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Contents

I Stochastic Calculus 1 4

1 Reminder 4

1.1 Metric Spaces, Extension theorem . . . . . . . . . . . 4

1.2 Random experiment, random variables: . . . . . . . . 5

1.3 Stochastic processes: . . . . . . . . . . . . . . . . . . 11

1.4 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 Conditional expectation: . . . . . . . . . . . . . . . . 13

2 Brownian motion 19

2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2 Existence of the Brownian Motion . . . . . . . . . . . 20

2.3 Some properties of the brownian motion . . . . . . . 27

2.4 Quadratic Variation: . . . . . . . . . . . . . . . . . . 31

3 Filtrations and Stopping times 34

3.1 Definitions and Properties . . . . . . . . . . . . . . . 34

3.2 Progressively measurable processes . . . . . . . . . . 36

4 Martingales 39

4.1 Definitions et properties: . . . . . . . . . . . . . . . . 39

4.2 Discrete Time Martingales . . . . . . . . . . . . . . . 41

4.3 Doob’s Inequality and Continuous Time Martingales 44

4.4 Uniform Integrability: . . . . . . . . . . . . . . . . . 50

4.5 Continuous Time Martingales . . . . . . . . . . . . . 53

5 Itˆo’s Integral 56

5.1 The space 2 : . . . . . . . . . . . . . . . . . . . . . 56

5.2 Itoˆ’s integral on H2 . . . . . . . . . . . . . . . . . . . 60

5.3 Semi-martingales and their quadratic variation: . . . 70

5.4 Itoˆ’s integral with respect to a semi-martingale . . . . 76

5.5 Itoˆ’s Formula . . . . . . . . . . . . . . . . . . . . . . 76

II Stochastic Calculus 2 80

6 Local martingales 80

7 a dA when A is of finite variation. 83

0 s s

8 a dM when M is a local martingale 84

0 s s

8.1 The covariation process of two semi-martingales . . . 84

8.2 The quadratic variation process of a local martingale 86

8.3 a dM when M is a martingale . . . . . . . . . . . 89

0 s s

8.4 a dM when M is a local martingale . . . . . . . . 92

R0 s s

9 Integral with respect to a semi-martingale 93

10 Itoˆ’s formula 94

11 First applications of Itoˆ’s formula 96

12 Girsanov’s theorem 99

12.1 Absolute continuity . . . . . . . . . . . . . . . . . . . 99

12.2 Girsanov . . . . . . . . . . . . . . . . . . . . . . . . . 101

12.3 Application to finance . . . . . . . . . . . . . . . . . 103

13 Stochastic di↵erential equations 105

13.1 Polish spaces and conditional probability . . . . . . . 105

13.2 The law of a continuous process . . . . . . . . . . . . 109

13.3 Main definitions for S.D.E. . . . . . . . . . . . . . . . 110

13.4 Pathwise uniqueness implies uniqueness in law . . . . 112

13.5 Tanaka equation . . . . . . . . . . . . . . . . . . . . 114

13.6 SDE with Lipschitz coecients . . . . . . . . . . . . 117

Part I

Stochastic Calculus 1

1 Reminder

1.1 Metric Spaces, Extension theorem

Exercise 1.1 1) Define the following notions: a metric d on a space

E, a convergent sequence, a Cauchy sequence.

2) Prove that a convergent sequence is always a Cauchy sequence.

3) Give an example of a Cauchy sequence in a metric space that

does not converge.

4) A metric space (E,d) is complete iif every Cauchy sequence is

convergent. As an important example, (R, . ) is a complete space

| |

(This results from the so called completeness axiom). Using the

completeness of (R, . ), prove that (Rn, . 1) is a complete space.

| | k k

5) Define A E is a closed set, A is an open set, the closure A of

⇢

A, A is dense in E.

Exercise 1.2 1) Define f : E D is continuous from (E,d ) to

(D,d ).

2)If x is a convergent sequence in E and f : E D is continu-

{ } !

ous, is f(x ) a convergent sequence in D? Why?

{ }

3)Find an example of a Cauchy sequence x and a continuous

{ }

function f such that f(x ) is not a Cauchy sequence?

{ }

4)Definef : E D isuniformlycontinuousfrom(E,d )to(D,d ).

E D

(compare with 1)

5) Prove that if x is a Cauchy sequence in E and f : E D is

{ } !

uniformly continuous, then f(x ) a Cauchy sequence in D?

{ }

6) Define f is a Lipschitz function. Prove that a Lipschitz function

is uniformly continuous.

Exercise 1.3 In a vector space E endowed with a scalar product

.,. , consider a point x and a subspace F. Let ↵ denote ↵ =

h i

inf x f : f F and let a F be such that x a ↵

n n

{k k 2 } { }⇢ k k !

as n .

! 1

1) Prove that a is a Cauchy sequence.

{ }

2) If F is complete (Hilbert space), a converges thus to a limit

{ }

a F such that x a = ↵

2 k k

3) Prove that x a is orthogonal to F, i.e. f F : x a,f = 0.

8 2 h i

The vector a is called the orthogonal projection of x on F.

Exercise 1.4 1) Let A E. Let f : A D. A function f : E

⇢ ! !

D is called an extension of f if it coincides with f on A.

2) Find an example of continuous function f : A D with no

continuous extension.

3) Prove the following Extension theorem:

If A is dense in E, if D is a complete space, if f : A D is

uniformly continuous, then f admits a unique continuous extension

on E.

1.2 Random experiment, random variables:

Arandomexperimentischaracterizedbyaprobabilityspace(⌦, ,P)

where ⌦ is the set of possible outcomes of the experiment, is the

-algebra that represent our level of knowledge of the experiment

and where P is a probability measure.

Definition 1.5 A random variable to (E, ) is a measurable map

X from (⌦, ) to (E, ).

A E

We will often consider real random variables where E = R and

= (The Borel -algebra on R) or random vectors where E = Rd

E BR

and =

BRd

X is called measurable (⌦, ) to (E, ) if and only if for all

A E

B , X 1(B) := ! X(!) B belongs to .

2E { | 2 } A

The random variable X is sending the probability P on(⌦, ) to

a probability P on (E, ):

B : P (B) := P(X 1(B))

8 2E

P is called the law of the r.v. X.

Exercise 1.6 Prove that P is indeed a probability on (E, )

Exercise 1.7 Prove that (X) := X 1(B) B is a -algebra

{ | 2E}

on ⌦. It is the smallest -algebra on ⌦ that makes X measurable.

Exercise 1.8 Prove that if is a family of -algebras sur

⇤

{F } 2

⌦, then is also a -algebra.

⇤

\ 2 F

In particular, if (⌦), we define ( ) as , where

⇤

G⇢P G \ 2 F

is the family of all -algebras that contain .

⇤

{F } 2 G

( ) is therefore the smallest -algebra that contains .

G G

Definition 1.9 A class (⌦) is called monotone iif

G⇢P

1) for all increasing sequence A (i.e. n : A

{ n }n 2N ⇢G 8 n ⇢

A ), we have A .

n+1 [n 2N n 2G

2) for all decreasing sequence A (i.e. n : A

{ n }n 2N ⇢G 8 n+1 ⇢

A ), we have A .

n \n 2N n 2G

Theorem 1.10 (monotone class theorem)

If is an algebra, if is a monotone class and if ,

A M A⇢M

alors ( ) .

A ⇢M

Exercise 1.11 Prove that two probability measures P 1 and P 2 on R

that coincide on intervals are equal on

In particular, show that two random variables with the same cumu-

lative distribution function have the same law.

Definition 1.12 The expectation of a r.v. X 0 is defined as:

E(X) = X(!)dP(!)

Z⌦

If ↵ [1, [, ↵(⌦, ,P) is defined as the space of r.v. X such

2 1 L A

that X ↵ := E( X ↵) < .

kL↵

| | 1

On 1(⌦, ,P), one defines E[X] as

L A

E(X) = E[X+] E[X ],

where X+ := max(X,0) and X := max( X,0)

Exercise 1.13 Let denote the set of (discrete) random variables

(i.e. such that X(⌦) is a set with finitely many elements).

1) If X , define E[X] by E[X] := rP( X = r ). Prove

2D r X(⌦) { }

that E[.] is linear on and uniformly continuous with respect to

X := sup X(!) : ! ⌦ .

k k1 {| | 2 }

2) Prove that E[.] can be extended by continuity to the set of bounded

r. v.

3) Define E[X] for to the set of positive r.v..

Exercise 1.14 Prove that X = 0 is equivalent to X = 0 P-a.s.

L↵

k k

(which means that P(X = 0) = 1)

. fails thus to be a norm on ↵(⌦, ,P). To overcome this

L↵

k k L A

diculty, one defines:

Definition 1.15 L↵(⌦, ,P) is the space of equivalence classes of

P-a.s.

r.v. in ↵(⌦, ,P) for the equivalence relation = .

L A

Theorem 1.16 For all ↵ [1, [, the space L↵(⌦, ,P) endowed

2 1 A

with the norm . is a Banach space.

L↵

k k

L2(⌦, ,P) endowed with the scalar product X,Y := E[X Y]

A h i ·

is an Hilbert space.

Exercise 1.17 Prove that if X L1(⌦, ,P), then

2 A

E(X) = xdP (x) :

ZRd

The expectation of a r.v. X just depends on its law P .

Definition 1.18 If X et Y are r. v. in L2(⌦, ,P), one defines

var[X] := E[(X E[X])2] and cov[X,Y] := E[(X E[X]) (Y

E[Y])]. If Z is a random vector, considered as a column matrix

whose entries are in L2(⌦, ,P), the covariance matrix var[Z] of Z

is defined as

var[Z] := E[(Z E[Z])(Z E[Z]) ].

Exercise 1.19 Prove that if v Rd, then var[ v,Z ] = v >var[Z]v.

2 h i

Infer from the previous result that var[Z] is a semi positive definite

matrix.

Exercise 1.20 (Jensen’s Inequality) Prove that if f : Rd R is

convex and if Y is a d-dimensional random vector, then E[f(Y)]

f(E[Y]).

From the previous result, infer that if ↵> then X

L↵

k k

X . In particular L↵(⌦, ,P) L(⌦, ,P)

k k A ⇢ A

Definition 1.21 A real random variable Z follows a standard nor-

mal law, which is denoted Z (0,1), if its density with respect to

⇠N

Lebesgue’s measure is:

1 z2

f (z) = exp( )

p2⇡ 2

Definition 1.22 A random variable Y is normally distributed with

expectation µ and variance 2 if Y = µ + Z where is standard

normal. This is denoted Y (µ,2). µ and 2 are respectively

⇠N

the expectation and the variance of Y.

Definition 1.23 The characteristic function of a random vector

Z in Rd is defined by: Z : Rd C : ↵ Z(↵) :

! !

(↵) = E(exp(i ↵ Z )).

·h | i

Theorem 1.24 The characteristic function characterizes the law:

if the r. v. X and Y have the same characteristic functions (i.e.

= ), then they have the same law: P = P .

X Y X Y

Exercise 1.25 ProvethatifZ (0,1)then (↵) = exp( ↵2/2).

⇠N

Compute (↵) when Y (µ,2).

⇠N

Definition 1.26 TwoeventsAandB of areindependentifP(A

A \

B) = P(A) P(B). This is denoted A B. Two sub -algebras

· ??

and of are independent, which is denoted , if B

B C A B??C 8 2

, C : B C. Two random variables X and Y on ⌦ are

B 8 2C ??

independent (X Y), if (X) (Y).

?? ??

Exercise 1.27 Prove that X Y if and only if ↵, :

?? 8

(↵,) = (↵) ()

(X,Y) X Y

Exercise 1.28 Prove that if X Y, then cov(X,Y) = 0. Find an

example of random vector (X,Y) such that cov(X,Y) = 0, but where

X and Y are not independent.

Exercise 1.29 Prove that if is a -algebra and if une algebra

B C

that satisfies B , C : B C, then ( ).

8 2B 8 2C ?? B?? C

Exercise 1.30 If Y N(µ ,2 ) and Y N(µ ,2 ) are two

1 1 1 2 2 2

⇠ ⇠

independent r. v., compute where S := Y +Y . Infer from this

S 1 2

that S N(µ +µ ,2 +2 ).

1 2 1 2

⇠

Definition 1.31 A sequence {X

n }n 2N

of random vectors in Rd con-

verges in law to X iif for all bounded continuous function f : Rd

R, we have the convergence of E[f(X n)] to E[f(X)], as n .

! 1

Law

Exercise 1.32 Prove that if X X, and P[X = a] = 0, then

P[X a] P[X a], as n .

 !  ! 1

Theorem 1.33 If X

is a sequence of random vectors in Rd, then

Law

X if and only if 8↵

Rd : Xn(↵)

X(↵).

The previous theorem can be used to prove the central limit the-

orem:

Theorem 1.34 (Central Limit Theorem) If X is a sequence

{ n }n 2N

of random variables in L2(⌦, ,P), independent, identically dis-

tributed, with expectation 0 and variance 1, then

i=1

converges in law to (0,1).

This theorem explains the frequent appearance of the normal law in

nature.

Exercise 1.35 Prove the central limit theorem.

Definition 1.36 A random vector Y in Rd is called gaussian if and

only if v Rd, v,Y is normally distributed.

8 2 h i

Exercise 1.37 ProvethatifY isagaussianvectorsuchwithE(Y) =

µ and with variance matrix var(Y) =⌃ , then (v) = exp[i v,µ

h i

vt⌃v] (The vectors v de Rd is considered in this formula as a col-

umn ematrix.) This indicates that the law of a gaussian vector is

completely determined once one knows µ and ⌃.

1.3 Stochastic processes:

Definition 1.38 A sctochastic process on (⌦, ,P), indexed by a

time set T R is a family X t t T of r.v. on (⌦, ,P) with values

⇢ { } 2 A

in the state space (E, ) of the process.

X is the state of the process at time t.

At fixed !, the map t X (!) is called the trajectory or the

sample path of the process corresponding to !.

We next define the following equivalence relations between pro-

cesses:

Definition 1.39 The processes X and Y on (⌦, ,P) are modifi-

modif

cations of each other (notation Y Y) if:

⌘

t 0,X (!) = Y (!) P a.s.

t t

Definition 1.40 The processes X and Y on (⌦, ,P) are indistin-

guishable if and only if the event:

! 0 t < , X (!) = Y (!)

t t

{ |8  1 }

is measurable and has a probability equal to 1.

Definition 1.41 The processes X and Y on (⌦, ,P) have the

same finite-dimensional laws iif :

0 t < t < ... < t < ,(X ,...,X ) and (Y ,...,Y )

8 

1 2 n

t1 tn t1 tn

have the same law.

Exercise 1.42 Prove that if X and Y are indistinguishable, then

they are modifications of each other.

Theconverseclaimisfalseasindicatesthefollowingexample: Let

⌦ = [0,1] and let P be the Lebesgue measure. Define, !,t [0,1]:

8 2

1 if t = !,

X (!) :=

( 0 otherwise.

and Y (!) := 0.

Then we have: P( ! X (!) = Y (!) )=P( t ) = 0 thus X = Y

t t t t

{ | 6 } { }

P-a.s. so X and Y are modifications of each other. But,

! t [0,1] : X (!) = Y (!) = ! t [0,1] : t = ! =

t t

{ |8 2 } { |8 2 6 } ;

Therefore P( ! t : X (!) = Y (!) ) = 0 = 1, and the processes X

t t

{ |8 } 6

and Y are not indistinguishable.

Exercise 1.43 Prove that if two processes have continuous trajec-

modif

tories, then X Y implies that X and Y are indistinguishable.

⌘

Exercise 1.44 Prove that if two processes are modifications of each

other then the have the same finite-dimensional laws.

1.4 Filtrations

Definition 1.45 A filtration is an incraesing family of sub-

t t T

{F } 2

-algebras of , which means that 0 s < t < .

s t

A 8  1F ⇢F ⇢A

represents the level of knowledge of the process at time t, i.e.

is the class of all events that one can identify as realized or not

at time t, based on the current information.

Definition 1.46 If is a filtration, the we denote:

• =( )

F1 t 0Ft

• = S

Ft+ ✏>0Ft+✏

• =( ).

Definition 1.47 A filtration is right-continuous (respectively

{F }

left-continuous) if = (resp. = ) t 0.

t t+ t t

F F F F 8

Definition 1.48 A filtration is complete if A such that

{F } 8 2F 1

P(A) = 0, A .

Definition 1.49 The process X is adapted to the filtration

t t 0

{F }

if and only if 0 t < , X is -measurable.

t t

8  1 F

Exercise 1.50 Let be a sub--algebra of and let be the class

F A N

of all sets in with p(A) = 0. := ( ) is called completion

A F F[N

¯ ¯

of . Prove that all -mesurable r.v. X is P-a.s. equal to an

F F

-mesurable r.v. X.

The filtration is called la completion of . Prove

t t T t t T

{F } 2 {F } 2

that any modification X of an -adapted process X is adapted to

0 t

t t T

{F } 2

1.5 Conditional expectation:

Exercise 1.51 If X is a r.v. in L2(⌦, ,P)), prove that C := E[X]

is the constant C that best approximates X in L2(⌦, ,P)).

A sub--algebra of represents a level of knowledge of the

B A

random experiment: At this level of information, one can decides of

anyeventB ifithasbeenrealizedornot. Onecanalsocompute

the value of any -measurable function. Extending the previous

exercise, it is then natural to define the conditional expectation as

follows:

Definition 1.52 Let and X L2( ). The random variable

B⇢A 2 A

X in L2( ) that best approximates X in L2 is denoted E[X ]: it

B |B

is called the conditional expectation of X given .

E[X ] is therefore the orthogonal projection of X on L2( ) .

|B B

Since L2( ) is an Hilbert space, this projection exists and is unique.

E[X ] is thus well defined. Remark E[X ] is defined in L2, as an

|B |B

P-a.s.

equivalence class of r.v. in 2 for the equivalence relation = .

The above uniqueness in L2 means that E[X ] is defined up to a

set of 0 probability.

The orthogonality relations are:

Z L2( ) : E[ZX] = E[ZE[X ]].

8 2 B |B

Exercise 1.53 Prove that, if X Y P a.s., then E[X ]

E[Y ] P a.s..

Exercise 1.54 If U L2( ), prove that U = E[X ] if and only if

2 B |B

B : E[U1 ] = E[X1 ].

B B

8 2B

Exercise 1.55 Prove the following theorem.

Theorem 1.56 1. ↵, RE[↵X+Y ]=↵E[X ]+E[X ].

8 2 |B |B |B

2. If , then E[X ] = E[E[X ] ].

B⇢C⇢A |B |C |B

3. If X is independent of , then E[X ] = E[X].

B |B

4. Z L2( ):

8 2 B

E[(X Z)2] = E[(X E[X ])2]+E[(E[X ] Z)2].

|B |B

5. Y L ( ): E[YX ] = YE[X ].

8 2 B |B |B

6. Let : R R be a convex function such that (X) L2( ),

! 2 A

then:

E[(X) ] (E[X] ]) a.s..

|B B

Proof:

1. Follows from the fact that an orthogonal projection on L2( )

is a linear operator.

2. Let Z L2( ), then E[ZE[X ]] = E[ZX]. Since ,

2 B |B B⇢C

one also have E[ZX] = E[ZE[X ]]. Therefore, Z L2( ),

|C 8 2 B

E[ZE[X B]] = E[ZE[X ]]. Since, by definition E[X ]

| |C |B 2

L2( ),thelastrelationistheorthogonalityrelationthatdefines

E[E[X ] ]. Hence E[X ] = E[E[X ] ].

|C |B |B |C |B

3. Since X independent of , Z L2( ) E[ZX] = E[X]E[Z] =

B 8 2 B

E[E[X]Z], therefore E[X ] = E[X]. Indeed, the constant

r.v. Y := E[X] is in L2( ) and satisfies for all Z L2( )

B 2 B

E[ZX] = E[ZY]. Y is therefore the conditional expectation

E[X ].

4. E[(X Z)2] = E[(X E[X ]+E[X ] Z)2]

|B |B

= E[(X E[X ])2]+E[(E[X ] Z)2]

|B |B

+2E[(X E[X ])(E[X ] Z)]

|B |B

But: (E[X ] Z) L2( ). It follows

|B 2 B

E[(X E[X ])(E[X ] Z)] = 0,

|B |B

and the result is proved.

5. Set H = YE[X ]. Since Y L ( ) and E[X ] L2( ), we

|B 2 B |B 2 B

have: H L2( ). Therefore, if Z L2( ), we get:

2 B 2 B

E[ZH] = E[ZYE[X ]] = E[Z(YX)] ( since ZY L2( )).

|B 2 B

Thus H = E[YX ] = YE[X ].

|B |B

6. We limit our proof to the case where is di↵erentiable. Since

is convex, we get y,z : (z) (y)+(y)(z y)

Therefore, !, one has :

(X(!)) (E[X ](!))+(E[X ](!))(X(!) E[X ](!))

|B |B |B

Taking the conditional expectation, we obtain then:

E[(X) ] E[((E[X ])+(E[X ])(X E[X ]) ]

|B |B |B |B |B

= (E[X ])+(E[X ])E[(X E[X ]) ]

|B |B |B |B

ButE[(X E[X ]) B] = E[X ] E[E[X ] ] = 0,andthus

|B | |B |B |B

E[(X) B] [E[X ]] P a.s.

| |B

Exercise 1.57 On the probability space (⌦, ,P), we consider the

class

:= B P(B) = 0 or P(B) = 1 .

B { 2A| }

1. Prove that is a -algebra.

2. Characterize the -mesurables functions.

3. For X L2(⌦, ,P), compute E[X ].

2 A |B

Exercise 1.58 If (X,Y) is a random vector whose density f

(X,Y)

satisfies x,y R : f (X,Y)(x,y) > 0, if := (Y);

8 2 B

1. Prove that E[X ] = g(Y), where

xf (x,y)dx

(X,Y)

g : R R : y g(y) := R

! ! f (x,y)dx

R R (X,Y)

2. If (X,Y) is a gaussian vector such that E[X] = E[Y] = 0,

var[X] = 1 = var[Y] and cov[X,Y] = ⇢ [0,1], compute

E[X ]. (see also exercise 1.62).

Exercise 1.59 If and X L2(⌦, ,P)

B⇢C⇢A 2 A

1. Prove that E[X ] = E[(E[X ]) ].

|B |C |B

2. Prove that the previous relation is not satisfied in general when

is not included in , considering := (X) and := (Y)

B C B C

in point 2) of the previous exercise.

Exercise 1.60 The law of a random vector X = (X ,...,X ) is

1 n

called exchangeable if for all permutation ⇡ of 1,...,n , the vectors

{ }

X and X have the same law, where X := (X ,...,X ).

⇡ ⇡ ⇡(1) ⇡(n)

If X has an exchangeable law and S := X +...+X , compute

1 n

E[X S] (E[X S] is a shorthand notation for E[X (S)]).

1 1 1

| | |

Exercise 1.61 If Z ,Z ,... is an i.i.d. sequence of (0,1) r.v., if

1 2

:= (Z ,...,Z ) and if X := Z +...+Z ,

n 1 n n 1 n

1. Prove that n m : E[X ] = X .

n m m

8 |F

2. Prove that n m : E[Y ] = Y , where Y := X2 n.

8 n |Fm m n n

3. Prove that n m : E[M ] = M , where M := exp(X

n m m n n

8 |F

n/2).

(As we shall see later, the previous relations indicates that the pro-

cesses X, Y and M are martingales.)

Exercise 1.62 A random vector X = (X ,...,X ) is called gaus-

1 n

sian if, v Rn, the random variable v,X is normally distributed.

8 2 h i

Let X de a gaussian vector with expectation µ and variance matrix

⌃.

1. ComputethecharacteristicfunctionofX: pour↵ Rn, X(↵) :=

E[exp(i ↵,X )].

h i

2. Assume next that n = m + l. The vector X can the be seen

as X = (U,V). Prove that U V if and only if cov(U ,V ) =

i j

0, i 1,...,m , j 1,...,l .

8 2{ } 8 2{ }

(Hint: one can use the following result: The random vectors

U and V are independent if and only if u Rm : v Rl :

8 2 8 2

(u,v) = (u) (v).)

(U,V) U V

3. Let be the vector space spanned by the random variables

X ,...,X and the constant 1. Let denote the -algebra

1 n 1

(X ,...,X ). Finally, let Y be the orthogonal projection

1 n 1

of X on . Prove that the vector (X ,...,X ,X Y) is

n 1 n 1 n

gaussian, and that X Y . Conclude that E[X ] = Y.

n n

?? B |B

4. With the previous result, compute E[X ] in exercise 1.58.2.

The conditional expectation E[X ] was defined above for ran-

dom variables X in L2(⌦, ,P). The well known Radon Nikodym’s

theorem is used in the following exercise to define the conditional

expectation of random variables X in L1(⌦, ,P). This theorem

states that if µ is a bounded signed measure and P a probability

onr (⌦, ), then the following claims are equivalent:

1) µ has a density Y with respect to P (i.e. Y L1(⌦, ,P) :

9 2 B

U L1(⌦, ,µ) : Udµ = E [U.Y])

8 2 B ⌦ P

2) µ is absolutely continuous with respect to P(i.e. B :

8 2B

P(B) = 0 µ(B) = 0.)

)

Exercise 1.63 Let X L1(⌦, ,P) and let be a sub--algebra of

2 A B

. Prove that the measure µ on (⌦, ) defined by µ(B) := E [X 1 ]

P B

A B ·

is absolutely continuous with respect to P. Prove that the density

Y of µ with respect to P, that exists according to Radon Nikodym’s

theorem, satisfies Z L (⌦, ,P): E[Z X] = E[Z Y] and Y is

8 2 B · ·

thus a natural candidate to define E[X ].

Exercise 1.64 Consider the random variables X ,...,X that are

1 n

independent and uniform on [0,1]. Let T := max X . Com-

i=1,...,n i

pute E[X T]. (Hint: T = T 11 + X 11 , where

{X1

{X1>T

T := max X . Compute the density of (X ,T ), and prove

0 i=2,...,n i 1 0

that for any measurable bounded function g : IR IR: E[X g(T)] =

g(s)

sn(n+1)ds.)

0 · 2

2 Brownian motion

2.1 Definition

Ashewasobservingwithamicroscopepollenparticlesinsuspension

in water, the english botanist Robert Brown noticed in 1827 that

these particles were moving in an erratic way.

This mouvement, called since then Brownian motion, was ex-

plained by Einstein (1905) as resulting from the collisions of water

molecules with the particle. He also listed the principal character-

istics of this movement. If B denotes the horizontal position of the

pollen particle at time t, Einstein suggested that

(1) ( t,s 0 : (B B ) (B ;0 u t).

t+s t u

8 ??  

(2) ( t,s 0 : (B B ) (0,s).

t+s t

8 ⇠N

(3) The trajectories of the process B are continuous.

Indeed the change of position of the particle (B B ) during

t+s t

a time interval [t,s] is the sum of the variations of position due

to each impact by a water molecule. One conceive easily that the

e↵ect of these impacts is null in average, and these impacts may

be considered as independent and identically distributed. Property

(1) follows thus from the independence of the impacts in the time

interval[t,s]withthepreviousimpacts. Thenormalityof(B B )

t+s t

statedinproperty(2)isduetothecentrallimittheorem: Thesumof

a large number of iid impacts is normally distributed. The variance

of this sum will be proportional to the number of impacts which is

in turn proportional to the length of the time interval. Property (3)

is obvious since B is the position of a physical particle.

The physical model of the Brownian motion described here above

seems to justify heuristically the existence of such a movement.

However this is far from being a proof. Indeed, since B is the posi-

tion of a physical particle, with a positive mass, it must not only be

continuous but also di↵erentiable. However, we will see later in this

course that a process satisfying properties (1),(2),(3) can not have

di↵erentiable trajectories. The mathematical proof of the existence

of a Brownian motion is due to Wiener in 1920.

2.2 Existence of the Brownian Motion

Definition 2.1 If is a filtration on (⌦, ), a process B

t t 0 t

{F } A

adapted to the filtration is an Brownian motion if:

t t 0 t

{F } F

• 1) B = 0 a.s.

• 2) 0 t < s < , B B is independent of and follows

s t t

8  1 F

a normal law with 0 expectation and variance s t.

• 3) The trajectories B(!) are continuous for P-almost every !.

In this section we construct explicitly a brownian motion on a

space (⌦, ,P). The only condition we impose on (⌦, ,P) is that

A A

there exists on that space an infinite sequence ⇠ of i.i.d. ran-

{ n }n 2N

dom variables that are (0,1).

ConsiderthespaceL2([0, [)ofmeasurablefunctionsf : [0, [

1 1 !

R such that kf kL2([0, 1[ := 01f2(x)dx < 1. The scalar product on

this space is hf,g

iL2([0, 1[

: R= 01f(x)g(x)dx. As well known there

exists an Hilbert basis of this space, i.e. a countable orthonormal1

family e L2([0, [) such that := spann( e ) is dense

{ i }i 2N ⇢ 1 U { i }i 2N

in L2([0, [).

1 ifi=j

1Afamily {ei}i

iscalledorthonormaliif 8i,j 2N: hei;ejiL2([0, 1[=

( 0 ifi=j

We next define a linear application F : L2(⌦) by the for-

mula:

F(g) := e ,g ⇠ .

i L2([0, [) i

h i 1

i=0

Observe that if g then g is a linear combination of finitely

many e ’s: g = N ↵ e , where ↵ =< e ,g > . Therefore

i i=0 i i i i L2([0, [

only finitely many terms in the sum defining F(g) are non 0, and F

is thus well defined.

Lemma 2.2 F is an isometry from to L2(⌦). It can thus be

extended by continuity into an isometry F¯ from L2([0, [) to L2(⌦).

Furthermore for all g L2([0, [), F¯ (g) is a centered normal

2 1

random variable with variance g 2.

k k

Proof:

If g , g = N ↵ e , where ↵ = e ,g . An easy com-

2U i=0 i i i h i iL2([0, 1[

putation shows that g 2 = N ↵2. Since the ⇠ are inde-

P k kL2([0, [) i=0 i i

pendent (0,1), one get as announced that F(g) = N ↵ ⇠

N P i=0 i i ⇠

(0, N ↵2). Therefore F(g) = g . Being a uni-

N i=0 i k kL2(⌦) k kL2([0, 1[) P

formly continuous map to a complete space, F can be extended by

continuity into F.

Let next g L2([0, [). Consider a sequence g in that

2 1 { } U

converges to g. Observe then that g g , as

n L2([0, [) L2([0, [)

kgn k2 Lk 2([0k

[)↵2 1 ! kgk k2 L2k

([0,

[)↵12

n goes to 1. So F(gn)(↵) = e 2 1

e 2 1 and

thus F¯ (g) (0, g 2 ) as announced.

⇠N k kL2([0, [)

We next define B := F(1 ).

s [0,s[

Lemma 2.3 The obtained process B satisfies properties 1) and 2)

in definition 2.1, with respect to the natural filtration := (B ,s

t s

F 2

[0,t]).

Proof: Obviously: B := F(0) = 0 P-a.s..

We next prove that B B is independent of . Observe

t+s t t

first that t ... t t t + s, the random vector B =

1 n

8    

(B ,...,B ,B B )isgaussian: indeed,if~v= (v ,v ,...,v ,v )

t1 tn t+s

t 1 2 n n+1

Rn+1, we get:

~vB,~ = n v B +v (B B )

h i i=1 i ti n+1 t+s t

= n v F¯ (1 )+v F¯ (1 )

Pi=0 i [0,ti[ n+1 [t,t+s[

= F¯ ( n v 1 +v 1 )

P i=0 i [0,ti[ n+1 [t,t+s[

since F¯ is linear. ThereforPe ~vB,~ follows a normal law, as it results

h i

from the previous lemma. This being true for all ~v, we infer that B

is indeed a gaussian vector.

We next prove that B B is independent of (B ,...,B ).

t+s

t t1 tn

SinceB isgaussian, itissucienttoprovethatcov(B ,B B ) =

ti t+s

0 (See exercise 1.62-2).

Now, we have :

cov(B ,B B ) = E[(B E(B ))(B B E(B B )]

ti t+s

t ti

ti t+s

t+s

= E[B (B B )] = B ,B B )

ti t+s t h ti t+s t iL2(⌦)

= F(1 ),F(1 ) = 1 ,1

h [0,ti[ [t,t+s[ iL2(⌦) h [0,ti[ [t,t+s[ iL2([0, [)

= 011 [0,ti[(x)1 [t,t+s[(x)dx = 0.

Thus (B B ) R (B ,v T). Finally, observe

t+s t T finite [0,t] v

??[ ⇢ 2

that (B ,v T) is an algebra that engenders and

Tfinite [0,t] v t

[ ⇢ 2 F

with exercise 1.29 we conclude that B B is independent of .

t+s t t

It remains to prove that B B (0,s). This follows from

t+s t

⇠N

¯ ¯ ¯

the fact that B B =F(1 ) F(1 )=F(1 ) is centered

t+s t [0,t+s[ [0,t[ [t,t+s[

and normaly distributed. We further have

1 2 = 1 12 (x)dx = s.

k ]t,t+s] k L2([0, [) ]t,t+s]

1 Z0

Hence B B (0,s).

t+s t

⇠N

It remains for us to analyse the continuity of the trajectopries of

¯ ¯

B. However B was defined above as F(10,t[) where F is a map

t [

to L2 and not to 2. Therefore, B is in fact defined up to a set

of 0-measure! Could we select for all t one member B˜ 2 of the

equivalence class B L2, in such a way that the resulting process

have all its trajectories continuous? To answer positively to this

question, we will need the following theorem:

Theorem 2.4 (Kolmogorov)

If X is a process on (⌦, ,P) such that there exists ↵> 0, > 0

and C > 0 such that : 0 s,t 1

8  

E[ X X ↵] C s t 1+,

s t

| |  | |

then there exists a modification X¯ of X satisfying [0, [:

8 2 ↵

¯ ¯ ↵

X X

s t

E sup | |,0 s,t 1,s = t < .

{ s t   6 } 1

✓ | | ◆

In particular, X is a process with P-almost every sample path

continuous on [0,1].

Proof: We set D := k ,k = 0,...,2n and D := D . De-

n {2n } [n 2N n

fineK := sup {|X ss tX t |,0



s,t

D,s 6= t }andK n := sup {|X ss tX t |,0



| | | |

s,t D ,s = t .

2 6 }

1) We first prove that E[K↵] < . Observe that K is an

1 { }

increasing sequence of positive functions that converges to K. Due

to the monotone convergence theorem of Beppo Levi, we infer then

that E[K↵] = lim E[K↵].

n n

If s < t are two points in D , set s the smallest point in D

n 0 n 1

such that s s and let t the biggest point in D such that t t.

0 0 n 1 0



Then s s t t and s t 2 n. Therefore:

0 0

   | |

X X X X

| s t | 2n X X +2n X X + | s 0 t 0|.

s t  | s s 0| | t 0 t | s t

| | | |

If s = t, the last term vanishes and is thus below K . If s = t,

0 0 n 1 0 0

then s t s t and the last term is then dominated by

0 0

| || |

X X X X

| |ss 00 t 0|t0|, and since s 0,t 0 2 D n 1, we get also that: | |s s0 t |t0|  K n 1.

Therefore, if R := max X X ,k = 0,...2n 1 , we find:

n k+1 k

{| 2n 2n| }

|X |ss tX |t |



2nR n + K n 1. This holds for all s 6= t

D n and we

conclude that

K 2 2nR +K . (1)

n n n 1

 ·

If ↵ 1, we find: E[K↵]1 = K 2 2nR +K

n ↵ k n kL↵  k · n n 1 kL↵ 

2 2n R + K . It results from the definition of R and

n L↵ n 1 L↵ n

· k k k k

from our hypothesis on X that:

2n 1

R ↵ = E[R↵] E[ X X ↵] 2nC2 n(1+) = C2 n.

k n kL↵ n  | k 2+ n1 2k n| 

k=0

Therefore: K 2 C1 2n( )+ K . Summing up these

n L↵ ↵ ↵ n 1 L↵

k k  · · k k

inequalities for n between 1 and N, we get thus:

N L↵

2 C↵1 2n( ↵) + K

L↵.

k k  · · k k

n=1

Since < 0, the geometric series appearing on the righthand

↵

side is then converging and thus

E[K↵]↵1 = lim K

n L↵

2 C↵1 1 2n( ↵) + K

0 L↵

< .

N k k  · · k k 1

!1 n=1

Let us next deal with the case ↵< 1: observe that if z [0,1],

then z↵ z. Thus for all x,y > 0 : ( x )↵ + ( y )↵ 1 and it

x+y x+y

follows that x↵ + y↵ (x + y)↵. Equation (1) yields then: K↵

n 

2↵ 2n↵R↵+K↵ . Since, aswehaveproved above: E[R↵] C2 n,

· n n 1 n 

wenextgetE[K↵] 2↵ C 2n(↵ )+E[K↵ ]and, since↵ < 0,

n  · · n 1

we infer that

E[K↵] = lim E[K↵] 2↵ C 2n(↵ ) +E[K↵]

n n  · · 0

!1 n=1

2)We next construct the process X: Set ⌦ := ! ⌦:

{ 2

K(!) < . According to the first part of the proof, P(⌦) = 1.

For ! ⌦, we define X (!) := 0. For ! ⌦, the map X(!) : t

0 t 0 .

62 2 2

D X (!) is Ho¨lder-continuous and thus uniformly continuous.

X(!) can thus be extended by continuity on the closure of D which

coincides with [0,1]. This extension is denoted X.(!). The resulting

process X is continuous and if J := sup

|X¯

X¯

t | : 0 s,t 1,s =

{ s t   6

t }, we get J = sup

{|X¯

tX¯

t | : s,t

D,s| 6= |

t }. If !

⌦ 0, then

¯ | |

t D: X (!) = X (!) and thus J(!) = K(!). Therefore J = K

t t

8 2

P-a.s. and thus E[J↵] = E[K↵] < .

It remains for us to prove that X is a modification of X. If

t [0,1], there exists a sequence t in D that converges to t.

2 { n }n 2N

But X = X P-a.s. since both variables coincide on ⌦. Due to

tn tn 0

¯ ¯

the continuity of X, we conclude that X converges P-a.s. to X .

tn t

OntheotherhandwehaveE[ X X ↵] 0, andthereexiststhus

tn|

a sub-sequence t of t such that X converge P-a.s. to

{ 0n}n 2N { n }n 2N t 0n

X . This implies that X = X P-a.s.

t t t

Exercise 2.5 Prove that, under the hypotheses of the previous the-

orem, there exists a set ⌦ ⌦ with P(⌦) = 1 such that: ! ⌦,

0 0 0

⇢ 8 2

the trajectory t [0,1] X (!) is -H¨older continuous for all in

2 !

[0,/↵ [. In other words:

[0,/↵ [, K() < : t,s [0,1] : X (!) X (!) K() t s

t s

8 2 9 1 8 2 | | | |

Exercise 2.6 ProvethatifX satisfiestheinequalityE[ X X ↵]

s t

| | 

C s t 1+, for all s,t 0, then there exists a modification X¯ of X

| |

with continuous trajectories on [0, [.

Let us then apply Kolmogorov’s theorem to the process B ob-

tained above: We have B B (0,s t). Thus B B has

s t s t

⇠N

the same law as ps tZ where Z (0,1). Therefore:

⇠N

E[ B

↵] = E[ps t↵ Z ↵] = s t ↵ 2C

↵

s t 1+

| | | | | |  | |

with C := E[ Z ↵] < and := ↵ 1. Since we must have > 0,

↵ | | 1 2

it must be the case that ↵> 2. We thus obtain a modification of

B with -Holder continuous for all < = 1 1. Since ↵ can be

↵ 2 ↵

taken arbitrarily big, we obtain the following result

Theorem 2.7 There exists a modification B of the process B with

locally -H¨older continuous trajectories for all in [0,1/2[.

¯ ¯

Exercise 2.8 Let bethecompletionof . ShowthatB isadapted

t t

F F

¯ ¯

to and is a -brownian motion.

t t

F F

Exercise 2.9 ShowthatifB isabrownianmotion, thencov(B ,B ) =

s t

s t, where s t is a notation for min(s,t).

^ ^

Show that if X is a continuous centered gaussian process - i.e. for

all finite family t ,...,t , the vector (B ,...,B ) is a centered

{

1 n

}

t1 tn

gaussian vector - and if s,t 0 : cov(B ,B ) = s t, then B is a

s t

8 ^

brownian motion on it’s natural filtration.

On ⌦ := ([0, [), we define the process X , ! ⌦

0 t 0 0

C 1 2 !

X (! ) := !(t). Let be it’s natural filtration.

t 0 t

Corollary 2.10 On (⌦ , ) there exists a unique probability mea-

sure ⇧ such that X is a -brownian motion. ⇧ is called Wiener

t t

measure.

Proof:

1) existence: Let (⌦, ,P) be the space on which we have con-

ˆ ˆ

structed the brownian motion B. We can consider B as an appli-

ˆ ˆ

cation that associates to ! ⌦the trajectory B(!) : t B (!).

. t

2 !

Thus, B is an application from ⌦ to ⌦ . It is trivially measurable

. 0

from to , since t : B is -measurable. We take ⇧ as the

A G1 8 A

image measure of P by B:⇧:= P . That is: C : ⇧(C) :=

. Bˆ

. 8 2G 1

P(Bˆ 1(C)).

Under ⇧, let’s show that X (! ) (0,t). Indeed,

t 0

⇠N

⇧( ! X (! ) A ) = ⇧(X 1(A)) = P(Bˆ 1(X 1(A)))

{

t 0

2 }

t . t

= P( ! Bˆ (!) X 1(A) )

{ |

}

= P( ! X (B(!)) A )

t .

{ | 2 }

= P( ! B (!) A )

{ | 2 }

The result follows since B is a brownian motion under P.

This argument can easily be generalized to show that X(! ) is a

brownian motion under ⇧.

2) uniqueness: If ⇧ and ⇧ are 2 measures on (⌦ , ) such that

0 0

X is a brownian motion, then, for all finite set J in R+,⇧and⇧

coincident on := (X ,t J). In fact, since X is a Brownian

J t

G 2

motion, we deduce that the random vector (X ,t J) is a centered

gaussian and cov(X ,X ) = t t. The law of (X ,t J) is hence

t t0

0 t

entirely characterized and is the same under ⇧ and ⇧.

In other words, := A ⇧(A) =⇧ (A) . So

J 0

G ⇢U { 2G 1| }

J finite J

[ G ⇢U

But is also a monotone class from the -additivity of ⇧ and

⇧. Then by the monotone class theorem ( ) . Since

0 [J fini GJ ⇢U

= ( ), by the definition of , we conclude that⇧and

G1 [J fini GJ U

⇧ coincide on .

2.3 Some properties of the brownian motion

Theorem 2.11 If B is a brownian motion, then :

1. c > 0 X = c 1B is a brownian motion.

t c2t

2. Y = tB is a brownian motion.

t t 1

3. > 0 Z = B B is a brownian motion.

t t+

Proof:

1. The trajectories of X are continuous, since the trajectories of

B are continuous. In addition X = c 1B = 0. We observe

t 0 0

that

X X = c 1(B B )=c 1(B B ) with t = c2t

t+s t c2(t+s) c2t t+s t 0

0 0 0

and s = c2s, so X X is independent from (B ,µ t) =

0 t+s t µ 0



(B ,µ c2t) = (X ,v t) so X X is independent

µ v t+s t

 

from (X ,v t). Since B B (0,c2s), it is clear

v c2(t+s) c2t

 ⇠N

that X X = c 1(B B ) (0,s). Hence X is a

t+s t c2(t+s) c2t t

⇠N

brownian motion.

2. Y is clearly a centered gaussian process (see exercise 2.2): For

all finite set J ]0, [, the random vector (Y ,t J) is the im-

⇢ 1 2

age of a centered gaussian (B ,t J) by a linear application.

In addition cov(Y ,Y ) = ts cov(B ,B ) = ts(1 1) = s t.

t s · 1 t 1 s t ^ s ^

So for all finite set J ]0, [, the random vectors (Y ,t J)

⇢ 1 2

and (B ,t J) have the same law, since they are 2 centered

gaussian vectors with the same covariance matrix. To use ex-

ercise 2.2 and conclude that Y is a brownian motion, we only

have to show that the trajectories of Y are continuous. By

the continuity of the trajectories of B, the ones of Y are triv-

ially continuous for t > 0 and we only have to show that

P( ! lim Y (!) = 0 ) = 1. Since t ]0, [ Y (!) is a

t 0 t t

{ | & } 2 1 !

continuous function, we have:

! limY (!) = 0 = ! lim Y (!) = 0

t t

{ |t 0 } { |q 0,q Q }

& & 2

If J isanincreasingsequenceoffinitesetssuchthattheir

{ n }n 2N

union is Q+, the set {! |lim

q &0,q

2QY t(!) = 0

}

is equal to

! k N: M N: n N: q J

]0,1/M]: Y q(!) 1/k .

{ |8 2 9 2 8 2 8 2 \ | | }

By monotonocity of the considered sequence of sets, the prob-

ability P( ! lim Y (!) = 0 ) can be written as

t 0 t

{ | & }

lim lim lim P( ! q J ]0,1/M]; Y (!) 1/k ).

n q

k M n { |8 2 \ | | }

!1 !1 !1

Since(Y ,q J ]0,1/M])hasthesamelawas(B ,q J ]0,1/M]),

q n q n

2 \ 2 \

we can conclude that

P( ! q J ]0,1/M]: Y (!) 1/k )

n q

{ |8 2 \ | | }

= P( ! q J ]0,1/M]: B (!) 1/k ),

n q

{ |8 2 \ | | }

and then

P( ! limY (!) = 0 ) = P( ! limB (!) = 0 ) = 1.

t t

{ |t 0 } { |t 0 }

& &

3. The process Z has obviously continuous trajectories and Z =

B B = 0.

Next Z Z = B B + B B = B B

t+s t t+s+ t+ t++s t+

which is independent from (B ,µ t+) (Z ,v t).

µ v

 

Finally, Z Z = B B (0,s). Thus Z is a

t+s t t++s t+ t

⇠N

brownian motion.

Theorem 2.12 (Zero-one law)

If B is a brownian motion and = (B ,s t), then

t s

F 

A : P(A) 0,1 .

8 2F 2{ }

Proof: Let 0 <✏< . Let us consider = (B B ,✏

✏, t ✏

F 

t ). Since t [0,] : B = lim B B , it follows that B

t ✏ 0 t ✏ t

 8 2 &

is ( )-measurable. Thus = ( ). If A ,

✏>0 ✏, ✏>0 ✏, 0+

[ F F [ F 2F

then ✏> 0, A which is independent from , hence, using

✏ ✏,

8 2F F

exercise 1.29, A is independent from ( ) = Ac. Thus

✏>0 ✏,

[ F F 3

A is independent from Ac, and then, P(A)P(Ac) = P(A Ac) = 0.

Finally P(A)(1 P(A)) = 0 P(A) = 0 ou P(A) = 1.

)

Corollary 2.13 If B is a brownian motion, then

P( ✏> 0, sup B > 0) = 1.

0 t ✏

 

Proof: Let A = sup B > 0 . We have:

✏ { 0 t ✏ t }

A = ✏> 0, sup B > 0 = A

t n 1 0+

{8 0 t ✏ } \ n 2F

 

then by the previous theorem P(A) = 0 or P(A) = 1.

But P(A) = limP(A ) limP(B > 0) = 1, so P(A) = 1.

n1 n1 2

Remark 2.14 Since B is also a brownian motion, we show in

addition that: P( ✏> 0,inf B < 0) = 1, and then

0 t ✏ t

8  

P( ✏> 0, sup B > 0 and inf B < 0) = 1.

t t

8 0 t ✏ 0 t ✏

   

Thus, before departing from 0, the generic trajectory of a brownian

motion oscillates infinitely many times between positive and nega-

tive values, as does the function t tsin(1/t). Between such os-

cillations, the brownian motion reaches 0, by continuity and thus

P( ! t : t 0 et B (!) = 0 ) = 1.

{ |9{ n }n 2N n & tn }

Corollary 2.15 P(sup B = ) = 1.

t 0 t 1

Proof: Let c > 0 and > 0, then

P(sup B c) = P(sup B )

t 0 t t 0 c t

= P(sup B )

t 0 c c2 2t

= P(sup X ),

t 0 t

where X is a brownian motion, as a result of the theorem 2.11. Tak-

ing the limit of the last line when 0, we obtain P(sup X >

& t 0 t

0) that is equal to 1 by the previous corollary. Hence c > 0,

P(sup B c) = 1,whichenablesustoconcludethatP(sup B =

t 0 t t 0 t

) = 1.

Remark 2.16 Since the generic trajectory of a brownian motion

is continuous, the previous corollary indicates that limsup B =

t t

. By symmetry, we also have liminf B = , and thus the

t t

1 %1 1

brownian motion (of dimension 1) oscillates infinitely many times

from to + : For all c R, the brownian motion B will then

1 1 2

take the value c infinitely many times. This is expressed saying the

brownian motion is a recurrent process.

Exercise 2.17 Let :]0, [ ]0, [. Show that there exists a con-

1 ! 1

stant c() [0, ] such that

2 1

P(limsup = c()) = 1.

(t)

t 0

Derive c() for the functions of the form (t) := t↵.

An important theorem by Levy indicates that c( ) = 1, where

(t) = 2tLog(Log(1/t)).

2.4 Quadratic Variation:

Let B a brownian motion and let s > t. For a finite sectioningof

theinterval[t,s](i.e.= t ,...,t , witht = t < t < t < ... <

1 n 1 2 3

{ }

t = s), we define the diameter of as: =maxi=n 1(t t ).

| |

i=1 i+1

Finally, let us consider the the random variable:

i=n 1

T = (B B )2.

s,t ti+1 ti

i=1

Theorem 2.18 If n is a sequence of partitions of [s,t] such

n N

{ } 2

that n 0 when n , then Tn s t in L2.

| |! ! 1 s,t !

Proof: Let us compute E[Tn]: since B B (0,t t ),

s,t ti+1 ti ⇠N i+1 i

B B can be written as pt t Z with Z (0,1) thus:

ti+1

ti i+1

i i i

⇠N

n 1 n 1

E[Tn ] = (t t )E[Z2] = (t t ) = t t = s t.

s,t i+1 i i i+1 i n 1

i=1 i=1

X X

Next observe that

var[T s ,tn] = i i= =n 11var[(B

ti+1

B ti)2]

= Pi i= =n 11var[(t

i+1

t i)Z i2]

= Pi i= =1n 1(t

i+1

t i)2var(Z i2)



Pi i= =1n 1(t

i+1

t i) |n |var(Z2)

= (s t)var(Z2) n n !1 0.

| | !

But var[Tn] = E[(Tn (s t))2] = Tn (s t) 2 .

s,t s,t k s,t kL2

Hence Tn s t in L2

s,t !

If f : [0, 1[

R, we note that V t,s(f) := sup

n i= 11 |f(t i+1)

f(t ) , where is in the class of finite sectioning of [t,s]. V (f) is

i P t,s

called variation of f on the interval [t,s]. Let’s recall the following

theorem that characterizes the class of functions of finite variation :

Theorem 2.19 V (f) < if and only if there exists 2 increasing

s,t

functions g and h : [s,t] R such that f = g h.

Proof: The condition is sucient since f = g h, V (f)

s,t



V (g)+V (h) and all increasing functions are of finite variation.

s,t s,t

The condition is also necessary since g(x) := f(s) + V (f) is

s,x

increasing: indeed, if x < x, V (f) = V (f)+V (f) V (f)

0 s,x s,x x,x s,x

0 0 0

(Proveit). Thefunctionh(x) := g(x) f(x) = f(s) f(x)+V (f)is

s,x

increasingaswell: ifx < x, h(x) h(x) = f(x) f(x)+V (f) 0

0 0 0 x,x

as f(x) f(x) V (f).

0 x,x

| | 0

Corollary 2.20 If B is a brownian motion on (⌦, ,P), there ex-

ists a subset ⌦ of ⌦ of probability P(⌦) = 1 such that ! ⌦ :

0 0 0

8 2

s > t 0 : V (B(!)) = .

t,s .

8 1

Proof: For all pair of rational numbers p < q, let’s take a sequence

n of partitions of [p,q] such that n n !1 0. By the theorem

p,q | p,q| !

2.18,

p,q converges in L2 to p q. By selecting a sub-sequence, we

p,q

can suppose the P-a.s. convergence. Thus, there exists a set ⌦

p,q

⇢

⌦of probability P(⌦ ) = 1 on which T p,q converges pointwise to

p,q p,q

p q. Since there is only a countable number of pairs of rational

numbers ⌦ := ⌦ has a probability of P(⌦) = 1.

0 \(p,q) 2Q2

p,q 0

Let ! ⌦. Let also s > t and let’s choose a pair of rational

numbers p < q in [t,s]. Since T p,q(!) converges to q p > 0, there

p,q

exists N such that n N, T p,q(!) > (q p)/2. Thus,

p,q

(q p)/2 < n i= 11(B ti+1(!) B ti(!))2



Pmax i |B ti+1(!) B ti(!)

| ·

n i= 11 |B ti+1(!) B ti(!)

max B (!) B (!) V (B(!))



ti+1

ti | ·Pp,q .

max B (!) B (!) V (B(!))



ti+1

ti |

s,t .

Since n tends to 0 and the trajectoryB(!) is uniformly contin-

| p,q| .

uous on [s,t], (max B (!) B (!) ) tends to 0 with n. This is

ti+1

only possible if V (B(!)) = .

s,t .

The previous corollary implies that the generic trajectory of the

brownian motion is not monotonic in any interval!

Corollary 2.21 If B is a brownian motion on (⌦, ,P), there ex-

ists a subset ⌦ of ⌦ of probability P(⌦) = 1 such that ! ⌦, the

0 0 0

8 2

trajectory B(!) is not H¨older continuous of order ↵> 1/2 in any

interval.

Proof: Let’s consider again the set ⌦ constructed in the previous

proof. Let ! ⌦, and let’s suppose the trajectory B(!) is Ho¨lder

0 .

continuous of order ↵> 1/2 on the interval [t,s]:

K < : t ,t [t,s] : B (!) B (!) K t t ↵

9 1 8

1 2

2 |

| |

If p,q are rational numbers such that t < p < q < s, then

T p ,qn p,q(!) = n i= 11(B ti+1(!) B ti(!))2



PK2 n i= 11 |t i+1 t i |2↵



K2 |Pn p,q|2↵ 1 n i= 11 |t i+1 t i

= K2 n 2↵ 1(p q)

| p,q| P

Since

p,q(!) q p > 0 and n 2↵ 1 0 (↵> 1/2), the

p,q ! | p,q| !

previous inequalities are only possible if K = .

Exercise 2.22 Show that with probability 1, the trajectories of the

brownian motion are not H¨older continuous of order 1/2 in any

interval.

3 Filtrations and Stopping times

3.1 Definitions and Properties

Definition 3.1 A random variable ⌧ :⌦

R¯+

is a stopping times

on the filtration

t t

0, i↵ t R+ ⌧ t t.

{F } 8 2 {  }2F

Example 3.2 If X is a continuous process, -adapted, if A is a

closed set in R, then: ⌧ A(!) := inf t X t(!) A , with the conven-

{ | 2 }

tion inf := , is a stopping time on .

; 1 F

Proof: Since A is closed and X continuous, we get:

! ⌧ A(!) t = ! inf d(X q(!),A) = 0,q Q+ .

{ |  } { |0 q t 2 }

 

Since x d(x,A)is a continuous function, it is also measurable on

(R, BR). !

d(X qn(!),A) is then Ft-measurable, as composition

of measurable functions and it follows that g :⌦ R defined

by g(!) := inf d(X q(!),A),q Q+ [0,t] is

measurable, as

{ 2 \ } F

infimum of a countable set of measurable functions. Therefore:

⌧ t = g 1( 0 ) .

A t

{  } { } 2F

⌧ is thus a stopping time.

Example 3.3 Let X be a process with right-continuous trajectories,

adapted to t. If O is an open set in R, then: ⌧

= inf t 0 X

F { | 2

O is a stopping time on the filtration

} F

Proof:

⌧ a = ! : n t Q+,t a+ 1,X O

{ O  } { 8 9 2  n t 2 }

= ! : t Q+,t a+ 1,X O

\n { 9 2  n t 2 }

= ! X (!) O .

\n [t 2Q \[0,a+ n1] { | t 2 }

But ! X (!) O .

{ | t 2 }2F t ⇢F a+ n1

Thus A := ! X (!) O and n m :

n [t 2Q \[0,a+ n1]{ | t 2 }2F a+ n1 8

A . Therefore ⌧ a = A F . It

n 2F a+ n1 ⇢F a+ m1 { O  } \n m n 2 a+ m1

follows that: ⌧ a = . Since this holds for all a,

{ O  }2 \ m Fa+ m1 Fa+

we have proved that ⌧ is a -stopping time.

O t+

Exercise 3.4 Explain why ⌧(!) := inf t X = max X is gen-

t s 0 s

{ | }

erally not a stopping time on the natural filtration of X.

Definition 3.5 If ⌧ is a -stopping time, we define

:= A , t 0 A ⌧ t .

⌧ t

F { 2F 1 8 \{  }2F }

Exercise 3.6 Prove that is a -algebra.

⌧

Proof: We have: ⌧ t = thus .

t ⌧

;\{  } ; 2F ; 2F

Let A . Let us prove that Ac . Indeed: Ac ⌧ t =

⌧ ⌧

2F 2F \{  }

⌧ t (A ⌧ t )c since ⌧ t and A ⌧ t

t t

{  }\ \{  } 2F {  }2F \{  }2

Let A , then n and t we have: A ⌧ t .

n ⌧ n t

{ }⇢F 8 8 \{  }2F

Therefore (A ⌧ t ) . So, forallt, ( A ) ⌧ t

n n t n n t

[ \{  } 2F [ \{  }2F

and thus A .

n n ⌧

[ 2F

is therefore a -algebra.

⌧

Theorem 3.7 If ⌧ and ⌧ are two -stopping times and if !

0 t

F 8

⌧(!) ⌧ (!), then .

0 ⌧ ⌧

 F ⇢F 0

Proof: If A , let us prove that A : If t 0, then

⌧ ⌧

2F 2F 0

⌧ t ⌧ t so A ⌧ t = A ⌧ t ⌧ t . Since

0 0 0

{  }⇢{  } \{  } \{  }\{  }

A , we get A ⌧ t and, as ⌧ is a stopping time, we

⌧ t 0

2F \{  }2F

have ⌧ t . Therefore A ⌧ t = A ⌧ t ⌧

0 t 0 0

{  }2F \{  } \{  }\{ 

t and, hence, A .

t ⌧

}2F 2F 0

Theorem 3.8 If ⌧ and ⌧ are two -stopping times, then:

0 t

⌧ ⌧ := max(⌧,⌧ ) and ⌧ ⌧ := min(⌧,⌧ ) are stopping times.

0 0

_ ^

Proof:

Let t 0, then ⌧ ⌧ t = ⌧ t ⌧ t since

0 0 t

{ _  } {  }\{  }2F

⌧ t and ⌧ t and since is a -algebra.

t 0 t t

{  }2F {  }2F F

Thus ⌧ ⌧ = max(⌧,⌧ ) is a stopping time.

0 0

Similarly, ⌧ ⌧ t = ⌧ t ⌧ t . Thus ⌧ ⌧ is

0 0 t 0

{ ^  } {  }[{  }2F ^

a stopping time.

Exercise 3.9 Prove that ⌧ is -measurable.

⌧

Preuve: This comes to prove that a 0 : ! ⌧(!) a .

⌧

8 { |  }2F

Let t 0, then:

⌧ a ⌧ t = ⌧ a t

a t t

{  }\{  } {  ^ }2F ^ ⇢F

Since this is true for all t, it follows that ⌧ a .

⌧

{  }2F

3.2 Progressively measurable processes

Let X be an -adapted process on a probability space (⌦, ,P)

F A

and let ⌧ be an -stopping time.

Definition 3.10 We denote X the map ! X (!) := X (!).

⌧ ⌧ ⌧(!)

The fact that X is adapted is not sucient to ensure that X is a

⌧

-measurable random variable. In this section, we aim to introduce

a sucient condition on X for the measurability of X .

⌧

Definition 3.11 A process X is progressively measurable if and

only if T 0,X : (⌦ [0,T]) R is T [0,T]-measurable.

8 ⇥ ! F B

Obviously, a progressively measurable process X with respect to

the filtration is in particular adapted to .

t t

F F

Theorem 3.12 If X is -progressively measurable, if ⌧ is an -

t t

F F

stopping time then X is measurable.

⌧ ⌧

Proof: We just have to prove that for all A : X 1(A) .

2B R ⌧ 2F ⌧

In other words: for all t 0, X 1(A) ⌧ t . Observe that

⌧

\{  }2F

X 1(A) ⌧ t = ! ⌧(!) t and X (!) A

⌧

\{  } { | 

⌧

2 }

= ! ⌧(!) t and X (!) A

⌧ t

{ |  ^ 2 }

= ! ⌧(!) t X 1(A).

{ |  }\ ⌧ ^t

If ⌧ is a stopping time, so is ⌧ t. Since ⌧ t is -measurable, it

⌧ t

^ ^ F ^

is thus measurable. Le us define g as follows:

g : (⌦ , ) (⌦ [0,t], ) : ! g(!) := (!,⌧ t(!)).

t t [0,t]

F ! ⇥ F B 7! ^

Since both components of g, ! and ⌧ t(!) are respectively mea-

surable from to and from to , g is measurable from

t t t [0,t] t

F F F B F

to . Let us next consider X as a measurable function on

t [0,t]

F B

⌦ [0,t]:

⇥

X : (⌦ ⇥[0,t],

B[0,t])

(R, BR) : (!,s)

X(!,s).

then X is the composition of the two functions X et g: X =

⌧ t ⌧ t

^ ^

g and X

⌧ ^t

is therefore measurable from (⌦, Ft) to (R, BR).

Therefore, A : X 1(A) . Since we also have ⌧ t

8 2B R ⌧ ^t 2F t {  }2

, we infer that X 1(A) .

Ft ⌧

⌧

Theorem 3.13 If X is a continuous -adapted process, then X is

progressively measurable.

Proof: Let us define the process Xn as

Xn(!) := X (!)1 .

t k k t

n {n n }

Xk 2N

Let us first prove that Xn est progressively measurable:

Let T be given. If t



T then X tn(!) = k k =n 0T 1 nk t

But X is measurable and 1 is measurable so

nk FT nk t

X nk1 nk t

that measurable.

T [0,T] N P

F B

By continuity of X, we have: lim Xn(!) = X (!) and, as Xn is

N n t t t

measurable, the same goes for X. In other words X is

T [0,T]

F B

progressively measurable.

Exercise 3.14 If ⌧ is a stopping time, if X is a continuous adapted

process, if X⌧(!) := X (!), prove that X⌧ is progressively mea-

t ⌧ t t

surable.

Exercise 3.15 If X is a continuous process such that X = 0, and

if 0 < a < b, what can you say about X (!) when ! is such that

⌧

[a,b]

⌧ (!) < ?

[a,b]

If X is a brownian motion, prove that P(⌧ (!) < ) = 1.

[a,b]

Exercise 3.16 If X is an -brownian motion, if a > 0, and if X

t ⇤

denotes X (!) := sup X (!),

s⇤ t [0,s] t

a) Prove that X is -measurable.

1⇤ F1

b) Prove that ⌧ s = X a .

{

}  } {

s⇤

}

c) Prove that X has the same law as psX and that ⌧ has

s⇤ 1⇤ a

{ }

the same law as a2/(X )2.

1⇤

d) Prove that ⌧ := inf t X = X and ⌧ := sup t 1 X = 0

{ |

t 1⇤

}

{  |

}

are not stopping times.

4 Martingales

4.1 Definitions et properties:

Definition 4.1 A martingale on a filtration t T (T discrete

F 2

ou continuous) is an adapted process such that:

1. t T X L1( )

t t

8 2 2 F

2. ts T t < s E[X ] = X as.

s t t

8 2 |F

Definition 4.2 Asub-martingale(respectivelysuper-martingale)on

a filtration ( ) (T continuous ou discrete) is an adapted pro-

t t T t

F 2 F

cess X such that:

1. t T X L1( )

t t

8 2 2 F

2. t,s T t < s E[X F ] X (respect. E[X F ] X ).

s t t s t t

8 2 | | 

Example 4.3 If B is a a brownian motion, then :

t t

1. B is a martingale .

2. X = B 2 t is a martingale.

t t

3. If ↵

R then M t↵ = exp(↵B t

↵ 22 t) is also a martingale.

Proof:

1. We know that B L1( ) and if s > t we have:

t t

2 F

E[B ] = E[B +(B B ) ] = B +E[B B ]

s t t s t t t s t

|F |F

since B B is independent from and as B B follows a

s t t s t

centeredgaussianlawE[B ]=B . Thatis,B isamartingale.

s t t t

2. We know that X = B 2 t L1( ) and if s > t

t t t

2 F

E(X ) = E(B 2 ) s

s t s t

|F |F

= E[(B +(B B ))2 ] s

t s t t

= E[B2 ]+E[(B B )2 ]+2E[B (B B ) ] s

t|Ft s

t |Ft t s

t |Ft

= B2 +E[(B B )2]+2B E[B B ] s

t s t t s t

= B2+s t+0 s = B2 t = X . Hence X is a martingale.

t t t t

3. It is obvious that M↵ L1( ) and if s > t we have:

t 2 Ft

E(M↵ ) = E[exp(↵B ) ]exp( ↵2 s)

s |Ft t |Ft 2

= E[exp(↵[B +(B B )]) ]exp(

↵2

t s t |Ft 2

= E[exp(↵(B B )) ]exp(↵B

↵2

s t |Ft t 2

= E[exp(↵ps tZ)]exp(↵B ↵2 s)

t 2

= ( i↵ps t)exp(↵B ↵2 s)

Z t 2

= exp((i↵p 2s t)2 )exp(↵B

↵ 22 s)

= exp(↵2( 2s t))exp(↵B

↵ 22 s)

= exp(↵B ↵2 t) = M↵. So M↵ is a martingale.

t 2 t t

Example 4.4 If Y L1( ) and if, for all t: X := E[Y ] then

t t

2 F1 |F

X is a martingale.

Indeed, if s > t, we have so:

t s

F ⇢F

E[X ] = E[E[Y ] ] = E[Y ] = X

s t s t t t

|F |F |F |F

Theorem 4.5 1. If X is a martingale and is a continuous and

convex function such that t (X ) L1( ) then (X ) is a

t t t

8 2 F

sub-martingale.

2. If X is a sub-martingale and if is a continuous, increasing

and convex function such that t (X ) L1( ) then (X ) is

t t t

8 2 F

a sub-martingale.

Proof:

1. By the Jensen inequality we have for s > t E[(X ) ]

s t

[E[X ]] = (X ).

s t t

2. If s > t E[(X ) ] [E[X ]] (X ) as X is a sub

s t s t t

|F |F

martingale and is increasing.

Example 4.6 IfX isamartingalethenY = X p isasub-martingale

t t t

| |

p 1 :

Indeed, the function x x p is continuous and convex.

7! | |

4.2 Discrete Time Martingales

Theorem 4.7 If the process H is bounded and -adapted

{Fn }n 2N

and if X is a -martingale, then the process Y defined by recur-

rence as: Y = 0 et Y = Y +H (X X ) is a martingale.

0 n+1 n n n+1 n

Proof: Since X is a martingale and H is -measurable, we have:

n n n

E[Y ] = E[Y +H (X X ) ]

n+1 n n n n+1 n n

|F |F

= Y +H (E[X ] X )

n n n+1 n n

= Y

Thus

E[Y ] = E[E[Y F ] ]

n+k n n+k n+k 1 n

|F | |F

= E[Y ]

n+k 1 n

···

= E[Y ]

n+1 n

= Y

If (X ) denotes the increment (X X ) of the process X,

n n+1 n

Y n can be written as: Y n = Y 0 + n t= 01H t(X t). This is a discrete

time version of the integral Y = Y + H dB that we will introduce

t P0 t t t

in the next chapter.

Theorem 4.8 If H is a bounded, adapted, positive process and

X an sub-martingale, then the process Y defined as: Y = 0 et

n 0

Y = Y +H (X X ) is a sub-martingale.

n+1 n n n+1 n

Proof: Proof identical to the previous one.

Theorem 4.9 (Stopping Theorem): If X is a -martingale, if

n n

⌧ are 2 bounded stopping times E(X )=X .

⌧ ⌧

 |F

Proof: Let B and let us define H := 1 1 . Then H is

⌧ n ⌧ n< B n

2F {  }

-measurable since B ⌧ n < = B ⌧ n n c

F \{  } \{  }\{  } 2

Let Y

= 0+ n n= =M

1H n(X

n+1

X n). Then

Y = H X +(H H )X +(H H )X + +H X .

M 0 0 0 1 1 1 2 2 M 1 M

···

Let’s define H = 1 1 and H⌧ = 1 1 . Then H =

n B n< n B n<⌧ n

{ } { }

H H⌧. We can thus write:

n n

Y = (1 H)X +(H H)X + +H X

M B 0 0 0 1 1 ··· M 1 M

(1 H⌧)X (H⌧ H⌧)X H⌧ X .

B 0 0 0 1 1 ··· M 1 M

But(H H )=1 (1 1 )=1 1 =1 1 .

n n+1 B {>n } {>n+1 } B {n+1 >n } B {=n+1 }

We have also that 1 H= 1 (1 1 )=1 1 . Hence:

B 0 B {>0 } B {=0 }

n=M n=M

Y = ( X 1 )1 ( X 1 )1 = (X X )1

M n =n B n ⌧=n B ⌧ B

{ } { }

n=0 n=0

X X

and since Y is a martingale:

E[Y ] = E[E[Y ]] = E[Y ] = 0.

M M 0 0

So B : E[1 (X X )] = 0 and with exercise 1.54, we

⌧ B ⌧

8 2F

conclude that: E[X ]=X .

⌧ ⌧

Remark 4.10 We have the same theorem for sub-martingales.

Theorem 4.11 (Maximal Inequality)

If X is a sub-martingale then > 0

n n=0,...,N n

{ } F 8

P(max(X ,X ,...,X ) ) E[X 1 ]

0 1 N



N {max(X0,X1,...,XN)

}

Proof: Let ⌧ := min n X avec min := N. Since X is

{ | } ;

a sub-martingale, it follows from the stopping theorem that X

⌧



E[X F ]. Thus

⌧

E[X 1 ] E[1 E[X F ]]

⌧ {max {Xn }>

} 

{max {Xn }

}

⌧

= E[X 1 ]

N max Xn

{ { } }

since 1 is measurable. When max X , we have

{max {Xn }

}

F⌧

{

}

bydefinitionof⌧ : X . Thus,X 1 1

⌧

⌧ {max {Xn }

}

{max {Xn }

}

and so

P(max X > ) E[X 1 ] E[X 1 ]

{

} 

⌧ {max {Xn }

} 

N {max {Xn }

}

Corollary 4.12 If X is a martingale in Lp for p 1 and X :=

n ⇤

max( X , X ,..., X ), then:

1 2 N

| | | | | |

pP(X ) E[ X p].

⇤ N

 | |

Proof: LetX beamartingale, thenY = X p isasub-martingale

n n n

| |

to which we can apply the previous theorem:

pP(max(Y ,...,Y ) p) E[Y 1 ] E(Y ).

1 N



N {max(Y1,...,YN) p

} 

Since X p = max(Y ,...,Y ) et X p = Y , we have, as stated

⇤ 1 N N N

| |

earlier: pP(X ) = pP(X p p) E[ Xp ].

⇤ ⇤  | N|

Theorem 4.13 (Doob’s Inequality): p > 1, if X is a

n n=1,...,N

8 { }

martingale in Lp then :

p p

E[ X p] E[X p] ( ) E[ X p]

N ⇤ N

| |   p 1 | |

with X := max( X ,..., X ).

⇤ 0 N

| | | |

Proof: We have X p X p, which implies E[ X p] E[X p].

N ⇤ N ⇤

| |  | | 

Also, the maximal inequality leads to: E[ X 1 ] E[1 ].

N X > X >

| | ⇤ ⇤

Multiplying this by p 2 and integrating on [0, [, we obtain:

1 E[ X 1 p 2]d 1 E[1 p 1]d

N X > X >

| | ⇤ ⇤

Z0 Z0

Since the functions integrated in the previous formula are positive,

we may apply Fubini’s theorem to conclude:

E[ |X

|(X p⇤)p 11 ] = E[ |X

011

⇤>p 2 d]

E[ 011

⇤>p 1d]

E[(X⇤)p

]

R p

Finally, if q is the conjugate of p (i.e. 1 + 1 = 1), we have, with

p q

Ho¨lder’s inequality:

E[ X

(X ⇤)p 1] (E[ X

p])p1 E[(X ⇤)(p 1)q] 1 q .

| |  | | ·

Since (p 1)q = p, we have

(E[ X N p])p1 (E[(X ⇤)p])1 q E[(X ⇤)p],

| | · p 1

which gives the announced result, after simplification:

(E[ X

p])p1

(E[(X

⇤)p])p1

| | p 1 ·

4.3 Doob’s Inequality and Continuous Time Martingales

Let the filtration on a probability space (⌦, ,P).

t t 0

{F } A

Definition 4.14 p( ) will denote the set of -martingales

t t

M {F } {F }

X of continuous trajectory such that X < , where

k k 1

X := sup X .

Lp t Lp

k k k k

t>0

If X p( ), we also define X := X , where

t Mp ⇤ Lp

2M {F } k k k k

X (!) := sup X (!) : s [0,t] .

t⇤

| 2 }

Exercise 4.15 Show that, if X p( ), then the application

2M {F }

t X is increasing. In particular X = lim X .

t Lp Lp t t Lp

! k k k k !1k k

Exercise 4.16 Showthat, ifX p( ), thenX is -measurable.

{Ft

}

t⇤ Ft

Exercise 4.17 Show that p( ) is a vector space and . is

t Lp

M {F } k k

a semi-norm on p( ). Show that X = 0 is equivalent to

t Lp

M {F } k k

modif

saying that X is a modification of 0: X 0.

⌘

Exercise 4.18 Show that if X and Y are continuous processes, then

modif

X Y if and only if X and Y are indistinguishable.

⌘

Proof: We already know that if X and Y are indistinguishable,

modif

then they are modifications of each other. Conversely, if X Y,

⌘

then for all t, P(A ) = 1, where A := ! : X (!) = Y (!) . Let

t t t t

{ }

A := A . Clearly P(A) = 1 and, if ! A, then for all

\q 2Q+ q

t Q+: X t(!) = Y t(!). By continuity of the trajectories, this

relation can be extended to all t 0. Thus, for P-almost all !, the

trajectories X(!) and Y(!) coincide: The processes X and Y are

. .

indistinguishable.

Since ( p, . ) is not a normed space, we had to consider the

L k k

space Lp of equivalence classes of p for the equivalence relation =

P as. In the same way, we introduce here the space Mp( ) of

{F }

equivalence classes in p( ) for the relation modif .

M {F } ⌘

Exercise 4.19 Showthat(Mp( ), . )isanormedvectorspace.

t Lp

{F } k k

Theorem 4.20 (Doob’s Inequality): p > 1, if X p( ),

8 2M {F }

then

X X X .

Lp Mp Lp

k k  k k  p 1k k

In other words, . and . are equivalent norms on Mp( ).

Lp Mp t

k k k k {F }

Proof: Let X p( ) and D an increasing sequence of

2M {Ft } { n }n 2N

sets whose union is Q+. Let t

:= maxD

and

X (!) := max X (!) : t D .

D⇤

n {|

| 2

}

It is trivial that t . Besides, since the trajectories of X are

% 1

continuous, X

⇤

= sup {|X

: t

Q+ }, and hence X

D⇤

n %

⇤

1 1

Theorem 4.13 indicates that

X X X .

tnkLp

 k

D⇤ nkLp

 p 1k

tnkLp

But, itfollowsfromexercise4.15that X = lim X and

kLp n

!1k

tnkLp

by monotone convergence: X = X = lim X ,

kMp

1⇤ kLp n

!1k

D⇤ nkLp

which completes the proof.

The following corollary will prove to be very important in our

future construction of the Itˆo integral.

Corollary 4.21 If is a complete filtration, then for all p > 1,

{F }

the space (Mp( ), . ) is a Banach space.

t Lp

{F } k k

Proof: We shall show that every Cauchy sequence Xn in

{ }n 2N

Mp( ) admits a limit X in Mp( ). Let’s consider such a

t t

{F } {F }

sequence Xn .

{ }n 2N

1) We will construct a “candidate” limit process X:

By the previous theorem, Xn is also a Cauchy sequence

{ }n 2N

with the norm . :

k k

✏> 0, N(✏) : m,n N(✏) : Xn Xm p ✏

8 9 8 k

kMp



Let’s fix n := N(2 (p+1)), and by recurrence

n := max(N(2 (p+1)(k+1)),1+n ).

k+1 k

The sequence n is therefore strictly increasing, since n > n

k k+1 k

N(2 (p+1)(k)), the sub-sequence {Xnk

}k 2N

satisfies

8k : kXnk+1 Xnk kp



2 (p+1)(k).

Let A := {!

⌦: 9L : 8k

L : kX .nk+1(!) X .nk(!)

k1 

2 k

}

where, for a function f(.) : [0, [ R, f(.) := sup f(t) .

1 ! k k1 t 2[0, 1[| |

We remark that if ! A, then the trajectories Xnk(!) consti-

2 .

tute a Cauchy sequence in the space2 ( ([0, [), . ) of continuous

C 1 k k1

functions on [0, [. Indeed, if k k,

kX .n k0(!) X .nk(!)

k1 

k01

kX .nj+1(!) X .nj(!)

k1 

1 2 j k ! !1 0.

j=k j=k

X X

The space ( ([0, [), . ) being complete, the sequence of tra-

C 1 k k1

jectories Xnk(!) converges in . to a continuous function X(!).

. k k1 .

Let’s define X(!) := 0 when !/ A. The resulting process X

has all its trajectories continuous.

2) Let’s now show that P(A) = 1:

Indeed, if we take Yk := Xnk+1 Xnk,

Ac = ! ⌦: L : k L : Yk(!) > 2 k = Ak,

{ 2 8 9 k . k1 } \L [k L

where Ak := ! ⌦: Yk(!) > 2 k . Thus, L,

{ 2 k . k1 } 8

P(Ac) P( Ak) P(Ak).

k L

 [ 

j=L

But Yk(!) = Yk (!) where the notation Yk was introduced in

k . k1 1⇤ ⇤

the definition 4.14. It follows from the Chebichev inequality that

P(Ak)2 kp



kYk

⇤

and the definition of the sub-sequence Xnk

2To be more accurate, . is not a norm on ([0, [) since continuous functions on

kk1 C 1

[0, 1[canbeunbounded. Oneshouldinsteadconsiderthesequence X .nk(!) X .n1(!)that

constitutesaCauchysequenceintheBanachspace b([0, [)ofboundedcontinuousfunctions

C 1

endowedwiththeuniformnorm.

indicates that Yk p = Yk p 2 (p+1)k. Also P(Ak) 2 k

⇤ kLp

kMp



and hence

P(Ac) 1 2 k L !1 0.

 !

k=L

Finally P(Ac) = 0.

3) We show that X is -adapted: P(Ac) = 0 implies that Ac

{F } 2

, for all t 0 since is a complete filtration. Also, the

t t

F {F }

process 1 (!)Xnk(!) is -adapted. And we have shown that

A t {Ft }

1 (!)Xnk(!) converges to X (!). Since the pointwise limit pre-

A t t

serves measurability, X is -measurable.

t t

4)NextweshowthatX isa -martingale: Since Xn Xm

{Ft } k t t kLp 

Xn Xm , the sequence of random variables Xn is a Cauchy se-

k kLp t

quence in Lp( ), which is a complete space. Xn converges therefore

Ft t

in Lp. Since a sub-sequence Xnk converges P-a.s. (on A) to X , we

t t

have shown that Xn converges to X in Lp. Since the conditional

t t

expectation is a continuous operator in Lp, we have that if s > t:

E[X ] = E[lim Xn ] = lim E[Xn ] = lim Xn = X

s |Ft

s|Ft

t t

!1 !1 !1

5) Finally, we show that Xn converges to X in . . Let ✏> 0,

k k

there exists N such that, if m,n N, then Xn Xm ✏. It

k k 

follows that if n N,

X Xn = lim Xm Xn lim Xn Xm ✏.

k t t kLp m k t t kLp  m k kLp 

!1 !1

So X Xn = sup X Xn ✏. The previous being true

k kLp t 0k t t kLp 

for all ✏> 0, we have shown that lim X Xn = 0.

n Lp

!1k k

Corollary 4.22 If X M2( ), then there exists a variable

2 {F }

X L2( ) such that X L2 X and X P-a.s. X when t .

t t

1 2 F1 ! 1 ! 1 ! 1

In particular, t : X = E[X ] and X = X .

t t L2 L2

8 1|F k k k 1k

Proof: To establish the L2-convergence of X and the existence of

X , we only have to show that, for every sequence t increas-

1 { n }n 2N

ing to , the sequence X converges in L2( ) (we remark

1 { tn}n 2N Ft

that all these sequences have the same limit, otherwise there would

be one without a limit!). L2 being complete, it’s enough to show

that X is a Cauchy sequence. If t t , X is the orthog-

{ tn}n 2N m n tn

onal projection of X on L2( ), X being a martingale. We then

tm Ftn

have the Pythagoras identity:

X X 2 = X 2 X 2

tnkL2

tmkL2

tnkL2

The function t X 2 is increasing and converges to X 2 as

! k t kL2 k kL2

t . We infer then that X X 2 0 as n,m :

! 1 k

tnkL2

! ! 1

X is therefore Cauchy sequence in L2.

{ tn}n 2N

From the convergence in L2 of X to the common limit X of all

sequences X ,itfollowsimmediatelythat X = lim X =

{

tn}

kL2 t

!1k

t kL2

X . From the continuity of the operator E[. ] with respect

L2 t

k 1k |F

to the L2-norm, we infer that:

E[X ] = lim E[X ] = X .

t s t t

1|F s |F

We have now to prove the P-a.s.-convergence of X to X . Let

us consider a sequence X that converges in L2 to X . We can

{

tn}

suppose, extracting if needed an appropriate sub-sequence from it,

that X converges also P-a.s. We now apply Doob’s inequality

{

tn}

to the martingale (Yn) , where Yn := X X :

s s 0 s tn+s tn

sup Yn 2 Yn = 2 Yn = 2 X X

k s 0 | s |kL2  k kL2 k 1kL2 k 1 tnkL2

Therefore the variables sup Yn converge to 0 in L2. Extract-

s 0| s |

ing again an appropriate sub-sequence, we may consider that they

converge P-a.s. to 0. If t t , we have:

X X X X + X X sup Yn + X X

| t 1|| t tn| | tn 1|

s 0

| s | | tn 1|

Since both terms in the right hand of this inequality converge P-a.s.

to 0, we have, as stated, the P-a.s.-convergence of X to X .

4.4 Uniform Integrability:

Definition 4.23 A family L1(⌦, ,P) is uniformly integrable

G⇢ A

lim sup E[ X 1 ] = 0.

c X X>c

!1 2G | | {| | }

Exercise 4.24 If g L1 show that the family := g is a U.I.

2 G { }

family

Proof: We have g 1 g L1 and lim g 1 = 0.

g>c c g>c

| | {| | } | |2 !1| | {| | }

So it follows from the dominated convergence theorem that:

lim E[ g 1 ] = 0. is thus a U.I. family

c g>c

!1 | | {| | } G

Exercise 4.25 Show that if is U.I and if g L1 then g is

G 2 G[{ }

U.I. In particular, finite families of L1 variables are U.I.

Proof: We have:

sup E[ X 1 ] sup E[ X 1 ]+E[ g 1 ]

X g X>c X X>c g>c

2G[{ } | | {| | }  2G | | {| | } | | {| |

but g and are U.I. families so the right hand side terms tend to

{ } G

0 when c tends to . Thus lim sup E[ X 1 ] = 0.

c X g X>c

1 !1 2G[{ } | | {| | }

Exercise 4.26 Show that if is a bounded subset of Lp( ) (p > 1)

G A

then is U.I.

Proof: Let M < such that g : E[gp] M. Applying

1 8 2G 

Ho¨lder’s inequality, with q such that 1/p+1/q = 1, we have g :

8 2G

E[ g 1 ] (E[ g p])1/p(E[1 ])1/q M1/p(P( g > c ))1/q.

g>c g>c

| | {| | }  | | {| | }  {| | }

By Chebichev’s inequality, we also have P( g > c )cp E[ g p]

{| | }  | | 

M. Also E[ g 1 ] M1/pM1/q = M . Hence

| | {|g |>c }  cp/q cp/q

supE[ g 1 ] !1 0.

g | | {|g |>c }  cp/q !

Theorem 4.27 is U.I if and only if satisfies the following 2

G G

properties:

1. is bounded in L1: M < : g E[ g ] M.

G 9 1 8 2G | | 

2. ✏> 0 : > 0 such that, A , if P(A) < then g

8 9 8 2A 8 2G

E[ g 1 ] ✏.

| | 

Proof: Supposethatclaims1)and2)aretrue. Then,fromChebichev’s

inequality, we get: f : M cP( f > c).

8 2G | |

Let ✏> 0 and consider the corresponding to 2). If c > M then

f : P( f > c) < M < . So, from property 2), we have:

8 2G | | c

E[ f 1 ] ✏. Thus sup E[ f 1 ] ✏. Hence, we have

f >c f f >c

| | {| | }  2G | | {| | } 

shown that lim sup E[ f 1 ] = 0.

c f f >c

!1 2G | | {| | }

Conversely, suppose is U.I. then c such that:

G 9

sup E[ f 1 ] 1.

f f >c

2G | | {| | } 

Therefore, f , E[ f ] = E[ f 1 ] + E[ f 1 ] 1 + c.

f >c f c

8 2G | | | | {| | } | | {| | } 

Thus 1) is true with M = 1+c.

Let ✏> 0 then c such that f : E[ f 1 ] ✏. Let’s

9 8 2G | | {|f |>c }  2

take := ✏ and let A such that P(A) . Then:

2c 2A 

✏

E[ f 1 ] E[ f 1 ]+E[ f 1 ] +c P(A) ✏.

A f >c f c A

| |  | | {| | } | | {| | }\  2 · 

Property 2) is satisfied as well.

Theorem 4.28 Let X be a sequence of r.v. in L1( ) such

{ n }n 2N A

that lim X = X P-a.s. Then, for X to converge to X in L1,

n n n

a necessary and sucient condition is X be U.I

{ n }n 2N

Proof: If lim X = X in L1 then X is bounded in L1 and

n n n

!1 { }

X L1 so X X is bounded in L1.

2 { }[{ }

Let ✏> 0 then N such that n N,E[ X X ] ✏.

9 8 | n |  2

Thefinitefamily = X ,X ,...,X ,X isU.I.Thereforethere

0 1 N

G { }

exists > 0 such that if P(A) < , then E[ X 1 ] < ✏ for n =

| n | A 2

0,...,N and E[ X 1 ] < ✏.

| | A 2

Therefore, for all n: E[ X 1 ] ✏. This relation is obvious if

n A

| | 

n N. Similarly, if n > N, we have:



E[ X 1 ] E[ X X 1 ]+E[ X 1 ]

n A n A A

| |  | | | |

E[ X X ]+E[ X 1 ]

n A

 | | | |

✏ + ✏

 2 2

= ✏.

The family X is thus U.I.

{ n }n 2N

Let’s show now that if the sequence, X is U.I., it converges

{ }

in L1: X is a bounded sequence in L1: M such that E[ X ]

n n

{ } 9 | | 

M. Let g := inf X . Then g constitutes an increasing

m n m n m

| | { }

sequence of r.v., and, since X converges to X P-a.s., we have P-

a.s:

lim g = liminf X = X .

m m n

!1 | | | |

By the monotone convergence theorem, E[g ] E[ X ] and as

% | |

g X , it follows: E[g ] E[ X ] M and thus E[ X ] M

m m m m

| |  | |  | | 

. Therefore X L1. Hence X X is U.I. and by 2) we have:

2 { }[{ }

✏> 0, > 0: A : P(A) < n : E[ X 1 ] ✏ and

8 9 8 2A ) 8 | n | A  3

E[ X 1 ] ✏.

| | A  3

Let c := 2M. Then the sequence 1 X X converges

{|X |

P-a.s. to 0 and is bounded by 2c. Then it converges in L1 by the

LDCT to 0, so N : n N :

9 8

✏

E[1 X X ] .

{|X |

|  3

If n N, then

E[ X X ] E[1 X X ]+E[ X 1 ]+E[ X 1 ],

| 

{|X |

| |

where A := X c X c . Since P(A) P( X

n n

{| | } [ {| | }  | |

c)+P( X c) 2M = ,itfollowsthatE[ X X ] ✏+✏+✏ = ✏.

| |  c | n |  3 3 3

We have thus shown that lim E[ X X ] = 0.

n n

!1 | |

Exercise 4.29 If is a family of -algebras and if X

s s S

{G } 2 ⇢A 2

L1( ), show that the family E[X ] is U.I.

s s S

A { |G } 2

Proof: Let X := E[X ]. Jensen’s inequality indicates that

s s

X E[ X ]. Also, since X > c , it follows that

s s s s

| | | ||G {| | }2G

E[ X 1 ] E[ X 1 ]. Since X is U.I.,

{|Xs |>c

}  | |

{|Xs |>c

} { }

✏> 0, > 0 : P(A) < E[ X 1 ] <✏.

8 9 ) | |

If c > E[ X ]/, then s S:

| | 8 2

P( X > c ) E[ X ]/c E[ X ]/c < ,

s s

{| | }  | |  | |

andthusE[ X 1 ] E[ X 1 ] ✏. Thefamily X

{|Xs |>c

}  | |

{|Xs |>c

}  {

s }s 2S

is thus U.I.

4.5 Continuous Time Martingales

Theorem 4.30 (Stopping Theorem): If X is a continuous martin-

gale and if ⌧ is a bounded stopping time : M ! , ⌧(!) M ,

9 8 

then:

X = E[X ] (2)

⌧ M ⌧

In particular: if ⌧ are 2 stopping times, and if is bounded,



then

X = E[X ].

⌧ ⌧

Proof: By the stopping theorem 4.9, the relation (2) is true if ⌧

takes finitely many values.

If ⌧ is a general stopping time, let us define

1 Mk

⌧ (!) := 1

n n {M(k n1)<⌧(!) M nk

}

k=0

Let us first prove that ⌧ is also a stopping time: Let t 0 and

let k be the largest integer k such that Mk t. Then

⇤ n 

⇤

⌧ t = ⌧ .

{

 } {  n }2F

M nk⇤

⇢F

⌧ is thus an -stopping time.

n t

Since ⌧ ⌧ when n , the continuity of the trajectories of

! ! 1

X enables us to conclude that X X P-a.s.

⌧n

⌧

⌧ takes at most n+1 values so, according to theorem 4.9, X =

n ⌧n

E[X ]. The family X is then U.I. as in exercise 4.29.

M |F⌧n

{

⌧n}n 2N

TheP-a.s. convergenceofX andtheU.I.natureofthesequence

⌧n

enable us to conclude, with the theorem 4.28 that X converges to

⌧n

X in L1.

⌧

If Z L ( ) L ( ) then: E[X Z] = E[X Z]. But,

1 F⌧

⇢

1 F⌧n ⌧n M

X X in L1 et Z L and so E[X Z] = lim E[X Z] =

⌧n

⌧

1 ⌧ n

⌧n

E[X Z]. X being -measurable and satisfying the previous in-

M ⌧ ⌧

equality Z L ( ), we conclude that : X = E[X ].

1 ⌧ ⌧ M ⌧

8 2 F |F

Let’s prove the second assertion: if ⌧ M, then

 

E[X ] = E[E[X ] ] = E[X ] = X ,

⌧ M ⌧ M ⌧ ⌧

|F |F |F |F

as X = E[X ] and .

M ⌧

|F F ⇢F

Theorem 4.31 If X is a continuous martingale, uniformly inte-

grable (i.e. the family X is U.I.), then X L1( ) such

t t 0

{ } 9 1 2 F1

that X converges P-a.s. to X and in L1 when t .

1 ! 1

In addition, for every stopping time ⌧: X = E[X ].

⌧ ⌧

1|F

Proof: This theorem generalizes the corollary 4.22. The proof is

not provided here.

Corollary 4.32 (GeneralStoppingTheorem)IfX isa -continuous

{F }

martingale that is U.I., then for all stopping times ⌧ : X =

⌧



E[X ].

⌧

Proof: The proof consists of replacing X by X in the end of the

theorem 4.30.

Remark 4.33 We cannot eliminate the hypothesis that X is U.I.

in the previous theorem, as shows the following example: Let B a

B.M. , = inf t B 1 . Since the brownian motion attains all

{ | }

points of R infinitely many times, we conclude that < P-a.s.

and therefore B = 1 P-a.s. Let ⌧ be the constant time ⌧ = 0, then

0 = B = E[B ] = 1.

⌧ ⌧

6 |F

Definition 4.34 If X is a process and ⌧ is a stopping time, X⌧

denotes the process X⌧ := X . X⌧ coincides thus with X till time

t ⌧ t

⌧, and, after that time, the process X⌧ becomes constant with value

X . X⌧ is sometimes referred to at the process X killed at time ⌧.

⌧

Theorem 4.35 IfX isan -martingaleand⌧ isan -stopping

t t

{F } {F }

time, then X⌧ is also an -martingale.

{F }

Exercise 4.36 Prove this theorem.

Exercise 4.37 Let B be a Brownian Motion, let a,b be two real

numbers with a < 0 < b. We set ⌧ := inf t > 0 : B = a , and

a t

{ }

⌧ := inf t > 0 : B = b . We also set ⌧ := ⌧ ⌧ .

b t a b

{ } ^

1) Prove that B⌧ is a U.I. martingale.

2) Prove that E[B ] = 0.

⌧

3) Compute p(⌧ <⌧ ).

a b

4) Compute E[⌧].

To conclude this section on continuous time martingales, let us

state without proof the following theorem on the ”regularization”

of the trajectories:

Theorem 4.38 If is a complete right-continuous filtration,

{F }

then every martingale X admits a modification X whose trajectories

are right continuous and the left limit exists for all t.

5 Itˆo’s Integral

5.1 The space 2 :

Let B be a brownian motion and a process on (⌦ , ). We would

t t t

like to define a process Y that is the integral:

Y = dB . (3)

t s s

The first idea that comes to mind is to work on a given trajectory:

we fix ! and try to define Y (!) := (!)dB (!). If f and g are

t 0 s s

functions from R to R, the Riemann- RStieltjes integral theory defines

f(s)dg(s) as the limit of Riemann sums ⌃f(s )(g(s ) g(s ))

0 i i+1 i

on partitions 0 = s < s < ... < s = t in [0,t] when the diameter

R 0 1 n

max s s of these partitions tends to 0. For this limit to exist,

i i+1 i

| |

we have to impose conditions on f and g.

Exercise 5.1 Show that if g is continuous and f := 1 then the

[a,b[

Riemann sums converge to g(t b) g(t a). Show also that if g

^ ^

is continuous, f(s)dg(s) is a linear map on the vector space

0 R

spanned by the functions 1 , a b.

R [a,b[



If we want to integrate more general functions f than the ones

of the previous exercise, for instance continuous functions, one has

to restrict the class of g-functions considered : Riemann-Stieltjes

theory suggests that g should be a function of finite variation in

order to guarantee the convergence of the Riemann-Stieltjes sums.

The trajectory t B (!) being generically of infinite variation, this

theory cannot be applied to our case.

To define the integral (3), we have to limit the class of processes

. The following is the first space we consider:

Definition 5.2 2 is the set of - progressively measurable pro-

H2 Ft

cesses such that:

2 := E[ 1 2(s)ds] < .

k kH 22 1

H2 is the quotient of 2 for the equivalence relation , where

2 H2 ⌘ ⌘ 0

if and only if 2 = 0.

k 0 kH 22

Exercise 5.3 Show that 2 is a vector space and that . is a

H2 k kH 22

semi-norm on this space. H2 is hence a normed vector space.

Remark 5.4 For ↵> 1, the following norms are sometimes consid-

ered k

kH 2↵

= E[( 012 sds)↵ 2] ↵1 , their corresponding spaces being

↵.

H2 R

Exercise 5.5 Let t < t and L2( ). We take (!) :=

1 2

Ft1 t

(!)1 (t). Show that 2 and compute .

[t1,t2[ 2H 2 k kH 22

Proof:

1) is progressively measurable, indeed:

Let T fixed, if T < t then :⌦ [0,T] 0 so is

1 T [0,T]

⇥ ! F B

measurable since it is a constant map.

IfT t then(!,t) = (!)1 (t),but (!)is -measurable

1 [t1,t2[ Ft1

andtherefore -measurableand1 (t)is measurable,hence

FT [t1,t2[ B[0,T]

(!,t) = (!)1 (t) is -measurable.

[t1,t2[ FT B[0,T]

2) 2 =E[ 12(s)ds] = E[ 112 (t) 2(!)dt]

k k H22 0 N 0 [t1,t2[

R R

= E[(t t ) 2(!)] = (t t )E( 2(!)) < .

2 1 2 1

Thus 2.

2H 2

Definition 5.6 Wedefine scasthevectorspacespannedby: (!)1 :

E {

[t1,t2[

t < t , L2( ) .

1 2

Ft1

}

Theorem 5.7 (H2, . ) is a Hilbert space.

2 k kH 22

Proof: H2 is a linear subspace of L2( ,µ) with µ =

2 F1 B[0, 1[

P , P being the probability measure on and the Lebesgue

measure on [0, [. Indeed:

2 = 2(!,t)dµ(!,t)

k kL2 ⌦ [0, [

⇥ 1

( 12(!,t)dt)dP(!)

⌦ 0

= RE( R012(!,t)dt).

L2( ,µ) being a HilRbert space, the first assertion follows

[0, [

F1 B 1

from the fact that H2 is closed in L2( ,µ):

N 2 F1 B[0, 1[

Let in L2( ,µ) with H2. Let us show that

n ! F1 B[0, 1[ nN 2 2

H2. For a fixed T, we have:

2 2 N

= E[ T ( (!) (!))2dt]

k n kL2( F1 B[0,T]) 0 n,t t



E[ R01( n,t t)2dt]

= 2 0.

k Rn kH 22 !

The restriction of to ⌦ [0,T] is the limit in L2( µ)

T [0,T]

⇥ F B

of restrictions of to ⌦ [0,T]. The restriction of to [0,T] is

n N

⇥

-measurable. ThisbeingtrueforallT, isprogressively

T [0,T]

F B

measurable and thus in H2.

N 2

Theorem 5.8 sc is dense in H2.

E 2

Proof: 1) If f L2([0, [) and n N, we define T n(f) by T n(f)

2 1 2

1 k=1(n kn

f(s)ds)1 [k,k+1[(t).

n n n

Let’s show that T (f) L2 and that T (f) 2 f 2 .

P R n 2 k n kL2  k kL2

kT n(f) k2 L2 = 01( 1 k=1(n kn n1 f(s)ds)1 [ nk,k+ n1[(t))2dt

= R 01 P1 k=1(n R kn

f(s)ds)21 [k,k+1[(t)dt

n n n

= R 1P(n n Rf(s)ds)21.

k=1 k 1 n

P R

Applying Jensen’s inequality: (E[f(U)])2 E[f2(U)] with U a uni-



form variable on [k 1, k], we get:

n n

T (f) 2 1 n f2(s)ds = 1 f2(s)ds = f 2 < .

n kL2

 k

kL2

Xk=1Zk n1 Z0

Let’s now show that T n(f) n !1 f in L2: the space K([0, [) of

! C 1

continuous functions on a compact support is dense in L2, thus if

L2([0, [) then ✏> 0, f ([0, [): f < ✏.

2 1 8 9 2C K 1 k kL2 3

But, the uniform continuity of f implies that T (f) converges

uniformly to f and thus T (f) f 0. Thus, N : n N :

n L2

k k ! 9 8

T (f) f < ✏. Therefore:

k n kL2 3

T () T ( f) + T (f) f + f

n L2 n n

k k  k k k k k k

✏ + ✏ + ✏ = ✏.

 3 3 3

This being true for all ✏, we conclude that T n(f) n !1 f in L2.

2) If H2 is such that = 0, t T, the process n defined

2 2 t 8

1 n

n(!,t) = T ((!,.))(t) := (n (!,s)ds)1 (t)

n [k,k+1[

k 1 n n

Xk=1 Z n

belongs to sc. Indeed, this sum has only finitely many non-zero

terms. But, if s k, (!,s) (!,s) is - measur-

 n ! Fnk ⌦B [0, nk]

able. Therefore, with Fubini’s theorem, ! n (!,s)ds is -

! k n1 Fnk

measurable. In addition

k k

n n

E[(n (!,s)ds)2] E[n (!,s)2ds)] < .

 1

k 1 k 1

Z n Z n

Thus n(!,t) sc.

Let us next prove that n in H2-norm as n : Setting

! 2 ! 1

Y n(!) := 01(n(!,t) (!,t))2dt, we observe that

R Y (!) = T ((!,.)) (!,.) 2 .

n k n kL2([0, [)

Also, !, Y (!) 0 when n . In addition

8 ! ! 1

Y (!) = T ((!,.)) (!,.)

n n L2([0, [)

k k 1

T ((!,.)) + (!,.)

p n L2([0, [) L2([0, [)

 k k 1 k k 1

2 (!,.)

L2([0, [)

 k k 1

Thus Y n(!)



4 01((!,t))2dt. The right hand side of this inequal-

ity has a finite expectation. We can therefore apply the LDCT to

conclude that n = E[Y ] 0.

H2 n

k k 2 !

3) If H2, we will show that 1 converges to in H2 when

2 2 [0,T[ 2

T tends to . The theorem will then be proven since, 1 can

[0,T[

be approximated by an arbitrarily close process in sc.

Since ( 1 )2 = 1 2 2 and that 1 2 T !1 0, we

[0,T[ [T, [ [T, [

1  1 !

only have to apply again the LDCT in the space L1( ,µ)

[0, [

F1 B 1

to conclude that 1 0.

[0,T] H2 N

k k 2 !

5.2 Itˆo’s integral on H2

We will now define the integral (3) of processes in H2. We could

have defined the variable Y = dB for a fixed time t, but such a

t 0 s s

definition would have given a variable in L2( ): Y would have been

R t t

defined up to a set of measure 0 and nothing would then guarantee

that the resulting process Y has a regular version. For this reason,

we will define the process Y as a whole and denote it I().

Definition 5.9 If sc, then for all ! the trajectory (!) is

in the space of the exercise 5.1. We can therefore define I()

trajectory by trajectory:

I() (!) := (!)dB (!),

t s s

the integral we consider is the one defined in exercise 5.1. In par-

ticular, if (!) = (!)1 (t), where t t and L2( ), we

t [t1,t2[ 1



Ft1

have

I() = (B B ).

t2 ^t

t1 ^t

Weremarkthat, if (!) = (!)1 (t), thedefinedprocess I()is

t [t1,t2[

-adapted and continuous. The integral of the exercise 5.1 being

{F }

linear on , the application I will be linear as well, and I is thus a

linear map from sc to the space of continuous and -adapted

E {F }

processes.

Exercise 5.10 If (!) = (!)1 (t), where t t and

t [t1,t2[ 1



L2( ), show that I() is a martingale and compute I() .

Ft1

kL2

Proof: Let s > t.

1) First suppose that t [t ,t ]: then L2( ), and t s =

1 2 t 1

2 2 F ^

t = t t t. So

1 1

^ 

E[I() ] = E[ (B B ) ] = (E[B ] B ).

s |Ft

t2 ^s

t1 ^s |Ft

t2 ^s |Ft

t1 ^t

Since B is a martingale and t s t = t t, we have

2 2

^ ^

E[I() ] = (B B ) = I() .

s |Ft

t2 ^t

t1 ^t t

2) If t < t then, let s t , and therefore

1 1



E[I() ] = E[0 ] = 0 = I() ,

s t t t

|F |F

let s > t , that is, from case 1):

E[I() ] = E[E[I() ] ] = E[I() ] = E[0 ] = 0 = I() .

s |Ft s |Ft1 |Ft t1|Ft |Ft t

3) If t > t , then I() = I() = I() , and since I() est

2 t t2 s t

-measurable, we have also E[I() ] = I() .

t s t t

F |F

Finally,

I() 2 = lim I() 2

k kL2 t !1k t kL2

= I() 2

t2kL2

= E[ 2(B B )2]

= E[ 2](t t ).

2 1

Comparingwiththeresultweobtainedinexercise5.5,weseethat

I() = . The following theorem generalizes this property:

L2 H2

k k k k 2

Theorem 5.11 I isalinearandisometricapplicationfrom( sc, . )

E k k 2

to (M2, . ).

k k

Proof: We know on one hand that I is linear. On the other hand

is of the form 1 , it follows from the previous exercise that

[t1,t2[

I() M2. By linearity, this property applies to every sc.

2 2E

We remark that if sc, then = n 1 with tk tk,

2E 0 k [tk 1,tk 2[ 1  2

L2(

Ftk

1) and [tk 1,tk 2[ \[tk 10,tk 20[=

;

si k

6= k 0. The computation

gives:

I() 2 = E[I()2 ]

k kL2

= E[( nI( 1 ) )2]

0 k [tk,tk[

1 2 1

= E[( n (B B ))2]

P0 k tk

= E[ n 2(B B )2]

P0 k tk 2 tk 1

+2E[ (B B )(B B )]

P k

= nE[ 2](tk tk)

0 Pk 2 1

since the intervals [tkP,tk[ and [tj,tj[ are disjoint if k < j, that is,

1 2 1 2

E[ (B B )(B B )] = 0. Besides:

k j tk

k k2

H 22

= E[ 01( n

[tk 1,tk

2[)2dt]

= E[

01( Pn

k21

[tk 1,tk

2[)dt]

= E[ 2(!)(tk tk)]

R k Pk 2 1

= E( 2)(tk tk)

Pk k 2 1

= I() 2 .

P k kL2

Hence, we conclude that I() = : I is an isometric appli-

L2 H2

k k k k 2

cation.

Corollary 5.12 If sc converges to H2 in . , then

{ n }⇢E 2 2 k kH 22

the sequence I( ) converges in M2.

{ }

If sc converges as well to , then

{

0n}⇢E

lim I( ) = lim I( ).

n 0n

n n

!1 !1

Proof: Indeed, if converges, it is a Cauchy sequence in H2 so,

{ n } 2

I being linear and isometric:

I( ) I( ) = I( ) = 0.

n m L2 n m L2 n m H2

k k k k k k 2 !

Thus, I( ) is a Cauchy sequence in M2 and, M2 being complete,

{ }

the limit lim I( ) exists. If is another sequence converg-

{

0n}

ingto, wecancreateathirdsequence thattakesalternatively

{

0n0

}

it’s values in the sequences and . Since converges to

{

} {

0n}

{

0n0

}

,thesequence I( ) convergesandallsub-sequencesof I( ) ,–

{

0n0

} {

0n0

}

I( ) and I( ) inparticular–, convergethustothesamelimit.

{

} {

}

Definition 5.13 If H2, there exists a sequence sc

2 2 { n }⇢E

that converges to . We define I() := lim . By the previous

n n

corollary, this limit does not depend on the chosen sequence .

{ }

I() is the Itˆo integral of the process .

Exercise 5.14 Show that I is linear and isometric, and that if

sc: I() = I().

Proof: Let and H2, let also and sc such that

0 2 2 { n } { 0n}⇢E

and 0n

0. Then 8↵,

R : (↵ n + 0n)

(↵+ 0)

and thus

I(↵+) = limI(↵ + )

0 n 0n

= ↵limI( )+limI( )

n 0n

¯ ¯

= ↵I()+I()

¯ ¯

therefore I is linear. I is isometric as well since:

I() = lim I( ) = lim = .

L2 n L2 n H2 H2

k k k k k k 2 k k 2

Finally, if sc, the constant sequence := is a sequence

in sc that converges to . Thus I() = limI( ) = I().

¯ ¯

Definition 5.15 We will now denote I() = 0· tdB t, I()

s b ¯ ¯

dB and dB = I() I() .

0 t t a t t b a R

R R ¯

We remark that I() was defined globally. Nothing indicates that,

for instance I() does not depend on the behavior of the process

after the time s, a property that would have been evident if the Itˆo

integral was defined trajectory by trajectory. The following exercise

shows that this is actually the case:

Exercise 5.16 If T is fixed, show that

H2 : I¯ () = I¯ (1 ) .

8 2 2 T [0,T[ 1

With the integral notations, this means:

dB = 1 (s) dB

s s [0,T[ s s

Z0 Z0

or equivalently T11 [0,T[(s) sdB

= 0.

Proof: Let’s dR efine the applications J : H2 L2( ) and K :

2 ! FT

H2 L2( ) as follows: J() := I¯ () and K() := I¯ (1 ) .

2 ! F1 T [0,T[ 1

J and K are linear as well since I is linear. Let’s show that they

are also continuous. Indeed:

¯ ¯

J() = I() I() = .

L2 T L2 L2 H2

k k k k  k k k k 2

¯ ¯

Similarly, K() = I(1 ) = I(1 ) = 1 ,

L2 [0,T[ L2 [0,T[ L2 [0,T[ H2

k k k 1k k k k k 2

therefore kK()

kL2

= E(

2 sds)



E( 012 sds) = k

22.

Let us take now q:= H2 J(q) = K() . By continuity

A {R 2 2| R }

and linearity of J and K, is a closed linear subspace of H2.

A 2

We observe that, if is of the form = 1 , then:

[t1,t2[

J() = I() = (B B )

T T ^t2

T ^t1

and,since[t ,t [ [0,T[= [T t ,T t [,wehave1 = 1 .

1 2

\ ^

2 [0,T[ [T ^t1,T ^t2[

Thus

K() = I(1 ) = (B B ) = J().

[0,T[

T ^t2

T ^t1

Hence the vector space contains all the of the form =

1 . contains therefore sc that is the vector space spanned

[t1,t2[

A E

by these . being closed, and sc being dense in H2, we have

A E 2

established that H2 : which is what we wanted to prove.

2 ⇢A

Exercise 5.17 Show that if H2, if A then:

2 2 2F t

1. (!) = 1 (!)1 (s) (!) H2.

s A [t, 1[ s 2 2

2. 0· sdB s = 1 A 01 [t, [ sdB s.

We aRctually show heRre that if Z L ( ) then

1 t

2 F

1 1

Z dB = Z dB

s s s s

Zt Zt

Proof:

1) First we show that is progressively measurable. Let T fixed:

If T < t then the restriction of to ⌦ [0,T] is identically 0 and

⇥

is thus -measurable.

T [0,T]

F ⌦B

If T t then 1 is -measurable and thus -measurable and

A t T

F F

1 is measurable, thus 1 1 est -measurable

[t, [ [0,T] A [t, [ t [0,T]

1 B 1 F ⌦B

and, since H2, (!) = 1 (!)1 (s) (!) is -

2 2 A [t, 1[ s Ft ⌦B [0,T[

measurable.

Let’s now compute :

k k 2

2 = E[ 1 2ds] = E[1 1 2ds] E[ 1 2ds] = 2 < .

k kH 22 s A s  s k kH 22 1

Z0 Zt Z0

Thus: H2.

2 2

2) To prove the second assertion, we consider

¯ ¯

J() := I(1 1 ) et K() := 1 I(1 ).

A [t, [ A [t, [

1 · 1

It is easy to see that if is of the form 1 , then J() = K().

[t1,t2[

We have trivially that J and K are linear and continuous appli-

cations from H2 to M2. We can therefore apply the result of the

previous exercise.

Exercise 5.18 If ⌧ is a discrete stopping time: ⌧(⌦) = t ,...,t

1 n

{ }

where t < ... < t , if H2, show that I¯ () =I¯ (1 ) . In

1 n 2 2 ⌧ [0,⌧[ 1

other words:

⌧

dB = 1 (s) dB .

s s [0,⌧[ s s

Z0 Z0

Next show that this relation is true for any stopping time ⌧.

Proof: Let us consider first a discrete stopping time ⌧. Then, since

⌧ = t is -measurable, we have, by the 2 previous exercises

{

}

Fti

that:

¯ ¯ ¯

I(1 ) = I() I(1 )

[0,⌧[ [⌧, [

1 1 1 1

= I¯ () I¯ ( n 1 1 )

1 i=1 {⌧=ti} [ti, 1[ 1

= I¯ () n 1 I¯ (1 )

Pi=1 {⌧=ti} [ti, 1[

= n 1 (I¯ () I¯ (1 ) )

i=1 {⌧=Pti} 1 [ti, 1[ 1

= n 1 I¯ (1 )

Pi=1 {⌧=ti} [0,ti[

= n 1 I¯ ()

Pi=1 {⌧=ti} ti

= I() .

P ⌧

If ⌧ is a general stopping time, there exists a sequence ⌧ of

{ }

discrete stopping times such that n : ⌧ ⌧ and ⌧ ⌧ (see the

n n

8 !

proofofthestoppingtheorem4.30). Bycontinuityofthetrajectories

I(), we have:

¯ ¯ ¯

I() = lim I() = lim I(1 )

⌧

⌧n

[0,⌧n[

!1 !1

To complete the proof, we have to show that the processes 1

[0,⌧n[

converge in H2 to 1 . But this follows from the LDCT applied

2 [0,⌧[

to the norm P on ⌦ [0, [: on one hand 1 converges

⌦ ⇥ 1

[0,⌧n[

pointwise to 1 and on the other hand, n : 1 .

[0,⌧[

8 |

[0,⌧n[

|| |

Exercise 5.19 If ⌧ is a stopping time and X a process, we recall

that the notation X⌧ stands for the process t X⌧ := X . Build

! t ⌧ ^t

on the previous proofs to show that

I¯ ()⌧ = I¯ (1 ).

[0,⌧[

Remark 5.20 It follows from the previous exercise that if ⌧ are



¯ ¯

2 stopping times, then the processes I(1 ) and I(1 ) coincide

[0,[ [0,⌧[

up to time : I¯ (1 ) = I¯ (1 ). We will make use of this

[0,[ [0,⌧[

remark to extend the definition of Itˆo’s integral I to a broader class

of processes.

Definition 5.21 A continuous process X is a -local martingale

{F }

if it is adapted to the filtration and there exists an increasing

{F }

sequence ⌧ of stopping times such that ⌧ P-a.s. and for

n n

% 1

all n: 1 X⌧n 2. We denote loc the set of continuous

{⌧n>0

} 2M M

local martingales and Mloc the quotient of loc by the equivalence

modif

relation .

⌘

Exercise 5.22 Show that every continuous martingale is a locale

martingale.

Remark 5.23 Contrary to what the terminology suggests, a local

martingale is generally not a martingale. The following exercise

illustrates this point.

Exercise 5.24 Let V a finite positive random variable such that

E[V] = . Let also B a brownian motion independent from V. We

take := (V,B ,s [0,t]) and X := V B .

t s t t

F 2 ·

1) Show that X is not in L1 if t > 0. Therefore X cannot be a

martingale.

2) Show that B is a -brownian motion.

{F }

3) Show that ⌧ := n1 is an -stopping time and that ⌧

n V n t n

 {F } %

4) Show that 1 X⌧n = 1 (V n) B . Conclude that

{⌧n>0 } s {V n } ^ · s ^n

X⌧n M2.

Exercise 5.25 Show that if X Mloc, and if ⌧ is a stopping time

such that X⌧ is a bounded process, then X⌧ is a martingale.

We cannot replace bounded with U.I. in the previous argument as

illustrates the example of the exercise 5.48

Definition 5.26 Hloc denotes the set of progressively measurable

processes a such that for all T [0, [:

2 1

a2dt < P-a.s.

t 1

For such a process, we will denote ⌧a the stopping time

⌧a := inf t a2ds n ,

n { | s }

with the usual convention inf = + . The process 1 a is there-

[0,⌧a[

; 1 n

fore in M2. In addition, if a Mloc, ⌧a constitutes an increasing

2 2 n

sequence of stopping times that tends P-a.s. to .

We want to define J() := 0· sdB

for a process in H 2loc. It is

fairly natural to impose that n: J() coincide with

⌧n

dB ,

R 8 ⌧n 0 s s

interpreted as the previously defined integral I(1 ) . We will

[0,⌧n[ 1R

actually require that the processes J() and I(1 ) coincide up

[0,⌧n[

to time ⌧. The following theorem indicates that such a J() exists.

Theorem 5.27 If is a complete filtration, then Hloc,

{Ft } 8 2 2

there exists a unique process J() in Mloc such that n : J()⌧n =

I(1 ).

[0,⌧n[

Proof: I¯ (1 ) is an element of M2, that is an equivalence class

[0,⌧n[

modif

for the relation . Let’s choose a representative Y of this class.

⌘

If n < m, the identity

I¯

(1 ) =

I¯

)⌧n

of the remark 5.20

[0,⌧n[ [0,⌧m[

translates in terms of Y and Y by Y

modif Y⌧n

. The exercise 4.18

n m n ⌘ m

tells us that Y and Y are indistinguishable. Thus P(A ) = 1,

n m n,m

where A := ! t 0 : Y (!) =

Y⌧n

(!) . Let A := A .

n,m { |8 n,t m,t } \n

We also have P(A) = 1.

LetB := ! : lim ⌧(!) = . WehavethereforeP(B) = 1,

{ n !1 n 1}

and so P(A) = 1, where A := A B. If ! belongs to A, we

0 0 0

define Y (!) as folows: there exists n such that t ⌧(!). We take

t  n

Y (!) := Y (!). This definition does not depend on the chosen n

t n,t

such that t ⌧(!). Indeed, if t ⌧(!), and for instance n < m,

 n  m

then Y (!) =

Y⌧n

(!) = Y (!) = Y (!).

n,t m,t m,t ^⌧n m,t

The process Y is therefore well defined on A. Thus we take

Y (!) := 0 if !/ A. The resulting process Y is continuous, and

t 0

if ! A we have Y (!) = Y (!) if t ⌧(!) and if t >⌧ (!),

2 0 n,t t  n n

Y (!) = Y (!) = Y (!). We have just shown that

Y⌧n

n,t n,⌧n(!) ⌧n(!) t

Y 1 .

n,t A

Also !: Y (!) = lim 1 Y (!). Since the filtration

t n A n,t t

8 !1 0 {F }

is complete and that P(A) = 1, 1 Y is -measurable and at

0 A n,t t

0 F

the limit, Y is -measurable as well: Y is adapted to . In

t t t

F {F }

addition, the relation

Y⌧n

= Y 1 implies that Y =

Y⌧n

P-

t n,t A 0 n,t t

a.s. Thus Y modif Y⌧n . Since Y 2, this relation indicates that

n n

⌘ 2M

Y loc. Finally, we define J() as the equivalence class on loc

2M M

for

modif

that contains Y. Thus we have n :

J()⌧n

I¯

(1 ).

⌘ 8

[0,⌧n[

It remains to show the uniqueness of J(): let Z loc such

that n : Y modif Z⌧n , then n : Y⌧nmodif Y modif Z⌧n . Since ⌧ ,

8 n ⌘ 8 ⌘ n ⌘ n % 1

modif

this implies clearly that Y Z.

⌘

Exercise 5.28 Show that the application J:Hloc Mloc: J()

2 ! !

is linear.

Show also that if H2, then J() = I¯ ().

2 2

Definition 5.29 J() is the Itˆo integral of the process Hloc.

2 2

We adopt the following integral notations: J() = dB and

t 0 s s

J() = 0· sdB s.

Exercise 5.30 We have defined the application J : Hloc Mloc

2 !

as the integral with respect to a given brownian motion B. If B1

and B2 are 2 independent brownian motions, we know that B3 :=

1 (B1 + B2) is also a brownian motion. To each one of these

brownian motions, corresponds therefore a di↵erent application J :

Hloc Mloc that we denote respectively by J , J and J . Show that

2 ! 1 2 3

J () = 1 (J ()+J ()). Or equivalently, with integral notations

3 p2 1 2

· dB3 = ( · dB1 + · dB2).

t t p2 t t t t

Z0 Z0 Z0

5.3 Semi-martingales and their quadratic variation:

In the previous section, we have defined the the Itˆo integral with

respect to a brownian motion. We will generalize this construction

in this section to define the Itˆo integral with respect to a semi-

martingale:

Definition 5.31 A semi-martingale is a continuous process X that

is adapted and can be written as X = M+A where M is a continuous

local martingale and A a continuous adapted process starting at 0

and of finite variation: With the notations of the theorem 2.19:

t < : V (A(!)) < P-as.

0,t .

8 1 1

Theorem 5.32 1) A continuous local martingale Y of finite varia-

tion is constant.

2) The representation of a semi-martingale X as X = M + A is

unique.

Proof: 1) Let Y be a continuous local martingale locale Y of finite

variationanddefineZ := Y Y andV (!) := V (Z(!)). Let⌧ :=

t t 0 t 0,t . n

inf t > 0 : max( Z ,V ) n . Since Z is of finite variation, the

t t

{ | | }

increasing sequence of stopping times ⌧ satisfies: lim ⌧ =

n n n

!1 1

P-a.s. and Z⌧n is thus a martingale bounded by n, and its variation

is also bounded by n.

Let = t ,...,t a sub-division of [0,T] and define :=

0 M k

{ }

Z t⌧ kn

Z t⌧ kn. Then Z T⌧n = M k= 01 k.

Therefore E[(Z T⌧n)2] = PE[ M k= 01 k2] + 2 M k= 01 M j= k+1 1E[ k j].

Since k < j, is -measurable and, as Z⌧n is a martingale,

k Ftj P P P

E[ ] = 0, hence E[ ] = 0. So, with := max k , we

j |Ftj k j ⇤ k

| |

have: E[(Z T⌧n)2] = E[ M k= 01 k2]



E[ ⇤ M k= 01 | k |]



E[ ⇤V T]



nE[ ] nE[max Z⌧n Z⌧n : s,t [0,T], s t n ]. Since

⇤  {| s P t | 2 P | || |}

the trajectories of Z⌧n are continuous, they are uniformly continu-

ous on the compact interval [0,T]. It follows that max Z⌧n Z⌧n :

{| s t |

s,t [0,T], s t converges pointwise to 0 when 0.

2 | || |} | |!

Since in addition Z⌧n Z⌧n 2n, we conclude with the LDCT that

| s t |

E[max Z⌧n Z⌧n : s,t [0,T], s t ] 0 when 0

{| s t | 2 | || |} ! | |!

and therefore E[(Z⌧n)2] = 0. This indicates that Z⌧n = 0,P-a.s.

T T

and thus Z = lim Z⌧n = 0,P-a.s., that is, Y = Y P-a.s., as

T n T T 0

claimed.

2)Ifasemi-martingaleX admits2decompositionsX = M+A =

M +A where M and M are local martingales and A and A are

0 0 0 0

processes of finite variation, then M M = A A. The left hand

0 0

side term of this equality is a local martingale of finite variation

since the right hand side term is of finite variation. By 1) we have

therefore M M = A A = A A = 0 P-a.s. if A and A start

t0 0t t 00 0 0

at 0.

A partition n of [0, [ is an increasing sequence n := tn

1 { k}k 2N

such that tn = 0 and tn when k . We suppose in what

0 k ! 1 ! 1

followsthatn n+1 andthat n := sup tn tn : k N 0

⇢ | | { k+1 k 2 }!

when n . For t [0, [, we will denote [t] := max s

! 1 2 1 { 2

n,s t . If X and Y are 2 processes, we take:

 }

Tn(X,Y) := (X X )(Y Y )+(X X )(Y Y ).

t tn k+1 tn k tn k+1 tn k t [t]n t [t]n

k:tn t

Xk+1

Definition 5.33 If there exists a continuous process A of finite

variation such that t : Tn(X,Y) converges in probability to A

8 t t

when n , this process is called covariation of X and Y and

! 1

is denoted X,Y := A . The process X,X , if it exits, is thus

t t

h i h i

increasing and called quadratic variation of X.

Exercise 5.34 If X is a continuous process and Y is a process of

finite variation, show that X,Y = 0.

h i

The lemmas that follow will be useful for the proof of the following

theorem:

Theorem 5.35

1) If X is a local martingale, then X,X exists.

h i

2) X,X is the unique increasing process such that X2 X,X is

h i h i

a local martingale.

3) If ⌧ is a stopping time, then X⌧,X⌧ = X,X ⌧.

h i h i

Lemma 5.36 If t n and m n then

Tn(X,X) Tm(X,X) = 2(X X )(X X )

t t tm j [tm j ]n tm j+1 tm j

j:tm

Lemma 5.37 If X is a continuous martingale bounded by M, and if

t n andm nthen Tn(X,X) Tm(X,X) 2 16p2M2 R ,

2 k t t kL2  k n kL2

where R := max (X X )2 : s s n .

n s s 0

{ 0 | || |}

In addition R 0 when n

n L2

k k ! ! 1

Poof: ThetermsappearingintheexpressionofTn(X,X) Tm(X,X)

t t

computed in the previous lemma have an expectation of 0 and are

not correlated. We thus have:

Tn(X,X) Tm(X,X) 2 = 4E[(X X )2(X X )2]

k t t kL2 j:tm j

4E[R (X X )2]

P n j:tm j

=4E[R Tm(X,X)]

nPt

4 R Tm(X,X) .

 k n kL2 k t kL2

In addition, E[Tm(X,X)2] =

= E[( (X X )2)2]

k:tm k+1t tm k+1 tm k

= E[ (X X )4]+2E[ (X X )2(X X )2]

Pk:tm k+1t tm k+1 tm k j

= E[ (X X )4]+2E[ (X2 X2 )(X X )2]

Pk:tm k+1t tm k+1 tm k Pj

= E[ (X X )4]+2E[ (X2 X2 )(X X )2]

Pk:tm k+1t tm k+1 tm k Pj:tm j+1

4M2E[Tm(X,X)]+4M2E[Tm(X,X)]

 P t t P

= 8M2E[(X X )2]

t 0

32M4



Finally, since X 2, the limit lim X exists almost surely

s s

2M !1

(Corollary 4.22). The trajectories of X are thus almost uniformly

continuous on [0, [ and R converges therefore P-a.s. to 0 since

n 0. Because R 4M2, by the LDCT, we have as stated,

| |! | |

R 0.

n L2

k k !

Lemma 5.38

1)If X is a bounded martingale, then Yn := X2 Tn(X,X) is a

t t t

martingale in 2.

2) Yn converges in ( 2, . ) to a continuous martingale Y.

M k k

3) Tn(X,X) converges thus in L2 to X2 Y that is a continuous and

t t t

increasing process. Hence, we have shown that X,X = X2 Y .

h it t t

4) X,X is the unique continuous process A, increasing and start-

h i

ing at 0 such that X2 A is a martingale.

t t

5) If ⌧ is a stopping time, then X⌧,X⌧ = X,X ⌧.

h i h i

Proof : 1) We have to show that if s t, then E[Yn ] = Yn.

s |Ft t

Let’s first consider the case in which s and t belong to the same

interval [tn,tn ] of the partition n: it follows from the definition

k k+1

of Tn(X,X) that

Yn Yn = X2 X2 Tn(X,X)+Tn(X,X)

s t s t s t

= X2 X2 (X X )2 +(X X )2

s t s tn k t tn k

= 2X (X X ),

tn s t

and thus since, X is -measurable (tn t), we deduce that:

tn k Ft k 

E[Yn ] Yn = E[Yn Yn ] = 2X E[X X ] = 0. The

s |Ft t s t |Ft tn k s t |Ft

general case s [s] = tn > tn > > tn > t tn = [t] follows

n k k 1 ··· l+1 l n

from the previous since E[Yn ] = Yn, E[Yn ] = Yn ,

s |Ftn k tn k tn k|Ftn k 1 tn k 1 ···

and E[Yn ] = Yn. Finally, because X2 is bounde d by M2 and,

tn l+1|Ft t

as we have shown in the previous proof, E[Tn(X,X)2] 32M4, we

t 

conclude that Yn 2.

2) Since Yn Ym = lim Tn(X,X) Tm(X,X) , we

k kL2 t !1k t t kL2

conclude from the previous lemma that Yn is a Cauchy sequence in

( 2, . ) and thus has a limit Y in that space.

M k k

3)So Tn(X,X) = X2 Yn converges in L2 to A := X2 Y that is

t t t t t t

a continuous process. Since if s t are 2 points of n and if m n,

then Tm(X,X) Tm(X,X), we conclude that t n A is

t  s 2 [n ! t

P-a.s. increasing. By continuity of the trajectories, A is therefore a

process P-a.s. increasing ( n is dense in [0, [). 4) We have just

[ 1

shown that Y = X2 A is a martingale. If A’ is another increasing

t t t

process starting at 0 such that X2 A is also a martingale, then

t 0t

A A is a martingale (as a di↵erence of 2 martingales) of finite

variation (as a di↵erence of 2 increasing processes) and is therefore

constant as follows from the theorem 5.32: A A = A A = 0.

0t 0

5) The process Y = X2 X,X being a martingale, the same

t t h it

goes for the process Y⌧ = (X⌧)2 X,X ⌧. The process X,X ⌧

t t h it h it

being increasing, continuous and starting at 0, we conclude with the

assertion 4) of this theorem, applied to the bounded martingale X⌧

that X⌧,X⌧ = X,X ⌧.

h it h it

Exercise 5.39 Prove the theorem 5.35 that is a corollary of the

previous lemma.

Exercise 5.40 1)Show that if X and Y are 2 processes such that

X,X , X,Y and Y,Y exist, then X + Y,X + Y exists and

h i h i h i h i

X +Y,X +Y = X,X + X,Y +2 Y,Y .

h i h i h i h i

2) In particular if X = M + A is a semi martingale, then X,X

h i

exists and X,X = M,M .

h i h i

Theorem 5.41 If B is a brownian motion. If a loc and X =

2H 2

. a dB , then X,X = t a2ds.

0 s s h it 0 s

RProof: The process A R:= t a2ds is continuous, increasing and

t 0 s

starting at 0. By the theorem 5.35, we only have to show that

X2 A is a local martingale. If ⌧ is the stopping time defined by

t t n

⌧ := inf t > 0 : A n , the process an := 1 a is in H2 and

n { t } t t ⌧n t 2

X⌧n = I(an). We have to show that (X⌧n)2 A⌧n is a martingale,

t t

that is, E[(X⌧n)2 (X⌧n)2 ] = E[A⌧n A⌧n ], when s > t. Since

s t |Ft s t |Ft

X⌧n M2, this is equivalent to showing that E[(X⌧n X⌧n)2 ] =

2 s t |Ft

E[ s (an)2dx ], or, C :

t x |Ft 8 2F t

s s

R E[1 ( andB )2] = E[1 (an)2dx].

C x x C x

Zt Zt

But it follows from the exercise 6.1 that

E[1 ( s andB )2] = E[(1 s andB )2]

C t x x C t x x

= E[( s 1 andB )2]

R t RC x x

= E[ s (1 an)2dx]

Rt C x

= E[1 s (an)2dx].

RC t x

Exercise 5.42 Show that if B is a brownian motion, and if a,b

loc, X = . a dB and Y = . b dB , then X,Y = t a b ds.

H2 0 s s 0 s s h it 0 s s

R R R

5.4 Itˆo’s integral with respect to a semi-martingale

. .

If X = M+A, the integral a dX is defined as the sum a dM +

0 t t 0 t t

. .

a dA . Since A is of finite variation, a dA can be defined tra-

0 t t R 0 t t R

jectory by trajectory in the sense of theLebesgue Stieltjes integral.

R R

To define Itˆo’s integral with respect to a semi-martingale, we only

have to define it with respect to the local martingale M.

By the stopping technique used above, the problem sums up to

defining the integral with respect to a bounded martingale M such

that M,M is also bounded.

h i

The construction of this integral is parallel to the one of the

integral with respect to a brownian motion: we define 2(M) as the

space of progressively measurable functions a such that a 2 :=

k kH 22(M)

E[ 01a2 sd hM,M, is] < 1. As previously, the integral I(a) is defined

trajectory by trajectory for a Esc. The resulting application I is

isometric from (Esc, . 2 ) to (M2, . ). Esc being dense in

k kH 22(M) k kL2

(H2(M), . 2 ), I can be extended to an isometric application I

2 k kH 22(M)

from (H2(M), . 2 ) to (M2, . ).

2 k kH 22(M) k kL2

Exercise 5.43 Do the construction by yourselves.

Exercise 5.44 IfX andY aresemi-martingalesandifR = a dX

t 0 s s

and Z = b dY ,

t 0 s s R

1) Show that R,R = t a2d X,X .

R h it 0 s h is

2) Show that R,Z = a b d X,Y

it R0 s s

5.5 Itoˆ’s Formula

Let’s start with a particular case of this formula: if n is an increas-

ing partition of [0, [. If X = M+A is a semi-martingale such that

M and A are bounded, then

X2 = X2 + X2 X2

t 0 k:tn k+1t tn k+1 tn k

= X2 + 2X (X X )+ (X X )2

0 Pk:tn k+1t tn k tn k+1 tn k k:tn k+1t tn k+1 tn k

= X2 + t andX +Tn(X,X)

0 P 0 s s t P

where an := R 2X 1 (s). By continuity of the tra-

s k:tn k+1t tn k [tn k,tn k+1[

jectories an 2X in H2(M) and we can trivially conclude that

s !P s 2

t andX t 2X dX . WealreadyknowthatTn(X,X) X,X .

0 s s ! 0 s s t ! h it

Thus,

R R

X2 = X2 + 2X dX + X,X

t 0 s s h it

and this result can be generalized for all semi-martingales by the

stopping technique.

Exercise 5.45 Show that if X and Y are semi-martingales then

X Y = X Y + X dY +Y dX + X,Y .

t t 0 0 0 s s s s h is

Proof:

We know that X Y = 1[(X +Y )2 (X Y )2]. We only have to

t t 4 t t t t

apply the previous result to semi-martingales X +Y and X Y.

The last exercise shows that the product of 2 semi-martingales is

also a semi-martingale. The following theorem shows that in fact, a

regular function of a semi-martingale is also a semi-martingale:

Theorem 5.46 If f : Rd R is 2, if Xj are semi-martingales

! C

and if X := (X1, ,Xd), then:

···

f(X ) = f(X )

t 0

+ t d @f (X )dXj (4)

0 j=1 @xj s s

0Pt

j,j 0

@x@ j2 @f

j0(X s)d hXj,Xj0 is.

R P

Proof: By the stopping technique of the previous proofs, we only

have to do the proof for semi-martingales X that that values on

a compact set K of Rd. Let the class of functions f : K R

V !

satisfying the theorem.

The constant functions, as well as the functions

gj : x = (x1, ,xd) gj(x) := xj

··· !

are in .

By linearity in f of the formula (4), if f and g are in and

↵, R then trivially ↵f +g .

2 2V

Let’s show that if f and g are in then h := f g also belongs

V ·

to V. Let F := f(X ), G := g(X ), H := g(X ), By the exercise

t t t t t t

5.45, we have:

dH = F dG +G dF +d F,G

t t t t t t

h i

Since f and g are in , we find:

dF = @ f(X ) dXj + 1 @ f(X ) d Xj,Xj0

t j j t · t 2 j,j 0 j,j 0 t · h it

dG = @ g(X ) dXj + 1 @ g(X ) d Xj,Xj0

t Pj j t · t 2Pj,j 0 j,j 0 t · h it

Since the terP ms in d Xj,Xj0 areP of 1-bounded variation , they are

h i

disregarded in the computation of d F,G . We find then:

h i

d F,G = @ f(X ) @ g(X ) d Xj,Xj0 .

t j t j t t

h i · 0 · h i

j,j

Also

dH = (G @ f(X )+F @ g(X )) dXj

t j t j t t j t · t

+1 Lj,j0 d Xj,Xj0

P2 j,j 0 t · h it

where Lj,j0 = (G @ f(XP) + F @ g(X ) + 2@ f(X ) @ g(X )).

t t j,j 0 t t j,j 0 t j t · j 0 t

Since @ h(X ) = G @ f(X ) + F @ g(X ) and @ h(X ) = Lj,j0, we

j t t j t t j t j,j 0 t t

conclude that dH satisfies the formula (4) and therefore h .

From the previous, we conclude that contains all polynomials.

Let ✏> 0. Every function f can be approximated in K by a

polynomial g such that f g ✏, j: @ f @ g ✏

,K j j ,K

k k1  8 k k1 

and j,j : @ f @ g ✏. Since the formula (4) is satisfied

0 j,j j,j ,K

8 k 0 0 k1 

by g, it will be satisfied by f at the limit, which is what we wanted

to provr.

Exercise 5.47 Using Itˆo’s formula, show that, if B is a brownian

motion, then, M = exp(↵B

↵2

t) is a martingale locale.

t t 2

Proof: Let f(x) := exp(x) and X := ↵B

↵2

t. Since X =

t t 2 t

t ↵dB + t ( ↵2/2)ds, we have X,X = t ↵2ds = ↵2t. Itˆo’s

0 s 0 h it 0

formula gives, with f(x) = f (x) = f (x):

R R 0 00 R

M = f(M )

t t

= f(X )+ t f (X )dX + 1 t f (X )d X,X

0 0 0 s s 2 0 00 s h is

= f(0)+ t M ↵dB t M ↵2 ds+ 1 t M ↵2ds

0R s s 0 sR2 2 0 s

= 1+ M ↵dB

0 R s s R R

Since the trajectorRies of B and consequently of M are continuous,

the variable M := sup M : 0 s T is, for all T < , a P-a.s.

T⇤

{

  } 1

finite variable. Since T M2↵2ds M 2 ↵2 T, the process M ↵

0 s  T⇤ · · t

belongs to H 2loc. It’s in Rtegral 0·M s↵dB

is thus a local martingale

and the same goes for M.

Exercise 5.48 LetB1,B2,B3 3independentbrownianmotions. Let

f(x ,x ,x ) := ( 3 (1+x )2) 1/2 and Z := f(B1,B2,B3).

1 2 3 i=1 i t t t t

1) Show that 3 @ f(x ,,x ,x ) = 0, for all (x ,x ,x ) =

P i=1 i,i 1 2 3 1 2 3 6

( 1, 1, 1).

2) Let ⌧ := inf t : Z n . Using Itˆo’s formula, show that Z⌧n

n t

{ }

is a positive martingale bounded by n.

3) Show that E[Z ] = 1/p3. Imply from this result that

⌧n

P(⌧ < ) 1/(np3)

1 

and that ⌧ P-ap. Z is therefore a local martingale.

% 1

4) Let := inf t : Z 1/m . Using the corollary 2.15, show

m t

{  }

that < P-a.s.. Show also that P-a.s..

m m

1 % 1

P-a.s.

Prove that Z 1/m,n P-a.s.. and that Z Z

⌧n ^m

2{ }

⌧n ^m

⌧n

when m . Imply from this that Z = n 1 P-a.s..

! 1

⌧n

{⌧n<

P-a.s.

5) Show that Z 0 when n tends to infinity.

⌧n

6) Using the normal density, show that there exists a constant

C < such that t E[Z2] C.

1 8 t 

7) If Z was a martingale, we would have Z 2. Show that by

the corollary 4.22, Z cannot be a martingale.

Z is therefore a bounded in norm L2 (so U.I.) local martingale

locale that is not a martingale.

Part II

Stochastic Calculus 2

6 Local martingales

In the course of Stochastic Calculus 1, we defined a dB for pro-

0 s s

cesses a in Hloc. The resulting process is a local martingale.

2 R

Definition 6.1 A local martingale is a continuous adapted process

X such that there exists a sequence ⌧ of stopping times such that

{ }

1) For all n⌧ ⌧ (increasing sequence)

n n+1



a.s.

2) ⌧ as n and

% 1 ! 1

3) For all n, 1 X⌧n is a U.I. martingale.

⌧n>0

{ }

A sequence of stopping times satisfying 1), 2) and 3) is said to

”reduce” the local martingale.

Exercise 6.2 Prove that a martingale is a local martingale.

Exercise 6.3 Give an example of local martingale which is not a

martingale

Exercise 6.4 Let ⌧ be a stopping time and X a local martingale.

Prove that X⌧ is a local martingale.

Exercise 6.5 Let X be a bounded local martingale. Prove that X

is a martingale.

OnecouldexpectthataU.I.localmartingalewouldbeamartingale.

In fact this is false as shown in exercise 11.7 hereafter.

Exercise 6.6 Let X be a local martingale and define ⌧ := inf t

{

0 : X n , with inf := . Prove that the sequence ⌧ reduces

t n

| | } ; 1 { }

the local martingale X.

Exercise 6.7 1) Prove that a process X whose initial value X is

a.s. finite, is a local martingale iif Y := X X is a local martin-

t t 0

gale.

2) If Y is a process starting at 0 (i.e. Y = 0), prove then that

1 Y⌧n = Y⌧n. For processes starting at 0, point 3) in the defi-

⌧n>0

{ }

nition of local martingale could be modified in ”For all n, X⌧n is a

U.I. martingale.”

Definition 6.8 A process X is of finite variation iif

a.s.

T : V (X) < .

[0,T] .

8 1

Theorem 6.9 A local martingale X of finite variation is constant:

a.s.

t : X = X .

t 0

Exercise 6.10 Prove this theorem:

1) First prove that a bounded martingale of bounded variation is

constant.

2) If f : [0,T] R is continuous and of finite variation, then prove

that the map t V (f) is continuous.

[0,t]

3) Define ⌧ := inf t : X n or V (X) n . Prove that ⌧ is

n t [0,t] . n

{ | | }

a stopping time.

4) Prove that the sequence ⌧ reduces the local martingale X.

{ }

5) Conclude that theorem 6.9 is true.

Definition 6.11 A continuous adapted process X is called a semi-

martingale iif X = M +A where M is a local martingale and A is

a.s.

a continuous process of finite variation starting at 0: (i.e. A = 0)

Exercise 6.12 Prove that the decomposition X = M+A of a semi-

martingale is unique.

Definition 6.13 loc is defined as the set of progressively measur-

T a.s.

able processes a such that for all T: a ds < .

0 | s | 1

Exercise 6.14 Let B is a brownian motion, a loc and b loc.

2H 2 2H 1

1)ProvethattheprocessAdefinedforall(!,t)byA (!) := b (!)ds

t 0 s

is an adapted process of finite variation. (Observe that this integral

exists for all (!,t): There is thus no diculty to define the process

A = b ds which is defined path by path. In contrast, the definition

0 s

of a dB was much more dicult since B is of infinite variation.)

0Rs s

t t

2) The process X := a dB + b ds is thus a semi-martingale.

t 0 s s 0 s

modif

Prove that X is a local martingale iif b = 0.

R R

Our aim in the following sections is to define a dX when X is

0 s s

a semi martingale. Since the decomposition X = M +A is unique,

we will define a dX as:

0 s s

R t t t

a dX := a dM + a dA

s s s s s s

Z0 Z0 Z0

The next section is devoted to the second integral. The first integral

will be analyzed in section 8.

7 a dA when A is of finite variation.

0 s s

A continuous process A of finite variation is the di↵erence of two

increasing continuous processes A+ and A starting at 0. For all !,

there correspond two positive measures µ+ and µ on [0, [ such

! ! 1

that, for all t, A+(!) = µ+([0,t]) and A (!) = µ ([0,t]). We the

t ! t !

define the positive measure µ := µ++µ , and the signed measure

| |! ! !

µ := µ+ µ .

! ! !

If a is progressively measurable, we may consider the map !

a (!)d µ (s). This integral makes senses for all ! since a posi-

0 | s | | |!

tive measurable function of s can always be integrated with respect

to a positive measure (with possibly infinite values).

Definition 7.1 1(A) is defined as the set of progressively measur-

able processes a such that a < , where:

H1(A)

k k 1 1

a := E[ a (!) d µ (s)].

H1(A) s !

k k 1 | | | |

We also define loc(A) as the set of progressively measurable pro-

cesses a such that for all T: T a (!) d µ (s) a <.s. . If a loc,

0 | s | | |! 1 2H 1

then a dA is defined as the process Y where

0 s s R

R t t t

Y (!) := a (!)dµ (s) = a (!)dµ+(s) a (!)dµ (s)

t s ! s ! s !

Z0 Z0 Z0

Note that both integrals on the right hand side exist and are a.s.

finite:

t t t

a.s.

a (!)dµ+(s) a (!) dµ+(s) a (!) d µ (s) < .

| s ! | | s | !  | s | | |! 1

Z0 Z0 Z0

Exercise 7.2 If a loc(A), prove that . a dA is a process of

2H 1 0 s s

finite variation.

Exercise 7.3 1) If a = (!)1 where ↵< and L1( ),

t [↵,[ ↵

2 F

compute a dA .

0 s s

2) If a was a continuous process, one could ”approximate” any pro-

cess a by a sequence of step processes an := n2 a 1 . Show

i=1 i [i,i+1[

n n n

that T andA is the Riemann-Stieltjes sums:

0 s s P

n a (A A ), where 0 = t < t < < t = T. One

i=0R ni i+ n1 ^T ni ^T 0 1 ··· n

could then prove that these Riemann-Stieltjes sums converge as n

goes to . This is the Riemann-Stieltjes definition of a dA . It

1 0 s s

works with continuous processes a, but not with any measurable pro-

cess in loc(A).

Our definition 7.1 is called Lebesgue-Stieltje definition of a dA .

0 s s

AnalternativeapproachtodefinetheLebesgue-Stieltjeintegral a dA

R 0 s s

would start by proving that sc is dense in 1(A). Defining then

E H1 R

. a dA as the limit of . andA , where a are step processes that

0 s s 0 s s n

converge to a. This would however be longer.

R R

8 a dM when M is a local martingale

0 s s

R .

Our definition a dB when B is a Brownian motion relies on the

0 s s

fact that the operator I is an isometry from sc to 2. In turn the

E M

proof of this result just follows two properties a Brownian motion:

1) B is a continuous martingale and 2) B2 t is a martingale.

t t

Thenextsectionisdevotedtotheanalysisaprocess M,M such

h i

that M2 M,M is a local martingale.

h i

8.1 The covariation process of two semi-martingales

Definition 8.1 A subdivision of R+ is an increasing sequence

t with t = 0 and lim t = . The diameter of is

{ n }n 2N 0 n !1 n 1 | |

defined as sup t t . The following notations are useful in the

n| n+1 n |

sequel of this section: will denote the interval [t ,t [ and, if X

k k k+1

is a process X := (X X ).

k tk+1

Definition 8.2 If X and Y are two processes, T(X,Y) will denote

the process

n⇤1

T(X,Y) := X Y +(X X )(Y Y ),

t k k t t n⇤ t t n⇤

k=0

where n := max k : t t .

⇤ k

{  }

Definition 8.3 A continuous adapted process A of finite variation

is called the covariation process of X and Y if and only if for all t:

T(X,Y) Probability A as 0.

t ! t | |!

If it exists, the covariation process of X and Y is denoted X,Y .

h i

If it exists, the covariation process X,Y is called de quadratic

h i

variation process of X.

Exercise 8.4 Prove the following claims:

1) X,Y = Y,X

h i h i

2) If X,Z and Y,Z exist, then ↵, R : ↵X +Y,Z exists

h i h i 8 2 h i

and ↵X +Y,Z = ↵ X,Z + Y,Z

h i h i h i

3) If X +Y,X +Y and X Y,X Y exist, then X,Y exists

h i h i h i

and

X,Y = ( X +Y,X +Y X Y,X Y )

h i 4 h ih i

Theorem 8.5 Let A be an adapted process of finite variation and

X a continuous process, then X,A = 0.

h i

Exercise 8.6 Prove this theorem. (Hint: using the technique of ex-

ercise 6.10, find an increasing sequence ⌧ of stopping times that

converges a.s. to such that Xn := 1 X⌧n is a bounded contin-

⌧n>0

uous process and An := 1 A⌧n is a process of bounded variation.

⌧n>0

Prove first that Xn,An = 0 and conclude that X,A = 0.

h i h i

In the next subsection (see theorem 8.9) we will prove that the

quadratic variation process M,M of any local martingale exists.

h i

As a corollary, we get:

Theorem 8.7 If X = M + A and X = M + A are two semi-

0 0 0

martingales then X,X exists and X,X = M,M

0 0 0

h i h i h i

Exercise 8.8 Prove this result, assuming theorem 8.9.

8.2 The quadratic variation process of a local martingale

The aim of this section is to prove the following theorem:

Theorem 8.9 If X is a local martingale, then X,X exists: It is

h i

the unique increasing process A starting at 0 such that X2 A is a

t t

local martingale.

The proof of this result is made of several steps.

Exercise 8.10 Prove that there is at most one increasing process A

starting at 0 such that X2 A is a local martingale.

t t

Exercise 8.11 1) If t = t , prove that

n 1

Y := X2 T(X,X) = X2 +2 X X

t t t 0 ti i

i=0

2) For the sequel of this exercise, we assume that X is a martingale

bounded by Kand starting at 0 (One could otherwise consider X :=

X X .) Prove then that Y is a martingale.

t 0

3) Conclude that Y 2. Indeed:

E[(Y)2] 4K2E[T(X,X)] = 4K2E[X2] 4K4.

t  t t 

4) If t = t = t = t , prove that

n 2 { i }i 2N ⇢ 0 { 0i}i 2N

n 1

T t(X,X) T t0(X,X) = 2(X

t0j

X ti) 0jX

Xi=0 j:t0j2X[ti,ti+1[

5) Conclude that E[(T(X,X) T0(X,X))2] 4E[S2T0(X,X)],

t t  t

where S := max X X : s,v [0, [, s v .

s v

{| | 2 1 | || |}

6) With Cauchy Swartz inequality, one gets thus

E[(T(X,X) T0(X,X))2] 4 E[S4] E[(T 0(X,X))2]

t t  t

q q

7) Since X is bounded, it is a U.I. martingale. With theorem 4.31,

a.s.

we thus have that X X . Infer from this that for almost every

! 1

!, the map t [0, [ X (!) is uniformly continuous. Prove next

2 1 !

that E[S4] 0 as 0.

! | |!

8) It follows from 1) and 3) that:

T0(X,X)) X2 + Y0 K2 +2K2

k t kL2  k tkL2 k t kL2 

Conclude then that if is a sequence of subdivisions with

{ n }n 2N

0, then Yn is a Cauchy sequence in 2. It has

| n |! { }n 2N M

therefore a limit Y.

9) Prove that Tn(X,X) L2 X2 Y . Then prove that X2 Y is

t ! t t t t

an increasing process and conclude that X,X = X2 Y .

h it t t

This proves the next theorem:

Theorem 8.12 If X is a bounded martingale, then X,X exists:

h i

It is the unique increasing process A starting at 0 such that X2 A

t t

is a martingale.

Exercise 8.13 Let X be a local martingale. There is no loss of

generality to assume X = 0 (otherwise consider X := X X and

0 t0 t

note that T(X ,X ) = T(X,X).) Consider the sequence ⌧ of

0 0 n

{ }

stopping times of exercise 6.6: ⌧ := inf t 0 : X n .

n t

{ | | }

1) Prove that X⌧n is a bounded martingale.

2) If m n, then Z := (X⌧m)2 X⌧m,X⌧m is a martingale. Z⌧n

t t h it

is thus also a martingale. Prove that X⌧n,X⌧n = X⌧m,X⌧m ⌧n.

h i h i

3) Conclude that there exists an increasing process A such that n :

X⌧n,X⌧n = A⌧n.

h i

4) Conclude that X2 A is a local martingale.

5) Prove that X,X exists and X,X = A

h i h i

Theorem 8.9 is thus prooved.

Exercise 8.14 Let B be an -Brownian Motion, let a be in H2

{Ft } 2

and let X denote X := a dB .

0 s s

1) If A , prove that, for all T > t:

t R

T T

a.s.

11 a dB = 11 a dB .

A s s A s s

Zt Zt

2) Prove that Y := X2 t a2ds is an -martingale.

t t 0 s {Ft }

3) Conclude that if a Hloc, then X,X := . a2ds

2 R2 h i 0 s

Exercise 8.15 Let X and Y be two local martingales and let :=

(X ,s t) and := (Y ,s t). We also set := ( ).

s t s t t t

 G  H F [G

In this exercise, we aim to prove the following claim:

If , then X , Y and X Y are -local martingales. In

t t t t t

F1?? G1 {H }

particular X,Y = 0.

h i

1) Let denote the set of finite unions n F G , where F

S [j=1 j \ j j 2F t

and G . Prove that is an algebra of sets such that ( ) = .

j t t

2G S S H

2) Prove that if R , then R = n F G , where F and

2S [j=1 j \ j j 2F t

G , and where F G F G = if i = j.

j t i i j j

2G \ \ \ ; 6

3) First consider the case where X and Y are martingales. If s t,

let denote the class of A such that :

C 2H

E[(X X )1 ] = E[(Y Y )1 ] = E[(X X )(Y Y )1 ] = 0

s t A s t A s t s t A

Prove that is a monotone class that contains . It contains there-

C S

fore . Conclude that X , Y and X Y are -martingales.

t t t t t t

H {H }

4) Consider then the case where X and Y are local martingales. and

reduce it by stopping to the previous case.

8.3 a dM when M is a martingale

0 s s

Let M be a bounded martingale such that M,M is bounded.

h i

Definition 8.16 Define 2(M) is the set of progressively measur-

able processes a such that a < , where

H2(M)

k k 2 1

a 2 := E[ 1 a2d M,M ]

k kH 22(M) s h is

Note that M,M is increasing, therefore the integral of the right

h i

hand side makes sense.

Definition 8.17 loc(M) is the set of progressively measurable pro-

cesses a such that T : T a2d M,M a <.s. .

8 0 s h is 1

Two processes a and b in loc(M) are equivalent iif

R H2

T : (a b )2d M,M a =.s. 0.

s s s

8 h i

H2(M) and Hloc(M) are the quotient of 2(M) and loc(M) with

2 2 H2 H2

respect to this equivalence relation

Exercise 8.18 Prove that sc is dense in H2(M)

E 2

Definition 8.19 If a sc, then for all ! ⌦, the map t a (!)

2E 2 !

belongs to . Therefore the following integral makes sense and the

process I(a : M) is defined as:

I(a : M) (!) := a (!)dM (!),

t s s

Alternatively, we could define

n n

I( 1 : M) (!) = (!)(M (!) M (!))

i [↵i,i[ t i i^t

↵i^t

i=1 i=1

X X

Exercise 8.20 If a sc,

1) Prove that I(a : M) 2.

2) Prove that I(a : M) = a

L2 H2(M)

k k k k 2

3) Prove that I( : M) has a unique continuous extension

I( : M) : H2(M) M2

· 2 !

4) Prove that I( : M) is linear and I(a : M) = a

L2 H2(M)

· k k k k 2

Definition 8.21 The process I(a : M) is also denoted a dM .

0 s s

Theorem 8.22 Let M and M be two bounded martingales with

bounded quadratic variation. If a H2(M), a H2(M ), if X :=

2 2 0 2 2 0

. .

a dM , X := a dM and if ⌧ is a stopping time, then

0 s s 0 0 0s s0

1) X⌧ = . 1 a dM

R 0 s<⌧R s s

{ }

2) X⌧ = . a dM⌧.

R0 s s

3) Y := X2 t a2d M,M is a martingale.

t Rt 0 s h is

4) X,X = a a d M,M .

R0 s 0s

0 is

Exercise 8.23 Proof of the previous theorem.

a) If ⌧ is a stopping time taking finitely many values, if a sc,

prove that 1 a sc, and that 1) holds in this case.

t<⌧ t

{ } 2E

b) Prove that 1) still holds when a 2(M).

2H 2

c) If ⌧ is a general stopping time, prove that there exists a sequence

a.s.

⌧ of stopping times taking finitely many values such that ⌧ ⌧.

n n

{ } !

H2(M)

d) Prove that 1 a 2 1 a and conclude that 1) holds.

{s<⌧n

}

{s<⌧

}

e) The maps a X⌧ = ( . a dM )⌧ and a . a dM⌧ are con-

! 0 s s ! 0 s s

tinuous from H2(M) to 2. They further coincide on sc. This

2 MR R E

proves 2).

f) If s t, the map a E[Y ] Y is continuous from H2(M)

! s |Ft t 2

to L1. It maps sc to 0. Prove it and infer that 3) holds.

g) If a sc, prove that X X a a d M,M is a martingale

2E t t0 0 s 0s h 0 is

and that 4) holds in this case.

h) If a H2(M) and a H2(M ), there exists sequences an

2 2 0 2 2 0 { }

and an in sc converging to a and a respectively in H2(M) and

{ 0 } E 0 2

H2(M ). Setting Xn := I(an : M) and X n := I(an : M ), prove

2 0 0 0 0

that XnX n converges in L1 to X X .

t t0 t t0

i) Assuming Kunita-Watanabe inequality here below (5), prove that

t anand M,M is converging in L1 to t a a d M,M .

0 s 0s h 0 is 0 s 0s h 0 is

j) Infer from h) and i) that X X a a d M,M is a martingale,

R t t0 0 s 0sR h 0 is

and thus that 4) holds for any a H2(M) and a H2(M ).

2R 2 0 2 2 0

Theorem 8.24 Let M and M be two local martingales, let a

H2(M) and a H2(M ). If s > t, we set M,M s := M,M

2 0 2 2 0 h 0 it h 0 is

M,M .

0 t

h i a.s.

1) M,M s M,M s M ,M s

|h 0 it|  h it h 0 0 it

a.s.

t t t

2) | 0 a sa 0sd hM,Mp0 is |  p 0 a2 sd hM,M is 0 a 0s2d hM 0,M 0 is

3) Kunita Wattanabe ineqquality: q

R R R

E[ a a d M,M ] a a (5)

s 0s

0 is

|  k

kH 22(M)

0 kH 22(M 0)

Exercise 8.25 Prove this theorem:

a) For R, define Y := M +M 0. Prove that p(A) = 1, where

A := ! : R : Y,Y s(!) 0 .

{ 8 2 h it }

b) For ! A, we thus have :

2 8

M,M s(!)+2 M,M s(!)+2 M ,M s(!) 0.

h it h 0 it h 0 0 it

This is only possible if 1) holds.

b) If a,a sc, prove 2) using a Cauchy Shwartz inequality.

c) Extend the result 2) for a H2(M) and a H2(M )

2 2 0 2 2 0

c) Claim 3) follows from 2) with another Cauchy Shwartz inequality

8.4 a dM when M is a local martingale

0 s s

Let MR be a local martingale and let a Hloc(M). Let ⌧ denote

2 2 n

⌧ := inf t 0 : M n or M,M n or a2d M,M n .

n { | t | h it s h is }

The a H2(M⌧n). Let Y denote Y := . a dM⌧n.

2 2 n n 0 s s

Exercise 8.26 1) Prove that ⌧ is an inRcreasing sequence of stop-

a.s.

ping times such that ⌧ .

! 1

2) If m n, prove that Y = Y⌧n. Conclude that Y converges a.s.

n m n,t

to a limit Y .

3) Prove that Y is a continuous adapted process and that Y⌧n = Y .

Conclude that Y is a local martingale.

Definition 8.27 The process Y thus obtained is denoted a dM .

0 s s

Theorem 8.22 has a local martingale version: R

Theorem 8.28 Let M and M be two local martingales. If a

Hloc(M), a Hloc(M ), if X := . a dM , X := . a dM and if ⌧

2 0 2 2 0 0 s s 0 0 0s s0

is a stopping time, then

R R

1) X⌧ = . 1 a dM

0 s<⌧ s s

{ }

2) X⌧ = . a dM⌧.

R0 s s

3) Y := X2 t a2d M,M is a local martingale.

t Rt 0 s h is

4) X,X = a a d M,M .

R0 s 0s

0 is

Exercise 8.29 Prove this theorem.

Exercise 8.30 If M is a local martingale, if a Hloc(M), if X =

2 2

. a dM then X is a local martingale. Consider then b Hloc(X)

0 s s 2 2

and set: Y := b dX .

R 0 s s

1) Compute Y,Y .

h i

2) Prove that a b Hloc(M).

s s 2 2

3) Prove that Y := b a dM

0 s s s

9 Integral with respect to a semi-martingale

Definition 9.1 If X is a semi-martingale with decomposition X =

M +A, if a loc(M) loc(A), we define . a dX by

2H 2 \H1 0 s s

. . . R

a dX := a dM + a dA .

s s s s s s

Z0 Z0 Z0

Since the decomposition ofX is unique, there is no ambiguity in this

definition. Note also that the resulting process a dX is afresh a

0 s s

. .

semi-martingale as a dM is a local martingale and a dA is a

0 s s R 0 s s

process of finite variation.

R R

Definition 9.2 A progressively measurable process a is called locally

bounded iif there exists an increasing sequence ⌧ of stopping times

{ }

that converges a.s. to , such that 1 a⌧n is bounded, for all n.

{⌧n>0

}

Exercise 9.3

1) Prove that any continuous adapted process is locally bounded.

2) Prove that a is locally bounded, then a loc(M) loc(A), for

2H 2 \H1

any semimartingale X = M +A.

The following result is a kind of dominated theorem for stochastic

integral.

Theorem 9.4 Let an be a sequence of progressively measurable

{ }

processes such that t : an a.s. a as n .

8 t ! t ! 1

If there exists a locally bounded process b that dominates the sequence

an (i.e. n, t : an b ), then t andX probability t a dX .

{ } 8 8 | t| t 0 s s ! 0 s s

R R

Exercise 9.5 Prove this theorem. Let X = M +A and set Yn :=

t andX and Y := t a dX . Let ⌧ a sequence of stopping times

0 s s 0 s s { k }

increasing a.s. to such that b 2(M⌧k) 1(A⌧k).

R 1R 2H 2 [H1

1) Prove that an a 0 and an a 0 as

kH 22(M⌧k)

! k

kH 11(A⌧k)

n 0.

2) Conclude that (Yn)⌧k probability Y⌧k.

t ! t

3) Observe that

P( Yn Y >⌘ ) P( Yn Y >⌘ &⌧ > t)+P(⌧ t),

| t t |  | t t | k k 

and conclude the proof of the theorem.

Exercise 9.6 Let X be a semi-martingale, let a and b be two locally

. .

bounded processes and set: Y := a dX and Z := b dY .

0 s s 0 s s

1)Prove that Z = a b dX .

0 s s s R R

2) Compute Z,Z , Z,Y and Z,X

h i h i h i

10 Itˆo’s formula

The aim of this section is the proof of theorem 10.3 hereafter. We

start by proving a particular case of this result.

Theorem 10.1 Let X and Y be two semi-martingales then

1) X2 = X2 + t 2X dX + X,X .

t 0 0 s s h it

t t

2)X Y = X Y + X dY + Y dX + X,Y .

t t 0 0R 0 s s 0 s s h it

R R

Exercise 10.2 Prove this theorem.

Hint: Let be a subdivision that contains t. With 1) in exercise

8.11, one gets: X t2

T t(X,X) = X 02 + 2 n i= 01X tiX i. Setting

X := 1 i=01 [ti,ti+1[X ti, it follows P

P t

X2 T(X,X) = X2 +2 XdX

t t 0 s s

Prove that t XdX converges in probability to t X dX , and de-

0 s s 0 s s

duce that claim 1) holds.

R R

Use polarity relations to prove claim 2).

Theorem 10.3 (Itˆo’s formula) Let X = (X1,...,Xd) be a d-

dimensional vector of semi-martingales taking values in an open set

U. Let f : U R be a 2-function on U and consider the process

! C

Y := f(X ). Then Y is a semi-martingale and

t t

d . 1 d d .

Y = Y + @ f(X )dXk + @ f(X )d Xk,Xl

0 k s s 2 k,l s h is

k=1Z0

k=1 l=1

X XX

(6)

In this formula,

@ f(x1,...,xd) := f(x1,...,xd),

k @xk

and

@ f(x1,...,xd) := f(x1,...,xd).

k,l @xk@xl

Exercise 10.4

1) Apply previous theorem to the case d = 1, and f(x) = x2.

2) Apply previous theorem to the case d = 2, and f(x,y) = xy

(Theorem 10.1 is thus a particular case of Itˆo’s formula)

Exercise 10.5 Prove theorem 10.3.

Hint: 1) Let K be a compact subset of U. First prove that formula

(6) holds for all K-valued semi-martingales X: Let denote the

class of 2 functions f : K R for which formula 6 holds for all

C !

K-valued semi-martingales X.

1-a) Prove that a constant functions f belongs to .

1-b) Prove that the function fi : x = (x1,...,xd) fi(x) := xi

belongs to .

1-c) Prove that f,g implies h , where h(x) := f(x).g(x).

2V 2V

1-d) Conclude that contains the polynomials in (x1,...,xd).

1-e) For all 2 function f on a compact set K Rd, there exists

C ⇢

a sequence p of polynomials such that f p 0, for all

n n K,

{ } k k 1 !

j, @ f @ p 0 and, for all j,k, @ f @ p 0

j j n K, j,k j,k n K,

k k 1 ! k k 1 !

as n . (Here f := sup f(x) : x K .) Conclude that

! 1 k k 1 {| | 2 }

formula (6) holds for all K-valued semi-martingales X and all 2

function on K.

2) As open subset of Rd, U is the union of an increasing sequence

K of compact sets K . Define ⌧ := inf t : X @K , where

n n n t n

{ } { 2 }

@K is the boundary of K . Prove that ⌧ is an increasing sequence

n n n

of stopping times that converges a.s. to . Conclude the proof of

theorem 10.3

11 First applications of Itoˆ’s formula

Exercise 11.1 Let M be a local martingale and let f be a 2 func-

tion de (x,t) satisfying the (time reversed) heat equation:

@ f(x,t)+ @ f(x,t) = 0

t xx

1) Prove that f(M , M,M ) is a local martingale.

t t

h i

2) With f(x,t) := x2 t, we recover the result of theorem 8.9 that

M2 M,M is a local martingale.

t h it

3) With f(x,t) := e↵x ↵ 22 t, we conclude that e↵Mt ↵ 22 hM,M it is a

local martingale. This process is sometimes called the Dol´eans-Dade

exponential process of M. The result holds for ↵ C.

Exercise 11.2 Let B = (B1,...,Bd) be a d-dimensional Brownian

motion (i.e. a vector of d independent Brownian motions). Let

u : Rd be a 2 function. Prove that the process u(B t) is a local

! C

martingale if and only if u(x) = 0. (Remember that u(x) :=

d @ u(x).)

i=1 ii

PThe next exercise underlines an important link between Partial

Di↵erential Equation and stochastic process.

Exercise 11.3 Let D be an open bounded domain of Rd. Consider

the following Cauchy PDE problem:

u(x) = 0 if x D

( u(x) = (x) if x @D

Let B be a d-dimensional Brownian motion starting at x and let ⌧

denote the hitting time of @D by B.

1) Prove that ⌧ is a.s. finite.

2) Consider the process X := u(B⌧). Prove that X is a bounded

t t

martingale

3) Prove that u(x) = E[(B )].

⌧

This last formula is important: it expresses the value u(x) of the

solution u to a Cauchy problem as an expectation of the limit con-

dition . This implies that u(x) can then be numerically computed

by Monte Carlo methods.

Exercise 11.4

1) Explain how these Montecarlo methods work.

2) Compare this method with the finite di↵erence methods.

In particular, what happens when the dimension d of the problem

increases?

These Montecarlo methods can be used to solve more general PDE

using stochastic di↵erential equations. (see exercise ??.)

Exercise 11.5 Let B be a d-dimensional Brownian motion starting

at x . Let ⌧ denote inf t : B = ✏ .

0 ✏ t

{ k k }

a.s.

1) If d = 2, prove that ⌧ is a.s. finite if ✏> 0. Prove that ⌧ =

✏ 0

2) What happens if d = 3?

Hint: Consider a function ( x ) such that Y := ( B ) is a local

t t

k k k k

martingale. Setting ⌧ := ⌧ ⌧ , where R > x , prove that ⌧ is a.s.

✏ R 0

^ k k

finite and that Y⌧ is a bounded martingale. Compute E[( B )].

⌧

k k

Exercise 11.6 Prove that a bounded harmonic (i.e. u = 0) func-

tion u is constant. Otherwise the process u(B ) would be a bounded

martingale. u(B ) would therefore a.s. converge as t . This is

! 1

impossible if u is not constant. (use the previous result)

Exercise 11.7 Let B be a 3-dimensional Brownian motion starting

at B = 0. Consider the process V := 1 .

0 6 t Bt

k k

1) Prove that V is a local martingale

2) Prove that E[V2] is bounded.

3) Conclude that V us a uniformly integrable local martingale.

4) Prove that V 0 as t and conclude that V is not a

! ! 1

martingale.

Theorem 11.8 (Levy’s characterization of the Brownian motion)

A continuous local martingale X starting at 0 is a Brownian motion

if and only if, for all t, X,X = t

h i

Exercise 11.9 Prove this theorem

Exercise 11.10 (Dambis Dubbins Schwartz Theorem) LetX

be a local martingale on a filtration starting at 0. Assume

t t 0

{F }

that X,X is a strictly increasing process with X,X = .

h i h i1 1

1) For s 0, define ✓ := inf t 0 : X,X s . Prove that ✓ is

s t s

{ h i }

an -stopping time.

{F }

2) Define := and Y := X . Prove that Y is a -local

Gs F✓s s ✓s s {Gs }s 0

martingale.

3) Prove that X,X is a -stopping time.

t s s 0

h i {G }

4) Prove that Y is a -Brownian motion.

s s s 0

{G }

(Hint: Z := X2 X,X is a -local martingale. De-

t t h it {Ft }t 0

duce from this that Z is a -local martingale. Therefore

✓s {Gs }s 0

Y,Y = s and apply Levy’s theorem above).

h i

5) Prove that X = Y . (A continuous local martingale is thus

t X,X t

h i

a Brownian motion up to a change of time)

12 Girsanov’s theorem

12.1 Absolute continuity

Definition 12.1

1)A probability Q on (⌦, ) is said to be absolutely continuous with

respect to a probability P (Notation: P >> Q) if and only if, for all

A , P(A) = 0 implies Q(A) = 0.

2) P and Q are two equivalent probabilities (Notation: P Q) if

⇠

and only if P >> Q and Q >> P.

Theorem 12.2 Let P and Q be two probabilities on (⌦, ). Then

P >> Q if and only if

✏> 0 : > 0 : A : P(A) Q(A) ✏ (7)

8 9 8 2A  ) 

Exercise 12.3 Prove this theorem.

Hint: Assume (7) is false. Then, there exists ✏> 0 such that for all

> 0 there exists an event A in with P(A ) and Q(A ) >✏ .

A 

Consider then the sequence , with := 1 , and define B :=

{ j } j 2j k

A and B := B . Prove that P(B) = 0 and Q(B) ✏.

[1j=k j \1k=0 k

Definition 12.4 A probability Q is said to have a density Y

L1(⌦, ,P) with respect to P if and only if for all A : Q(A) =

A 2A

E [1 Y].

P A

In that case we write dQ = Y.

Exercise 12.5

1) Prove that the density Y of Q with respect to P on (⌦, ) is

unique in L1(⌦, ,P) (i.e. unique up to a set of P-measure 0).

2) If is a sub--algebra of , and if Y is a density of Q with

B A

respect to P on (⌦, ), and if Y := E[Y ], then for all A :

A |B 2B

Q(A) = E [1 Y] = E [1 Y ]. So Y and Y are two candidates

P A P A 0 0

to be the density of Q with respect to P on (⌦, ), but only Y is

-measurable. The notation dQ is thus misleading since the density

B dP

depends on the -algebra considered. When required, we will use the

following notation: dQ = Y and dQ = Y

dP|A dP|B 0

Exercise 12.6 Prove that Q has a density Y with respect to P if

and only if X L1(⌦, ,Q) : XY L1(⌦, ,P) and

8 2 A 2 A

E [X] = E [XY].

Q P

Theorem 12.7 (Radon Nikodym) If P and Q are two probabil-

ities on (⌦, ), then P >> Q if and only if Q has a density with

respect to P.

Exercise 12.8 Prove theorem 12.2 using the fact that the family

dQ is U.I.

{dP}

Exercise 12.9 If P >> Q prove the following claims:

P a.s. Q a.s.

1) X n X implies that X n X

! !

P probability Q probability

2) X X implies that X X

n n

! !

If P and Q are two probabilities on (⌦, ), if X is a process,

then X,X is defined as the limit in probability of T(X,X),

h it t

as 0. So, a priori, X,X could depend on the probability

| |! h i

P or Q considered. In the same way, a dX is defined as the

0 s s

limit in probability of t andX , where an sc, and the notion of

0 s s R 2E

integral could thus depend on the probability P or Q considered.

However, the previous exercise indicates that the convergence in P-

probability is equivalent to the convergence in Q-probability when

P Q. Therefore X,X and a dX are univocally defined,

⇠ h it 0 s s

independently of the equivalent probability P or Q considered.

100

Theorem 12.10 Let Q be an absolutely continuous probability with

respect to P on (⌦, ), and let Y be the density of Q with respect to

P. If is a sub-algebra of , and if X L1(⌦, ,Q), then

B A 2 A

E [YX ]

E [X ] = |B

|B E [Y ]

Exercise 12.11 Prove this theorem.

Q a.s.

Note in particular that E [Y ] > 0 so that the quotient by

E [Y ] makes sense Q-a.s.

12.2 Girsanov

Consider two equivalent probabilities P and Q on a filtered space

(⌦, ). Let Y denote dQ.

{Ft }t 0 dP

Exercise 12.12

1) Prove that Y > 0 P-a.s. and Q-a.s.

2) Prove that dP = 1.

dQ Y

Theorem 12.13 If P Q, and if y := E [Y ] is a continuous

t P t

⇠ |F

P-martingale then

1) X is Q-a.s. of finite variation iif X is P-a.s. of finite variation.

2) X is a Q-martingale iif Xy is a P-martingale

3) X is a Q-local martingale iif Xy is a P-local martingale

4) X is a Q-semi martingale iif X is a P-semi martingale.

Exercise 12.14 Prove this theorem.

Hint: 1) is trivial, 2) follows from theorem 12.10, 3) follows from

2). To get claim 4); observe that, if X is a Q- semi-martingale, then

X = M +A where M is a Q-local martingale and A is a process of

Q-a.s. finite variation starting at 0. Next observe that X = My+A.

101

Since both My and y are P-local martingales, it follows from Itˆo’s

formula that My and thus X are P semi-martingales.

Theorem 12.15 If P Q, and if y := E [Y ] is a continuous

t P t

⇠ |F

P-martingale then

1) There exists a P- local martingale N such that y

= y 0eNt 1 2hN,N it.

In particular dy = y dN .

t t t

2) M is a Q-local martingale iif M+ . 1 d M,y is a P-local mar-

0 ys h is

tingale.

3) M is a Q-local martingale iif M+ N,M is a P-local martingale.

h i

4) B is a Q-Brownian motion iif B + N,B is a P-Brownian mo-

h i

tion.

Exercise 12.16 Prove this theorem.

1) Due to Itˆo’s formula, L := Log[y ] is a P-semi martingale. So

t t

L = N + A where N is a P-local martingale, and A is of finite

variation. Then y = ELt. Apply Itˆo’s formula and infer 1) from

the fact that y is a P-local martingale.

2) V := My is a P-local martingale and so is W := . 1 dV . Next

0 ys s

dW = dM + Mtdy + 1d M,y

t t yt t yt h it R

3) Since dy = y dN , we have M,N = t 1 d M,y and claim 3)

t t t h it 0 ys h is

follows.

4) Follows from Levy’s theorem 11.8.

Remark that in the previous results, we only used the fact that the

density process dQ = y is a strictly positive continuous martin-

dP|Ft t

gale, but we didn’t use the fact that dQ exists. The martingale

dP|F1

y could fail to be U.I. Here is an example:

Exercise 12.17

1) Let B be a P-Brownian motion and a = 0 be a constant. Consider

the equation

dy = y a dB and y = 1 (8)

t t t 0

102

Prove that y := eaBt a2t/2 is a solution to that equation.

2) Let y˜ be another solution to (8). Let V denote V = y˜. Apply

Itˆo’s formula to infer that dV = 0. Conclude that the stochastic

di↵erential equation (8) has a unique solution.

3) Let Q denote the probability on with density y with respect

T T T

to P. Note that if A and A , then Q (A) = Q (A).

t2 t1 t2

There exists thus a function Q defined on the algebra by:

T>0 T

[ F

Q(A) := Q (A) if A .

T T

4) Prove that y can’t be a U.I. martingale. (Hint: Prove first that

a.s.

liminf y = 0. If y was a U.I. martingale we thus would have

t t

a.s.

y = lim y = 0, and thus Y = E[y ] = 0.)

t t t t

1 !1 1|F

12.3 Application to finance

The values at time t of a risk-less asset (asset 0) and of a risky asset

(asset 1) are respectively denoted

Sˆ0

and

Sˆ1

. Their evolution is

t t

described by the following equation:

dSˆ0

Sˆ0rdt Sˆ0

= 1

t t 0 (9)

( dSˆ t1 = Sˆ t1(µdt+dB t)

where B is a Brownian motion.

Exercise 12.18 As well known, the first equation indicates that

Sˆ0 = ert.

1) Find ↵ and such that the process

Sˆ1

Sˆ1e↵Bt+t

satisfies the

t 0

second equation in (9).

2) Prove that the process found above is the unique solution of that

equation.

An investment policy is characterized by two adapted processes V

and ✓: ✓ represents the number of shares of asset 1 in the port-

folio at time t. V represents the value of the portfolio at time t.

103

Assume next that the investment policy changes of positions only

on a discrete grid of times: 0 = t < t < < t = T. Assume

0 1 N

···

also that the total value of the portfolio remains unchanged at a

change of position (Self financed portfolio). V is then a continuous

process. The total value invested in the risky asset just after time

t (i.e. at time t+) is then ✓ Sˆ1. The total value invested in asset

i i t+

0 is then: Vˆ ✓ Sˆ1. The number of shares of asset 0 at time t+ is

ti t+

ti i

then

Vˆ ti✓ t+

Sˆ t1

i. Since the positions remain unchanged till time t ,

Sˆ0 i+1

we get that

Vˆ

Vˆ ti✓ t+

Sˆ t1 iSˆ0

+✓

Sˆ1

. Defining V :=

Vˆ

t and

ti+1 Sˆ t0

ti+1 t+

ti+1 t Sˆ t0

S1 := Sˆ t1 , this last relation leads to

t Sˆ0

ti+1

V = V ✓ S1 +✓ S1 = V + ✓ dS1

ti+1 ti t+ i ti t+ i ti+1 ti s s

Zti

Therefore

V = V + ✓ dS1 (10)

t 0 s s

Last relation that was established for piecewise constant processes

✓ is taken as definition for self-financing policies even if ✓ change of

position in continuous time. A self financed portfolio is therefore

completely defined by V and ✓.

A european contingent claim is a financial product whose value

at time T is a random variable L. As an example, a european call

on asset 1 with strike K and exercise date T is a contract that

allows its owner to buy one unit of asset 1 at date T at a prefixed

price K. Clearly, if

Sˆ1

> K, the owner will exercise his option,

buying one unit of asset 1 at price K and reselling it immediately

on the market at price

Sˆ1.

Doing so, the benefit of the owner is thus

Sˆ1

K. In case

Sˆ1

K, there is no profit to make and therefore

T T 

Lˆ = (Sˆ1 K)+, where (x)+ denotes (x 0).

T _

The question is then how to find the value at date 0 of such

104

a contingent claim and how to hedge the risk involved in such a

contract. To answer these questions, we introduce the notion of

replication portfolio: It is a self financed portfolio (V ,✓) such that

ˆ ˆ

V = L. Obviously, if there exists such a replication portfolio for

ˆ ˆ

L, the price of this contingent claim at time t should be V , and

the hedging policy for a bank issuing the contingent claim would be

ask that replication price at date 0, and invest this money in the

replication portfolio. Selling this portfolio at time T, the bank will

have in hand exactly the amount needed to honor the contract L.

The risk of the bank is thus completely hedged.

So, the question is now how to find a replication portfolio for L?

NotethatifthereexistsaMartingaleequivalentprobabilityQunder

which S1 is a martingale, relation (10) indicates that V is a Q-local

martingale, and if L := Lˆ /Sˆ0 is not too big (say L L2(⌦, ,Q)),

T 2 FT

then V is a Q-martingale. Therefore V = E [V ] = E [L].

0 Q T Q

Exercise 12.19

1) Find and equivalent martingale measure Q for asset 1 satisfying

equation (9).

2) Compute the price of a european call on asset 1 with expiry T

and strike K. (Hint: Use Girsanov’s theorem)

13 Stochastic di↵erential equations

13.1 Polish spaces and conditional probability

We have seen that E[X Y] = (Y), where (y) := E[X Y = y].

| |

With this notation, we meant that once we get to know that Y = y,

we change our mind about the random experiment and we use the

conditional probability p(. Y = y) to compute the expectation of

X. This conditional probability p(. Y = y) makes a perfect math-

105

ematical sense when the even Y = y has a non zero probability.

{ }

We could also define a similar conditional probability in particular

cases, when p( Y = y ) = 0. For instance, when (X,Y) has a prob-

{ }

ability density with respect to the Lebesgue measure, the law of X

conditionally to Y = y was defined trough the conditional density

f (x,y)

f (x) := (X,Y) . It is therefore tempting to define the notion

X |Y=y fY(y)

of regular conditional probability kernel:

Definition 13.1 If Y is a random variable from (⌦, ,P) to (E, ),

A E

a regular conditional probability kernel given Y is a map:

µ : (y,A) E µ (A) [0,1]

2 ⇥A! 2

satisfying

1) y E, the map µ : [0,1] is a probability on (⌦, )

8 2 A! A

2) A the map µ(A) : y µ (A) is measurable from (E, ) to

. y

8 2A ! E

([0,1], )

[0,1]

a.s.

3) A : E[1 Y](!) = µ (A)

A Y(!)

8 2A |

In this definition, µ is then the conditional probability on (⌦, )

given that Y = y .

{ }

Exercise 13.2 Prove that conditions 2) and 3) are equivalent to:

For all bounded random variable X on (⌦, ,P), the map : y

A !

(y) := E [X] is measurable from (E, ) to ([0,1], ). Futher-

µy

B[0,1]

a.s.

more, E[X Y] = (Y).

Remark 13.3 NotethatE [X] canalsobedenotedE [X] = X(!)dµ (!),

µy µy ⌦ y

and the previous relation yields the following ”desintegration” for-

mula:

X(!)dP(!) = E[X]

⌦

= E[E[X Y]]

= E[(Y)]

= (y)dP (y)

E Y

= X(!)dµ (!)dP (y)

RE ⌦ y Y

R R

106

where P denotes the law of Y. In other words, the lottery P that

selects ! is equivalent to the following compound lottery: first select-

ing y with the lottery P , then selecting ! with the corresponding µ .

Y y

This ”desintegration” into a compound lottery is clearly not possible

when the regular conditional probability kernel does not exist.

We next explain why such a regular kernel could fail to exist: Given

a probability space ⌦( , ,P) and a random variable Y measur-

able from (⌦, ) to (E, ), for all A , the conditional ex-

A E 2A

pectation E[1 Y] is a well defined element of L1(⌦,(Y),P), i.e.

it is an equivalence class of random variables in 1(⌦,(Y),P).

For all A , let us then pick an arbitrary element U in this

equivalence class E[1 Y]. Since it is (Y)-measurable, we know

that, !, U (!) = (Y(!)) for an apropriate measurable function

A A

: (E, E)

(R, BR).

We could then define µ (A) := (y). This map µ would clearly

y A

satisfy condition 2) and 3) in definition 13.1. There is however

a problem with condition 1). Indeed, if A ,A are two disjoint

1 2

elements of , we get 1 = 1 +1 and therefore

A1 [A2 A1 A2

a.s.

U = U +U .

A1 A2 A1 A2

[

As a consequence, the set

E := y E : µ (A A ) = µ (A )+µ (A )

A1,A2

{ 2

y 1

[

2 y 1 y 2

}

satisfies P (E ) = 1. The set of y E such that µ is additive

Y A1,A2

is thus

\A1,A2 2A:A1 \A2=

;

A1,A2

But this is the intersection of an uncountable family of E and

A1,A2

could therefore be non measurable, or have a probability P strictly

less than 1.

107

Unfortunately, without additional conditions on (⌦, ,P), there

is no argument that allows to conclude that there exists an appro-

priate selection of U in the equivalence class E[1 Y] leading to an

A A

additive µ , for all y E.

Definition 13.4 Aspace(⌦, ,P)iscalledapolishprobabilityspace

if there exists a metric d on ⌦, such that

1) (⌦,d) is a complete space

2) There exists a countable dense subset in (⌦,d)

3) is the Borel -algebra corresponding to that metric.

An obvious example of polish probability space is (R, ,P) with

an arbitrary probability P.

The importance of this notion is due to the following theorem,

that will not be proven in this course:

Theorem 13.5 If (⌦, ,P) is a polish probability space then for all

measurable random variable Y from (⌦, ,P) to (E, ), the regular

A E

conditional probability kernel given Y exists.

We could also define the regular conditional probability kernel

given , where is a sub -algebra of setting (E, ) = (⌦ , ) in

B B A E B

the previous definition and considering the identity map Y : !

Y(!) := !.

Definition 13.6 If is a sub -algebra of , a regular conditional

B A

probability kernel given is a map:

µ : (! ,A) ⌦ µ (A) [0,1]

0 !

2 ⇥A! 0 2

satisfying

1) ! ⌦, the map µ : [0,1] is a probability on (⌦, )

0 !

8 2 0 A! A

2) A the map µ(A) : ! µ (A) is measurable from (⌦, )

. 0 !

8 2A ! 0 B

to ([0,1], )

[0,1]

a.s.

3) A : E[1 ](! ) = µ (A)

A 0 !

8 2A |B 0

108

13.2 The law of a continuous process

We will use similar notations as in section when defining the Wiener

measure (see corrolary 2.10). The space ⌦ will denote ⌦ :=

0 0

([0,T]). On that space, we define the canonical process X with

time set [0,T] as X : ! ⌦ X (! ) := ! (t). Let be its

t 0 0 t 0 0 t

2 ! G

natural filtration: := (X ,s t).

t s

G 

Acontinuous( )-adaptedprocessZ onaprobabilityspace(⌦, ,P)

F A

induces a map Z :⌦ ⌦ , where Z (!) is the continuous function

• ! •

Z (!) : t [0,T] Z (!).

• 2 !

Exercise 13.7 ProvethatZ ismeasurablefrom(⌦, )to(⌦ , ).

T 0 T

• F G

Hint: First prove that Z 1((X )) . It follows that Z 1( )

t t

⇢F D ⇢

• •

, where := (X ). Since = ( ), conclude that :

T t [0,T] t T

F D [ 2 G D

Z 1( ) = (Z 1( )) .

T T

G D ⇢F

• •

Definition 13.8 The law of the process Z is then the probability P

on (⌦ , ) defined by B : P (B) = P( ! : Z (!) B )

0 T T Z

G 8 2G { • 2 }

As it follows from Theorem 13.5, the next result will allow us to

deal with the conditional law of a continuous process.

Theorem 13.9 (⌦ , ,P ) is a polish space.

0 T Z

proof: Indeed, (⌦ , . ) is a complete normed space and the

k k1

polynomials with rational coecients form a countable dense sub-

set in ⌦ . Let be the Borel -algebra on ⌦ . Since, for all

0 B⌦0 0

t [0,T], the map X

is continuous from (⌦ 0, . ) to (R, . ), it

2 k k1 | |

is measurable from to . Therefore (X ) and thus

B⌦0 BR t ⇢B ⌦0

= ( (X )) .

GT [t 2[0,T] t

⇢B

⌦0

On the other hand, if f ⌦ , the closed ball B with center f

0 f,✏

and radius ✏ is, due to the continuity of f

B := ! ⌦ : ! f ✏ = C ,

f,✏ { 0 2 0 k 0 k1  } \t 2[0,T] \Q t

109

where C := ! ⌦ : ! (t) f(t) ✏ . Next, observe that

t 0 0 0

{ 2 | | }

C = X 1([f(t) ✏,f(t)+✏]). Therefore C (X ) , and thus,

t t

⇢G

as a countable intersection of such C , we conclude that B .

t f,✏ T

Finally, if denotes the class of open sets in (⌦ , . ), any

O k k1

O is a countable union of such B and will then belong to .

f,✏ T

2O G

As a consequence , and thus := ( )

O⇢G

T B⌦0

O ⇢G

The law of a multidimensional continuous process Y = (Y1,Y2)

canbedefinedinthesameway: P ishereaprobabilityon ([0,T])2.

Exercise 13.10 If V = (X,Y,Z) is a continuous 3-d process on

(⌦, ,P) where X is a semi-martingale that satisfies: Z = Y dX .

A 0 s s

If V = (X ,Y ,Z ) is another 3-d process defined on another prob-

0 0 0 0 R

ability space (⌦, ,P ). Further assume that V and V have the

0 0 0 0

same law (P = P ). Prove that Z = Y dX

V V 0 0 0 s0 s0

(Hint: If = t ,...,t is a subdiviR sion of [0,T], define Y :=

0 N

{ }

N i= 01Y ti1 [ti,ti+1[ and Y 0 := N i= 01Y t0 i1 [ti,ti+1[. Define next W :=

. YdX and W := . Y dX .

P0 s s 0 0 s0 Ps0

Provethattheprocesses(X,Y,Z,W)and(X ,Y ,Z ,W )have

R R 0 0 0 0

the same law. Therefore P( Z W >✏ ) = P ( Z W >✏ ).

| t t | 0 | t0 t0 |

Letting go to 0, conclude that (X,Y,Z,W) and (X ,Y ,Z ,W )

0 0 0 0

| |

. .

have the same law, where W := Y dX and W := Y dX .

0 s s 0 0 s0 s0

Conclude that 1 = P(Z = W ) = P (Z = W ).)

t t R0 t0 t0 R

13.3 Main definitions for S.D.E.

Let a and b be two functions R R+ R. The evolution of a

⇥ !

physical real valued quantity y is often described by a di↵erential

equation dy(t) = b(y(t),t). This equation may as well be written

as dy(t) = b(y(t),t)dt. The evolution of y is often perturbed by a

random error, and the perturbed equation becomes

dy = a(y ,t)dB +b(y ,t)dt,

t t t t

110

where B is a Brownian motion. We refer to this equation as equa-

tion e(a,b). In the model, the fact that the random term is of

the form a(y ,t)dB indicates that the perturbation is essentially

t t

N(0,a(y ,t)dt), but in fact the Brownian motion itself is never ob-

served. So B is not an external data of the problem. This is the

reason why a solution to e(a,b) is defined as:

Definition 13.11 A solution to e(a,b) is a pair of processes (y,B)

where B is a brownian motion with respect to the natural filtration

(y,B) that satisfies:

. .

y = y + a(y ,t)dB + b(y ,t)dt

0 t t t

Z0 Z0

A solution (y,B) to e(a,b) is called strong iif y is adapted to the

filtration ( ) , where := (y , B ), otherwise, the solu-

t t 0 t 0 s s [0,t]

F F { } 2

tion is called weak.

Exercise 13.12 Let (y,B) be a solution to e(a,b) defined on a prob-

ability space (⌦, ,P). Let (y ,B ) be a process defined on another

0 0

probability space (⌦, ,P ). Assuming that (y,B) and (y ,B ) have

0 0 0 0 0

the same law, prove that (y ,B ) is also a solution to e(a,b).

0 0

For the sake of simplicity, we will deal in this course with one

dimensional Stochastic Di↵erential Equations, but all the concepts

could be generalised to the multidimensional framework: in this

case, y is a process with values in Rk ⇥1, B is an l dimensional Brow-

nian Motion with values in in Rl ⇥1. a is then a map

a : (y,t) (Rk ⇥1 R) a(y,t) Rk ⇥l

2 ⇥ ! 2

Similarly:

b : (y,t) (Rk ⇥1 R) b(y,t) Rk ⇥1

2 ⇥ ! 2

The term a(y ,t)dB has then to be interpreted as a matrix product.

t t

111

As we will see, there are two di↵erent notions of uniqueness for

the solution of an equation e(a,b).

Definition 13.13 Equation e(a,b) satisfies pathwise uniqueness if,

for any two solutions (y,B) and (y ,B) of e(a,b) defined on the same

probability space (⌦, ,P), with the same Brownian Motion B, we

P a.s

have the implication y

= y

implies y and y

are indistinguish-

able.

The second notion is natural if we consider a Stochastic di↵er-

ential equation as a perturbed deterministic equation: we are then

interested in the probability distribution of the position y of the

process, and more generally the law of the process y.

Definition 13.14 Equation e(a,b) satisfies uniqueness in law if, for

any two solutions (y,B) and (y ,B ) of e(a,b) (defined possibly on

0 0

Law

di↵erent probability spaces), we have the implication y = y im-

0 00

plies y and y have the same law: P = P .

0 y y

13.4 Pathwise uniqueness implies uniqueness in law

Theorem 13.15 If pathwise uniqueness holds for e(a,b), then any

solution to e(a,b) is a strong solution.

Proof: Assume that pathwise uniqueness holds for e(a,b) and let

(y,B) be a solution to that equation. The law ⇢ of (y,B) is thus

a probability measure on ( ([0,T])2, ) where := (y ,B ,s

T T s s

C G G 

T). Since ( ([0,T])2, ) is a polish space, we may consider the

C G

regularconditionalprobabilitykernel⇢ on( ([0,T])2, )given

(y0,B)

(y ,B). Let µ denote the marginal probability distribution of

0 (y0,B)

⇢ on y ([0,T]). Let also µ denote the marginal distribution

(y0,B)

of ⇢ on (y 0,B) R ([0,T]). According to Remark 13.3, we may

2 ⇥C

”desintegrate” the lottery ⇢ into a compound lottery: On the first

112

stage lottery, one selects (y ,B) with the lottery µ, and then, on the

second stage lottery, one select y with the conditional probability

µ . Thiswillyieldapair(y,B)withthesamelaw⇢astheinitial

(y0,B)

solution. According to exercise 13.12, (y,B) is thus also a solution

to e(a,b).

Suppose now that after selecting (y ,B) with the lottery µ, we

select two independent y and y with the same conditional lottery

µ . Let ⌫ denote the law of the resulting 3-d process (y,y ,B).

(y0,B) 0

Observe that both (y,B) and (y ,B) are ⇢-distributed. Therefore,

they are both solutions to e(a,b), with the same Brownian motion.

⌫ a.s.

Furthermore, y

= y 0. It results from the pathwise uniqueness

property of e(a,b) that y and y are indistinguishable. Therefore

⌫(y = y ) = 1 and thus ⌫ (y = y ) = 1 for µ-a.e (y ,B). In

0 (y0,B) 0 0

other words, conditionally to (y ,B), we select y and y with two

0 0

independent lotteries µ and we obtain with probability 1 the

(y0,B)

same result. This is only possible if µ is deterministic: it must

(y0,B)

be a Dirac measure concentrated on a trajectory F that depend on

(y ,B): µ := .

0 (y0,B) F(y0,B)

We thus have proved that there exists a map

F : (y 0,B) (R ([0,T])) ([0,T])

2 ⇥C !C

such that 1 = ⌫( y = F(y ,B) ) = ⇢( y = F(y ,B) ).

0 0

{ } { }

This implies that y is a deterministic function of (y ,B ,s T)

T 0 s



and is thus (y ,B ,s T)-measurable. (y,B) is thus a strong

0 s



solution to e(a,b).

Theorem 13.16 Ifpathwiseuniquenessholdsfore(a,b), thene(a,b)

satisfies uniqueness in law.

Proof: Assume that pathwise uniqueness holds for e(a,b) and let

(y,B) and (y˜,B) be two solutions to that equation, possibly on two

Law

di↵erent probability spaces with the same initial law: y = y˜ .

0 0

113

Let ⇢ and ⇢˜ denote the laws of (y,B) and (y˜,B) respectively. As

in the previous proof, define µ and µ˜, the marginals law of (y ,B)

and (y˜ 0,B) on R ([0,T]). Define also the regular conditional

⇥C

laws µ and µ˜ of y given (y ,B) and, respectively, y˜ given

(y0,B) (y˜0,B˜) 0

(y˜ ,B).

Since B and B are both Brownian motions, their law is the

Wiener measure W. Since y and y˜ have the same law ✓ and are

0 0

respectively independent of B and B respectively, we conclude that

µ = ✓ W = µ˜.

⌦

As in the previous proof, we next consider a two stage lottery.

At stage 1, one selects (y ,B) with the lottery µ = µ˜. On stage 2,

we select y and y with two independent lotteries µ and µ˜ .

0 (y0,B) (y0,B)

Let ⌫ denote the law of the resulting process (y,y ,B).

Clearly the law of (y,B) is ⇢ and that of (y ,B) is ⇢˜. According

to exercise 13.12, (y,B) and (y ,B) are solution to e(a,b). Further-

⌫ a.s.

more, y

= y 00.

Since pathwise uniqueness holds for e(a,b), we conclude that

⌫( y = y ) = 1 and thus the law ⇢ of (y,B) coincides with the

{ }

law ⇢˜ of (y ,B). It follows that y and y˜ have the same law, and

e(a,b) satisfies uniqueness in Law.

13.5 Tanaka equation

For x R, we set s(x) := 1 if x 0 and s(x) := 1 if x < 0.

Tanaka equation is equation e(s,0). As we will see, this equation

illustrates that the converse of previous theorem does not hold:

Exercise 13.17

1) Prove that e(s,0) has at least one solution (y,B) with y = 0.

2) Prove that pathwise uniquenes does not hold for e(s,0).

Proof: Indeed,lety beaBrownianmotionanddefineB := s(y )dy .

0 t t

114

Since y is a Brownian motion, we infer that B is a martingale. Fur-

thermore B,B = t s(y )2dv = t 1dv = t. It results then from

h it 0 v 0

Levy’s characterisation theorem 11.8 that B is a Brownian Motion.

R R

t t

Next dB = s(y )dy . Therefore s(y )dB = s(y )s(y )dy =

t t t 0 v v 0 v v v

1dy = y , and (y,B) is thus a solution to e(s,0).

0 v t R R

Define next y¯:= y. We now prove that (y¯,B) is also a solution

of e(s,0) with y¯ = 0. It follows from the definition of s(x) that

s( x) = s(x) if x = 0. Since y is a Brownian motion, p(y = 0) =

6 6

a.s

1 if v > 0. Therefore s( y ) = s(y ).

v v

Theprocessess(y¯)and s(y)arethusidenticalinHloc andthere-

t t

fore s(y¯ )dB = s(y )dB = y = y¯.

0 v v 0 v v t t

(y¯,B) is then a solution to Tanaka equation, as is (y,B). Fur-

R R

thermore y¯ = y = 0. Would pathwise uniqueness hold for e(s,0),

0 0

a.s.

we would have y = y¯. But this is impossible since y = y¯ if v > 0.

v v

Exercise 13.18 ProvethatuniquenessinlawholdsforTanakaequa-

tion.

Proof: Let(y,B)beasolutiontoe(s,0), andlet✓ belawofy . B is

(y,B)

then a Brownian motion on the filtration := (y ,B ,s t).

Ft s s 

Therefore, the process V := . s(y )dB is an (y,B)-local martin-

0 t t F

gale. Furthermore, since s(x)2 = 1, we get as in the previous exer-

cise that V,V = t s(y )2ds = t, and V is the a Brownian motion

h it 0 s

on the filtration (y,B).

In particular, the law of the process V is the Wiener measure W.

(y,B)

On the other hand, V = V V and thus y V. The law

t t 0 ?? F0 0 ??

of y = y +V is the completely determined by the law ✓ of y . Two

0 0

Law

solutions (y,B) and (y ,B ) satisfying y = y have thus the same

0 0 0 00

Law

law: y = y .

Exercise 13.19 Prove that Tanaka equation has no strong solution

115

(y,B) with y = 0.

Proof: Assume there exists a strong solution (y,B) to e(s,0) sat-

isfying y = 0. According to the definition of strong solution, y

must then be adapted to the filtration B := (B ,s t) and thus

Ft s 

y := (y ,s t) B.

Ft s  ⇢F t

On the other hand, we prove now that B must be measurable

on Ft| | := ( |y s |,s



t) . Indeed, we get that dy t = s(y t)dB t implies

that B = s(y )dy .

t 0 s s

Let next be an increasing sequence of even ( ( x) =

R { n }n 2N n

n(x)), positive, convex 2 functions n : R R, such that forall

C !

x: lim (x) = x and lim d (x) = sg(x), where

n !1 n | | n !1 dx n

1 if x > 0

sg(x) = 1 if x < 0

< 0 if x = 0

a.s. >

: a.s.

Since y = 0 when t > 0, we infer that s(y ) = sg(y ) and thus

t t t

B = sg(y )dy . Due to Itˆo’s formula, we have

t 0 s s

R t 1 t

(y )dy = (y ) (y )ds

0n s s n t

0n0 s

Z0 Z0

Since and thus also are even, (y ) = ( y ) and (y ) =

n 0n0 n t n

0n0 s

( y ), the right hand side is therefore y -adapted.

0n0

F| |

Since (y ) sg(y ), (y )dy converges in probability to

0n s ! s 0 0n s s

B = t sg(y )dy . B must therefore be y -adapted as well.

t 0 s s R F| |

If there were a strong solution (y,B) to e(s,0) satisfying y = 0,

R 0

we would have: y B |y |. But this is impossible: indeed, it

Ft ⇢F t ⇢F t

would imply on one hand that E[y t |Ft| |] = y t.

On the other hand, since y and y are both Brownian mo-

tions, (y t, |y |) has the same law as ( y t, |y |). So for all A

t| |,

E[y t1 A] = E[ y t1 A]. Hence, 8A

t| | : E[y t1 A] = 0 and therefore

a.s.

E[y t |Ft| |] = 0. But y t 6= 0.

116

13.6 SDE with Lipschitz coecients

Definition 13.20

1) An equation e(a,b) is said to be Lipschitz iif

K : x,y R : t 0 : a(x,t) a(y,t) K x y

9 8 2 8 | | | |

and b(x,t) b(y,t) K x y

| | | |

2) e(a,b) is said to be bounded iif

K : t 0 : a(0,t) K and b(0,t) K.

9 8 | | | |

3) e(a,b) is said to be locally Lipschitz iif

n : K : x,y [ n,n] : t 0 : a(x,t) a(y,t) K x y

n n

8 9 8 2 8 | | | |

and b(x,t) b(y,t) K x y

| | | |

we aim to prove in this section that

Theorem 13.21

1) If e(a,b) is locally Lipschitz, then it satisfies pathwise uniqueness.

2) If e(a,b) is Lipschitz and bounded, if B is a Brownian motion in-

dependent of a random variable ⇠, then there exists a unique process

y such that y = ⇠ and (y,B) is solution to e(a,b).

(y,B) is a strong solution to e(a,b).

Let ⇠ be a random variable in L2, let B be a Brownian motion

independentof⇠,andlet denote(⇠,(B ) ). Foraprogressively

t s s t

F 

measurable process y, we define the process T(y) by

. .

T(y) := ⇠ + a(y ,s)dB + b(y ,s)ds

s s s

Z0 Z0

Note that (y,B) is solution to e(a,b) and y = ⇠ is equivalent to the

claim y = T(y).

For a given function ↵ : [0, [ R+, we define the ↵-norm of a

1 !

process y by

y := ↵(t)E[ y 2]dt.

↵ t

k k s Z0 | |

117

The proof of the above theorem relies on the next result:

2K2 (t+t2

)

Theorem 13.22 Let ]0,1[. If ↵(t) = e 2 2 , then for all

processes y1,y2: T(y1) T(y2) y1 y2

↵ ↵

k k  k k

Proof: First observe that T(y1) T(y2) = t L dB + t M ds,

t t 0 s s 0 s

whereL := a(y1,s) a(y2,s)andM := b(y1,s) b(y2,s). Therefore

s s s s s R s R

kT(y1) T(y2) k2

↵

= 01↵(t)E 0t L sdB

+ 0t M sds dt



R ⇣ R t 2R t ⌘ 2



01↵(t)2E

L sdB

M sds dt



R 1↵(t)2E ⇣ Rt L2ds+t⌘ t M⇣ 2Rds dt ⌘

 0 0 s 0 s

h i

R R R

where the last inequality follows from the Cauchy Schwartz inequal-

t t t

ity: M ds M2ds 12ds. Due to the Lipschitz prop-

| 0 s | 0 s 0

erty, L K y1 qy2 and Mq K y1 y2 , therefore

|R s | | s Rs| | sR | | s s|

T(y1) T(y2) 2 1↵(t)2E t (1+t)K2 y1 y2 2ds dt

k k↵  0 0 | s s|

= R1 1↵(t)h2R(1+t)K2E[ y1 y2 2]ditds

0 s | s s|

Observing that 1↵(t)2(R1+Rt)K2dt = ↵(s)2 we get as announced

T(y1) T(y2) 2 2 1 ↵(s)E y1 y2 2 ds = 2 y1 y2 2

k k↵  | s s| k k↵

⇥ ⇤

Exercise 13.23 Let e(a,b) be a locally Lipschitz equation with two

solutions (y1,B) and (y2,B), with y1 a =.s. y2, let ⌧ denote

0 0 n

⌧ := inf t 0 : y1 n or y2 n .

n { | t| | t| }

2Kn2 (t+t2

)

1) Prove that, with ↵(t) := e 2 2 ,

y1,⌧n y2,⌧n y1,⌧n y2,⌧n

↵ ↵

k k  k k

2) Conclude that y1 m =odif y2. Claim 1) in theorem 13.21.

118

Exercise 13.24 Let e(a,b) be a Lipschitz bounded equation, Let B

be a Brownian motion independent of a random variable ⇠ L2. Let

denote := (⇠,(B ) ). Let ↵ be as in theorem 13.22, and

t t s s t

F F 

let denotes the set of -progressively measurable processes y such

V F

that y < .

↵

k k 1

1) Define y y iif y y = 0. Set V := / . Prove that

0 0 ↵

⌘ k k V ⌘

(V, . ) is a complete space.

↵

k k

2) If y0 denotes the constant process t : y0 := ⇠. Prove that y0 .

8 t 2V

3) e(a,b) is Lipschitz and bounded. Prove then that

a(x,t) K(1+ x ) and b(x,t) K(1+ x ).

| | | | | | | |

4) Infer from 3) that y1 := T(y0) belongs to

5) Define recursively yn := T(yn 1). Prove that yn+1 yn

↵

k k 

n y1 y0 . Conclude then that n : yn .

↵

k k 8 2V

6) Prove that yn is a Cauchy sequence in V, and has then a

{ }n 2N

limit y V. Prove that (y ,B) is a solution to e(a,b).

1 1

Exercise 13.25 Let denote the set of maps

Y : (x,!,t) (R ⌦ [0, [) Yx(!) R

2 ⇥ ⇥ 1 ! t 2

such that

a) For all x: Yx is a progressively measurable process.

b) For all (!,t), the map x Yx(!) is continuous

! t

c) Y < , where Y := sup Yx , where ↵ is defined as

k kW 1 k kW xk k↵

in the previous exercise.

1) Prove that ( , . ) is a Banach space.

W k kW

2) If Y , we define T(Y) as the element Z satsifying for

2W 2W

all x:

. .

Zx := x+ a(Yx,s)dB + b(Yx,s)ds

s s s

Z0 Z0

3) Prove that T has a unique fixed point Y.

119

4) Prove that for all x, (Yx,B) is the unique solution of e(a,b) that

satisfies Yx = x.

5) Let ⇠ be a random variable independent of the Brownian motion

⇠(!)

B. Consider the process U : (!,t) U (!) := Y (!). Prove that

! t t

(U,B) is the unique solution of e(a,b) satisfying U = ⇠.

Exercise 13.26 For the two following claims, if they are true, prove

it. Otherwise give a counterexample:

1) If U L1(⌦, ,p), for all pair of sub--algebras ( , ),

2 A B C

E[E[U ] ] = E[E[U ] ]

|B |C |C |B

(We do not assume any inclusion between and .)

B C

2) If U L1(⌦, ,p), for all pair of stopping times (,⌧):

2 A

E[E[U ] ] = E[E[U ] ]

⌧ ⌧

|F |F |F |F

(( ) is a filtration, but we do not assume any inequality between

t t 0

and ⌧.)

proof: Claim 1) is clearly false if no inclusion is assumed: An easy

example is as follows: Set := (X) and = (Y), where (X,Y)

B C

is a centred gaussian vector with covariance matrix:

1 ⇢

⇢ 1 !

with 0 <⇢< 1. Then E[X Y] = ⇢Y and E[Y X] = ⇢X. Taking

| |

U := X, we get E[E[U ] ] = E[E[X X] Y] = E[X Y] = ⇢Y =

|B |C | | | 6

E[E[U ] ] = E[E[X Y] X] = E[⇢Y X] = ⇢2X.

|C |B | | |

Claim 2) is true. Indeed, as we shell prove: E[E[U ] ] =

⌧

|F |F

E[U ] and similarly E[E[U ] ] = E[U ].

⌧ ⌧ ⌧

|F ^ |F |F |F ^

We first prove that ⌧ . Indeed, ⌧ = ⌧ =

{  }2F {  } { ^

. Since ⌧ is -measurable, it is also -measurable, as is

⌧

} ^ F ^ F

. As announced: ⌧ .

{  }2F

120

We next claim that, if A , then A ⌧ . Indeed,

⌧

2F \{  }2F

for all t > 0:

A ⌧ ⌧ t = A ⌧ t ⌧ t

\{  }\{  } \{  }\{  }\{  }

Since A ⌧ , it follows that A ⌧ t .

\{  }2F \{  }\{  }2F

We also have ⌧ t , since ⌧ is a stopping time. Therefore

{  }2F

A ⌧ ⌧ t and thus A ⌧ . Finally,

t ⌧

\{  }\{  }2F \{  }2F

since both A and ⌧ belong to , we conclude that A

{  } F \{ 

⌧ .

⌧

}2F ^

If B , then B ⌧< . Indeed, from the previous

⌧ ⌧

2F \{ }2F ^

argument with A := ⌦, we infer that ⌧< = ⌧ c .

⌧

{ } {  } 2F ^

It follows that B ⌧< . Next, t 0, we have:

⌧

\{ }2F 8

B ⌧< t = B ⌧< ⌧ t t

\{ }\{  } \{ }\{  }\{  }

Since both B ⌧< ⌧ t and t belong to , so

\{ }\{  } {  } F

does B ⌧< t and thus B ⌧< . Therefore

\{ }\{  } \{ }2F

B ⌧< = .

⌧ ⌧

\{ }2F \F F ^

Let next Y denote Y := 1 E[U ] + 1 E[U ]. It

⌧ ⌧< ⌧

{  } |F { } |F

follows from the previous discussion that Y is -measurable.

⌧

F ^

Next if C we get

⌧

2F ^

E[1 Y] = E[1 1 E[U ]]+E[1 1 E[U ]]

C C ⌧ C ⌧< ⌧

{  } |F { } |F

= E[E[1 1 U ]]+E[E[1 1 U ]]

C ⌧ C ⌧< ⌧

{  } |F { } |F

= E[1 1 U]+E[1 1 U]

C ⌧ C ⌧<

{  } { }

= E[1 U]

We conclude therefore that Y = E[U F ]

⌧

| ^

121