). 84

(ii) The map σ → ξ is linear. (iii) ξ(t, w) and ξ 2 (t, w) −

Rt

aσ2 ds are Martingales.

0

(iv) If σ is bounded, ξ ∈ I[0, aσ2 ].

85 Proof.

(i) It is easily seen that for simple functions the stochastic integral is linear. Therefore (ξn − ξm )(t, w) =

Zt

(σn − σm )(s, w)dY(s, w).

0

Since ξn − ξm is an almost surely continuous martingale ! 1 P sup |ξn (t, w) − ξm (t, w)| ≥ ǫ ≤ 2 E[(ξn − ξm )2 (T, w)]. ǫ 0≤t≤T This is a consequence of Kolmogorov inequality (See Appendix). Since Zt 2 (ξn − ξm ) − a(σn − σm )2 ds 0

is a Martingale, and a is bounded,

(*)

 T  Z    2 2 E[(ξn − ξm ) (T, w)] = E  (σn − σm ) a ds .   0

 T  Z  1   ≤ const 2 E  (σn − σm )2 ds .  ǫ  0

Therefore Lt E[(ξn − ξm )2 (T, w)] = 0.

n,m→∞

Thus (ξn −ξm ) is uniformly Cauchy in probability. Therefore there exists a progressively measurable ξ such that ! Lt P sup |ξn (t, w) − ξ(t, w)| ≥ ǫ = 0, ∀ǫ > 0, ∀T. n→∞

0≤t≤T

It can be shown that ξ is almost surely continuous.

11. Stochastic Integration

86

If (σn ) and (σ′n ) are two sequences of simple functions approxi- 85 mating σ, then (∗) shows that E[(ξn − ξn′ )2 (T, w)] → 0. Thus Lt ξn = Lt ξn′ , n

n

i.e. ξ is independent of (σn ). (ii) is obvious. (iii) (*) shows that ξn → ξ in L and therefore ξn (t, ·) → ξ(t, ·) in L1 for each fixed t. Since ξn (t, w) is a martingale for each n, ξ(t, w) is a martingale. (iv) ξn2 (t, w) −

Rt

aσ2n is a martingale for each n.

0

Since ξn (t, w) → ξ(t, w) in L2 for each fixed t and ξn2 (t, w) → ξ 2 (t, w) in

L1

for each fixed t.

For ξn2 (t, w) − ξ 2 (t, w) = (ξn − ξ)(ξn + ξ) and using H¨older’s inequality, we get the result. Similarly, since σn → σ in L2 ([0, t] × Ω),

σ2n → σ2 in L1 ([0, t] × Ω), and because a is bounded aσ2n → aσ2 in L1 ([0, t]×Ω). This shows Rt that ξn2 (t, w) − aσ2n ds converges to 0

2

ξ (t, w) −

Zt 0

aσ2 ds

87 for each t in L1 . Therefore 2

ξ (t, w) −

Zt

aσ2 ds

0

is a martingale.

86

(v) Let σ be bounded. To show that ξ ∈ I[0, σ2 ] it is enough to show that t 2

θξ(t,w)− θ2

e

R

aσ2 ds

0

is a martingale for each θ, the other conditions being trivially satisfied. Let t 2

θξn (t,w)− θ2

Zn (t, w) = e

R

aσ2n ds

0

We can assume that |σn | ≤ C if |σ| ≤ C (see the proof of Lemma 3).   t Zt   2 Z (2θ)   Zn = exp 2θξn (t, w) − aσ2n ds + θ2 aσ2n ds .   2 0

0

Thus (**)

  t  2θξn (t,w)− (2θ)2 R aσ2n ds  2   0 E(Zn ) ≤ const E e  = const 

since Zn is a martingale for each θ. A subsequence Zni converges to t 2

θξ(t,w)− θ2

e

R

aσ2 ds

0

almost everywhere (P). This together with (**) ensures uniform integrability of (Zn ) and therefore 2

θξ(t,w)− θ2

e

Rt 0

aσ2 ds

11. Stochastic Integration

88

is a martingale. Thus ξ is an Ito process, ξ ∈ I[0, aσ2 ].



Definition . With the hypothesis as in the above theorem we define the stochastic integral ξ(t, w) =

Zt

σ(s, w)dY(s, w).

0

87

Exercise. Show that d(X + Y) = dX + dY. Remark . If σ is bounded, then ξ satisfies the hypothesis of the previous theorem and so one can define the integral of ξ with respect to Y. Further, since ξ itself is Itˆo, we can also define stochastic integrals with respect to. Examples. 1. Let {β(t) : t ≥ 0} be a Brownian motion; then β(t, w) is progressively measurable (being continuous and Ft -measurable). Also, Z Zt Zt Z Zt t β2 (s)ds dP = β2 (s)dP ds = sds = 2 Ω

0

Hence

0

Zt

Ω

0

β(s, w)dβ(s, w)

0

is well defined. Rt 2. Similarly β(s/2)dβ(s) is well defined. 0

3. However

Zt

β(2s)dβ(s)

0

is not well defined, the reason being that β(2s) is not progressively measurable.

89 Exercise 5. Show that β(2s) is not progressively measurable. 88 (Hint: Try to show that β(2s) is not Fs -measurable for every s. To show this prove that Fs , F2s ). Exercise 6. Show that for a Brownian motion β(t), the stochastic integral Z1

β(s, ·)dβ(s, ·)

0

is the same as the left integral

L

Z1 0

defined earlier.

β(s, ·)dβ(s, ·)

12. Change of Variable Formula WE SHALL PROVE the

89

Theorem . Let σ be any bounded progressively measurable function and Y be an Ito process. If λ is any progressively measurable function such that  t  Z    E  λ2 ds < ∞, ∀t,   0

then (*)

Zt

λdξ(s, w) =

0

Zt

λ(s, w)σ(s, w)dY(s, w),

Zt

σ(s, w)dY(s, w).

0

where ξ(t, w) =

0

Proof. Step 1. Let λ and σ be both simple, with λ bounded. By a refinement of the partition, if necessary, we may assume that there exist reals 0 = s0 , s1 , . . . , sn , . . . increasing to +∞ such that λ and σ are constant on [s j , s j+1 ), say λ = λ j (w), σ = σ j (w), where λ j (w) and σ j (w) are Fs j measurable. In this case (*) is a direct consequence of the definition. 91

12. Change of Variable Formula

92

Step 2. Let λ be simple and bounded. Let (σn ) be a sequence of simple bounded functions as in Lemma 3. Put Zt ξn (t, w) = σn (s, w)dY(s, w) 0

By Step 1, Zt

(**)

λdξn =

0

Zt

λσn dY(s, w).

0

90

Since λ is bounded, λσn converges to λσ in L2 ([0, t] × Ω). Hence, Rt Rt by definition, λσn dY(s, w) converges to λσdY in probability. 0

0

Further, Zt

λdξn (s, w) = λ(s0 , w)[ξn (s1 , w) − ξn (s0 , w)] + · · ·

0

+ · · · + λ(sk , w)[ξn (t, w) − ξn (sk−1 , w)], where s0 < s1 < . . . . . . is a partition for λ, and ξn (t, w) converges to ξ(t, w) in probability for every t. Therefore Zt

λdξn (s, w)

Zt

λdξ(s, w).

0

converges in probability to

0

Taking limit as n → ∞ in (**) we get Zt 0

λdξ(s, w) =

Zt 0

λσdY(s, w).

93 Step 3. Let λ be any progressively measurable function with Zt E( λ2 ds) < ∞, ∀t. 0

Let λn be a simple approximation to λ as in Lemma 3. Then, by Step 2, (***)

Zt

λn (s, w)dξ(s, w) =

0

Zt

λn (s, w)σ(s, w)dY(s, w).

0

By definition, the left side above converges to Zt

λ(s, w)dξ(s, w)

0

in probability. As σ is bounded λn σ converges to λσ in L2 ([0, t] × Ω). 91 Therefore   Zt Zt     P  sup | λn σdY(s, w) − λσ dy(s, w)| ≥ ǫ   0≤t≤T 0

0

  t  Z   ||a||∞ 1/ǫ 2 E  (λn σ − λσ)2 ds   0

(see proof of the main theorem leading to the definition of the stochastic integral). Thus Zt λn σdY(s, w) 0

converges to Zt

λσ dY(s, w)

0

in probability. Let n tend to + in (***) to conclude the proof.

94

12. Change of Variable Formula 

13. Extension to Vector-Valued Itˆo Processes Definition. Let (Ω, F , P) be a probability space and (Ft ) an increasing 92 family of sub σ-algebras of F . Suppose further that (i)

a : [0, ∞) × Ω → S +d

is a probability measurable, bounded function taking values in the class of all symmetric positive semi-definite d × d matrices, with real entries; (ii)

b : [0, ∞) × Ω → Rd

is a progressively measurable, bounded function; (iii)

X : [0, ∞) × Ω → Rd

is progressively measurable, right continuous for every w and continuous almost everywhere (P); Z(t, ·) = exp[hθ, X(t, ·)i −

Zt

hθ, b(s, ·)ids

0

(iv)

−

1 2

Zt

hθ, a(s, ·)θids]

0

is a martingale for each θ ∈ Rd , where hx, yi = x1 y1 + · · · + xd yd , x, y ∈ Rd . 95

13. Extension to Vector-Valued Itˆo Processes

96

Then X is called an Itˆo process corresponding to the parameters b and a, and we write X ∈ I[b, a] Note.

1. Z(t, w) is a real valued function.

2. b is progressively measurable if and only if each bi is progressively measurable. 3. a is progressively measurable if and only if each ai j is so.

93

Exercise 1. If X ∈ I[b, a], then show that (i) (ii)

Y=

d X i=1

Xi ∈ I[bi , aii ], θi Xi ∈ I[hθ, bi, hθ, aθi],

where θ = (θ1 , . . . , θd ). (Hint: (ii) (i). To prove (ii) appeal to the definition). Remark . If X has a multivariate normal distribution with mean µ and covariance (ρi j ), then Y = hθ, Xi has also a normal distribution with mean hθ, µi and variance hθ, ρθi. Note the analogy with the above exercise. This analogy explains why at times b is referred to as the “mean” and a as the “covariance”. Exercise 2. If {β(t) : t ≥ 0} is a d-dimensional Brownian motion, then β(t, w) ∈ I[0, I] where I = d × d identity matrix. Rt As before one can show that Y(t, ·) = X(t, ·) − b(s, w)ds is an Itˆo 0

process with parameters 0 and a.

Definition . Let X be a d-dimensional Ito process. σ = (σ1 , . . . , σd ) a d-dimensional progressively measurable function such that  t   Z   E  hσ(s, ·), σ(s, ·)i > ds   0

97 is finite or, equivalently,  t  Z    2 E  σi (s, ·)ds < ∞,  

(i = 1, 2, . . . d).

0

94

Then by definition Zt

hσ(s, ·), dX(s, ·)i =

0

d Z X

t

i=1 0

σi (s, ·)dXi (s, ·).

Proposition . Let X be a d-dimensional Itˆo process X ∈ I[b, a] and let σ be progressively measurable and bounded. If ξi (t, ·) =

Zt

σi dXi (s, ·),

0

then ξ = (ξ1 , . . . , ξd ) ∈ I[B, A], where B = (σ1 b1 , . . . , σd bd ) and Proof.

Ai j = σi σ j ai j .

(i) Clearly Ai j is progressively measurable and bounded. Since a ∈ S +d , A ∈ S +d .

(ii) Again B is progressively measurable and bounded. (iii) Since σ is bounded, each ξi (t, ·) is an Itˆo process; hence ξ is progressively measurable, right continuous, continuous almost everywhere (P). It only remains to verify the martingale condition. Step 1. Let θ = (θ1 , . . . , θd ) ∈ Rd . By hypothesis, (*)

E(exp[(θ1 X1 + · · · +

θd Xd )|ts

···

Zt s

(θ1 b1 + · · · + θd bd )du

13. Extension to Vector-Valued Itˆo Processes

98 1 − 2

Zt X

θi θ j ai j ds]|Fs ) = 1.

0

95

Assume that each σi is constant on [s, t], σi = σi (w) and Fs measurable. Then (*) remains true if θi are replaced by θi θi (w) and since σi ’s are constant over [s, t], we get Zt X Zt d E(exp[ θi σi (s, ·)dXi (s, ·) − θi bi σi (s, ·)ds i=1

0

−

1 2

Zt 0

0

X

θi θ j σi (s, ·)σ j (s, ·)ai j ds]| s )

 s  Zs Zs Z X d    exp  θi σi (s, ·)dXi (s, ·) − hθ, Bidu − 1 hθ, Aθidu .   0

i=1

0

0

Step 2. Let each σi be a simple function.

By considering finer partitions we may assume that each σi is a step function,

finest partition

i.e. there exist points s0 , s1 , s2 , . . . , sn , s = s0 < s1 < . . . < sn+1 = t, such that on [s j , s j+1 ) each σi is a constant and s j -measurable. Then (**) holds if we use the fact that if C1 ⊃ C2 . E(E( f |C1 )|C2 ) = E( f |C2 ).

99

96

Step 3. Let σ be bounded, |σ| ≤ C. Let (σ(n) ) be a sequence of simple functions approximating σ as in Lemma 3. (**) is true if σi is replaced by σ(n) i for each n. A simple verification shows that the expression Zn (t, ·), in the parenthes is on the left side of (**) with σi replaced by σi(n) , converges to

= Exp



Zt 0

Z(t, ·) = Zt X X θi σi (s, ·)ds − θi bi σi (s, ·)ds− i

0

−

1 2

Zt 0

X i, j

i

 θi θ j σi σ j ai j ds

as n → ∞ in probability. Since Zn (t, ·) is a martingale and the functions σi , σ j , a are all bounded, sup E(Zn (t, ·)) < ∞. n

This proves that Z(t, ·) is a martingale.



Corollary . With the hypothesis as in the above proposition define Z(t) =

Zt

hσ(s, ·), dX(s, ·)i.

0

Then Z(t, ·) ∈ I[hσ, bi, σaσ∗ ]

where σ∗ is the transpose of σ.

Proof. Z(t, ·) = ξ1 (t, ·) + · · · + ξd (t, ·).



Definition. Let σ(s, w) = (σi j (s, w)) be a n × d matrix of progressively measurable functions with   t Z    E  σ2i j (s, ·)ds < ∞.   0

13. Extension to Vector-Valued Itˆo Processes

100

If X is a d-dimensional Itˆo process, we define  t  Z  X d Zt   σi j (s, ·)dX j (s, ·).  σ(s, ·)dX(s, ·) = 0

j=1 0

i

Exercise 3. Let

Z(t, w) =

Zt

σ(s, ·)dY(s, ·),

0

where Y ∈ I[0, a] is a d-dimensional Itˆo process and σ is as in the above definition. Show that Z(t, ·) ∈ I[0, σaσ∗ ] is an n-dimensional Ito process, (assume that σ is bounded). Exercise 4. Verify that  t  Z    E(|Z(t)|2 ) = E  tr(σaσ∗ )ds .   0

Exercise 5. Do exercise 3 with the assumption that σaσ∗ is bounded. Exercise 6. State and prove a change of variable formula for stochastic integrals in the case of several dimensions. (Hint: For the proof, use the change of variable formula in the one dimensional case and d(X + Y) = dX + dY).

97

14. Brownian Motion as a Gaussian Process SO FAR WE have been considering Brownian motion as a Markov pro- 98 cess. We shall now show that Brownian motion can be considered as a Gaussian process. Definition. Let X ≡ (X1 , . . . , XN ) be an N-dimensional random variable. It is called an N-variate normal (or Gaussian) distribution with mean µ ≡ (µ1 , . . . , µN ) and covariance A if the density function is ! 1 1 1 −1 ∗ exp − [(X − µ)A (X − µ) ] 2 (2π)N/2 (det A)1/2

where A is an N × N positive definite symmetric matrix. Note.

1. E(Xi ) = µi .

2. Cov(Xi , X j ) = (A)i j . Theorem . X ≡ (X1 , . . . , XN ) is a multivariate normal distribution if and only if for every θ ∈ RN , hθ, Xi is a one-dimensional Gaussian random variable. We omit the proof. Definition. A stochastic process {Xt : t ∈ I} is called a Gaussian process if ∀t1 , t2 , . . . , tN ∈ I, (Xt1 , . . . , XtN ) is an N-variate normal distribution. 101

14. Brownian Motion as a Gaussian Process

102

Exercise 1. Let {Xt : t ≥ 0} be a one dimensional Brownian motion. Then show that (a) Xt is a Gaussian process.

99

(Hint: Use the previous theorem and the fact that increments are independent) (b) E(Xt ) = 0, ∀t, E(X(t)X(s)) = s ∧ t.

Let ρ : [0, 1] = [0, 1] → R be defined by ρ(s, t) = s ∧ t. Define K : L2R [0, 1] → L2R [0, 1] by K f (s) =

Z1

ρ(s, t) f (t)dt.

0

Theorem . K is a symmetric, compact operator. It has only a countable number of eigenvalues and has a complete set of eigenvectors. We omit the proof. Exercise 2. Let λ be any eigenvalue of K and f an eigenvector belonging to λ. Show that (a) λ f ′′ + f = 0 with λ f (0) = 0 = λ f ′ (1). (b) Using (a) deduce that the eigenvalues are given by λn = 4/(2n + 1)2 π2 and the corresponding eigenvectors are given by √ fn = 2 Sin 1/2[(2n + 1)πt]n = 0, 1, 2, . . . . Let Z0 , Z1 , . . . , Zn . . . be identically distributed, independent, normal random variables with mean 0 and variance 1. Then we have Proposition . Y(t, w) =

∞ P

√ Zn (w) fn (t) λn

n=0

converges in mean for every real t.

103 100

Proof. Let Ym (t, w) =

m P

√ Zi (w) fi (t) λi . Therefore

i=0

2

E{(Yn+m (t, ·) − Yn (t, ·)) } = E(||Yn+m (·) − Yn (·)||2 ≤

n+m X n+1

n+m X

fi2 (t)λi ,

n+1

λi → 0. 

Remark. As each Yn (t, ·) is a normal random variable with mean 0 and n P variance λi fi2 (t), Y(t, ·) is also a normal random variable with mean i=0

zero and variance

∞ P

i=0

λi fi2 . To see this one need only observe that the

limit of a sequence of normal random variables is a normal random variable. Theorem (Mercer). ρ(s, t) =

∞ X i=0

λi fi (t) fi (s), (s, t) ∈ [0, 1] × [0, 1].

The convergence is uniform. We omit the proof. Exercise 3. Using Mercer’s theorem show that {Xt : 0 ≤ t ≤ 1} is a Brownian motion, where X(t, w) =

∞ X

√ Zn (w) fn (t) λn .

n=0

This exercise now implies that Z1 0

X 2 (s, w)ds = (L2 − norm of X)2

14. Brownian Motion as a Gaussian Process

104

=

X

λn Zn2 (w),

since fn (t) are orthonormal. Therefore −λ

E(e

R1

X 2 (s,·)ds

−λ

) = E(e

0

∞ P

n=0

λn Zn2 (w)

)=

101 ∞ Y

2

E(e−λλn Zn )

n=0

(by independence of Zn ) =

∞ Y

2

E(e−λλn Z0 )

n=0

as Z0 , Zn . . . are identically distributed. Therefore −λ

E(e

R1

)=

∞ Y

√ 1/ (1 + 2λλn )

=

∞ Y

√ 1/ 1 +

X 2 (s,·)ds

0

n=0

n=0

8 8λ (2n + 1)2 Π2

√ √ = 1/ (cosh) (2λ). R1 APPLICATION. If F(a) = P( X 2 (s)ds < a), then 0

Z∞

−λa

e

0

= E(e−λ

R1 0

dF(a) =

Z∞

−∞

X 2 (s)ds

e−λa dF(a)

√ √ ) = 1/ (cosh) (2λ).

!

15. Equivalent For of Itˆo Process LET (Ω, F , P) BE A probability space with (Ft )t≥0 and increasing fam- 102 ily of sub σ-algebras of F such that σ(U Ft ) = F . Let t≥0

(i) a : [0, ∞)×Ω → S d+ be a progressively measurable, bounded function taking values in S d+ , the class of all d × d positive semidefinite matrices with real entries; (ii) b : [0, ∞) × Ω → Rd be a bounded, progressively measurable function; (iii) X : [0, ∞) × Ω → Rd be progressively measurable, right continuous and continuous a.s. ∀(s, w) ∈ [0, ∞) × Ω. For (s, w) ∈ [0, ∞) × Ω define the operator L s,w =

d d X ∂2 ∂ 1X ai j (s, w) + b j (s, w) . 2 i, j=1 ∂xi ∂x j j=1 ∂x j

For f , u, h belonging to C0∞ (Rd ), C0∞ ([0, ∞) × Rd ) and Cb1,2 ([0, ∞) × Rd ) respectively we define Y f (t, w), Zu (t, w), Ph (t, w) as follows: Y f (t, w) = f (X(t, w)) −

Zt 0

105

(L s,w ( f )(X(s, w))ds,

15. Equivalent For of Itˆo Process

106

Zu (t, w) = u(t, X(t, w)) −

! ∂u + L s,w u (s, X(s, w))ds, ∂s

Zt 0

Ph (t, w) = exp[h(t, X(t, w)) − 1 − 2 103

Z

Zt

! ∂h + L s,w h (s, X(s, w)ds− ∂s

0

t

0

ha(s, w)∇x h(s, X(s, w)), ∇x h(s, X(s, w))ids].

Theorem . The following conditions are equivalent. Rt Rt (i) Xθ (t, w) = exp[hθ, X(t, w)i − hθ, b(s, w)ids − hθ, a(s, w)θids] 0

is a martingale relative to (Ω, Ft , P), ∀θ ∈

0 d R .

(ii) Xλ (t, w) is a martingale ∀λ in Rd . In particular Xiθ (t, w) is a martingale ∀θ ∈ Rd . (iii) Y f (t, w) is a martingale for every f ∈ C0∞ (Rd ) (iv) Zu (t, w) is a martingale for every u ∈ C0∞ ([0, ∞) × Rd ). (v) Ph (t, w) is a martingale for every h ∈ Cb1,2 [(0, ∞) × Rd ). (vi) The result (v) is true for functions h ∈ C 1,2 ([0, ∞) × Rd ) with linear growth, i.e. there exist constants A and B such that |h(x)| ≤ A|x| + B. ∂2 h ∂h ∂h , , and − which occur under the integral ∂t ∂xi ∂xi ∂x j sign in the exponent also grow linearly. The functions

Remark. The above theorem enables one to replace the martingale condition in the definition of an Itˆo process by any of the six equivalent conditions given above. Proof. (i) (ii). Xλ (t, ·) is Ft -measurable because it is progressively measurable. That E(|Xλ (t, w)|) < ∞ is a consequence of (i) and the fact that a is bounded.

107 Xλ (t, w) is continuous for fixed t, s, w, (t > s). Xλ (s, w) Morera’s theorem shows that φ is analytic. Let A ∈ Fs . Then Z Xλ (t, w) dP(w) Xλ (s, w) φ

The function λ − →

A

is analytic. By hypothesis, Z Xλ (t, w) dP(w) = 1, ∀λ ∈ Rd . Xλ (s, w) A

Thus

R Xλ (t, w) dP(w) = 1, ∀ complex λ. Therefore A Xλ (s, w) E(Xλ (t, w)|Fs ) = Xλ (s, w),

proving (ii). (ii) ⇒ (iii). Let   Zt   Zt 1   hθ, a(s, w)θids , θ ∈ Rd . A(t, w) = exp −i hθ, b(s, w)ids +   2 0

0

By definition, A is progressively measurable and continuous. Also dA | (t, w)| is bounded on every compact set in R and the bound is indt dependent of w. Therefore A(t, w) is of bounded variation on every interval [0, T ] with the variation ||A||[0,T ] bounded uniformly in w. Let M(t, w) = Xiθ (t, w). Therefore sup |M(t, w)| ≤ e1/2 T sup |hθ, aθi|.

0≤t≤T

0≤t≤T

By (ii) M(t, ·) is a martingale and since

! E sup |M(t, w)| ||A||[0,T ] (w) < ∞, ∀T, 0≤t≤T

1 M(t, ·)A(t, ·) − 2

Zt 0

M(s, ·)dA(s, ·)

104

15. Equivalent For of Itˆo Process

108

is a martingale (for a proof see Appendix), i.e. Y f (t, w) is a martingale when f (x) = eihθ,xi . Let f ∈ C0∞ (Rd ). Then f ∈ F (Rd ) the Schwartz-space. Therefore by the Fourier inversion theorem Z f (x) = fˆ(θ)eihθ,xi dθ. Rd

105

On simplification we get Y f (t, w) =

Z

fˆ(θ)Yθ (t, w)dθ

Rd

where Yθ ≡ Ye ihθ, xi. Clearly Y f (t, ·) is progressively measurable and hence Ft -measurable. Using the fact that E(|Yθ (t, w)|) ≤ 1 + t d|θ| ||b||∞ +

d2 2 |θ| ||a||∞ , 2

the fact that F (Rd ) ⊂ L1 (Rd ) and that F (Rd ) is closed under multiplication by polynomials, we get E(|Y f (t, w)|) < ∞. An application of Fubini’s theorem gives E(Y f (t, w)|Fs ) = Y f (s, w), if t > s. This proves (iii). (iii) ⇒ (iv). Let u ∈ C0 ([0, ∞) × Rd ). Clearly Zu (t, ·) is progressively measurable. Since Zu (t, w) is bounded for every w, E(|Zu (t, w)|) < ∞. Let t > s. Then E(Zu (t, w) − Zu (s, w)|F s ) = = E(u(t, X(t, w) − u(s, X(s, w)|F s) − E(

Zt

(

∂u + Lσ,w u)(σ, X(σ, w)dσ|F s) ∂σ

s

= E(u(t, X(t, w) − u(t, X(s, w))|F s) + E(u(t, X(s, w) − u(s, X(s, w))|F s)− Zt ∂u − E( ( + Lσ uw )(σ, X(σ, w))dσ|F s ) ∂σ s

109 Zt

(Lσ,w u)(t, X(σ, w))dσ|F s)+

Zt

(

∂u (σ, X(s, w))dσ|F s )− ∂σ

Zt

(

∂u + Lσu , w)(σ, X(σ, w))dσ|F s ), ∂σ

= E(

s

+ E(

s

− E(

by (iii)

s

= E(

Zt

[Lσ,w u(t, X(σ, w)) − Lσ,w u(σ, X(σ, w))]dσ|F s)

s

+ E(

Zt

[

∂u ∂u (σ, X(s, w)) − (σ, X(σ, w))]dσ|F s) ∂σ ∂σ

s

= E(

Zt

(Lσ,w u(t, X(σ, w)) − Lσ,w u(σ, X(σ, w))]dσ|F s)

s

− E(

Zt s

dσ

Zσ

Lρ,w

∂u (σ, X(ρ, w))dρ|F s) ∂σ

s

106

The last step follows from (iii) (the fact that σ > s gives a minus sign). Zt Zt ∂ = E( dσ Lσ,w u(ρ, X(σ, w))dρ|Fs ) ∂ρ 0

σ

Zt Zσ ∂u − E( dσ Lρ,w (σ, X(ρ, w))dρ|Fs ) ∂σ s

s

=0 (by Fubini). Therefore Zu (t, w) is a martingale. Before proving (iv) ⇒ (v) we show that (iv) is true if u ∈ Cb1,2 ([0, ∞) × Rd . Let u ∈ Cb1,2 .

15. Equivalent For of Itˆo Process

110

that

(*) Assume that there exists a sequence (un ) ∈ C0∞ [[0, ∞) × Rd ] such un → u,

107

∂un ∂u ∂un ∂2 u ∂u ∂un → , → → , ∂t ∂t ∂xi ∂xi ∂xi ∂x j ∂xi ∂x j

uniformly on compact sets. Then Zun → Zu pointwise and sup(|Zun (t, w)|) < ∞. n

Therefore Zu is a martingale. Hence it is enough to justify (*). For every u ∈ Cb1,2 ([0, ∞) × Rd ) we construct a u ∈ Cb1,2 ((−∞, ∞) × Rd ) ≡ Cb1,2 (R × Rd ) as follows. Put    u(t, x), if t ≥ 0, u(t, x) =   C1 u(−t, x) + C2 u(− t , x), if t < 0; 2

∂˜u ∂u matching , at t = 0 and uˆ (t, x) and u(t, x) at t = 0 and u˜ (t, x) and ∂t ∂t u(t, x) at t = 0 yields the desired constants C1 and C2 . In fact C1 = −3, C2 = 4. (*) will be proved if we obtain an approximating sequence for u˜ . Let S : R be any C function such that if |x| ≤ 1,    1, if |x| ≤ 1, S (x) =   0, if |x| ≥ 2. ! |x|2 Let S n (x) = S where |x|2 = x21 + · · · + x2d+1 . Pur un = S n u˜ . n This satisfies (*). (iv) ⇒ (v). Let h ∈ Cb1,2 ([0, ∞) × Rd ). Put u = exp(h(t, x)) in (iv) to conclude that h(t,X(t,w))

M(t, w) = e

−

Zt 0

108

is a martingale. Put

h(s,X(s,w))

e

"

# ∂h 1 + L s,w h + h∇x h, a∇x hids ∂s 2

111  t   Z 1 ∂h   (s, w) + L s,w − (s, w) + ha(s, w)∇ x h, ∇ x hids . A(t, w) = exp −    ∂s 2 0

A((t, w)) is progressively measurable, continuous everywhere and ||A||[0,T ] (w) ≤ C1 ∈ C2 T

dA where C1 and C2 are constants. This follows from the fact that | | is dt uniformly bounded in w. Also sup |M(t, w)| is uniformly bounded in 0≤t≤T

w. Therefore E( sup |M(t, w)| ||A||[0,T ] (w)) < ∞. 0≤t≤T

Hence M(t, ·)A −

Rt 0

M(s, ·)dA(s, ·) is a martingale. Now

" # dA(s, w) ∂h 1 =− (s, w) + L s,w h(s, w) + ha∇x h, ∇x hi A(s, w) ∂s 2 Therefore M(t, w) = eh(t,X(t,w)) +

Zt

eh(s,X(s,w))

dA(s, w) . A(s, w)

0

M(t, w)A(t, w) = Ph (t, w) + A(t, w)

Zt

eh(s,X(s,w))

dA(s, w) A(s, w)

0

Zt

M(s, ·)dA(s, ·) =

0

Zt

eh(s,X(s,w)) dA(s, w)

0

+

Zt 0

dA(s, w)

Zs

eh(σ,X(σ,w))

dA(σ, w) A(σ, w)

0

Use Fubini’s theorem to evaluate the second integral on the right above and conclude that Ph (t, w) is a martingale.

15. Equivalent For of Itˆo Process

112

(vi) ⇒ (i) is clear if we take h(t, x) = hθ, xi. It only remains to prove that (v) ⇒ (vi). (v) ⇒ (vi). The technique used to prove this is an important one and we shall have occasion to use it again. 109

Step 1. 0 Let h(t, x) = θ1 x1 + θ2 x2 + · · · θd xd = hθ, xi for every (t, x) ∈ [0, ∞) × Rd , θ is some fixed element of Rd . Let   Zt Zt   1   Z(t) = exp hθ, Xt i − hθ, bids − hθ, aθids   2 0

0

We claim that Z(t, ·) is a supermartingale.

Let f : R → R be a C ∞ function with compact support such that f (x) = x in |x| ≤ 1/2 and | f (x)| ≤ 1, ∀x. Put fn (x) = n f (x/n). Therefore | fn (x)| ≤ C|x| for some C independent of n and x and fn (x) converges to x. d P Let hn (x) = θi fn (xi ). Then hn (x) converges to hθ, xi and |hn (x)| ≤ i=1

C ′ |x| where C is also independent of n and x. By (v),   ! Zt Zt   1 ∂h   n Zn (t) = exp hn (t, Xt ) − + L s,w h ds − ha∇x hn , ∇x hn ids   ∂s 2 0

0

is a martingale. As hn (x) converges to hθ, xi, Zn (t, ·) converges to Z(t, ·) pointwise. Consequently E(Z(t)) = E(lim Zn (t)) ≤ lim E(Zn (t)) = 1 and Z(t) is a supermartingale. Step 2. E(exp B sup |X(s, w)|) < ∞ for each t and B. For, let Y(w) = 0≤s≤t

sup |X(s, w)|, Yi (w) = sup |Xi (s, w)| where X = (X1 , . . . , Xd ). Clearly

0≤s≤t

0≤s≤t

Y ≤ Y1 + · · · + Yd . Therefore E(eBY ) ≤ E(eBY 1eBY 2 . . . eBY d).

113 110

The right hand side above is finite provided E(eBY i) < ∞ for each i as can be seen by the generalised Holder’s inequality. Thus to prove the assertion it is enough to show E(eBY i) < ∞ for each i = 1, 2, . . . d with aB′ different from B; more specifically for B′ bounded. Put θ2 = 0 = θ3 = . . . = θd in Step 1 to get u(t) = exp[θ1 X1 (t) −

Zt

1 θ1 b1 (s, ·)ds − θ12 2

0

Zt

a11 (s, ·)ds]

0

is a supermartingale. Therefore ! 1 1 P sup u(s, ·) ≥ λ ≤ E(u(t)) = , ∀λ > 0. λ λ 0≤s≤t (Refer section on Martingales). Let c be a common bound for both b1 and a11 and let θ1 > 0. Then (∗) reads ! 1 2 1 P sup exp θ1 X1 (s) ≥ λ exp(θ1 ct + θ1 ct) ≤ . 2 λ 0≤s≤t Replacing λ by 2

eλθ1 e−ctθ1 −1/2ctθ1 we get !

2

P sup exp θ1 X1 (s) ≥ exp λθ1 ≤ e−λθ1 +θ1 ct+1/2θ1 ct , 0≤s≤t

i.e.

! 2 P sup X1 (s) ≥ λ ≤ e−λθ1 +θ1 ct+1/2θ1 ct , ∀θ1 > 0. 0≤s≤t

Similarly ! 2 P sup −X1 (s) ≥ λ ≤ e−λθ1 +θ1 ct+1/2θ1 tc , ∀θ1 > 0. 0≤s≤t

As (

) ( ) {Y1 (w) ≥ λ} sup X1 (s) ≥ λ ∪ sup −X1 (s) ≥ λ , 0≤s≤t

0≤s≤t

15. Equivalent For of Itˆo Process

114 we get

111 2

P{Y1 ≥ λ} ≤ 2e−λθ1 +θ1 ct+1/2θ1 ct , ∀θ1 > 0. Now we get 1 E(exp BY1 ) = B

Z∞

exp(Bx)P(Y1 ≥ x)dx

2 B

Z∞

1 exp(Bx − xθ1 + θ1 ct + θ12 ct)dx 2

(since Y1 ≥ 0)

0

≤

0

< ∞,

if

B < θ1

This completes the proof of step 2. Step 3. Z(t, w) is a martingale. For |Zn (t, w)| = Zn (t, w)   ! Zt Zt   ∂h 1   n = exp hn (Xt ) − + L s,w hn dx − ha∇x hn , ∇x hn ids   ∂s 2 0 0   Zt     ≤ exp hn (Xt ) − L s,w hn    0

(since a is positive semidefinite and ∂hn /∂s = 0).

Therefore |Zn (t, w)| ≤ A exp(B sup |X(s, w)|) (use the fact that 0st

∂hn ∂2 hn |hn (s)| ≤ C|x| and , are bounded by the same constant). The ∂xi ∂xi ∂x j result now follows from the dominated convergence theorem and Step 2.  112

Remark. In Steps 1, 2 and 3 we have proved that (v) ⇒ (i). The idea of the proof was to express Z(t, ·) as a limit of a sequence of martingales proving first that Z(t, ·) is a supermartingale. Using the supermartingale inequality it was then shown that (Zn ) is a uniformly integrable family proving thereby that Z(t, ·) is a martingale.

115 Step 4. Let h(t, x) ∈ C 1,2 ([0, ∞)×Rd ) such that h(t, x),

∂h ∂h (t, x), (t, x), ∂s ∂xi

∂2 h (t, x) are all dominated by α|x| + β for some suitable scalars α and ∂xi ∂x j β. Let φn be a sequence of real valued C ∞ functions defined on Rd such that    1 on |x| ≤ n φn =   0 on |x| ≥ 2n

and suppose there exists a common bound C for φn ,

∂φn ∂2 φn , (∀n). ∂xi ∂xi ∂x j

Let hn (t, x) = h(t, x)φn (x). By (v) Zhn (t, w) is a martingale. The conditions on the function h and φn ’s show that ! |Zhn (t, w)| ≤ A exp B sup |X(s, w)| 0≤s≤t

where A and B are constants. By Step 2, (Zhn ) are uniformly integrable. Also Zhn (t, ·) converges pointwise to Ph (t, ·) (since hn → h pointwise). By the dominated convergence theorem Ph (t, ·) is a martingale, proving (vi).

16. Itˆo’s Formula Motivation. Let β(t) be a one-dimensional Brownian motion. We have 113 seen that the left integral  t   Z    (*) L 2 β(s, ·)dβ = [β2 (t, ·) − β2 (0, ·) − t]   0

Formally (*) can be written as

dβ2 (t) = 2β(t)dβ(t) + dt. For, on integrating we recover (*). Newtonian calculus gives the result: d f (β(t)) = f ′ (β(t))dβ(t) +

1 ′′ f (β(t))dβ2 (t) + · · · 2

for reasonably smooth functions f and β. If β is of bounded variation, only the first term contributes something if we integrate the above equaP tion. This is because dβ2 = 0 for a function of bounded variation. P For the Brownian motion we have seen that dβ2 → a non zero value, P 3 but one can prove that dβ , . . . converge to 0. We therefore expect the following result to hold: d f (β(t)) ≈ f ′ (β(t))dβ(t) +

1 ′′ f (β(t))d2 β(t). 2

We show that for a one-dimensional Brownian motion X (dβ)3 , sup(dβ)4 , . . . 117

16. Itˆo’s Formula

118 all vanish.

 n  X X  X 3 3  E( (dβ) ) = E  [β(ti+1 ) − β(ti )]  = E[(β(ti+1 ) − β(ti )]3 i=0

i=0

=

n X

0 = 0,

i=1

114

because β(ti+1 ) − β(ti ) is a normal random variable with mean zero and variance ti+1 − ti . Similarly the higher odd moments vanish. Even moments of a normal random variable with mean 0 and variance σ2 are connected by the formula µ2k+2 = σ2 (2k + 1)µ2k , k > 1. So

n X X (dβ)4 = (β(ti+1 ) − β(ti ))4 . i=0

Therefore n X X E( (dβ)4 ) = E([β(ti+1 ) − β(ti ))4 ] i=0 n X

=3

i=0

(ti+1 − ti )2 ;

the right hand side converges to 0 as the mesh of the partition goes to 0. Similarly the higher order even moments vanish. More generally, if β(t, ·) is a d-dimensional Brownian motion then we expect d f (β(t)) ≈ ∇ f (β(t)) · dβ(t) +

1 X ∂2 f dβi dβ j . 2 i, j ∂xi ∂x j

119 However

P

dβi dβ j = 0 if i , j (see exercise below). Therefore

1 d f (β(t)) ≈ ∇ f (β(t)) · dβ(t) + ∆ f (β(t))dβ2 (t). 2 The appearance of ∆ on the right side above is related to the heat equation. Exercise 4. Check that dβi dβ j = δi j dt. 115 (Hint: For i = j, the result was proved earlier. If i , j, consider a partition 0 = t0 < t1 < . . . < tn = t. Let ∆k βi = βi (tk ) − βi (tk−1 ). Then  n 2 X X X   E  ∆k βi ∆k β j  = E(∆2k βi ∆2k β j ) + 2 E[(∆k βi )(∆1 β j )]; k=1

k

k,l

the right side converges to 0 as n → ∞ because ∆k βi and ∆ℓ β j are independent for k , ℓ). Before stating Itˆo’s formula, we prove a few preliminary results. Lemma 1. Let X(t, ·) ∈ I[b, 0] be a one-dimensional Itˆo process. Then X(t, ·) − X(0, ·) =

Zt

b(s, ·)ds

a.e.

0

Proof. exp[θX(t, ·) − θX(0, ·) − θ Therefore E(exp[θ(X(t, ·) − X(0, ·)) − θ

Rt 0

Zt

b(s, ·)ds] is a martingale for each θ.

b(s, ·)ds]) = constant = 1, ∀t.

0

Let W(t, ·) = X(t, ·) − X(0, ·) −

Zt

b(s, ·)ds.

0

Then E(exp θW(t, ·)) = Moment generating function of w = 1, ∀t. Therefore θ(t, ·) = 0 a.e.



16. Itˆo’s Formula

120 116

Remark . If X(t, ·) ∈ I[0, 0] then X(t, ·) = X(0, ·) a.e.; i.e. X(t, ·) is a trivial process. We now state a theorem, which is a particular case of the theorem on page 103. Theorem . If h ∈ C 1,2 ([0, ∞) × Rd ) such that (i) |h(x)| ≤ A|x| + B, ∂h ∂h ∂2 h ∀x ∈ [0, ∞) × Rd , for constants A and B (ii) , also grow , ∂t ∂xi ∂xi ∂x j linearly, then

exp[h(t, β(t, ·) −

Zt 0

! Zt ∂h 1 1 |∇h|2 (s, β(s, ·))ds] + ∆h (s, β(s, ·) − ∂s 2 2 0

is a martingale. Ito’s Formula. Let f ∈ C01,2 ([0, ∞) × Rd ) and let β(t, ·) be a d-dimensional Brownian motion. Then f (t, β(t)) − f (0, β(0)) =

Zt

∂f (s, β(s, ·))ds+ ∂s

0

+

Zt

1 h∇ f (s, β(s, ·)), dβ(s, ·)i + 2

0

Zt

∆ f (s, β(s, ·))ds.

0

where ∆≡

∂2 ∂2 + · · · + . ∂x21 ∂x2d

Proof. Step 1. Consider a (d + 1)-dimensional process defined by X0 (t, ·) = f (t, β(t, ·)), X j (t, ·) = β j (t, ·).

121 We claim that X(t, ·) ≡ (X0 , X1 , . . . , Xd ) is a (d + 1)-dimensional Itˆoprocess with parameters " ! # ∂f 1 b= + ∆ f (s, β(s, ·)), 0, 0, . . . 0 d terms ∂s 2 and

117

  a00 a01 . . . a0d  a   10  a =  .  .  . Id×d    ad0

where

a00 = |∇x f |2 (s, β(s, ·)), ! ∂ a0 j = f (s, β(s, ·)), ∂x j a j0 = a0 j . For, put h = λ f (t, x) + hθ, xi, x = (x1 , . . . , xd ) ∈ Rd , in the previous theorem. Then ∂h ∂f ∂h ∂f = λ , ∆h = λ∆ f, =λ + θ j. ∂s ∂s ∂x j ∂x j Therefore we seen that exp[λ f (t, β(t, ·)) + hθ, β(t, ·)i − λ

Zt

! ∂f 1 + ∆x f (s, β(s, ·)ds ∂s 2

0

1 − λ2 2

Zt

1 |∇ f | (s, β(s, ·))ds − |θ|2 t − λhθ, 2 2

0

Zt

∇( f (s, β(s, ·)))dsi]

0

is a martingale.   Consider (λ, θ)a λθ . We have " # " # P a00 λ + dj=1 a0 j θ j λ a = , θ ρ+θ

  a10    ρ = λ  . . .  .   ad0

16. Itˆo’s Formula

122 Therefore

" # d d X X λ 2 (λ, θ)a = a00 λ + λ a0 j θ j + a0 j θ j + θ j=1

j=1

∂f = λ2 |∇ f |2 + 2λ θ j + |θ|2 . ∂x j 118

Thus (*) reads exp[λ f (t, β(t, ·) + hθ, β(t, ·)i − λ

Zt

1 b0 (s, ·)ds − 2

0

Zt

hα, aαids]

0

is a martingale where α = (λ, θ) ∈ Rd+1 . This proves the claim made above. Step 2. Derine σ(s, ·) = (1, −∇x f (s, β(s, ·))) and let Z(t, ·) =

Zt

hσ(s, ·), dX(s, ·)i where X ≈ (X0 , X1 , . . . , Xd )

0

is the (d + 1)-dimensional Itˆo process obtained in Step 1. Since f ∈ Cb1,2 , Z(t, ·) is an Itˆo process with parameters hσ, bi and σaσ∗ : ∂f + "∂s a aσ∗ = 00 ρ∗

hσ, bi =

1 ∆ f, 2# " # ρ 1 I −∇x f

  0 " #   0 a − hρ, ∇x f i = 00 ∗ =  .  ρ − ∇x f  ..    0

Therefore σaσ∗ = 0. Hence by Lemma 1, Z(t, ·) − Z(0, ·) −

Zt 0

hσ, bids = 0(a.e.),

123

Z(t, ·) =

Zt

dX0 (s) −

Zt

h∇ f (s, β(s, ·)), dβ(s, ·)i

0

0

= f (t, β(t)) − f (0, β(0)) −

Zt

h∇ f ( f, β(s, ·)), dβ(s, ·)i

0

119

Hence Z(0) = 0. Thus f (t, β(t)) − f (0, β(0)) −

Zt

h∇ f (s, β(s, ·))dβ(s, ·)i−

0

−

Zt

! ∂f 1 + ∆x f (s, β(s, ·))ds = 0 a.e. ∂s 2

0

This estabilished Itˆo’s formula.  Exercise. (Itˆo’s formula for the general case). Let φ(t, x) ∈ Cb1,2 ([0, ∞) × Rd ). If X(t, ·) is a d-dimensional Itˆo process corresponding to the parameters b and a, then the following formula holds: φ(t, X(t, w)) − φ(0, X(0, w)) Zt Zt Zt X ∂φ 1 ∂2 = (s, X(s, x))ds + h∇x φ, dXi + ai j ds. ∂s 2 ∂xi ∂x j 0

0

0

This is also written as dφ(s, X(s, w)) = φ s ds + h∇x φ, dXi +

1X ∂2 φ ai j dx. 2 ∂xi ∂x j

To prove this formula proceed as follows.

16. Itˆo’s Formula

124

(i) Take h(t, x) = λφ(t, x) + hθ, xi in (vi) of the theorem on the equivalence of Itˆo process to conclude that Y(t, ·) = (φ(t, X(t, ·)), X(t, ·)) is a (d + 1)-dimensional Ito process with parameters ! ∂φ ′ b = + L s,w φ, b ∂t 120

and

ha∇x φ, ∇x φi, a∇x φ 1×1 1 × d A = a∇x φ a d×1 d × d

(ii) Let σ(t, x) = (1, −∇x φ(t, x)) and Z(t, ·) =

Zt

hσ(s, X(s, ·)), dY(s, ·)i.

0

The assumptions on φ imply that Z is an Itˆo process corresponding to ! ∂φ 1 X ∂2 φ ′ ∗ (hσ, b i, σAσ ) ≡ + ,0 . ai j ∂t 2 ∂xi ∂x j (iii) Use (ii) to conclude that Z(t, ·) =

Zt

hσ, b′ ids

a.e.

0

This is Itˆo’s formula. (iv) (Exercise) Verify that Itˆo’s formula agrees with the formula obtained for the case of Brownian motion. Note. Observe that Itˆo’s formula does not depend on b.

125 Examples.

1. Let β(t) be a one-dimensional Brownian motion. Then d(et φ(β(t)) = et dφ(β(t)) + φ(β(t))d(et ) = et φ′ (β(t))dβ(t) + φ(β(t))et dt+ 1 + φ′′ (β(t))et dt. 2 Rt

2. To evaluate d(e e V(β(s,·))ds u(t, β(t))) where V is a smooth function, 121 put

0

# " # 0 b = 1 V((t, ·)) b2

# 1 0 a= . 0 0

Let X1 (t, ·) = β(t, ·), X = (X1 , X2 ).

X2 (t, ·) =

Zt

V(β(s, ·))ds, b =

"

"

Then exp[θ1 X1 (t, ·) − θ2 X2 (t, ·) − θ1

Zt

b1 (X(s, ·))ds

Zt

haθ, θids]

0

− θ2

Zt

1 b2 (X(s, ·))ds − 2

0

0

exp[θ1 X1 (t, ·) −

θ122 2

t];

the right side is a martingale. Therefore (X1 , X2 ) is a 2-dimensional Itˆo process with parameters b and a and one can use Itˆo’s formula to write  Rt    V(β(s,·))ds 0 d e u(t, β(t)) = d eX2 (t) u(t, β(t)) Rt

= e 0 V(β(s,·)) Rt

+e 0 V(β(s,·))dt

∂ u(t, β(t))dt+ ∂t

Rt ∂ u(t, β(t))d(t) + e 0 V(β(s,·))ds u(t, β(t))dX2 ∂x

16. Itˆo’s Formula

126

1 Rt ∂ + e 0 V(β(s,·))ds u(t, β(t))dt 2 ∂t Rt ∂ 1 ∂2 ∂ = e 0 V(β(s,·))dt u(t, β(t))dt + u(t, β(s))d(t) + u(t, β(t))dt ∂t ∂t 2 ∂x2 +u(t, β(t))V(β(t, ·)dt]. 122

3. Let σi , fi , i = 1, 2, . . . k be bounded and progressively measurable relative to a one-dimensional Brownian motion (Ω, Ft , P). Write Xi (t, ·) =

Zt

σi (s, ·)dβ(s, ·) +

0

Zt

fi (s, ·)ds.

0

Rt Then Xi (t, ·) is an Itˆo process with parameters ( fi (s, ·)ds, σ2i ) and 0

(X1 , . . . , Xk ) is an Itˆo process with parameters  t  Zt Z    B =  f1 (s, ·)ds, . . . , fk (s, ·)ds ,   0

0

A = (Ai j ) where Ai j = σi σ j δi j .

If φ ≡ φ(t, X1 (t) . . . , Xk (t)), then by Itˆo’s formula ∂φ ∂φ ∂φ ds + dX1 + · · · + dXk ∂s ∂x1 ∂xk 1X ∂2 φ + σi σ j δi j ds 2 ∂xi ∂x j ∂φ ∂φ ∂φ = ds + dX1 + · · · + dXk ∂s ∂x1 ∂xk k 1 X 2 ∂φ + σ ds 2 i=1 i ∂xi

dφ =

Exercise. Take σ1 = 1, σ2 = 0, f1 = f2 = 0 above and verify that if φ = eX2 (t,·) u(t, β(t)), then one gets the result obtained in Example 2 above.

127 We give below a set of rules which can be used to calculate dφ in practice, where φ is as in Example 3 above. √ 1. With each dβ associate a term (dt) 2. If φ = φ(t, X1 , . . . , Xk ), formally differentiate φ using ordinary 123 calculus retaining terms upto the second order to get (*) dφ =

1 X ∂2 φ ∂φ ∂φ ∂φ dt + dX1 + · · · + Xk + dXi dX j ∂t ∂x1 ∂xk 2 ∂xi ∂x j

3. Formally write dXi = fi dt + αi dβi , dX j = σ j dβ j + f j dt. 4. Multiply dXi dX j and retain only the first order term in dt., For dβi dβ j substitute δi j dt. Substitute in (*) to get the desired formula. Illustration of the use of Itˆo Calculus. We refer the reader to the section on Dirichlet problem. There it was shown that Z u(x) = u(y)π(x, dy) = E(u(X(τ))) G

satisfies ∆u = 0 in a region G with u = u(X(τ)) on the boundry of G (here τ is the first hitting time).

The form of the solution is given directly by Itˆo’s formula (without having recourse to the mean value property). If u = u(X(t)) satisfies ∆u = 0 then by Itˆo’s formula du(X(t)) = h∇u, dXi.

16. Itˆo’s Formula

128 Therefore u(X(t)) = u(X(0)) +

Zt

h∇u(X(s)), dX(s)i

0

124

Assuming ∇u to be bounded, we see that u(X(t)) is a martingale. By the optional stopping theorem E(u(X(τ)) = u(X(0)) = u(x). Thus Itˆo’s formula connects solutions of certain differential equations with the hitting probabilities.

17. Solution of Poisson’s Equations LET X(t, ·) BE A d-dimensional Brownian motion with (Ω, Ft , P) as 125 1 usual. Let u(x) : Rd → R be such that ∆u = f . Assume u ∈ Cb2 (Rd ). 2 Rt 1 Then L s,w u ≡ ∆u = f and we know that u(X(t, ·)) − f (X(s, ·))ds is 2 0 a (Ω, Ft , P)-martingale. Suppose now that u(x) is defined only on an 1 open subset G ⊂ Rd and ∆u = f on G. We would like to consider 2 Z(t, ·) = u(X(t, ·)) −

Zt

f (X(s, ·))ds

0

and ask whether Z(t, ·) is still a martingale relative to (Ω, Ft , P). Let τ(w) = inf{t, X(t, w) ∈ ∂G}. Put this way, the question is not well-posed because Z(t, ·) is defined only upto time τ(w) for u is not defined outside G. Even if at a time t > τ(w)X(t, w) ∈ G, one needs to know the values of f for t > τ(w) to compute the integral. To answer the question we therefore proceed as follows. Let At = [w : τ(w) > t]. As t increases, At have decreasing measures. We shall give a meaning to the statement ‘Z(t, ·) is a martingale on At ’. Define Z(t, ·) = u(X(τ ∧ t, ·)) −

Zτ∧t 0

129

f (X(s, ·))ds.

17. Solution of Poisson’s Equations

130 Therefore

126

   on At Z(t), Z(t, ·) =   Z(τ, ·), on (At )c .

Since Z(t, ·) is progressively measurable upto time τ, Z(t, ·) is Ft measurable. Theorem . Z(t, ·) is a martingale. Proof. Let Gn be a sequence of compact sets increasing to G such that Gn ⊂ G0n+1 . Choose a C ∞ function φn such that φn = 1 on Gn and 1 support φn ⊂ G. Put un = φn u and fn = ∆un . Then 2 Zn (t, ·) = un (X(t, ·)) −

Zt

fn (X(s, ·))ds

0

is a martingale for each n. Put τn = inf{t : X(t, ·) < Gn } Then Zn (τn ∧ t, ·) is also a martingale (See exercise below). But Zn (τn ∧ t) = Z(τn ∧ t). Therefore Mn (t, ·) = Z(τn ∧ t, ·) is a martingale. Observe that τn ≤ τn+1 and since Gn  G we have τn  τ. Therefore Z(τn ∧t) → Z(τ∧t) (by continuity); also |Mn (t, ·)| ≤ ||u||∞ + || f ||∞ t. Therefore Z(τ ∧ t) = Z(t, ·) is a martingale.  Exercise. If M(t, ·) is a (Ω, Ft , P)-martingale, show that for my stopping time τ, M(τ ∧ t, ·) is also a martingale relative to (Ft ). [Hint: One has to show that if t2 > t1 , Z Z M(τ ∧ t2 , w)dP(w) = M(τ ∧ t1 , w)dP(w), ∀A ∈ Ft1 . A

A

131 The right side = Z M(t1 , w)dP(w) + A∩(τ>t1 )

The left side Z =

127

M(τ, w)dP(w).

A∩(τ
M(t2 , w)dP(w) +

A∩(τ>t1 )

Z

Z

M(τ, w)dP(w).

A∩(τ
Now use optional stopping theorem]. Lemma . Let G be a bounded region and τ be as above. Then E x (τ) < ∞, ∀x ∈ G, where E x = E Px . Proof. Without loss of generality we assume that G is a sphere of radius R2 − |x|2 1 ≥ 0 and satisfies ∆u = −1 in G. By R. The function u(x) = d 2 the previous theorem Zτ∧t u(X(τ ∧ t, ·) + ds 0

is a martingale. Therefore E x (u(X(τ ∧ t, ·))) + E x (τ ∧ t) = u(X(0)) = u(x). Therefore E x (τ ∧ t) ≤ u(x) (since u ≥ 0). By Fatou’s lemma, on letting t → ∞, we obtain E x (τ) ≤ u(x) < ∞. Thus the mere existence of 1  a u satisfying ∆u = 1 helps in concluding that E x (τ) < ∞. 2 Theorem . Let u ∈ Cb2 (G) and suppose that u satisfies (*)

1 ∆u = f in G, 2 u = g on ∂G. Rτ Then u(x) = E x [g] − E x [ f (X(s, ·))ds] solves (*). 0

132

17. Solution of Poisson’s Equations

Remark . The first part of the solutin u(x) is the solution of the homogeneous equation, and the second part accounts for the inhomogeneous term. 128

Proof. Define Z(t, ·) = u(X(τ ∧ t)) −

τ∧t R 0

f (X(s, ·))ds. Then Z is a mar-

tingale. Also |Z| ≤ ||u||∞ + τ|| f ||∞ . Therefore, by the previous Lemma, Z(t, ·) is a uniformly integrable martingale. Therefore we can equate the expectations at time t = 0 and at time t = ∞ to get Zτ u(x) = E x (g) − E x [ f (X(s, ·))ds]. 0



18. The Feynman-Kac Formula WE NOW CONSIDER the modified heat equation ∂u 1 + ∆u + V(x)u(t, x) = 0, 0 ≤ t ≤ T, ∂t 2 where u(T, x) = f (x). The Feynman-Kac formula says that the solution for s ≤ T is given by (*)

(**)

RT

u(s, x) = E s,x (e

V(X(s))dS

s

f (X(T ))).

Observe that the solution at time s depends on the expectation with respect to the process starting at time s. Note . (**) is to be understood in the following sense. If (*) admits a smooth solution then it must be given by (**). We shall not go into the conditions under which the solution exists. Let Rt

Z(t, ·) = u(t, X(t, ·))e

s

V(X(σ,·)dσ)

,

t ≥ s.

By Ito’s formula (see Example 2 of section 16), we get Z(t, ·) = Z(s, ·) +

Zt

Rt

e

s

V(X(σ,·)dσ)

s

h∇u(λ, X(λ)), dX(λ)i,

provided that u satisfies (*). Assume tentatively that V and ∇u are bounded and progressively measurable. Then Z(t, ·) is a martingale. Therefore E s,x (Z(T, ·)) = Ez,x (Z(s, ·)), 133

129

18. The Feynman-Kac Formula

134 or

R

E s,x (u(T, X(T )))e

T s

V(X(σ,·))dσ

= u(s, x).

This proves the result. 130

We shall now remove the condition ∇u is bounded and prove the uniqueness of the solution corresponding to (*) under the assumption that V is bounded above and α

|u(t, x)| ≤ eA+|x| ,

α < 2,

on

[s, T ).

In particular, the Feynman-Kac formula extends the uniqueness theorem for the heat equation to the class of unbounded functions satisfying a growth condition of the form given above. Let φ be a C ∞ function such that φ = 1 on |X| ≤ R, and φ = 0 outside |x| > R + 1. Put uR (t, x) = u(t, x)φ, Rt

ZR (t, x) = uR (t, x)e

s

V(X(σ)dσ)

.

By what we have already proved, ZR (t, ·) is a martingale. Let τR (ω) = inf{t : t ≥ sω(t) ∈ S (0; R) = {|x| ≤ R}}. Then ZR (t ∧ τR , ·)) is also a martingale, i.e. R t∧τ

uR (t ∧ τR , X(t ∧ τR , ·))e

s

R

V(X(σ))dσ

is a martingale. Equating the expectations at time t = s and time t = T and using the fact that uR (t ∧ τR , X(t ∧ τR , ·)) = u(t ∧ τR , X(t ∧ τR , ·))), we conclude that Rτ

u(s, x) = E s,x [u(τR ∧ T, X(τR ∧ T, ·))e R

= E s,x [X(τR ∧T ) f (X(T ))e

T s

s

R ∧T

V(X(s))ds RτR

+ E s,x [X(τR ≤T ) u(τR , X(τR ) e s

V(X(s))ds

]+

V(X(s))ds

]

]

135 131

Consider the second term on the right: Rτ

|E s,x [X(τR ≤T ) u(τR , X(τR ))e Rα

s

R

V(X(s))ds

]|

≤ A′ e P[R ≤ T ]

(where A′ is a constant given in terms of the constants A and T and the bound of V) α = A′ eR P[ sup |X(σ)| ≥ R]. s≤σ≤T

2

P[ sup |X(σ)| ≥ R] is of the order of e−c(T )R and since α < 2, the s≤σ≤T

second term on the right side above tends to 0 as R → ∞. Hence, on letting R → +∞, we get, by the bounded convergence theorem, RT

u(s, x) = E s,x [ f (X(T ))e

s

V(X(s))ds

]

Application. Let β(t, ·) be a one-dimensional Brownian motion. Recall 4 (Cf. Reflection principle) that P{ sup |β(s)| ≤ 1} is of the order of e − π 0≤s≤t π2 t . The Feynman-Kac formula will be used to explain the occurance 8 π2 π2 λ2 of the factor in the exponent. First observe that = where λ is 8 8 2 the first positive root of Cos λ = 0. Let τ(w) = inf{t : |β(t)| ≥ 1}. Then P{ sup |β(s, ·)| ≤ 1} = P{τ ≥ t}. 0≤s≤t

Let φ(x) = E x [eλτ ], λ < 0. We claim that φ satisfies (*)

1 ′′ φ + λφ = 0, |x| < 1, 2 φ = 1, |x| = 1.

132

18. The Feynman-Kac Formula

136

Assume φ to be sufficiently smooth. Using Itˆo’s formula we get 1 d(eλt φ(β(t)) = eλt φ′ (β(t))dβ(t) + [λφ(β(t)) + φ′′ (β(t))]eλt dt. 2 Therefore λt

e φ(β(t)) − φ(β(0)) =

Zt

eλs φ′ (β(s))dβ(s)+

0

+

Zt

1 [λφ(β(s)) + φ′′ (β(s))]eλs ds, 2

0

i.e. λt

e φ(β(t)) − φ(β(0)) −

Zt

1 [λφ(β(s)) + φ′′ (β(s))]eλs ds 2

0

is a martingale. By Doob’s optional sampling theorem we can stop this martingale at time τ, i.e. λ(t∧τ)

e

Zt∧τ 1 φ(β(t ∧ τ)) − φ(β(0)) − [λφ(β(s)) + φ′′ (β(s))]eλs ds 2 0

is also a martingale. But for s ≤ t ∧ τ, 1 λφ + φ′′ = 0. 2 Thus we conclude that φ(β(t ∧ τ))eλ(τ∧t) is a martingale. Since λ < 0 and φ(β(t ∧ τ)) is bounded, this martingale is uniformly integrable. Therefore equating the expectation at t = 0 and t = ∞ we get (since φ(β(τ)) = 1) φ(x) = E x [eλτ ]. 133

By uniqueness property this must be the solution. However (*) has a solution given by √ Cos( (2λx)) (x) = √ . Cos( (2λ))

137 Therefore E0 [eλτ ] =

(1)

1 √ (λ < 0), Cos( (2λ))

If F(t) = P(τ ≥ t), then Z∞

eλt dF(t) = E0 (eλτ ).

0

A theorem on Laplace transforms now tells us that (1) is valid till 1 π2 we cross the first singularity of √ . This occurs at λ = . By 8 Cos( (2λ)) the monotone convergence theorem E0 [eτπ R∞

2 /8

] = +∞

π2 π2 and diverges for λ ≥ . 8 8 0 R∞ π2 is the supremum of λ for which eλt dF(t) converges, i.e. sup Thus 8 0 [λ : E0 (eλτ )] exists, i.e. the decay rate is connected to the existence or the non existence of the solution of the system (*). This is a general feature and prevails even in higher dimensions. Hence

eλt dF(t) converges for λ <

19. An Application of the Feynman-Kac Formula. The Arc Sine Law. LET β(t, ·) BE THE one-dimensional Brownian motion with β(0, ·) = 0. 134 Define Zt 1 ξt (w) = X[0,∞) (β(s, w))ds. t 0

ξt (w) is a random variable and denotes the fraction of the time that a Brownian particle stays above the x-axis during the time interval [0, t]. We shall calculate P[w : ξt (w) ≤ a] = Ft (a) 1 Brownian Scaling. Let Xt (s) = √ β(ts). Then Xt is also a Brownian t motion with same distribution as that of β(s). We can write

ξt (w) =

Z1

X[0,∞) (Xt (s, w))ds.

0

The ξt (w) = time spent above the x-axis by the Brownian motion Xt (s) in [0, 1]. Hence Ft (a) is independent of t and is therefore denoted 139

19. An Application of the Feynman-Kac Formula....

140

by F(a). Suppose we succeed in solving for F(a); if, now, ξt∗ (w)

=

Zt

X[0,∞) (β(s))ds = tξt ,

0

then the amount of time β(s, ·) > 0 in [0, t] is t (amount of time Xt (s) > 0 in [0, 1]). Hence we can solve for P[ξt∗ (w) ≤ a] = Gt (a). Clearly the solution of Gt is given by Gt (a) = F(a/t). It is clear that

135

   0 if F(a) =   1 if

a ≤ 0, a ≥ 1.

Hence it is enough to solve for F(a) in 0 ≤ a ≤ 1. Let uλ (t, x) = E x [e−(λ

Rt 0

X[0,∞) (β(s,w))ds)

Then −ξ(w))

u1 (t, 0) = E[e

]=

Z1

]

e−tx dF(x).

0

Also note that uλ (t, x) is bounded by 1, if λ ≥ 0. By the Feynman1 1 Kac formula (appropriately modified in case ∆ is replaced by − ∆) 2 2 u1 (t, x) satisfies ∂u 1 ∂2 u = − u, x > 0, ∂t 2 ∂x2 1 ∂2 u = , x < 0, 2 ∂x2

(*)

and u(0, x) = 1. Let φα (x) = α

Z∞ 0

u(t, x)e−αt dt,

α > 0,

where

u = u1 ,

141

=

Z∞

u(t, x)(−de−αt ).

0

A simple integration by parts together with (*) gives the following system of ordinary differential equations for φα : 1 + (α + 1)φα = α, x > 0, − φ′′ 2 α 1 − φ′′ + αφα = α, x < 0. 2 α These have a solution √ √ α + Aex (2(α+1)) + Be−x (2(α+1)) , φα (x) = α+1 √ √ = 1 + Cex (2α) + De−x (2α) , x < 0.

x > 0,

136

However u is bounded by 1 (see definition of uλ (t, x)). Therefore φα is also bounded by 1. This forces A = D = 0. We demand that φα and dφα should match at x = 0. (Some justification for this will be given dx later on). Lt φα (x) = Lt φα (x) x→0+

x→0−

gives

α + B = C + 1. 1+α √ √ dφα Similarly we get −B (2(α + 1)) = C (2α) by matching . Solvdx ing for B and C we get √ α √ √ , B= (1 + α)( α + (α + 1)) C=√ Therefore

1 √ √ . (1 + α)( α + (α + 1))

√ α α φα (0) = +B= √ , α+1 (α + 1)

19. An Application of the Feynman-Kac Formula....

142 i.e.

Z∞

E[e−tξt αe−αt ]dt = √

0

√

α . (α + 1)

Using Fubini’s theorem this gives E[

√ α α ]= ( ), α+ξ α+1

or √ 1 1 E[ ]= ( ), 1 + ξ/a 1 + (1/α)

i.e.

Z1 0

dF(x) 1 . =√ 1 + γx (1 + γ)

137

This can be inverted to get dF(x) =

2 dx √ . π (x(1 − x))

(Refer tables on transforms or check directly that 2 π

Z1 0

1 dx 1 √ =√ 1 + βx (x(1 − x)) (1 + β)

by expanding the left side in powers of (β). Therefore F(a) = Hence Gt (a) =

2 π

2 π

arcsin

√ ( a),

0 ≤ a ≤ 1.

√ arcsin ( ( at )), 0 ≤ a ≤ t, i.e.

P[ξt ≤ a] =

2 π

arcsin

√ a ( ( )), t

0 ≤ a ≤ t.

This result goes by the name of arc sine law for obvious reasons. We now give some justification regarding the matching conditions used above.

143 The equation we solved was 1 αφ − φ′′ + Vφ = f 2 where φ was bounded V ≥ 0. Suppose we formally use Itˆo’s formula to calculate Rt  d φ(β(t)e− 0 (α+V)(β(s)ds) ) = e−

Rt 0

(α+V)(β(s,·))ds

[− f (β(s, ·))dt +

dφ (β(t))dβ(t)] dX

(see Example 2 of Itˆo’s formula). Therefore −

(Z(t, ·) = φ(β(t, ·))e

Rt

(α+V)(β(s,·))ds 0

+

Z

0

t

f β(s, ·) exp(−

Zs

(α + V)dσ)ds

0

is a martingale. Since φ, f are bounded and V ≥ 0, |Z(t, ·)| ≤ ||φ||∞ + || f ||∞

Z∞

138

e−αs ds ≤ ||φ||∞ + C|| f ||∞ .

0

Therefore Z(t, ·) is uniformly integrable. Equating the expectations at time 0 and ∞ gives (*)

φ(0) = E0

Z∞

R

[ f (β(s, ·))e−αs−

s 0

V(β(σ)dσ)

]ds.

0

This is exactly the form obtained by solving the differential equations. In order to use Itˆo’s formula one has to justify it. If we show that Itˆo’s formula is valid for functions having a discontinuity in the second derivative, (*) will be a legitimate solution and in general there is no reason why the second derivatives (or higher derivatives) should be dφ matched. This partially explains the need for matching φ and only. dx

19. An Application of the Feynman-Kac Formula....

144

Proposition . Let β(t, ·) denote a one-dimensional Brownian motion. Suppose φ ∈ Cb1 and satisfied 1 αφ − φ′′ + Vφ = f, 2 Then φ(β(t)) −

Zt

f (β(s))ds

0

is a martingale. Proof. Let (φǫ ) ∈ Cb2 such that αφǫ − 21 φ′′ ǫ + Vφǫ + Vφǫ = fǫ and such that (i) φǫ converges to φ uniformly on compact sets, (ii) φ′ǫ converges to ′′ φ′ uniformly on compact sets, (iii) φ′′ ǫ converges pointwise to φ except at 0. We may suppose that the convergence is bounded.  139

Claim.

Rt

fǫ (β(s))ds converges to

0

Rt

f (β(s))ds a.e. As fǫ (β(s)) converges

0

to f (β(s)) except when β(s) = 0, it is enough to prove that (*) P[w: Lebesgue measure (s : β(s) = 0) > 0]= 0. Let X{0} denote the indicator function of {0}. Then E

Zt

X{0} (β(s))ds =

0

Zt

EX{0} (β(s))ds = 0.

0

Thus (*) holds and establishes the claim. Now φǫ (β(t)) −

Zt

fǫ (β(s))ds

0

is a uniformly bounded martingale converging to

φ(β(t)) −

Zt 0

f (β(s))ds.

145 Therefore φ(β(t)) −

Zt 0

is a martingale.

f (β(s))ds

20. Brownian Motion with Drift LET Ω = C[0, ∞; Rd ], F = BOREL σ-FIELD of Ω, {X(t, ·)} ≡ Brow- 140 nian motion, Ft = σ[X(s, ·) : 0 ≤ s ≤ t], P x ≡ probability measure on Ω corresponding to the Brownian motion starting at time 0 at x. F = σ( U Ft ). Let b : Rd → Rd be any bounded measurable funcit≥0

ton. Then the map (s, w)| → b(w(s)) is progressively measurable and Zt Zt 1 Z(t, ·) = exp[ hb(X(s, ·)), dX(s, ·)i − |b(X(s, ·))|2 ds] 2 0

0

is a martingale relative to (Ω, Ft , P x ). Define Qtx on Ft by Z t Q x (A) = Z(t, ·)dP x , A

i.e. Z(t, ·) is the Radon-Nikodym derivative of Qtx with respect to P x on Ft . Proposition .

(i) Qtx is a probability measure.

(ii) {Qtx : t ≥ 0} is a consistent family on ∪ Ft , i.e. if A ∈ Ft1 and t2 ≥ t1 then Qtx 1(A) = Qtx 2(A).

t≥0

Proof. Qtx being an indefinite integral, is a measure. Since Z(t, ·) ≥ 0, Qtx is a positive measure. Qtx (Ω) = E x (Z(t, ·)) = E x (Z(0, ·)) = 1. This proves (i). (ii) follows from the fact that Z(t, ·) is a martingale. 147

20. Brownian Motion with Drift

148 If A ∈ Ft , we define

Q x (A) = Qt(A) x . 141

The above proposition shows that Q x is well defined and since (Ft ) S is an increasing family, Q x is finitely additive on Ft .  t≥0

S Exercise. Show that Q x is countably additive on the algebra Ft . t≥0 S Then Q x extends as a measure to F = σ( Ft ). Thus we get a t≥0

family of measures {Q x : x ∈ Rd } defined on (Ω, F ). Proposition . If s < t then

Q x (Xt ∈ A|Fs ) = QX(s) (X(t − s) ∈ A) a.e. Definition . If a family of measures {Q x } satisfies the above property it is called a homogeneous Markov family. Proof. Let B ∈ Fs . Therefore B ∩ Xt−1 (A) ∈ Ft and by definition, Z Q x ((X(t) ∈ A) ∩ B)) = Z(t, ·)dP x B∩Xt−1 (A)

E Px (Z(t, ·)χ(w) B χA (X(t, ·))) Z(t, ·) Z(s, ·)χ(w) = E Px (E Px ( B χA (X(t, ·))|F s ) Z(s, ·) Z(t, ·) = E Px ([χB Z(s, ·))E Px ( χA (χ(t, ·))]|Fs ) Z(s, ·) (since B ∈ Fs and Z(s, ·) is Fs -measurable) (1)

Z(t, ·) χA (χ(t, ·))|Fs )] . . . Z(s, ·) Zt Zt 1 Qx Px = E [χB E (exp[ hb, dXi − |b|2 ]χA (X(t, ·))|Fs )] 2 = E Qx [χB E Px (

s

0

149

Qx

= E [χB E

PX(s)

Zt−s Zt−s 1 |b|2 ]XA (X(t − s))] (exp[ hb, dXi 2 0

0

(by Markov property of Brownian motion) Qx

= E (χB E

Qt−s X(s)

(χA (χ(t − s)))

(since

dQt−s X(s) dPX(s)

= Z(t − s, ·)

= E Qx (XB E QX(s) χA (X(t − s, ·))  142 The result follows from definition. Let b : [0, ∞] × Rd → Rd be a bounded measurable function, P s,x the probability measure corresponding to the Brownian motion starting at time s at the point x. Define, for t ≥ s, Zt Z s,t (w) = exp[ hb(σ, X(σ, w)), dX(σ, w)i s

1 − 2

Zt

|b(σ, X(σ, w))|2 dσ]

s

(i) Z s,t is a martingale relative to (Fts , P s,x ). R (ii) Define Qts,x by Qts,x (A) = Z s,t dP s,x , ∀A ∈ Fts .

Exercise.

A

Show that

Qts,x

is a probability measure on Fts .

(iii) Qts,x is a consistent family. (iv) Q s,x defined on U Fts by Q s,x |Fts = Qts,x is a finitely additive set t≥s

function which is countably additive.

(v) The family {Q s,x : 0 ≤ s < ∞, x ∈ Rd } is an inhomogeneous Markov family, i.e. Q s,x (X(t, ·) ∈ A|Fσs ) = Qσ,X(σ,·) (X(t, ·) ∈ A), ∀s < σ < t, A ∈ Fts .

20. Brownian Motion with Drift

150

[Hint: Repeat the arguments of the previous section with obvious 143 modifications]. Proposition . Let τ be a stopping time, τ ≥ s. Then Q s,x [X(t, ·) ∈ A|Fτs ] = Qτ,Xτ (·) (X(t, ·) ∈ A) on τ(w) < t, = χA (X(t, ·)) on τ(w) ≥ t.

Proof. Let B ∈ Fτs and B ⊂ {τ < t} so that B ∈ Fts . E Qs,x (χB χA (Xt )) = E Ps,x (Z s,t χB χA (Xt )) = E Ps,x [E Ps,x (Zτ,t Z s,τ χB χA (Xt )|Fτs )] (since Z satisfies the multiplicative property) = E Ps,x (Z s,τ χB E Ps,x (Zτ,t χA (Xt )|Fτs )] (since Z s,τ is Fτs -measurable) = E Ps,x [Z s,τ XB E Pτ ,Xτ (Zτ,t χA (Xt ))]

(*)

(by strong Markov property). Now dQ s,x = Z s,t . dP s,x Fts

144

so that the optional stopping theorem, dQ s,x = Z s, on {τ < t}, ∀x. dP s,x Fτs Putting this in (*) we get

E Qs,x [XB χA (Xt )] = E Ps,x [Z s,τ χB E Qτ,Xτ (χA Xt ) ]. Observe that χB E Qτ,Xτ (χA (Xt )) is Fτs -measurable to conclude the first part of the proof. For part (ii) observe that Xt−1 (A) ∩ {τ ≥ t} ∩ {τ ≤ k}

151 is in Fks if k > s, so that Xt−1 (A) ∩ {τ ≥ t} ∈ Fτs . Therefore E Qs,x (Xt ∈ A ∩ (τ ≥ t))|Fτs ) = χA (Xt )χ{τ≥t} , or E Qs,x [(Xt ∈ A)|Fτs ] = χA (Xt ) if

τ ≥ t. 

Proposition . Let b : [0, ∞) × Rd → Rd be a bounded measurable function, f : Rd → R any continuous bounded function. If ∂u 1 + ∆u + hb(s, x), ∆ui = 0, ∂s 2 u(t, x) = f (x)

0 ≤ s ≤ t,

has a solution u, then u(s, x) =

Z

f (Xt )dQ s,x .

Ω

Remark . b is called the drift. If b = 0 and s = 0 then we recover the 145 result obtained earlier. With the presence of the drift term, the result is the same except that instead of P s,x one has to use Q s,x to evaluate the expectation. Proof. Let Y(σ, ·) =

Zσ s

1 hb(θ, X(θ, ·)), dX(θ, ·)i − 2

Zσ

|b(θ, X(θ, ·)|2 dθ.

s

Step 1. (X(σ, ·)−X(s, ·), Y(σ, ·)) is a (d+1)-dimensional Itˆo process with parameters 1 (0, 0, . . . , 0, − |b(σ, X(σ, ·))|2 ) and 2

20. Brownian Motion with Drift

152 d terms

"

I bd×1 a = d×d ∗ b1×d b

#

Let λ = (λ1 , . . . , λd ). We have to show that µ exp[µY + 2

Zσ

|b(σ, X(σ, ·)|2 dσ + hλ, X(σ, ·) − X(s, ·)i−

s

−

1 2

Zσ

hλ, λi + 2µhλ, b + µ2 |b(σ, ·)|2 dσ]

s

is a martingale, i.e. that exp[hλ, X(σ, ·) − X(s, ·)i + µ

Zσ s

1 hb, dXi − 2

Zσ s

|λ + bµ|2 dρ].

is a martingale; in other words that Zσ Zσ 1 exp[ hλ + bµ, dXi − |λ + bµ|2 dρ] 2 s

146

s

is a martingale. But this is obvious because Z(σ, ·) =

Zσ s

hλ + µb, dXi

is a stochastic integral and hence an Itˆo process with parameters (0, |λ + µb|2 ). (Refer to the section on vector-valued Itˆo process). Step 2. Put φ(σ, X(σ, ·), Y(σ, ·)) = u(σ, X(σ, ·))eY(σ,·) . By Itˆo formula, dφ = eY

∂u 1X π2 φ dt + eY h∇u, dXi + ueY dY + ai j , ∂t 2 ∂zi ∂z j

where z = (x, y), or dφ = eY [

∂u µ 1 dt + h∇u, dXi + uhb, dXi − |b|2 dt + ∇udt + hb, ∇uidt+ ∂t 2 2

153 1 + u|b|2 dt] 2 Y = e [h∇u, dXi + uhb, dXi]. Therefore φ is an Itˆo process and hence a martingale. Therefore E(φ(t, ·)) = E(φ(s, ·)) u(s, x) = E

P s,x

R

[( f (X(t))e

t hb,dXi− 12 s

Rt 0

|b|2 dθ

]

= E Qs,x [ f (X(t))], which proves the theorem.  Alternate Proof.

147

Exercise . Let Y(σ, ·) be progressively measurable for σ ≥ s. Then Y(σ, ·) is a martingale relative to (Q s,x , Fts ) if and only if Y(σ)Z s,σ is a martingale relative to (P s,x , Fts ). Now for any function θ which is progressively measurable and bounded, Zt Zt 1 exp[ hθ, dXi − |θ|2 dσ] 2 s

s

is a martingale relative to (Ω, Fts , P s,x ). In particular let θ be replaced by θ + b(σ, w(σ)). After some rearrangement one finds that Xt is an Itˆo process with parameters b, I relative to Q s,x . Therefore u(t, Xt ) −

Zt s

! ∂u 1 + hb, ∇ui + ∇u dσ ∂σ 2

is a martingale relative to Q s,x . But ∂u 1 + hb, ∇ui + ∇u = 0. ∂σ 2

20. Brownian Motion with Drift

154 Therefore

E Qs,x (u(t, X(t)) = u(s, X). We have defined Q s,x by using the notion of the Radon-Nikodym derivative. We give one more relation between P and Q. Theorem . Let T : C([s, ∞), Rd ) → C([s, ∞), Rd ) be given by TX = Y

where

Y(t) = X(t) −

Zt

b(σ, X(σ))dσ.

s

(b is as before). Then Q s,x T −1 = P s,x . 148

Proof. Define Y(t, w) = X(t, T w) where X is a Brownian motion. We prove that Y is a Brownian motion with respect to Q s,x . Clearly Y is progressively measurable because T is (Ft − Ft )-measurable for every t, i.e. T −1 (Ft ) ⊂ Ft and X is progressively measurable. Clearly Y(t, w) is continuous ∀ w. We have only to show that Y(t2 ) − Y(t1 ) is Q s,x independent of Fts1 and has distribution N(0; (t2 − t1 )I) for each t2 > t1 ≥ s. But we have checked that 1 exp[hθ, Xt − xi − |θ|2 (t − s) − 2

Zt

hθ, bidσ]

s

is a martingale relative to Q s,x . Therefore 1 E Qs,x (exphθ, Yt2 − Yt1 i|Fts1 ) = exp( |θ|2 (t2 − t1 )), 2 showing that Yt2 − Yt1 is independent of Fts1 and has normal distribution N(0; (t2 − t1 )I). Thus Y is a Brownian motion relative to Q s,x . Therefore Q s,x T −1 = P s,x . 

21. Integral Equations Definition . A function b : Rd → Rd is said to be locally Lipschitz if 149 given any x0 ∈ Rd there exists an open set U0 containing x0 such that b|U0 is Lipschitz. Exercise 1. b is locally Lipschitz iff b|K is Lipschitz for every compact set K i.e. iff b|K is Lipschitz for every closed sphere K. Exercise 2. Every locally Lipschitz function is continuous. Theorem . Let b : Rd → Rd be locally Lipschitz and X : [0, ∞) → Rd continuous. Then (i) the equation Y(t) = X(t) +

Zt

b(Y(s))ds

(∗)

0

has a continuous solution near 0, i.e. there exists an ǫ > 0 and a continuous function Y : [0, ǫ] → Rd such that the above equation is satisfied for all t in [0, ǫ]. (ii) (Uniqueness) If Y1 , Y2 are continuous solutions of the above equation in [0, T ], then Y1 = Y2

on

[0, T ].

Proof. (ii) (Uniqueness) Let f (t) = |Y1 (t) − Y2 (t)|. As Y1 , Y2 are continuous, there exists a k > 0 such that |Y1 (t)|, |Y2 (t)| ≤ k for all t in 155

21. Integral Equations

156

150

[0, T ]. Choose C such that |b(x) − b(y)| ≤ C|x − y| for |x|, |y| ≤ k and Rt n C sup f (t). Then f (t) ≤ C and f (t) ≤ C f (s)ds so that f (t) ≤ (ct) n! for 0≤t≤T

0

all n = 1, 2, 3, . . .. Thus Y1 (t) = Y2 (t), proving uniqueness. (i) (Existence) We can very well assume that X(0) = 0. Let a = inf{t : |X(t)| ≥ 1}, M > sup{|b(x)| : |x| ≤ 2},

α = inf{a,

1 }, M

C , 0, a Lipschitz constant, so that |b(x) − b(y)| ≤ C|x − y| for all |x|, |y| ≤ 2. Define the iterations Y0 , Y1 , . . . by Y0 (t) = X(t),

Yn+1 (t) = X(t) +

Zt

b(Yn (s))ds

0

for all t ≥ 0. By induction, each Yn is continuous. By induction again, |Yn (t) − X(t)| ≤ Mt for all n, 0 ≤ t ≤ α. Again, by induction |Yn+1 (t) − (Ct)n+1 Yn (t)| ≤ M C (n+1)! for 0 ≤ t ≤ α. Again, by induciton |Yn+1 (t) − Yn (t)| ≤ M (Ct)n+1 C (n+1)!

for 0 ≤ t ≤ α. Thus Yn (t) converges uniformly on [0, α] to a continuous function Y(t) which is seen to satisfy the integral equation.  Remark . Let X : [−δ, ∞) → Rd be continuous where δ > 0. Then a similar proof guarantees that the equation (*) has a solution in [−ǫ, ǫ] for some ǫ > 0. Define B(X) = sup{t : (∗) has a solution in [0, t]}. The theorem above implies that 0 < B(X) ≤ ∞. B(X) is called the exploding time. Remark . If b is, in addition, either bounded or globally Lipschitz, B(X) = ∞ for every continuous X : [0, ∞) → Rd . 151

Example. Let b(y) = y2 , X(t) = 0. The equation Y(t) = x0 +

Zt 0

b(y(s))ds

157 with x0 > 0 has a solution Y(t) =

1 x0

1 1 , ∀t < ; x0 −t

the solution explodes at t = x−1 0 . Proposition . If B(w) < ∞,

then

Lt |y(t)| = +∞.

t↑B(w)

Proof. Suppose that lim |y| = R < ∞. Let (tn ) be a sequence increast→B(w)

ing to B(w) such that |y(tn )| ≤ R + 1, ∀n. Let τn = inf{t ≥ tn : |y(t) − y(−n )| ≥ 1}. Then 1 = |y(τn ) − y(tn )|

≤ w(τn ) − w(tn )| + (τn − tn ) sup |b(λ)| . . . , (1) λ ∈ S (y(tn ), 1).

Since (tn ) is bounded, we can choose a constant M such that |w(tn ) − w(t)| <

1 2

if

|t − tn | ≤ M.

Then using (1), τn − tn ≥ inf{M, (2 sup |b(λ)|)−1

where

λ ∈ S (Y(tn ); 1)

Therefore τn − tn ≥ inf(M, (2 sup |b(λ)|)−1 , λ ∈ S (0; R + 2)) = α(say)∀ n. 152

21. Integral Equations

158

Chose n such that τn > B(w) > tn . Then y is bounded in [tn , B(w)] and hence it is bounded in [0, B(w)). From the equation y(t) = X(t) +

Zt

b(y(s))ds

0

one then gets that

Lt y(t) exists. But this is clearly a contradiction

t→B(w)

since in such a case the solution exists in [0, B(w) + ǫ) for suitable ǫ, contradicting the definition of B(w). Thus lim |y(t)| = +∞

t→B(w)

and hence lim |y(t)| = +∞.

t→B(w)

 Corollary . If b is locally Lipschitz and bounded, then B(X) = ∞ for all X in C([0, ∞), Rd ). Proof. Left as an exercise.



Proposition . Let b : Rd → Rd be locally Lipschitz and bounded. Define T : C([0, ∞), Rd ) → C([0, ∞), Rd ) by T X = Y where Y(t) = X(t) +

Zt

b(Y(s))ds.

0

Then T is continuous.

153

Proof. Let X, X ∗ : [0, ∞) → Rd be continuous, K > 0 be given. Let Y0 , Y1 , . . . , Y0∗ , Y1∗ , . . . be the iterations for X, X ∗ respectively. Then |Yn (t) − X(t)| ≤ K||b||∞ for 0 ≤ t ≤ K, n = 0, 1, 2, 3, . . ., so that we can find R such that |Yn (t)|, Yn∗ (t)| ≤ R for 0 ≤ t ≤ k, n = 0, 1, 2, . . ., Let C ≥ 1 be any Lipschitz constant for the function b on |x| ≤ R. Then |Yn (t) − Yn∗ (t)| ≤ sup |X(t) − X ∗ (t)| · (1 + Ct + 0≤t≤K

(Ct)2 + 2!

159 + ··· +

(Ct)n ) n!

for

0 ≤ t ≤ K,

n = 0, 1, 2, 3, . . . .

A b is bounded, Yn converges uniformly to Y on [0, K]. Letting n → ∞, we get sup |(T X)t − T X ∗ )t| ≤ eck sup |X(t) − X ∗ (t)|, . . . (2)

0≤t≤K

0≤t≤K

where c depends on sup |X(t)|, sup |X ∗ (t)|. The proof follows by (2). 0≤t≤K

0≤t≤K



22. Large Deviations LET Pǫ BE THE Brownian motion starting from zero scaled to Brown- 154 ∆ ian motion corresponding to the operator ǫ . More precisely, let 2 ! A Pǫ (A) = P √ ǫ where P is the Brownian motion starting at time 0 at the point 0. Interpretation 1. Let {Xt : t ≥ 0} be Brownian motion with X(0) = x. Let Y(t) = X(ǫt), ∀t ≥ 0. Then Pǫ is the measure induced by the process Y(t). This amounts to stretching the time or scaling time. √ Interpretation 2. Let Y(t, ·) = ǫX(t, ·). In this case also Pǫ is the measure induced by the process Y(t, ·). This amounts to ‘looking at the process from a distance’ or scaling the length. Exercise. Make the interpretations given above precise. (Hint: Calculate (i) the probability that X(ǫt) ∈ A, and (ii) the probability √ that ǫX(t, ) ∈ A). Problem. Let 1 I(w) = 2

Z1

|w(t)| ˙ 2 dt

0

if w(0) = 0, w absolutely continuous on [0, 1]. Put I(w) = ∞ otherwise. 161

22. Large Deviations

162 We would like to evaluate

Z

e

F(w) ǫ

dPǫ (w)

Ω

155

for small values of ǫ. Here F(w) : C[0, 1] → R is assumed to the a bounded and continuous function. Theorem . Let C be any closed set in C[0, 1] and let G be any open set in C[0, 1]. Then Z lim sup ∈ log Pǫ (C) ≤ − I(w), ǫ→0

w∈C

lim inf ǫ log Pǫ (G) ≥ − inf I(w). w∈G

ǫ→0

Here Pǫ (G) = Pǫ (π−1 G) where π : C[0, ∞) → C[0, 1] is the canonical projection. Significance of the theorem . If 1. dPǫ = e then Pǫ(A) =

Z

−I(w) ǫ

e

,

−I(w) ǫ

dPǫ

A

is asymptotically equivalent to 1 exp[− inf I(w)]. ǫ w∈A 2. If A is any set such that inf I(w) = inf I(w),

w∈A0

w∈A

then by the theorem Lt log Pǫ (A) = inf I(w).

ǫ→0

w∈A

163 156

Proof of the theorem. Lemma 1. Let w0 ∈ Ω with I(w0 ) = ℓ < ∞. If S = S (w0 ; δ) is any sphere of radius δ with centre at w0 then lim ∈ log Pǫ (S ) ≥ −I(w0 ). ǫ→0

Proof. P(S ) = =

Z

Z

χ(w) S (w0 ,δ) dPǫ 0) χ(w−w S (0;) dPǫ

! w = χS (0;δ) (λw)dP √ , where λ(w) = w − w0 , ǫ Z √ = χS (0;δ) (λ( ǫw))dP(w)   1 Z Z1 Z  √ 1   |w˙ 0 |2 dσ dP(w) = χS (0,δ) ( ǫw) exp  hw0 , dXi −   2 Z

0

0

 1  Z Z  √   = χS (0;δ) ( ǫw) exp  hw˙ 0 , dXi − I(w0 ) dP(w)   0

=

Z

  Z1   1  1  χS (0;δ) (w) exp − hw˙ 0 , dXi − I(w0 ) dPǫ (w)  ǫ  ǫ 0

=e

−I(w0 ) ǫ

1 Pǫ (S (0; δ)) Pǫ (S (0; δ))

Z

S (0;δ)

  Z1   1   hw˙ 0 , dXi dPǫ exp −  ǫ 

 Z  −I(w0 ) 1  1 ǫ ≥e Pǫ (S (0; δ))e  −  ǫ Pǫ (S (0; δ))

0

Z1

S (0;δ) 0

by Jensen’s inequality,

=e =e

−I(w0 ) ǫ −I(w0 ) ǫ

   hw˙ 0 , dXidPǫ  

Pǫ (S (0, δ))e0 (use Pǫ (w) = Pǫ (−w) if w ∈ S (0; δ)) Pǫ (S (0, δ)).

22. Large Deviations

164 157

Therefore Pǫ (S (w0 ; δ)) ≥ e

−I(w0 ) ǫ

δ P(S (0; √ )) ǫ

or, δ ǫ log Pǫ (S (w0 ; δ)) ≥ −I(w0 ) + ǫ log P(S (0; √ )); ǫ let ǫ → 0 to get the result. Note that the Lemma is trivially satisfied if I(w0 ) = +∞.  Proof of Part 2 of the theorem. Let G be open, w0 ∈ G; then there exists δ > 0 with S (w0 , δ) ⊂ G. By Lemma 1 lim ∈ log Pǫ (G) ≥ lim ∈ log Pǫ (S (w0 ; δ)) ≥ −I(w0 ).

ǫ→0

ǫ→0

Since w0 is arbitrary, we get lim ∈ log Pǫ (G) ≥ − inf{I(w0 ) : w0 ∈ G}. For part 1 we need some more preliminaries. Lemma 2. Let (wn ) ∈ C[0, 1] be such that wn → w uniformly on [0, 1], I(wn ) ≤ α < ∞. Then I(w) < α, i.e. I is lower semi-continuous. Proof. n Step 1. w is absolutely continuous. Let {(x′i , x′′ i )}i=1 be a collection of mutually disjoint intervals in [0, 1]. Then n X i=1

|wm (x′i ) − wm (x′′ i )| ≤

n X i=1|

′′

Zxi ′ 1/2 |x′′ [ |wm |2 ]1/2 i − xi | x′i

(by H¨older’s inequality)

165  1/2 x′′ 1/2 X  X n Zi n     2 ′′ ′ ≤  |wm |   |xi − xi |   i=1 x′ i

(again by H¨older)

i=1

X √ ′ 1/2 ≤ (2α)( |x′′ . i − xi |)

158

Letting m → ∞ we get the result. Step 2. Observe that wm (0) = 0S 0 w(0) = 0. Therefore 1 wn (x + h) − wn (x) 2 | =| | h h ≤

1 h

 x+h 2 Zx+h  Z  1   wn dt|2 ≤ 2  |wn |dt  h  x

x x+h Z x

|wn |2 dt.

Hence Z1−h Zh Z1−h 1 wn (x + h) − wn (x) 2 | dx ≤ dx |(w˙ n (x + t))|2 dt | h h 0

0

≤

1 h

Zh

dt

0

1 ≤ 2 2

Zh

0 1−h Z

|w˙ n (x + t)|2 dx

0

dt = 2α

0

letting n → ∞, we get Z1−h w(x + h) − w(x) 2 | | dx ≤ 2α. h 0

Let h → 0 to get I(w) ≤ α, completing the proof.

22. Large Deviations

166

 159

Lemma 3. Let C be closed and put C δ =

S

S (w; δ); then

w∈C

lim ( inf I(w)) = inf I(w). w∈C

δ→0 w∈C δ

Proof. If δ1 < δ2 , then C δ1 ⊂ C δ2 so that inf I(w) is decreasing. As w∈C δ

C δ ⊃ C for each δ,

lim ( inf I(w)) ≤ inf I(w)

δ→0 w∈C δ

w∈C

Let ℓ = lim ( inf I(w)). Then there exists wδ ∈ C δ such that I(wδ ) → δ→0 w∈C δ

ℓ, and therefore (I(wδ )) is a bounded set bounded by α (say). Claim. |wδ (t1 ) − wδ (t2 )| = |

Zt2 t1

Z wδ dt| ≤ |(|t1 − t2 |)( |wδ |2 )1/2 √

√

≤ (2α|t1 − t2 |). The family (wδ ) is therefore equicontinuous which, in view of the fact that wδ (0) = 0, implies that it is uniformly bounded and the claim follows from Ascoli’s theorem. Hence every subfamily of (wδ ) is equicontinuous. By Ascoli’s theorem there exists a sequence δn → 0 such that wδn → w uniformly on [0, 1]. It is clear that w ∈ C. By lower semicontinuity of I(w), lim inf I(w) ≥ inf

δ→0 w∈C δ

w∈C

completing the proof.  160

Proof of Part 1 of the theorem. Let X be continuous in [0, 1]. For each n let Xn be a piecewise linear version of X based on n equal intervals, i.e.

167 Xn is a polygonal function joining the points (0, X(0)), (1/n, X(1/n)), . . . , (1, X(1)). Pǫ (||Xn − X|| ≥ δ), (|| · || = || · ||∞ )   !  [ j − 1 δ  ≤ P  sup · sup |Xr (t) − Xr | ≥ √  , n 2 d n i≤ j≤n j−1 ≤t≤ j n

n

where

≤ ndPǫ ( sup |X(t) − X(0)| ≥ 0≤t≤1/n

X = (X1 , . . . , Xd ).

δ √ (Markov property; here X is one2 d

dimensional).    δ  ≤ ndPǫ  sup |Xt | ≥ √  (since X(0) = 0) 2 d 0≤t≤1/n δ ≤ 2nd Pǫ ( sup Xt ≥ √ ) 2 d 0≤t≤1/n δ = 2dn P( sup Xt ≥ √ ) 2 ǫd 0≤t≤1/n δ = 4dn P(X(1/n) ≥ √ ) (by the reflection principle) 2 ǫd Z∞ 1 2 √ = 4dn e−ny /2 dy 2π/n √ √ = 4d

δ n/2 ǫd Z∞

√ √ δ n/2 ǫd

2 1 √ e−x /2 dx 2π

Now, for every a > 0,

a

Z∞ a

Thus

−x2 /2

e

dx ≤

Z∞

2 /2

xe−x

2 /2

dx = e−a

.

a

161

22. Large Deviations

168

Pǫ (||Xn − X|| ≥ δ) ≤

√ 2 ǫ −nδ2 /(8∈d) 4dn e−nδ /(8∈d) = C (n) √ e , √ √ 1 δ δ n/2 ǫd

where C1 depends only on n. We have now Pǫ (Xn ∈ C δ ) ≤ Pǫ (I(Xn ) ≥ ℓδ ) where ℓδ = inf{I(w)w ∈ C δ }.   ! n−1  j   1 X j + 1 n|X −X |2 ≥ ℓδ  = P  2 j=0 n n ! 2ℓδ 2 2 2 = P Y1 + Y2 + · · · + Ynd ≥ , ǫ √ where Y1 = n(X1 (1/n) − X1 (0)) etc. are independent normal random variables with mean 0 and variance 1. Therefore, 2ℓδ 2 ) P(Y12 + · · · + Ynd ≥ ǫ Z 2ℓδ −(y2 +···+y2 )1/2 nd = e 1 dy1 . . . dynd . ǫ y21 +···+y2nd

= C(n) √

Z∞

e−r

2 /2

rnd−1 dr,

(2ℓδ /ǫ)

using polar coordinates, i.e. 2 P(Y12 + Y22 + · · · + Ynd

2ℓδ ≥ ) = C ′ (n) ǫ

Z∞

nd

e−s s 2 −1 ds

(ℓδ /ǫ)

(change the variable from r to s = Z∞

r2 ). An integration by parts gives 2

e−s sk ds = e−α (αk +

α

k! αk−2 + · · · ). (k − 1)!

Using this estimate (for n even) we get 2 P((Y12 + · · · + Ynd )≥

2ℓδ ℓδ nd ) ≤ C2 (n)e−ℓδ /ǫ ( ) 2 −1 , ǫ ǫ

169 where C2 depends only on n. Thus, Pǫ (C) ≤ Pǫ (||Xn − X|| ≥ δ) + Pǫ (Xn < C δ ) ! nd2 −1 √  ǫ  −nδ2 /(8∈d) −ℓδ /ǫ ℓδ + C2 (n)e ≤ C1 (n) e δ ǫ ! nd2 −1 √  ǫ  −nδ2 /(8∈d) −ℓδ /ǫ ℓδ ≤ 2 max[C1 (n) e , C2 (n)e δ ǫ √  ǫ  −nδ2 /(8∈d) ∈ log Pǫ (C) ≤ ǫ log 2 + ǫ max[log(C1 (n) e δ ℓδ nd log C2 (n)e−ℓδ /ǫ ( ) 2 −1 ] ǫ 162

Let ǫ → 0 to get

(

) −nδ2 −ℓδ lim ∈ log Pǫ (C) ≤ max , . 8d 1

Fix δ and let n → ∞ through even values to get lim ∈ log Pǫ (C) ≤ −ℓδ . Now let δ → 0 and use the previous lemma to get Z lim ∈ log Pǫ (C) ≤ − I(w). ǫ→0

w∈C

Proposition . Let ℓ be finite; then {w : I(w) ≤ ℓ} is compact in Ω. Proof. Let (wn ) be any sequence, I(wn ) ≤ ℓ. Then √ |wn (t1 ) − wn (t2 )| ≤ (ℓ|t1 − t2 |) and since wn (0) = 0, we conclude that {wn } is equicontinuous and uniformly bounded.  Assumptions. Let Ω be any separable metric space, F = Borel σ-field on Ω. For every ǫ > 0 let Pǫ be a probability measure. Let I : Ω → [0, ∞] be any function such that

22. Large Deviations

170 (i) I is lower semi-continuous. (ii) ∀ finite ℓ, {w : I(w) ≤ ℓ} is compact.

163

(iii) For every closed set C in Ω, lim sup ∈ log Pǫ (C) ≤ − inf I(w). w∈C

ǫ→0

(iv) For every open set G in Ω lim inf ∈ log Pǫ (G) ≥ − inf I(w). w∈G

ǫ→0

Remark . Let Ω = C[0, 1], Pǫ the Brownian measure corresponding to 1 R1 2 |w| dt if w(0) = 0 and ∞ otherwise, then all the scaling ǫ. If I(w) = 20 the above assumptions are satisfied. Theorem . Let F : Ω → R be bounded and continuous. Under the above assumptions the following results hold. (i) For every closed set C in Ω Z F(w) lim sup ǫ log exp dPǫ ≤ sup(F(w) − I(w)). ǫ→0 ǫ w∈C C

(ii) For every open set G in Ω Z F(w) lim inf ∈ log exp dPǫ ≥ sup(F(w) − I(w)). ǫ→0 ǫ w∈G G

In particular, if G = Ω = C, then Z F(w) lim ∈ log exp dPǫ = sup(F(w) − I(w)). ǫ→0 ǫ w∈Ω Ω

171 Proof. Let G be open, w0 ∈ G. Let δ → 0 be given. Then there exists a neighbourhood N of w0 , F(w) ≥ F(w0 ) − δ, ∀w in N. Therefore Z Z F(w0 )−δ F(w) F(w) exp dPǫ ≥ exp dPǫ ≥ e ǫ Pǫ (N). ǫ ǫ G

N

164

Therefore ǫ log

Z

exp

F(w) dPǫ ≥ F(w0 ) − δ + ǫ log Pǫ (N). ǫ

G

Thus lim log

Z

exp

F(w) dPǫ ≥ F(w0 ) − δ + lim ǫ log Pǫ (N). ǫ

G

≥ F(w0 ) − δ − inf I(w) ≥ F(w0 ) − I(w0 ) − δ. w∈N

Since δ and w0 are arbitrary (w0 ∈ G) we get Z F(w) lim ∈ log exp dPǫ ≥ sup(F(w) − I(w)). ǫ w∈G G

This proves Part (ii) of the theorem.



Proof of Part (i). Step 1. Let C be compact; L = sup(F(w) − I(w)). If L = −∞ it follows w∈G

easily that lim sup ∈ log

ǫ→0

Z

eF/ǫ dPǫ ≤ −∞.

C

(Use the fact that F is bounded). Thus without any loss, we may assume L to be finite. Let w0 ∈ C; then there exists a neighbourhood N of w0 such that F(w) ≤ F(w0 ) + δ and by lower semi-continuity of I, I(w) ≥ I(w0 ) − δ, ∀w ∈ N(w0 ).

22. Large Deviations

172

By regularity, there exists an open set Gw0 containing w0 such that Gw0 Gw0 N(w0 ). Therefore 165 ! Z F(w) F(w0 ) + δ exp dPǫ exp Pǫ (Gw0 ). ǫ ǫ G w0

Therefore lim sup ∈ log

ǫ→0

Z

exp

F(w) dPǫ ≤ F(w0 ) + δ + ǫ lim Pǫ (Gw0 ) ǫ→0 ǫ

Gw0

≤ F(w0 ) + δ − inf I(w) w∈G w0

F(w0 ) + δ − I(w0 ) + δ ≤ L + 2δ.

Let Kℓ = {w : I(w) ≤ ℓ}. By assumption, Kℓ is compact. Therefore, for each δ > 0, there exists an open set Gδ containing Kℓ ∩ C such that Z F(w) lim sup ∈ log e ǫ dPǫ ≤ L + 2δ. ǫ→0

Gδ

Therefore lim sup ∈ log

ǫ→0

Z

Z

Z

e

F(w) ǫ

dPǫ ≤ L + 2δ,

G δ ∩C

e

F(w) ǫ

dPǫ ≤ eM/ǫ P(C ∩ Gcδ ).

e

F(w) ǫ

dPǫ ≤ M + lim sup ǫ log Pǫ (Cδc ∩ C)

G cδ ∩C

Therefore lim sup ∈ log

ǫ→0

G cδ ∩C

ǫ→0

≤ M − inf

w∈C∩G cδ

I(w).

173 166

Now Gcδ ⊂ Kℓc ∩ C c . Therefore C ∩ Gcδ ⊂ C ∩ Kℓc

if w ∈ C ∩ Gcδ , w < Kℓ . Therefore I(w) > ℓ. Thus Z lim sup ∈ log eF(w)/ǫ dPǫ ≤ M − ℓ ≤ L ≤ L + 2δ. ǫ→0

G cδ ∩C

This proves that lim sup ∈ log

ǫ→0

Z

exp

F(w) dPǫ ≤ L + 2δ. ǫ

C

Since C is compact there exists a finite number of points w1 , . . . , wn in C such that n [ C⊂ Gwi i=1

Therefore lim ∈ log

Z C

F(w) exp dPǫ ≤ lim ǫ log ǫ S

≤ lim(ǫ log n Max 1≤i≤

Z

Z

eF(w)/ǫ dPǫ

n i=1 G wi

exp

F(w) dPǫ ) ǫ

G wi

≤ L + 2δ. Since δ is arbitrary. Z F(w) lim ∈ log exp dPǫ ≤ sup(F(w) − I(w)). ǫ w∈C C

The above proof shows that given a compact set C, and δ > 0 there 167 exists an open set G containing C such that Z F(w) lim ∈ log exp dPǫ ≤ L + 2δ. ǫ G

22. Large Deviations

174 Step 2. Let C be any arbitrary closed set in Ω. Let L = sup(F(w) − I(w)). w∈C

Since F is bounded there exists an M such that |F(w)| ≤ M for all w. Choose ℓ so large that M − ℓ ≤ L. Since δ is arbitrary Z F(w) lim sup ∈ log exp dPǫ ≤ sup(F(w) − I(w)) ǫ→0 ǫ w∈C C

We now prove the above theorem when Pǫ is replaced by Qǫx . Let Pǫx be the Brownian motion starting at time t = 0 at the space point x corresponding to the scaling ǫ. Precisely stated, if τǫ : C([0, ∞); Rd ) → C([0, ∞); Rd ) ′′ is the map given by (τǫ w)(t) = w(ǫt), then Pǫx = P x τ−1 ǫ . Note T 1 τǫ = T ǫ def

and T ǫ is given by T ǫ w = y where y(t) = w(ǫt) +

Zt

b(y(s))ds.

0

Hence −1 ǫ −1 P x T ǫ−1 = P x (T 1 , τǫ )−1 = P x τ−1 ǫ T1 = Px T1 ;

either of these probability measures is denoted by Qǫx . Theorem . Let b : Rd → Rd be bounded measurable and locally Lipschitz. Define Z1 1 I(w) : |X(t) − b(X(t))|2 dt 2 0

168

If w ∈ C([0, ∞); Rd ), w(0) = x and x absolutely continuous. Put I(w) = ∞ otherwise. If C is closed in C[(0, 1]; Rd ), then lim ∈ log Qǫx (C) ≤ − inf I(w).

ǫ→0

w∈C

175 If G is open in C([0, 1]; Rd ), then lim ∈ log Qǫx (G) ≥ − inf I(w). w∈G

ǫ→0

As usual Qǫx (C) = Qǫx π−1 (C) where π : C([0, ∞); Rd ) → C([0, 1]; Rd ) is the canonical projection. Remark. If b = 0 we have the previous case. Proof. Let T be the map x(·) → y(·) where y(t) = x(t) +

Zt

b(y(s))ds.

0

Then Qǫx = Pǫx (T −1 ). If C is closed Qǫx (C) = Pǫx (T −1 C). The map T is continuous. Therefore T −1 (C) is closed. Thus lim sup ∈ log Q x (C) = lim sup ∈ log Pǫx (T −1 C)

ǫ→0

≤−

(*)

ǫ→0

inf

w∈T −1 (C)

= − inf

w∈C

1 2

1 2

Z1

Z1

|X|2 dt

(see Exercise 1 below)

0

|T −1 w|2 dt.

0

Now T −1

y(·) −−−→ y(t) −

Zt 0

169

b(y(s))ds.

22. Large Deviations

176 Therefore (T −1 y) = y − b(y(s)). Therefore

lim sup ∈ log Qǫx (C) ≤ − inf I(w). w∈C

ǫ→0

The proof when G is one is similar.



Exercise 1. Replace Pǫ by Pǫx and I by I x where 1 I x (w) = 2

Z1

|w|2 , w(0) = x, w absolutely continuous,

0

= ∞ otherwise. Check that (*) holds, i.e. lim sup ∈ log Pǫx (C) ≤ − inf I x (w), if C is closed, w∈C

ǫ→0

and lim inf ∈ log Pǫx (G) ≥ − inf I x (w). w∈G

ǫ→0

set in Rn ,

Let G be a bounded open with a smooth boundary Γ = ∂G. d d ∞ Let b : R → R be a smooth C function such that (i) hb(x), n(x)i0, ∀x ∈ ∂G where n(x) is the unit inward normal. (ii) there exists a point x0 ∈ G with b(x0 ) = 0 and |b(x)| > 0, ∀x in G − {x0 }. 170

(iii) for any x in G the solution ξ(t) = x +

Zt

b(ξ(s))ds,

0

of the vector field starting from x converges to x0 as t → +∞. Remark.

(a) (iii) is usually interpreted by saying that “x0 is stable”.

177 (b) By (i) and (ii) every solution of (iii) takes all its values in G and ultimately stays close to x0 . Let ǫ > 0 be given; f : ∂G → R be any continuous bounded function. Consider the system Lǫ uǫ =

1 ∆uǫ + b(x) · ∆uǫ = 0 in G 2 uǫ = f on ∂G.

We want to study lim uǫ (x). Define ǫ→0

I0T (X(t))

1 = 2

ZT

|X(t) − b(X(t))|2 dt; X : [0, T ] → Rd

0

whenever X is absolutely continuous, = ∞ otherwise. Remark. Any solution of (iii) is called an integral curve. For any curve X on [0, T ], I0T gives a measure of the deviation of X from being an integral curve. Let VT (x, y) = inf{I0T (X) : X(0) = x; X(T ) = y} and V(x, y) = inf{VT (x, y) : T > 0}. V has the following properties. (i) V(x, y) ≤ V(x, z) + V(z, y) ∀x, y, z. (ii) Given any x, ∃δ → 0 and C > 0 such that for all y with |x − y| ≤ δ. V(x, y) ≤ C|x − y| Proof. Let X(t) = Put

t(y − x) + x. |y − x| T = |y − x|, X(0) = x, X(T ) = y,

171

22. Large Deviations

178

I0T (X(t))

1 = 2

ZT 2 y − x − b(X + S (y − x)) ds. T T 0

Then I0T

1 ≤ 2

ZT

! |y − x|2 2 + ||b||∞ ds, 2 T2

0

where ||b||∞ =

sup

b(λ),

|λ−x|≤|y−x|

or, I0T ≤ (1 + ||b||2∞ )|y − x|. As a consequence of (ii) we conclude that     2  V(x, y) ≤ 1 + sup |b(λ)|  |y − x|, |λ−x≤|y−x|

i.e. V is locally Lipschitz. The answer to the problem raised is given by the following.



Theorem . lim uǫ (x) = f (y0 )

ǫ→0

where y0 is assumed to be such that y0 ∈ ∂G and V(x0 , y0 ) < V(x, y), ∀y ∈ ∂G, y , y0 . 172

We first proceed to get an equivalent statement of the theorem. Let be the Brownian measure corresponding to the starting point x, and corresponding to the scaling ǫ. Then there exists a probability measure Qǫx such that dQǫx Ft = Z(t) dPǫx Pǫx

179 where Z(t, ·) = exp

Zt

1 hb∗ (X(s)), dX(s)i − 2

0

Zt

b∗ (X(s))ds;

0

b∗ is any bounded smooth function such that b∗ = b on G. Further we have the integral representation Z uǫ (x) = f (X(τ))dQǫx ∂G

where τ is the exit time of G, i.e. τ(w) = inf{t : w(t) < G}. Z |uǫ (x) − f (y0 )| = | ( f (X(τ)) − f (y0 ))dQǫx | ≤|

∂G

Z

( f (X(τ)) − f (Y0 ))dQǫx |+

N∩∂G

+|

Z

N c ∩∂G

( f (X(τ)) − f (Y0 ))dQǫx |

(N is any neighbourhood of y0 ).

≤

Qǫx (X(τ)

∈ N ∩ ∂G) sup | f (λ) − f (y0 )|+

λ∈N∩∂G ǫ +2|| f ||∞ Q x (X(τ) ∈ N c

∩ ∂G). 173

Since f is continuous, to prove the theorem it is sufficient to prove the Theorem . lim Q x (X(τ) ∈ N c ∩ ∂G) = 0

ǫ→0

for every neighbourhood N of y0 .

22. Large Deviations

180 Let N be any neighbourhood of y0 . Let V = V(x0 , y0 ), V ′ =

inf

y∈N c ∩∂G

V(x, y).

By definition of y0 and the fact that N c ∩∂G is compact, we conclude that V ′ > V. Choose η = η(N) > 0 such that V ′ = V + η. For any δ > 0 let D = S (x0 ; δ) = {y : |y − x0 | < δ}, ∂D = {y : |y − x0 | = δ}. Claim. We can choose a δ2 such that 3η , ∀x ∈ ∂D2 , y ∈ N c ∂G. 4 η (ii) V(x, y0 ) ≤ V + , ∀x ∈ ∂D2 . 4 (i) V(x, y) ≥ V +

Proof.

(i) V(x0 , y) ≥ V + η, ∀y ∈ N c ∂G. Therefore V + η ≤ V(x0 , y) ≤ V(x0 , x) + V(x, y) ≤ C|x − x0 | + V(x, y).

η Choose C such that C|x − x0 | ≤ . Thus 4 V+ 174

3η η ≤ V(x, y) if C|x − x0 | ≤ , ∀y ∈ N c ∂G. 4 4

C depends only on x0 . This proves (i). η (ii) |V(x0 , y0 ) − V(x, y0 )| ≤ V(x0 , x) ≤ C|x0 − x| ≤ if x is close to x0 . 4 Thus η η V(x, y0 ) ≤ V(x0 , y0 ) + = V + 4 4 if x is close to x0 . This can be achieved by choosing δ2 very small.  Claim (iii). We can choose δ1 < δ2 such that for points x1 , x2 in ∂D1 there is a path X(·) joining x1 , x2 with X(·) ∈ D2 − D1 , i.e. it never penetrates D1 ; and η I(X) ≤ . 8

181 Proof. Let C = sup{|b(λ)|2 : |λ − x0 | ≤ δ2 }. Choose X(·) to be any path on [0, T ], taking values in D2 with X(0) = x1 ; X(T ) = x2 and such that |X| = 1 (i.e. the path has unit speed). Then I0T (X)

≤

ZT

2

(|X| + C)dt ≤ CT +

0

ZT

|X|dt

0

= (c + 1)T = (C + 1)|x2 − x1 |. η Choose δ1 small such that (C + 1)|x2 − x1 | ≤ . 8 Let Ωδ1 = {w : w(t) ∈ G − D1 , ∀t ≥ 0}, i.e. Ωδ1 consists of all  trajectories in G that avoid D1 . Claim (iv). inf inf I T (X(·)) T >0 X∈Ωδ1 ,X(0)∈∂D2 0

≥V+

3η 4

X(T ) ∈ N c ∩ ∂G 175

Proof. Follows from Claim (i) and (ii).



Claim (v). inf

inf

T >0 X∈Ωδ1 ,X(0)∈∂D2 X(T )=y0

I0T (X(·)) ≤ V +

3η . 8

η Proof. By (ii) V(x, y0 ) ≤ V + ∀x ∈ ∂D2 , i.e. 4 η I0T (X(·)) ≤ V + . T >0 X(0)=x,X(T )=y0 4 inf

inf

η Let ǫ > 0 be arbitrary. Choose T and X(·) such that I0T (X) ≤ V + +ǫ 4 with X(0) = x, X(T ) = y0 , X(·) ∈ G. If X ∈ Ωδ1 define Y = X. If X < Ωδ1 define Y as follows:

22. Large Deviations

182

Let t1 be the first time that X enters D1 and t2 the last time that it gets out of D1 . Then 0 < t1 ≤ t2 < T . Let X ∗ be a path on [0, s] such η that (by Claim (iii)) I0s (X ∗ ) ≤ with X ∗ (0) = X(t1 ) and X ∗ (s) = X(t2 ). 8 Define Y on [0, T − (t2 − t1 ) + s] [T − (t2 − t1 ) + s, ∞) by Y(t) = X(t) on [0, t1 ] = X ∗ (t − t1 ) on [t1 , t1 + s]

= X(t − t1 − s + t1 ), on [t1 + s, T − (t2 − t1 ) + s] = X(t2 ), for t ≥ T − (t2 − t1 ) + s.

Then I0T −t2 +t1 +s

1 = 2

Zt1

1 |X − b(X(s))| ds + 2 2

|X ∗ − b(X ∗ (s))|2 ds

0

0

+

1 2

ZT t2

|X(s) − b(X(s))|2 ds

≤V+ 176

Zs

η η +ǫ+ 4 8

by choice of X and X ∗ . As Y ∈ Ωδ1 , we have shown that inf inf I T (X(·)) T >0 X∈Ωδ1 ,X(0)∈∂D1 0 X(T )=y0

≤V+

3η + ǫ. 8

Since ǫ is arbitrary we have proved (v).



Lemma . I0T is lower semi-continuous for every finite T . Proof. This is left as an exercise as it involves a repetiti on of an argument used earlier.  Lemma . Let Xn ∈ Ωδ1 . If T n → ∞ then I0T n (Xn ) → ∞. This result says that we cannot have a trajectory which starts outside of a deleted ball for which I remains finite for arbitrary long lengths of time.

183 Proof. Assume the contrary. Then there exists a constant M such that I0T n (X0 ) ≤ M, ∀n. Let T < ∞, so that MT = sup I0T (Xn ) < ∞. n

Define XnT = Xn |[0,T ] .



Claim. {XnT }n=1 is an equicontinuous family.

177

Proof. |XnT (x2 ) −

XnT (x1 )|2

=|

Zx2

XnT (t)dt|2

x1

Zx2

≤ |x2 − x1 |2

|XnT |2 dt

x1

2

≤ 2|x2 − x1 |

Zx2

|XnT

−

b(XnT )|2 ds

+

ZT

b(XnT )2 ds]

0

≤ 2|x2 − x1 |2 [2MT + T ||b||2∞ ]. Thus, {XnT }n is an equicontinuous family. Since G is bounded, {XnT }n is uniformly bounded. By Arzela-Ascoli theorem and a “diagonal procedure” there exists a subsequence Xnk and a continuous function to X uniformly on compact subsets of [0, ∞). As Xnk (·) ∈ G−D1 , X ∈ G−D1 . Let m ≥ n. I0T n (Xm ) ≤ M. Xn → X uniformly on [0, T n ]. By lower semicontinuity I0T (X) ≤ M. Since this is true for every T we get on letting T tend to ∞, that Z∞ 1 |X − b(X(s))|2 ds ≤ M. 2 0

Thus we can find a sequence a1 < b1 < a2 < b2 < . . . such that Iabnn (X(·))

1 = 2

Zbn

an

|X(t) − b(X(t))|2 dt

22. Large Deviations

184

converges to zero with bn − an → ∞. Let Yn (t) = X(t + an ). Then I0bn −an (Yn ) → 0 178

with bn − an → +∞,

Y n ∈ Ω δ1 .

Just as X was constructed from Xn , we can construct Y from Yn such that Yn → Y uniformly on compact subsets of [0, ∞). I0bn −an (Y) ≤ inf I0bn −an (Ym ) = 0 m≥n

(by lower semi-confirmity of I0T ). Thus I0bn −an (Y) = 0, ∀n, showing that Z∞

Y(t) − b(Y(t))|2 dt = 0

0

Thus Y satisfies Y(·) = x +

Rt

b(Y(s))ds with Y(t) ∈ G − ∂D1 , ∀t.

0

Case (i). Y(t0 ) ∈ G for some t0 . Let Z(t) = Y(t+t0 ) so that Z is an integral curve starting at a point of G and remaining away from D1 contradicting the stability condition. Case (ii). Y(t0 ) < G for any t0 , i.e. Y(t) ∈ ∂G for all t. Since Y(t) = b(Y(t))hY(t), n(Y(t))i is strictly positive. But Y(t) ∈ ∂G and hence hY(t), n(Y(t))i = 0 which leads to a contradiction. Thus our assumption is incorrect and hence the lemma follows.  Lemma . Let x ∈ ∂D2 and define E = {X(t) exits from G before hitting D1 and it exits from N}

F = {X(t) exists from G before hitting D1 and it exits from N c } Then Qǫx (F) 3η 1 ≤ exp − + 0 Qǫx (E) 8ǫ ǫ 179

!!

→ 0 uniformly in x(x ∈ ∂D2 ).

Significance. Qǫx (E) and Qǫx (F) are both small because diffusion is small and the drift is large. The lemma says that Qǫx (E) is relatively much larger than Qǫx (F).

185 Proof. Qǫx (E) ≥ Qǫx {X(t) exists from G before hitting D1 , and exists in 1 N before time T }, = Q x (B) ≥ exp[− inf I0T (X(·))] where the infimum ǫ is taken over the interior of B, " ! !# 1 3η 1 ≥ exp − V + +0 . ǫ 8 ǫ Similarly, Qǫx (F)

"

! !# 1 3η 1 ≤ exp − V + +0 . ǫ 4 ǫ

Therefore " !# Qǫx (F) 1 3η ≤ exp − + 0 → 0 as Qǫx (E) 8ǫ ǫ

ǫ → 0.

We now proceed to prove the main theorem. Let τ0 = 0, τ1 = first time ∂D1 is hit, τ2 = next time ∂D2 is hit, τ3 = next time ∂D1 is hit, .................. and so on. Observe that the particle can get out of G only between the time intervals τ2n and τ2n+1 . Let En = {between τ2n and τ2n+1 the path exits from G for the first time and that it exits in N}, Fn = {between τ2n 180 and τ2n+1 the path exits from G for the first time and that it exists in N c }. Qǫx (X(τ) ∈ N) + Q x (X(τ) ∈ N c ) = 1. Also Qǫx (X(τ) ∈ N c ) =

∞ X

Qǫx (X(τ) ∈ N) =

∞ X

Qǫx (Fn ),

n=1

n=1

Qǫx (En ),

22. Large Deviations

186

≤

X

∞ X

Qǫx (Fn ) =

n=1

Qǫx

E [χ(τ>τ2n ) sup x∈∂D2

n=1

≤ 0(ǫ)

∞ X

∞ X

n=1 ǫ Q x (F)]

ǫ

E Qx (Qǫx (Fn |Fτ2n )) (by the Strong Markov property)

ǫ

E Qx [χ(τ>τ2n ) inf Qǫx (E)]

(as

x∈∂D2

n=1

≤ 0(ǫ)

∞ X n=1

Qǫx (F) → 0) Qǫx (E)

Qǫx (En ) = 0(ǫ)Q x (X(τ) ∈ N c ).

Therefore Qǫx (χ(τ) ∈ N) → 1,

Q x (X(τ) ∈ N c ) → 0. 

Exercise. Suppose b(x) = ∇u(x) for some u ∈ C 1 (G ∪ ∂G, R). Assume that u(x0 ) = 0 and u(x) < 0 for x , x0 . Show that V(x0 , y) = −2u(y). [Hint: For any trajectory X with X(0) = x0 , X(T ) =

y, I0T (X)

1 = 2

ZT 0

181

2

|X + ∇u(X)| dt − 2

ZT

∇u(X) · X(t)dt ≥ −2u(y)

0

so that V(x0 , y) ≥ −2u(y). For the other inequality, let u be a solution of duX(s) X(t) + ∇u(X(t) = 0 on [0, ∞) with X(0) = y. Show that because 0 ds for X(s) , 0 and x0 is the only zero of u, limitX(t) = x0 . Now conclude t→∞ that V(x0 , y) ≤ −u(y)].

23. Stochastic Integral for a Wider Class of Functions WE SHALL NOW define the stochastic integral for a wider class of 182 functions. Let θ : [0, ) × Ω → Rd be any progressively measurable function such that for every t Zt

|θ(s, w)|2 ds < ∞,

a.e.

0

Define, for every finite L ≥ 0,      θ(s, w),     θL (s, w) =        0,

if if

Rs 0

|θ(t, w)|2 dt < L < ∞,

0

|θ(t, w)|2 dt ≥ L.

Rs

We can write θL (s, w) = θ(s, w)χ[0,L) (φ(s, w)) where

φ(s, w) =

Zs

|θ(t, w)|2 dt

0

is progressively measurable. Hence θL (s, w) is progressively measur187

23. Stochastic Integral for a Wider Class of Functions

188

able. It is clear that

RT 0

|θL (s, w)|2 ds ≤ L, a.e. ∀T . Therefore ZT E( |θL (s, w)|2 ds) ≤ L. 0

Thus the stochastic integral ξL (t, w) =

Rt 0

hθL (s, w), dX(s, w)i is well de-

fined. The proofs of the next three lemmas follow closely the treatment of Stochastic integration given earlier. 183

Lemma 1. Let τ be a bounded, progressively measurable, continuous function. Let τ be any finite stopping time. If θ(s, w) = 0, ∀(s, w) such Rt that 0 ≤ s ≤ τ(w) then hθ(s, w), dX(s, w)i = 0 for 0 ≤ t ≤ τ(w). 0

Proof. Define θn (s, w) = θ( [ns] n , w). θn is progressively measurable and by definition of the stochastic integral of θn , Zt

hθn (s, w), dX(s, w)i = 0,

∀t,

0 ≤ t ≤ τ(w).

0

Now  t  Z    E  |θn (s, w) − θ(s, w)|2 ds   0

 t  ! Z  [ns]   = E  |θ w − θ(s, w)|2 ds → 0 as n → ∞   n 0

and

Zt 0

hθn (s, w), dX(s, w)i →

Zt 0

hθ(s, w), dX(s, w)i

189 in probability. Therefore Zt

hθ, dXi = 0 if 0 ≤ t ≤ τ(w).

0

 Lemma 2. If θ is progressively measurable and bounded the assertion of Lemma 1 still holds. Proof. Let 1 θn (t, w) = n

Zt

θ(s, w)ds.

(t−1/n)V0

Then  T  Z    E  |θn (t, w) − θ(t, w)|2 dt → 0 (Lebesgue’s theorem).   0

θn is continuous and boundd, θn (t, w) = 0 for 0 ≤ t ≤ τ(w). By lemma 1 184 Zt

hθn (s, w), dX(s, w)i = 0

0

if 0 ≤ t ≤ τ(w). This proves the result.



Lemma 3. Let θ be progressively measurable such that, for all t,  t  Z    E  |θ(s, w)|2 ds < ∞.   0

If θ(s, w) = 0 for 0 ≤ s ≤ τ(w), then Zt 0

hθ(s, w), dX(s, w)i = 0 for

0 ≤ t ≤ τ(w).

23. Stochastic Integral for a Wider Class of Functions

190 Proof. Define

   θ, θn =   0,

Then Zc

if |θ| ≤ n, if |θ| > n.

hθn , dXi = 0, if 0 ≤ t ≤ τ(w),

(Lemma 2) and

0

Rt E( |θn − θ|2 ds) → 0. The result follows.



0

Lemma 4. Let θ be progressively measurable such that ∀t, Zt

|θ(s, w)|2 ds < ∞

a.e.

0

Then Lt ξL (t, w) exists a.e. L→∞

185

Proof. Define   Zs         2 ; |θ(σ, w)| dσ ≥ L s : τL (w) = inf         0

clearly τL is a stopping time. If L1 ≤ L2 , τL1 (w) ≤ τL2 (w) and by assumptions of the lemma τL ↑ ∞ as L ↑ ∞. If L1 ≤ L2 ,

θL1 (s, w) = θL2 (s, w) for 0 ≤ s ≤ τL1 (w).

Therefore by Lemma 3, ξL2 (t, w) = ξL1 (t, w) if 0 ≤ t ≤ τL1 (w). Therefore as soon as L is large enough such that t ≤ τL (w), ξL (t, w) remains constant (as a function of L). Therefore Lt ξL (t, w) exists a.e.  L→∞

191 Definition. The stochastic integral of θ is defined by Zt

hθ(s, w), dX(s, w)i = Lt ξL (t, w). L→∞

0

Exercise. Check that the definition of the stochastic integral given above coincides with the previous definition in case  t  Z    E  |θ(s, w)|2 ds < ∞, ∀t.   0

Lemma . Let θ be a progressively measurable function, such that Zt

|θ(s, w)|2 ds < ∞, ∀t.

0

If ξ(t, w) denotes the stochastic integral of θ, then

186

  T  Z  L2  P sup |ξ(t, ·)| ≥ ǫ ≤ P  |θ|2 ds ≥ L + 2 .  ǫ  0≤t≤T !

0

Proof. Let τL = inf{t :

Rt 0

θ(s, w). Also

|θ|2 ds ≥ L}. If T < τL (w), then θL (s, w) =

ξL (t, w) = ξ(t, w) for 0 ≤ t ≤ T.  Claim. ( ) w : sup |ξ(t, w)| ≥ ǫ 0≤t≤T ( ) w : sup |ξL (t, w)| ≥ ǫ ∪ {w : τL (w) ≤ T } 0≤t≤T

23. Stochastic Integral for a Wider Class of Functions

192

For, if w is not contained in the right side, then sup |ξL (t, w)|2 < ǫ

and

0≤t≤T

|τL (w)| > T.

If τL > T , ξL (t, w) = ξ(t, w) ∀t ≤ T . Therefore sup |ξL (t, w)| = sup |ξ(t, w)|

0≤t≤T

0≤t≤T

Therefore w < left side. Since     ZT         2 {w : τL (w) > T } =  w : |θ| ds ≥ L          0

we get

P sup |ξ(t, ·)| ≥ ǫ 0≤t≤T  T Z

 ≤ P  

0

!

 !   2 |θ| ds ≥ L + P sup |ξL (t, ·)| ≥ ǫ  0≤t≤T

  T  Z  L2  ≤ P  |θ|2 ds ≥ L + 2 ,  ǫ  0

187

by Kolmogorov’s inequality. This proves the result. Corollary . Let θn and θ be progressively measurable functions such that (a)

Rt 0

|θn (s, w)|2 ds < ∞,

(b) Lt

n→∞

Rt 0

Rt 0

|θ(s, w)|2 ds < ∞, ∀t;

|θn (s, w) − θ(s, w)|2 ds = 0 in probability.

If ξn (t, w) and ξ(t, w) denote, respectively the stochastic integrals of θn and θ, then sup |ξn (t, w) − ξ(t, w)| converges to zero in probability. 0≤t≤T

193

Proof. Let τn,L (w) = inf{t :

Rt 0

|θn |2 ds ≥ L}; replacing θ by θn − θ and ξ

by ξn − ξ in the previous lemma, we get P sup |ξn (t, ·) − ξ(t, ·)| ≥ ǫ 0≤t≤T

!

 T  Z  L2   ≤ 2 + P  |θn (s, ·) − θ(s, ·)|2 ds ≥ L .   ǫ 0

Letting n → ∞, we get

lim P sup |ξn (t, ·) − ξ(t, ·)| ≥ ǫ

n→∞

0≤t≤T

!

L2 . ǫ2

As L is arbitrary we get the desired result.



Proposition . Let θ be progressively measurable such that Zt

|θ(s, w)|2 ds < ∞,

∀t and ∀ w.

0

Then

188

 t  Zt Z  1   Z(t, ·) = exp  hθ(s, ·), dX(s, ·)i − |θ(s, ·)|2 ds   2

(*)

0

0

is a super martingale satisfying (a) E(Z(t, ·)) ≤ 1; (b) Lt E(Z(t, ·)) = 1. t→0

Proof. Let (θn ) be a sequence of bounded progressively measurable functions such that Zt 0

|θn − θ|2 ds → 0 ∀t,

∀w.

23. Stochastic Integral for a Wider Class of Functions

194

(For example we may take θn = θ if |θ| < n, = 0 otherwise). Then (∗) is a martingale when θ is replaced by θn . This martingale satisfies E(Zn (t, ·)) = 1, and Zn (t, ·) → Z(t, ·) pointwise (a) now follows from Fatou’s lemma: ! lim E(Z(t, ·)) ≥ E lim Z(t, ·) t→0

t→0

= E(1) = 1. Therefore Lt E(Z(t, ·)) = 1. This proves (b). t→0



24. Explosions Exercise. Let R > 0 be given. Put bR = bφR where φR = 1 on |x| ≥ R, 189 φR = 0 if |x| ≥ R + 1; φR is C ∞ . Show that bR = b on |x| ≤ R, bR is bounded on Rd and bR is globally Lipschitz. Let ΩT = {w ∈ Ω : B(w) > T }. Let S T = ΩT → C[0, T ] be the map Rt S T w = y(·) where y(t) = w(t) + b(y(s))ds on [0, T ]. Unless otherwise 0

specifie b : Rd → Rd is assumed to be locally Lipschitz. Define the measure QTx on (Ω, T ) by QTx (A) = P x {w : S T w ∈ A, B(w) > T }, where P x is the probability measure corresponding to Brownian motion. Theorem . QTx (A)

=

Z

∀A ∈ FT ,

Z(T, ·)dP x ,

A

where

  T ZT  Z 1  2  |b(X(s, ·))| ds . Z(T, ·) = exp  hb, dXi −   2 0

0

Remark . If b is bounded or if b satisfies a global Lipschitz condition then B(w) = ∞, so that ΩT = Ω and QTx are probability measures. Proof. Let 0 ≤ R < ∞. For any w in Ω, let y be given by y(t) = w(t) +

Zt 0

195

b(y(σ))dσ.

24. Explosions

196

Define σR (w) = inf{t : |y(t)| ≥ R and let bR be as in the Exercise. Then the equation yR (t) = w(t) +

Zt

bR (yR (σ))dσ

0

190

has a global solution. Denote by S R : Ω → Ω the map w → yR . If QR,x is the measure induced by S R , then  t  Zt Z  dQR,x 1   2  |bR | ds . Ft = Zr (t) = exp  hbR , dXi −  dP x 2 0

0

Let τR (w) = inf{t : |w(t)| > R}. τR is a stopping time satisfying τR S R = σR . By the optional stopping theorem. dQR,x (1) FτR ∧T = ZR (τR ∧ T ) = Z(τR ∧ T ). dP x



Claim. QR,x ((τR > T ) ∩ A) = QTx ((τR > T ) ∩ A), ∀A in FT . Proof. Right side = P x {w : B(w) > T, S T (w) ∈ A ∩ (τR > T )} = P x {w : B(w) > T, y ∈ A, sup |y(t)| < R} 0≤t≤T

= P x {w : y is defined at least upto time T, y ∈ A, sup |y(t)| > R} 0≤t≤T

= P x {w : yR ∈ A, sup |yR (t)| < R} 0≤t≤T

= P x {w : S R (w) ∈ A, τR S R (w) > T }

= QR,x {(τR > T ) ∩ A}

(by definition). As Ω is an increasing union of {τR > T } for R increasing, QTx (A) =

lt

R→+∞

QTx ((τR > T ) ∩ A), ∀A in FT ,

197 = lt QR,x ((τR > T ) ∩ A) (by claim) R→∞   τR ∧T τZ R ∧T Z   Z 1   2  |b| ds dP x = lt exp  hb, dXi −   R→∞ 2 = =

Z A Z

(τR ∧T )∩A

(by (1))

0

0

 T  ZT Z  1   exp  hb, dXi − |b|2 ds dP x   2 0

0

Z(T )dP x .

A

 191 Theorem . Suppose b : Rd → Rd is locally Lipschitz; let L =

∆ + b.∇. 2

(i) If there exists a smooth function u : Rd → (0, ∞) such that u(x) → ∞ as |x| → ∞ and Lu ≤ cu for some c > 0 then P x {w : B(w) < ∞} = 0, i.e. for almost all w there is no explosion. (ii) If there exists a smooth bounded function u : Rd → (0, ∞) such that Lu ≥ cu for some c > 0, then P x {w : B(w) < ∞} > 0, i.e. there is explosion. Corollary . Suppose, in particular, b satisfies |hb(x), xi| ≤ A + B|x|2 for some constants A and B; then P x (w : B(w) < ∞) = 0. Proof. Take u(x) = 1 + |x|2 and use part (1) of the theorem.



∆ Proof of theorem. Let bR be as in the Exercise and let LR = + bR · ∇; 2 then LR u(x) ≤ cu(x) if |x| ≤ R. Claim. u(X(t))e−ct is a supermartingale upto time τR relative to QRx ,   t  Zt  Z  1    d u(X(t))e−ct exp  hbR , dXi − |bR |2 ds    2 0

0

24. Explosions

198 Rt Rt −ct hbR ,dXi− 21 |bR |2 ds

e

0

0

|2

×

|bR 1 dt] + bR udt + |bR |2 udt} 2 2 Zt Zt 1 = exp(−ct + hbR , dXi − |bR |2 ds) 2

×{−cudt + h∇u, dXi + u(x)[hbR , dXi −

0

0

·[LR − c)u + h∇u, dXi + uhbR , dXi]. 192

Therefore −ct

Rt

e0

hbR ,dXi− 21

Rt 0

|bR |2 ds

u(X(t))e E   Zt Zs Zs     − exp −cs + hbR , dXi − |bR |2 ds · (LR − c)u(X(s))ds   0

0

0

is a Brownian stochastic integral. Therefore   τR ∧t τR ∧t Z Z   1   2  |bR | ds − u(X(τR ∧ t)) exp −c(τR ∧ t) + hbR , dXi −   2 0 0   τR ∧t Z Zs Zs   1   2  − exp −cs + hbR , dXi − |bR | ds (LR − c)u(X(s))ds   2 0

0

0

is a martingale relative to P x , FτR ∧t . But bR (x) = b(x) if |x| ≤ R. Therefore   τR ∧t τR ∧t Z Z   1   u(X(τR ∧ t)) exp −c(τR ∧ t) + hb, dXi − |b|2 ds −   2 0

0

  τR ∧t Z Zs Zs     − exp −cs + hb, dXi − |b|2 ds (LR − c)u(X(s))ds   0

0

199

193

is a martingale relative to FτR ∧t . But (LR − c)u ≤ 0 in [0, τR ]. Therefore τR ∧t τR ∧t Z Z 1 u(X(τR ∧ t)) exp(−c(τR ∧ t) + hb, dXi − |b|2 ds) 2 0

0

is a supermartingale relative to FτR ∧t , P x . Therefore u(X(τR ∧ t)e−c(τR ∧t) is a supermartingale relative to QRx (optional sampling theorem). Therefore R E Qx (u(X(tR ∧ t))e−c(τR ∧t) ≤ u(x); letting t → ∞, we get, using Fatou’s lemma, R

E xQ (u(X(τR )e−cτR ) ≤ u(x). Therefore R

E Qx (e−cτR ) ≤ Thus E Px (e−cσR ) ≤

u(x) . inf |u(y)|

|y|=R

u(x) inf |u(y)|

|y|=R

(by change of variable). Let R → ∞ to get Lt

R→∞

P x {w : B(w) < ∞} = 0.

R

e−cσR dP x = 0, i.e.

Sketch of proof for Part (ii). By using the same technique as for Part (i), show that u(X(t))e−ct is a submartingale upto time τR relative to QRx , so that E Px (e−cσR ) ≥

u(x) u(x) ≥ > 0; sup |u(y)| ||u||∞

|y|=R

let R → ∞ to get the result.

194

∂2

1 ∂ + x3 , there is explosion. (Hint: take 2 ∂x ∂x −1 2 u = etan (x ) and show that Lu ≥ u). Exercise. Show that if L =

25. Construction of a Diffusion Process Problem . Given a : [0, ∞) × Rd → S d+ , bounded measurable and b : 195 [0, ∞) × Rd → Rd bounded measurable, to find (Ω, Ft , P, X) where Ω is a space, {Ft }t≥0 an increasing family of σ-algebras on Ω, P a probability measure on the smallest σ-algebra containing all the Ft ’s. and X : [0, t) × Ω → Rd , a progressively measurable function such that X(t, w) ∈ I[b(t, Xt ), a(t, Xt )]. Let Ω = C[0, ∞); Rn ), β(t, ·) = n-dimensional Brownian motion, Ft = σ{β(s) : 0 ≤ s ≤ t}, P the Brownian measure on Ω and a and b as given in the problem. We shall show that thee problem has a solution, under some special conditions on a and b. Theorem . Assume that there exists σ : [0, ∞) × Rd → Md×n (Md×n = set of all d × n matrices over the reals) such that σσ∗ = a. Further let X i, j

X i, j

X j

|σi j (t, x)| ≤ C,

X j

|b j (t, x)| ≤ C,

|σi j (t, x1 ) − σi j (t, x2 )| ≤ A|x1 − x2 |, |b j (t, x1 ) − b j (t, x2 )| ≤ A|x1 − x2 |. 201

25. Construction of a Diffusion Process

202 Then the equation

(1)

ξ(t, ·) = x +

Zt

hσ(s, ξ(s, ·)), dβ(s, ·)i +

0

196

Zt

b(s, ξ(s, ·))ds

0

has a solution. The solution ξ(t, w) : [0, ∞) × Ω → Rd can be taken to be such that ξ(t, ·) is progressively measurable and such that ξ(t, ·) is continuous for a, a.e. If ξ, η are progressively measurable, continuous (for a.a.e) solutions of equation (l), then ξ = n for a.a.w. Proof. The proof proceeds in several steps.



Lemma 1. Let Ω be any space with (Ft )t≥0 an increasing family of σalgebras. If 0 ≤ T ≤ ∞ then there exists a σ-algebra A0 ⊂ A = B[0, T ) × FT such that a function f : [0, T ] × Ω → R is progressively measurable if and only if f is measurable with respect to A0 . Proof. Let A0 = {A ∈ A : χA is progressively measurable}. Clearly [0, T ] × Ω ∈ A0 , and if A ∈ A0 , Ac ∈ Ac . Thus A0 is an algebra. As increasing limits (decreasing limits) of progressively measurable functions are progressively measurable, A0 is a monotone class and hence a σ-algebra.  Let f : [0, T ] × Ω → R be progressively measurable; in fact, f + = f + 1 = f| − f − |f| ,f = . Let g = f + . Then 2 2 gn =

n2n X i−1 i=1

197

2n

χg−1 [ i−1n , 2

i ) 2n

+ nχg−1 [n,)

is progressively measurable. Hence nVgn is progressively measurable, i.e. nχg−1 [n,∞) is progressively measurable. Similarly φg−1 [ i−1n , in ) is pro2 2 gressively measurable, etc. Therefore, by definition, gn is measurable with respect to A0 . As g = f + is the pointwise limit of gn , f + is measurable with respect to A0 . Similarly f − is A0 -measurable. Thus f is A0 -measurable.

203 Let f : [0, T ] × Ω → R be measurable with respect to A0 . Again, if g = f+ n2n X i−1 gn = χ −1 i−1 i + nχg−1 [n,∞) 2n g [ 2n , 2n ) i=1 i−1 i is A0 -measurable. Since g−1 [n, ∞), . . . g−1 [ n , n ) ∈ A0 . So gn 2 2 is progressively measurable. Therefore g is progressively measurable. Hence f is progressively measurable. This completes the proof of the Lemma. To solve (1) we use the standard iteration technique. Step 1. Let ξ0 (t, w) = x, ξn (t, w) = x +

Zt

hσ(s, ξn−1 (s, w)), dβ(s, w)i +

0

Zt

b(s, ξn−1 (s, w))ds.

0

By induction, it follows that ξn (t, w) is progressively measurable. Step 2. Let ∆n (t) = E(|ξn+1 (t) − ξn (t)|2 ). If 0 ≤ t ≤ T , ∆n (t) ≤ C ∗

Rt 0

∆n−1

(s)ds and ∆0 (t) ≤ C ∗ t, where C ∗ is a constant depending only on T . Proof. ∆0 (t) = E(|ξ(t) − x|2 )   t Zt   Z   2 = E | h(s, x), dβ(s, x)i + b(s, x)ds|    0

0

  t   Z   2 ≤ 2E | hσ(s, x), dβ(s, x)i|  +   0

  t   Z   2 + 2E | b(s, x)ds|   

(use the fact that |x + y|2

0

≤ 2(|x|2 + |y|2 ) ∀x, y ∈ Rd )

25. Construction of a Diffusion Process

204

 t   t  Z   Z     ∗  2   = 2E  T r σσ ds = 2E | b(s, x)ds|  .     0

0

198

or

 t   t  Z   Z     ∗  2    ∆0 (t) ≤ 2E  tr σσ ds + 2E t |b(s, x)| ds     0

0

(Cauchy-Schwarz inequality)

≤ 2nd C 2 (1 + T )t.

∆n (t) = E(|ξn+1 (t) − ξn (t)|2 )  t  Z  = E | hσ(s, ξn (s, w)) − σ(s, ξn−1 (s, w))dβi+  0

+

Zt 0

   b(s, ξn (s, w)) − b(s, ξn−1 (s, w))ds|2  

 t   Z    ≤ 2E | hσ(s, ξn (s, w)) − (s, ξn−1 (s, w)), dβ(s, w)i|2  +   0

+ 2E(|

Zt

(b(s, ξn (s, w)) − b(s, ξn−1 (s, w))ds|2 )

0

Zt ≤ 2E( tr[(σ(s, ξn (s, w)) − σ(s, ξn−1 (s, w))]× 0 ∗

× [σ (s, ξn (s, w)) − σ∗ (s, ξn−1 (s, w))]ds]+   t   Z   + 2E t |b(s, ξn (s, w)) − b(s, ξn−1 (s, w))|2 ds   0

≤ 2dn A

2

Zt 0

2

∆n−1 (s)ds + 2tA n

Zt 0

∆n−1 (s)ds

205

2

≤ 2dn A (1 + T )

ZT

∆n−1 (s)ds.

0

199

This proves the result. Step 3. ∆n (t) ≤



(C ∗ t)n+1 ∀n in 0 ≤ t ≤ T , where (n + 1)!

C ∗ = max{2nd C 2 (1 + T ),

and

A2 (1 + T )}.

Proof follows by induction on n. Step 4. ξn |[0,T ]×Ω is Cauchy in L2 ([0, T ] × Ω, B([0, T ] × Ω), µ × P), where µ is the Lebesgue measure on [0, T ]. Proof. ∆n (t) ≤

(C ∗ t)n+1 implies that (n + 1)! ||ξn+1 − ξn ||22 ≤

(C ∗ T )n+2 . (n + 2)!

Here || · ||2 is the norm in L2 ([0, T ] × Ω). Thus ∞ X n=1

||ξn+1 − ξn ||2 < ∞,

proving Step (4). 

Step 5. (4) implies that ξn |[0,T ]×Ω is Cauchy in L2 ([0, T ] × Ω, A0 , µ × P) where A0 is as in Lemma 1. Thus ξn |[0,T ]×Ω converges to ξT in L2 ([0, T ] × Ω) where ξT is progressively measurable. Step 6. If ξn |[0,T 2 ]×Ω ξT 2 in L2 ([0, T 2 ] × Ω) and ξn |[0,T 1 ]×Ω ξT 1

in

L2 ([0, T 1 ] × Ω),

then ξT 2 |[0,T ]×Ω = ξT 1 a.e. on [0, T 1 ] × Ω, T 1 < T 2 . 1 This follows from the fact that if ξn → ξ in L2 , a subsequence of (ξn ) converges pointwise a.e. to ξ.

200

25. Construction of a Diffusion Process

206

Step 7. Let ξ be defined on [0, ∞) × Ω by ξ|[0,T ]×Ω = ξ T . We now show that ξ(t, w) = x +

Zt

hσ(s, ξ(s, ·)), dβ(s, ·)i +

0

Zt

b(s, ξ(s, ·))ds.

0

Proof. Let 0 ≤ t ≤ T . By definition, ξn (t, w) = x +

Zt

hσ(s, ξn−1 (s, ·)), dβ(s, ·)i +

0

Zt

b(s, ξn−1 (s, ·))ds.

0

 t 2  Z      E  h(σ(s, ξn (s, ·)) − σ(s, ξ(s, w))), dβ(s, w)i    0

= E(

ZT

tr[(σ(s, ξn (s, w)) − σ(s, ξ(s, w)))(σ(s, ξn (s, w)) − σ(s, ξ(s, w)))∗ ds

0

≤ dn A

ZT Z 0

|ξn (s, w) − ξ(s, w)|2 ds → 0

as n → ∞

Ω

(by Lipschitz condition on σ). Therefore Zt

hσ(s, ξn−1 (s, w), dβ(s, w)i →

0

201

Zt

hσ(s, ξ(s, w)), dβ(s, w)i

0

in L2 (Ω, P). Similarly, Zt

b(s, ξn (s, w))ds →

0

Zt

b(s, ξ(s, w))ds,

in

0

Thus we get (*)

ξ(t, w) = x +

Zt 0

hσ(s, ξ(s, w)), dβ(s, w)i +

L2 .

207

+

Zt

b(s, ξ(s, w))ds

a.e. in

t, w.

0

 Step 8. Let ξ(t, w) ≡ the right hand side of (∗) above. Then ξ(t, w) is almost surely continuous because the stochastic integral of a bounded progressively measurable function is almost surely continuous. The re∞ S sult follows by noting that [0, ∞) = [0, n] and a function on [0, ∞) is n=1

continuous iff it is continuous on [0, n], ∀n.

Step 9. Replacing ξ by ξ in the right side of (∗) we get a solution Zt

ξ(t, w) = x +

hσ(s, ξ), dβi +

0

Zt

b(s, ξ(s, w))ds

0

that is a.s. continuous ∀t and a.e. Uniqueness. Let ξ and η be two progressively measurable a.s. continuous functions satisfying (1). As in Step 3, Zt Zt ∗ E(|ξ(t, w) − x| ) ≤ 2(E( tr σσ ds) + 2E(t b|2 ds) 2

0

0

ZT ZT ∗ ≤ 2E( tr σσ ds + 2E(T |b|2 ds), 0

if 0 ≤ t ≤ T

0

< ∞. 202

Thus

E(|ξ(t, w)|2 )

is bounded in 0 ≤ t ≤ T . Therefore

φ(t) = E(|ξ(t, w) − η(t, w)|2 )

≤ 2E(|ξ(t, w)|2 ) + 2E(|η(t, w)|2 )

25. Construction of a Diffusion Process

208

and so φ(t) is bounded in 0 ≤ t ≤ T . But 2

φ(t) ≤ 2dn A (1 + T )

Zt

φ(s)ds

0

as in Step 2; using boundedness of φ(t) in 0 ≤ t ≤ T we can find a constant C such that φ(t) ≤ Ct

and

φ(t) ≤ C

Zt

φ(s)ds,

0 ≤ t ≤ T.

0

By iteration φ(t) ≤

(Ct)n (CT )n ≤ . Therefore n! n! φ=0

on

[0, T ],

i.e. ξ(t, w) = η(t, w) a.e. in [0, T ]. But rationals being dense in R we have ξ = η a.e. and ∀t. It is now clear that ξ ∈ I[b, a]. Remark. The above theorem is valid for the equation ξ(t, w) = x0 +

Zt t0

hσ(s, ), dβi +

Zt

b(s, ξ)ds,

t0

∀t ≥ t0 .

This solution will be denoted by ξt0 ,x0 . 203

Proposition . Let φ : C[(0, ∞); Rn ) → C([t0 , ∞); Rd ) be the map sending w to ξt0 ,x0 , P the Brownian measure on C([0, ∞); Rn ). Let Pt0 ,x0 = Pφ−1 be the measure induced on C([t0 , ∞); Rd ). Define X : [t0 , ∞) × C)[t0 , ∞); Rd ) by X(t, w) = w(t). Then X is an It¨o process relative to (C([t0 , ∞); Rd ), t0 , Pt0 ,x0 ) with parameters [b(t, Xt ), a(t, Xt )]. The proof of the proposition follows from

209 Exercise. Let (Ω, F t , P), (Ω, Ft , P) be any two measure spaces with X, Y progressively measurable on Ω, Ω respectively. Suppose λ : Ω → Ω is such that λ is (F t , Ft )-measurable for all t, and Pλ−1 = P. Let X(t, w) = Y(t, λw), ∀w ∈ Ω. Show that (a) If X is a martingale, so is Y. (b) If X ∈ I[b(t, Xt ), a(t, Xt )] then Y ∈ I[b(t, Yt ), a(t, Yt )].

P Lemma . Let f : R2 → R be (Ω, P)-measurable, a sub - σ - algebra P of F . Let X : Ω → R and Y : Ω → R be such that X is -measurable P and Y is -independent. If g(w) = f (X(w), Y(w)) with E(g(w)) < ∞, then X E(g| )(w) = E( f (x, Y)| x=X(w) , i.e.

E( f (X, Y)|P )(w)

=

Z

f (X(w), Y(w′ ))dP(w′ ).

Ω

Proof. Let A and B be measurable subsets in R. The result is trivially 204 verified if f = XA×B. The set A = {F ∈ R : the result is true for XF } is a monotone class containing all measurable rectangles. Thus the Lemma is true for all characteristic functions. The general result follows by limiting procedures. 

26. Uniqueness of Diffusion Process IN THE LAST section we proved that ξ(t, w) = x0 +

Zt

205

hσ(s, ξ(s, w)), dβ(s, w) +

t0

Zt

b(s, (s, w)ds

t0

has a solution under certain conditions on b and σ where σσ∗ = a. The measure Pt0 ,x0 = Pξt−1 was constructed on (C([t0 , ∞); Rd ), Ft0 ) so that 0 ,x0 the map X(t, w) = w(t) is an Itˆo process with parameters b and a. We now settle the uniqueness question, about the diffusion process. Theorem . Let (i) a : [0, ∞) × Rd → S d+ and b : [0, ∞) × Rd → Rd be bounded measurable functions; (ii) Ω = C([0, ); Rd ); (iii) X : [0, ∞) × Ω → Rd be defined by X(t, w) = w(t); (iv) Xt = σ{X(s) : 0 ≤ s ≤ t}; (v) P be any probability measure on   [   = σ  Xt  t≥0

211

26. Uniqueness of Diffusion Process

212

such that P{X(0) = x0 } = 1 and X is an Itˆo process relative to (Ω, Xt , P) with parameters b(t, Xt ) and a(t, Xt ); (vi) σ : [0, ∞) × Rd → Md×n be a bounded measurable map into the set of all real d × n matrices such that σσ∗ = a on [0, ∞) × Rd . 206

Then there exists a generalised n-dimensional Brownian motion β P on (Ω, t , Q) and a progressively measurable a.s. continuous map ξ : [0, ∞) × Ω → Rd satisfying the equation (1)

ξ(t, w) = x0 +

Zt 0

hσ(s,

Z

(s, w)), dβ(s, w)i +

Zt

b(s, ξ(s, w))ds

0

with Qξ −1 = P, where ξ : Ω → Ω is given by (ξ(w))(t) = ξ(t, w). Roughly speaking, any Itˆo process can be realised by means of a diffusion process governed by equation (1) with σσ∗ = a. Proof. Case (i). Assume that there exist constants m, M > 0 such that mI ≤ a(t, x) ≤ MI and σ is a d × d matrix satisfying σσ∗ = a. In this P case we can identify (Ω, t , Q) with (Ω, Ft , P). Since D(t, ·) is an Itˆo process, exphθ, X(t)i −

Zt

1 hθ, b(s, X(s, ·))ids − 2

0

Zt

hθ, a(s, X(s, ·))θids

0

is a (Ω, Ft , P)-martingale. Put

Y(t, w) = X(t, w) −

Zt

b(s, X(s, w))ds − x0 .

0

Clearly Y(t, w) is an Itˆo process corresponding to the parameters [0, a(s, X s )],

213 so that 1 exphθ, Y(t, w)i − 2

Zt

hθ, a(s, X(s, ·))θids

0

is a (Ω, Ft , P)-martingale. The conditions m ≤ a ≤ M imply that σ−1 exists and is bounded. Let η(t) =

Zt

−1

σ dY =

0

Zt

σ−1 (s, X(s, ·))dY(s, ·),

0

so that (by definition of a stochastic integral) η is a (Ω, Ft , P)-Itˆo pro- 207 cess with parameters zero and σ−1 a(σ−1 )∗ = 1. Thus η is a Brownian motion relative to (Ω, Ft , P). Now by change of variable formula for stochastic integrals, Zt 0

σdη =

Zt

σσ−1 dY

0

= Y(t) − Y(0) = Y((t), since Y(0) = 0. Thus X(t) = x0 +

Zt

σ(s, X(s, ·))d +

0

Zt

b(s, X(s, ·))ds.

0

Taking Q = P we get the result. Case (ii). a = 0, b = 0, x0 = 0, σ = 0 where σ ∈ Md×n . Let (Ω∗ , Ft∗ , P∗ ) be an n-dimensional Brownian motion. Define X (Ω, , Q) = (Ω × Ω∗ , Ft × Ft∗ , P × P∗ ). t

If β is the n-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ ), we define β on Ω by β(t, w, w∗ ) = β(t, w∗ ). It is easy to verify that β is an P n-dimensional Brownian motion on (Ω, t , Q). Taking ξ(t, w, w∗) = x0 we get the result. 

26. Uniqueness of Diffusion Process

214

Before we take up the general case we prove a few Lemmas. Lemma 1. Let σ : Rn → Rd be linear σσ∗ = a : Rd → Rd ; then 208 there exists a linear map which we denote by σ−1 : Rd → Rn such that σ−1 aσ−1∗ = πNσ⊥ , where π denotes the projection and Nσ null space of σ. Proof. Let Rσ = range of σ. Clearly σ : Nσ⊥ → R is an isomorphism. Let τ : Rσ → Nσ⊥ be the inverse. We put ⊥ σ−1 = τ ⊕ 0 : Rσ ⊕ R⊥ σ → Nσ ⊕ Nσ .

 Lemma 2. Let X, Y be martingales relative to (Ω, Ft , P) and (Ω, F t , P) respectively. Then Z given by Z(t, w, w) = X(t, w)Y(t, w) is a martingale relative to (Ω × Ω, Ft × F t , P × P). Proof. From the definition it is clear that for every t > s Z Z Z(t, w, w)d(P × P)|Fs ×Fs = Z(s, w, w)d(P × P) A×A

A×A

if A ∈ Fs and A ∈ F s . The general case follows easily.



As a corrollary to Lemma 2, we have Lemma 3. Let X be a d-dimensional Itˆo process with parameters b and a relative to (Ω, Ft , P) and let Y be a d-dimensional Itˆo process relative to (Ω, F t , P) relative to b and a. Then Z(t, w, w) = (X(t, w), Y(t, w))h is ai (d + d)-dimensional Itˆo process with parameters B = (b, b), A = a0 0a

relative to (Ω × Ω, Ft × F t , P × P).

215 209

Lemma 4. Let X be an Itˆo process relative to (Ω, Ft , P) with parameRt ters 0 and a. If θ is progressively measurable such that E( |θ|2 , ds) < 0

Rt

∞, ∀t and θaθ∗ is bounded, then hθ, dXi ∈ I[0, θaθ∗ ]. 0

Proof. Let θn be defined by    θi , θni =   0,

if |θ| ≤ n, otherwise;

Rt Then hθn , dXi ∈ I[0, θn aθn∗ ]. Therefore 0

Xn (t) = exp(λ

Zt

λ2 hθn , dXi − 2

Zt

hθn , aθn ids

0

0

is a martingale converging pointwise to  t  Zt  Z  2 λ   hθ, aθids . X(t) = exp λ hθ, dXi −   2 0

0

Rt

To prove that hθ, dXi is an Itˆo process we have only to show that 0

Xn (t) is uniformly integrable. Without loss of generality we may assume that λ = 1. Let [0, T ] be given   t  Zt   Z     E(Xn2 (t, w)) = E exp 2 hθn , dXi − hθn , aθn ids    0

0

  t Zt   Z   = E exp 2 hθn , dXi − 2 hθn , aθn ids   0

+

Zt 0

   hθn , aθn ids . 

0

26. Uniqueness of Diffusion Process

216

≤ eT sup hθn , aθn i. 0≤t≤T

210

But hθ, aθ∗ i is bounded and therefore hθn , aθn i is uniformly bounded in n. Therefore (Xn ) are uniformly integrable. Thus X(t, ·) is a martingale. Case (iii). Take d = 1, and assume that Zt

a−1 χ(a>0) ds < ∞, ∀t

0

with a > 0; let σ = +ve squareroot of a. Define 1/σ = 1/σ if σ > 0, and 1/σ = 0 if σ = 0. Let Y(t) = X(t) − x0 −

Zt

b(s, X(s))ds.

0

Denote by Z the one-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ ) where Ω∗ = C([0, ∞), R). Now Y ∈ I[0, a(s, X(s, ·))], Z ∈ I[0, 1]. By Lemma 3,

!! a 0 (Y, Z) ∈ I (0, 0); . 0 1

If ∗

η(t, w, w ) =

Zt 0

1 χ(σ>0) , χσ=0 , d(Y, Z)i h σ(s, X(s, ·))

!

then Lemma 4 shows that η ∈ I[0, 1]. Therefore η is a one-dimensional Brownian motion on Ω = (Ω × Ω∗ , Ft × Ft∗ , P × P∗ ). Put Y(t, w, w∗ ) = Y(t, w) and

X(t, w, w∗) = X(t, w);

217 211

then Zt

σdη =

Zt

=

Zt

0

1 σ χ(σ>0) dY + σ

0

Zt

σχ(σ=0) dZ

0

χ(σ>0) dY.

0

Since  t  2   t Z Z          2    E  χ(σ=0) dY   = E  σ χ(σ=0) ds = 0,       0

0

it follows that

Zt

σdη =

0

Zt

dY = Y(t) = Y(t, w, w∗ ).

0

Thus, ∗

X(t, w, w ) = x0 +

Zt

σ(s, X(s, w, w∗ )dη+

Zt

b(s, X(s, w, w∗ )ds

0

+

0

with X(t, w, w∗) = X(t, w). Now −1

(P × P∗ )X (A) = (P × P∗ )(A × Ω∗ ) = P(A). Therefore (P × P∗ )X

−1

=P

or

QX

−1

= P.

26. Uniqueness of Diffusion Process

218 Case (iv). (General Case). Define

Y(t, ·) = X(t, ·) − x0 −

Zt

b(s, X(s, ·))ds.

0

Therefore Y ∈ I[0, a(s, X(s, ·))] relative to (Ω, Ft , P). Let Z be the 212 n-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ ) where Ω∗ = C([0, ∞); Rn ). " " ## a(s, X s ), 0 (Y, Z) I (0, 0); 0 I Let σ be a d × n matrix such that σσ∗ = a on [0, ∞) × Rd . Let η(t, w, w ) =

Zt

=

Zt

∗

−1

σ (s, X(s, w))dY(s, w) +

0

Zt

rNσ (s, Z(s, w∗))dZ(s, w∗ )

0

h(σ−1 (s, X(s, w)), πNσ (s, Z(s, w∗ ))), d(Y, Z)i.

0

Therefore η is an Ito process with parameters zero and ! ∗! a 0 σ−1 −1 A = (σ , πN ) 0 I πNσ∗ = σ−1 a(σ−1 )∗ + πNσ πNσ∗ . = πNσ + πNσ

(for any projection PP∗ = PP = P)

= IRn . Therefore η is n-dimensional Brownian motion on (Ω, Ft , P) = (Ω × Ω∗ , Ft × Ft∗ , P × P∗ ). Zt σ(s, X(s, w))dη(s, w, w∗ ) 0

219

=

Zt

=

Zt

σσ dY +

=

Zt

πRσ dY, since σσ−1 = πRσ and σπNσ = 0,

=

Zt

σ(s, X(s, w, w∗ ))dη(s, w, w∗ ), where X(s, w, w∗) = X(s, w),

0

−1

0

Zt

σπNσ dZ.

0

0

(1 − πRσ )dY.

0

 213 Claim.

Rt

πRσ dY = 0.

0

For

 t 2  Zt Z   Zt     E  πRσ dY   = aπRσ ds = σσ∗ πRσ ds    0

0

=

Zt

0

σ(0)ds = 0.

0

Therefore we obtain Zt

∗

σ(s, X(s, w))dη(s, w, w ) =

0

Zt

dY = Y(t) − Y(0) = Y(t)

0

putting Y(t, w, w∗) = Y(t, w), one gets ∗

X(t, w, w ) = x0 +

Zt 0

σ(s, X(s, w, w∗ ))dη(s, w, w∗ )

26. Uniqueness of Diffusion Process

220

+

Zt

b(s, X(s, w, w∗ ))ds.

0

As in Case (iii) one shows easily that (P × P∗ )X

−1

= P.

This completes the proof of the theorem. Corollary . Let a : [0, ∞) × Rd → S d+ , and b : [0, ∞) × Rd → Rd be bounded, progressively measurable functions. If for some choice of a Lipschitz function σ : [0, ∞) × Rd → Md×n , σσ∗ = a then the Itˆo process corresponding to [b, a) is unique. 214

To state the result precisely, let P1 and P2 be two probability measures on C([0, ∞); Rd ) such that X(t, w) = w(t) is an Itˆo process with parameters b and a. Then P1 = P2 . Proof. By the theorem, there exists a generalised n-dimensional BrowP nian motion βi on (Ωi , it , Qi ) and a map ξi : Ωi → Ω satisfying (for i = 1, 2) (t,w)

i

= x0 +

Zt

σ(s, ξi (s, w))dβi (s, w) +

0

Zt

b(s, ξi (s, w))ds.

0

and Pi = Qi ξi−1 . Now σ is Lipschitz so that ξi is unique but we know that the iterations converge to a solution. As the solution is unique the iterations converge to ξi . Each iteration is progressively measurable with respect to i t = σ{βi (s); 0 ≤ s ≤ t} so that ξi is also progressively

measurable with respect to Fti . Thus we can restate the result as follows: There exists (Ωi , Fti , Qi ) and a map ξi : Ωi → Ω satisfying ξi (t, w) = x0 +

Zt 0

σ(s, ξi (s, w))dβi (s, w)

221

+

Zt

b(s, ξi (s, w))ds,

0

and Pi = Qi ξi−1 . (Ωi , Fti , Qi , βi ) can be identified with the standard Brownian motion ∗ (Ω , Ft∗ , Q, β). Thus P1 = Qξ −1 = P2 , completing the proof. 

27. On Lipschitz Square Roots Lemma . Let f : R → R be such that f (x) ≥ 0, f (x) ∈ C 2 and | f ′′ (x)| ≤ 215 A on (−∞, ∞); then √ p | f ′ (x)| ≤ f (x) 2A.

Proof.

0 ≤ f (y) = f (x) + (y − x) f ′ (x) + ≤ f (x) + Z f ′ (x) +

(y − x)2 ′′ f (ξ) 2

Z 2 ′′ f (ξ) 2

where Z = y − x, or f (y) ≤ f (x) + Z f ′ (x) +

AZ 2 . Therefore 2

AZ 2 + Z f ′ (x) + f (x) ≥ 0, ∀Z ∈ R 2 | f ′ (x)|2 ≤ 2A f (x). So | f ′ (x)| ≤

√

2A f (x). 

Note. If we take f (x) = x2 , we note that the constant is the best possible. Corollary . If f ≥ 0, | f ′′ | ≤ A, then √ √ √ | ( f (x1 )) − ( f (x2 )) ≤ (A/2)|x1 − x2 |. 223

27. On Lipschitz Square Roots

224

√ Proof. Let ǫ > 0, then ( f (x) + ǫ) is a smooth function. √ f ′ (x) ( f (x) + ǫ)′ ( ( f (x) + ǫ))′ = √ = √ . 2 ( f (x) + ǫ) 2 ( f (x) + ǫ) Therefore √ √ √ |( ( f (x) + ǫ))′ | ≤ (2A/2) ≤ (A/2), or

√ √ √ | ( f (x1 ) + ǫ) − ( f (x2 ) + ǫ)| ≤ (A/2)|x1 − x2 |.

216

Let ǫ → 0 to get the result. We now consider the general case and give conditions on the matrix a so that σ defined by σσ∗ = a is Lipschitz.  Theorem . Let a : Rn → S d+ be continuous and bounded C 2 -function such that the second derivative is uniformly bounded, i.e. ||D s Dr ai j || ≤ d ). If σ : Rn → S d+ is M, where M is independent of i, j, r, s; (Dr ≡ dxr the unique positive square root of a, then ||σ(x1 ) − σ(x2 )|| ≤ A|x1 − x2 |, ∀x1 , x2 , A = A(M, d). Proof. Step 1. Let A ∈ S d+ be strictly positive such that ||I − A|| < 1. Then √

√ A = (I − (I − A)) ∞ X Cr = (I − A)r , r! r=0

√ so that on the set {A : ||I − A|| < 1} the map A → A is C ∞ (in fact analytic). Now assume that A is any positive definite matrix. Let λ1 , . . . , λn be the eigen values so that λ j > 0, j = 1, 2 . . . n. Therefore I − ǫA is

225 aymmetric with eigen values 1 − ǫλ j . By choosing ǫ sufficiently small we can make ||I − ǫA|| = max{1 − ǫλ1 , . . . , 1 − ǫλn } < 1. Fixing such an ǫ we observe that √

1 √ 1 √ A = √ (ǫA) = √ (I − (I − ǫA)). ǫ ǫ

So the map A → definite matrices.

√

217

A is smooth on the set of symmetric positive

√ Step 2. Let n = 1, σ(t0 ) = (a(t0 ) where a(t0 ) is positive definite. Assume a(t0 ) to be diagonal so that σ(t0 ) is also diagonal. X σi j (t)σ jk (t) = aik (t). j

Differentiating with respect to t at t = t0 we get X X σ′i j (t0 )σ jk (t0 ) = a′ik (t0 ) σi j (t0 )σ′jk (t0 ) + j

j

or

√

√ aii (t0 )σ′ik (t0 ) + akk (t0 )σ′ik (t0 ) = a′ik (t0 )

or σ′ik (t0 )

a′ik (t0 ) =√ √ . (aii (t0 )) + (akk (t0 ))

Since the second derivatives are bounded by 4M and aii −2ai j +a j j ≥ 0, we get √ √ |a′ii (t) + 2a′i j (t) + a′j j (t)| ≤ (8M) (aii (t) + 2ai j (t) + a j j (t)) √ √ √ ≤ (8M) 2 (aii + a j j )(t) or (1)

√ √ √ |a′ii (t) + 2a′i j (t) + a′j j (t)| ≤ 4 M( aii + a j j ).

27. On Lipschitz Square Roots

226 Since a is non-negative definite,

√ √ |a′ii (t)| ≤ (2M) (aii (t)), ∀i. substituting this in (1), we get √ √ √ |a′i j (t)| ≤ 4 M( aii + a j j ), and hence

√ |σ′i j (t0 )| ≤ 4 M.

Step 3. Let a(t0 ) be positive definite and σ its positive definite square root. There exists a constant unitary matrix α such that αa(t0 )α−1 = b(t0 ) is a diagonal positive definite matrix. Let λ(t0 ) be the positive square root of b(t0 ) so that λ(t0 ) = ασ(t0 )α−1 . Therefore σ′ (t0 ) = (α−1 λ′ α)(t0 ) where (σ′ (t0 ))i j = σ′i j (t0 ) and a′′ (t0 ) = (α−1 b′′ α)(t0 ). Since α is unitary. ||λ|| = ||σ||, ||a′′ || = ||b′′ ||, ||λ′ || = ||σ′ ||. By hypothesis, ||b′′ || = ||a′′ || ≤ C(d) · M. Therefore √ ||λ′ || ≤ 4 (MC(d)),

i.e.

√ ||σ′ || ≤ 4 (MC(d)). Thus ||σ(t1 ) − σ(t2 )|| ≤ |t1 − t2 |C(M, d).

Step 4. Let a : R → S d+ and σ be the unique non-negative definite square root of a. For each ǫ > 0 let aǫ = a + ǫI, σǫ = unique positive square root of aǫ . Then by step 3, ||σǫ (t1 ) − σǫ (t2 )|| ≤ C(M, d)|t1 − t2 |.

218

227

219

If a is diagonal then it is obvious that σǫ → σ as ǫ → 0. In the general case reduce a to the diagonal form and conclude that σǫ → σ. Thus ||σ(t1 ) − σ(t2 )|| ≤ C(M, d)|t1 − t2 |. Step 5. Let a : Rn → S d+ and σ2 = a, with ||Dr D s ai j || ≤ M, ∀x, i, j; r, s × ǫRn . Choose x1 , x2 ∈ Rn . Let x1 = y1 , y2 , . . . , yn+1 = x2 be (n + 1) points such that yi and yi+1 differ almost in one coordinate. By Step 4, we have ||σ(yi ) − σ(yi+1 )|| ≤ C|yi − yi+1 |.

(*)

The result follows easily from the fact that ||x||1 =

n X i=1

|xi | and

||x||2 = (x1 + · · · + xn )1/2

are equivalent norms. This completes the proof of the theorem. 

28. Random Time Changes LET

X ∂ 1X ∂2 L= ai j + bj 2 i, j ∂xi ∂x j ∂x j j

with a : Rb → S d+ and b : Rd → Rd bounded measurable funcitons. Let X(t, ·), given by X(t, w) = w(t) for (t, w) in [0, ∞) × C([0, ∞) : Rd ) be an Itˆo process corresponding to (Ω, Ft , Q) with parameters b and a where Ω = C([0, ∞); Rd ). For every constant c > 0 define    1 X  2 X ∂ ∂  ai j bj Lc ≡ c  +  2 i, j ∂xi ∂x j ∂x j j

Define Qc by Qc = PT c−1 where (T c w)(t) = w(ct). Then one can show that X is an Itˆo process corresponding to (Ω, Ft , Qc ) with parameters cb and ca [Note: We have done this in the case where ai j = δi j ]. Consider the equation ∂u = Lc u with ∂t

u(0, x) = f (x).

∂u This can be written as = Lu with u(0, x) = f (x) when τ = ct. ∂τ Thus changing time in the differential equation is equivalent to stretching time in probablistic language. So far we have assumed that c is a constant. Now we shall allow c to be a function of x. 229

220

28. Random Time Changes

230

Let φ : Rd → R be any bounded measurable function such that 0 < C1 ≤ φ(x) < C2 < ∞,

∀x ∈ Rd

and suitable constants C1 and C2 . If " X # X 1 ∂ ∂2 L≡ ai j + bj 2 ∂xi ∂x j ∂x j 221

we define # " X X ∂ 1 ∂2 ai j + bj . Lφ = φL ≡ φ 2 ∂xi ∂x j ∂x j In this case we can say that the manner which time changes depends on the position of the particle. Define T φ : Ω → Ω by (T φ w)(t) = w(τt (w)) where τt (w) is the solution of the equation Zτt

ds = t. φ(w(s))

0

1 1 ≤ t ≤ τt . When φ ≡ c a C1 C2 constant, then τt = ct and T φ coincides with T c . As Zλ 1 ds 0 < C1 ≤ φ ≤ C2 < ∞, φ(w(s)) As C1 ≤ φ ≤ C2 it is clear that τt

0

is continuous and increases strictly from 0 to ∞ as λ increases, so that τt exists, is unique, and is a continuous function of t for each fixed w. Some properties of T φ . (i) If l is the constant function taking the value 1 then it is clear that T l = identity.

231 222

(ii) Let φ and ψ be two measurable funcitons such that 0 < a ≤ φ(x), ψ(x) ≤ b < ∞, ∀x ∈ Rd . Then T φ ◦ T = T φψ = T ψ ◦ T φ . Proof. Fix w. Let τt be given by Zτt

1 ds = t. φ(w(s))

0

Let w∗ (t) = w(τt ) and let σt be given by Zσt 0

1 ds = t. φ(w∗ (s))

Let w∗∗ (t) = w∗ (σt ) = w(τσt ). Therefore ((T ψ ◦ T φ )w)(t) = (T φ w∗ )(t) = w∗ (σt ) = w∗∗ (t) = w(τσt ).

Hence to prove the property (ii) we need only show that Zτσt

1 1 ds = t. φ(w(s)) ψ(w(s))

0

Since

Zτt

1 ds = t, φ(w(s))

dt 1 = dτt φ(w(τt ))

0

and

dt 1 1 = = dσt ψ(w∗ (σt )) ψ(w(τσt )) Therefore dτσt dτσt dσt = − = φ(w(τσt ))φ(w∗ (σt )) dt dσt dt = φ(w(τσt )ψ(w(τσt ))

28. Random Time Changes

232

= (φψ)(w(τσt )). 223

Thus

Zτσt

1 ds = t. (φψ)(w(s))

0

This completes the proof.



(iii) From (i) and (ii) it is clear that T φ−1 = T φ−1 where φ−1 =

1 . φ

(iv) (τt ) is a stopping time relative to τt . i.e.     Zλ       1   w : ds ≥ r ∈ λ for each λ ≥ 0.       φ(w(s))     0

(v) T φ (w)(t) = w(τt w) = Xτt (w). Thus T φ is (Ft − Fτt )-measurable, i.e. T φ−1 (Ft ) ⊂ Fτt . Since X(t) is an Itˆo process, with parameters b, a, ∀ f ∈ C0∞ (Rd ), Rt f (X(t)) − (L f )(X(s))ds is a martingale relative to (Ω, Ft , P). By the 0

optional sampling theorem

f (Xτt ) −

Zτt

(L f )(X(s))ds

0

is a martingale relative to (Ω, Fτt , P), i.e.

f (Xτt ) −

Zt 0

(L f )(X(τ s ))dτ s

233 is a martingale relative to (Ω, Fτt , P). But f (X(τt )) −

Zt

dτ s = φ. Therefore dt

(L f )(Xτs )φ(Xτs )ds

0

is a martingale. Put Y(t) = Xτt and appeal to the definition of Lφ to conclude that f (Y(t)) −

Zt

224

(Lφ f )(Y(s))ds

0

is a martingale. Y(t, w) = Xτt (w) = (T φ w)(t). Let F t = σ{Y(s) : 0 ≤ s ≤ t}. Then F t = T φ−1 (Ft ) ⊂ Fτt . Thus f (Y(t)) −

Zt

(Lφ f )(Y(s))ds

0

is a martingale relative to (Ω, F t , P). Define Q = PT φ−1 so that f (X(t)) −

Zt

(Lφ f )(X(s))ds

0

is an (Ω, Ft , Q)-martingale, i.e. Q is an Itˆo process that corresponds to the operator φL. Or, PT φ−1 is an Itˆo process that corresponds to the operator φL. We have now proved the following theorem. Theorem . Let Ω = C([0, ∞); Rd ); X(t, w) = w(t); X ∂2 ∂ 1X L= ai j + bj . 2 i, j ∂xi ∂x j ∂x j Suppose that X(t) is an Itˆo process relative to (Ω, Ft , P) that corresponds to the operator L. Let 0 ≤ C1 ≤ φ ≤ C2 where φ : Rd → R 225 is measurable. If Q = PT φ−1 , then X(t) is an Itˆo process relative to (Ω, Ft , Q) that corresponds to the operator φL.

28. Random Time Changes

234

As 0 < C1 ≤ φ ≤ C2 , we get 0 < 1/C2 ≤ 1/φ < 1/C1 with T φ−1 ◦ T φ = I. We have thus an obvious corollary. Corollary . There exists a probability measure P on Ω such that X is an Itˆo process relative to (Ω, Ft , P) that corresponds to the operator L if and only if there exists a probability measure Q on Ω such that X is an Ito process relative to (Ω, Ft , Q) that corresponds to the operator φL. Remark . If C2 ≥ φ ≥ C1 > 0 then we have shown that existence and uniqueness of an Itˆo process for the operator L guarantees existence and uniqueness of the Itˆo process for the operator φL. The solution is no longer unique if we relax the strict positivity on C1 as is illustrated by the following example. 1 ∂2 Let φ ≡ a(x) = |x|α ∧ 1 where 0 < α < 1 and let L = a . Define 2 ∂x δ0 on {C([0, ∞); R)} by    1, if θ ∈ A, ∀Aǫ, δ0 (A) =   0, if θ < A, where θ is the zero function on [0, ∞).

Claim. δ0 is an Itˆo process with parameters 0 and a. For this it is enough to show that, ∀ f ∈ C0∞ (R) f (X(t)) −

Zt

(L f )(X(s))ds

0

226

is a martingale, using a(0) = 0, it follows easily that Z Zt A

(L f )(X(σ))dσdδ0 = 0

0

∀ Borel set A of C([0, ∞); R) and Z A

R

f (X(t))dδ0 = 0 if θ < A and

A

f (X(t))dδ0 = f (0)

235 if θ ∈ A, and this is true ∀t, showing that X(t, w) = w(t) is an Itˆo process relative to δ0 corresponding to the operator L. Next we shall define T a (as in the theorem); we note that T a cannot be defined everywhere (for example T a (θ) is not defined). However T a is defined a.e. P where P = P0 is the Brownian motion.  t  Z  Zt Z∞ −y 1 1 1   P √ E  ds = e 2s dy ds < ∞ α α  |X(s)|  y (2πs) 0

0

0

since 0 < α < 1. Thus by Fubini’s theorem, Zt 0

1 ds < ∞ |w(s)|α

a.e.

Taking t = 1, 2, 3 . . ., there exists a set Ω∗ such that P(Ω∗ ) = 1 and Zt 0

1 ds < ∞, |w(s)|α

Observe that

Zt 0

∀t,

∀w ∈ Ω∗

1 ds < ∞ |w(s)|α

implies that Zt 0

1 ds < ∞, |w(s)|α ∧ 1

for

227

Zt

=

Z

[0,t]{|w(s)|α >1}

ds = |w(s)|α ∧ 1 0 Z ds + |w(s)|α ∧ 1

{|w(s)|α ≤1}[0,t]

ds ,... |w(s)|α

28. Random Time Changes

236

α

≤ m{(|w(s)| > 1)[0, t]} +

Zt 0

1 ds < ∞ |w(s)|α

(m = Lebesgue measure) Thus T a is defined on the whole of Ω∗ . Using the same argument as in the theorem, it can now be proved that X is an Itˆo process relative to Q corresponding to the operator L. Finally, we show that Q{θ} = 0. Q{θ} = PT a−1 {θ}. Now: T a−1 {θ} = empty. For, let w ∈ T a−1 {θ}. Then w(τt ) = 0, ∀t, w ∈ Ω∗ . Since |τt − τ s | ≤ |t − s|, one finds that τt is a continuous function of t. Further τ1 > 0, and w = 0 on [0, τ1 ] gives Zτ1 0

1 ds = ∞. |w(s)|α ∧ 1

This is false unless T a−1 {θ} = empty. Thus Q{θ} = 0 and Q is different from δ0 .

29. Cameron - Martin Girsanov Formula LET US REVIEW what we did in Brownian motion with drift. Let (Ω, Ft , P) be a d-dimensional Brownian motion with P{w : w(0) = x} = 1. Let b : Rd → Rd be a bounded measurable function and define   t Zt Z  1   |b|2 ds . Z(t) = exp  hb, dxi −   2 0

0

Then we see that Z(t, ·) is an (Ω, Ft , P)-martingale. We then had a probability measure Q given by the formula dQ = Z(t, ·). dP Ft

We leave it as an exercise to check that in effect X is an Itˆo process relative to Q with parameters b and I. In other words we had made a transition from the operator ∆/2 to ∆/2 + b · ∇. We now see whether such a relation also exists for the more general operator L. Theorem . Let a : Rd → S d+ be bounded and measurable such that a ≥ CI for some C > 0. Let b : Rd → Rd be bounded, Ω = ([0, ∞); Rd ), X(t, w) = w(t), P any probability measure on Ω such that X is an Itˆo 237

228

29. Cameron - Martin - Girsanov Formula

238

process relative to (Ω, Ft , P) with parameters [0, a]. Define Qt on Ft by the rule dQt dP

229

F

t

 t  Zt Z  1   −1 −1 = Z(t, ·) = exp  ha b, dXi − hb, a bids .   2 0

0

Then (i) {0t }t ≥ 0 is a consistent family. (ii) there exists a measure Q on σ(||Ft ): Q = Qt . Ft

(iii) X(t) is an Itˆo process relative to (Ω, Ft , Q) with parameters [b, a], i.e. it corresponds to the operator X ∂ ∂2 1X ai j bj + . 2 i, j ∂xi ∂x j ∂x j j

Proof.

Rt (i) Let A(t) = ha−1 b, dXi. Then A ∈ I[0, hb, a−1 bi]. 0

Therefore Z(t) is a martingale relative to (Ω, Ft , P) hence {Qt }t≥0 is a consistent family. (ii) Proof as in the case of Brownian motion. (iii) We have to show that

exp[hθ, X(t, ·)i − hθ,

Zt

1 bdsi − 2

0

is a martingale relative to (Ω, Ft , Q).

Zt 0

hθ, aθids]

239 Now for any function θ which is progressively measurable and bounded

Zt Zt 1 hθ, aθids] exp[ hθ, dXi − 2 0

230

0

is an (Ω, Ft , P)-martingale. Replace θ by θ(w) = θ + (a−1 b)(χ(s, w)), where θ now is a constant vector. Then Zt Zt 1 −1 exp[ hθ + a b, dXi − hθ + a−1 b, aθids 2 0

0

is an (Ω, Ft , P)-martingale, i.e. 1 exp[hθ, X(t)i − 2

Zt

1 hθ, aθids − 2

0

Zt

1 ha−1 b, aθi − 2

0

Zt

hθ, bi]

0

is an (Ω, Ft , Q)-martingale, and ha−1 b, aθi = ha∗ a−1 b, θi = haa−1 b, θi

(since a = a∗ )

= hb, θi.

Thus 1 exp[hθ, X(t)i − 2

Zt

hθ, aθids −

0

Zt

hθ, bids]

0

is an (Ω, Ft , Q)-martingale, i.e. X is an Itˆo process relative to (Ω, Ft , Q) with parameters [b, a]. This proves the theorem.  We now prove the converse part. Theorem . Let L1 =

1X ∂2 i, jai j 2 ∂xi ∂x j

29. Cameron - Martin - Girsanov Formula

240 and L2 ≡

231

X 1X ∂ ∂2 ai j + bj , 2 i, j ∂xi ∂x j ∂x j j

where a : Rd → S d+ is bounded measurable such that a ≥ CI for some C > 0; b : Rd → Rd is bounded and measurable. Let Ω = C([0, ∞); Rd ) with Ft as usual. Let θ be a probability measure on σ(∪Ft ) and X a progressively measurable function such that X is an Itˆo process relative to (Ω, Ft , Q) with parameters [b, a] i.e. X corresponds to the operator L2 . Let Zt Zt 1 hb, a−1 bids]. Z(t) = exp[− ha−1 b, dXi + 2 0

0

Then (i) Z(t) is an (Ω, Ft , Q)-martingale. (ii) If Pt is defined on Ft by dPt = Z(t), dQ Ft Then there exists a probability measure P on σ(∪Ft ) such that P = Pt Ft

(iii) X is an Itˆo process relative to (Ω, Ft , P) corresponding to parameters [0, a], i.e. X corresponds to the operator L1 . Proof.

(i) Let A(t) =

Zt

h−a−1 b, dXi.

0

Then A(t) is an Itˆo process with parameters [h−a−1 b, bi, ha−1 b, bi].

241 Thus Zt

exp[A(t) −

1 h−a b, bids − 2 −1

Zt

ha−1 b, bids]

0

0

is an (Ω, Ft , Q)-martingale, i.e. Z(t) is an (Ω, Ft , Q) martingale. (ii) By (i), Pt is a consistent family. The proof that there exists a 232 probability measure P is same as before. Since X is an Itˆo process relative to Q with parameters b and a, Zt Zt Zt 1 hθ, aθids] exp[ hθ, dXi − hθ, bids − 2 0

0

0

is a martingale relative to Q for every bounded measurable θ. Replace θ by θ(w) = θ − (a−1 b)(X(s, w)) where θ now is a constant vector to get Zt

exp[hθ, X(t)i −

−1

ha b, dXi −

0

−

Zt

hθ, bi +

0

1 2

Zt

Zt

ha−1 b, bids−

0

hθ − a−1 b, aθ − bids]

0

is an (Ω, Ft , Q) martingale, i.e. exp[hθ, Xi −

Zt

−1

ha b, dXi −

0

1 − 2

Zt

Zt

hθ, bids +

0

1 hθ, aθids − 2

0

Zt

1 ha b, bids + 2 −1

0

+

Zt 0

ha−1 b, aθids]

Zt

ha−1 b, bids−

Zt

hθ, bids+

0

0

29. Cameron - Martin - Girsanov Formula

242

is an (Ω, Ft , Q) martingale. Let θ ∈ Rd , so that 1 exp[hθ, Xi − 2

Zt

1 hθ, bids − 2

0

Zt

1 hθ, aθids + 2

0

Zt

ha−1 b, aθids]Z(t)

0

is an (Ω, Ft , Q)-matringale and ha−1 b, ai = hb, θi 233

Therefore 1 exp[hθ, Xi − 2

(since a = a∗ ).

Zt

hθ, aθids]Z(t)

0

is an (Ω, Ft , Q) martingale. dP Using the fact that = Z(t), we conclude that dQ Ft 1 exp[hθ, Xi − 2

Zt

hθ, aθids]

0

is a martingale relative to (Ω, Ft , P), i.e. X ∈ I[0, a] relative to (Ω, Ft , P). This proves the theorem.



Summary. We have the following situation L1 , Ω, Ft , Ω = C([0, ∞); Rd ), L2 , Ω, Ft .   P a probability measure   X is an Itˆo process relative   to a probability measure Q such that X is an Itˆo Pro-    cess relative to P corre-  =⇒  corresponding to L2 . Q is    dQ sponding to the operator   given by = Z(t, ·) L1 . dP Ft X is an Itˆo process relative to P corresponding to L1 dP 1 where F = dQ t Z(t, ·)

     

⇐=

  X is an Itˆo process relative  to Q corresponding to L . 2   

243 Thus existence and uniqueness for any system guarantees the existence and uniqueness for the other system. Application. (Exercise). Take d = 1, a : R → R bounded and measurable with 0 < C1 ≤ a ∂2 ∂ a < C2 < ∞. Let L = + b . Show that there exists a unique 2 2 ∂x ∂x probability masure P on Ω = C([0, ∞); R) such that X(t) is Itˆo relative to P corresponding to L. (X(t, w) ≡ w(t)) for any given starting point.

234

30. Behaviour of Diffusions for Large Times LET L2 = ∆/2 + b · ∇ WITH b : Rd → Rd measurable and bounded 235 on each compact set. We assume that there is no explosion. If P x is the d-dimensional Brownian measure on Ω = C([0, ∞); Rd ) we know that there exists a probability measure Q x on Ω such that  t  Zt Z  1 dQ x  2  |b| ds = exp  hb, dXi −  dP x t 2 0

0

Let K be any compact set in Rd with non-empty interior. We are interested in finding out how often the trajectories visit K and whether this ‘frequency’ depends on the starting point of the trajectory and the compact set K. Theorem . Let K be any compact set in Rd having a non-empty interior. Let K E∞ = {w : w revisits K for arbitrarily large times}

= {w : there exists a sequence t1 < t2 < · < ∞ with tn → ∞ such that w(tn ) ∈ K}

Then, K either Q x (E∞ ) = 0, ∀x, and ∀K,

or

K Q x (E∞ ) = 1, ∀x, and ∀K.

245

30. Behaviour of Diffusions for Large Times

246

Remark. 1. In the first case lim |X(t)| = +∞ a.e. Q x , ∀x, i.e. almost t→+∞ all trajectories stay within K only for a short period. These trajectories are called transient. In the second case almost all trajectories visit K for arbitrary large times. Such trajectories are called recurrent.

236

2. If b = 0 then Q x = P x . For the case d = 1 or d = 2 we know that the trajectories are recurrent. If d ≥ 3 the trajectories are transient. Proof. Step 1. We introduce the following sets. E0K = {w : X(t, w) ∈ K for some t ≥ 0},

EtK0 = {w : X(t, w) ∈ K for some t ≥ t0 }, 0 ≤ t0 < ∞. Then clearly K E∞ =

∞ \

EnK =

n=1

\

EtK0 .

t0 ≥0

Let K ψ(x) = Q x (E∞ ), F = χE∞K .

K E Qx (F|Ft ) = E Qx (χE∞K |Ft ) = QX(t) (E∞ )

by the Markov property, = ψ(X(t)) a.e. Q x . Next we show that ψ(X(t)) is a martingale relative to Q x . For, if s < t, E Qx (ψ(X(t))|Fs ) = E Qx (E Qx (F|Ft )|Fs ) = E Qx (F|Fs ) = ψ(X(s)).

247 237

Equating the expectations at time t = 0 and time t one gets Z ψ(x) = ψ(X(t))dQ x =

Ω Z

ψ(y)q(t, x, y)dy,

where q(t, x, A) = Q x (Xt ∈ A), ∀A Borel in Rd . We assume for the present that ψ(x) is continuous (This will be shown in Lemma 4 in the next section). By definition 0 ≤ ψ ≤ 1. Step 2. ψ(x) = 1, ∀x or ψ(x) = 0, ∀x. Suppose that ψ(x0 ) = 0 for some x0 . Then Z 0 = ψ(x0 ) = ψ(y)q(t, x0 , y)dy. As q > 0 a.e. and ψ ≥ 0 we conclude that ψ(y) = 0 a.e. (with respect to Lebesgue measure). Since ψ is continuous ψ must vanish identically. If ψ(x0 ) = 1 for some x0 , we apply the above argument to 1 − ψ to conclude that ψ = 1, ∀x. We now show that the third possibility 0 < ψ(x) < 1 can never occur. Since K is compact and ψ is continuous, 0 < a = inf ψ(y) ≤ sup ψ(y) = b < 1. y∈K

y∈K

From an Exercise in the section on martingales it follows that ψ(X(t)) → χE∞K

a.e.

Qx

as t → +∞.

Therefore lim ψ(X(t))(1 − ψ(X(t))) = 0 a.e. Q x . Now t→∞

K ψ(x0 ) = Q x0 (E∞ ) = Q x0 {w : w(t) ∈ K for arbitrary large time}

≤ Q x0 {w : a ≤ ψ(X(t, w)) ≤ b for arbitrarily large times} ≤ Q x0 {w : (1 − b)a ≤ ψ(X(t))[1 − ψ(X(t)] ≤ b(1 − a) for arbitrarily large times}

30. Behaviour of Diffusions for Large Times

248 = 0.

238

Thus ψ(x) = 0 identically, which is a contradiction. Thus for the given compact set K, either

K Q x (E∞ ) = 0, ∀x, K Q x (E∞ ) = 1, ∀x.

or

◦

K0 Step 3. If Q x (E∞ ) = 1 for some compact set K0 (K 0 , ∅) and ∀x, then K Q x (E∞ ) = 1, ∀ compact set K with non-empty interior. K0 We first given an intuitive argument. Suppose Q x (E∞ ) = 1, i.e. almost all trajectories visit K0 for arbitrarily large times. Each time a trajectory hits K0 , it has some chance of hitting K. Since the trajectory visits K0 for arbitrarily large times it will visit K for arbitrarily large times. We now give a precise arguent. Let

τ0 = inf{t : X(t) ∈ K0 } τ1 = inf{t ≥ t0 + 1 ...

...

...

...

X(t) ∈ K0 }

τn = inf{t ≥ tn−1 + 1 : X(t) ∈ K0 } ...

...

...

...

Clearly τ0 < τ1 < . . . < and τn ≥ n. Q x (EnK ) ≥ Q x {X(t) ∈ K for t ≥ τn } ∞ [ [τ j , τ j + 1]} ≥ Q x {X(t) ∈ K for t ∈ j=n

         \ = 1 − Qx  X(t) < K for t ∈ [τ , τ + 1]  j j       j≥n          \ X(τ + 1) ∈ K ≥ 1 − Qx   j       j≥n

249 239

We claim that \ Q x ( X(τ j + 1) < K) = 0, j≥n

K ) = 1, completing the so that Q x (EnK ) = 1 for every n and hence Q x (E∞ proof of the theorem. Now ◦

q(1, x, K) ≥ q(1, x, K) > 0,

◦

∀x, K

interior of

K.

It is clear that if xn → x, then ◦

◦

lim q(1, xn , K) ≥ q(1, x, K). ◦

Let d = inf q(1, x, K). Then there exists a sequence xn in K0 such x∈K0

◦

that d = Lt q(1, xn , K). K0 being compact, there exists a subsequence n→∞ yn of xn with yn → x in K0 , so that ◦

◦

◦

d = lim q(1, x, K) = lim q(1, yn , K) ≥ q(1, x, K) > 0. n→∞

Thus inf q(1, x, K) ≥ d > 0.

x∈K0

Now   N Y  X(τ j + 1) < K|FτN  Q x  j=n

=

N−1 Y j=n

χ(X(τ j + 1) < K)Q x (X(τN + 1) ∈ K|FτN ) because

τ j + 1 ≤ τN

=

for

j < N,

N−1 Y (χX(τ j +1) < K)QX(τN ) (X(1) < K) by the strong j=n

30. Behaviour of Diffusions for Large Times

250

Markov property, =

N−1 Y

q(1, X(τN ), K c ) χ(X(τ j +1)
j=1

240

Therefore   N \  Q x  X(τ j + 1) < K  j=n

N \ = E (Q x ( X(τ j + 1) < K|τN ))) Qx

j=n

  N−1 Y   Qx  c = E  (χ[X(τ j +1)
Since K0 is compact and X(τN ) ∈ K0 ,

q(1, X(τN ), K c ) = 1 − q(1, X(τN ), K) ≤ 1 − d Hence     N  N−1  \ \    X(τ j + 1) < K  . Q x  X(τ j + 1) < K  ≤ (1 − d)Q x  j=n

j=n

Iterating, we get   N  \ Q x  X(τ j + 1) < K  ≤ (1 − d)N−n+1 , ∀N. j=n

241

Let N → ∞ to get

   \ Q x  X(τ j + 1) < K  = 0, j=n

since 0 ≤ 1 − d < 1. Thus the claim is proved and so is the theorem.



251 ◦

K ) = 1 if and only if Corollary . Let K be compact, K , ∅. Then Q x (E∞ Q x (E0K ) = 1, ∀x. K ) = 1; then Q (E K ) = 1 because E K E K . SupProof. Suppose Q x (E∞ x 0 ∞ 0 pose Q x (E0K ) = 1, then

Q x (EnK ) = E Qx (E Qx (χEnK |Fn )) = E Qx (QX(n) (E0K )) = E Qx (1) = 1, ∀n. K ) = 1. Therefore Q x (E∞



K ) = 0 then it need not imply that Remark. If Q x (E∞

Q x (E0K ) = 0. Example . Take b = 0 and d = 3. LEt K = S 1 = {x ∈ R3 such that |x| ≤ 1}. Define    1, for |x| ≤ 1, ψ(n) =    1 , for |x| ≥ 1. |x|

K ) = 0. In fact, P (E K ) = ψ(x) (Refer P x (E0K ) , constant but P x (E∞ x 0 Dirichlet Problem).

31. Invariant Probability Distributions Definition. Let {P x } x∈Rd be a family of Markov process on Ω = C([0, ∞); Rd ) indexed by the starting points x, with homogeneous transition probability p(t, x, A) = P x (Xt ∈ A) for every Borel set A in Rd . A probability measure µ on the Borel field of Rd is called an invariant distribution if, ∀A Borel in Rd . Z p(t, x, A)dµ(x) = µ(A). Rd

We shall denote dp(t, x, y) by p(t, x, dy) or p(t, x, y)dy if it has a density. Proposition . Let L2 = ∆/2 + b · ∇ with no explosion. Let Q x be the associated measure. If {Q x } has an invariant measure µ then the process is recurrent. Proof. It is enough to show that if K is a compact set with non-empty interior then K Q x (E∞ )=1 for some x. Also Q x (EtK ) ≥ Q x (Xt ∈ K) = q(t, x, K). Therefore Z Z µ(K) = q(t, x, K)dµ(x) ≤ Q x (EtK )dµ(x). 253

242

31. Invariant Probability Distributions

254

K ) as t → ∞. Now, 0 ≤ Q x (EtK ) ≤ 1 and Q x (EtK ) decreases to Q x (E∞ Therefore by the dominated convergence theorem Z K µ(K) ≤ Q x (E∞ )dµ(x).

243 Sn If the process were transient, then Q x (E∞ ) = 0, ∀n, where S n = d {x ∈ R : |x| ≤ n}, i.e. µ(S n ) = 0, ∀n. Therefore µ(Rd ) = 0, which is false. Thus the process is recurrent. The converse of this proposition is not true as is seen by the following example. 1 ∂2 Let L = so that we are in a one-dimensional situation (Brow2 ∂x2 nian motion). Then Z −(x−y)2 1 1 √ e 2t dy ≤ √ λ(K), p(t, x, K) = (2πt) (2πt) K

where λ denotes the Lebesgue measure on R. If there exists an invariant distribution µ, then Z Z 1 λ(K) µ(K) = p(t, x, K)dµ(x) ≤ √ λ(K) dµ(x) = √ (2πt) (2πt) Letting t → ∞, we get µ(K) = 0 ∀ compact K, giving µ = 0, which is false.  Theorem . Let L = ∆/2 + b · ∇ with no explosion. Assume b to be C ∞ . Define the formal adjoint L∗ of L by L∗ = ∆/2 − ∇ · b (i.e. L∗ u = 1 ∆u − ∇ · (bu)). Suppose there exists a smooth function φ(C 2 - would 2 do) suchR that L∗ φ = 0, φ ≥ 0, inf φdx = 1. If one defines µ by the rule µ(A) = φ(y)dy, then µ is an invariant distribution relative to the family A

{Q x }.

244

Proof. We assume the following result from the theory of partial differential equations.

255 If f ∈ C0∞ (G) where G is a bounded open set with a smooth boundary ∂G and f ≥ 0, then there exists a smooth function UG : [0, ∞)×G → [0, ∞) such that ∂UG = LUG on (0, ∞) × G, ∂t UG (0, x) = f (x) on {0} × G, ∀x ∈ G.

UG (t, x) = 0,

Let t > 0. As UG , φ are smooth and G is bounded, we have Z Z Z ∂ ∂ UG (t, x)φ(x)dx = φLUG dx UG φds = ∂t ∂t G

G

G

Using Green’s formula this can be written as # Z Z Z " ∂ 1 ∂UG ∂ ∗ UG (t, x)φ(x)dx = UG L φ − φ − UG dS + ∂t 2 ∂n ∂n G G ∂G Z + hb · niUG (t, x)φ(x)dS ∂G

Here n is assumed to be the inward normal to ∂G. So, Z Z ∂ 1 ∂UG UG (t, x)φ(x)dx = − (x) (t, x)dS ∂t 2 ∂n G

∂G

(Use the equation satisfied by φ and the conditions on UG ). Now UG (t, x) ≥ 0, ∀x ∈ G, UG (t, x) = 0, ∀x in ∂G, so that ∂UG (t, x) ≥ 0. ∂n This means that ∂ ∂t

245

Z G

UG (t, x)φ(x)dx ≤ 0, ∀t > 0,

31. Invariant Probability Distributions

256 i.e.

R

UG (t, x)φ(x)dx is a monotonically decresing function of t. There-

G

fore

Z

Z

UG (t, x)φ(x)dx ≤

G

G Z

=

G Z

=

UG (0, x)φ(x)dx f (x)φ(x)dx f (x)φ(x)dx.

Rd

Next we prove that if U : [0, ∞)×Rd → [0, ∞) is such that ∀t > 0 and U(0, x) = f (x), then Z Z U(t, x)φ(x)dx ≤ f (x)φ(x)dx. Rd

∂U ∂t

= LU,

Rd

The solution UG (t, x) can be obtained by using Itˆo calculus and is given by Z UG (t, x) = f (X(t))χ{τG>t } dQ x . We already know that U(t, x) = Therefore Z Now

U(t, x)φ(x)dx = "

Z

f (X(t))dQ x .

"

f (X(t))φ(x)DQ x dx.

f (X(t))χ{τG >t} dQ x φ(x)dx Z Z UG (t, x)φ(x)dx ≤ f (x)φ(x)dx. Rd

257 246

Letting G increase to Rd and using Fatou’s lemma, we get " Z f (X(t))φ(x)dQ x dx ≤ f (x)φ(x)dx  This proves the assertion made above. Let Z µ(A) = φ(X)dx, ν(A) =

Z

A

Q x (Xt ∈ A)dµ(x) =

Z

q(t, x, A)dµ(x).

Let f ∈ C0∞ (G), f ≥ 0, where G is a bounded open set with smooth boundary. Now Z " f (y)dν(y) = f (y)q(t, x, y)dµ(x)dy " = f (X(t))dQ x dµ(x) Z = U(t, x)dµ(x) Z = U(t, x)φ(x)dx Z Z ≤ f (x)φ(x)dx = f (x)dµ(x). Thus, ∀ f ≥ 0 such that f ∈ C0∞ , Z Z f (x)dν(x) ≤ f (x)dµ(x). This implies that ν(A) ≤ µ(A) for every Borel set A. (Use mollifier 247 s and the dominated convergence theorem to prove the above inequality for χA when A is bounded). Therefore ν(Ac ) ≤ µ(Ac ), or 1 − µ(A) ≤ 1 −

31. Invariant Probability Distributions

258

µ(A), since µ, ν are both probability measures. This gives µ(A) = ν(A), i.e. Z µ(A) = q(t, x, A)dµ(x), ∀t, i.e. µ is an invariant distribution. We now see whether the converse result is true or not. Suppose there exists a probability measure µ on Rd such that Z Q x (Xt ∈ A)dµ(x) = µ(A), ∀A Borel in Rd and ∀t.

R The question we have in mind is whether µ(A) = φdx for some A R smooth φ satisfying L∗ φ = 0, φ ≥ 0, φ(x)dx = 1. To answer this we proceed as follows. R By definition µ(A) = q(t, x, A)dµ(x). Therefore " f (X(t))dQ x dµ(x) " = f (y)q(t, x, y)dy dµ(x) Z (1) = f (y)dµ(y)∀ f ∈ C0∞ (Rd )|| f ||∞ ≤ 1. Since X is an Itˆo process relative to Q x with parameters b and I, f (X(t)) −

Zt

(L f )(X(s))ds

0

248

is a martingale. Equating the expectations at time t = 0 and time t we obtain   t  Z   E Qx ( f (X(t)) = f (x) + E Qx  (L f )(X(s))ds   0

Integrating this expression with respect to µ gives

"

f (X(t))dQ x dµ(x) =

Z

f (x)dµ(x)

" Zt Rd

0

(L f )(X(s))ds dQ x dµ.

259 Using (1), we get 0=

Z Z Zt

Rd Ω

(L f )(X(s))dQ x ds dµ(x)

0

Applying equation (1) to the function L f we then get 0=

Z Zt

Rd

=t

Z

(L f )(y)dµ(y)ds

0

(L f )(y)dµ(y),

∀t > 0.

Rd

Thus 0=

Z

(L f )(y)dµ(y),

∀ f ∈ C0∞ (Rd ).

Rd

In the language of distributions this just means that L∗ µ = 0. From the theory of partial differential equations it then follows that there exists a smooth function φ such that ∀A Borel in Rd , Z µ(A) = φ(y)dy A

with L∗ φ = 0. As µ ≥ 0, φ ≥ 0 and since Z µ(Rd ) = 1, φ(x)dx = 1. Rd

249

We have thus proved the following (converse of the previous) theorem. Theorem . Let µ be an invariant distribution with respect to the family {Q x } with b : Rd → Rd being C ∞ . Then there exists a φ ∈ L′ (Rd ), φ ≥ 0, φ smooth such that Z L∗ φ = 0, φ(y)dy = 1

31. Invariant Probability Distributions

260 and such that µ(A) =

Z

φ(y)dy,

∀A

Borel in Rd .

A

Theorem (Uniqueness). Let φ1 , φ2 be smooth on Rd such that Z Z φ1 , φ2 ≥ 0, 1 = φ1 dy = φ2 dy, L∗ φ1 = 0 = L∗ φ2 . Rd

Rd

Then φ1 = φ2 . Proof. Let f (x) = φ1 (x) − φ2 (x), Z µi (A) = φi (x)dx,

i = 1, 2.

A

Then µ1 , and µ2 are invariant distributions. Therefore Z Z q(t, x, y)φi (x)dx = q(t, x, y)dµi (x) = φi (y),

(a.e.),

Taking the difference we obtain Z q(t, x, y) f (x)dx = f (y),

i = 1, 2.

a.e.

Now Z

Z Z | f (y) dy = | q(t, x, y) f (x)dx|dy " ≤ q(t, x, y)| f (x)|dx dy Z Z = | f (x)|dx q(t, x, y)dy Z = | f (x)|dx.

261 250

(*)

Thus "

Z Z | f (x)|q(t, x, y)dx dy = | q(t, x, y) f (x)dx|dy ∀t.

We show that f does not change sign,R i.e. f ≥ 0 a.e. or f ≤ 0 a.e. The result then follows from the fact that f (x)dx = 0. Now Z Z | q(1, x, y) f (x)dx| ≤ q(1, x, y)| f (x)|dx and (∗) above gives Z Z " | q(1, x, y) f (x)dx|dy = q(1, x, y)| f (x)|dx dy. Thus |

Z

q(1, x, y) f (x)dx| =

Z

q(1, x, y)| f (x)|dx a.e. y,

i.e. |

Z

q(1, x, y) f (x)dx +

E−

=

Z

E+

Z

q(1, x, y) f (x)dx|

E−

q(1, x, y) f (x)dx −

Z

q(1, x, y) f (x)dx a.e. y,

E−

where E + = {x : f (x) > 0},

E − = {x : f (x) < 0},

E 0 = {x : f (x) = 0}.

Squaring both sides of the above equality, we obtain      Z Z  q(1, x, y) f (x)dx  q(1, x, y) f (x)dx = 0, (**)     E+

a.e.

y.

E−

251

31. Invariant Probability Distributions

262

Let A be a set of positive Lebesgue measure; then p(1, x, A) = P x (X(1) ∈ A) > 0. Since Q x is equivalent to P x on Ω we have Q x (X(1) ∈ A) = q(1, x, A) > 0. Therefore q(1, x, y) > 0 a.e. y for each x. By Fubini’s theorem q(1, x, y) > 0 a.e. x, y. Therefore for almost all y, q(1, x, y) > 0 for almost all x. Now pick a y such that (∗∗) holds for which q(1, x, y) > 0 a.e. x. We therefore conclude from (∗∗) that either Z q(1, x, y) f (x)dx = 0, in which case f ≤ 0 a.e., E+

or

Z

q(1, x, y) f (x)dx = 0,

in which case

E−

f ≥0

a.e.

Thus f does not change its sign, which completes the proof.



Remark. The only property of the operator L we used was to conclude q > 0. We may therefore expect a similar result for more general operators. R Theorem . Let L∗ φ = 0 where φ ≥ 0 is smooth and φ(x)dx = 1. Let K be any compact set. Then Z sup |q(t, x, y) − φ(y)|dy → 0 as t → +∞. x∈K

Lemma 1. Let b be bounded and smooth. For every f : Rd → Rd that is bounded and measurable let u(t, x) = E Qx ( f (X(t)). Then for every fixed t, u(t, x) is a continuous function of x. Further, for t ≥ ǫ > 0, |u(t, x) −

252

Z

1 |x − y|2 u(t − ǫ, y) √ exp − dy| 2ǫ (2πǫ)d √ ≤ || f ||∞ (ect (ecǫ − 1)),

where c is a constant depending only on ||b||∞ .

263 Proof. Let (T t f )(x) = E Qx ( f (X(t)) = E Px ( f (X(t))Z(ǫ, t))+

(1)

+ E Px ( f (X(t))(Z(t) − Z(ǫ, t))),

where   t Zt Z  1   |b|2 ds , Z(t) = exp  hb2 , dxi −   2 0 0  t  Zt Z  1   |b(X(s))|2 ds . Z(ǫ, t) = exp  hb, dxi −   2 ǫ

ǫ

E Px ( f (X(t))Z(ǫ, t)) = E Px (E PX ( f (X(t))Z(ǫ, t)|ǫ )) = E Px (E PX ǫ)( f (X(t − ǫ))Z(t − ǫ))) (by Markov property),

Px

= E (u(t − ǫ, X(ǫ)). (2)

=

Z

" # |(x − y)|2 1 u(t − ǫ, y) √ exp − dy. 2ǫ ( (2πǫ))d

Now (E Px (|Z(t) − Z(ǫ, t)|)) =

= E Px (|Z(ǫ)Z(ǫ, t) − Z(ǫ, t)|))2

= E Px (Z(ǫ, t)Z(ǫ) − 1|))2

≤ (E Px ((Z(ǫ) − 1)2 ))(E Px (Z 2 (ǫ, t)))

(by Cauchy Schwarz inequality),

Px

≤ E (Z 2 (ǫ) − 2Z(ǫ) + 1)E Px (Z 2 (ǫ, t))

≤ E Px (Z 2 (ǫ) − 1)E Px (Z 2 (ǫ, t)), (since E Px (Z(ǫ)) = 1), Zt Zt Zt 2 Px 2 Px 2 |b| ds + |b|2 ds)) ≤ E (Z (ǫ) − 1)E (exp(2 hb, dXi − 2 ǫ

ǫ

ǫ

31. Invariant Probability Distributions

264 ≤ E Px (Z 2 (ǫ) − 1)ect , 253

using Cauchy Schwarz inequality and the fact that Px

E (exp(2

Zt

22 hb, dXi − 2

Zt

|b|2 ds)) = 1.

ǫ

ǫ

Thus E Px (|Z(t) − Z(ǫ, t)||2 ≤ (ecǫ − 1)ect where c depends only on ||b||∞ . Hence (3)

|E Px ( f (X(t))(Z(t) − Z(ǫ, t))| ≤ || f ||∞ E Px (|Z(t) − Z(ǫ, t)|) √ ≤ || f ||∞ ((ecǫ − 1)ect ). Substituting (2) and (3) in (1) we get " # Z 1 −|x − y|2 |u(t, x) − u(t − ǫy) · √ exp dy 2ǫ ( (2πǫ)d ) √ ≤ || f ||∞ ((ecǫ − 1)ect )

Note that the right hand side is independent of x and as ǫ → 0 the right hand side converges to 0. Thus to show that u(t, x) is a continuous function of x (t fixed), it is enough to show that " # Z 1 −|x − y|2 u(t − ǫ, y) √ exp dy 2ǫ ( (2πǫ)d 254

is a continuous function of x; but this is clear since u is bounded. Thus for any fixed tu(t, x) is continuous.  Lemma 2. For any compact set K ⊂ Rd , for r large enough so that K ⊂ {x : |x| < r},

x → Q x (τr ≤ t)

is continuous on K for each t ≥ 0, where τr (w) = inf{s : |w(s)| ≥ r}.

265 Proof. Q x (τr ≤ t) depends only on the coefficient b(x) on |x| ≤ r. So modifying, if necessary, outside |x| ≤ r, we can very well assume that |b(x)| ≤ M for all x. Let τǫr = inf{s : s ≥ ǫ, |w(s)| ≥ r}.

Q x (τǫr ≤ t) = E Qx (u(X(ǫ))),

where u(x) = Q x (τr ≤ t − ǫ) As b and u are bounded, for every fixed ǫ > 0, by Lemma 1, Q x (τr ≤ t) is a continuous function of x. As |Q x (τǫr ≤ t) − Q x (τr ≤ t)| ≤ Q x (τr ≤ ǫ), to prove the lemma we have only to show that limit sup Q x (τr ≤ ǫ) = 0 ǫ→0 x∈K

Now Q x (τr ≤ ǫ) =

Z

Z()dP x

{τ ≤ǫ}

r Z √ ≤ ( (Z(ǫ))2 dP x )1/2 · P x (τr ≤ ǫ),

by Cauchy-Schwarz inequality. The first factor is bounded because b is 255 bounded. The second factor tends to zero uniformly on K because sup P x (τr ≤ ǫ) ≤ P( sup |w(s)| > δ) x∈K

0≤s≤ǫ

where δ = inf |(x − y)|. y∈K |x|=r.



Lemma 3. Let K be compact in Rd . Then for fixed t, Q x (τr ≤ t) monotically decreses to zero as r → ∞ and the convergence is uniform on K.

31. Invariant Probability Distributions

266

Proof. Let fr (x) = Q x (τr ≤ t). As {τr ≤ t} decreases to the null set, fr (x) decreases to zero. As K is compact, there exists an r0 such that for r ≥ r0 , fr (x) is continuous on K, by Lemma 2. Lemma 3 is a consequence of Dini’s theorem.  Lemma 4. Let b : Rr → Rd be smooth (not necessarily bounded). Then E Qx ( f (X(t))) is continuous in x for every fixed t, f being any bounded measurable function. Proof. Let br be any bounded smooth function on Rd such that br ≡ b on |x| ≤ r and Qrx the measure corresponding to br . Then by Lemma 1, E Qx ( f (X(t))) is continuous in x for all r. Further, r

|E Qx ( f (X(t))) − E Qx ( f (X(t)))| ≤ 2|| f ||∞ · Q x (τr ≤ t). 256

The result follows by Lemma 3.



Lemma 5. With the hypothesis as the same as in Lemma 1, (S 1 ) is an equicontinuous family, where S 1 = { f : Rd → R, f bounded measurable, || f ||∞ ≤ 1} Proof. For any f in S 1 , let U(x) = U(t, x) ≡ E Qx ( f (X(t))) and " # Z −|x − y|2 1 exp Uǫ (x) = Uǫ (t, x) = U(t − ǫ, y) √ dy. 2ǫ ( (2πǫ)d ) By Lemma 1, |U(x) − Uǫ (x)| ≤ (((ecǫ − 1)ǫ ct ))1/2

|U(x) − U(y)| ≤ |U(x) − Uǫ (x)| + |Uǫ (y) − U(y)| + |Uǫ (x) − Uǫ (y)| √ ≤ 2 ((ecǫ − 1)ect ) + |Uǫ (x) − Uǫ (y)|. The family {Uǫ : f ∈ S 1 } is equicontinuous because every U occuring in the expression for Uǫ is bounded by 1, and the exponential factor is uniformly continuous. Thus the right hand side is very small if ǫ is small and |x − y| is small. This proves the lemma. 

267 Lemma 6. Let b be smooth and assume that there is no explosion (b is not necessarily bounded). Then (S 1 ) is an equi-continuous family ∀t > 0. Proof. Let r > 0 be given. Define br ∈ C ∞ such that br = 0 on |x| > r+1, br = b on |x| ≤ r, br : Rd → R. By Lemma 2, we have that r

{E Qx ( f (X(t))) : f ∈ S 1 } is equicontinuous, where Qrx is the probability measure corresponding 257 to the function br . r

E Qx ( f (X(t))χ{τr >t} )E Qx ( f (X(t))χ{τr >t} ).

(1) Therefore

r

|E Qx ( f (X(t))) − E Qx ( f (X(t))|

= |E Qx ( f (X(t))χ{τr >t} ) + E Qx ( f (X(t))χ{τr ≤t} ) r

r

− E Qx ( f (X(t))χ{τr >t} ) − E Qx ( f (X(t)))χ{τr ≤t} ) r

= |E Qx ( f (X(t)χ{τr ≤t} ) − E Qx ( f (X(t))χ{τr ≤t} )| r

≤ || f ||∞ (E Qx (χ{τr ≤t} ) + E Qx (χ{τr ≤t} )

≤ l[E Qx (χ(τr ≤t) ) + E Qx (χ(τr ≤t) )](use (1) with f = 1)

= 2E Qx (χ(τr ≤t) ). Thus

r

sup sup |E Qx ( f (X(t)) − E Qx ( f (X(t))| ≤ 2 sup(χ{τr ≤t} ). x∈K || f ||∞ ≤1

x∈K

By Lemma 3, sup E 0x (τr ≤ t) → 0 x∈K

for every compact set K as n → ∞, for every fixed t. The equicontinuity of the family (S 1 ) now follows easily. For fixed 258 r x0 , put ur (x) = E Qx ( f (X(t))) and u(x) = E Qx ( f (X(t))) and let K = s[x0 , 1] = {x : |x − x0 | ≤ 1}. Then |u(x) − u(x0 )| ≤ |u(x) − ur (x)| + |u(x0 ) − ur (x0 )| + |ur (x) − ur (x0 )|

31. Invariant Probability Distributions

268

≤ 2 sup E Qy (χ(τr ≤|t) ) + |ur (x) − ur (x0 )| y∈K

By the previous lemma {ur } is an equicontinuous family and since sup E Qy (χ(τr ≤1) ) → 0, {u : || f ||∞ ≤ 1} is equicontinuous at x0 . This y∈K

proves the Lemma.



Lemma 7. T r ◦ T s = T t+s , ∀s, t ≥ 0. Remark. This property is called the semigroup property. Proof. T r (T s f )(x) " = f (z)q(s, y, z)q(t, x, y)dy dz. Thus we have only to show that Z q(t, x, y)q(s, y, A)dy = q(t + s, x, A). q(t + s, x, A) = E Qx (X(t + s) ∈ A)

= E Qx (X(t + s) ∈ A|t ))

= E Qx (E Qx X(t)(X(s) ∈ A))), by Markov property

Qx

= E (q(s, X(t), A)) Z = q(t, x, y)q(s, y, A)dy, 259

which proves the result. As a trivial consequence we have the following. Lemma 8. Let ǫ > 0 and let S 1 be the unit ball in B(Rd ). Then is equicontinuous.

 S

t≥ǫ

T t (S 1 )

269 Proof.

S

t≥ǫ>0

T t (S 1 ) = T (

S

T t (S 1 )) (by Lemma 7) T ǫ (S 1 ).

t≥0

The result follows by Lemma 6.



Lemma 9. Let u(ttx) = E Qx ( f (X(t))) with || f ||∞ ≤ 1. Let ǫ > 0 be given and K any compact set. Then there exists a T 0 = T 0 (ǫ, K) such that ∀T ≥ T 0 and ∀x1 , x2 ∈ K, |u(T, x1 ) − u(T, x2 )| ≤ ǫ. Proof. Define q∗ (t, x1 , x2 , y1 , y2 ) = q(t, x1 , y1 )q(t, x2 , y2 ) and let Q(x1 ,x2 ) be the measure corresponding to the operator 1 L = (∆x1 + ∆x2 ) + b(x1 ) · ∇x1 + b(x2 ) · ∇x2 2 i.e., for any u : Rd × Rd → R, 1 X ∂2 u X ∂u + bi (x1 , . . . , xd ) 2 + 2 2 i=1 ∂xi ∂xi i=1 2d

Lu =

+

t

d X

bi (xd+1,...,x2d )

i=1

∂u . ∂xi+d

Then Q(x1 ,x2 ) will be a measure on C([0, ∞); Rd × Rd ). We claim that Q(x1 ,x2 ) = Q x1 × Q x2 . Note that C([0, ∞); Rd × Rd ) = C([0, ∞); Rd ) × C[(0, ∞); Rd ) and since C([0, ∞); Rd ) is a second countable metric space, the Borel 260 field of C([0, ∞)Rd × Rd ) is the σ-algebra generated by B = (C([0, ∞); Rd )) × B(C[0, ∞); Rd ). By going to the finite-dimensional distributions one can check that P(x1 ,x2 ) = P x1 × P x2 .  t  Zt Z  dQ(x1 ,x2 ) 1  (1) (1) 2  = exp hb , dX i − |b | ds   × 1 dP(x1 ,x2 ) Ft 2 0

0

31. Invariant Probability Distributions

270

 t  Zt Z  1  (2) (2) 2  × exp  hb , dX2 i − |b | ds ,   2 0

where

0

b(1) (x1 . . . xd ) = b(x1 . . . , xd ) · b(2) (xd+1 . . . x2d ) = b(xd+1 , . . . , x2d ), so that Q(x1 ,x2 ) = Q x1 × Q x2 . It is clear that if φ defined an invariant measure for the process Q x , i.e. Z Z φ(x)dx = φ(y)Qy(Xt ∈ A)dy, A

then φ(y1 )φ(y2 ) defines an invariant measure for the process Q(x1 ,x2 ) . Thus the process Q(x1 ,x2 ) is recurrent. Next we show that u(T − t, X1 (t)) is a martingale (0 ≤ t ≤ T ) for any fixed T on C([0, T ]; Rd ). E Qx (u(T − t, X(t)|Fs )) Z = [ u(T − t, y)q(t − s, x, dy)]x=X(s) " =[ f (z)q(T − t, y, dz)q(t − s, x, dy)]x=X(s) Z = [ f (z)q(T − s, x, dz)]x=X(s) = u(T − s, X(s)),

s < t.

261

It now follows that u(T − t, X1 (t)) is a martingale on C([0, ∞); Rd ) × C([0, ∞); Rd ). Hence u(T − t, X1 (t)) − u(T − t, X2 (t)) is a martingale relative to Q(X1 ,x2 ) . Let V = S (0, δ/2) ⊂ Rd × Rd with δ < 1/4. If (x1 , x2 ) ∈ V, then |x1 − x2 | ≤ |(x1 , 0) − (0, 0)| + |(0, 0) − (0, x2 )| < δ. 

271 Claim 1. Q(x1 ,x2 ) (τV ≤ T ) → 1 as T → ∞, where τV is the exit time from Rd − V. Proof. If w is any trajectory starting at some point in V, then τV = 0 ≤ T , ∀T . If w starts at some point outside V then, by the recurrence property, w has to visit a ball with centre 0 and radius δ/2; hence it must get into V at some finite time. Thus {τV ≤ T } ↑ to the whole space as T ↑ ∞. Next we show that the convergence is uniform on compact sets. If x1 , x2 ∈ K, (x1 , x2 ) ∈ K × K (a compact set). Put gT (x1 , x2 ) = Q(x1 ,x2 ) (τV ≤ T ). Then gT (x1 , x2 ) ≥ 0 and gT (x1 , x2 ) increases to 1 as T tends to ∞. gT (x1 , x2 ) = Q(x1 ,x2 ) (τV ≤ T ) Q(x1 ,x2 ) (τ1V ≤ T ),

where

262

τ1V

= inf{t ≥ 1 : (x1 , x2 ) ∈ V}.

Therefore gT (x1 , x2 ) ≥ E Q (x1 , x2 )(E Q (x1 , x2 )((τ1V ≤ T )|1 )) = E Q (x1 , x2 )(Q(X1 (1),X2 (1)) {τ1V ≤ T )}) = E Q (x1 , x2 )(ψT (X1 (1), X2 (1))),

where ψT is a bounded non-negative function. Thus, if hT (x1 , x2 ) = Q(x1 ,x2 ) (τ1V ≤ T ) =

= E Q (x1 , x2 )(ψT (X1 (1), X2 (1))),

then by Lemma 4, hT is continuous for each T , gT ≥ hT and hT increases to 1 as T → ∞. Therefore, hT converges uniformly (and so does gT ) on compact sets. Thus given ǫ > 0 chose T 0 = T 0 (ǫ, K) such that if T ≥ T 0 , sup sup Q(x1 ,x2 ) (τV ≥ T − 1) ≤ ǫ.

x2 ∈K x1 ∈K

31. Invariant Probability Distributions

272

By Doob’s optional stopping theorem and the fact that u(T − t, X1 (t)) − u(t − t, X2 (t)) is a martingale, we get, on equating expectations, |u(T, x1) − u(T, x2 )|

= |E Q(x1 ,x2 ) [u(T − 0, X1 (0) − u(T − 0, X2 (0)]|

= |E Q(x1 ,x2 ) [u(T − (τv ∧ (T − 1)), X1(T − (τv ∧ (T − 1))− − u(T − (τv ∧ T (−1)), X2(T − (τv ∧ (T − 1)]| Z | [u(1, X1(1)) − u(1, X2(1))]dQ(x1,x2 ) + {τv ≥T −1}

Z

+

[u(T − τv , X1 (T − τv )) − u(T − τv ), X2 (T − τv ))dQ(x1 ,x2 ) |.

{τv <(T −1)}

263

Therefore |u(T, x1) − u(T, x2)| Z ≤ |[u(1, X1(1)) − u(1, X2 (1))]|dQ(x1,x2 ) + {τv ≥(T −1)}

+|

Z

[u(T − τv , X1 (T − τv )) − u(T − τv , X2 (T − τv ))dQ(x1 ,x2 ) |

{τv <(T −1)}

≤ 2ǫ + |

Z

[u(T − τv , X1 (T − τv )) − u(T − τv , X2 (T − τv ))]dQ(x1 ,x2 ) |,

{τv <(T −1)}

since u is bounded by 1. The second integration is to be carried out on the set {T − v ≥ 1}. S Since T t (S 1 ) is equicontinuous we can choose a δ > 0 such that t≥1

whenever x1 , x2 ∈ K such that |x1 − x2 | < δ |u(t, x1 ) − u(t, x2 )| ≤ ǫ,

∀t ≥ 1.

264

Thus |u(T, x1 ) − u(T, x2 )| ≤ 3ǫ whenever x1 , x2 ∈ K and T ≥ T 0 . This proves the Lemma. 

273 Corollary to Lemma 9. sup x1 ,x2 ∈K

R

|q(t, x1 , y)|dy converges to 0 as t → ∞.

Proof. Since the dual of L1 is L∞ , we have Z |q(t, x1 , y) − q(t, x2 , y)|dy Z = sup | [q(t, x1 , y) − q(t, x2 , y)] f (y)dy| || f ||∞ ≤1

and the right side converges to 0 as t → ∞, by Lemma 9. We now come to the proof of the main theorem stated before Lemma 1. Now Z |q(t, x, y) − φ(y)|dy Z Z = |q(t, x, y) − φ(x1 )q(t, x1 , y)dx1 |dy (by invariance property) Z Z Z 1 1 = | q(t, x, y)φ(x )dx − φ(x1 )q(t, x1 , y)dx1 |dy Z (since φ(x1 )dx1 = 1) " ≤ |q(t, x, y) − q(t, x1 , y)|φ(x1 )dx1 dy (since φ ≥ 0) Z Z 1 1 = φ(x )dx |q(t, x, y) − q(t, x1 , y)|dy

Since

Z

1

φ(x )dx = Lt

n→∞ |x1 |≤n

choose a compact set L K such that Z

Z

1

φ(x1 )dx1

Z

R

φ(x1 )dx1 ,

φ(x1 )dx1 < ǫ. Then

Rd −L

|q(t, x, y) − q(t, x1 , y)|dy

265

31. Invariant Probability Distributions

274 =

Z

1

Z

1

φ(x )dx

L

+

Z

1

1

φ(x )dx

Z

|q(t, x, y) − q(t, x1 , y)|dy

"

1

1

Z

|q(t, x, y) − q(t, x1 , y)|dv + 2ǫ.

Rd −L

≤

|q(t, x, y) − q(t, x1 , y)|dy+

L

φ(x )dx

Chose t0 such that whenever t ≥ t0 , Z |q(t, x, y) − q(t, x1 , y)|dy ≤

1+

R

ǫ φ(x1 )dx1

L

∀x, x1 in L. (Corollary to Lemma 9). Then Z |q(t, x, y) − φ(y)|dy ≤ 3ǫ, if t ≥ t0 ∀x ∈ K completing the proof of the theorem.



32. Ergodic Theorem Theorem . Let f : Rd → R be bounded and measurable with || f ||∞ ≤ 1. 266 If φ is an invariant distribution for the family {Q x }, x ∈ Rd then Z Qx lim E ( f (X(t1 )) f (X(t2 ))) = [ f (y)φ(y)dy]2 t1 →∞ 0≤t2 −t1 →∞

Proof. E Qx [( f (X(t1 ) f (X(t2 ))] = E Qx (E Qx [ f (X(t1 )) f (X(t2 ))|Ft1 ]) = E Qx ( f (X(t1 ))(E Qx [ f (X(t2 ))|Ft1 ])) Z Qx = E ( f (X(t1 )) f (y)q(t2 − t1 , X(t1 ), y))dy), t2 > t1 (1)

=

Z

f (z)q(t1 , x, z)dz

Z

(by Markov property), f (y)q(t2 − t1 , z, y)dy

does any bounded an measurable f . By theorem of § 31, Z sup | f (y)[q(t, x, y) − φ(y)]dy| → 0 x∈K

as t → +∞. We can therefore write (1) in the form E Qx [( f (X(t1 )) f (X(t2 ))] = Z Z Z = ( f (z)q(t1 , x, x)dz) fφ + f (z)q(t1 , x, z)A(t2 − t1 , z)dz, 275

32. Ergodic Theorem

276

267

where A(t2 −t1 , z) converges to 0 (uniformly on compact sets as) t2 −t1 → +∞. To prove the theorem we have therefore only to show that Z f (z)q(t1 , x, z)A(t2 − t1 , z)dz → 0 R R as t1 → +∞ and t2 − t1 → ∞ (because f (z)q(t1 , x, z)dz → f φ). Now Z | f (z)q(t1 , x, z)A(t2 − t1 , z)dz| Z ≤ || f ||∞ q(t1 , x, z)|A(t2 − t1 , z)|dz Z (2) ≤ q(t1 , x, z)|A(t2 − t1 , z)|dz Let K be any compact set, then Z Z Z q(t1 , x, z)dz = χK q(t1 , x, z)dz → χK φ(z)dz K

at t1 → ∞. Given ǫ > 0, let K be compact so that Z | χK c φ(z)dz| ≤ ǫ; R then | χK c q(t1 , x, z)dz| ≤ 2ǫ if t1 ≫ 0. Using (2) we therefore get Z | f (z)q(t1 x, z)A(t2 − t1 , z)dz| Z Z ≤ q(t1 , x, z)|A(t2 − t1 , z)|dz + q(t1 , x, z)|A(t2 − t1 , z)|dz ≤

K Z

Kc

q(t1 , x, z)|A(t2 − t1 , z)|dz + 2

K

≤

Z K

Z

q(t1 , x, z)dz,

Kc

since |A(t2 − t1 , z)| ≤ 2, q(t1 , x, z)|A(t2 − t1 , z)|dz + 2ǫ, if t1 ≫ 0.

277 268

The theorem now follows from the fact that lim sup |A(t2 − t1 , z) = 0.

t2 −t1 →∞ z∈K

 Weak Ergodic Theorem.    Zt   Q x  1 f (X(s))ds − f (x)φ(x)dx| > ǫ  = 0. lim E | t→∞   t 0

Proof.

  Z   Zt 1   f (X(s))ds − E Qx | f (x)φ(x)dx| > ǫ    t 0

  Z  Zt  1 Qx  1  ≤ 2 E | f (X(s))ds − f (y)φ(y)dy|2  ,  t  ǫ 0

by Tchebychev’s inequality. We show that the right side → 0 as t → ∞. Now   Z  Zt  1   f (X(s))ds − f φ|2  E Qx |  t  0

1 = E [| 2 t Qx

−2

1 t

Zt Zt

Z f (X(σ1 )) f (X(σ2 ))dσ1 dσ2 + ( f φdy)

0

0

Zt

f (X(σ))dσ

Z

f φdy]

0

Also Qx

sup |E [ f (X(t)) − x∈K

Z

f (y)φ(y)dy]|

32. Ergodic Theorem

278 = sup |

Z

Z

q(t, x, y) f (y)dy − f (y)φ(y)dy| Z ≤ || f ||∞ sup |q(t, x, y) − φ(y)|dy; x∈K

x∈K

269

the right hand side tends to 0 as t tends to +∞. Consider Zt Z 1 |E ( f (X(σ))dσ − f (y)φ(y)dy)| t 0  T  Zt Z   Z 1 1  Qx  f (X(σ))dσ − f (y)φ(y)dy + f (X(σ))dσ , 0 ≤ T ≤ t, = |E   t t Qx

0

≤

1 | t

ZT

0

E Qx f (X(σ))dσ − T

0

Z

f (y)φ(y)dy|+

   Zt  t − T  Z 1   f (y)φ(y)dy |. + |E Qx  f (X(σ))dσ − t  t T

Given ǫ > 0 choose T large so that Z |E Qx ( f (X(σ)) − f (y)φ(y)dy| ≤ ǫ,

(σ ≥ T ).

Then |E

Qx

≤|

1 Z t

1 t

0 T Z

t

f (X(σ))dσ −

Z

E Qx [ f (X(σ))] −

0

T t

f (y)φ(y)dy| ≤ Z

f (y)φ(y)dy]| +

≤ 2ǫ provided t is large. Thus 1 lim E [ t→+∞ t Qx

Zt 0

f (X(σ))dσ] =

Z

f φdy.

t−T ǫ t

279 To prove the result we have therefore only to show that   #2  Zt Zt  "Z 1  Qx    lim E  2 f (X(σ1 )) f (X(σ2 ))dσ1 dσ2  = f (y)φ(y)dy t→+∞  t 0

0

270

POR is the region σ2 ≥ t0 , σ1 − σ2 ≥ t0 . Let I   !  Zt Zt  Z 1  = E Qx  f (X(σ1 )) f (X(σ2 ))dσ1 dσ2  − f φdy t  0 0 !2  Z  Z  2  Qx = 2 f (y)φ(y)dy  dσ1 dσ2 E ( f (X(σ1 )) f (X(σ2 ))) − t 0 ≤ σ2 ≤ σ1 ≤ t.

Then 2 |I| ≤ 2 t

Z ∆PQR

Qx

|E ( f (X(σ1 )) f (X(σ2 ))) −

Z

!2 f (y)φ(y)dy |dσ1 dσ2

32. Ergodic Theorem

280 +

2 · 2|| f ||2∞ 2 t

[area of OAB − area of PQR]

By the Ergodic theorem the integrand of the first term on the right can be made less than ǫ/2 provided t0 is large (see diagram). Therefore  !2   2 (t − 2t0 )  ǫ 2 4 2  t |I| ≤ · 2 area of PQR + 2 || f ||∞  −  2 t 2 2 t ǫ 2|| f ||2∞ + [4tt0 − 4t02 ]. 2 t2 <ǫ ≤

271

if t is large. This completes the proof of the theorem.



33. Application of Stochastic Integral LET b BE A bounded function. For every Brownian measure P x on 272 Ω = C([0, ∞); Rd ) we have a probability measure Q x on (Ω, F ). Problem. Let q(t, x, A) = Q x (Xt ∈ A) · q(t, x, ·) is a probability measure on Rd . We would like to know if q(t, x, ·) is given by a density function on Rd and study its properties. Step (i). q(t, x, ·) is absolutely continuous with respect to the Lebesgue measure. For, p(t, x, A) = P x (X(t) ∈ A) is given by a density function. Therefore p(t, x, ·) ≫ md (Lebesgue measure). Since Q x ≪ P x on Ft ,

q(t, z, ·) ≤ Md on Ft . Step (ii). Let q(t, x, y) ≥ 0 be the density function of q(t, x, ·) and write p(t, x, y) for the density of p(t, x, ·). Let 1 < α < ∞. Put q(t, x, y) . r(t, x, y) = p(t, x, y) Z Z qα dy = rα pα dy Rd

=

Z

α 1/α

r p

P

α−1 α

dy ≤

281

Z

!1/α r pdy × α2

33. Application of Stochastic Integral

282

× 273

Z

p

!α−1/α dy

Z

q(t, x, y)dy

α+1

Step (iii). Q x (X(t) ∈ A) = = = Therefore

Z

Z

r(t, x, y)p(t, x, y)dy r(t, x, y)P x (Xt ∈ dy).

dQ x t = r(t, x, y) dP x t

Therefore !1/α Z dQ x t α α2 r pdy = dP x t α2 ,Px dQ x α , since Ftt ⊂ Ft , ≤ dP x t α2 ,Px 2

{E Px [Z(t)α ]}1/α Zt Zt α2 Px 2 |b|2 ds)]}1/α = {E [exp(α hb, dXi − 2 Px

= {E [exp(α

2

0

0

Zt

Zt

0

α4 hb, dXi − 2

α4 − α2 |b| ds + 2 2

0

Zt

|b|2 ds)]}1/α ,

0

i.e., Z

α2

r pdy

!1/α

  1/α  t Zt    4    2 4 Z  α − a α  Px     2 2    ≤ E exp ct + α hb, dXi − |b| ds               2  2   0

0

where c is such that |b|2 ≤ c. Using Schwarz inequality we then get !1/α " !#1/α Z α4 − α2 α2 r pdy ≤ exp ct . 2

283 274

Hence Z

qα dy ≤ exp

"

α4 − α2 ct 2

#!1/α Z

!α−1/α Pα+1 dy

Significance. Pure analytical objects like q(t, x, y) can be studied using stochastic integrals.

33. Application of Stochastic Integral

284

Appendix Language of Probability

275

Definition. A probability space is a measure space (Ω, B, P) with P(Ω) = 1. P is called a probability measure or simply a probability. Elements of B are called events. A measurable function X : (Ω, B) → Rd is called d-dimensional random variable. Given the random variable X, define F : Rd → R by F((a1 , . . . an )) = P{w : Xi (w) < ai , for i = 1, 2, . . . , d} where X = (X1 , X2 , . . . , Xd ). Then F is called the distribution funcR tion variable X. For any random variable X, X dP = R of the random R ( X1 dP, . . . , Xd dP), if it exists, is called mean of X or expectation of R X and is denoted by E(X). Thus E(X) = XdP = µ. E(X n ), where X n = (X1n , X2n , . . . , Xdn ) is called the nth moment about zero. E((X − µ)n ) is called the nth central moment. The 2nd central moment is called variance and is denoted by σ2 we have the following. Tchebyshev’s Inequality. Let X be a one-dimensional random variable with mean and variance µ. Then for every ǫ > 0, P{w : |X(w) − µ| ≥ ǫ} ≤ σ2 /ǫ 2 .

276

Generalised Tchebyshev’s Inequality. Let f : R → R be measurable such that f (u) = f (−u), f is strictly positive and increasing on (0, ∞). Then for any random variable X : Ω → R, P(w : |X(w)| > ǫ) ≤

E( f (X)) f (ǫ)

for every ǫ > 0. For any random variable X : Ω → Rd , φ(t) = E(eitX ) : Rd → C is called the characteristic function of X. Here t = (t1 , . . . , td ) and tX = t1 X1 + t2 X2 + · · · + td Xd .

285 Independence. Events E1 , . . . , En are called independent if for every {i1 , . . . , ik } ⊂ {1, 2, . . . , n} we have P(Ei1 ∩ . . . ∩ Eik ) = P(Ei1 )P(Ei2 ) . . . P(Eik ). An arbitrary collection of events {Eα : α ∈ I} is called independent if every finite sub-collection is independent. Let {Fα : α ∈ I} be a collection of sub-σ-algebras of B. This collection is said to be independent if for every collection {Eα : α ∈ I}, where Eα ∈ Fα , of events is independent. A collection of random variables {Xα : α ∈ I} is said to be independent if {σ(Xα ) : α ∈ I} is independent where σ(Xα ) is the σ-algebra generated by Xα . Theorem . Let X1 , X2 , . . . , Xn be random variables with F X1 , . . . , F Xn as their distribution functions and let F be distribution function of X = (X1 , . . . , Xn ), φX1 , . . . , φXn the characteristic functions of X1 , . . . , Xn and φ that of X = (X1 , . . . , Xn ). X1 , . . . , Xn are independent if and only if F((a1 , . . . , an )) = F X1 (a1 ) . . . F Xn (an ) for all a1 , . . . , an , iff φ((t1 , . . . , tn )) = φX1 (t1 ) . . . φXn (tn ) for all t1 , . . . , tn . 277 Conditioning. Theorem . Let X : (Ω, B, P) → Rd be a random variable, with E(X) finite, i.e. if X = (X1 , . . . , Xd ), E(Xi ) is finite for each i. Let C be a subd σ-algebraR of B. Then R there exists a random variable Y : (Ω, C ) → R such that YdP = XdP for every C in C . C

C

If Z is any random variable with the same properties then Y = Z almost everywhere (P).

Definition . Any such Y is called the conditional expectation of X with respect to C and is denoted by E(X|C ). If X = χA , the characteristic function of A in B, then E(χA |C ) is also denoted by P(A|C ). Properties of conditional expectation. 1. E(1|C ) = 1.

33. Application of Stochastic Integral

286

2. E(aX + bY|C ) = aE(X|C ) + bE(Y|C ) for all real numbers a, b and random variables X, Y. 3. If X is a one-dimensional random variable and X ≥ 0, then E(X|C ) ≥ 0. 4. If Y is a bounded C -measurable real valued random variable and X is a one-dimensional random variable, then E(Y X|C ) = Y E(X|C ). 5. If D ⊂ C ⊂ B are σ-algebras, then E(E(X|C )|D) = E(X|D). 278

6.

R

Ω

|E(X|D)|d(P|D) ≤

R

E(X|C )d(P|C ).

Ω

Exercise 1. Let (Ω, B, P) be a probability space, C a sub-σ-algebra of B. Let X(t, ·)Y(t, ·) : Ω → R be measurable with respect to B and C respectively where t ranges over the real line. Further let E(X(t, ·)|C ) = Y(t, ·) for each t. If f is a simple C -measurable function then show that Z Z X( f (w), w)d(P|C ) = Y( f (w)w)dP C

C

for every C in C . [Hint. Let A1 , . . . , An be a C -measurable partition such that f is constant on each Ai . Verify the equality when C is replaced by C ∩ Ai .] Exercise 2. Give conditions on X, Y such that exercise 1 is valid for all bounded C -measurable functions and prove your claim. The next lemma exhibits conditioning as a projection on a Hilbert space. Lemma . Let (Ω, B, P) be any probability space C a sub-σ-algebra of B. Then

287 (a) L2 (Ω, C , P) is a closed subspace of L2 (Ω, B, P). (b) If π : L2 (Ω, B, P) → L2 (Ω, C , P) is the projection,then π( f ) = E( f |C ). (a) is clear, because for any f ∈ L1 (Ω, C , P) Z Z f d(P|C ) = f dP

Proof.

Ω

Ω

(use simple function 0 ≤ s1 ≤ . . . ≤ f , if f ≥ 0) and L2 (Ω, C , P) 279 is complete. (b) To prove this it is enough to verify it for characteristic functions because both π and f → E( f |C ) are linear and continuous. Let A ∈ B, C ∈ C then π(χC ) = χC . As π is a projection Z Z χA π(χC )d(P|B), π(χA )χC d(P|B) = i.e.

Z

π(χA )d(P|B) =

C

Z

XA d(P|B).

C

Since π(χA ) is C -measurable, Z Z π(χA )d(P|B) = π(χA )d(P|C ) C

C

Therefore Z C

π(χA )d(P|C ) =

Z

χA d(P|B), ∀C in C .

C

Hence π(χA ) = E(χA |C ). 

33. Application of Stochastic Integral

288 Kolmogorov’s Theorem.

280

Statement. Let A be any nonempty set and for each finite ordered subset (t1 , t2 , . . . , tn ) of A [i.e. (t1 , . . . , tn ) an ordered n-tuple with ti in A], let P(t1 ,...,tn ) be a probability on the Borel sets in Rdn = Rd × Rd × · · · Rd . Assume that the family P(t1 ,...,tn ) satisfies the following two conditions (i) Let τ : {1, 2, . . . , n} → {1, 2, . . . , n} be any permutation and fτ : Rdn → Rdn be given by fτ ((x1 , . . . , xn )) = (xτ(1) , . . . , xτ(n) ). We have P

(E) (tτ(1) ,...,tτ(n) )

= P(t1 ,...,tn ) ( fτ−1 (E))

for every Borel set E of Rdn . In short, we write this condition as Pτt = Pt τ−1 . (ii) P (E) = P(t1 ,t2 ,...,tn ,tn+1 ,...tn+m ) (E × Rdm ) for all Borel sets E of Rdn (t1 ,...,tn )

and this is true for all t1 , . . . , tn , tn+1 , . . . , tn+m of A. Then, there exists a probability space (Ω, B, P) and a collection of random variable {Xt : t ∈ A} : (Ω, B) → Rd such that P (E) = P{w : (Xt1 (w), . . . , Xtn (w)) ∈ E} (t1 ,...,tn )

for all Borel sets E of Rdn .

281

Proof. Let Ω = π{Rdt : t ∈ A} where Rdt = Rd for each t. Define Xt : Ω → Rd to be the projection given by Xt (w) = w(t). Let B0 be the algebra generated by {Xt : t ∈ A} and B the σ-algebra generated by {Xt : t ∈ A}. Having got Ω and B we have to construct a probability P on (Ω, B) satisfying the conditions of the theorem. Given t1 , . . . , tn define π(t1 ,...,tn ) : Ω → Rd × Rd × · · · × Rd (n times)

289 by π(t1 ,...,tn ) (w) = (w(t1 ), . . . , w(tn )). It is easy to see that every element of B0 is π−1 (t1 ,...,tn ) (E) for suitable dn t1 , . . . , tn in A and a suitable Borel set E of R . Define P on B0 by P(π−1 (t1 ,...,tn ) (E)) = P(t1 ,...,tn ) (E). Conditions (1) and (2) ensure that P is a well-defined function on B0 and that, as P(t1 ,...,tn ) are measures, P is finitely additive on B0 .  Claim . Let C1 ⊃ C2 ⊃ . . . ⊃ Cn ⊃ . . . be a decreasing sequence in B0 with limit P(Cn ) ≥ δ > 0. Then ∩Cn is non-empty. Once the claim is n→∞ proved, by Kolmogorov’s theorem on extension of measures, the finitely additive set function P can be extended to a measure P on B. One easily sees that P is a required probability measure. Proof of the Claim. As Cn ∈ B0 , we have Cn = π−1(n)

(n) (t1 ,...,tk(n) )

(En ) for suitable ti(n) in A

and Borel set En in Rdk(n) . Let (n) T n = (t1(n) , . . . , tk(n) )

and

(n) An = {t(n) , . . . , tk(n) }.

We can very well assume that An is increasing with n. Choose a compact 282 subset En′ of En such that PT n (En − En′ ) ≤ δ/2n+1 . ′ ′ n+1 . If C ′′ = C ′ ∩C ′ ∩. . .∩C ′ If Cn′ = π−1 n n T n (E n ), then P(C n −C n ) ≤ δ/2 1 2 ′′ ′ ′′ then Cn ⊂ Cn ⊂ Cn , Cn is decreasing and

P(Cn′′ ) ≥ P(Cn ) −

n X i=1

P(Ci − Ci′ ) ≥ δ/2.

We prove ∩Cn′′ is not empty, which proves the claim. Choose wn in Cn′′ . As πT 1 (wn ) is in the compact set E1 for all n, choose a subsequence (1) n(1) 1 , n2 , . . . of 1, 2, . . . such that πT 1 (wnk (1))

33. Application of Stochastic Integral

290

converges as k → ∞. But for finitely many n(1) k ’s, πT 2 (ωnm (1)) is in (1) ′ the compact set E2 . As before choose a subsequence n(2) k of nk such that πT 1 (ωnk (2)) converges as k → ∞. By the diagonal process obtain a subsequence, w∗n of wn such that πT m (w∗n ) converges as n → ∞ for all m. Thus, if t is in ∞ [ Am , then limit w∗n (t) = xt n→∞

m=1

exists. Define w by w(t) = 0 if t ∈ A easily sees that w ∈ 283

Martingales.

∞ T

n=1

∞ S

m=1

Am , w(t) = xt if t ∈

∞ S

Am . One

m=1

Cn′′ , completing the proof of the theorem.

Definition. Let (Ω, F , P) be a probability space, (T, ≤) a totally ordered set. Let (Ft )t∈T be an increasing family of sub-σ-algebras of F . A collection (Xt )t∈T of random variables on Ω is called a martingale with respect to the family (Ft )t∈T if (i) E(|Xt |) < ∞, ∀t ∈ T ; (ii) Xt is Ft -measurable for each t ∈ T ; (iii) E(Xt |Fs ) = X s a.s. for each s, t in T with t ≥ s. (Markov property). If instead of (iii) one has (iii)′ E(Xt |Fs ) ≥ (≤)X s a.s., then (Xt )t∈T is called a submartingale (respectively supermartingale). From the definition it is clear that (Xt )t∈T is a submartingale if and only if (−Xt )t∈T is a supermartingale, hence it is sufficient to study the properties of only one of these. T is usually any one of the following sets [0, ∞), N, Z, {1, 2, . . . , n}, [0, ∞]

or

N ∪ {∞}.

291 Examples. (1) Let (Xn )n=1,2... be a sequence of independent random variables with E(Xn ) = 0. Then Yn = X1 + · · · + Xn is a martingale with respect to (Fn )n=1,2,... 284 where Fn = σ{Y1 , . . . , Yn } = σ{X1 , . . . , Xn }. Proof. By definition, each Yn is Fn -measurable. E(Yn ) = 0. E((X1 + · · · + Xn + Xn+1 + · · · + Xn+m )|σ{X1 , . . . , Xn })

= X1 + · · · + Xn + E((Xn+1 + · · · + Xn+m )|σ{X1 , . . . , Xn }) = Yn + E(Xn+1 + · · · + Xn+m ) = Yn .

 (2) Let (Ω, F , P) be a probability space, Y a random variable with E(|Y|) < ∞. Let Ft ⊂ F be a σ-algebra such that ∀t ∈ [0, ∞) Ft ⊂ Fs

if t ≤ s.

If Xt = E(Y|Ft ), Xt is a martingale with respect to (Ft ). Proof.

(i) By definition, Xt is Ft -measurable.

(ii) E(Xt ) = E(Y) (by definition) < ∞. (iii) if t ≥ s, E(Xt |Fs ) = E(E(Y|Ft )|Fs ) = E(Y|Fs ) = X s  Exercise 1. Ω = [0, 1], F = σ-algebra of all Borel sub sets of Ω, P = Lebesgue measure. Let Fn =-algebra generated by the sets h 1 h 1 2  h 2n − 1 i 0, n n , n ; . . . , ,1 . 2 2 2 2n

33. Application of Stochastic Integral

292

Let f ∈ L′ [0, 1] and define  n n −1 Zj/2 Z1 2X  n Xn (w) = 2  χ j−1 j f dy + χ[ 2n n−1 ,1] 2  j=1 [ 2n , 2n ) j−1/2n

2n −1/2n

Show that (Xn ) is a martingale relative to (Fn ).

285

   f dy 

Exercise. Show that a submartingale or a supermartingale {X s } is a martingale iff E(X s ) = constant. Theorem . If (Xt )t∈T , (Yt )t∈T are supermartingales then (i) (aXt + bYt )t∈T is a supermartingale, ∀a, b ∈ R+ = [0, ∞). (ii) (Xt ∧ Yt )t∈T is a supermartingale. Proof.

(i) Clearly Zt = aXt + bYt is Ft -measurable and E(|Zt |) ≤ aE(|Xt |) + bE(|Yt |) < ∞. E(aXt + bYt |Fs ) = aE(Xt |Fs ) + bE(Yt |Fs ) ≤ aX s + bY s = Z s ,

if t ≥ s.

(ii) Again Xt ∧ Yt is Ft -measurable and E(|Xt ∧ Yt |) < ∞, E(Xt ∧ Yt |Fs ) ≤ E(Xt |Fs ) ≤ X s . 286

Similarly E(Xt ∧ Yt |Fs ) ≤ E(Yt |Fs ) ≤ Y s ,

if

t ≥ s.

Therefore E(Xt ∧ Yt |Fs ) ≤ X s ∧ Y s .  Jensen’s Inequality. Let X be a random variable in (Ω, B, P) with E(|X|) < ∞ and let φ(x) be a convex function defined on the real line such that E(|φ0 X|) < ∞. Then φ(E(X|C )) ≤ E(φ0 X|C ) a.e. where C is any sub-σ-algebra of B.

293 Proof. The function φ being convex, there exist sequences a1 , a2 , . . . an , . . . , b1 , b2 , . . . of real numbers such that φ(x) = sup(an x + bn ) for each x. n

Let Ln (x) = an x + bn . Then Ln (E(X|C )) = E(Ln (X)|C ) ≤ E(φ(X)|C ) for all n so that φ(E(X|C )) ≤ E(φ(X)|C ).  Exercise. (a) If {Xt : t ∈ T } is a martingale with respect to {Ft : t ∈ T } and φ is a convex function on the real line such that E(|φ(Xt )|) < ∞ for every t, then {φ(Xt )} is a sub martingale. (b) If (Xt )t∈T is a submartingale and φ(x) is a convex function and nondecreasing and if E(|φ0 Xt |) < ∞, ∀t then {φ(Xt )} is a submartingale. (Hint: Use Jensen’s inequality). Definition . Let (Ω, B, P) be a probability space and (Ft )t∈[0,∞) an in- 287 creasing family of sub-σ-algebras of F . Let (Xt )t∈[0,∞) be a family of random variables on Ω such that Xt is Ft -measurable for each t ≥ 0. (Xt ) is said to be progressively measurable if X : [0, t] × Ω → R

defined by

X(s, w) = X s (w)

is measurable with respect to the σ-algebra B[0, t] × Ft for every t. Stopping times. Let us suppose we are playing a game of chance, say, tossing a coin. The two possible outcomes of a toss are H (Heads) and T (Tails). We assume that the coin is unbiased so that the probability of getting a head is the same as the probability of getting a tail. Further suppose that we gain +1 for every head and lose 1 for every tail. A game of chance of this sort has the following features. 1. A person starts playing with an initial amount N and finishes with a certain amount M.

33. Application of Stochastic Integral

294

2. Certain rules are specified which allow one to decide when to stop playing the game. For example, a person may not have sufficient money to play all the games, in which case he may decide to play only a certain number of games.

288

It is obvious that such a game of chance is fair in that it is neither advantageous nor disadvantageous to play such a game and on the average M will equal N, the initial amount. Furthermore, the stopping rules that are permissible have to be reasonable. The following type of stopping rule is obviously unreasonable. Rule. If the first toss is a tail the person quits at time 0 and if the first toss is a head the person quits at time t = 1. This rule is unreasonable because the decision to quit is made on the basis of a future event, whereas if the game is fair this decision should depend only on the events that have already occured. Suppose, for example, 10 games are played, then the quitting times can be 0, 1, 2, . . . , 10. If ξ1 , . . . , ξ10 are the outcomes (ξi = +1 for H, ξi = −1 for T ) then the quitting time at the 5th stage (say) should depend only on ξ1 , . . . , ξ4 and not any of ξ5 , . . . , ξ10 . If we denote ξ = (ξ1 , . . . , ξ10 ) and the quitting time τ as a function of ξ then we can say that {ξ : τ = 5 depends only ξ1 , . . . , ξ4 }. This leads us to the notion of stopping times. Definition. Let (Ω, F , P) be a probability space, (Ft )t∈[0,∞) an increasing family of sub-σ-algebras of F . τ : Ω → [0, ∞] is called a stopping time or Markov time (or a random variable independent of the future) if {w : τ(w) ≤ t} ∈ Ft

for each

t ≥ 0.

Observe that a stopping time is a measurable function with respect to σ(∪Ft ) ⊂ F . 289

Examples.

1. τ = constant is a stopping time.

2. For a Brownian motion (Xt ), the hitting time of a closed set is stopping time. T Exercise 2. Let Ft+ ≡ Fs ≡ Ft . Def s>t

295 [If this is satisfied for every t ≥ 0, Ft is said to be right continuous]. If {τ < t} ∈ Ft for each t ≥ 0, then τ is a stopping time. (Hint: {τ ≤ t = ∞ T {τ < t + 1/n} for every k). n=k S We shall denote by F∞ the σ-algebra generated by Ft . If τ is a t∈T

stopping time, we define

Fτ = {A ∈ F∞ : A ∩ {τ ≤ t} ∈ Ft , ∀t ≥ 0} Exercise 3.

(a) Show that Fτ is a σ-algebra. (If A ∈ Fτ , Ac ∩ {τ ≤ t} = {t ≤ t} − A ∩ {τ ≤ t}).

(b) If τ = t (constant) show that Fτ = Ft . Theorem . Let τ and σ be stopping times. Then (i) τ + σ, τvσ, τ ∧ σ are all stopping times. (ii) If σ ≤ τ, then Fσ ⊂ Fτ . (iii) τ is Fτ -measurable. (iv) If A ∈ Fσ , then A ∩ {σ = τ} and A ∩ {σ ≤ τ} are in Fσ∧τ ⊂ Fσ ∩ Fτ . In particular, {τ < σ}, {τ = σ}, {τ > σ} are all in 290 Fτ ∩ Fσ . (v) If τ′ is Fτ -measurable and τ′ ≥ τ, then τ′ is a stopping time. (vi) If {τn } is a sequence of stopping times, then lim τn . lim τn are also stopping times provided that Ft+ = Ft , ∀t ≥ 0. (vii) If τn ↓ τ, then Fτ = Proof.

∞ T

n=1

Fτn provided that Ft+ = Ft , ∀t ≥ 0.

(i) {σ + τ} > t} = {σ + τ > t, τ ≤ t, σ ≤ t} ∪ {τ > t} ∪ {σ > t}; {σ + τ > t, σ ≤ t} = τ ≤ A

33. Application of Stochastic Integral

296 [

=

r∈Q 0≤r≤t

{σ > r > t − τ, τ ≤ t, σ ≤ t}

(Q = set of rationals) {σ > r > t − τ, τ ≤ t, σ ≤ t} = {t ≥ σ > r} ∩ {t ≥ τ > t − r} = {σ ≤ t} ∩ {σ ≤ r}c ∩ {τ ≤ t} ∩ {τ ≤ t − r}c .

The right side is in Ft . Therefore σ + τ is a scopping time. {τVσ ≤ t} = {τ ≤ t} ∩ {σ ≤ t}

{τ ∧ σ > t} = {τ > t} ∩ {σ > t} (ii) Follows from (iv). (iii) {τ ≤ t}{τ ≤ s} = {τ ≤ t ∧ s} ∈ Ft∧s ⊂ Fs , ∀s ≥ 0. (iv) A ∩ {σ < τ} ∩ {σ ∧ τ ≤ t} = [A ∩ {σ ≤ t < τ}] U[A ∩ U {σ ≤< τ} ∩ {τ ≤ t}] ∈ Ft . r∈Q 0≤r≤t

291

A ∩ {σ ≤ τ} ∩ {σ ∧ τ ≤ t} = A ∩ {σ ≤ τ} ∩ {σ ≤ t}.

It is now enough to show that (σ ≤ τ) ∈ Fσ ; but this is obvious because (τ < σ) = (σ ≤ τ)c is in Fσ∧τ ⊂ Fσ . Therefore A ∩{σ ≤ τ} ∈ Fσ∧τ and (iv) is proved. (v) {τ′ ≤ t} = {τ′ ≤ t} ∩ {τ ≤ t} ∈ Ft as (τ′ ≤ t) ∈ Fτ . Therefore τ′ is a stopping time. (vi) lim τn ≡ sup inf τk n k≥n

= sup inf inf{τn , τn+1 , . . . , τn+ℓ }. n

ℓ

By (i), inf{τn , τn+1 , . . . , τn+ℓ } is a stopping time. Thus we have only to prove that if τn ↑ τ or τn ↓ τ where τn are stopping times, ∞ T then τ is a stopping time. Let τn ↑ τ. Then {τ ≤ t} = {τn ≤ t} n=1

so that τ is a stopping time. Let τn ↓ τ. Then ∞ \ {τ ≥ t} = {τn ≥ t}. n=1

297 By Exercise 3, τ is a stopping time. That lim τn is a stopping time is proved similarly. (vii) Since τ ≤ τn , ∀n, Fτ ⊂

∞ T

n=1

Fτn . Let A ∈

A ∩ (τn < t) ∈ Ft , ∀n. A ∩ (τ < t) =

∞ T

∞ T

n=1

Fτn . Therefore

(A ∩ (τm < t)) ∈ Ft .

m=1

Therefore A ∈ Fτ .



Optional Sampling Theorem. (Discrete case). Let {X1 , . . . , Xk } be a martingale relative to {F1 , . . . , Fk }. Let {τ1 , . . . , τ p } be a collection of stopping times relative to {F1 , . . . , Fk } such that τ1 ≤ τ2 ≤ . . . ≤ τ p 292 a.s. and each τi takes values in {1, 2, . . . , k}. Then {Xτ1 , . . . , Xτ p } is a martingale relative to {Fτ1 , . . . , Fτ p } where for any stopping time τ, Xτ (ω) = Xτ(w) (ω). Proof. It is easy to see that each Xτi is a random variable. In fact Xτm = k P Xi χ{τm =i} . Let τ ∈ {1, 2, . . . , k}. Then i=1

E(|Xτ |) ≤

k Z X j=1

|X j |dP < ∞.

Consider (Xτ j ≤ t) ∩ (τ j ≤ s) = that

\ (Xℓ ≤ t) ∈ Fs . ℓ≤s

Then (Xτ j ≤ t) is in Fτ j , i.e. Xτ j is Fτ j -measurable. Next we show E(Xτ j |Fτk ) ≤ Xτk ,

(*)

(*) is true if and only if Z Z Xτ j dP ≤ Xτk dP A

if

j ≥ k.

for every

A ∈ Fτk .

A

The theorem is therefore a consequence of the following



33. Application of Stochastic Integral

298

Lemma . Let {X1 , . . . , Xk } be a supermartingale relative to {F1 , . . . , Fk }. If τ and σ are stopping times relative to {F1 , . . . , Fk } taking values in {1, 2, . . . , k} such that τ ≤ σ then Z Z Xτ dP ≥ Xσ dP for every A ∈ Fτ . A

293

A

Proof. Assume first that σ − τ ≤ 1. Then Z

(Xτ − Xσ )dP =

k X

=

k X

A

Z

j=1 [A∩(τ= j)∩(τ<σ)]

Z

j=1 [A∩(τ= j)]

(Xτ − Xσ )dP

(X j − X j+1 )dP

A ∈ Fi . Therefore A ∩ (τ = j) ∈ F j . By supermartingale property Z (X j − X j+1 )dP ≥ 0. [A∩(τ= j)]

Therefore

Z

(Xτ − Xσ )dP ≥ 0.

A

Consider now the general case τ ≤ σ. Define τn = σ ∧ (τ + n). Therefore τn ≥ τ. τn is a stopping time taking values in {1, 2, . . . , k}, τn+1 ≥ τn , τn+1 − τn ≤ 1, τk = σ. R R Therefore Xτn dP ≥ Xτn+1 dP, ∀A ∈ Fτn . If A ∈ Fτ then A ∈ A

A

Fτn , ∀n. Therefore Z Z Z Xτ1 dP ≥ Xτ2 dP ≥ . . . ≥ Xτk dP, A

A

A

∀A ∈ Fτ .

299 Now τ1 − τ ≤ 1. τ ≤ τ1 . Therefore Z Z Z Xτ dP ≥ Xτ1 dP ≥ Xσ dP. A

A

A

This completes the proof.



N.B. The equality in (*) follows by applying the argument to {−X1 , . . . , −Xk }. Corollary 1. Let {X1 , X2 , . . . , Xk } be a super-martingale relative to {F1 , . . . , Fk }. If τ is any stopping time, then E(Xk ) ≤ E(Xτ ) ≤ E(X1 ). Proof. Follows from the fact that {X1 , Xτ , Xk } is a supermartingale relative to {F1 , Fτ , Fk }.  Corollary 2. If {X1 , X2 , . . . , Xk } is a super-martingale relative to {F1 , . . . , Fk } and τ is any stopping time, then E(Xτ ) ≤ E(|X1 |) + 2E(Xk− ) ≤ 3 sup E(|Xn |) 1≤n≤k

where for any real x, x− =

|x| − x . 2

|Xk | − Xk Proof. Xk− = , so 2E(Xk− ) = E(|Xk |) − E(Xk ). 2 By theorem {Xτ ∧0, Xk ∧0} is a super-martingale relative to {Fτ , Fk }. Therefore E(Xk ∧ 0|Fτ ) ≤ E(Xτ ∧ 0). Hence E(Xk− ) ≥ E(Xτ− ) =

E(|Xτ |) − E(Xτ ) . 2

294

33. Application of Stochastic Integral

300 Therefore

E(|Xτ |) ≤ 2E(Xk− ) + E(Xτ )

≤ 2E(Xk− ) + E(X1 ) ≤ 3 sup E(|Xn |). 1≤n≤k



295

Theorem . Let (Ω, F , P) be a probability space and (Ft )t≥0 on increasing family of sub-σ-algebras of F . Let τ be a finite stopping time, and (Xt )t≥0 a progressively measurable family (i.e. X : [0, ∞) × Ω → R defined by X(t, w) = Xt (w) is progressively measurable). If Xτ (w) = Xτ(w) (w), then Xτ is Fτ -measurable. Proof. We show that {w : X(τ(w), w) ≤ t, τ(w) ≤ s} ∈ Ft for every t. Let Ω s = {w : τ(w) ≤ s}; Ω s ∈ Fs and hence the σ-algebra induced by Fs on Ω s is precisely {A ∩ Ω s : A ∈ Fs } = {A ∈ Fs : A ⊂ Ω s }. Since τ(w) is measurable, w → (τ(w), w)

of Ω s → [0, s] × Ω s

is (Fs , B[0, s] × Fs )-measurable. Since X is progressively measurable, X

[0, s] × Ω s − → R is measurable. Therefore {w : X(τ(w), w) ≤ t, τ(w) ≤ s} ∈ σ-algebra on Ω s . Therefore Xτ is Fτ measurable. The next theorem gives a condition under which (Xt )t≥0 is progressively measurable.  Theorem . If Xt is right continuous in t, ∀w and Xt is Ft -measurable, ∀t ≥ 0 then (Xt )t≥0 is progressively measurable. Proof. Define ! [nt] + 1 [nt] + 1 Xn (t, w) = X ,w · ↓ t. n n

301 Then Lt Xn (t, w) = X(t, w) (by right continuity)

n→∞

296

Step 1. Suppose T is rational, T = m/n where m ≥ 0 is an integer. Then {(t, w) : 0 ≤ t < T, Xn (t, w) ≤ α}   −1  [    h i i + 1  Xi+1  = , X (−∞, α]       n n n 0≤i≤m−1

Thus if T = m/n, Xn |[0,T )×Ω is B[0, T ] × FT -measurable. Now km T = . Letting k → ∞, by right continuity of X(t) one gets X|[0,)×Ω is kn [0, T ] × FT -measurable. As X(T ) is FT -measurable, one gets X|[0,T ]×Ω is [0, T ] × FT -measurable. Step 2. Let T be irrational. Choose a sequence of rationals S n increasing to T . {(t, w) : 0 ≤ t ≤ T, X(t, w) ≤ α} ∞ [ = {(t, w) : 0 ≤ t ≤ S n , X(t, w) ≤ α} ∪ {T } × XT−1 (−∞, α] n=1

The countable union is in B[0, T ]×FT by Step 1. The second member is also in B[0, T ] × FT as X(T ) is FT -measurable. Thus X|[0,T ]×Ω is B[0,T ] × FT -measurable when T is irrational also.  Remark. The technique used above is similar to the one used for proving that a right continuous function f : R → R is Borel measurable. Theorem . Let {X1 , . . . , Xk } be a supermartingale and λ ≥ 0. Then R Xk dP (1) λP( sup Xn ≥ λ) ≤ E(X1 ) − 1≤n≤k

  sup Xn <λ 1≤n≤k

≤ E(X1 ) + E(Xk− ).

297

33. Application of Stochastic Integral

302 (2) λP( inf Xn ≤ −λ) ≤ − 1≤n≤k

R

Xk dP

{inf Xn ≤−λ}

≤ E(Xk− ). Proof. Define

τ(w) = inf{n : Xn ≥ λ} if = k,

if

sup Xn < λ.

sup Xn ≥ λ,

n

Clearly τ ≥ 0 and τ is a stopping time. If τ < k, then Xτ (w) ≥ λ for each w. Z Z E(Xτ ) = Xτ dP + Xτ dP (sup Xn <λ)

(sup Xn ≥λ)

≥ λP(sup Xn ≥ λ) +

Z

Xk dP.

(sup Xn <λ)

Therefore E(X1 ) ≥ λP(sup Xn ≥ λ) +

Z

Xk dP,

(sup Xn <λ)

λP(sup Xn ≥ λ) ≤ E(X1 ) −

Z

Xk dP ≤ E(X1 ) + E(Xk− )

(sup Xn <λ)

The proof of (2) is similar if we define    inf{n : Xn ≤ −λ}, if inf Xn ≤ −λ, τ(w) =   k, if inf Xn > −λ. 298



Kolmogorov’s Inequality (Discrete Case). Let {X1 , . . . , Xk } be a finite sequence of independent random variables with mean 0. Then ! 1 2 P sup (|X1 + · · · + Xn | ≥ λ) ≤ 2 E((X1 + X2 + · · · + Xk ) ) λ 1≤n≤k

303 Proof. If S n = X1 + · · · + Xn , n = 1, 2, . . . , k, then {S 1 , . . . , S k } is a martingale with respect to {F1 , . . . , Fk } where Fn = σ{X1 , . . . , Xn }. Therefore S 12 , . . . , S k2 is a submartingale (since x → x2 is convex). By the previous theorem, λ2 P{inf −S n2 ≤ −λ2 } ≤ E((−S 2K )− ) Therefore P{sup |S n | ≥ λ} ≤ =

E((−S k2 )− ) λ2

=

E(S k2 ) λ2

1 E((X1 + X2 + · · · + Xk )2 ). λ2 

Kolmogorov’s Inequality (Continuous case). Let {X(t) : t ≥ 0} be a continuous martingale with E(X(0)) = 0. If 0 < T < ∞, then for any ǫ>0 n  1 P w : sup |X(s, w)| ≥ ǫ ≤ 1 E((X(T ))2 ). ǫ 0≤s≤T Proof. For any positive integer k define Y0 = X(0), ! T  T  2T Y1 = X k − X(0), Y2 = X k − X k , . . . , Y2 k 2 2 2 ! ! k k 2T (2 − 1) =X −X T . 2k 2k By Kolmogorov inequality for the discrete case, for any δ > 0.    nT    1  P  sup |X k | > δ ≤ 2 E((X(T ))2 ). δ 2 0≤n≤2k  nT  By continuity of X(t), Ak = {w : sup |X k | > δ} increases to 299 2 0≤n≤2k { sup |X(s)| > δ} so that one gets 0≤s≤T

(1)

! 1 P sup |X(s)| > δ ≤ 2 E((X(T ))2 ). δ 0≤s≤T

33. Application of Stochastic Integral

304 Now

!

1 P sup |X(s)| ≥ ǫ ≤ limit P sup |X(s)| > ǫ − m→∞ m 0≤s≤T 0≤s≤T ≤ limit m→∞

!

1 E((X(T ))2 ), (ǫ − 1/m)2

by (1).

= 1/ǫ 2 E((X(T ))2 ). This completes the proof.



Optional Sampling Theorem (Countable case). Let {Xn : n ≥ 1} be a supermartingale relative to {Fn : n ≥ 1}. Assume that for some X∞ ∈ L1 , Xn ≥ E(X∞ |Fn ). Let σ, τ be stopping times taking values in N ∪ {∞}, with σ ≤ τ. Define Xτ = X∞ on {σ = ∞} and Xτ = X∞ on {σ = ∞}. Then E(Xτ |Fσ ) ≤ Xσ . Proof. We prove the theorem in three steps. Step 1. Let X∞ = 0 so that Xn ≥ 0. Let τk = τ ∧ k, σk = τ ∧ k. By optional sampling theorem for discrete case E(Xτk ) ≤ E(Xk ) ≤ E(X1 ). By Fatou’s lemma, E(Xτ ) < ∞. Again by optional sampling theorem for the discrete case, E(Xτk |Fσk ) ≤ Xσk . . . , (0). Let A ∈ Fσ . Then A ∩ {σ ≤ k} ∈ Fσk , and by (0) Z Z Z Z Xτ dP ≤ Xτk dP ≤ Xσk dP ≤

300

A∩(τ≤k)

A∩{τ≤k}

Letting k → ∞, (1)

Z

A∩(σ≤k)

Xτ dP ≤

A∩(τ,∞)

Z

A∩(σ≤k)

Xσ dP.

A∩(σ,∞)

Clearly (2)

Z

A∩(τ=∞)

Xτ dP =

Z A

X∞ dP =

Z

A∩(σ=∞)

Xσ dP

Xσ dP.

305 By (1) and (2), inf X dP ≤ A

R

Xσ dP, proving that

A

E(Xτ |Fσ ) ≤ Xσ . Step 2. Suppose Xn = E(X∞ |Fn ). In this case we show that Xτ = E(X∞ |Fτ ) for every stopping time so that E(Xτ |Fσ ) = Xσ . If A ∈ Fτ , then Z Z Xτ dP = X∞ dP for every k. (τ≤k)

(1)

A∩(τ≤K)

Letting k → ∞, Z

A∩(τ,∞)

(2)

Z

Z

Xτ dP =

X∞ dP,

A∩(τ,∞)

Xτ dP =

Z A

A∩(τ=∞)

X∞ dP =

Z

X∞ dP

A∩(τ=∞)

The assertion follows from (1) and (2). Step 3. Let Xn be general. Then

301

Xn = Xn − E(X∞ |Fn ) + E(X∞ |Fn ). Apply Step (1) to Yn = Xn − E(X∞ |Fn ) and Step (2) to Zn = E(X∞ |Fn ) to complete the proof.  Uniform Integrability. Definition . Let (Ω, B, P) be any probability space, L1 = L1 (Ω, B, P). A family H ⊂ L1 is calledR uniformly integrable if for every ǫ > 0 there exists a δ > 0 such that |X|dP < ǫ for all X in H. (|X|≥δ)

33. Application of Stochastic Integral

306

Note. Every uniformly integrable family is a bounded family. Proposition . Let Xn be a sequence in L1 and let Xn → X a.e. Then Xn → X in L1 iff {Xn : n ≥ 1} is uniformly integrable. Proof. is left as an exercise.



As {Xn : n ≥ 1} is a bounded family, by Fatou’s lemma X ∈ L1 . Let ǫ > 0 be given. By Egoroff’s theorem there exists a set F such that P(F) < ǫ and Xn → X uniformly on F. Z Z |Xn − X|dP ≤ ||Xn − X||∞,Ω−F+ |Xn − X|dP ≤ ||Xn − X||∞,Ω−F + Z

≤ ||Xn − X||∞,Ω−F + +

Z

F

|Xn |dP +

F

|X|dP

Z

|Xn |dP +

|X|dP+

F∩(|X|≥δ)

Z

|Xn |dP + Z

Z F

F∩(|Xn |≥δ)

F∩{|Xn |≤δ}

≤ ||Xn − X||∞,Ω−F +

Z

XdP

F∩(|X|≤δ)

|Xn |dP +

(|Xn |≥δ)

Z

|X|dP + 2δǫ

(|X|≥δ)

302

The result follows by uniform integrability of {X, Xn : n ≥ 1}. Corollary . Let C be any sub-σ-algebra of B. If Xn → X a.e. and Xn is uniformly integrable, then E(Xn |C ) → E(X|C ) in L1 (Ω, C , P). Proposition . Let H ⊂ L1 . Suppose there exists an increasing convex function G : [0, ∞) → [0, ∞) such that limit t→∞

G(t) = ∞ and t

sup E(G(|X|)) < ∞.

X∈H

Then the family H is uniformly integrable.

307 Example . G(t) = t2 is a function satisfying the conditions of the theorem. Proof. (of the proposition). Let M = sup E(G(|X|)). X∈H

Let ǫ > 0 be given. Choose δ > 0 such that G(t) M ≥ t ǫ Then for X in H Z Z ǫ |X|dP ≤ M (|X|≥δ)

for

t ≥ δ.

G(|X|)dP ≤

(|X|≥δ)

ǫ M

Z

G(|X|)dP ≤ ǫ

G

 Remark. The converse of the theorem is also true. Exercise . Let H be a bounded set in L∞ , i.e. there exists a constant M such that ||X||∞ ≤ M for all X in H. Then H is uniformly integrable. Up Crossings and Down Crossings. Definition . Let a < b be real numbers; let s1 , s2 , . . . , sk be also given reals. Define i1 , i2 , . . . , ik as follows.    inf{n : sn < a}, i1 =   k, if no si < a;    inf{n > i1 : sn > b}, i2 =   k, if sn ≤ b for each n > i1 ;    inf{n > i2 : sn < a}, i3 =   k, if sn ≥ a for each n > i2 ;

and so on

303

308

33. Application of Stochastic Integral

Let t1 = si1 , t2 = si2 , . . .. If (t1 , t2 ), (t3 , t4 ), . . ., (t2p−1 , t2p ) are the only non-empty intervals and (t2p+1 , t2p+2 ), . . . are all empty, then p is called the ******** of the sequence s1 , . . . , sk for the interval [a, b] and is denoted by U(s1 , . . . , sk ; [a, b]). Note. U (the up crossing) always takes values in {0, 1, 2, 3, . . .}. Definition. For any subset S of reals define U(S ; [a, b]) = sup{U(F; [a, b]) : F is a finite subset of S } The number of down crossings is defined by D(S ; [a, b]) = U(−S ; [−b, −a]). For any real valued function f on any set S we define U( f, S , [a, b]) = U( f (S ), [a, b]). If the domain of S is known, we usually suppress it. 304

Proposition . Let a1 , a2 , . . . be any sequence of real numbers and S = {a1 , a2 , . . .}. If U(S , [a, b]) < ∞ for all a < b, then these sequence {an } is a convergent sequence. Proof. It is clear that if T ⊂ S then U(T, [a, b]) ≤ U(S , [a, b]). If the sequence were not convergent, then we can find a and b such that lim inf an < a < b < lim sup an . Choose n1 < n2 < n3 . . .; m1 < m2 < . . . such that ani < a and ami > b for all i. If T = {an1 , am1 , an2 , am2 , . . .}, then U(S ; [a, b]) ≥ U(T ; [a, b]) = ∞; a contradiction.  Remark. The converse of the proposition is also true. Theorem . (Doob’s inequalities for up crossings and down crossings). Let {X1 , . . . , Xk } be a submartingale relative to {F1 , . . . , Fk } a < b. Define U(w, [a, b]) = U(X1 (w), . . . , Xk (w); [a, b]) and similarly define D(w, [a, b]). Then (i) U, D are measurable functions;

309 (ii) E(U(·, [a, b])) ≤

E((Xk − a) + 1) − E((X1 − a)+ ) ; b−a

(iii) E(D(·, [a · b])) ≤ E((Xk − b)+ )/(b − a). Proof.

(i) is left as an exercise.

(ii) Define Yn = (Xn − a)+ ; there are submartingales. Then clearly Yn ≤ 0 if and only if Xn ≤ a and Yn ≥ b − a iff Xn ≥ b, so that UY1 (w), . . . , Yk (w); [0, b − a]) = U(X1 (w), . . . , Xk (w); [a, b]) 305

Define τ1 = 1    inf{n : Yn = 0} τ2 =   k, if each Yn = 0    inf{n > τ2 : Yn > b − a, τ3 =   k, if Yn < b − a for each n > τ2 ;

τk+1 = k.

As {Y1 , . . . , Yk } is a submartingale, by optional sampling theorem Yτ1 , . . . , Yτk+1 is also a submartingale. Thus E(Yτ2 − Yτ1 ) + E(Yτ4 − Yτ3 ) + · · · ≥ 0.

(1) Clearly

[(Yτ3 − Yτ2 ) + (Yτ5 − Yτ4 ) + · · · ](w) ≥ (b − a) ∪ (Y1 (w), . . . Yk (w); [0, b − a]) = (b − a) ∪ (w, [a, b]).

Therefore (2)

E(Yτ3 − Yτ2 ) + E(Yτ5 − Yτ4 ) + · · · ≥ (b − a)E(U(·, [a, b])).

33. Application of Stochastic Integral

310 By (1) and (2),

E(Yk − Y1 ) ≥ (b − a)E(U(·, [a, b])) giving the result. (iii) Let Yn = (Xn − a)+ so that D(Y1 (w), . . . Yk (w); [0, b − a]) = D(X1 (w), . . . , Xk (w); [a, b]) 306

Define τ1 = 1;    inf{n : Yn ≥ b − a}, τ2 =   k, if each Yn < b − a;    inf{n > τ2 : Yn = 0}, τ3 =   k, if each Yn > 0 for each n > τ2 ;

τk+1 = k.

By optional sampling theorem we get 0 ≥ E(Yτ2 − Yτ3 ) + E(Yτ4 − Yτ5 ) + · · · . Therefore 0 ≥ (b − a)E(D(Y1 , . . . , Yk ; [0, b − a])) + E((b − a) − Yk ). Hence E(D(·, [a, b])) ≤ E((Xk − a)+ − (b − a))/(b − a) ≤

E((Xk − b)+ ) , for (c − a)+ − (b − a) ≤ (c − b)+ (b − a) for all c. 

311 Corollary . Let {X1 , . . . , Xk } be a supermartingale. U, D as in theorem. Then (i) E(D(·, [a, b])) ≤ (ii) E(U(·, [a, b])) ≤

E(X1 ∧ b) − E(Xk ∧ b) . b−a E((Xk − b)− ) . b−a

(i) E(D(·, [a, b])) = E(U(−X1 (w), . . . , −Xk (w), [−b, −a])

Proof.

≤ ≤

E((−Xk + b)+ − (−X1 + b)+ ) , b−a E((b ∧ Xk ) − (b ∧ X1 )) , b−a

by above theorem,

since for

since for all a, b, c, (b − c)+ − (b − a)+ ≤ (b ∧ a) − (b ∧ c). (ii) E(U(·, [a, b])) = = E(D(−X1 (w), . . . , −Xk (w); [−b, −a])) E((−Xk + a)+ ) , by theorem, b−a E((Xk − b)− ) ≤ , b−a (since (−Xk + a)+ ≤ (Xk − b)− ,

≤

307



Theorem . Let {Xn : n = 1, 2, . . .} be a supermartingale relative to {Fn : n = 1, 2, . . .}. Let (Ω, F , P) be complete. (i) If sup E(Xn− ) < ∞, then Xn converges a.e. to a random variable n

denoted by X∞ . (ii) if {Xn : n ≥ 1} is uniformly integrable, then also X∞ exists. Further, {Xn : n = 1, 2, . . . , n = ∞} is a supermartingale with the natural order. (iii) if {Xn : n ≥ 1} is a martingale, then {Xn : n ≥ 1, n = ∞} is a martingale.

33. Application of Stochastic Integral

312 Proof.

(i) Let U(w[a, b]) = U(X1 (w), X2 (w), . . . , [a, b]). By the corollary to Doob’s inequalities theorem, E(U(·, [a, b]) ≤ sup E((Xn − b)− ) < ∞ n

308

for all a < b. Allowing a, b to vary over the rationals alone we find that the sequence Xn is convergent a.e. (ii) Sup E(Xn− ) ≤ sup E(|Xn |) < ∞ so that X∞ exists. As Xn → X∞ in n

n

L1 we get that {Xn : n ≥ 1, n = ∞} is a supermartingale. (iii) follows from (ii).  Proposition . Let {Xt : t ≥ 0} be a supermartingale relative to {Ft : t ≥ 0}. I = [r, s], a < b and S any countable dense subset. Let U(w, S ∩ I, [a, b]) = U(·, {Xt (w) : t ∈ S ∩ I}, [a, b]). Then E(U(·, S ∩ I, [a, b)) ≤

E((X s − b)− ) . b−a

Proof. Let S ∩ I be an increasing union of finite sets Fn : then E(U(·, Fn , [a, b])) ≤

E((Xmax Fn − b)− ) E((X s − b)− ) ≤ . b−a b−a

The result follows by Fatou’s lemma.



Exercise . If further Xt is continuous i.e. t → Xt (w) is continuous for each w, then prove that E(U(·, I, [a, b])) ≤

E((X s − b)− ) b−a

Theorem . Let (Ω, F , P) be complete and {Xt : t ≥ 0} a continuous supermartingale. 309

(i) If sup E(Xt− ) < ∞, then Xt converges a.e. to a random variable t≥0

X∞ .

313 (ii) If {Xt : t ≥ 0} is uniformly integrable then also X∞ exists and {Xt : t ≥ 0, t = ∞} is a supermartingale. (i) E(U(·, [0, n], [a, b])) ≤ E((Xn − b)− )/(b − a) so that

Proof.

limit E(U(·, [0, n], [a, b])) ≤ sup n→∞

0≤s

E((X s − b)− ) b−a

for all a < b. Thus {Xt (w) : t > 0} converges a.e. whose limit in denoted by X∞ which is measurable. (ii) As E(Xt− ) ≤ E(|Xt |) by (i) X∞ exists, the other assertion is a consequence of uniform integrability.  Corollary . Let {Xt : t ≥ 0} be a continuous uniformly integrable martingale. Then {Xt : 0 ≤ t ≤ ∞} is also a martingale. Exercise. Let {Xt : t ≥ 0} be a continuous martingale such that for some Y with 0 ≤ Y ≤ 1 E(Y|Ft ) = Xt show that Xt → Y a.e. Lemma . Let (Ω, F , P) be a probability space, F1 ⊃ F2 ⊃ F3 . . . be sub-σ-algebras. Let X1 , X2 , . . . be a real valued functions measurable with respect to F1 , . . . , Fn , . . . respectively. Let (i) E(Xn−1 |Fn ) ≤ Xn (ii) sup E(Xn ) < ∞. n

Then {Xn : n ≥ 1} is uniformly integrable.

310

Proof. By (i) E(Xn ) is increasing. By (ii) given ǫ > 0, we can find n0 such that if n ≥ n0 then E(Xn ) ≤ E(Xn0 ) + ǫ. For and δ > 0, Z n ≥ n0 |Xn |dP = E(Xn ) +

Z

(|Xn |≥δ)

(Xn ≤−δ)

−Xn dP −

Z

(Xn <δ)

Xn dP

33. Application of Stochastic Integral

314 ≤ ≤ǫ+

Z

Z

−Xn0 dP −

(Xn ≤−δ)

Xn0 dP −

(Xn ≥δ)

Z

Z

Xn0 dP + E(Xn )

by (i)

(Xn <δ)

(because E(Xn ) ≤ E(Xn0 ) + ǫ)

Xn0 dP

(Xn ≤−δ)

≤ǫ+

Z

|Xn0 |dP

(|Xn |≥δ)

Thus to show uniform integrability we have only to show P(|Xn | ≥ δ) → 0 uniformly in n as δ → ∞. Now E(|Xn |) = E(Xn + 2Xn− )

≤ E(Xn ) + 2E(|X1 |) by (i) ≤ M < ∞ for all n by (ii)

The result follows as P(|Xn | ≥ δ) ≤ M/δ.

311



Optional Sampling Theorem. (Continuous case). Let {Xt : t ≥ 0} be a right continuous supermartingale relative to {Ft : t ≥ 0}. Assume there exists an X∞ ∈ L′ (Ω, F , P) such that Xt ≥ E(X∞ |Ft ) for t ≥ 0. For any stopping time τ taking values in [0, ∞], let Xτ = X∞ on {τ = ∞}. Then (i) Xτ is integrable. (ii) If σ ≤ τ are stopping times, then E(Xτ |Fσ ) ≤ Xσ . Proof. Define [2n σ] + 1 [2n σ] + 1 , τ = . n 2n 2n Then σn , τn are stopping times, σn ≤ τn , σ ≤ σn , τ ≤ τn . σn , τn take values in Dn = {∞, 0, 1/2n , 2/2n , . . . , 1/2n , . . .} so that we have E(Xτn |Fσn ) ≤ Xσn . Thus if, A ∈ Fσ ⊂ Fσn , then Z Z (*) Xτn dP ≤ Xσn dP σn =

A

A

315 As σ1 ≥ σ2 ≥ . . ., by optional sampling theorem for the countable case, we have E(Xσn−1 |Fσn ) ≤ Xσn . Further Fσ1 ⊃ Fσ2 ⊃ . . . ;

E(Xσn |F0 ) ≤ X0 .

By the lemma {Xσn }, {Xτn } are uniformly integrable families. By right continuity Xσn → Xσ pointwise and Xτn → Xτ pointwise. Letting n → ∞ in (*) we get the required result.  Lemma (Integration by Parts). Let M(t, ·) be a continuous progres- 312 sively measurable martingale and A(t, w) : [0, ∞) × Ω → R be of bounded variation for each w. Further, assume that A(t, w) is Ft -measurable for each t. Then Y(t, ·) = M(t, ·)A(t, ·) =

Zt

M(s, ·)dA(s, ·)

0

is a martingale if E( sup |M(s, ·)| ||A(·)||t ) < ∞ 0≤s≤t

for each t, where ||A(w)||t is the total variation of A(s, w) in [0, t]. Proof. By hypothesis, n X i=0

M(s, ·)(A(si+1 , ·) − A(si , ·))

converges to Zt

M(u, ·)dA(u, ·) in L1 as n → ∞

s

and as the norm of the partition s = s0 < s1 < . . . < sn+1 = t converges to zero. Hence it is enough to show that E([M(t, ·)A(t, ·) −

n X i=0

M(si+1 , ·)(A(si+1 , ·) − A(si , ·))]|Fs )

33. Application of Stochastic Integral

316

= M(s, ·)A(s, ·). But the left side = E(M(sn+1 , ·)A(sn+1 , ·)− −

n X i=0

M(si+1 , ·)(A(si+1 , ·) − A(si , ·))|Fs )

= M(s, ·)A(s, ·). 313

Taking limits as n → ∞ and observing that sup |(si+1 − si )| → 0,

0≤i≤n

we get E(M(t, ·)A(t, ·) −

Zt

M(u, ·)dA(u, ·)|Fs )

0

= M(s, ·)A(s, ·) −

Zs

M(u, ·)dA(u, ·).

0



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´ [11] LEVY, P. Processus Stochastic et Mouvement Brownien, GaulhierVillars, Paris (1948). 315

[12] MEYER, P.A. Probability and Potentials, Blaisdell, Wallham (1966). [13] NEVEU, J. Discrete Parameter Martingales, North Holland, Amsterdam (1975). [14] PARTHASARATHY, K.R. Probability Measures on Metric Spaces, Academic Press, New York (1967). [15] PROTTER, M.H. and H.F. WEINBERGET. Maximum Principles in Differential Equations, Printice Hall, Englewood Cliffs (1967). [16] STROOCK, D.W. and S.R.S. VARADHAN. Diffusion Process in Several Dimension, Springer-Verlag, New York (to appear) [17] WILLIAMS, D. Brownian motions and diffusions as Markov Processes. Bull. Lond. Math. Soc. 6 (1974) 257 - 303.