‫مجلة جامعة كرميان‬

Journal of Garmin University

‫طؤظاري زانكؤي طةرميان‬

Equivalent Martingale Measure on d -Stochastic Process By Bouhra Youssif Hussein Dept. of Mathematics, College of Education, AL-Qadisiya University E-mail: [email protected] Abstract Suppose that (Xt) tN is adapted sequence of d-dimensional bounded random variables on probability space (, F, P). In this paper we obtain condition which is necessary and sufficient for the existence of unique probability measure Q equivalent to P on dstochastic process. Keyword :- Stochastic process, probability measure, equivalent measure, martingale

1-Introducation. In recent years the mathematical theory of stochastic integration (stochastic process) has become of interest because of its several areas of mathematical stochastic integrals with respect to martingales which were first discussed by Winner (1923). The extension of this definition is due of square martingale. In their fundamental paper, Harrison and Kreps (1981), discussed the fundamental theorem and introduced the concepts of equivalent martingale measure. Absence of arbitrage alone was not sufficient to obtain an equivalent martingale measure for the stochastic process. The triple (,F,P) is called probability space, where  be a non-empty set , F is a -field on , and P is a probability measure. The process X, Sometimes denoted (Xt)tI is supposed to be d-valued, although all proofs work with a d-dimensional process as well. The theory of d-dimensional stochastic integration is a little more subtle than the one dimensional theory but no difficulties arise, this process adapted to filter {Ft}tI. The probability measure Q defined on F is equivalent to P if Q and P contain the same null sets. We say the equivalent probability measure Q is equivalent martingale measure to (Xt)tI if (Xt)tI is martingale with respect to Q, in other word if X is integral with respect to Q and for all tI, i.e. E((XtXt-1 F)=0. 649

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Vol.1 No.11 (January, 2017)

Journal of Garmin University

‫مجلة جامعة كرميان‬

‫طؤظاري زانكؤي طةرميان‬

2.1 Basic concepts. Let (, F, P) be any probability space. We consider (Xt) tN be d-valued bounded random variables on (, F, {F t} tN, P). Theorem (2.1),[1]:”Kolmogorov theorem” Let (E,d) be a complete space, and let Ux be an E-valued process for all x dyadic rational in n. Suppose that for all x,y we have d(Ux,Uy) is a random variable and that there exists strictly positive constants G,,C, such that E{(d(Ux,Uy))} x7C -yG+ then for almost all  the function xUx can be extended uniquely to a continuous function from n to E.

Definition (2.2), [5]: Let  be a collection of subsets of a set of , and let A. The restriction (or trace) of  on A is the collection of all sets of the form AB, where B, and it is denoted by A or A, i.e. A= A= {AB: B} A is a collection of subsets of A. The -field (A) generated by A sometime denoted by A (A), i.e. (A)=A (A) Theorem (2.3), [5]: Let  be a collection of subsets of a set , and let A : (1) A () is the -field of subsets of A. (2) (A) = A (A). (3) If  is closed under finite intersection and A, then: A= {B : BA}. (4) If  is a -field on , then A is a -field on A. Definition (2.4), [6]: Let (,F) be a measurable space. A set function :F  is said to be 650

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Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬ (1) Finitely additive, if n

n

i 1

i 1

μ(  Ai )   μ(Ai ) Whenever A1,A2…An are disjoint sets in F. (2) -additive, (or countable additive), if 



i 1

i 1

μ(  Ai )   μ(Ai ) Whenever {An} is a sequence of disjoint sets in F.

Definition (2.5), [6]: Let  be a finitely additive set function on a measurable space (,F) with (A) >- for all AF. We say that  is (1) Continuous from below at AF, if (An)(A) Whenever {An} is a sequence of sets in F with AnA such that (An)<- for some n. (2) Continuous from above at AF, if (An)(A) Whenever {An} is a sequence of sets in F with AnA such that (An)< for some n. (3) Continuous at AF, if it is continuous both from below and from above at A.

3. Equivalences In this section, we discuss the relation between no arbitrage and the condition how is necessary and sufficient to get equivalent measure in discrete time. Definition (3.1), [1]: We say that the process (Xt) tN satisfies the no arbitrage condition if For t=1, 2… N and each Ft-measurable bounded d-valued function h such that h (), Xt () 0 a.e. [P], we have h (), Xt () =0 a.e. [P]. Proposition (3.2):

651

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Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬ N

Let M0= {  hi (), Xt () : hi be Ft-measurable bounded d-valued function for i 1

1 I N}. Then M0 is convex.

Proof: Let x, yM0 and let 0  1 N

N

i 1

i 1

Then x=  hi (), Xt (), y=  hi (), Xt () When hi, hi are F t-measurable bounded d-valued functions N

N

i 1

i 1

x+(1- )y =  hi(),Xt()+(1- )  hi(),Xt() N

N

i 1

i 1

=  hi (),Xt()+  (1- )hi(),Xt() N

Then x+(1- )y =  hi()+(1- )hi(),Xt()M0 i 1



Thus M0 is convex. In this following we need the following proposition. Proposition (3.3): Let X be the set of all bounded

Ft-measurable d-valued functions. Then the

process (Xt) tN satisfies the no-arbitrage condition if and only if M0X+= {0}. Proof: Suppose the process (Xt) tN satisfies the no-arbitrage condition, the function h (), Xt () are element of M0, If M0X+ {0} this contradicts the process (Xt) tN satisfies the no-arbitrage condition. Suppose M0X+= {0} If N=1 Since M0= {Xi (),hN (): hN is Ft-measurable bounded d-valued function} N

Let fN=  Xi (), hi (), 1 i N i 1

652

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Vol.1 No.11 (January, 2017)

‫مجلة جامعة كرميان‬

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

Suppose that fN 0. We show that fN =0. If fN-1=0 N 1

We have fN =  Xi (), hi ()+XN(),hN() i 1

=0+XN (),hN () =0 by case N=1. If fN-10 Let A= {fN-1<0} A FN-1 with strictly positive measure Then f=XN (),hN ().IA  f 0 and f0 This contract to the case N=1.



Proposition (3.4): Let {Xt} tN be an d-valued stochastic process, and then the following conditions are equivalent: (1)

P(A{:Xt+1()>Xt()}>0



P(A{:Xt+1()0

for

any

t,

AFt. (2) For t=1, 2… N the process (Xt) tN satisfies the no-arbitrage condition. Proof: (1) (2): suppose (1) is true for any AF For t=1, 2… N and each Ft-1-measurable bounded d-valued function h such that h(),Xt()  0 a.e.[P] i.e. h (),Xt+1()-Xt()  0 a.e. [P] To prove h (),Xt+1()-Xt() =0 Suppose not, i.e. there exists hL 0  such that  h(),Xt+1()-Xt() >0 Then  h(),Xt+1() >  h(),Xt() P{ :  h(),Xt+1() >  h(),Xt() } > 0 i.e. P{ :  h(),Xt+1()-Xt() > 0} > 0 Then P{ :  h(),Xt+1()-Xt() < 0} > 0 653

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‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬

Thus  h (),Xt+1()-Xt() < 0 for some hL 0  but  h (),Xt+1()-Xt()  0 this contradiction. Therefore for t=1,2,…,N and each Ft-1-measurable bounded d-valued function h such that  h(),X()=0 a.e.[P]. (2) (1): Suppose (2) is true Suppose P (A {  : Xt+1()> Xt()}) > 0 for any tN, AFt. To prove P (A {  : Xt+1()< Xt()}) > 0 for any tN, AFt. Let h be a positive Ft-1-measurable function such that the set

{ : h()h(A)}

satisfies P{{:h()h(A)}{ : Xt+1()>Xt()}}  0 Then A  { : Xt+1()>Xt()}= the set {:  h(),Xt+1()}{: h(),Xt()}  t=0,1,…,N  h (),Xt+1()   h(),Xt()  .  h (),Xt+1() -  h(),Xt()  0  h (),Xt+1() - Xt()=0 by (2) then P {{ : h ()A} { : Xt+1


4. Main results Let (Xt) tN be a stochastic process on a probability space (,F, P). Let F = (X0,X1…) , X0=0 The following theorem is the main result.

Theorem (4.1): Let X () = {Xt():} be a closed set in the Tichonov topology on N. The following conditions are equivalent: (1) P(A (Xt+1 > Xt)) > 0  P(A (Xt+1 < Xt)) >0 for tN, AFt. (2) There exists a measure Q on F for which (Xt)tN is a martingale with respect to { F t} and QF t  PF t , tN. Proof: Before proof the part one of this theorem we need the following notations and the following lemmas: 654

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Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬ Notations:

Let (Yt) tN be a stochastic process on a probability space (,F, P). By Yt we also denoted the canonical representation on the space ([0,,(),PY. Thus: (1) Yt() =Yt(1,2,…)=t  N. (2) n(A(n))={(1,2,…)N ; (1,2,…,n) A(n), A(n)(n)} (3) n={n(A(n)) ; (A(n))(n)} (4) =  ζ n  ζ 1  ζ 2  ... nN Lemma (4.2): Let T is any set closed in n in the Tichonov topology, Pn is

a probability on n  T,

nN satisfying: Pn+1(n+1(AnR) T) =Pn(n+1(An) T)

(*)

Where R (n). Then there exists an uniquely a probability measure P define on (T) =T () such that Pn ζ  T =P(nTAn), An(n). n Proof: Let n=Tn  =  (T  ζ n ) =T  (  ζ n )=T  

(by not.3)

nN

nN

Let B then B can representation of the form B=Tn(An), An(n). We defined Q: by Q(B) =Pn(Tn(An)) Since, Pn is unique  nN therefore Q is unique  B. To prove (1) Q is finitely additive on . Let Bi,  i=1, 2,…,n n Then Q(Bi)=Pn(T (  (ζ n (A n )) i ) i 1

n

=Pn(  T  (ζ n (A n )) i ) i 1

655

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Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬ n

n

i 1

i1

=  Pn (T  (ζ n (A n )) i ) =  Q(Bi ) this implies Q is finitely additive on . (2) Q is norm on . Let B (i) Q(B)  0 (Pn is non negative function) (ii) If Q(B)=0  Pn(T (n(An)))=0 implies (T (n(An)))=  therefore B= If B= We have B= (T (n(An)))= implies Pn(T (n(An)))=0 then Q(B)=0 (iii) Let , and B Q(B)= Pn((T (n(An))))= Pn(T (n(An))). (iv) Let B1,B2 Thus B1= T (n(An)), An(n), B2= T (n( A n )), A n (n). Q(B1+B2)= Pn((T (n(An)))(T (n( A n )))). = Pn(T (n(An)))+Pn(T (n( A n ))) (Pn is additive) = Q(B1)+Q(B2)  nN. (3) To prove Q is -additive on . Let B1B2… Assume Bn=n(An) and suppose

Lim (Bn)0 this implies Q(Bi)   >0.

n 

To show that Bi.

~

We consider Pn (Cn) =Pn(TCn) for Cnn. i.e. we put Cn=n(TnAn) Tn={(1,2,…,n ):(1,2,…,n,n+1,…)T for some n+1,n+2,…. If C1C2…

~

then Pn (Cn)   >0 656

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‫مجلة جامعة كرميان‬

‫طؤظاري زانكؤي طةرميان‬

there exists   C i iI

Since T is closed  T Then (T (n(An))=  B n nN

Therefore Q is continuous Thus Q is countable additive on  Q(B)=Pn((T (n(An))) =Pn+1((T (n(AnR)))  Q ζ n T ((n(An))= Pn((T (n(An))) ,  n(An).



the Lemma proved.

Lemma (4.3):Let Yt,tN+ be a stochastic process on a probability space (,F,P) where F =(Y1,Y2,…) and Y() is closed in [0, and if P *Y ζ n  P Y ζ n For some probability measure P *Y on (), then there exists a uniquely defined probability measure P* on F satisfying:P *Y (A)=P*(Y-1(A)) ,  A() Proof: We take T=Y() and defined Pn(CnT)= P *Y (Cn) for Cnn. If CnT= Cn T for Cn, Cn  Implies (Cn Cn )T= then P *Y (Cn Cn )=0 we have Pn((Cn Cn )T)=0 therefore P *Y (Cn Cn )=0 i.e. Pn(CnT)=Pn( Cn T) thus Pn is well define. Also, by definition of Pn, Pn satisfy the condition (*).

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Journal of Garmin University

‫مجلة جامعة كرميان‬

by Lemma(3.3.2), there exists a probability measure on T() and Pn=P ζ n T , the measure P* is uniquely defined by P*(Y-1(A))= P *Y (A)=P(AY()) for A() The measure P on ()T corresponds to measure

**

P Y (A)= P*(Y-1(A))=P(AT) on () . But from Pn=P ζ  T we get n P *Y* (Cn)=P *n (CnT)

(1)

*

P Y (Cn)=P *n (CnT

(2)

From (1) and (2) P *Y* (Cn)= P *Y (Cn). To prove P* is countable additive on () if AY()=  A()  P *Y (A)= P*(Y-1(A))=P(AY())=0  P* is countable additive on (). The Lemma is proved.



Proof Theorem (4.1): (1) (2) d d 1 d 2 1 2 Let Yt=(Y1t , Y 2 t … Y t ) =(X t -X t 1 , X t -X t 1 … X t -X t 1 )

d =(X1t ,X 2 t ,…,X t ).

Let W = {: Y(1)() >0}  P (W) =P {: Y(1)() >0} 0 by (Lemma 3.3.2), we discuss a canonical representation (RN,(),PY) for the process Yt=(Y 1t ,…, Y dt ), there exists a measure P *Y on () satisfying P *Y  ζ  PY ζ , tN. t t To show this, we use mathematical induction a sequence of probability P (t) on () satisfying P (A) ζ =P(B) ζ where B(X0,…,Xt+1) and t t 658

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Journal of Garmin University

‫مجلة جامعة كرميان‬ EP(t)[Xt+1-Xt ζ ] =0

 tN.

t

ζ i.e. E t [Yt+1] =0  tN. P(n)

If t=0 , let P(0)=PY To prove PY=P (1) Define 1()=1 if P{W}=0 Otherwise n

 ei Xi (ω)

if

Y(1)(),e > 0

1

if

Y(1)(),e=0

if

Y(1)(),e < 0

i 1

1()=

n

 ei Yi (ω)

i 1

Where Xi(),Yi() are uniquely determined on a set (Y(1)0) n

n

 ei Xi (ω) Y (1) (t)-  ei Yi (ω) Y (1) (t)=0

i 1 n

i 1

[3]

n

ζ

 ei Xi (ω) .E P0 (I{:Y(1)(),ei >0})-  ei Yi (ω) .

i 1

E

i 1

Y

ζ0 ζ (I{:Y(1)(),ei < 0})= E 0 I{:Y(1)(),ei 0}) PY PY

[4]

With 0={,n} 1 if A and I()A={: Y(1)(),ei}= 0 if A  P({: i=0}t)=P({: i=1}t) P(0) ζ =P(1) ζ 0 0 ζ ζ ζ E P0(1) [Yi(1)]= E P0(0) [Yi(1)]= E P0

Y.φ1

[Yi(1)]=0

i i  Y (1) .1dPY=  Y (1) .dPY=0

Ω

659

Ω

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Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬

Assume, the theorem is true for t=n-1. To prove it for t=n . Define n+1()=1 if Pn(: Yn+1() >0)=0

n

 ei Xi (ω) if

i 1

n+1()=

1

Yn+1(),e > 0

if Yn+1(),e=0

n

 ei Yi (ω) if Yn+1(),e < 0

i 1

where Xi(),Yi() are uniquely determined almost everywhere on a set (Yn+10) by n

ζ

n

 ei Xi (ω) . E Pn(n) (I{:Yn+1(),ei >0})-  ei Yi (ω) .

i 1

i 1

ζ ζ E Pn(n) (I{:Yn+1(),ei < 0})= E n I{:Yn+1(),ei 0} Pn P((Y1,,…,Yn+1)  t)=P((Y1,,…,Yn+1)  t)  P(n+1)  ζ t =P(n)  ζ t = P *Y  ζ t  PY ζ t  P(n+1)  ζ t  PY ζ t  Pn+1=PY.n+1  dPn+1/dPY=n+1. (Which is Randon-Nikodem)

*

The measure P Y =Product measure, is uniquely (from the () is smallest

-field).

ζ and, E n (Yn+1)=0 Pn  P *Y (Ai)= P *i (Y i1 (Ai)) , A() P*is a probability on F it =(Y1i , Y i2 ,…) , i=1,…,d for F it =( Y10 ,…, Y it ) this implies F it =(Yi)( it ) [email protected] 660

Vol.1 No.11 (January, 2017)

‫طؤظاري زانكؤي طةرميان‬

Journal of Garmin University

‫مجلة جامعة كرميان‬  Ft= F 1t  F 2t … F dt

1 d 1 2 =(Y 11 (1t )Y  2 ( t )…Y n ( t ))

 P* P F t on Ft= F1t … F dt F

 E P *t Yn+1=0  Xt=(X 1t ,…,X n t )

is martingale with respect to product measure, and

-field

(X i0 ,…,X it )=(Y1i ,…,Y in ). (2) (1) Suppose there exist a measure Q on F which (Xt)tN is a martingale with respect to { F t}tN. Then the process (Xt)tN satisfies no-arbitrage condition (1) is true by proposition (3.2.4).

References [1]E.Czkwanianc and A.Pazkiewicz,"on martingale measure for stochastic process with discrete time",Probability and Potentia,New York,Oxford,1987. [2] E.jouini and H.Kallal ,"Martingale and Arbitrage in securities markets with transaction costs",Journal of mathematical Economics, (1999). [3]J.M.Harrian and .M.Kreps,"Martingales and stochastic inferals in the theory continuous trading",stochastic processand Application,vol.11,pp.215-260(1981) [4]N.Gaussal,"The

martingale

property

of

prices

when

arbitrage

opportunities

exist",university of Paris I,Ph,D,2002 [5]R.B.Ash,"Real analysis and probability", Academic process,New York,1992. [6]S.T.Taylor,"Introducation

to

measure

and

integration",Combridge

University

press,1966.



661

[email protected]

Vol.1 No.11 (January, 2017)