S R S Varadhan Tata Institute of Fundamental Research Lectures on Diffusion Problems and Partial Differential Equations 1981

7/27/2019 S R S Varadhan Tata Institute of Fundamental Research Lectures on Diffusion Problems and Partial Differential Equ

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Lectures on

Diffusion Problems and Partial Differential

Equations

By

S. R. S. Varadhan

Notes by

P. MuthuramalingamTara R. Nanda

Tata Institute of Fundamental Research, Bombay1989


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Author

S. R. S. VaradhanCourant 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 with-

out 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


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iv Contents

15 Equivalent For of Ito Process 105

16 Itos Formula 117

17 Solution of Poissons 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


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Contents v

Language of Probability 284

Kolmogorovs Theorem 288

Martingales 290

Uniform Integrability 305

Up Crossings and Down Crossings 307

Bibliography 317


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vi Contents

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


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1. The Heat Equation

LET US CONSIDER the equation 1

(1) ut 1

2u = 0

which describes (in a suitable system of units) the temperature distribu-

tion of a certain homogeneous, isotropic body in the absence of any heat

sources within the body. Here

u u(x1, . . . , xd, t); ut ut

; u =

di=1

2

ux2

i

,

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

Ltt0

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

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


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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

,3

(6) u(t, x) =1

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

2

2(t + s)

and for t > s, y Rd,

(7) u(t, x) =1

[2(t

s)]d/2

exp |x y|2

2(t

s)

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) = u(t, x) where stands for the Fourier transform in thespace variables only (in this case), i.e.

v(t, x) =

Rd

u(t,y)ei xydy.

Using equation (1), one easily verifies that

(8) vt(t, x) =1

2 |x|2v(t, x)

with

(9) v(0, x) = f(x).

The solution of equation (8) is given by

(10) v(t, x) = f(x)et|x|2 /2.

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


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5

v(t, x) =

t0

w(t, x, s)ds

where

w(t, x, s) =

Rd

g(s,y)1

[2(t, s)]d/2exp

|x y|

2

2(t s)

dy.

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

smooth and has compact support. vt 12

v = Ltst

w(t, x, s) and now use

part (b) of Exercise (1).

Remark 1. We can assume g has compact support because in evaluating

vt 1

2v the contribution to the integral is mainly from a small neigh-

bourhood of the point (t, x). Outside this neighbourhood

1[2(t s)]d/2 exp

|x y|22(t s)

satisfies

ut 1

2u = 0.

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, ) =

t0

g(s, )exp 12

(t s)||2d,

orv

t=

v

t= g(t, ) +

1

2v =

g(t, ) +

1

2v

.

Thereforev

t 1

2v = g.


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6 1. The Heat Equation

Exercise 4. Solve wt 1

2w = g on [0, ) Rd with w = f on {0} Rd 6

(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 sat-isfying

ut

u2

0 in (0, T] Rd and u(0, x) 0, x Rd.

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

2> 0 in (0, T] Rd.

Claim . v cannot attain a minimum for t0 (0, T]. Assume (to get acontradiction) 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

vt v2 (t0, x0) 0.Thus, ifv has any minimum it should occur at t0 = 0.

Step 2. Let > 0 be arbitrary. Choose such that

h(t, x) = |x|2 + t

satisfies7ht h

2= d > 0 (say = 2d).


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8 1. The Heat Equation

Then

u(0, x) = 0, ut =u

2, u 0,

i.e. u satisfies

ut 1

2

2u

x2= 0, 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(2t)d/2

exp |x|2

2tfor t > 0. Then

p(t, ) p(s, ) = p(t+ s, ).

Proof. Let f be any bounded continuous function and put

u(t, x) =

Rd

f(y)p(t, x y)dy.

Then u satisfies

ut 1

2u = 0, u(0, x) = f(x).

Let

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

Then

vt 12

v = 0, v(0, x) = u(s, x).

This has the unique solution

v(t, x) =

u(s,y)p(t, x y)dy.

Thus

Rd

f(y)p(t+ s, x y)dy = f(z)p(s,y z)p(t, x y)dz dy.


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4. Construction of Wiener

Measure

ONE EXAMPLE WHERE the Kolmogorov construction yields a proba- 17

bility measure concentrated on a nice class 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

Ft1 . . . tk(A) =

A

p(0, x, t1, x1)p(t1, x1, t2, x2) . . .

p(tk1, xk1, 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 k may be thought of as representing particles performing

Brownian movement; {Xt : 0 t < } then represents the trajectoriesof these particles in the space Rd as functions of time and Bcan be con-

sidered as a representation of the observations made on these particles.

Exercise 2. (a) Show that Ft1...tk defined above is a probability mea-

sure on Rd Rd (k times).

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23

Step 3. We want to show that Px((D)) = 1. By Exercise 4(b) this is

equivalent to showing that Ltj

P(N,1/j

D(f) 1

k) = 1 for all N and k.

The lemmas which follow will give the desired result.

Lemma (Levy). Let X1, . . .Xn be independent random variables, > 0

and > 0 arbitrary. If

P(|Xr + Xr+1 + + X| ) r, such that1 r n, then

P( sup1jn |X1 + + Xj| 2) 2.

(see Kolmogorovs theorem) for every j = 1, 2, . . . , for every N = 22

1, 2, . . . and for every k = 1, 2, . . .. (Hint: Use the fact that the pro-

jections are continuous).

(b) Show that (D) =

N=1

k=1

j=1

{N,1j

D(f) 1

k} and hence (D) is

measurable in

{Rdt : t

D

}.

Let t1...tk : (D) Rd Rd Rd (k times) be the projectionsand let

Et1...tk = 1t1...tk

(B(Rd) k times

B(Rd)).

Put

E = {Et1...tk : 0 t1 < t2 < . . . < tk < ; ti D}.

Then, asEt1...tk Es1...s1 E1...m ,

where

{t1 . . . tk, s1 . . . s1} {1 . . . m},E is an algebra. Let (E) be the -algebra generated by E.

Lemma . LetB be the (topological) Borel -field of (D). Then B is

the -algebra generated by all the projections

{ti...tk : 0 t1 < t2 < . . . < tk, ti D}.


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Lemma . Let{X(t)} be a Brownian motion, I [0, ) be a finite inter-val, F I D be finite. Then

Px

Supt,F

|X(t) X()| 4 C(d) |I|24

,

where |I| is the length of the interval and C(d) a constant depending onlyon d.

Remark. Observe that the estimate is independent of the finite set F.

Proof. Let F =

{ti : 0

t1 < t2 < . . . < tk

= Px(T,hD

> )

(T, , h) = C h4

Th + 1 .

Note that(T, , h) 0 as h 0 for every fixed T and .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

Ij

[0, T] (j = 1, 2, . . . , r).

Clearly there are [ Th

] + 1 of them. If|t s| h then t, s Ij for some26j, 1 j r. Write D =

n=1

Fn where Fn Fn+1 and Fn is finite. Then

Px

sup|ts|h

t,s[0,T]D

|X(t) X(s)| >

= Px

n=1

sup|ts|h

t,sDFn

|X(t) X(s)| >

= supn

Pxsupj supt,sFn(|XIj (t) X(s)| > )


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29

everything else that affects it) through a vector x. The symmetry of the

physicl laws governing this motion tells us that any property exhibited29

by the first process should be exhibited by the second process and viceversa. Mathematically this is expressed by

Theorem . Px = PT1

x .

Proof. It is enough to show that

Px(Tx1t1...tk

(A1 Ak)) = P(1t1...tk(A1 Ak))

for every Ai Borel in Rd. Clearly,

Tx1t1...tk

(A1 Ak) = 1t1...tk(A1 x Ak x).

Thus we have only to show thatA1x

. . .

Akx

p(0, x, t1, x1) . . . p(tk1, xk1, tk, xk)dx1 . . . dxk

= A1

. . .Ak

p(0, 0, t1, x1) . . . p(tk1, xk1, tk, xk)dx1 . . . dxk,

which is obvious.

Exercise. (a) If (t, ) is a Brownian motion s tarting at (0, 0) then1

(t) is a Brownian motion starting at (0, 0) for every > 0.

(b) IfX is a d-dimensional Brownian motion and Y is a d-dimensio-nal Brownian motion then (X, Y) is a d+ d dimensional Brownianmotion provided that X and Y are independent.

(c) IfXt = (X1t , . . . ,X

dt ) is a d-dimensional Brownian motion, then X

jt

is a one-dimensional Brownian motion. (j = 1, 2, . . . d).

(w) = inf{t : |Xt(w)| +1}= inf

{t :

|w(t)

| 1}

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


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41

i.e.41

2P(B)

P

n

i=1 Ai ,or

P

max1in

Xi > a

2P{Xn > a}

Lemma 2. Let Yi, . . . , Yn be independent random variables. Put Xn =n

k=1Yk and let = min{i : Xi > a}, a > 0 and = if there is no such i.

Then for each > 0,

(a) P{ n 1,Xn X } P{ n 1,Xn a} +n1j=1

P(Yj > ).

(b) P{ n1,Xn > a+2} P{ n1,XnX > }+n1j=1

P{Yj > }

(c) P

{Xn > a + 2

} P

{

n

1,Xn > a + 2

}+ P

{Yn > 2

}.

If, further, Y1, . . . , Yn are symmetric, then

(d) P{max1in

Xi > a,Xn a} P{Xn > a + 2} P{Yn 2}

2n1j=1

P{Yj > }

(e) P{max1

i

n

Xi > a} 2P{Xn > a + 2} 2n

j=1P{Yj > }

Proof. (a) Suppose w { n 1,Xn X } and w { n 1,Xn a}. Then Xn(w) > a and Xn(w) + X(w)(w) or,X(w)(w) > a + .

By definition of(w), X(w)1(w) a and therefore,

Y(w)(w) = X(w)(w) X(w)1(w) > a + a =

if(w) > 1; if(w) = 1, Y(w)(w) = X(w)(w) > a + > .

Thus Yj(w) > for some j n 1, i.e. 42


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=

nk=1

P{ = k}P{Xn Xk > } n1j=1

P(Yj > ) (by symmetry)

= P{ n 1,Xn X } n1j=1

P(Yj > )

P{ n 1,Xn X > } n1j=1

P(Yj > )

P{ n 1,Xn > a + 2} 2n1j=1

P{Yj > } (by (b))

P{Xn > a + 2} P{Yn > 2} 2n1

j=1 P{Yj > } (by (c)) 43This proves (d).

(e) P{max1in

Xi > a} = P{max1in

Xi > a,Xn a} + P{max1in

Xi > a,Xn > a}= P{max

1inXi > a,Xn a} + P{Xn > a}

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

2 n1j=1

P{Yj > } (by (d))

Since P{Xn > a + 2} P{Xn > a} and

P{Yn > 2} P{Yn > } 2P{Yn > },

we get

P{max1inXi > a} 2P{Xn > a + 2} 2

nj=1

P(Yj > )

This completes the proof. 44

Proof of the reflection principle.

By Lemma 1

p = P

max1jn

X jt

n

> 2P(X(t) > a).


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We let n tend to through values 2, 22, 23, . . . so that we get

2P{X(t) > a + 2

} 2n P

{X(t/n) >

}P

max1jn

X(t/n) > a

2P{X(t) > a},

or

2P{X(t) > a} 2P{X(t) a} P

max0st

X(t) > a

2P

{X(t) > a

},

on letting n + first and then letting 0. Therefore,

P

max0st

X(s) > a

= 2P{X(t) > a}

= 2

a

1/

(2t)ex2 /2tdx.

AN APPLICATION. Consider a one-dimensional Brownian motion. Aparticle starts at 0. What can we say about the behaviour of the particle

in a small interval of time [0, )? The answer is given by the following

result.

P(A) P{w : > 0, t, s in [0, ) such that Xt(w) > 0 andXs(w) < 0} = 1.

INTERPRETATION. Near zero all the particles oscillate about their 46starting point. Let

A+ = {w : > 0 t [0, ) such that Xt(w) > 0},A = {w : > 0 s [0, ) such that Xs(w) < 0}.

We show that P(A+) = P(A) = 1 and therefore P(A) = P(A+ A) = 1.

A+ n=1

sup0t1/n

w > 0 = n=1

m=1

sup0t1/n

w(t) 1/m


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46 7. Reflection Principle

Therefore

P(A+

) Ltn supm P sup0t1/n w(t) 1/m 2 Lt

nmsup 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 ofR. Then

P0{Xt A,Xs < a s such that0 s t}

=

A

1/

(2t)ey2/2tdy

A

1/

(2t)e(2ay)2 /2tdy

Proof. Let (w) = inf{t : w(t) a}. By the strong Markov property ofBrownian motion,

P0{B(X( + s) X() A)} = P0(B)P0(X(s) A)for every set B in Ft. This can be written as

E(X(X(+s)X()A)|F) = P0(X(s) A)Therefore47

E(X(X(+(w))X()A)|F) = P0(X((w)) A)for every function (w) which is F-measurable. Therefore,

P0(( t) ((X( + (w)) X()) A) =

{t}

P0(X((w)) A)dP(w)

In particular, take (w) = t (w), clearly (w) is F-measurable.Therefore,

P0(( t)((X(t) X()) A)) = {t}

P0(X((w) A)dP(w)).


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47

Now X((w)) = a. Replace A by A a to get

(*) P0(( t) (X(t) A)) = {t}

P0(X((w) A a)dP(w))

Consider now

P2a(X(t) A) = P0(X(t) A 2a)

= P0(X(t) 2a A) (by symmetry ofx)= P0(( t) (X(t) 2a A)).

The last step follows from the face that A (, a) and the conti-nuity of the Brownina paths. Therefore

P2a(X(t)

A) = {t} P0(X((w)) a A)dP(w), (using )= P0(( t) (X(t) A)).

Now the required probability

P0{Xt A,Xs < a s 0 s t} = P0{Xt A} P0{( t) (Xt A)}

=A

1/(2t)ey2 /2tdy A

1/(2t)e(2ay)2 /2tdy.

The intuitive idea of the previous theorem is quite clear. To obtain 48

the paths that reach A at time twithout hitting the horizontal line x = a,

we consider all paths that reach A at time tand 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 tafter hitting x = a we consider

a typical path X(w).


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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 t is thesame 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 t, thisparticle has to hit x = a).

We shall use the intuitive approach in what follows, the mathemati-49

cal analysis being clear, thorugh lengthy.

Theorem . Let X(t) be a one-dimensional Brownian motion, A

(

1, 1)

any Borel subset ofR. Then

P0

sup

0st|X(s)| < 1,X(t) A

=

A

(t,y)dy,

where

(t,y) =

n=(1)n/

(2t)e(y2n)

2 /2t.


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=

A

1/

(2t)ey2 /2tdy P[(E1 F1) A0],

where

A0 = {X(t) A} =A

1/

(2t)ey2 /2tdy P[(E1 A0) (F1 A0)].

Use the fact that P[A B] = P(A) + P(B) P(A B) to get

(t,A) =A

1/

(2t)ey2 /2tdyP[E1A0]P[F1A0]+P[E1F1A0],

as E1 F1 = E2 F2. Proceeding successively we finally get

(t,A) = A 1/

(2t)ey2 /2tdy+

n=1(1)n P[En

A0]+

n=1(1)nP[Fn

A0]

We shall obtain the expression for P(E1 A0) and P[E2 A0], theother terms can be obtained similarly.

E1 A0 consists of those trajectries that hit x = 1 at some time t and then reach A at time t. Thus P[E1 A0] is given by theprevious theorem by

A

1/

(2t)e(y2)2 /2tdy.

E2 A0 consists of those trajectories that hit x = 1 at time 1, hit51x = 1 at time 2 and reach A at time t(1 < 2 < t).

According to the previous theorem we can reflect the trajectory upto

2 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 1 and ends up in A at time t.We can now reflect the trajectory


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51

upto time 1 (the dotted curve should be reflected) about x = 3 toobtain the required probability as if the trajectory started at x = 4.Thus,

P(E2 A0) = A

e(y+

4)2

/2t/(2t)dy.

Thus

(t,A) =

n=

(1)nA

1/

2te(y2n)2 /2tdy

= A (t,y)dy.The previous theorem leads to an interesting result: 52

P

sup

0st|X(s)| < 1

=

11

(t,y)dy

Therefore

P

sup0st

|X(s)| 1 = 1 P sup0st

|X(s)| < 1


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= 1 1

1

(t,y)dy,

(t,y) =

n=

(1)n/(2t)e(y2n)2 /2t

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,

(t,y) 1/(2t) ey2/2t e(y2)2/2t e(y+2)2 /2t .Therefore

11

(t,y)dy 4/(2t)e1/2t.

Case (ii). t is large. In this case we use Poissons summation formula

for (t,y):

(t,y) =

k=0

e(2k+1)2 2t/8Cos{(k+ 1)/2y},

to get1

1 (t,y)dy 4/e2t/8

for large t. Thus, P( > t) = 4/e2t/8.53

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 isgoverned by the factor e

2t/8. This is connected with the solution of a

certain differential equation as shall be seen later on.


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54 8. Blumenthals Zero-One Law

is clearly dense in L2(,B, P).

Proof of zero-one law. Let55

Ht = L2(,Ft, P),H = L

2(,B, P),H0+ =t>0

Ht.

Clearly H0+ = L2(,F0+, P).

Let t : H Ht be the projection. Then tf 0+ff in H.To prove the law it is enough to show that H0+ contains only constants,

which is equivalent to 0+f = constant f in H. As 0+ is continuousand linear it is enough to show that 0+ = const of the Lemma 2:

0+ = Ltt0

t = Ltt0

E(|t) by Lemma 1,= Lt

t0E((t1, . . . , tk)|Ft).

We can assume without loss of generality that t < t1

< t2

< . . . < tk

.

E((t1, . . . , tk)|Ft) =

(y1, . . . ,yk)1/

(2(t1 t))e|y1Xt(w)|2/2(t1t) . . .

. . . 1/

(2(tk tk1))e|ykyk1 |2

2(tktk1 ) dy1 . . . dyk.

Since X0(w) = 0 we get, as t 0,

0+ = constant.

This completes the proof.

APPLICATION. Let 1A = {w :1

0

|w(t)|/t < }. Then A F0+.

For, if 0 < s < 1, then1

s|w(t)|/t < . Therefore w A or not according

ass

0

|w(t)|/tdt converges or not. But this convergence can be asserted56


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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(limXt = ) = 1; consequently P(lim Xt < ) = 0.

(b) P(limXt

=

) = 1; consequently P(lim X

t>

) = 0.

(c) P(limXt = ); lim Xt = ) = 1.

SIGNIFICANCE. By (c) almost every Brownian path assumes each

value infinitely often.

Proof.

{limXt = } =

n=1

(lim Xt > n)

=

n=1

( limrational

X > n) (by continuity of Brownian paths)

First, note that

P0

sup0st

X(s) n = 1 P0 sup0st

X(s) > n

57


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59

Remark . If d 3 we shall see later that P( Ltt

|Xt| = ) = 1. i.e.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) =

ni=1

|w(ti) w(ti1)| (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) =

ni=1

|(w(ti) w(ti1)|2.

If

An(w, Qn) sup1in |w(ti) w(ti1)|,

then

An(w, Qn)Vn(w, Qn) Un(w, Qn).By continuity Lt

nAn(w, Qn) = 0.

Claim. Ltn

E[(Un(w, Qn) (b a))2] = 0.

Proof.

E[(Un (b a))2]

= E

nj=1

[(Xtj Xtj1 )2 (b a/n)]

2

E[((Z2j b a/n))2], Zj = Xtj Xtj1 ,= nE[(Z21 b a/n)2]


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64 10. Dirichlet Problem and Brownian Motion

Proof. The distributions {Ft1,...,tk} defining Brownian motion are invari-62ant under rotations. Thus S(0, ) is a rotationally invariant probabilitymeasure. The result follows from the fact that the only probability mea-sure (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) = Ex(f(XG )). 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 Px(A) is measurable for every Borel set A.

Let C = {A B : x Px(A) is measurable}

It is clear that 1

t1 ,...,tk(B) C

, Borel set B in Rd

Rd

. AsC is a monotone class C = B.

(ii) Let S be any sphere with centre at x, and S G. Let = Sdenote the exit time through S. Clearly G. By the strongMarkov property,

u(X) = E(f(XG )|F).

Now

u(x) = Ex(f(XG )) = Ex(E(f(XG ))|F)

= Ex(u(X)) =

S

u(y)S(x, dy)

=1

|S| Su(y)dS;

|S

|= surface area ofS.

63


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As Ans are increasing, Bns are decreasing and Bn F1/n; so thatB F0+. We show that P(B) > 0, so that by Bluementhals zero-onelaw, P(B) = 1, i.e. P(A) = 0.

Py(B) = limn Py(Bn) limn Py{w : w(0) = y, w(

1

2n) Ch {y}}

Thus

Py(b) lim

Ch{y}

1/

(2/2n)d exp(|z y|2/2/2n)dz

=

C

1/

(2)e|y|2 /2dy,

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 Px(C) is upper semi-continuous.

For, denote by XC the indicator function of C. As C is closed (in

a metric space) a sequence of continuous functions fn decreasing toXC such that 0 fn 1. Thus Ex(fn) decreases to Ex(XC) = Px(C).Clearly x Ex(Fn) is continuous. The result follows from the fact thatthe infimum of any collection of continuous functions is upper semi-

continuous.

Step 3. Let > 0,65

N(y; ) = {z G : |z y| < },B = {w : w(0) G,XG (w) G N(y; )},

i.e. B consists of trajectories which start at a point ofG and escape for

the first time through G at a point not in N(y; ). IfC = B, then

C {

w : w(0) = y}

A {

w : w(0) = y}

where A is as in Step 1.


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67

For, suppose w C{w : w(0) = y}. Then there exists wn B suchthat wn w uniformaly on compact sets. Ifw A{w : w(0) = y} thereexists > 0 such that w(t) G t in (0, ]. Let = inf0t d(w(t), G N(y, )). Then > 0. If tn = G(wn) and tn does not converge to 0,then there exists a subsequence, again denoted by tn, such that tn k >0 for some k (0, 1). Since wn(k) G and wn(k), w(k) G, acontradiction. Thus we can assume that tn converges to 0 and also that

tnn, But then(*) |wn(tn) w(tn)| .

However, as wn converges to w uniformly on [0, ],

wn(tn) w(tn) w(0) w(0) = 0contradicting (*). Thus w A{w : w(0) = y}.

Step 4. limxy,xG

Px(B) = 0.

For, 66

limxy

Px(B) limxy

Px(C) Py(C) (by Step 2)

= Py(C {w : w(0) = y}) Py(A) (by Step 3)= 0.

Step 5.

|u(x) f(y)| = |

f(XG (w))dPx(w)

f(y)dPx(w)|

B

|f(XG (w)) f(y)|dPx (w) + |

B

(f(XG (w)) f(y))dPx(w)|

B

|f(XG (w)) f(y)|dPx (w) + 2||f||Px(B)

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


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Remark. The theorem is local.

Theorem . Let G = {y Rd : < |y| < R}, f any continuous functionon G = {|y| = } {|y| = R}. If u is any harmonic function in G withboundary values f , then u(x) = Ex(f(XG )).

Proof. Clearly G has the exterior cone property. Thus, if

v(x) = Ex(f(XG )),

then v is harmonic in G and has boundary values f (by the previoustheorem). 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| = is of spe-cial interest. Denote by R,0

,1the corresponding solution of the Dirichlet

problem.

Exercise. (i) Ifd = 2 then67

UR,0,1

(x) =logR log |x|logR log , x G.

(ii) Ifd 3 thenU

R,0,1

(x) =|x|n+2 Rn+2n+2 Rn+2 .

Case (i): d = 2. Then

logR log |x|logR log = U

R,0,1

(x).

Now,

Ex(f(XG )) =

|y|=

G(x, dy) = Px(|XG | = ),

i.e.

logR log |x|logR log = Px(|XG | = )


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69

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

Fix R and let

0; then 0 = Px

(the particle hits 0 before hitting

|y| = R).Let R take values 1, 2, 3, . . ., then 0 = Px (the particle hits 0 before

hitting any of the circles |y| = N). Recalling that

Px(lim |Xt| = ) = 1,

we get

Proposition . A two-dimensional Brownian motion does not visit apoint.

Next, keep fixed and let R , then,

1 = Px(|w(t)| = 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 recurrenceproperty.

Case (ii): d 3. In this case

Px(w : w hits |y| = before it hits |y| = R)= (1/|x|n2 1/Rn2)/(1/n2 1/Rn2).

Letting R

we get

Px(w : w hits |y| = ) = (/|x|)n2

which lies strictly between 0 and 1. Fixing and letting |x| , wehave

Proposition . If the particle start at a point for away from 0 then it has

very little chance of hitting the circle |y| = .If

|x|

, then

P(w hits S) = 1 where S = {y Rd : |y| = }.


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Let

V(x) = (/|x|)n2 for |x| .In view of the above result it is natural to extend V to all space

by putting V(x) = 1 for |x| . As Brownian motion has the Markovproperty

Px{w : w hits S after time t}

=

V(y)1/

(2t)d exp |y|2/2t dy 0 as t +.

Thus P(w hits S for arbitrarily large t) = 0. In other words, P(w :69limt

|w(t)| ) = 1. As this is true > 0, we get the followingimportant result.

Proposition . P(limt

|w(t)| = ) = 1,i.e. for d 3, the Brownian particle wander away to +.


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11. Stochastic Integration

LET {Xt : t 0} BE A one-dimensional Brownian motion. We want 70first to define integrals of the type

0

f(s)dX(s) for real functions f

L1[0, ). IfX(s, w) is of bounded variation almost everywhere then wecan give a meaning to

0

f(s)dX(s, w) = g(w). However, since X(s, w)

is not bounded variation almost everywhere, g(w) is not defined in the

usual sense.

In order to define g(w) = 0

f(s)dX(s, w) proceed as follows.

Let f be a step function of the following type:

f =

ni=1

aiX[ti,ti+1), 0 t1 < t2 < . . . < tn+1.

We naturally define

g(w) =

0

f(s)dX(s, w) =

ni=1

ai(Xti+1 (w) Xti (w))

=

ni=1

ai(w(ti+1) w(ti)).

g satisfies the following properties:

(i) g is a random variable;

71


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72 11. Stochastic Integration

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

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 inde-pendent increments, i.e. we have

E

0

f dX

= 0, E|

0

f dX|2 = ||f||22.

71

Exercise 1. If

f =

ni=1

aiX[ti,ti+1), 0 t1 < . . . < tn+1,

g =

mi=1

biX[si,si+1), 0 s1 < . . . < sm+1,

Show that

0

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

0

f dX(s, w) +

0

gdX(s, w)

and

0(f)dX(s, w) =

0f dX(s, w),

R.

Remark. The mapping f

0

f dX is therefore a linear L2R

-isometry of

the space S of all simple functions of the type

ni=1

aiX[ti,ti+1), (0 t1 < . . . < tn+1)

into L2(,B, P).


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73

Exercise 2. Show that S is a dense subspace ofL2[0, ).Hint: Cc[0, ), i.e. the set of all continuous functions with compact sup-port, is dense in L

2

[0, ). Show that S contains the closure ofCc[0, ).

Remark. The mapping f

0

f dX can now be uniquely extended as

an isometry of L2[0, ) into L2(,B, P).Next we define integrals fo the type 72

g(w) =

t

0 X(s, w)dX(s, w)Put t = 1 (the general case can be dealt with similarly). It seems

natural to define

(*)

10

X(s, w)dX(s) = Ltsup |tjtj1|0

nj=1

X(j)(X(tj) X(tj1))

where 0 = t0 < t1 < . . . < tn = 1 is a partion of [0, 1] with tj1 j tj.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 j, we may obtain

different limits. To consider an example we choose j = tj and then

j = tj1 and compute the right hand side of (). Ifj = tj1,n

j=1

Xj (Xtj Xtj1 ) =n

j=1

Xtj1 (Xtj Xtj1 )

=1

2

nj=1

(Xtj ) (Xtj1 ) 1

2

nj=1

(Xtj Xtj1 )

1

2[X2(1) X2(0)] 1

2as n , and sup |tj tj1| 0,

arguing as in the proof of the result that Brownian motion is not of

bounded variation. Ifj = tj,

Ltn

Sup |tjtj1 |0

nj=1

Xtj (Xtj Xtj1 ) = 1/2X(1) 1/2X(0) + 1/2.


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Thus we get different answers depending on the choice of j and73

hence one has to be very careful in defining the integral. It turns out

that the choice of j = tj1 is more appropriate in the definition of theintegral and gives better results.

Remark. The limit in () should be understood in the sense of conver-gence probability.

Exercise 3. Let 0 a < b. Show that the left integral (j = tj1) isgiven by

L

b

a

X(s)dX(s) = X2

(b) X2

(a) (b a)2

and the right integral (j = tj) is given by

R

bs

X(s)dX(s) =X2(b) X2(a) + (b a)

2.

We now take up the general theory of stochastic integration. To

motivate the definitions which follow let us consider a d-dimensional

Brownian motion {(t) : t 0}. We have

E[(t+ s) (t) A|Ft] =A

1/

(2s)e|y|2 /2sdy.

Thus

E(f((t + s) (t))|Ft] =

f(y)1/

(2s)e|y|2/2s

dy.

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

E[eiu((t+ s) (t))|Ft] =

eiu.y1/

(2s)e|y|2 /2sdy

= es|u|2

2 .

Thus74


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75

E[eiu.(t+s) |Ft] = eiu.(t)es|u|2 /2,

or,

E[eiu.(t+s)+(t+s)|u|2 /2|Ft] = eiu.(t)+t|u|2 /2.Replacing iu by we get

E[e.(s)|s|2 /2 | Ft] = e.(t)t||

2/2

, s > t, .

It is clear that e.(t)t||2 /2 is Ft-measurable and a simple calculation

gives

E(e.(t)||2

t/2|) < .We thus have

Theorem . If {(t) : t 0} is a d-dimensional Brownian motion thenexp[.(t) ||2t/2] is a Martingale relative to Ft, the -algebra gener-ated by ((s) : s t).

Definition. Let (,B, P) be a probability space (Ft)t

0 and increasing

family of sub--algebras ofF with F = (t0Ft).

Let

(i) a : [0, ) [0, ) be bounded and progressively measurable;

(ii) b : [0, ) R be bounded and progressively measurable;

(iii) X : [0,

)

R be progressively measurable, right continuous

on [0, ), w , and continous on [0, ) almost everywhereon ;

(iv) Zt(w) = eX(t,w)

t0

b(s,w)ds 22

t0

a(s,w)ds

75

be a Martingale relative to (Ft)t0.

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.


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Example. If {(t) : t 0} is a Brownian motion, then X(t, w) = t(w)is an Ito process corresponding to parameters 0 and 1. (i) and (ii) are

obvious. (iii) follows by right continuity of t and measurability of trelative to Ft and (iv) is proved in the previous theorem.

Exercise 4. Show that Zt(w) defined in (iv) is Ft-measurable and pro-

gressively measurable.

[Hint:

(i) Zt is right continuous.

(ii) Use Fubinis theorem to prove measurability].

Remark. If we put Y(t, w) = X(t, w) t

0

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

gressively measurable and Y(t, w) is an Ito process corresponding to

the parameters 0, a. Thus we need only consider integrals of the typet

0 f(s, w)dY(s, w) and definet

0

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

t0

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

t0

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

(Note that formally we have dY = dX dbt).Lemma . If Y(t, w)

I[0, a], then76

Y(t, w) and Y 2(t, w) t

0

a(s, w)ds

are Martingales relative to (Ft).

Proof. To motivate the arguments which follow, we first give a formal

proof. Let

Y(t) = eY(t,w) 2

2

t0

a(s,w)ds

.


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77

Then Y(t) is a martingale, . ThereforeY 1

is a Martingale, .

Hence (formally),

lim0

Y 1

= Y|=0is a Martingale.

Step 1. Y(t, ) Lk(,F, P), k = 0, 1, 2, . . . and t. In fact, for everyreal , Y(t) is a Martingale and hence E(Y) < . Since a is boundedthis means that

E(eY(t,)) 0.

Step 2. Let X(t) = [Y(t, ) t

0

ads]Y(t) =d

dY(t, ). 77

Define

A() =

A

(X(t, ) X(s, ))dP(w)

where t > s, A Fs. Then2

1

A()d =

21

A

[X(t, ) X(S, )]dP(w)d.

Since a is bounded, sup||

E([Y(t, )]k) < , and E(|Y|k) < , k; wecan use Fubinis theorem to get

21

A()d = A

21

[X(t, ) X(s, )]ddP(w).


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or

21

A()d =A

Y2 (t, ) Y1 (t, )dP(w) A

Y1 (s, ) Y1 (s, )dP(w).

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

21

A()d = 0.

This is true 1 < 2 and since A() is a continuous function of ,we conclude that

A() = 0, .

In particular, A() = 0 which means that

A

Y(t, )dP(w) = A

Y(s, )dP(w), A Fs, t > s,

i.e., Y(t) is a Martingale relative to (,Ft, P).78

To prove the second part we put

Z(t,

) =

d2

d2

Y(t)

and

A() =

A

{Z(t, ) Z(s, )}dP(w).

Then, by Fubini,

21

A()d = A

21

Z(t, ) Z(s, )d dP(w).


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79

or,

21

A()d = A(2) A(1)

= 0 if A Fs, t > s.

Therefore

A() = 0, .In particular, A() = 0 implies that

Y2(t, w) t

0

a(s, w)ds

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

Definition. A function : [0, ) R is called simple if there existreals s0, s1, . . . , sn, . . .

0 s0 < s1 < . . . < sn . . . < ,

sn increasing to + and

(s, w) = j(w)

if s [sj, sj+1), where j(w) is Fsj -measurable and bounded. 79Definition. Let : [0,

)

R be a simple function and Y(t, w)

I[0, a]. We define the stochastic integral ofwith respect to Y, denotedt

0

(s, w)dY(s, w)),

by

(t, w) =

t0

(s, w)dY(s, w)


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=

kj=1

j1(w)[Y(sj, w) Y(sj1, w)] + k(w)[Y(t, w) Y(sk, w)].

Lemma 2. Let : [0, ) R be a simple function and Y(t, w) I[0, a]. Then

(t, w) =

t0

(s, w)dY(s, w) I[0, a2].

Proof. (i) By definition, is right continuous and (t, w) is Ft-

measurable; hence it is progressively measurable. Since a is pro-

gressively measurable and bounded

a2

: [0, ) [0, )is progressively measurable and bounded.

(ii) From the definition of it is clear that (t, ) is right continuous,80continous almost everywhere and Ft-measurable therefore is

progressively measurable.

(iii) Zt(w) = e[(t,w)

2

2

t

0 a2 ds]is clearly Ft-measurable . We show that

E(Zt) < , t and E(Zt2 |Ft1 ) = Zt1 ift1 < t2.

We can assume without loss of generality that = 1 (if 1 we

replace by ). Therefore

Zt(w) = e[(t,w) t

0

a2ds]

.


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Therefore

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

2

t2t1

a2ds)|Ft1 )

as Y I[0, a].

(*) E(exp[(Y(t2, w) T(t1, w)) 2

2

t2t1

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

and since k(w) is Ft1 -measurable () remains valid if is replaced byk. 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 (t, w) I[0, a2]. Corollary . (i) (t, w) is a martingale; E((t, w)) = 0;82

(ii) 2(t, w) t

0

a2ds

is a Martingale with

E(2(t, w)) = E(

t0

a2(s, w)ds.

Proof. Follows from Lemma 1.

Lemma 3. Let (s, w) be progressively measurable such that for each

t,

E(

t0

2(s, w)ds) < .


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83

Then there exists a sequence n(s, w) of simple functions such that

limnE

t

0

|n(s, w) (s, w)|2ds = 0.Proof. We may assume that is bounded, for if N = for || Nand 0 if || > N, then n , (s, w) [0, t] . N is progressivelymeasurable and |n |2 4||2. By hypothesis L([0, t] : ).

Therefore E(t

0 |n|ds) 0, by dominated convergence. Further,we can also assume that is continuous. For, if is bounded, define

h(t, w) = 1/h

t(th)v0

(s, w)ds.

n is continuous in t and Ft-measurable and hence progressively mea-

surable. Also by Lebesgues theorem

h(t, w) (t, w), as h 0, t, w.

Since is bounded by C, h is also bounded by C. Thus 83

E(

t0

|h(s, w) (s, w)|2ds) 0.

(by dominated convergence). If is continuous, bounded and progres-

sively measurable, then

n(s, w) =

[ns]

n, w

is progressively measurable, bounded and simple. But

Ltn

n(s, w) = (s, w).


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Thus by dominated convergence

E

t0

|n |2ds 0 as n .

Theorem . Let(s, w) be progressively measurable, such that

E(

t0

2(s, w)ds) <

for each t > 0. Let(n) be simple approximations to as in Lemma 3.

Put

n(t, w) =

t0

n(s, w)dY(s, w)

where Y I[0, a]. Then

(i) Ltn

n(t, w) exists uniformly in probability, i.e. there exists an al-

most surely continuous (t, w) such that

Ltn

P

sup

0

t

T

|n(t, w) (t, w)|

= 0

for each > 0 and for each T. Moreover, is independent of the

sequence (0).

(ii) The map is linear.84

(iii) (t, w) and2(t, w) t

0a2ds are Martingales.

(iv) If is bounded, I[0, a2].


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85

Proof. (i) It is easily seen that for simple functions the stochastic

integral is linear. Therefore

(n m)(t, w) =t

0

(n m)(s, w)dY(s, w).

Since n m is an almost surely continuous martingale

P

sup

0tT |n(t, w)

m(t, w)

|

1

2E[(n

m)

2(T, w)].

This is a consequence of Kolmogorov inequality (See Appendix).

Since

(n m)2 t

0

a(n m)2ds

is a Martingale, and a is bounded,

E[(n m)2(T, w)] = E

T0

(n m)2a ds .(*)

const 12

E

T

0

(n m)2ds .

ThereforeLt

n,mE[(n m)2(T, w)] = 0.

Thus (n m) is uniformly Cauchy in probability. Therefore thereexists a progressively measurable such that

Ltn

P

sup

0

t

T

|n(t, w) (t, w)|

= 0, > 0, T.

It can be shown that is almost surely continuous.


92/323


If (n) and (n) are two sequences of simple functions approxi- 85

mating , then () shows that

E[(n n)2(T, w)] 0.

Thus

Ltn

n = Ltn

n,

i.e. is independent of (n).

(ii) is obvious.

(iii) (*) shows that n in L and therefore n(t, ) (t, ) in L1 foreach fixed t. Since n(t, w) is a martingale for each n, (t, w) is a

martingale.

(iv) 2n(t, w) t

0

a2n is a martingale for each n.

Since n(t, w) (t, w) in L2

for each fixed t and

2n(t, w) 2(t, w) in L1 for each fixed t.

For 2n(t, w) 2(t, w) = (n )(n + ) and using Holders in-equality, we get the result.

Similarly, since

n in L2([0, t] ),2n 2 in L1([0, t] ),

and because a is bounded a2n a2 in L1([0, t]). This showsthat 2n(t, w)

t0

a2nds converges to

2(t, w) t

0

a2ds


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87

for each t in L1. Therefore

2(t, w) t

0

a2ds

is a martingale. 86

(v) Let be bounded. To show that I[0, 2] it is enough to showthat

e(t,w) 2

2

t0

a2ds

is a martingale for each , the other conditions being trivially sat-

isfied. Let

Zn(t, w) = en (t,w)

2

2

t0

a2nds

We can assume that |n| C if || C (see the proof of Lemma3).

Zn = exp2n(t, w) (2)

2

2

t0

a2nds + 2

t0

a2nds .

Thus

(**) E(Zn) const E

e2n(t,w) (2)

2

2

t0

a2nds

= const

since Zn is a martingale for each . A subsequence Zni converges

to

e(t,w) 2

2

t0

a2ds

almost everywhere (P). This together with (**) ensures uniform

integrability of (Zn) and therefore

e(t,w) 2

2

t0

a2ds


94/323


is a martingale. Thus is an Ito process, I[0, a2].

Definition. With the hypothesis as in the above theorem we define the

stochastic integral

(t, w) =

t0

(s, w)dY(s, w).

87

Exercise. Show that d(X + Y) = dX + dY.

Remark. If is bounded, then satisfies the hypothesis of the previ-

ous theorem and so one can define the integral of with respect to Y.

Further, since itself is Ito, we can also define stochastic integrals with

respect to.

Examples. 1. Let {(t) : t 0} be a Brownian motion; then (t, w) isprogressively measurable (being continuous and Ft-measurable).

Also,

t0

2(s)ds dP =

t0

2(s)dP ds =

t0

sds =t

2

Hencet

0

(s, w)d(s, w)

is well defined.

2. Similarlyt

0

(s/2)d(s) is well defined.

3. Howevert

0

(2s)d(s)

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

measurable.


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89

Exercise 5. Show that (2s) is not progressively measurable. 88

(Hint: Try to show that (2s) is not Fs-measurable for every s. To show

this prove that Fs F2s).

Exercise 6. Show that for a Brownian motion (t), the stochastic integral

10

(s, )d(s, )

is the same as the left integral

L

10

(s, )d(s, )

defined earlier.


96/323


97/323

12. Change of Variable

Formula

WE SHALL PROVE the 89

Theorem . Let be any bounded progressively measurable function

and Y be an Ito process. If is any progressively measurable function

such that

Et

0

2ds < , t,then

(*)

t0

d(s, w) =

t0

(s, w)(s, w)dY(s, w),

where

(t, w) =

t0

(s, w)dY(s, w).

Proof.

Step 1. Let and be both simple, with bounded. By a refinement

of the partition, if necessary, we may assume that there exist reals 0 =

s0, s1, . . . , sn, . . . increasing to +

such that and are constant on

[sj, sj+1), say = j(w), = j(w), where j(w) and j(w) are Fsj -

measurable. In this case (*) is a direct consequence of the definition.

91


98/323

92 12. Change of Variable Formula

Step 2. Let be simple and bounded. Let (n) be a sequence of simple

bounded functions as in Lemma 3. Put

n(t, w) =

t0

n(s, w)dY(s, w)

By Step 1,

(**)

t0

dn =

t0

ndY(s, w).

90

Since is bounded, n converges to in L2([0, t] ). Hence,

by definition,t

0

ndY(s, w) converges tot

0

dY in probability.

Further,

t0

dn(s, w) = (s0, w)[n(s1, w) n(s0, w)] +

+ + (sk, w)[n(t, w) n(sk1, w)],where s0 < s1 < . . . . . . is a partition for , and n(t, w) converges to

(t, w) in probability for every t. Therefore

t0

dn(s, w)

converges in probability to

t0

d(s, w).

Taking limit as n in (**) we gett

0

d(s, w) =

t

0

dY(s, w).


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93

Step 3. Let be any progressively measurable function with

E(

t0

2ds) < , t.

Let n be a simple approximation to as in Lemma 3. Then, by Step

2,

(***)

t0

n(s, w)d(s, w) =

t0

n(s, w)(s, w)dY(s, w).

By definition, the left side above converges to

t0

(s, w)d(s, w)

in probability. As is bounded n converges to in L2([0, t] ). 91

Therefore

P

sup0tT |t

0

ndY(s, w) t

0

dy(s, w)|

||a||1/2E

t0

(n )2ds

(see proof of the main theorem leading to the definition of the stochasticintegral). Thus

t0

ndY(s, w)

converges tot

0 dY(s, w)

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


100/323

94 12. Change of Variable Formula


101/323

13. Extension to

Vector-Valued Ito Processes

Definition. Let (,F, P) be a probability space and (Ft) an increasing 92

family of sub -algebras ofF. Suppose further that

(i) a : [0, ) Sd+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, ) Rd

is a progressively measurable, bounded function;

(iii) X : [0, ) Rd

is progressively measurable, right continuous for every w and continu-

ous almost everywhere (P);

Z(t, ) = exp[,X(t, ) t

0

, b(s, )ds

12

t0

, a(s, )ds](iv)

is a martingale for each Rd

, where

x,y = x1y1 + + xdyd, x, y Rd.

95


102/323

96 13. Extension to Vector-Valued Ito Processes

Then X is called an Ito 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 progres-

sively 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

Xi I[bi, aii],(i)

Y =

di=1

iXi I[, b, , a],(ii)

where

= (1, . . . , d).

(Hint: (ii) (i). To prove (ii) appeal to the definition).

Remark. If X has a multivariate normal distribution with mean and

covariance (i j), then Y = ,X has also a normal distribution withmean , and variance ,. Note the analogy with the above exer-cise. This analogy explains why at times b is referred to as the mean

and a as the covariance.

Exercise 2. If {(t) : t 0} is a d-dimensional Brownian motion, then(t, w) I[0,I] where I = d d identity matrix.

As before one can show that Y(t, ) = X(t, ) t

0

b(s, w)ds is an Ito

process with parameters 0 and a.

Definition . Let X be a d-dimensional Ito process. = (1, . . . , d) a

d-dimensional progressively measurable function such that

Et

0

(s, ), (s, ) > ds


103/323

97

is finite or, equivalently,

Et

0

2i (s, )ds < , (i = 1, 2, . . . d).94

Then by definition

t0

(s, ), dX(s, ) =d

i=1

t0

i(s, )dXi(s, ).

Proposition . Let X be a d-dimensional Ito process X I[b, a] and let be progressively measurable and bounded. If

i(t, ) =t

0

idXi(s, ),

then = (1, . . . , d) I[B,A],

where

B = (1b1, . . . , dbd) and Ai j = ijai j.

Proof. (i) Clearly Ai j is progressively measurable and bounded.

Since a Sd+, A Sd+.

(ii) Again B is progressively measurable and bounded.

(iii) Since is bounded, each i(t, ) is an Ito process; hence is pro-gressively measurable, right continuous, continuous almost ev-

erywhere (P). It only remains to verify the martingale condition.

Step 1. Let = (1, . . . , d) Rd. By hypothesis,

E(exp[(1X1 + + dXd)|ts t

s

(1b1 + + dbd)du(*)


104/323


12

t0

ijai jds]|Fs) = 1.

Assume that each i is constant on [s, t], i = i(w) and Fs-95

measurable. Then (*) remains true ifi are replaced by ii(w) and since

is are constant over [s, t], we get

E(exp[

t0

d

i=1ii(s, )dXi(s, )

t0

ibii(s, )ds

12

t0

iji(s, )j(s, )ai jds]|s)

exp

s

0

di=1

ii(s, )dXi(s, ) s

0

,Bdu 1s

0

,Adu .

Step 2. Let each i be a simple function.

By considering finer partitions we may assume that each i is a step

function,

finest partition

i.e. there exist points s0, s1, s2, . . . , sn, s = s0 < s1 < . . . < sn+1 = t,

such that on [sj, sj+1) each i is a constant and sj -measurable. Then (**)

holds if we use the fact that ifC1

C2.

E(E(f|C1)|C2) = E(f|C2).


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99

Step 3. Let be bounded, || C. Let ((n)) be a sequence of sim-ple functions approximating as in Lemma 3. (**) is true ifi is re-96

placed by (n)

i for each n. A simple verification shows that the expres-sion Zn(t, ), in the parenthes is on the left side of (**) with i replacedby

(n)

i, converges to

Z(t, ) =

= Exp t

0

i

ii(s, )ds t

0

i

ibii(s, )ds

12

t0

i,j

ijijai jds

as n in probability. Since Zn(t, ) is a martingale and the functionsi, j, a are all bounded,

supn

E(Zn(t, )) < .

This proves that Z(t, ) is a martingale. Corollary . With the hypothesis as in the above proposition define

Z(t) =

t0

(s, ), dX(s, ).

Then

Z(t, ) I[, b, a]where is the transpose of .

Proof. Z(t, ) = 1(t, ) + +d(t, ). Definition. Let (s, w) = (i j(s, w)) be a n d matrix of progressivelymeasurable functions with

Et

0

2i j(s, )ds < .


106/323


If X is a d-dimensional Ito process, we define 97

t

0

(s, )dX(s, )i

=

dj=1

t0

i j(s, )dXj(s, ).

Exercise 3. Let

Z(t, w) =

t0

(s, )dY(s, ),

where Y I[0, a] is a d-dimensional Ito process and is as in the abovedefinition. Show thatZ(t, ) I[0, a]

is an n-dimensional Ito process, (assume that is bounded).

Exercise 4. Verify that

E(|Z(t)|2

)=

Et

0

tr(a)ds .Exercise 5. Do exercise 3 with the assumption that a 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 di-

mensional case and d(X + Y) = dX + dY).


107/323

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

(1, . . . , N) and covariance A if the density function is1

(2)N/21

(det A)1/2exp

1

2[(X)A1(X)]

where A is an N N positive definite symmetric matrix.

Note. 1. E(Xi) = i.

2. Cov(Xi,Xj) = (A)i j.

Theorem . X (X1, . . . ,XN) is a multivariate normal distribution if andonly if for every RN, ,X is a one-dimensional Gaussian randomvariable.

We omit the proof.

Definition. A stochastic process {Xt : t I} is called a Gaussian processift1, t2, . . . , tN I, (Xt1 , . . . ,XtN) is an N-variate normal distribution.

101


108/323

102 14. Brownian Motion as a Gaussian Process

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 : [0, 1] = [0, 1] R be defined by

(s, t) = s

t.

Define K : L2R

[0, 1] L2R

[0, 1] by

K f(s) =

10

(s, t)f(t)dt.

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 be any eigenvalue ofKand f an eigenvector belonging

to . Show that

(a) f + f = 0 with f(0) = 0 = f(1).

(b) Using (a) deduce that the eigenvalues are given by n = 4/(2n +1)22 and the corresponding eigenvectors are given by

fn =

2Sin1/2[(2n + 1)t]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) =

n=0Zn(w)fn(t)nconverges in mean for every real t.


109/323

103

Proof. Let Ym(t, w) =m

i=0Zi(w)fi(t)

i. Therefore100

E{(Yn+m(t, ) Yn(t, ))2} =n+mn+1

f2i (t)i,

E(||Yn+m() Yn()||2 n+mn+1

i 0.

Remark. As each Yn(t, ) is a normal random variable with mean 0 andvariance

ni=0

i f2

i(t), Y(t, ) is also a normal random variable with mean

zero and variance

i=0i f

2i

. To see this one need only observe that the

limit of a sequence of normal random variables is a normal random vari-

able.

Theorem (Mercer).

(s, t) =

i=0

i fi(t)fi(s), (s, t) [0, 1] [0, 1].

The convergence is uniform.

We omit the proof.

Exercise 3. Using Mercers theorem show that {Xt : 0 t 1} is aBrownian motion, where

X(t, w) =

n=0

Zn(w)fn(t)

n.

This exercise now implies that

10

X2(s, w)ds = (L2 norm ofX)2


110/323

104 14. Brownian Motion as a Gaussian Process

=

nZ

2n (w),

since fn(t) are orthonormal. Therefore 101

E(e

10

X2(s,)ds) = E(e

n=0nZ

2n (w)

) =

n=0

E(enZ2n )

(by independence ofZn)

=

n=0

E(enZ20 )

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

E(e

10

X2(s,)ds) =

n=0

1/

(1 + 2n)

=

n=01/

1 +

8 8

(2n + 1)22

= 1/

(cosh)

(2).

APPLICATION. IfF(a) = P(1

0

X2(s)ds < a), then

0

eadF(a) =

eadF(a)

= E(e1

0X2(s)ds ) = 1/

(cosh)

(2).


111/323

15. Equivalent For of Ito

Process

LET (,F, P) BE A probability space with (Ft)t0 and increasing fam- 102ily of sub -algebras ofF such that (UFt

t0) = F. Let

(i) a : [0, ) S+d

be a progressively measurable, bounded func-

tion taking values in S+d

, the class of all ddpositive semidefinitematrices with real entries;

(ii) b : [0, ) Rd be a bounded, progressively measurablefunction;

(iii) X : [0, ) Rd be progressively measurable, right continu-ous and continuous a.s. (s, w) [0, ) .

For (s, w)

[0,

)

define the operator

Ls,w =1

2

di,j=1

ai j(s, w)2

xixj+

dj=1

bj(s, w)

xj.

For f, u, h belonging to C0

(Rd), C0

([0, ) Rd) and C1,2b

([0, ) Rd) respectively we define Yf(t, w), Zu(t, w), Ph(t, w) as follows:

Yf(t, w) = f(X(t, w)) t

0

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

105


112/323

106 15. Equivalent For of Ito Process

Zu(t, w) = u(t,X(t, w)) t

0

u

s+ Ls,wu

(s,X(s, w))ds,

Ph(t, w) = exp[h(t,X(t, w)) t

0

h

s+ Ls,wh

(s,X(s, w)ds

12

t0

a(s, w)xh(s,X(s, w)), xh(s,X(s, w))ds].

Theorem . The following conditions are equivalent.103

(i) X(t, w) = exp[,X(t, w) t

0

, b(s, w)ds t

0

, a(s, w)ds]

is a martingale relative to (,Ft, P), Rd.(ii) X(t, w) is a martingale in Rd. In particular Xi(t, w) is a mar-

tingale Rd.(iii) Yf(t, w) is a martingale for every f

C

0

(Rd)

(iv) Zu(t, w) is a martingale for every u C0 ([0, ) Rd).

(v) Ph(t, w) is a martingale for every h C1,2b [(0, ) Rd).(vi) The result (v) is true for functions h C1,2([0, ) Rd) with

linear growth, i.e. there exist constants A and B such that |h(x)| A|x| + B.

The functions ht

, hxi

, and 2hxixj

which occur under the integral

sign in the exponent also grow linearly.

Remark. The above theorem enables one to replace the martingale con-

dition in the definition of an Ito process by any of the six equivalent

conditions given above.

Proof. (i) (ii). X(t,

) is Ft-measurable because it is progressively mea-

surable. That E(|X(t, w)|) < is a consequence of (i) and the fact thata is bounded.


113/323

107

The function X(t, w)

X(s, w)is continuous for fixed t, s, w, (t > s).

Moreras theorem shows that is analytic. Let AFs. Then

A

X(t, w)

X(s, w)dP(w)

is analytic. By hypothesis, 104A

X(t, w)

X(s, w)dP(w) = 1, Rd.

Thus

A

X(t, w)

X(s, w)dP(w) = 1, complex . Therefore

E(X(t, w)|Fs) = X(s, w),

proving (ii). (ii) (iii). Let

A(t, w) = exp it

0

, b(s, w)ds + 12

t

0

, a(s, w)ds , Rd.By definition, A is progressively measurable and continuous. Also

|dAdt

(t, w)| is bounded on every compact set in R and the bound is in-dependent of w. Therefore A(t, w) is of bounded variation on every in-

terval [0, T] with the variation ||A||[0,T] bounded uniformly in w. LetM(t, w) = Xi(t, w). Therefore

sup0tT

|M(t, w)| e1/2 T sup0tT

|, a|.

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

E

sup

0tT|M(t, w)| ||A||[0,T](w)

< , T,

M(t, )A(t, ) 12

t0

M(s, )dA(s, )


114/323


is a martingale (for a proof see Appendix), i.e. Yf(t, w) is a martingale

when f(x) = ei,x.

Let f C0 (Rd

). Then f F(Rd

) the Schwartz-space. Thereforeby the Fourier inversion theorem

f(x) =

Rd

f()ei,xd.

105

On simplification we get

Yf(t, w) =Rd

f()Y(t, w)d

where Y Yei, x. Clearly Yf(t, ) is progressively measurable andhence Ft-measurable.

Using the fact that

E(|Y(t, w)|) 1 + t d|| ||b|| +d2

2 ||2 ||a||,

the fact that F(Rd) L1(Rd) and that F(Rd) is closed under multi-plication by polynomials, we get E(|Yf(t, w)|) < . An application ofFubinis theorem gives E(Yf(t, w)|Fs) = Yf(s, w), ift > s. This proves(iii).

(iii) (iv). Let u C0([0, ) Rd).Clearly Z

u(t,

) is progressively measurable. Since Z

u(t, w) is boun-

ded for every w, E(|Zu(t, w)|) < . Let t > s. Then

E(Zu(t, w) Zu(s, w)|Fs) =

= E(u(t,X(t, w) u(s,X(s, w)|Fs) E(t

s

(u

+ L,wu)(,X(, w)d|Fs)

= E(u(t,X(t, w) u(t,X(s, w))|Fs) + E(u(t,X(s, w) u(s,X(s, w))|Fs)

E(t

s

( u

+ Luw)(,X(, w))d|Fs)


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109

= E(

ts

(L,wu)(t,X(, w))d|Fs)+

+ E(

ts

(u

(,X(s, w))d|Fs)

E(t

s

(u

+ Lu, w)(,X(, w))d|Fs), by (iii)

= E(

t

s

[L,w

u(t,X(, w))

L,w

u(,X(, w))]d|F

s)

+ E(

ts

[u

(,X(s, w)) u

(,X(, w))]d|Fs)

= E(

ts

(L,wu(t,X(, w)) L,wu(,X(, w))]d|Fs)

E(t

s

d

s

L,wu

(,X(, w))d|Fs)

106

The last step follows from (iii) (the fact that > s gives a minus

sign).

=

E(

t

0

d

t

L,wu(,X(, w))d|F

s)

E(t

s

d

s

L,wu

(,X(, w))d|Fs)

= 0

(by Fubini). Therefore Zu(t, w) is a martingale.

Before proving (iv) (v) we show that (iv) is true ifu C1,2b ([0, ) Rd. Let u C1,2

b.


116/323


(*) Assume that there exists a sequence (un) C0 [[0, )Rd] suchthat

un u, unt

ut

, unxi

uxi

, unxixj

2

uxixj

uniformly on compact sets.107

Then Zun Zu pointwise and supn

(|Zun (t, w)|) < .Therefore Zu is a martingale. Hence it is enough to justify (*).

For every u C1,2b

([0, ) Rd) we construct a u C1,2b

((, ) Rd) C1,2

b(R Rd) as follows. Put

u(t, x) =

u(t, x), ift 0,C1u(t, x) + C2u( t2 , x), ift < 0;matching

u

t,

u

tat t = 0 and u(t, x) and u(t, x) at t = 0 and u(t, x) and

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,

S(x) =

1, if |x| 1,0, if |x| 2.Let Sn(x) = S

|x|2n

where |x|2 = x2

1+ + x2

d+1. Pur un = Snu.

This satisfies (*).

(iv) (v). Leth C1,2

b([0, ) Rd).

Put u = exp(h(t, x)) in (iv) to conclude that

M(t, w) = eh(t,X(t,w)) t

0

eh(s,X(s,w))

h

s+Ls,wh +

1

2xh, axhds

is a martingale.Put108


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111

A(t, w) = exp

t0

h

s(s, w) + Ls,w (s, w) + 1

2a(s, w)xh, xhds

.

A((t, w)) is progressively measurable, continuous everywhere and

||A||[0,T](w) C1 C2T

where C1 and C2 are constants. This follows from the fact that |dA

dt| is

uniformly bounded in w. Also sup0tT

|M(t, w)| is uniformly bounded in

w. ThereforeE( sup

0tT|M(t, w)| ||A||[0,T](w)) < .

Hence M(t, )A t

0

M(s, )dA(s, ) is a martingale. Now

dA(s, w)

A(s, w)=

h

s(s, w) + Ls,wh(s, w) +

1

2axh, xh

Therefore

M(t, w) = eh(t,X(t,w)) +

t0

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

A(s, w).

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

t0


A(s, w)

t0

M(s, )dA(s, ) =t

0


+

t0

dA(s, w)

s0

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

A(, w)

Use Fubinis theorem to evaluate the second integral on the right

above and conclude that Ph(t, w) is a martingale.


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(vi) (i) is clear if we take h(t, x) = , x. It only remains to provethat (v) (vi).

(v) (vi). The technique used to prove this is an important one andwe shall have occasion to use it again.

Step 1. 0 Let h(t, x) = 1x1 + 2x2 + dxd = , x for every (t, x) 109[0, ) Rd, is some fixed element ofRd. Let

Z(t) = exp

,Xt

t0

, bds 12

t0

, ads

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

Let f : R R be a C function with compact support such thatf(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 ofn and x and fn(x) converges tox.

Let hn(x) =d

i=1 ifn(xi). Then hn(x) converges to , x and |hn(x)| C|x| where C is also independent ofn and x. By (v),Zn(t) = exp

hn(t,Xt) t

0

hn

s+ Ls,wh

ds 1

2

t0

axhn, xhnds

is a martingale. As hn(x) converges to , x, Zn(t, ) converges to Z(t, )pointwise. Consequently

E(Z(t)) = E(limZn(t)) lim E(Zn(t)) = 1

and Z(t) is a supermartingale.

Step 2. E(exp B sup0st

|X(s, w)|) < for each t and B. For, let Y(w) =sup

0st|X(s, w)|, Yi(w) = sup

0st|Xi(s, w)| where X = (X1, . . . ,Xd). Clearly

Y

Y1 +

+ Yd. Therefore

E(eBY) E(eBY1eBY2 . . . eBYd).


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113

The right hand side above is finite provided E(eBYi) < for each i110as can be seen by the generalised Holders inequality. Thus to prove the

assertion it is enough to show E(eBY

i) < for each i = 1, 2, . . . d withaB different from B; more specifically for B bounded.

Put 2 = 0 = 3 = . . . = d in Step 1 to get

u(t) = exp[1X1(t) t

0

1b1(s, )ds 1

221

t0

a11(s, )ds]

is a supermartingale. Therefore

P

sup0st

u(s, )

1

E(u(t)) =1

, > 0.

(Refer section on Martingales). Let c be a common bound for both b1and a11 and let 1 > 0. Then () reads

P

sup

0

s

t

exp 1X1(s) exp(1ct+1

221ct)

1

.

Replacing by

e1 ect11/2ct21

we get

P

sup

0stexp 1X1(s) exp 1

e1 +1ct+1/221 ct,

i.e.

P

sup0st

X1(s) e1 +1ct+1/221 ct, 1 > 0.Similarly

P

sup

0stX1(s)

e1 +1ct+1/221 tc, 1 > 0.

As

{Y1(w) } sup0st

X1(s) sup0st

X1(s) ,


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we get 111

P{Y1 } 2e1 +1ct+1/221

ct, 1 > 0.Now we get

E(exp BY1) =1

B

0

exp(Bx)P(Y1 x)dx (since Y1 0)

2B

0

exp(Bx x1 + 1ct+1

221ct)dx

< , if B < 1This completes the proof of step 2.

Step 3. Z(t, w) is a martingale. For

|Zn(t, w)| = Zn(t, w)

= exp hn(Xt) t

0 hn

s+ Ls,whn dx

1

2

t

0 axhn, xhnds exp

hn(Xt) t

0

Ls,whn

Documents

S R S Varadhan Tata Institute of Fundamental Research Lectures on Diffusion Problems and Partial Differential Equations 1981