Brownian Motion and Stochastic Calculus

Xiongzhi ChenUniversity of Hawaii at ManoaDepartment of Mathematics

July 5, 2008

Contents

1 Preliminaries of Measure Theory 11.1 Existence of Probability Measure . . . . . . . . . . . . . . . . 5

2 Weak Convergence of Probability Measures 11

3 Martingale Theory 17

Brownian Motion and Stochastic CalculusChapter 0: Preparations

1 Preliminaries of Measure Theory

De�nition 1 F � P () is said to be an algebra if (1) 2 F (2) A;B 2 Fimplies A

SB 2 F (3) A 2 F implies AC 2 F . F is said to be a semi-

algebra or semi-ring is (1) ;? 2 F (2) A;B 2 F implies AB 2 F (3)A;B 2 F implies AnB =

Pnk=1Ak for some Ak 2 F

De�nition 2 �-Class: If A;B 2 � implies AB 2 �

De�nition 3 �-Class: if (1) 2 �;(2) A;B 2 � with A � B impliesA�B 2 �; (3) An 2 �; An " A implies A 2 �

De�nition 4 Let L � R be such that � 2 L implies �+; �1 2 L. L-class:(1)1 2 L (2) L is linearly closed (3) �n 2 L; 0 � �n " �; � bounded or � 2 Limplies � 2 L

Proposition 5 (i)M being �-Class & �-Class impliesM is �-algebra (ii)If �-Class � contains the �-Class �; then � � � (�)

1

Proof. (i) It su¢ ces to showM is closed under countable union.(ii) Then intersecton F 0 of all �-classes that contain the �-Class � is

� (�) : De�neF1 =

�A : AB 2 F 0;8B 2 �

Then F1 is a �-class and F1 � �: So F1 � F 0; which means

AB 2 F 0;8A 2 F 0; B 2 �

Similarly, de�neF2 =

�B : AB 2 F 0;8A 2 F 0

Then F2 is also a �-class and F2 � �: So F2 � F 0; which means

AB 2 F 0;8A;B 2 F 0

and � � � (�) = F 0

Proposition 6 If L-class L contains the c.f.� �A of each member A of the�-Class �;then L contains all � (�)-measurable functions in L

Proof. The setG = fA � : �A (!) 2 Lg

is a �-Class � and since � � � it�s clear that � � � (�) ; that is

�A (!) 2 L;8A 2 � (�)

Let � 2 L be nonnegative, � (�)-measurable and

�nk =

�k

2n� � (!) < k + 1

2n

�2 � (�)

Then 0 � �n " � and � 2 L implies � 2 L: Finally, from the fact that � is� (�)-measurable i¤ �+; �� are � (�)-measurable it�s clear that � 2 L

Theorem 7 (Hahn Decomposition) Suppose (;S) is a measurable spacewith � being a signed measure which does not assume �1 and +1 at thesame time, then there exists a partition A;B 2 S of such that

� (A) = inf f� (E) : E 2 Sg

and� (B) = sup f� (E) : E 2 Sg

from which two measures are induced

�� (E) := �� (AE) ; �+ (E) = � (BE) ;8E 2 S

and� = �+ � ��

2

Theorem 8 (Radon-Nikodym) Let � be a �-�nite measure and � a signedmeasure on (;S) : Assume that � is absolutely continuous with respect to�: Then there is a Borel measurable function g : ! R such that

� (A) =

ZAgd�;8A 2 S

and if h is another such function, then g = h [�] (From Rober B. Ash "Realanalysis and probability)

Theorem 9 (Lebesgue Decomposition) Let � be a �-�nite measure and� a �-�nite signed measure on (;S) : Then � has unique decomposition as�1 + �2, where �1 � � and �2 ? �

Proof. (Trick) Let m = �+ �: Then �� m and �� m

Theorem 10 (Representation) Let (X; �) be a metric space and � 2(Cb (X))

� with � (�) � 0:Then

� (F ) = inf f� (f) : f � �F ; f 2 Cb (X)g ;8F 2 ClX

where ClX denotes all closed subsets of X

� (G) = sup f� (F ) : F 2 ClXg

for any G 2 OpX : De�ne

�� = infG2OpX

f� (G) : A � Gg ;8A � X

Theorem 11 (Extension of Independent Classes) Let (;S; �) be a prob-ability space and fCi; i 2 Ig is a family of �-classes of S each of which con-tains : Suppose for 8Cj 2 Cj ; j 2 J � I and J �nite

P�T

j2J Cj�=Qj2J P (Cj) (1)

Then the family of �-algebras f� (Ci) = Bi; i 2 Ig is independent. Moreover,for

B0i = fBSN : B 2 Bi; N 2 S; P (N) = 0g

it�s also true that the familynB0i; i 2 I

ois independent.

Proof. Case 1: I is �nite. Fix i 2 I and de�ne

D =nD 2 S : P

�DTj2J Cj

�= P (D)

Qj 6=i P (Cj) ;8Cj 2 Cj ; j 6= i; j 2 I

o

3

Then Ci � D for this i 2 I: To show that Bi � D; it su¢ ces to show D is a�-class. Obviously 2 D: Suppose D1; D2 2 D and D1 � D2 and denoteTj2J Cj = C; then

P (D2)Qj 6=i P (Cj) = P (D2C) = P ((D1 + (D2 �D1))

TC)

= P (D1C) + P ((D2 �D1)TC)

= P (D)Qj 6=i P (Cj) + P ((D2 �D1)

TC)

that is, P (D2 �D1)Qj 6=i P (Cj) = P

�(D2 �D1)

Tj2J Cj

�and D2 �D1 2

D: Finally, ifDn 2 D andDn " D then P (DnTC) = limn!1 P (Dn

TC) =

limn!1 P (Dn)P (C) = P (limn!1Dn)P (C), that is, limn!1Dn 2 D:Consequently Bi = � (Ci) = � (Ci) � D and fBi; Cj ; j 6= ig satis�es (1).Recursively and after rearrangement, de�ne

Dk+1 =(D 2 S : P

�DTj>k+1Cj

Tl�k Cl

�= P (D)

Qj 6=k+1 P (Cj) ;

8Cj 2 Cj ; j > k + 1; Cl 2 Bl; l � k

)

then Bk+1 � Dk+1 and inductively�B1; � � � ;Bk;Dk+1; � � � ; CjIj

satis�es (1).

Thus by induction, case 1 is provedCase 2: I is in�nite. Take any �nite J � I: Then from case 1 the

assertion is true.Further, if B

0i = Bi

SN 2 B0i; B 2 Bi; N 2 S; P (N) = 0: ThenQ

J Bj �QJ B

0i � (

QJ Bj)

S(QJ Nj)

and when J is �nite

P (QJ Bj) � P

�QJ B

0i

�� P (

QJ Bj) +

Pj P (Nj) =

QJ P (Bj)

Consequently

P�Q

J B0j

�=QJ P

�B0j

�which completes the proof.

Notation 12 C =�BI �

QT�I k : BI 2

QI Ai; I � T; jIj <1

is an al-

gebra, where Ai is an �-algebra on and T is (non-void) index set

Notation 13 Let ���BS �

QT�S : BS 2

QS At; S � T;

�:= AS If jSj <

1; thenQS At �= AS :

Theorem 14 Let f(;At) ; t 2 Tg be a family of measurable spaces andS � T be �nite. Then B =

SS�T (

QS At) is an algebra on

QT t := T :

Moreover F =SS�T AS =

QT At := � (C) is a �-algegra on T ; where S

ranges over all countable subsets of T

4

Proof. There is some �nite S; S1; S2 � T such that B 2 AS ; B1 2 AS1 ; B2 2AS2 if B;B1; B2 2 B: Thus BC 2 AS : and B1

SB2; B1

TB2 2 AS1[S2 ;which

means B is an algebra.F is just the maximal element of the chain. Let B 2 F : Then there is

some countable S � T such that B 2 AS : So BC 2 AS : For Bn 2 F ; thereis a sequence of countable Sn � T such that Bn 2 ASn : Let S� =

Sn Sn:

Then S� is countable andSnBn 2 AS� : Thus F is a �-algebra.

Since for any �nite I � T it�s clear that AI �SS�T AS = F where S is

countable, then

C =�BI �

QT�I k : BI 2

QI Ai; I � T; jIj <1

is a sub-algebra of F : So � (C) =

QT At := AT � F : But on the other hand,

for any B 2 F its corresponding AS � AT ; thus F � AT : Whence AT = F

Corollary 15 (Countably Generated) If B is a countably generated �-algebra of

QT At := AT ; then there is a countable subset S � T such that

B � AS : Particularly, every member of the �-algebra generated by any r.v.� 2 (

QT t;

QT At) depends only on countably many coordinates.

Proof. Let B = � (An; n � 1) �QT At: Then each An 2 ASn for some

countable Sn 2 T: Let S =Sn Sn: ThenAn 2 AS ;8n � 1 and B = � (An; n � 1) �

ASLet A 2

QT At and B =

�A;AC ;?;

: Then A depends only on count-

ably many coordinates from the above assertion. Since the Borel �eld Bon R is countably generated, that is, B = � (Bn; n � 1) for some speci�cBn � R; then for any � 2

QT At it�s clear that ��1 (B) := � (� 2 Bn; n � 1)

is a countably generated sub-�-algebra ofQT At: Thus � depends only on

countably many coordinates.

1.1 Existence of Probability Measure

De�nition 16 (Consistency Condition) Let � = f�t (�)gt2T be a ran-dom process, where �t (�) : (;S; P ) ! (I;�) and each �t (�) is measurable.For any n 2 N and ti 2 T; i = 1; � � � ; n the joint p.d.f. of

��t1 (�) ; � � � ; �tn (�)

�is the induced probability measure pt1;���tn (�) on (In;�n) given by

pt1;���tn (B) := P��! 2 :

��t1 (�) ; � � � ; �tn (�)

�2 B

�where B 2 �n:

pt1;���tn (�) is called the marginal measure of � at time ti 2 T; i = 1; � � � ; nand it has the following properties for any B 2 �m

(K.1) Compatibility: pt1;���tn;��� ;tn+m (B � Im) = pt1;���tn (B)

5

(K.2) Symmetry: pt1;���tn (�B) = p�(t1;���tn) (B) for any � 2 Sn;where

�B :=��x�(1); � � � ; x�(n)

�: (x1; � � � ; xn) 2 B

Theorem 17 (Kolmogorov) Let I be a complete, seperable measurable,metric space, i.e., a Polish space and suppose the family of probability mea-sures

fpt1;���tn (�) : n 2 N; ti 2 T; i = 1; � � � ; ng

satisfy the consistency condition, then exists a unique probability measureon (;F) such that

P (f! = (!t : t 2 T ) : (!t1;��� ;!tn) 2 Bg) = pt1;���tn (B) (2)

for all n 2 N and distinct ti 2 T; i = 1; � � � ; n and B 2 In;where = ITand F = � (C)

Proof. De�ne the cylinder

Ct1;��� ;tn (B) :=�! = (!t : t 2 T ) 2 IT : (!t1;��� ;!tn) 2 B

for each B 2 �n and distinct ti and de�ne

C = fCt1;��� ;tn (B) : B 2 �n; ti; 1 � i � n 2 Ng

Then (16) and (16) ensure that (2) well de�nes the unique, �nitely additiveset function P (�) on C. Since C is a semi-ring, the validity of the �-additivityof P (�) on C implies that P (�) is the unique probability measure on � (C)satisfying (2)

In order to show the �-additivity of P (�) ;it su¢ ces to show: Cn 2 C; Cn #?;then

limn!1

P (Cn) = 0 (3)

Since the sequence fCngn depends on at most coutably many coordinates,say, (t1; � � � ; tn; � � � ) and low dimensional cylinder corresponds to a highdimensional cylinder, we can let

Cn = f! = (!t; t 2 T ) : (!t1;��� ;!tn) 2 Bng

Suppose (3) is false, that is,

P (Cn) � " > 0;8n 2 N

we�ll show thatTnCn 6= ?: If each Cn is compact, then obviously

TnCn 6=

?: (the intersection of a sequence of non-empty monotone compact sets isnonempty) Since Bn is an n-dimensional Borel set, there exists compact setKn � Bn such that

pt1;��� ;tn (Bn �Kn) <"

2n+1(4)

6

LetDn = f! = (!t; t 2 T ) : (!t1;��� ;!tn) 2 Kng (shrink a little bit) and de�ne

~Dn =Qnk=1Dn =

Qns=1 f! = (!t; t 2 T ) : (!t1;��� ;!ts) 2 Ksg

=�!; (!t1 ; � � � ; !tn) 2

Qns=1

�Ks � In�s

�Then P

�~Dn

�< "

2 > 0 since by (4)

P (Cn �Dn) �"

2n+1

(Cn �Dn � Bn �Kn) and

P�Cn � ~Dn

��

nXk=1

P (Cn �Dk) �nXk=1

P (Ck �Dk)

�nXk=1

"

2k+1� "

2

(Cn � Ck for k � n)Thus

P�~Dn

�� P (Cn)� P

�Cn � ~Dn

�� "

2> 0

andTnCn �

Tn~Dn 6= ?:Finally, C being a semi-algebra implies the exten-

sion of P to � (C) is unique.

Remark 18 In the above proof the key is to show the continuity of P at? and the property of Polish space is exploited to show the existence of anon-empty intersection. In case I does not have any topological structure,if there exists a sequence

�!n =��!1; �!2; � � � ; �!n; !nn+1; !nn+2 � � �

�2 Cn (5)

then there is a diagonalized point

�! = (�!1; � � � ; �!n; �!n+1; � � � ) 2TnCn

(Note that each Cn is a cylinder). Since generally for any

!n = (!n1 ; !n2 ; � � � ; !nm; � � � ) 2 Cn

de�ne

An (!n) := f! = (!1; � � � ; !m; � � � ) : !i = !ni ; 1 � i � ng

it�s clear that An (!n) � Cn (Cn being a cylinder). Thus �! 2 An (!n) � Cn

7

Theorem 19 (Generalized Tulcea) Let fFngn�0 be a non-decreasing se-quence of sub-�-algebras of the measurable sapce (;F) such that Fn " F andlet fQ (0; �)g be a family of probability measures on F0 and fQ (n; !; �) ; n � 1gbe a family of conditional probability measures such that Q (n; !; �) is a prob-ability measure on (;Fn) for �xed ! and that Q (n; �;A) is Fn�1-measurablefunction for �xed A 2 Fn. Moreover, suppose fQ (0; �)g ; fQ (n; !; �) ; n � 1gsatis�es (the following consistency conditions)

(T.1) Q (n; !;A) = 1 for 8! 2 A 2 Fn�1 (i.e., E (1AjFn�1) (!) = 1A (!)for 8! 2 A 2 Fn�1M)

(T.2) For any sequence !(n) 2 ; n � 1;TNn=1An

�!(n)

�6= ?;8N � 1 =)

T1n=1An

�!(n)

�6= ?

Then there is a unique probability measure P on (;F) such that(1) P jF0 = Q (0; �) (i.e., P (A) = Q (0; A) ;8A 2 F0)(2) E (1AjFn�1) (!) = 1A (!) for 8! 2 A 2 Fn�1M (=) P (A) =RQ (n; !;A)P (d!))

Proof. Step 1: LetP (A) = Q (0; A) ;8A 2 F0

and inductively for A 2 Fn de�ne

P (A) =

ZQ (n; !;A)P (d!)

since Q (n; �;A) 2 Fn�1: From (T.1) it�s clear that for any A 2 Fn�1 � FnZQ (n; !;A)P (d!) =

ZAP (d!) +

ZACQ (n; !;A)P (d!)

=

ZQ (n� 1; !;A)P (d!) +

Z1Ac

�1�Q

�n; !;AC

��P (d!)

=

ZQ (n� 1; !;A)P (d!)

that is, P (A) is identical when de�ned on Fn�1 or Fn:Thus we have a well-de�ned �nitely additive non-negative set function on

C = fC : 9n � 0; C 2 Fng

Step 2: Show P (�) is countably additive on CChoose a sequence Cn 2 C with Cn # ? but P (Cn) � " > 0 (n � 1), it

su¢ ces to show TnCn 6= ?

WLOG, suppose Cn 2 Fn:We have to �nd !(n) 2 Cn such thatT1n=1An

�!(n)

�6=

? for any N � 1: Then from (T.2) it�s clear that �! 2 Cn:

8

LetQ (�1; n;!;A) := P (A)

andQ (n; n+ 1;!;A) := Q (n+ 1; !;A)

and (m-step transition probability which is also the measure of the n-section)

Q (n; n+m;!;A) (6)

: =

ZQ (n; n+m� 1;!; d!1)Q (n+m;!1;A)

(i.e., integralQ (n+m;!1;A) (Fn+m�1-measurable for �xedA) w.r.t. Q (n; n+m� 1;!; d!1)(a probability measure on Fn+m�1 for �xed !). Obviously, Q (n; n+m;!;A) 2Fn for A 2 Fn+m

De�ne

F kn =n! : Q (k; n;!;Cn) �

"

2n+1

o2 Fk � Fn�1 (7)

for k = 1; 2; � � � and n = k + 1; k + 2; � � �From

Q (k; n+ 1;!;Cn+1) =

ZQ (k; n;!; d!1)Q (n+ 1; !1;Cn+1)

�ZQ (k; n;!; d!1)Q (n+ 1; !1;Cn)

= Q (k; n;!;Cn)

it�s clear that F kn � F kn+1 for 8n > kNow let�s show by induction that there exists !(k) 2

�Tn>k F

kn

�TAk�1

�!(k�1)

�(k = 1; 2; � � � ) and !(0) 2

Tn>0 F

0n : Since

Q (k; n;!;Cn) =

ZQ (k + 1; n;!; d!1)Q (k; k + 1;!; d!1) (8)

� Q�k; k + 1;!;F k+1n

�+

Z(Fk+1n )

C

"

2k+2Q (k; k + 1;!; d!1)

� Q�k; k + 1;!;F k+1n

�+

"

2k+2

(�rst inequality: successive induction) Then for k = �1; we have from (8)

" � P (Cn) = Q (�1; n;!;Cn)� Q

��1; 0;!; F 0n

�+"

2= P

�F 0n�+"

2

andP�T

n�0 F0n

�= limn!1

P�F 0n�� "� "

2="

2> 0

9

Thus there is some !(0) 2Tn�0 F

0n such that

Q�0; n;!(0); Cn

�� "

2;8n � 0

Suppose !(k) 2�T

n>k Fkn

�TAk�1

�!(k�1)

�; then from (8), (7) it�s clear

that for !(k) 2 F kn

Q�k; k + 1;!(k);F k+1n

�� Q

�k; n;!(k); Cn

�� "

2k+2

� "

2k+1� "

2k+2=

"

2k+2

Consequently

Q�k; k + 1;!(k);

TF k+1n

�= limn!1

Q�k; k + 1;!(k);F k+1n

�� "

2k+2> 0

since !(k) 2 Ak�!(k)

�then for 8A 2 Fk with A � Ak

�!(k)

�we have

Q�k; k + 1;!(k); A

�= 1 =) Q�

�k; k + 1;!(k); A

�= 1

where Q� denotes the outer measure induced by Q; then

Q��k; k + 1;!(k);

�Tn>k+1 F

k+1n

�TAk

�!(k)

��> 0

and there must be

!(k+1) 2�T

n>k+1 Fk+1n

�TAk

�!(k)

�(9)

Finally, from (9) we have

!(k) 2 Ak�1�!(k�1)

�and since Ak�1

�!(k�1)

�2 Fk�1 � Fk it�s clear that

Ak

�!(k)

�� Ak�1

�!(k)

�� Ak�1

�!(k�1)

�Thus

T0�n�k An

�!(n)

�6= ? (for �nite k � 0). Whence from (T.2), it�s clear

that 9�! 2Tk�0Ak

�!(k)

�: Notice that

1Ck

�!(k)

�= Q

�k; k + 1;!(k); Ck

�� Q

�k; k + 1;!(k); Ck+1

�>

"

2k+1> 0

it�s clear that !(k) 2 Ck and from Ck 2 Fk we �nally have

�! 2 Ak�!(k)

�� Ck;8k � 0

which completes the proof.

10

2 Weak Convergence of Probability Measures

De�nition 20 A completely regular T1-space is called Tychono¤ space

Notation 21 For a topological space (X; �) the Borel algebra is always de-noted by B (X) ;i.e., the �-algebra generated by all open sets of (X; �) ; whichis the same if X is a metric space compatible with this topology. P (X) de-notes all probability measures on (X;B (X))

Notation 22 Cb (X) =�f 2 C0 (X) : kfk1 <1

Theorem 23 Every mesure on any seperable, complete metric space is reg-ular.

Proof. Use classical methods

Theorem 24 (Tychono¤) If X is a T1-space, there the following are equiv-alent

1. X is regular and has a countable base (for its toplogy)

2. X is homeomorphic to a subspace of the cube Q!

3. X is metrizable and seperable.

Theorem 25 A space is Tychono¤ i¤ it�s homeomorphic to a subset of thecube

Proof. Construct a countable family of continuous function that seperatespoints and points and closed sets

De�nition 26 Let �n; � 2 P (X) Then there are usually three types of con-vergence on P (X)

1. Uniform:limn!1

supA2B(X)

j�n (A)� � (A)j = 0

i¤

limn!1

supf2C(X);kfk1�1

j�n (f)� � (f)j = limn!1

supf2C(X);kfk1�1

����Z fd�n �Zfd�

���� = 02. Strong:

limn!1

j�n (A)� � (A)j = 0;8A 2 B (X)()

limn!1

j�n (f)� � (f)j = 0;8f 2 B (X)

11

3. Weak:limn!1

j�n (f)� � (f)j = 0;8f 2 Cb (X)

which is denoted by �n �

Theorem 27 Let �n; � 2 P (X) Then the following are equivalent

1. �n �

2. �n (f)! � (f) ;8f 2 Cbu (X)

3. limn!1�n (C) � � (C) ;8closed C � X

4. limn!1�n (C) � � (C) ;8open G � X

5. �n (A)! � (A) ;8�-continuity set A � X

Proof. Only show (2)=)(3), (3) or (4) =) (5), (5)=) (1)For closed C � X let

fm (x) =1

1 +m� (x;C)# 1C (x) ;8x 2 X

Obviously fm 2 Cbu (X) and

� (C) = limm!1

Zfm (x) d� = lim

m!1limn!1

Zfn (x) d�n � limn!1�n (C)

(3) or (4) =) (5): Let A be a �-continuity set, then � (A) = ���A�=

� (A�) :Then

� (A) = � (A�) � limn!1�n (A�) � limn!1�n (A)

and� (A) = �

��A�� limn!1�n

��A�� limn!1�n (A)

Hence� (A) = lim

n!1�n (A)

(5)=) (1): The idea is to use step function fn de�ned on �-continuitysets to uniformly approximate f 2 Cb (X) : Since � is a probability measure,the set

Jn (a) =

�a 2 [0; 1] : � (fx 2 X : f (x) = ag) � 1

n

�has cardinality at most n and

J (a) =Sn Jn (a) = fa 2 [0; 1] : � (fx 2 X : f (x) = ag) > 0g

12

is at most countable. Consequently, for 8n > 0;M > kfk ;there 9a(n)1 ; � � � ; a(n)2n 2R such that

a(n)k 2

��k � 1n

� 1�M;

�k

n� 1�M

�and

��nx 2 X : f (x) = a

(n)k

o�= 0;81 � k � 2n

ThusA(n)k =

nx 2 X : a

(n)k�1 � f (x) < a

(n)k

oare all �-continuity sets. For 8f 2 Cb (X) ; de�ne

fn (x) =

2nXk=1

a(n)k 1

A(n)k

(x)

Then obviously

supx jfn (x)� f (x)j � max1�k�2n

���a(n)k � a(n)k�1��� < 2M

n! 0

and Zfm (x) d�n =

2mXk=1

a(m)k �n

�A(n)k

�!Zfm (x) d�

Finally from limmRfm (x) d�n =

Rf (x) d�n uniformly for n;it�s clear that

limn

Zf (x) d�n = lim

nlimm

Zfm (x) d�n = limm

limn

Zfm (x) d�n

= limm

Zfm (x) d� =

Zf (x) d�

De�nition 28 A set A is called a �-continuity set i¤ ���A�= � (A�)

Theorem 29 There is a metric ~� de�ned on P (X) such that �n � i¤~� (�n; �)! 0

Proof. By Tychono¤ Embedding theorem, that is, any seperable metricspace could be homeomorphically embeded into the compact space [0; 1]N ;then the closure �X is also compact and any function f 2 Cbu (X) can beextended to ~f 2 C

��X�. Consequently C

��X�is seperable with, say, ffngn

being one of its countable dense subset. De�ne

~� (�1; �2) =

1Xj=1

1

2j

��R fjd�1 � R fjd�2��1 +

��R fjd�1 � R fjd�2��and (P (X) ; ~�)

13

Remark 30 Another metric d could be de�ned on P (X) as

d (�1; �2) = infn� : �1 (A) � �2

�A��+ �; �2 (A) � �1

�A��+ �;8closed A � X

owhere

A� = fx 2 X : � (x;A) � �g

The metric d is called Levy-Prohorov metric and is equivalent to ~�

Theorem 31 Let X be a compact metric space, then P (X) is sequentiallycompact under w.r.t weak topology

Proof. Since X is a compact metric space, then Cb (X) is seperable (withsupremum norm). Let ffngn be a dense subset of Cb (X) and for 8fn and�j 2 P (X) ; n; j � 1

j�j (fn)j =����Z fnd�j

���� � kfnkExtract a diagonal sequence such that

limn

Zfnd�jr exists;8n � 1

for some��jr: Since for 8f 2 Cb (X) ;there 9nk such that

supx2X jfnk (x)� f (x)j ! 0

then ����Z fd�jr �Zfd�js

�����

����Z (f � fnk) d�jr

����+ ����Z (f � fnk) d�js

����+ ����Z fnkd�jr �Zfnkd�js

����Since for �xed nk there 9N such that����Z fnkd�jr �

Zfnkd�js

���� < "

2

for all r; s > N:It�s clear that����Z fd�jr �Zfd�js

���� < "

2

i.e., there exists

� (f) = limr

Zfd�jr ;8f 2 Cb (X)

14

i.e.� 2 (Cb (X))�

By the Rieze representation theorem, there is some �ntie measure � on B (X)such that

� (f) =

Zfd�

Obviously

� (X) = � (1) = limn

Z1d�n = 1

implies� 2 P (X) ; �n �

Remark 32 If X is not compact, then Cb (X) may not be seperable. TakeR and de�ne for any subset E � Z

fE (x) =

8<:1 if x 2

�n+ 1

3 ; n+23

�; n 2 E

linear if x 2�n; n+ 1

3

�;�n+ 2

3 ; n�

0 otherwise

Obviously for each F � Z with F 6= E

kfE � fF k = 1

and ffE : E � Zg is an uncountable set.

De�nition 33 f�tgt � P (X) is said to be tight i¤ for 8" > 0; there9compact K" � X such that

�t (X �K") < ";8t

Theorem 34 (Prohonov) Let X be a seperable metric space and f�n 2 P (X)gn�1be tight. Then f�ngn�1 has a weakly convergent subsequence, i.e., f�ngn�1is sequentially compact under weak topology.

Proof. Since X is hemeomorphic to a subset of [0; 1]Z+

;there is an equiva-lent metric on X such that �X is compact under this metric. Consequently,taken as probability measures on P

��X�(how?); f�ng has a subsequence

�nk � 2 P��X�

To show � 2 P (X) ; it su¢ ces to show

���X �X

�= 0

15

To this end, take 0 < "j # 0 and select compact sets fKj "g such that

�n (Kj) > 1� "j

Thus

1 � � (X) � limj� (Kj) � lim

jlimk�nk (Kj) � limj (1� "j) = 1

and0 � �

��X �X

�� 1� � (X) = 0

i.e.,� 2 P (X)

Theorem 35 Let X be a seperable, complete metric space and A � P (X)be compact w.r.t to the weak topology. Then for any " > 0 there existscompact set Ks � X such that � (Ks) � 1� " for any � 2 A

Proof. For any � 2 P (X) and open Gn " X; it�s clear that � (X �Gn) # 0:Moreover, since X �Gn is closed, then

limm�m (XnGn) � � (XnGn)

for all �m � and any �xed n: On the other hand, there 9�m 2 A suchthat

sup�2A

� (XnGn) = limm�m (XnGn) ;8n (10)

Since A is compact, we can assume �m �0: Then

sup�2A

� (XnGn) � limm�m (XnGn) � �0 (XnGn)! 0 (11)

as n!1: The seperabilty of X implies

X =S1j=1 S

�xj;r; 2

�r�where S (x;R) is the open sphere centered at x with radius R: Let

G (n; r) =Snj=1 S

�xj;r; 2

�r�then G (n; r) " X and for any " > 0 there is some N ("; r) such that (from(11))

� (XnG (n; r)) < "

2r;8� 2 A;8n>N ("; r)

Thus settingG" =

T1r=1G (N ("; r) ; r) ;K" = G"

16

It�s clear that K" is totally bounded (and hence is compact since X is com-plete) and

� (K") � 1� � (XnG") � 1�1Xr=1

� (XnG (N ("; r) ; r))

� 1�1Xr=1

"

2r= 1� "

3 Martingale Theory

Lemma 36 Suppose fxn; y : n � 1g is a closed submartingale sequence, thenfor any c > 0

c� (fsupxn � cg) �Zfsupxn�cg

y� (d!)

Proof. Since�supxn > c

0 =Xm

�xk � c0; k < m; xm > c0

=X

Am

ThenZfsupxn�c0g

y� (d!) =

ZPAm

y� (d!) =XZ

Am

y� (d!)

=XZ

Am

E (yjxk; k � m)� (d!)

�XZ

Am

xm� (d!) = c0���supxn > c

0�Now let c0 ! c;it�s clear that the assertion holds.

Lemma 37 Suppose x; z are non-negative r.v.�s such that E (xr) ; E (zr) <1 and for any � > 0

�� (fz (!) � �g) �Zfz(!)��g

x� (d!)

then

E (zr) ��

r

r � 1

�rE (xr) ; r > 1

17

Proof. Choose nondecreasing function � (�) ; � > 0 such that � (0) = 0 andE (j� (z)j) <1 then

E (� (z)) =

Z 1

0� (�) d� (fz (!) < �g)

= �Z 1

0� (�) d� (fz (!) � �g)

�Z 1

0� (fz (!) � �g) d� (�) �

Z 1

0

d� (�)

�

Zfz(!)��g

x� (d!)

=

Zx� (d!)

Z z(!)

0

d� (�)

�

Now take � (�) = �r;then

E (zr) � r

r � 1E�xzr�1

�� r

r � 1E (xr)1=r E (zr)1�1=r

i.e.,

(E (zr))r ��

r

r � 1

�rE (xr)E (zr)r�1

Remark 38 Let q = rr�1 and p = r: Then

1p +

1q = 1

Lemma 39 Suppose g (x) is convex on R and E (j� (!)j) <1; E (jg (�)j) <1; then

g (E (�jB)) � E (g (�) jB) ; a:s:

Proof. Let Q be the set of all rationals and for any �i 2 Q let

G (�i; !) = � (� (!) � �ijB)

where the rhs is a version of E��f�(!)��ig (!) jB

�. Then there is some

measurable set A with � (A) = 0 such that any G (�i; !) is right-continuous,non-decreasing wrt to �i for ! =2 A and

lim�i!1

G (�i; !) = 1

Now extend G : R� ! R by letting

G (�; !) =

�lim�i#�G (�i; !) if ! =2 A

F (�) if ! 2 A

where F (�) is any d.f. independent of !: Thus for each ! 2 ; G (�; �) is ad.f. on R with the induced measure G (B;!) for any measurable B:Let

M = fB 2 � (R) : G (B;!) 2 B �� () g

18

Then obviously M is a �-class that includes the �-class consisted of all(�1; �]: Consequently M =� (R) and G (B;!) is measurable in ! for anyB 2 � (R) :

Further, let

L = ff 2 � (R) : E (jf (� (!))j) <1g

and

L =

�f 2 � (R) : E (f (� (!)) jB) =

ZRf (x)G (dx; !)

�Obviously L is a L-class that contains all the c.f.�s of the sets (�1; �]: ThusL � � (R) and speci�cally for f (x) = g (x) and f (x) = x it�s clear that

g

�ZRxG (dx; !)

��ZRg (x)G (dx; !)

which is a consequence of Jessen�s inequality.

Remark 40 it�s clear that

E (f (�)) =

ZRf (x)F� (dx)

by integral transformation formula if E (jf (�)j) <1: Similarly in

E (f (� (!)) jB) =ZRf (x)G (dx; !)

the G (dx; !) is called the conditional d.f. w.r.t B (or regular c.d.f)

De�nition 41 � : ! N̂ is called a stopping time wrt the �ltration fFngi¤

f� = ng 2 FnMoreover,

F� = fE 2 F1 : E \ f� = ng 2 Fn; n � 1g

is called the �-�eld prior to � ;where

F1 =W1n=1Fn

(is the lattice theory de�nition)

Remark 42 If fXn;Fng is adaptive, then X1 = lim supXn is F1-measurableand consequently X� is F�-measurable is � is adapted.

19

Lemma 43 Let fXn;Fng be an adaptive (sub)martingale and �; � be bounded,stopping times with � � �;then

E (X�jF�) = X� (� X�)

Proof. Let G 2 F� and Gn = G \ f� = ng : Clearly for any k � j

Gj \ f� > kg 2 FkBy the de�nition of submartingaleZ

Gj\f�>kgXkdP �

ZGj\f�>kg

Xk+1dP

then ZGj\f��kg

XkdP �ZGj\f�=kg

XkdP +

ZGj\f�>kg

Xk+1dP

Thus ZGj\f��kg

XkdP �ZGj\f��k+1g

Xk+1dP �ZGj\f�=kg

X�dP

Suppose m is supremum for �. Summing up the above inequality w.r.t. kfrom j to m yieldsZ

Gj\f��jgX�dP �

ZGj\f��m+1g

Xm+1dP �ZGj\fj���mg

X�dP

i.e., ZGj

X�dP �ZGj

X�dP

FinallyZPmj=1Gj

X�dP =

mXj=1

ZGj

X�dP �mXj=1

ZGj

X�dP =

ZPmj=1Gj

X�dP

Theorem 44 (Doob) Let fXn;Fng be an adaptive martingale. Then forp > 1 max1�j�n

jXj j p

= E

��max1�j�n

jXj j�p�

� qpE (jXnjp) = qp kXnkp

Proof. It�s the same as (37) but with a di¤erent proof. Note that

E (jXj) =

ZjX (!)j dP (!) =

Z

Z 1

0Ifx�jX(!)jgdxdP (!)

=

Z 1

0dx

ZIfx�jX(!)jgdP (!) =

Z 1

0P (jX (!)j � x) dx

=

Z 1

0(1� F (x)) dx

20

Then

E

��max1�j�n

jXj j�p�

=

Z 1

0P

��max1�j�n

jXj j�p� x

�dx

=

Z 1

0dx

ZIfx�fmax1�j�njXj jgpgdP =

ZdP

Z 1

0Ifx�fmax1�j�njXj jgpgdx

=

Z

�max1�j�n

jXj j�pdP =

ZdP

Z max1�j�njXj j

0pxp�1dx

=

ZdP

Z 1

0Ifx�max1�j�njXj jgpx

p�1dx

= p

Z 1

0xp�1P

��x � max

1�j�njXj j

��dx � p

Z 1

0xp�2E

�jXnj Ifx�max1�j�njXj jg

�dx

= pE

jXnj

Z max1�j�njXj j

0xp�2dx

!= qE

jXnj

�max1�j�n

jXj j�p�1!

� q (E (jXnjp))1=p�E

�max1�j�n

jXj j�p�1=q

Brownian Motion and Stochastic CalculusChapter 1: Martingales, Stopping times, Filtrations

Problem 45 Let Y be a modi�cation of X and suppose that both processeshave a.s. right-continuous sample paths. Then X and Y are indistinguish-able.

Solution 46 Since for any t � 0

P (Xt = Yt) = 1

Choose Q = fri : i � 1g. Then

P�S

ri(Xri 6= Yri)

��X

riP (Xri 6= Yri) =

Xri(1� P (Xri = Yri)) = 0

that is,

P�\

ri(Xri = Yri)

�= 1

Since for any t � 0;there is a sequence fri;tgi � Q such that ri;t # t; then

limiXri;t = Xt; lim

iYri;t = Yt; a:s:8! 2

by the hypothesis and

1 � P (Xt = Yt;8t � 0) = P�T

t�0 (Xt = Yt)�

= P

�Ti

�limiXri;t = lim

iYri;t

��� P

�Tri(Xri = Yri)

�= 1

21

that is, X and Y are indistinguishable.

De�nition 47 Let (X;S) ; (Y; T ) be two measurable spaces. The productionmeasure space (X � Y;S T ) is de�ned to be

S T = � (S � T ) = � (fA�B 2 S � T g)

where each A�B 2 S � T is called a measurable rectangle.

Lemma 48 Every section of a measurable set is a measurable set (PaulHalmos)

Proof. LetE = fE 2 X � Y : Ex 2 S; Ey 2 T g

Then obviouslyS � T � E

and E is a �-ring. Thus E � S T

Corollary 49 If Xt (!) 2 B ([0;1))F ; then Xt (!) 2 B ([0;1)) for �xed! 2

Proof. By (). The trajectory for a �xed ! 2 is a section of Xt (!) andhence is measurable w.r.t B ([0;1)); that is, sincen

(t; !) : Xt (!) 2 A 2 B�Rd�o

= A1 �A2 2 B ([0;1))F

then any �xed ! 2 ;nt : Xt (!) 2 A 2 B

�Rd�o

= A1 2 B ([0;1))

(Fubini�s Theorem is too much for this corollary).

Problem 50 Let X be a process, every sample path of which is RCLL. LetA be the event that X is continuous on [0; t0): Show that A 2 FXt0

Proof. For any t 2 (0; t0) ;let

En;kt =

�! 2 :

���Xt (!)�Xt� 1k(!)��� < 1

n

�and

A� =Tn�1

Sm�1

Tk�mE

n;kt

Since Xt (!) is RCLL for each ! 2 ;it�s clear that A = A�: Note that forany k > 0;

Xt� 1k(!) 2 FX

t� 1k

; Xt (!) 2 FXt

22

thenXt (!)�Xt� 1

k(!) ;

���Xt (!)�Xt� 1k(!)��� 2 FXt

since�FXttis increasing. Consequently,

A� = limnlimkEn;kt 2 FXt � FXt0

Problem 51 Let X be a process whose sample paths are RCLL almostsurely, and let A be the event that X is continuous [0; t0): Show that A canfail to be in FXt0 ; but if fFt; t � 0g is a �ltration satisfying F

Xt � Ft; t � 0

and Ft0 contains all P -null sets of F ;then A 2 Ft0

Problem 52 Let X be a process with sample paths are LCRL almost surely,and let A be the event that X is continuous [0; t0] : Let X be adapted to aright-continuous �itration fFtg : Show that A 2 FXt0 :

Theorem 53 (Crossing) 000

23