stoch diff eqns tifr68

8/10/2019 Stoch Diff Eqns Tifr68

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Lectures onTopics in Stochastic Di ff erential Equations

ByDaniel W. Stroock

Tata Institute of Fundamental ResearchBombay

1982


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Lectures on

Topics in Stochastic Di ff erential Equations

ByDaniel W. Stroock

Lectures delivered at the

Indian Institute of Science, Bangaloreunder the

T.I.F.R.I.I.SC. Programme in applications of

Mathematics

Notes by

Satyajit Karmakar

Published for the

Tata Institute of Fundamental Research, Bombay

Springer-Verlag

Berlin Heidelberg New York

1982


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Author

Daniel W. StroockDepartment of Mathematics

University of Colorado

Boulder, Colorado 80309

U.S.A.

c Tata Institute of Fundamental Research, 1981

ISBN 3-540-11549-8 Springer-Verlag, Berlin. Heidelberg. New York ISBN 0-387-11549-8 Springer-Verlag, New York. Heidelberg. Berlin

No part of this book may be reproduced in anyform by print, microlm or any other means with-out written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 400 005

Printed by N. S. Ray at the Book Centre Limited,Sion East, Bombay 400 022 and published by H. Goetze,

Springer-Verlag, Heidelberg, West Germany

Printed in India


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Preface

THESE NOTES are based on ve weeks of lectures given during De-cember 1980 and January 1981 at T.I.F.R. in Bangalore. My purpose inthese lectures was to provide some insight into the properties of solu-tions to stochastic di ff erential equations. In order to read these notes,one need only know the basic It o theoryof stochastic integrals. Forexample, H.P. McKeans Stochastic Integrals (Academic Press, 1969)contains all the background material.

After developing a little technical machinery, I have devoted ChapterI to the study of solutions of S.D.E.s as a function of the initial point.This topic has been discussed by several authors; the treatment givenhere is based on a recent paper by H. Kunita (to appear in the Proceed-ings of the 1980 L.M.S. conference at Durham). In fact, the only way inwhich the present treatment di ff ers from this is that I have been a littlemore careful about the integrability estimates.

Chapter II is devoted to the study of solutions as a function of time.A large part of this material is adapted from my work with S.R.S. Varad-han (On the support of di ff usion processes with applications to thestrong maximum principle, Proc. 6th Berkeley Symp. on Math. Stat.and Prob., Vol.III (1970)). The presentation of this material has beengreatly aided by the incorporation of ideas introduced by Y. Takahashiand S. Watanabe in their paper The probability functionals of di ff usionprocesses (to appear in the Proceedings of the 1980 L.M.S. conferenceat Durham). To intorduce the reader to their work, in the last section

of Chapter II, I have derived a very special case of the general resultderived in the paper by Takahashi and Watanabe.

v


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vi

I would like to thank S. Karmakar for his e ff orts in guiding these

notes to completion. Also, I am grateful to R. Karandikar for his usefulcomments on these notes. Finally, it gives me pleasure to express mygratitude to K.G. Ramanathan and the Tata Institute for the opportunityto visit India.


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Contents

1 Solutions to Stochastic Di ff erential Equations.... 11 Basic Inequalities of Martingale Theory . . . . . . . . . 12 Solutions to Stochastic Di ff erential Equation.... . . . . . 43 Diff erentiation with respect to x . . . . . . . . . . . . . 194 An application to Partial Di ff erential Equations . . . . . 25

2 The Trajectories of Solutions to Stochastic.... 291 The Martingale Problem . . . . . . . . . . . . . . . . . 292 Approximating Di ff usions by Random Evolutions . . . . 323 Characterization of Supp( p x), the non-degenerate case . 464 The Support of P x L, the Degenerate case . . . . . . . 575 The Most Probable path of a brownian motion with drift 73

vii


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Chapter 1

Solutions to Stochastic

Diff

erential Equations as aFunction of the StartingPoint

1 Basic Inequalities of Martingale Theory1

The most important inequality in the martingale theory is the Doobs in-equality which states that if ( X (t ), F t , P) is a right continuous, integrablesub-martingale, then

P(Sup o t T X (t ) ) < 1

E [ X (T ), Sup o t T X (T ) ], > 0 (1.1)

for all T > 0.As a consequence of ( 1.1 ), one has, assuming that X (.) 0:

E Sup o t T X (t ) p 1/ p

p p 1

E [ X (T ) p]1/ p, 1 < p < . (1.2)

The passage from ( 1.1 ) to (1.2) goes as follows:

1


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2 1. Solutions to Stochastic Di ff erential Equations....

Let X (T ) = Sup 0 t T X (t ). Then using ( 1.1) we have

E [( X (T ) ) p] = p

0

p 1 P( X (T ) )d

p

0

p 2 E [ x(T ), X (T ) ]d

= p

0

p 2d

0

P( X (T ) , X (T ) )d .

= p p 1

0 E [ X (T ) ) p 1 , X (T )] ]d .

= p

p 1 E [( X (T ) ) p 1 , X (T )]

p

p 1 E [( X (T ) ) p]1 1/ p, E [( X (T )) p]1/ p

where we have used H olders inequality to obtain the last line. Nowdividing both sides by E [( X (T ) ) p]1 1/ p, we get the required result.

The basic source of continuous martingales is stochastic integrals.2Let ( t , F t , P) be a l-dimensional Brownian motion and (.) be a F .-progressively measurable R -valued function satisfying

E

T

0

| (t )|2dt <

for all T > 0. Set

X (t ) =t

0

(s)d (s).

Then ( X (t ), F t , P), and ( X 2(t ) t

0 (s)2ds , F t , P) are continuous mar-


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1. Basic Inequalities of Martingale Theory 3

tingales. In particular,

E [ X 2(T )] = E

t

0

(t )2dt for all T > 0.

By (1.2), this means that

E [( X (T ) )2] 2 E [

T

0

(t )2dt ]1/ 2 .

The following theorem contains an important generalization of thisobservation.

Theorem 1.3 (Burkholder) . Let ( (t ), F t , P) be a 1-dimensional Brow-nian motion and (.) be a F.-progressively measurable R n Rd -valued function satisfying

E

T

0

Trace a (t )dt < , T > 0

where a (.) = (.) (.) . Set X (t ) =t

0 (s)d (s). Then for 2 p < and T > 0 ,

E [Sup o t T | X (t )| p]1/ p C p E [

T

0

Trace a (t )dt ) p/ 2 ]1/ p

where 3

C p = p p+ 1

2( p 1) p 1

Proof. By Itos formula:

| X (t ) p

| = p

t

0 | X (s)| p 2

(s) X (s)d (s)


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+ 12

t

0 p| X (s)| p 2

(Trace a(s)+

( p 2)< X (s), a(s) X (s) >

| X (s)|2 ds

Thus,

E [| X (T )| p] =12

E [

T

0

p| X (t )| p 2 Trace a(t ) + ( p 2)< X (t ), a (t ) X (t ) >

| X (t )|2 dt ]

p( p 1)

2 E [

T

0

p| X (t )| p 2 Trace a(t )dt ]

p( p 1)

2 E [Sup 0 t T | X (t )|

p 2

T

0Trace a(t )dt ]

p( p 1)

2 E [Sup 0 t T | X (t )|

p]1 2/ p E [(

T

0

Trace a(t )dt ) p/ 2]2/ p

and so E [Sup 0 t T | X (t )| p] < (

p p 1

) E [| X (T )| p]

p p+ 1

2( p 1) p 1 E [Sup 0 t T | X (t )|

p]1 2/ p E [(

T

0

Trace a(t ) p/ 2 ]2/ p

Now dividing both sides by E [Sup 0 t T | X (t )| p]1 2/ p we get the re-

quired result.

2 Solutions to Stochastic Di ff erential Equation as aFunction of (t , x)

Let : Rn Rn Rd and b : Rn Rn be measurable functions4satisfying

|| ( x) ( y)|| H .S . L| x y|, x, y Rn

|b( x) b( y)| L| x y|, x, y Rn (2.1)where L < and ||.|| H .S . denotes the Hilbert Schmidt norm 1 .

1 || A|| H .S . is the Hilbert-Schmidt norm of the matrix A. That is, || A||2 H .S . = Trace AA .


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2. Solutions to Stochastic Di ff erential Equation.... 5

Also assume that ( (t ), F t ( p) is a d dimensional Brownian motion

Theorem 2.2. For each x Rn there exists a unique solution (., x) to

(t , x) = x +t

0

( (s, x))d (s) +t

0

b( (s, x))ds , t 0. (2.3)

In fact, for each 2 p < and T > 0 there are A p(T ) < suchthat

E [Sup 0 t T | (t , x) (t , y)| p] A p(T )(| x y| p) (2.4)

Proof. The existence of solution is proved by the Picard iteration. De-ne 0(.) = x and

n+ 1 (t ) = x +t

0

( n(s))d (s) +t

0

b( n )( s)ds

Then for 2 p < ,

E [Sup 0 t T | n+ 1(t ) n(t )| p]1/ p

E [|

t

0

b( n (t )) b( n 1 (t )dt | p]1/ p

+ E [|

T

0

( ( n(s)) ( n 1(s)))d (s)| p]1/ p

L

T

0

E [| n (t ) n 1(t )| p]1/ pdt

+ C P LE [(

T

0

| n(t ) n 1(t )|2dt ) p/ 2 ]1/ p

L

T

0 E [| n (t ) n 1(t )| p

]1/ p

dt


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+ C P LT (P 2)/ (2P)

E [

T

0 | n(t ) n 1 (t )| p

dt ]1/ p

and so5 E [Sup 0 t T | n(t ) n 1 (t )|

p]

2 p 1 L p(T p 1 + C p pT p 1/ 2)

T

0

E [Sup 0 t T | n(s) n 1(s)| p]dt .

From this, it follows by induction that

E [Sup 0 t T | n+ 1 (t ) n (t )| p] 0 and 1 p < .To prove uniqueness, note that if (.) is a second solution and R =6

inf {t 0 : |(.) x| R}, then

E [|(t R) (t R)|2 ]

2 L2 E [

t R

0

|(s) (s)|2ds ] + 2 L2 tE [

t R

0

|(s) (s)|2ds ]

and so (t R) = (t R) (a .s., P) for all t and R. Since (.) and (.)are P a .s. continuous and P(Sup 0 t T | (t )| < ) = 1 for all T > 0,uniqueness is now obvious.

Finally, E [Sup 0 t T | (t , y) (t , x)|

p]1/ p

| x y| + C P LE [

T

0 | (t , y) (t , x)|2dt )

p/ 2]

1/ p


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+ L

T

0 E [| (t , y) (t , x)| p]1/ p

dt

and so

E [Sup 0 t T | (t , y) (t , x)| p] 3 p 1 | x y| p + 3 p 1 L p(C p pT

p 1/ p)T

0

E [Sup 0 t T | (s, y) (s, x)| p] dt

Therefore

E [Sup 0 t T | (t , y) (t , x)| p] 3 p 1 | x y| p exp(3 p 1 L p

T

0 (C p p t

( p 1)/ 2 ++ t p 1 ) dt ).

This proves our theorem.The importance of (2.4) is that it allows us to nd a version of (t , x)

which is a.s. continuous with respect to ( t , x). To see how this is done,we need the following real-variable lemma.

Lemma 2.5. Let P and be strictly increasing continuous function on 7[0, ] such that P (0) = (0) = 0 and ( ) = . Also suppose that Lis a normed linear space and that f : Rd L is strongly continuous on B(a , r )( { x Rd : | x a | < r }). Then

B(a ,r )

B(a ,r )

|| f ( x) f ( y)||

P(| x y|)dxdy B

implies that

|| f ( x) f ( y)|| 8

| x y|

0

1 4d + 2

2u2d P(du), x, y B(a , r )

where

= inf x B(a ,r ) inf 1


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Now let = (2d + / 2) p. Then, by (2.8):

Sup N E [ B(0 ,r )

B(0 ,r )

]|| X ( N )( x) X ( N )( y)||

| x y|

p

dxdy C r 2d

where C < depends on C and d . By (2.5 ), this means that for L > 0

Sup N P (|| X ( N )( x) X ( N )( y)|| K L1/ p| x y|/ 2 p, x, y B(0, r ))

1 C r 2d

L ,

where K depends on d , p and . Notice that10

Sup x, y B(0 , R) x y

|| X ( N )( x) X ( N )( y)||| x y|/ 2 p

is a non-decreasing function of N . Hence

P Sup N Sup x, y B(0 , R) x y

|| X ( N )( x) X ( N )( y)||| x y|/ 2 p

K L1/ p 1 C r 2d

L .

Since X (n) k 2n

= X k 2n

for all N 0 and k Z d , x( N )(.) converges

a.s., uniformly on B(0, r ) to a continuous function X (.) which coincides

with X ( x) at x = k / 2 N

B(0, r ). Since

E [|| X ( x) X ( y)|| p] 0 as | x y| 0,

it is clear that X ( x) = X ( x) a.s. for all x B(0, r ). Finally, since r wasarbitrary, the proof is complete.

Exercise 2.9. Let f ( x) = xVa. Then f ( x) = [a , ]( x) and f ( x) = a ( x)(Diracs function). Thus by It os formula, if ( t , F t , P) is a one-dimensional Brownian motion:

12

t

0 a ( (s))ds = (t )Va t

0 [a , ]( (s))d (s)


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That is, we would suspect that

lim0

14

t

0

[a ,a+ ]( (s))ds = (t )Va t

0

[a , ]( (s))d (s).

To check this, dene 11

a (t ) = (t )Va t

0

X [a , ]( (s))d (s).

By the technique which we have been using, show that there is aversion L a (t ) of a (t ) which is a.s. continuous in (t , x)a[0 , ] R. Check that for f C 0 ( R)

R

f (a) La (t )ds = F ( (t )) t

0

F ( (s))d (s)

where F ( x) = ( xVa) f (a)da.From this conclude that

R

f (a) La (t )da = 12

t

0

f ( (s))ds

for all such f s. The identication of L a (t ) as

lim0

14 X [a ,a+ ]( (s))ds

is now easy

Exercise 2.10. Again let ( t , F t , P) be a 1-dimensional Brownian mo-tion. Given t > 0 and > 0 , let N (t ) be the number of times that | (.)|goes from to 0 during [0 , t ]. (That is, N n if and only if there exist 0 < u1 < v1 < un < vn t such that | (u j)| = and | (v j)| = 0 for 1 j n). Show that

N (t ) 2 L0 (t ) a. s. 0.


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The idea is the following: First show that

| (t )| =t

0

sgn (s)d (s) + 2 L0 (t ).

Next dene 0 = 0 and for n 112

n = inf {t n 1 : | (t )| = }n = inf {t n : | (t )| = 0}.

Then

1

| (n t )| | ( n t )| = N (t ) + (| (t )| )

1

X [ n ,n ](t ).

At the same time:

1

| (n t )| | ( n t )| = | (t )|

1

n t

n 1 t

sgn (s)d (s) 2 L0(t ).

Thus

N (t ) 2 L0(t ) = | (t )|

1 X [n 1 , n ](t )

1 X [ n ,n ](t )

+

1

n t

n 1 t

sgn (s)d (s).

From this check that

E [| N (t ) 2 L0 (t )|2 ] ct , 0 < 1;

and therefore that

1n2

N 1/ n2 (t ) 2 L0(t ) a.s as n .


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Finally, note that if 1/ (n + 1)2 < 1/ n2 , then

1(n + 1)2

N 1/ n2 (t ) N (t ) 1/ n2 N 1/ (n+ 1)2 (t ),

and so N (t ) 2 L0(t ) a.s. as 0.

Lemma 2.11. Let ( (t ), F t , P) be a d dimensional Brownian motion 13and suppose that (.) and b (.) are as in (2.1 ). Then there is a choice of (t , x) solving ( 2.3 ) such that

(t , x) (t , x)

From now on, (t , x) refers to the map in ( 2.11 ). We next want todiscuss the mapping x (t , x) for xed t > o. We will rst show thata.s. the maps x (t , x) are 1-1 and continuous for all t 0.

Lemma 2.12. Let T > 0 and p R be given. Then there is a C p(T ) < such that

E [| (t , x) (t , y)| p] C p(T )| x p | p, t [0, t ] and x , y Rn . (2.13)

Proof. Set f ( z) = | z| p for z Rn {0}. Then

f zi

= p| z|

p 2

zi

and2 f

zi z j= p( p 2) | z| p 4 zi z j + i j p| z| p 2

Now let x y be given and for 0 < < | x y| let

= inf {t 0 : | (t , x) (t , y)| }

and dene z (t ) = (t , x) (t , y).

Now by It o s formula: 14


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| z (t )| p

n

i= 1

t

0 (bi( (s, x)) b

i( (s, y)))

f xi ( z (s))ds

12

n

i= 1

t

o

(d

l= 1

il( (s, x) (s, y)) jl ( (s, x) (s, y)))

2 f xi x j

( z (s))ds

is a martingale. Notice that for 0 s :

|

n

i= 1 (bi

( (s, x)) bi

( (s, y))) f xi ( z (s)) |

| p| L| z | p 1n

i= 1

| z,i(s)| 1/ 2 L|P || z (s)| p.

Also for 0 s :

n

i, j= 1

d

= 1

( i ( (s, x)) i ( (s, y)))(

i ( (s, x)))

j ( (s, y))

2 f xi x j

( z (s))

dL2 p( p 2) | z (s)| p 2n

i, j= 1

z,i(s) z, j(s) + dL2 p| z (s)| p

(dnL 2 p( p 2) + dL2 p)| z (s)| p.

Thus

E [| z (t )| p.] | x y| p + Ad ,n( p)

t

o

E [| z (t )| p]ds ,

and so E [| (t , x) (t , y)| p] | x y| pe Ad ,n( p)t

Letting 0, we see that

E [| (t , x) (t , y)| p] | x y| pe Ad ,n( p)t


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where = inf {t > o : (t , x) = (t , y)}. Taking p = 1, we conclude that15

P( < ) = 0 and therefore that ( 2.13 ) holds. Lemma ( 2.12 ) show thatfor each x y : (t , x) (t , y), t > 0 a.s. We now have to show thatexceptional set does not depend on x and y.

Lemma 2.14. Let

(t , x, y) = 1

| (t , x) (t , y)|, t > o and x y.

Then is a.s. a continuous function of (t , x, y) for t > 0 and x y.

Proof. In view of ( 2.6) and the fact that is a.s. continuous on {(t , x, y) :

(t , x) (t , y)}, we need only to show that for some p > 2(2n + 1), onehas

E [|(t , x, y) (t , x , y )| p] C p,T ()( | x x | p+ | y y | p+ |t t | p/ 2) (2.15)

for all T > 0, > 0, 0 t , t T and | x x | | y y | .But

|(t , x, y) (t , x , y )| p

= || (t , x) (t , y)| | (t , x ) (t , y )|| (t , x) (t , y)| | (t , x ) (t , y )|

| p

2 p 1 |(t , x, y)| p |(t , x , y )| p(| (t , x) (t , x )| p)+

+ | (t , y) (t , y )| p).

Thus

E [|(t , x, y) (t , x , y )| p]

2 p 1 E [|(t , x, y)|4 p]1/ 4 E [|(t , x , y )|4 p]1/ 4

( E [| (t , x, ) (t , x , )|2 p]1/ 2 + [| (t , y) (t , y )|2 p]1/ 2)

The rst factors are easily estimated by (2.7), so along as | x x

|

| y 16 y | .


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Moreover, by the argument used to derive ( 2.4) , it is easy to check

that

E [| (s, x) (t , y)|2 p] C p(T )( |s t | p) + | x y|2 p),0 s, t T , x, y Rn. (2.16)

Thus (2.15 ) has been proved.

Exercise 2.17. Prove ( 2.16 ) for all 1 p < .We now want to show that a.s. the map x (t , x) is onto for all

t 0. The idea is that this is certainly true when t = 0 and that the map (t , .) is homotopically connected to (0, .). In order to take advantageof these facts, we must show that they continue to hold on the one-point compactication of R n .

Lemma 2.18. For each T > 0 and p R there is a C p(T ) < suchthat

E [(1 + | (t , x)|2) p] C p(T )(1 + | x|2 ) p, 0 t T .

Proof. Let f ( z) = (1 + | z|2 ) p. Then

f / zi = 2 p(1 + | z|2 ) p 1 ziand

2 f

zi z j= 2 p( p 1)(1 + | z|2 ) p 2 zi z j + 2 pi j (1 + | z|2) p 1 .

Hence

E [(1 + | (t , x)|2) p] (1 + | x|2 ) p

= E [

t

o

n

i= 1

bi( (s, x) f zi

( (s, x))ds ] + 12 E [

t

o

n

i, j= 1

i ( (s, x))

j ( (s, x) 2 f zi z j

( (s, x))ds ]

C

t

o E [(1 + | (s, x)|2) p

]ds ,


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since17

max1 i n

|bi( z)| < max1 i n

|bi(0) | + L| z| < 2 max1 i n

|bi(0) |VL(1 + | z|2)1/ 2

and similarly

max1 i n1 n

| i ( z)| < 2 max1 i n1 n

| i (0) |VL(1 + | z|2 )1/ 2 .

Hence the desired result follows by Gronwalls inequality.

Lemma 2.19. Let Rn = Rn {} denote the one-point compacticationof Rn . Dene

(t , x) =1

1+ | (t , x)| , x

Rn

.0 , x = .

Then is a.s. continuous on [0, ] Rn into R 1 .

Proof. Since is a.s. continuous on [0 , ] Rn, if suffi ces to show thatfor each T > 0 and > 0 there is a.s. an R > 0 such that (t , x) < if | x| R and 0 t T . Choose p > 2(2n + 1). Then

|(t , y) (s, x)| p | (t , y)| p |(s, x)| p| (t , y) (s, x)| p

and so 18

E [|(t , y) (s, x) p |] E [|(t , y)|4 p]1/ 4 E [|(s, x)|4 p]1/ 4 E [| (t , y) (s, x)|2 p]1/ 2

Since (1 + | x|) (1 + | x|2)1/ 2 12

(1 + | x|), we see from ( 2.15) that

E [|(t , y)|4 p]1/ 4 E [|(s, x)|4 p]1/ 4 K p(T )(1 + | x|) p(1 + | y|) p

On the other hand, by ( 2.15 ),

E [| (t , x) (s, y)|2 p]1/ 2 K p(T )(|t s| p/ 2 + | x y| p)

for 0 s, t T . Thus

E [|(s, x) (t , y) p|] K p (T )( | x y|(1 + | x|)(1 + | y|)) p + |t s|

p/ 2

(1 + | x|)(1 + | y|) p]


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We now dene

(t , x) = (t , x/ | x|2 ) for x 0,

and(t , 0) = 0.

Since is a.s. continuous on [0 , ] Rn , we will be done once weshow that admits an a.s. continuous version. Noting that if | y| | x| >0:

| x| x|2 y| y|2

|

(1 + | x|| x| )(1 + | y|| y| )

=

1| y|2

| y x |2 x y|

1| x|| y| (1 + | x|)(1 + | y|)

=

x y |

y2

x x x| + x y | x y|

(1 + | x|)(1 + | y|)

=

x y

2

y x2

1 + x y | x y|

(1 + | x|)(1 + | y|)

=

1| y| (| y| | x|)(| y| + | x|) +

x y | x y|

(1 + | x|)(1 + | y|)

3| x y|.

19

We see that

E [|(s, x) (t , y)| p] K p (T ).[| x y| p + |t s| p/ 2 ]

if 0 s, t T and x, y 0. Since by ( 2.18)(u, z) 0 in L p for each u > 0 as | z| , this inequality con-

tinuous to hold even when x or y is 0. This, by ( 2.6 ), admits an a.s.continuous version. This proves the lemma.

Theorem 2.20. Let (.) and b (.) be as in ( 2.1) and let (t , x) be as in(2.11 ). Then a.s. (t , .) is a homeomorphism of R n onto R n for all t 0.


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3. Di ff erentiation with respect to x 19

Proof. We have seen that a.s. (t .) determines a one-to-one continuous

map (t , .) of Rn

into itself for all t 0. Moreover, (0 , .) is certainlyonto. Thus, by standard homotopy theory, a.s. (t , .) is onto all t 0.Also, by the invariance domain theorem (t , .) 1 must be continuous.

Finally, since a.s. (t , ) = for all t 0, we see that a.s. (t , .) isa homomorphism of Rn onto Rn for all t 0.

3 Diff erentiation with respect to x

We now want to di ff erentiate (t , x) with respect to x. 20

Lemma 3.1. Let (t , X ) , t 0 and X R D , be an a.s. continuous

R M

valued process such that (., X ) is F progressively measurable for each X R D. Further assume that for T > 0 , R > 0 and 2 p there is a C p(T , R) < such that

E [Sup 0 t T | (t , X )| p] < C p(T , R)(1 + | X |) p, | X | R,

and

E [Sup 0 t T | (t , X ) (t , Y )| p] C p(T , R)| X Y | p, | X |V |Y | R.

Let : R M R N R N Rd and b : R M R N R N be C functions satisfying

Sup (,) R M R N || i, ||H.S.V | b j

, | < ,

for 1 j N, and assume that (, 0) and b(, 0) and all their deriva-tion slowly increasing. Finally, let f : R D R N be a smooth functionwith bounded rst derivatives. Then for each X R D the equation

(t , X ) = f ( X ) +t

o

( (s, X ), (s, X ))d (s) +t

o

b( (s, X ), (s, X ))ds , t 0

has precisely one solution (., X ). Moreover, for T, R > 0 and 2 p there exists C p(T , R) < such that

E [Sup 0 t T |(t , X )| p] C p(T , R)(1 + | X |)

p, | X | R,


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and

E [Sup 0 t T |(t , X ) (t Y )| p] < C p(T , R)| X Y |

p, | X |V |Y | R.

21

In particular, admits a version which is a.s. continuous in (t , X ).

Proof. The techniques used to prove this lemma are similar to thoseintroduced in section 2. The details are left as an exercise.

Lemma 3.2. Let everything be as in lemma ( 3.1 ). Assume, in addition,that a.s. (t , .) C 1( R D) for all t 0 and that for all T, R > 0 and 2 p

max1 D

E [Sup 0 t T | X

(t , X )| p] C p(T , R)(1 + | X |) p, | X | R,

and

max1 D

E [Sup 0 t T | X

(t , X ) X

(t , X )| p] C p(T , R)| x y| p, | X |V |Y | R.

Then a.s. (t , .) C 1( Rn) for all t 0 and for T, R > 0 and 2 p there exists C p(T , R) such that

max1 D

E [Sup1

T

|

X | p] C

p(T , R)(1 + | X |) p, | X | R,

and

max1 D

E [Sup 0 t T | X

(t , X ) X

(t , Y )| p] C p(T , R)| x y| p, | X |V |Y | R.

In particular, for each 1 D, X

admits an a.s. continuous

version.

Proof. Choose and x 1 D and let e denote the unit vector in the -th direction. For h R\{0}, dene

h(t , X ) = (t , X + he ) (t , X ),


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h (t , X ) = (t , X + he ) (t , X )

and22

h f ( X ) = f ( X + he ) f ( X ).

Then h(., X ) is determined by the equation:

h (t , X ) = h f ( X )

+

t

o M

i= 1

(

1

o

i

( (s, X ) + h (s, X ), (s, X )

+ h(s, X ))d ) h i(s, X )d (s)

+

t

o

N

i= 1

(

1

o

j

( (s, X ) + h (s, X ), (s, X )

+ (s, X ))d ) h j(s, X )d (s)

+

t

o

M

i= 1

(

1

o

b j

( (s, X ) + h (s, X ), (s, X )

+ h(s, X ))d ) h i(s, X )ds

+

t

o

N

j= 1

(

t

o

b j

( (s, X ) + h (s, X ), (s, X )

+ h(s, X ))d ) h j(s, X )ds

Thus if

(t , X , h) =

(0) (t , x, h) (1) (t , x, h) (2) (t , x, h)

(3) (t , x, h) (4) (t , x, h)

=

(t , X ) h (t , X )

1h

h (t , X )

(t , X ) h(t , X )


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where 1

h

h t , X )

X (t , X ) for h = 0, and

(, ) = M

i= 1

1

o

i

( (0) + (1) , (3) + (4) )d i(2)

+ N

i= 1

1

o

j

(( (0) + (1) , (3) + (4) )d j)

and b(, ) is dened analogously, then the process (t , X , h) determinedby

(t , X , h) = h f h

( X ) +t

o

( (s, X , h), (s, X , h))d (s)

+

t

o

b( (s, X , h), (s, X , h))ds,

where h f / h( X ) f X ( X ) if h = 0, has the property that23

(., X , h) = h(., X )/ h a.s.

for each h R\{0} and X R D.

Since, by ( 3.1), has an a.s. continuous version, we conclude that/ X exists a.s. and has an a.s. continuous version. Also, by ( 3.1) ,/ X satises the asserted moment inequalities.

Theorem 3.3. Let : Rn Rn Rd and b : Rn Rn be C functionssuch that (| |/ x ) and (| |b/ x ) are bounded for each | | 1. Let (t , x) be an a.s. continuous version of the solution of the solution to(2.3 ). Then a.s. (t , .) C ( Rn) for all t 0 and (| | / x) is a.s.continuous in (t , x) for all . Moreover, for each T, R > 0 , m 0 and 2 p < , there is a C p(T , R, m) such that

Sup| X | R | | m E [ Sup0 t T || |

x (t , x)| p

]C p(T , R, m). (3.4)


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Finally, for given m 1, let (., x) denote || |

x (., x) : | | m 1

and let (., x) denote || | x

(., x) : | | = m . Then (., x) is relatedto (., x) by a stochastic di ff erential equation of the sort described inlemma (3.2).

In particular, if 24

J (t , x) = (( i

x j(t , x)))1 i, j n (3.5)

then J (t , x) is determined by the equation

J (t , x)=

I +

d

k = 1

t

o S (k )( (s, x)) J (s, x))d k

(s)

+

t

o

B( (s, x)) J (s, x)ds, t 0, (3.6)

where

S (k ) = (( ik x j

))1 i, j n and B = ((bi

x j))1 i, j n .

Proof. The facts that a.s. (t , .) C ( Rn) for all t 0 and that ( | | )/( x ) is a.s. continuous for all is derived by induction on | |, using

(3.2) at each stage. The induction procedure entails showing at eachstage that derivatives of order m are related to those of order m 1by an equation of the sort described in lemma ( 2.3 ), and and so thedescribed equation are a consequence of ( 3.2 ).

Lemma 3.7. Suppose that (.), S k (.)(1 < k < d ) , and B(.) areF progressively measurable R N R N valued functions such that

max1


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where A o R N R N is invertible. Then a.s. A (t ) is invertible for all25

t 0 and A 1

(.) satises

A 1(t ) = A 10 d

k = 1

t

o

A 1(u)S k (u)d k (u)

+

t

o

(d

k = 1

A 1(u)S 2k (u) A 1 (u) B(u))du t 0.

Proof. The theorem is proved by dening M (.) by the equation which A 1(.) is supposed to satisfy and then using It os formula to check that

d ( M (t )) A(t )) = d ( A(t ) M (t )) = 0.

Exercise 3.8. In order to see how one can guess the equation satised by A 1(.) , suppose that A 1(.) exists and denote it by M (.). Assume that

dM (t ) =d

k = 1

M (t )S k (t )d k (t ) + M (.) B(.)dt

and use It os formula to nd out what

S k

(.) and

B

(.) must be.

Theorem 3.9. Let everything be as in theorem (3.3 ). Then a.s. J (t , x) isinvertible for all (t , x) [0, ] Rn and J 1(., x) determined by

J 1(t , x) = I d

k = 1

t

o

J 1(s, x)S k ( (s, x))d k (s) (3.10)

+

t

o

(d

k = 1

J 1(s, x)S 2k ( (s, x)) J 1(s, x) B( (s, x)))ds .

In particular, a.s. (t , .) is a C di ff eomorphism of R n onto R n .26


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4. An application to Partial Di ff erential Equations 25

4 An application to Partial Di ff erential Equations

Let (t , x) be as in section ( 2). Then it is well-known ( c f . Chapter 4of [S &V ]) that (t , x) is strong Markov in the sense that if is a niteF .-stopping time and f B( Rn) (the bounded measurable functions on Rn) then for all x Rn :

E [ f ( ( + t , x))|F ] = u f (t , (, x)) a.s., t 0 (4.1)

whereu f (t , x) = E [ f ( (t , x))] , (t , x) [0, ] Rn . (4.2)

Lemma 4.3. Given x Rn and R > 0 there exist A 1( x, R) < and

0 < A2( x, R) < depending only on n, Sup y B( x, R) ( y) H .S . and Sup y B( x, R) |b( y)| such that

P ( r h) A1( x, R) exp( A2( x, R)r 2/ h)

for 0 r R and h > 0 , where

r = inf {t 0 : | (t , x) x| r }.

Proof. Given Rn, dene

X (t )=

exp[ < , (t , x)

x

t

o b( (s, x)ds >

12

t

o

< , ( (s, x) > d s]

Then by It os formula, ( X (t R), F t , p) is a martingale for all R > 0.set

b R = Sup y B( x, R) |b( y)|, a R Sup y B( x, R) ( y) H .S

and dene h R = b R/ 2. For S n 1 (the unit sphere in Rn 1 ,) > 0, 270 < r R and 0 < h h R:

P (Sup 0 t h < , (t , x) x > r ) < P (Sup 0 t h < , (t R, x)


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x

t R

o b( (s, x))ds ) r 2 )

< P (Sup 0 t h X (t R) exp(r 2

2h

2 a R))

exp(r 2

+ 2h

2a R)

Taking = r / (2ha R), we obtain

P (Sup 0 t h < , (t , x) x > r ) exp( r 2 / (8ha R))

and so by choosing successively to point along n coordinate axes:

P (Sup 0 t h max1 i n| i(t , x) xi | r ) 2n exp( r 2 / 8ha R)

Clearly the required estimate follows from this.

Lemma 4.4. Let (.) and b (.) be as in theorem ( 3.3 ) and dene (t , x)accordingly. Given f C ( R

n) (the C -functions f such that the D f are slowly increasing for all ), dene u f (t , x) by (4.1 ). Then u f C ([0 , ) Rn) and for each T, R > 0 and m 0 there is a C (T , R, m) < such that

max2 + || m Sup 0

t T | | R

|

t | |u

f x (t , x)| C (T , R, m)max| | m

| | f x u (4.5)

Finally, u f is the unique u C ([0 , ) Rn ) C 1,2 ((0 , ) Rn ) suchthat

ut

= Lu, t 0 (4.6)

limt 0

u(t ., ) = f (.)

where28

L=

1

2

n

i, j= 1( )i j

( x)

2

xi x j+

n

i= 1 bi

( x)

xi


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4. An application to Partial Di ff erential Equations 27

Proof. Diff erentiating E [ f ( (t , x))] with respect to x, one sees by induc-

tion that| |u f x

(t , x) =

E | | f x

( (t , x)))P , ( (| |) (t , x)) , (4.7)

in| | x

(t , x) : | | | | . Using It o s formula and the fact that

| | x

(., x) : | | | | satises a stochastic di ff erential equation withcoeffi cients in C , one sees that

t | |u f x (t , x)

=

| || |+ 2 E

| | f

x (t , x) (,, )

(

(| |) (t , x)) , (4.8)

where the ,, C ( R D| |). From ( 4.8) and moment estimates in ( 3.3 ),

it is clear that u f C ([0 , ) Rn) and that ( 4.5 ) holds.

We next show that u f satises (4.6). Clearly

limt 0

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

Moreover, by (4.1)

u f (t + h, x) = E [ f ( (h + t , x))] = E [u f (t , (h, x))] .

Let = inf {s 0 : | (s, x) x| 1|}. Then

E [u f (t , (h, x))] = E [u f (t , (h , x))]+ E [u f (t , (h, x)), h].

29

Since u f (t , .) is slowly increasing, it follows from ( 4.3 ) that

1/ hE [u f (t , (h, x)), h] 0 as h 0.

On the other hand, by Ito s formula

1h E [u f (t , (h , x))] u(t , x)


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=

1h E [

h

0 Lu f (t , (s, x))ds ] Lu f (t , x) as h 0

since P ( > 0) = 1. Thus we have proved that

u f t

= Lu f , t 0.

Finally, we must show that if u C ([0 , ) Rn) C 1,2((0 , ) Rn

satises ( 4.6 ). Then u = u f . To this end, let

R = inf {t 0 : | (t , x) x| R}.

Then by I t o s formula for xed T > 0 we know that

(u(T t R, (t R, x)), F t , P)

is a martingale, 0 t T . In particular,

u(T , x) = E [u(T T R, (T R, x)].

since R as R and u is slowly increasing,

E [u(T T R, (T R, x)] E [ f (t , ( x(T ))] as R .

Exercise 4.9. Under the condition of theorem ( 3.3 ), show that for each30

m 0 and 2 p < there exist A m, p , Bm, p < and m, p > 0(depending on m , p, n and the bounds on and b and their derivatives)such that

E Sup 0 t m | | m || | x

(t , x)|2 p/ 2

Am, pe Bm, pT (1 + | x|)m, p . (4.10)

Calculate from this that if f C ( Rn), then u f C ([o, T ] R

n)for all T > o. Check also that for a given m 1 and 2 p < , m, pdoes not depend on (.) and b(.) while Am, p and Bm, p depend only onthe bounds on

| | x (.) and

| |b x (.) for | | m.


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Chapter 2

The Trajectories of Solutionsto Stochastic Di ff erentialEquations

1 The Martingale Problem31

In Chapter 1 we have discussed di ff usions from It o s point of view,that is as solutions to stochastic di ff erential equations. For the purposeof certain applications it is useful to adopt a slightly di ff erent point of

view.Let = C ([0 , ), Rn] and think of as a Polish space with themetric of uniform convergence on compact sets. Denote by M the Boreleld over . Given t > o and , let x(t , ) be the position in Rn

of at time t and set M t = ( x(s) : 0 s t ). If L is a diff erentialoperator of the from

L = 12

n

i, j= 1a i j ( x)

2

xi x j+

n

i= 1

bi( x) xi

, (1.1)

where a(.) and b(.) are continuous functions with values in S n (the set of

symmetric, non-negative denite real n n matrices) and Rn

respectivelyand if x Rn , we say that P solves the martingale problem for L starting

29


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30 2. The Trajectories of Solutions to Stochastic....

from x (Notation P L at x) if P is a probability measure on ( , M ),

P( x(o) = x) = 1 and for all f C 0 ( R

n) : ( X f (t ), M t , P) is a martingale,

where

X f (t ) f ( x(t ))

t

0

L f ( x(s))ds.

If for each x Rn there is precisely one such P, we say that themartingale problem for L is well-posed.

Theorem 1.2. Let (.) and b (.) be as in (I, 3.3) and let L be gives as32(1.1 ) with a (.) = (.). Then the martingale problem for L is well- posed. In fact, for each x Rn , the unique P x L at x coincides withthe distribution of (., x) is the solution to (I, 2.3).

Proof. The rst step is the simple remark that, by It o s formula, if (., x)is given by (I, 2.3), then P L at x, where P denotes the distributionof (., x). Thus for each x Rn the martingale problem for L admits asolution.

The next step is to show that if P L at x and if f C ([o, T ) Rn )for some T > o, then

( f (t T , x(t )) t T

o

(/ s + L) f (s, x(s))ds , t , P)

is a martingale. The proof of this fact runs as follows: Given o s o, choose R C 0 ( B(0, 2 R)) so that 0 R 1 and R 1 on B(o, R). Then the function

F R(t , y) = R( y)u f (T t , y) C 0 ([0 , T ] Rn)

for any T > 0. Hence

(F R(t T , x(t T )) t

o

(/ s + L)F R(s, x(s))ds , t , P)

is a martingale. Thus by Doobs stopping time theorem,

(u f (T t R T , x(t R T )), t , P)

is a martingale, where

R = inf {t o : | x(t ) x(o)| R}.

Since R as R and u f is bounded, it follows that

(u f (T t T , x(t T )), t , P)

is a martingale. In particular, if 0 s T then

E p[ f ( x(T )) | s ] = E p[u f (o, x(T )) | s ]= u f (T s, x(s)) (a .s., P).

working by induction on N , it follows easily that if 0 t 1 < t N and 34 f 1 , . . . , f N C ( Rn), then

E p[ f 1( x(t 1 )) . . . f N ( x(t N ))] = E [ f 1( (t 1, x)) . . . f N ( (t N , x))]

Clearly this identies P as the distribution of (., x).


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Exercise 1.3. Using exercise (I, 4.9), show that theorem (1.2 ) continues

to hold if (.) , b(.) C 3( R

n) and (

| |/ x

)(.) and (

| |b/ x

)(.) are

bounded for 1 | | 3. The point is that under these conditions, onecan show that the u f dened by (I, 4.1 ) is in C 1,2([0 , ] Rn) and stillsatises (I, 4.5). Actually, theorem (1.2 ) is true even if (.) and b (.) just satisfy (I, 2.1); however, the proof in this case is quite di ff erent (cf.chapter 5 and 6 of [S & V]).

Exercise 1.4. Let { m(.)}m= 1 and {bm(.)}m= 1 be sequences of co-e fficients

satisfying the hypotheses of theorem (I, 3.3). Assume that m(.) (.)and that b m(.) b(.) uniformly on compact sets where (.) and b (.)again satisfy the hypotheses of (I, 3.3). For m 1 and x Rn let P m

x L

m at x, where L

m is dened for

m(.) and b

m(.); and for x Rn

let P x L at x where L is dened for (.) and b (.). Show that if x m x,then P m xm p x on , where means convergence in the sense of weak convergence of measures. The idea is the following.

In the rst place one can easily check that 35

Sup m 1 E P m xm [| x(t ) x(s)|4 ] C (T )|t s|2 , 0 s t T

for each T > o.Combining this with the fact that { xm}1 is relatively compact, one

can use (I, 2.7 ) to see that for each > 0 there is a compact K such that P m x m(K ) 1 for all m 1. By Prohorovs theorem (cf.Chapter 1 of [S & V]) this means that {P m xm }m= 1 is relatively compact in the weak topology. Finally, one can easily check that if {P m

xm} is any

weakly convergent subsequence of {P m xm } and Pm xm

P, then P L at x;and so P m xm P x.

2 Approximating Di ff usions by Random Evolutions

We start with an example:Let 1 , 2 , . . . , n . . . be independent unit exponential random vari-

ables on some probability space (i.e. P(1 > s1 , . . . , n > sn) = exp

(n

j= 1s j) for all ( s1 , . . . , sn) [0, )n). Dene T o o, and T n =

n

j= 1 j,


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2. Approximating Di ff usions by Random Evolutions 33

n 1; and consider N (t ) = max {n 0 : T n t }. Since P(( n > 1) :

n = o) = o, it is clear that a.s. the path t N (t ) is right continuous,non-decreasing, piecewise constant and jumps by 1 when it jumps.

Lemma 2.1. For 0 s < t and m 0:

P ( N (t ) N (s) = m| ( N (u), u s)) = (t s)m

m! e (t s)a .s. (2.2)

That is, N (t ) N (s) is independent of ( N (u) : 0 u s) and is aPoisson random variable with intensity (t s).

Proof. Note that if t 0 and h > 0, 36

P( N (t + h) N (t ) 1/ ( N (u) : u t ))= P ( N (t + h) N (t ) 1/ (T 1 , . . . , T N (t ), N (t )))= P ( N (t )+ 1 t + h T N (t )/ (T 1 , . . . , T N (t ), N (t )))= P ( N (t )+ 1 t + h T N (t )/ N (t )+ 1 > t T N (t ))= 1 P( N (t )+ 1 > t + h T N (t )/ N (t )+ 1 > t T N (t ))= 1 e h

= h + (h)

Similarly

P ( N (t + h) N (t ) 2/ ( N (u) : u t ))= P ( N (t )+ 1 + N (t )+ 2 t + h T N (t )/ N + 1 > t T N (t ))

=

h

0

e u (1 e (h u))du = (h).

Because of the structure of the paths N (.), if o s < t and un,k

s + k n

(t s)(n 1 and 0 k n), then we now see that P( N (t ) N (s) =

m| ( N (u) : u s))

= limn A { 1,...,n}

| A|= mP ( N (un,k )

N (un,k 1)

= X A(k ), 1

k

n

| ( N (u) : u

s))


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= limn

(nm)(hn + (hn) (hn))m(1 hn (hn))n m

= (t s)m

m! e (t s)

where we have used hn t s

nNext dene (t ) = ( 1) N (t ) , t 0. Clearly (.) is right continuous,37

piece -wise constant and (t ) (t ) implies (t ) = (t ).

Lemma 2.3. If f : R1 { 1, 1} R1 is a function such that f (., ) C 1b ( R

), {1, 1} , then

( f (

t

0 (u)du , (t ))

t

0[ (s) f (

t

0 (u)du , (s))

+ K f (

s

0

(u)du , (s))]ds , F t , p)

is a martingale, where F t = ( (u) : 0 u t ), f ( x, ) = d dx

f ( x, )and K f ( x, ) = f ( x, ) f ( x, ).

Proof. If we can prove the result when f does not depend on x, then thegeneral case follows immediately by the argument given in the secondstep of the proof of theorem (II, 4.2). But from (2.2) it is easy to see that

d dt

E [ f ( (t )) |F s = E [K f ( (t )) |F s , 0 s t ,

and so the result holds for f not depending on x.Consider the process

x (t ) = t

0

(s)ds , (2.4)

where > 0, and dene

X (t ) = x (t ) +

2 (t ). (2.5)


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Lemma 2.6. If C 0 ( R1) , then there is an F . progressively measur-

able process L ,(.) such that | L,(.)| < C ||||C 3b ( R1) and

(( x (t )) 2

2

t

0

( x (s))ds 3t

0

L, (s)ds, F t , P

is a martingale. 38

Proof. By (2.3) with f ( x, ) = ( x + 2

):

(( X (t ))

t

0 (s)( X (s))ds

t

0

( ( X (s)) (s))( X (s)))ds , F t , P)

is a martingale. By Taylors theorem

( X (s) (s)) ( X (s))

= (s) ( X (s)) + 2

2 ( X (s))

2

6 ( X (s) (s) (s)

where 0 < (s) < . Thus we can take

L, = 16

( X (s) (s) (s)).

We now see that for C 0( R1):

(( X ( t 2

)) 12

t

0

( X ( 22

))ds t

0

L,( s2

)ds , F t / 2 , P)

is a martingale. Thus if we let p

be the distribution on

(n =

1) of X (./ 2), then, since | X (./ 2 ) X (./ 2)| / 2, it is reasonable to


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suppose that P as 0, where 1

2

d 2

dx2 at 0 (i.e. is 1-

dimensional Wiener measure), in the weak topology. If we know that{P : > 0} were compact, then this convergence would follow imme-diately from the observation that if n 0 and P n P , then

(( x(t )) 12

t

0

( x(s))ds , t , P)

is a martingale for all C 0 ( R1).

In order to handle the compactness question, we state the following39theorem.

Theorem 2.7. Let {P : > 0} be a family of probability measureson ( , ) and let A () , C 0 ( R

n) , be a non-negative number such that A() < C ||||C k ( Rn) for some C < and k 0. Assume that for each >0 there is a M.-progressively measurable function X : [0, ) Rn

such that

(i) lim R

Sup > 0 P (| X (0) | R) = 0

(ii) X (.) is right continuous and has left limits (P a.s.),

(iii) lim

P (Sup 0 t T | x(t ) X (t ) | ) = 0 for each T > 0 and > 0 ,

(iv) (( X (t )) + A()t , t , P ) is a submartingale for each > 0.

If {n}1 is any sequence of positive numbers tending to zero, then{P n : n 1} has a weakly convergent sub-sequence.

Comment on the proof: We will not give the proof here because itis somewhat involved. However, in the case when X (.) = x(.) for all > 0, the proof is given in Chapter 1 of [S &V ] (cf. Theorem 1.4.6) and

the proof of the general case can be easily accomplished by modifyingthe lemma 1.4.1 given there. The modication is the following.


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Lemma. Let t 1 and t 2 be any pair of points in [0, T]. such that |t 2 t 1 |


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+ K f ( A(s), (s))]ds , F t P) (2.8)

is a martingale; where

K f ( x, ) = S d 1

( f ( x, ) f ( x, ))d

and d denotes normalized uniform measure on S d 1

Lemma 2.9. For each > 0 , let (.) be a right continuous R N -valued,F.-progressively measurable function such that

| (.)| ()

where () 0 as 0. Set

A(t ) =t

0

(s)ds .

Then for any h C 1,0( R N S d 1 ), T > 0 and > 0:

lim0

P( Sup0 t T |t /

0

(h( A (s), (s)) h( A (s)))ds |) = 0

where h( x) S d 1 h( x, )d .Proof. Clearly we may assume that h(0 , 0) = 0. Set

(t ) = t

0

(h( A (s), (s)) h( A (s)))ds

and (t ) = (t ) + h( A (t ), (t )).

Clearly it su ffi ces to prove that42

Sup 0 t T | (t / )| 0 in probability, as 0.


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43

Thus it su ffi ces to show that E [2( (T / ))] 0 as 0. Finally,

|2 ( (T / )) M 2,2 (T / )| C (2)( + ())T ,

and so E [2 ( (T / )) C (2 )( + ())T ,

since E [ M ,2 (T / )] = E [ M ,2 (0)] = 0.

Theorem 2.10. Let F : Rn

S d 1

Rn

be a smooth bounded functionhaving bounded derivatives and satisfying

S d 1

F ( x, )d = 0 for all x Rn.

Let b : Rn Rn be a smooth bounded function having bounded derivatives. Dene

a i j ( x) = S d 1

S d 1

(F i( x, ) F i( x, ))(F j( x, ) F j( x, ))d d

and

ci( x) = S d 1

n

j= 1

F j( x, )F i

x j( x, )d ;

and dene x (., x) by

x (t , x) = x + t

0

F ( x (s, x), (s))ds + 2t

0

b( x (s, x), (s))ds , t 0,

for > 0 and x Rn

. Let P () x on ( , ) be the distribution of x (./ 2) and set 44


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+ 3 R,( x, )

where 45| R,( x, )| C ||||C 3b ( Rn)

Thus if

a i j ( x, ) = S d 1

(F i( x, )F i( x, ))(F j( x, ) F j( x, ))d

and

i( x, ) = bi( x) +n

j= 1

(F jF i

x j)( x, ),

then, since

F ( x, )d = 0,

( X (t )) 2t

0

< ( x (s), (s)), grad x( X (s)) > d s

2

2

t

0

n

i, j=

1

a i j ( x (s), (s)) 2 xi x j

( X (s))ds

3t

0

L,( x(s), (s))ds

is an F .- martingale with respect to P , where46

| L,( x, )| C ||||C 3b ( Rn )

By (2.7) this proves that {P : > 0} is relatively compact. Also, itshows that there is an L,( x, ) satisfying

| L,( x, )| C ||||C 3b ( Rn )


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such that

( x (t )) 2t

0

< ( x (s), (s)), grad x( X (s)) > d s

2

2

t

0

n

i, j= 1

a i j ( x (s), (s)) 2 xi x j

( x (s))ds

3t

0

L,( x (s), (s))ds + (( X (t )) ( x (t )))

is an F -martingale with respect to P . Since|( X (t )) ( x (t )) | C

and, by ( 2.8 ), we know that

Sup 0 t T |2

t / 2

0

[ L( x (s)) < ( x (s)), (s)), grad x( x(s)) >

12

n

i, j= 1

a i j ( x (s), (s)) 2 xi x j

( x(s))]ds |

0 in probability, it is now clear that n

0 and P n P, then P Lat 0.

Corollary 2.11. Let : Rn Rn Rd and b : R N Rn be bounded smooth functions having bounded derivatives of all orders and dene

L = 12

d

k = 1

(n

i= 1

ik ( x) xi

)2 +n

i= 1

bi( x) xi

. (2.12)

For > 0 and x Rn , let x (., x) be the process determined by

x(t , x) = x +d 2

1/ 2

t

0

( x (s, x)) (s)ds + 2t

0

b( x (s, x))ds , t 0,

(2.13)


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and denote by P x the distribution on ( ,) of x (./ 2 , x). Then as 0, P xconverges weakly to the unique P x L at x. In particular, if S ( x; , b)denotes the set of paths (.) of the form

(t ) = x +t

0

((s))(s)ds +t

0

b((s))ds, (2.14)

where : [0, ] Rd is a locally bounded, right continuous function47 possessing left limits, then

Supp (P x) S ( x; , b)

Proof. The second part follows immediately from the rst, since (2.13 )shows explicitly that

Supp( P x) S ( x; , b)

for each > 0.To prove the convergence result, rst observe that, by (1.2), the mar-

tingale problem for L well-posed. Next, dene

F i( x, ) = (d / 2)1/ 2d

k = 1

ik ( x) k , 1 i n.

Then F ( x, ) satises the hypotheses of theorem ( 2.10 ). Moreover,

S d 1

(F i( x, ) F i( x, ))(F j( x, ) F j( x, ))d

= d 2

k ,

ik ( x) j ( x)

S d 1

( k k )( )d

= d 2

k ,

ik ( x) j ( x)( k +

1d

k , ).

Thus in the notation of ( 2.10 ):

a i j ( x) =

i ( x) j ( x).


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Next:48

S d 1

n

j= 1

F i( x, )F i

x j( x, )d

= d 2

n

j= 1

d

k , = 1

jk ( x) i x j

( x) S d 1

k d

= 12

n

j= 1

d

k = 1

jk ( x) ik x j

( x)

Thus in the notation of ( 2.10 )

ci( x) = 12

d

k = 1

n

j= 1

jk ( x) ik x j

( x)

But this means that

12

n

i, j= 1

a i j ( x) 2

xi x j+

n

i= 1

ci( x) xi

= 12

d

k = 1

(n

i= 1

jk ( x) x j

)2,

and so the L is given in ( 2.12 ) is the one associated with F ( x, ) andb( x, ) via the prescription given in ( 2.10 ).

Exercise 2.15. Let (.) and b (.) be as in the preceding and suppose that (.) and b(.) are a second pair of such functions. Assume that Range( ( x)) Range ( ( x)) for all x Rn and that b( x) b( x) Range( ( x)) for all x Rn. Show that

S ( x; , b) S ( x; , b)

for all x Rn

Remark 2.16. An alternative description of S ( x; , b) can be obtained

as follows:Let 49


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X k =n

i= 1

ik ( x)

xi, 1 < k < d ,

and denote by Lie ( X 1 , . . . , X d ) the Lie algebra of vector elds generatedby { X 1, . . . , X d }. Then S ( x; , b) is the closure of the integral curves,starting at x, of vector elds Z + Y , where Z Lie { X 1 , . . . , X d } and Y =

n

i= 1bi( x)

xi

. The derivation of this identication rests on the elementary

facts about how integral curves change under the Lie bracket operationand linear combinations.

3 Characterization of Supp ( p x), the non-degenerate

caseIn Corollary ( 2.11 ), we showed that if L is given by ( 2.12 ) and P x Lat x, then Supp( P x) S ( x; , b). In the case when (.) > 0, thisinclusion does not provide any information. What we will show in thepresent section is that, in fact, if (.) > 0, then Supp ( P x) = x {w : x(0 , w) = 0}. Actually, we are going to derive a somewhatmore general result.

Lemma 3.1. Given > 0 , dene

u (t , x) =

n=

[ t ( x y 4n) t ( x y (4n + 2))]dy (3.2)

for t > 0 and x ( , ) , where

t ( z) = 1

(2t )1/ 2e z

2 / 2t .

Then

u C ((0 , ) R1),

u t

= 12

2u x2

in(0, ) R1,

limt 0

u (t , x) = 1 for x ( , ), u(t , x) > 0 for (t , x) (o, ) ( , ) ,50

u(t , ) = 0 for t > 0 and 2udx 2

(t , x) 0 for (t , x) (0, ) ( , ).


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3. Characterization of Supp ( p x), the non-degenerate case 47

Finally, if ( (t ), F t , P) is a one dimensional Brownian motion, T > 0 and

x ( , ) , then

u (T , x) = P ( x + (t )) ( , ) for t [0 , T ]). (3.3)

Proof. LetS = U ((4n 1), (4n + 1))

andS = 2 + S = {2 + z : z S }.

Then

u(t , x) = S

t ( x y)dy S

t ( x y)dy.

This proves thatu C ((0 , ) R1)

and thatut

= 12

2u x2

for t > 0. Also, if x ( , ), then x S and x S .Hence

limt 0

S

t ( x y)dy = 1 and limt 0

S

t ( x y)dy = 0.

Thus

lim0

S

u (t , x) = 1 for x ( , ).

Next observe that 51

u (t , ) = S +

t ( y)dy S +

t ( y)dy = 0.

Since u (t , .) is clearly even, this shown that u(t , ) = 0 for t > 0.We next prove ( 3.3 ). To this end, let

x = inf {t 0 : x + (t ) ( , )}.


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Then, by It o s formula applied to u (T t , x):

u (T , x) = E [u (T x T , x + ( x T )= P ( x > T ) = P ( x + (t ) ( , ) for t [0, T ]).

From ( 3.3) it is clear that

ut

(t , x) 0,

and so2u x2

= 2ut

0.

Finally, we must show that u (t , x) > 0 for all t > 0 and x ( , ).To this end, Suppose that xo ( , ) and that u (T , xo ) = 0 for someT > 0. Then we would have that

E [e z xo ]

is an entire function of z C . On the other hand a given 0 < 0 dene

= inf {t : |(t ) ( )| }.

If b(s) 0 for T s T and a (s) A for T s T then

P( T ) (1 u ( AT , 0))P ( T ).

Proof. Let F : [0, ) R1 E R1 have the properties that foreach ( t , x) [0, ) R1 F (t , x) is F t measurable and for each q E ,F (t , x, q) C 1,2b ([0, ) R

1 . Then by Doobs stopping time theorem plus 53elementary properties of conditional expectations (cf 1.5.7 in [ S &V ]),

E [F (T , (T )) F ( T , ( T )) |F ]

= E [

T

T (F

s

+ 1

2

a(s)2 F

x2 )( s, (s)ds |F ]

(a.s., P ) on { < }In particular, with F (t , x) = u ( A(T T t ), x ( ))( 0 if = ),

we have

P( > T , T ) = E [u ( A(T T ), ( T ) ( )), T ]= E [F (T , (T )), T ]= E [F ( T , ( T )), T ]

+ E [

T

T (F

s +

1

2a(s)

2 F

x (s, (s))ds, T ].


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since F ( T , ( T )) = u ( A(T ), 0) u ( AT , 0) on { T } and

(F s

+ a(s)

22 F x2

)( s, (s))

= ( Aus

+ a(s)

2u x2

)( A(T s), (s) ( ))

= 12

( A a(s))2u x2

( A(T s), (s) ( )) 0

for T s T , we see that

P ( > T , T ) u ( AT , 0)P ( T ).

Hence

P( T ) = P ( T , T )= P ( T ) P( T , > T )< (1 u ( AT , 0))P ( T ).

54

Lemma 3.5. Let ( (t ), F t , P) be a d-dimensional brownian motion, (.)a bound F.-progressively measurable R n Rd valued function, b (.)a bounded F.-progressively measurable R n- valued function and c (.) abound F .-progressively measurable R d -valued function such that c (t ) 0 for t T. Dene

R = exp[

0

< c(s), d (s) > 12

0

|c(s)|2 ds ]. (3.6)

Then R > 0 (a.s., P) E p[ R] = 1 and if dQ = RdP, then for every f C 1,2b ([0, ), R

n )

( f (t , (t )))

t

0 ( f s + Ls f + < (s)c(s), grad f >)( s, (s))ds , F t , Q)


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is a martingale, where

Ls 12

n

i, j= 1

( (s))i j 2

xi x j+

n

i= 1

bi(s) xi

.

Proof. It is clear that R > 0 (a.s., P ). Furthermore, by It os formula, if

R(t ) = exp[

t

0

< c(s), d (s) > 12

t

0

|c(s)|2 ds ].

then

R(t ) = 1 +t

0 R(s) < c(s), d (s) >, t 0. (3.7)

Thus E P [ R] = E P [ R(T )] = 1. Finally, another application of It os 55formula shows that

( R(t ) f (t , (t )) t

0

( R(s)( f s

+ Ls f + < (s)c(s), grad f >)( s, (s))ds , F t , P )

is a martingale. Since, from ( 3.7)

E P [ R|F t ] = R(t ) (a , s., P)

for all t 0, it follows immediately that

( f (t , (t )) t

0

( f s

+ Ls f + < (s)c(s), grad f >)( s, (s))ds , F t , Q)

is a martingale.

Theorem 3.8. Let ( (t ), F t , P) be a d-dimensional Brownian motion and let (.) be a bounded F.-progressively measurable R n Rd -valued func-tion and (.) a bounded F.-progressively measurable R d -valued func-tion. Set

(t ) =

t

0 (s)d (s) +t

0 (s) (s)ds


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and assume that

Trace (s)|(s) (s)|

[, )(| (s)|)( 0 if (s) = 0)

is uniformly bounded for each > 0. Then for each T > 0 and > 0 ,

P (Sup 0 t T | (t )| < ) > 0.

Proof. Set (t ) = | (t )|2 and dene

= inf {t 0 : |(t )| 2}

and56 = inf {t 0 : |(t ) ( )| }.

Then

P(Sup 0 t T | (t )| (3)1/ 2 ) = P (Sup 0 t T |(t )| 3) P ( T ).

Thus we must show that P ( T ) < 1. Without loss of generality,we will assume that (S ) 0 if

| (s)| (3)1/ 2 .

To prove that P ( T ) < 1, rst note that

(t ) = 2

t

0

< (s) (s), d (s) > +t

0

Trace (s)ds

+ 2

t

0

< (s) (s), (s) > d s

Now dene

c(s) = (s) 12

( Trace (s)| (s) (s)|2

(s) (s)) [0 ,T ](s) [ / 2, ]((s)).


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Then c(.) is uniformly bounded and c(s) 0 for s T . Set

R = exp[

0

< c(s), d (s) > 12

0

|c(s)|2 ds ]

and dQ = RdP . Since Q and P are mutually absolutely continuous

Q( T ) < 1 iff P( T ) < 1.

But, by ( 3.5 ),

( f (t , (t ))

t

0 ( f s +

12 a(s)

2 f x2 + b(s)

f x ) (s, (s)ds , F t , Q)

is a martingale for all f C 1,2b ([0, ) R1), where 57

a(s) = 4| (s) (s)|2

and b(s) = Trace (s) + 2 < (s), c(s) > + < (s) (s), (s) >. Inparticular, a(.) and b(.) are bounded and b(s) = 0 for T s T .Thus, by ( 3.4) , Q( T ) < 1.

Corollary 3.9. Let ( (t ), F t , P) be a d-dimensional Brownian motion,(.) a bounded F . progressively measurable R n Rd -valued functionand v (.) a bounded F . progressively measurable R n-valued function. Assume that (.) I for some > 0. set

(t ) = x +t

0

(s) d (s) +t

0

v(s)ds, t 0.

Then for every C 1b [0, ), Rn) satisfying (0) = x, every > 0

and every T > 0:

P (Sup 0 t T | (t ) (t )| < ) > 0.


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Proof. Given (.), set

(t ) = (t ) (t )Then

(t ) =t

0

(s)d (s) +t

0

(s) (s)ds

where (s) = (s) ( (s)) 1(v(s) (s)).

Note that58

| (s) (s)|2 = < (s), (s) (s) > | (s)|2,

and soTrace (s)| (s) (s)|

[, ](| (s)|)

is uniformly bounded for each > 0.Thus, by theorem ( 3.8 ),

P (Sup 0 t T | (t )| < ) > 0

for each > 0 and T > 0.

Corollary 3.10. Let : Rn Rn Rd and b : Rn Rn be C b -functions

and set L =

12

n

i, j= 1

( )i, j( x) 2

xi x j+

n

i= 1

bi( x) xi

.

Assume that ( x) > 0 for each x Rn , let P x L at x. Then

Sup p(P X ) = { C ([0 , ), Rn) : (0) = X }

Proof. As we already know, P x is the distribution of the process (., x)given by

(t , x) = x +

t

0 ( (s, x)) s (s) +t

0 b( (s, x))ds


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where ( (t ), F t , P) is any d -dimensional Brownian motion. In particular,

we may assume that

(, ) = ( 1(.), . . . , d (.))

where ( (t ), F t , P) is a (d + n) dimensional Brownian motion. Now let 59 C 1b ([0 , ), R

n) with (0) = x be given. Set

K = Sup 0 t T |(t )| + 2

and dene

(t ) = x +t

0 ( (s, x))d (s) +

t

0b( (s, x))ds , t 0.

where : Rn Rn Rd + n is given by

( y) = ( ( y), [K , )( y) I ).

Then (t ) = (t , x) for 0 t K inf {s 0 : | (s, x)| K }. Inparticular

P (Sup 0 T )= P (Sup 0 t T | (t ) (t )| < ).

But ( y) = ( y) + (K , )(Y ) I I where

= inf {< , (Y ) > / | |2 : |Y | K and Rn \{0}} > 0}

Thus, by corollary ( 3.9) ,

P (Sup 0 t T | (t ) (t )| < ) > 0.

We have therefore proved that

{ C 1b ([0 , ) : Rn); (0) = X } Sup( P x).

60

Since Sup( P X ) is closed in , this completes the proof.


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Corollary 3.11 (The Strong Maximum Principle) . Let L be as in (3.10 )

and let G be an open subset of R1 R

n. Suppose that u C

1,2(G) satises

ut

+ Lu 0

in G and that (t 0 , x) G has the property that u (t 0 , x) u(t , x) for all(t , x) G. Then u(t 1 , x1) = u(t 0 , x) for all (t 1 , x1) G(t 0 , x0). whereG(t 0 , x0) is the closure of the points (t 1 , (t 1 t 0 )) such that t 1 t 0 , C ([0 , ) : Rn) , (0) = x0 and (t , (t t 0) G for t 0 t t 1 . In particular, if U is a connected open set in R n and u C 2(U ) satises Lu 0 in U , then u is constant in U if u attains its maximum in U .

Proof. By replacing G by {(t t 0 , x) : (t , x) G}and u(t , x) by u(t t 0 , x).We will assume that t 0 = 0. Furthermore, by approximating G frominside with relatively compact open regions, we will assume that u C 1,2b (G).

Note that by It os formula and Doobs stopping time theorem E P

0 x[u(t , (t ))] u(0, x0 )

= E P0 x[

t

0

(us

+ Lu)( s, x(s))ds] 0,

where P x0 L at x0

and = inf {t 0 : (t , x(t ) G)}Thus61

E P x0 [u(t , X (t )) u(0, x0 )] 0, t 0.

Now suppose that C ([0 , ) : Rn , (0) = x and (t , (t )) G for0 t t 1 . If u(t 1 , (t 1 )) < u(0, x0 ). Then we could nd an > 0 suchthat dist. (( s, (s)), GC ) > for all 0 s t and u(t 1 , x) u(0, x) for | x (t 1 )| < . Thus we would have

E P0 x[u(t 1 , x(t 1 )) u(0, x0 )]

P x0 (Sup 0 t t 1 | x(t ) (t )| < ) < 0,

which is a contradiction


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4. The Support of P x L, the Degenerate case 57

Remark 3.12. The preceding result can be extended in the following

way:Let a = R1 Rn Rn Rn and b : R1 Rn Rn be measurable

functions such that a is symmetric and uniformly positive denite oncompact sets and a and b are bounded on compact sets. Set

Lt = 12

n

i, j= 1a i j (t , x)

2

xi x j+

n

i= 1

bi(t , x) xi

.

If u C 1,2(G) satises

u

t

+ Lu 0

in G and u attains its maximum at ( t 0 , x0) G, then u(t 1 , x1 ) = u(t 0 , x0 ) 62for all ( t 1 , x1) G(t 0 , x). The proof can be constructed along the samelines as we have just used, the only missing ingredient is a more sophis-ticated treatment of the existence theory for solutions to the martingaleproblem (cf [Berk. Symp.] for the details).

4 The Support of P x L, the Degenerate case

We are going to show that if

L = 12

d

= 1( i ( x)

xi

)2 +n

i= 1

bi( x) xi

, (4.1)

and if P x L at x, then

supp( P x) = S ( x; , b). (4.2)

Since we already know that supp ( P x) S ( x; , b), it suffi ces toshow that if

(t ) = x +

t

0 ((s))(s)ds +t

0 b((s))ds , t 0, (4.3)


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where C 20 ((0 , ); Rd ), then

P x(Sup o t T | x(t ) (t ) | ) > 0 (4.4)

for all > 0 and T > 0. Actually, what we are going to prove is slightlymore rened result than (4.4). Namely, suppose that ( (t ), F t , P) is ad dimensional Brownian motion and that

(t , x) = x +t

0

( (s, x))d (s) +t

0

b( (s, x))ds , t 0, (4.5)

where

b( x) = b( x) + 12

n

j= 1

d

= 1 j ( x) x j

( x). (4.6)

63

Then P x is the distribution of (., x). We will show that

lim0

P(Sup 0 t T | | Sup 0 t T | (t ) (t ) | ) = 1 (4.7)

for all > 0 and T > 0. Since

P(Sup 0 t T | (t ) (t ) | ) > 0

for > 0 and T > 0, this will certainly prove ( 4.4 ). The proof of ( 4.7 )relies on a few facts about Brownian motion.

Lemma 4.8. Let ( (t ), F t , P) be a d-dimensional Brownian motion.Then there exist A > 0 and B > 0 , depending on d, such that

P(Sup 0 t T | (t ) |< ) A exp BT 2

, T > 0 and > 0. (4.9)

Proof. First note that if

(T , ) = P (Sup 0 t T | (t ) |< ,

then (T , ) = (T / 2 , 1).


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The reason for this is that for any > 0, (.) has the same distri-

bution as (2, .) (cf. exercise ( 4.11 ) below). Thus we need only check

that (t , 1) Ae Bt , t > 0.

Next observe that

P (Sup 0 t T | (t ) |< ) P (Sup 0 t T max1 d | (t ) |< / d 1/ 2 )

= P (Sup 0 t T | 1 (t ) |< / d 1/ 2 )d

since the (.) s are mutually independent and each has the same distri- 64bution. Thus we will restrict our attention to the case when d = 1.

To prove that (T , 1) Ae Bt when d = 1, we will show that if f C ([ 1, 1])

E [ f ( x + (T )), Sup 0 t T | x + (t ) |< 1]

=

m= 0

am( f )e( m22 / 8)T sin( m/ 2)( x + 1)) , (4.10)

T > 0 and x ( 1, 1),

where

am( f ) =1

0

f ( y) sin( m/ 2)( y + 1))dy/1

0

sin 2((m/ 2)( y + 1))dy.

Given ( 4.10 ), we will have

P (Sup 0 t T | (t ) |< 1) = a 1(1) e( 2 / 8)T +

m= 1

a 2m+ 1(1)( 1)m+ 1e( (2m+ 1)

2 2 / 8)T

Since a1(1) > 0 and | am(1) | 1 for all m, it is clear form this that

limT

e(2 / 8)T P(Sup 0 t T | (t ) |< 1) = a1 (1) > 0,

and so estimate will be established. To prove ( 4.10 ), rst note that

sin m2

( x + 1) : m 1


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is an orthonormal basis in L2 (( 1, 1)) (cf. exercise ( 4.11 ) below).

Next observe that for, f C 0 (( 1, 1)), 65

m= 0

am( f ) sin( m/ 2( x + 1))

is uniformly and absolutely convergent to f and that if u(T , x) is givenby the right hand side of ( 4.10 ) then

uT

= 12

2u x2

, T > 0 and x R1

limT 0

u(T , x) = f , 1 < x < 1

u(T , 1) = 0.

Hence, ( u(T t T , x + (t )), F t , P) is a martingale and so

u(T , x) = E [ f ( x) + (T )), x > T ]

where x = inf {t 0 :| x + (t ) | 1}. Thus (4.10 ) holds for f C 0 (( 1, 1)). Using obvious limit procedure, it is now easy ot see that(4.10 ) continues to hold for all f ([ 1, 1]) so long as T > 0.

Exercise 4.11. Fill the missing details in the preceding proof. In partic-ular, use the martingale problem characterization of Brownian motionto check that (./ 2 ) has the same distribution as (.) for any R/ {0}.Second, show that

{sin(m2

( x + 1) :, m 1}

is an orthogonal basis in L 2 (( 1, 1)) , that

Sup m | am( f ) | c || f || L2(( 1,1)) ,

and that 66

m= 1

am( f ) sin(m2

( x + 1))

is absolutely and uniformly convergent to f if f C 0 (( 1, 1)) . All these facts are easy consequences of the elementary theory of Fourier series.


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Lemma 4.12. Let ( B(t ), F t , P) be a 1-dimensional Brownian motion.

Then P a.s., there is precisely one solution to the equation

(t ) = 2

t

0

| (s) |1/ 2 dB(s) + 2t , t 0.

Moreover, the unique solution (.) is non-negative and B (.)-measu-rable.

Proof. For 0 < 1 and n 1, consider the equation

n(t , )=

+

2

t

0 n( n(s, ))dB(s)+

2t ,

where { n}1 c ( R1), Sup n 1 Sup | x| 1 | n( x) | 1 and n( x) = | x |1

/ 2

for | x | / n. Since n(.) is uniformly Lipschitz continuous, n(., ) isuniformly determined and B(.)-measurable. Moreover, if

n = inf {t 0 : n(t , ) / n},

then n+ 1(t , ) = b(t , ), 0 t n (a.s., P). Next note that n (a.s.,P) as n . To see this, dene

n, R = inf {t 0 :| n(t , ) | R}

for R > . Then

limn

P(n n, R) = 0 for all R > 0

67

Indeed, choose u C 2b ( R1) so that

u( x) = log( nx/ )2 / log( nR/ )2

for / n x R. Then, by It os formula

P (n > n, R) = E [u( n (n n, R))] = u()


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= log( n2 )/ log( nR/ )2 1

as n . Next note that by Doobs inequality, for any T > 0

P Sup 0 t T | n(t , ) 2t | R)

12

E [| n(T , ) 2T |]

c(T )

R

where c(T ) is independent of n. Hence

Sup n 1 P( n, R T ) 0

as R , and so we conclude that

lim P(n T ) = 0

for all T > 0.We have now proved that there is for each > 0, P-a.s. a unique

continuous (., ) such that (t , ) = n(t , ), 0 t n , and that n (a.s., P ) as n . Clearly (., ) is B(.)-measurable and non-negative(a.s., P ). Also

(t , ) = + 2

| ( (s, )) |1/ 2 dB(s) + 2t , t 0.

68

We next show that

P (Sup 0 t T | (t , ) (t ) | ) 0

as , 0 for each T > 0 and > 0. To this end, note that

0 < () = Sup x, y 1| x y|

(| x |1/ 2 | y|1/ 2) C (1/ 2 ), > 0.

Thus we can nd {k } (0, 1) so thatk 1

k 1

2() d = k .


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Choose {k } C 0 ((k , k 1)) so that 0

k (.)

2

k 2(.)

and

0

k ()d = 1.

Set

k () =| |

0

dt

t

0

k (s)ds.

Then, by It os formula:

E [k ( (T , ) (T , ))]

k ( ) +12

E [

T

0

2(| (t , ) (t , ) |)k ( (t , ) (t , ))dt ]

| | + T / k .

Since k ( x) | x | as k , we now see that

E [| (T , ) (T , ) |] | | .

But ( (t , ) (t , ), F t , P) is a martingale; and so, by Doobs in- 69equality

P (Sup 0 t T | (t , ) (t , ) | )

| |

0.

We now see that there exist n 0 such that (., n) (.) (a.s.,P) uniformly on nite intervals. Clearly (.) is B(.)-measurable, P - a.s.non-negative and satises

(t ) = 2

t

0 | (s) |1/ 2

dB(s) + 2t , t 0


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(cf. Exercise ( 4.13) below). Finally, if (t ) were a second process satis-

fying the same equation, then, using the functions k constructed above,we would nd that

(T ) = (T ) (a.s.,P)

for all T > 0.

Exercise 4.13. Show that in fact (., ) (., ) (a.s., P) if 0 < .The idea is to employ a variation on the ideas used prove

P Sup 0 t T | (t , ) (t , ) | 0 as , 0.

Lemma 4.14. Let ( (t ), F t , P) be a d dimensional Brownian motionwhere d 2. Given 1 i j d, dene

Li j(t ) =t

0

i( 2i +

2 j)

1/ 2d j

t

0

i( 2i +

2 j )

1/ 2d i

here we take i

( 2i + 2 j)

1/ 2 =

j( 2i +

2 j)

1/ 2 = 0 if 2i +

2 j

= 0

70

Then L i j (t ), F t , P is a 1-dimensional Brownian motion and the pro-cess L i j (.) is independent of the process 2i (.) +

2 j(.).

In particular, if (.) is a bounded F .-progressively measurable R-valued process, then for M 1 and > 0:

P (Sup 0 t T |

t

0

(u) | (u) | dLi j (u) |> M | Sup 0 t T | (t ) |< )

2 exp M 2

2T || ||2u (4.15)

Proof. Without loss of generality, we will assume that i = 1 and j = 2.Let L(t ) = L12 (t ). To see that ( L(t ), F t , P) is a 1-dimensional Brownianmotion, note that by Itos formula

( f ( L(t ))

t

012 f

( L(s))ds , F t , P)


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is a martingale for all f C 2b ( R1).

(Remember that P (

0 {0}( 21 (s) + 22(s))ds = 0) = 0 since

E

T

{0}( 21 (s) + 22(s))ds =

T

dt {0}

12t

e y2 / 2t dy = 0

for all 0 < < T < .)We next show that L(.) is independent of 21(.) +

22(.). Dene

B(t ) =t

0 1

21 + 22

d 1 +t

0 2

21 + 22

d 2

(with the same convention when 21 + 22

= 0).Again use It os formula to show that 71

( f ( L(t ), B(t )) 12

t

0

f ( L(s), B(s)ds , F t , P)

is a martingale for all f C 2b( R2). Thus (( L(t ), B(t )), F t , P) is a 2-

dimensional Brownian motion, and so L(.) is independent of B(.).Finally, by I t o s formula, if (t ) = 21 (t ) +

22(t ) then

(t ) = 2t

0

( 1(s)d 1 (s) + 2 (s)d 2 (s)) + 2t

= 2

t

0

| (s) |1/ 2 dB(s) + 2t

and so, by ( 4.12 ), (.) is B(.)-measurable. Hence L(.) is independent of 21(.) + 22(.).


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Finally, since L(.) is independent of (.) and they both are indepen-

dent of d

3 2 j(.), we can argue as in (I. 4.3) to prove that

P Sup 0 t T |

t

0

(s) | (s) | dL(s) | | (.) |

2 exp 2

2 || ||2uT

0 | (s) |2

where || ||u denotes the uniform norm.In particular72

P Sup 0 t T |

t

0

(s) | (s) | dL(s) | M Sup 0 t T | (t ) |

2 exp M 2/ 2T || ||2u

Exercise 4.16. Let ( (t ), F t , P) be a 2-dimensional Brownian motion.Prove that for each > 0 , P(| (tV ) |> 0 , t 0) = 1 , where =inf {t 0 : (t ) | }. Next show that P ( 0 as 0) = 1. Concludethat P (| (t ) |> 0 for all t > 0) = 1. Thus P-a.s., (t ) = arg (t ) iswell dened for all t > 0). In order to study (t ) , t > 0 , write z(t ) = 1(t ) + i 2 (t ). Given an analytic function f on C show that d f ( z(t )) = f ( z(t ))dz(t ).

In particular, show that for xed t 0 > 0

log z(t ) log z(t 0 ) =

t

t 01

| z(s) | dB(s) +

t

t 0dL(s)| z(s) | , t t 0,


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76/93


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t

0 k

(u) < ( (u)) grad f ( (u)), d (u) >

t

0

k (u) L f ( (u))du

t

0

[ k (u)( ( (u)) grad f ( (u)) + (u)( ( (u)) grad f ( (u))k ]du .

All except the second term on the right clearly tend to zero as || (.)||0T 75 0. Moreover, the second term is covered by ( 4.18 ) with = 2. Thus

lim0

P(||

.

0

f ( (u))d ( k (u))||0T || (.)||0T < ) = 0

for all > 0 and 1 k , d . This proves our rst assertion upon takingk = .

To prove the second assertion, note that

t

0

f ( (u)) k (u)d (u) = 12 f ( (u))d ( k (u) (u)) + 12

t

0

(u)| (u)|dL k , (u),

where Lk , (.) is as in (4.14 ) and

(.) = ( k (.)2 + (.)2)1/ 2

| (.)| f ( (u)).

Thus the rst term is of the sort just treated and the second one iscovered by ( 4.14) .

Theorem 4.20. Let ( (t ), F t , P) be a d dimensional Brownian motionand let (., x) be given by ( 4.5 ). Given C 20 ((0 , ); R

d ) , dene (.) by(4.3) . Then for each > 0 and T > 0:

lim0

P(Sup 0 t T | (t , x) (t )| Sup 0 t T || (t ) (t )| ) = 1.


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= exp( < (T ), (T ) >

T

0 < (s), (s) > d s 12

T

0 |(s)|2

ds ).

77

Observe that the ratio in the rst factor tends to

exp( (T )|2 +t

0

< (s), (s) > d s + 12

T

0

|(s)|2ds ) as 0

while the ratio in the second factor tends to

exp( |(T )|

2

t

0 < (s), . . .

(s), > d s 1

2

T

0 |(s)|

2

ds ;

and so the products tends to 1.To handle the case when 0 and b depends on ( t , x), let (t ) =

(t , x). Then

i(t ) = xi + i ( (t )) (t )

t

0

(s) i, j( (s)) jk ( (s))d

k (s)

t

0 (s)( L i )( (s))ds

1

2

t

0 i, j( (t ))

j ( (s))ds

+

t

0

bi(s, (s))ds

= xi +t

0

bi(s, (s))ds i(t ),

where

i(t ) =

i ( (t ))

(t )

k

t

0 i, j,

jk ( (s))

(s)d

k (s)


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12

t

0 ik , j,

ik ( (s))d (

k

(s))2

t

0

(s)( Li

)( (s))ds .

We have used in these expressions the convention that repeated in-78dices are summed and the notation

ik , j = ik x j

By Lemma (4.19 ),

P (|| (.)||0T || )(.)||0T ) 0 as 0

for each > 0 and T > 0. Hence

lim0

)P (|| (.) x .

0

b(s, (s))ds ||oT || (.)||0T = 0

But, since |b(t , x) b(t , y)| c| x y|, it is easily seen from this that

lim0

)P(|| (.) (.)||oT || (.)||0T ) = 0,

where

(t ) = x +t

0

b(s, (s))ds , t 0.

Corollary 4.21. Let G be an open set in R Rn and suppose that u C 1,2(G) satises

ut

+ Lu 0

in G, where L is given by ( 4.1 ). If (t 0 , xo) G and u (t 0 , x0) = maxG u,then u (t , x) = u(t 0 , xo) at all points (t , x) G L(t 0 , xo) , where G L(t 0 , xo)is the closure in G of the points (t 1 , (t 1 t 0 )), t 1 t 0 , where S ( xo ; , b) satises (t , (t t 0 )) G for t 0 t t 1. In fact, if

u(t 0 , x) = maxG L(t 0 , xo )

u,

then u u(t 0 , x) on G L(t 0, xo).79


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5. The Most Probable path of a brownian motion with drift 73

Proof. Given ( 4.20 ), the proof is precisely the same as that of ( 3.11 ).

Remark 4.22. W

stoch diff eqns tifr68

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