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Journal of Babylon University/Pure and Applied Sciences/ No.(3)/ Vol.(19): 2011 On Equivalent (Super/Sub) Martingale Measure Boushra Youssif Hussein Dept. of Mathematics, College of Education ,AL-Qadisiya University Abstract In this paper we proved if we have two process, Z and Z′ such that Z′ is super martingale and Z is sub-martingale, we can find a measure Q equivalent to given measure P such that there exist an other process Z* lying between them (Z ≤Z*≤ Z′) and Z* is martingale with respect to measure Q and {Ft}.Also we proved under these condition there exist a corresponding between the set of all equivalent measure and the set of all convex functional. ' ' الخالصة Q نستطيع إن نجد قياس مكافئ, مارتنجل جزئيZ مارتنجل أعظم وZ بحيثZ,Z في هذا البحث إذا كان لدينا عمليتين { وأيضا برهنا تحتFt} وQ مارتنجل بالنسبة إلى القياسZ* وZ ≤Z*≤ Z' تقع بينهماZ* بحيث توجد عملية أخرىP إلى قياس معلوم .هذه الشروط وجود تقابل بين مجموعة كل القياسات المكافئة ومجموعة كل الداليات المحدبة 1.Introduction A probability space is a triple (,F,P) where is non-empty set, F is -field and P is probability measure.[N, R.Ash ]. The function Z: R is called random variables, if Z is measurable function.[ R.Ash] A stochastic process is a collection of random variables Z={Z(t) :t I} [ F] A stochastic process Z={Z(t) : Ft : t I} is said adopted to the filtration { Ft}t I if Z(t) is Ft –measurable random variables [ R.Ash] We say that the measure Q equivalent to P iff they have the same null set i.e. Q(A)=0 P(A) =0 . A process Z′ is said to be super- martingale with respect to Q and adapted to { F t} if P{ Z′(t) | F s} ≤ Z′(s), and we say that Z is a sub-martingale if P{Z(t) | F s} ≥ Z(s), ts. A process Z is said to be martingale if P{Z(t) | F s}=Z(t) , t s. Let X be a set of all random variables on (,F,P)with norm topology . And X+={xX: p(x 0) =1 and p(x0)=0} If we have a, b are real numbers, we denote a+=( max (0,a)) a- =(- min (0,a)) We consider t T, T [0, T] be a filtration, and let Z′ (t) = ( Z1′(t),…,Z′k(t)) Z (t) = (Z1(t),…, Z k(t)) are two K-dimensional positive process, which are adapted to { Ft}tT, and E (Zk2(t) , E(Z′k2(t) tT and K=1,…..,k . We assume that Z, Z′ and the filtration {Ft}tT are right continuous . 2.Multi-Free lunch:we start this section by the following definition:- 844 Definition (2.1),[F ]: A simple trading strategy is a pair (H,H′ )of non-negative and non-decreasing kdimensional processes such that 0≤t0≤…≤tN=T for which (H(t,w), H (t,w)) is constant over [tn-1,tn) n=1,…,N. We assume: (1) (H, H ) is a dapted to { Ft}tT , and (2) E((Hk Z′k)2(t))< ∞ , E(( Hk Zk)2(t))< ∞ tT and K=1,2,…,k (3) Z1(t)= Z′1 (t)=1 , tT. Definition (2.2),[R.C ]: A simple strategy (H, H ) is self-financing if n=1,…, N we have: (H (tn)-H (tn-1)) Z (tn) ≤ (( H (tn) - H (tn-1)) Z (tn) The set of all self-financing simple strategy is denoted by . Let M={xX: (H- H )+(T). Z (T)- (H- H )-(T).Z (T) ≥ x for some (H, H )} And :MR defined by: (x)=inf{ Lim {Hα(0)Z(0)- H α (0)Z(0)} where (H, H ), (Hα- H α )+(T). Z′ (T)- (H- H )-(T).Z(T) ≥xα and xαx Definition (2.3),[J,W]: Let (,F, P) be a probability space. A multi-free lunch is a net (x, m) in XM such that: (1) xx*X+. (2) a net (H, H α ) such that (H- H α )(T)≥x and (H- H )+(T) Z′ (T)- (H- H )-(T)Z(T)X+. (3) (m)=inf {{(H(0)Z(0)- H α (0) Z′ (0)} ≤ 0. Proposition (2.4): Let (,F,P) be a probability space, then M is convex cone and is convex functional. Proof: Let x,yM (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)≥ x (J- J )+(T) Z′ (T)- (J- J )- (T)Z(T)≥y for some (H, H ) for some (J, J ) By additive [(H- H ) +(T) + (J- J )+(T)] Z′ (T)- [(H- H )- (T) + (J- J )- (T)]Z(T) ≥ x+y [max {0,(H- H )(T)}+max{0,(J- J )(T)}] Z′ (T)-[(-min{0,(H- H )}+(-min{0,(JJ )(T)})]Z(T) ≥ x+y Then [max {0,(H- H )(T)+ (J- J )}] Z′ (T)- [(-min {0, (H- H ) + (J- J ) (T))] Z (T) ≥x+y and ((H- H ) (T) + (J- J ))+ Z′ (T)-((H- H ) +(J- J ))- (T)Z(T)≥x+y also ((H+J)-( H + J ))+(T) Z′ (T)- ((H+J)-( H + J ))-(T)Z(T)≥x+y therefore (- θ )+(T) Z′ (T)- (- θ )-(T)Z(T)≥x+y 845 Journal of Babylon University/Pure and Applied Sciences/ No.(3)/ Vol.(19): 2011 where =H+J , θ = H + J Thus x+yM. Let >0, xM Then there exist (H, H ) such that (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)≥ x [(H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)]= (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) = max {0, H- H } Z′ (T) - (-min {0, H- H }Z(T) =max {0, H- H } Z′ (T) - (-min {0, H- H } Z (T) = (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T). Since (H, H ) then [(H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)]= (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T). Also (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) ≥x. therefore xM Thus M is convex cone. Now, to prove is convex functional. Let x, yM Then (H, H ) , (J, J ) such that (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)≥ x (J- J )+(T) Z′ (T)- (J- J )- (T)Z(T)≥y α (x)=inf { Lim {(H(0)Z(0)- H (0) Z′ (0))}} (y)=inf{ Lim {(J(0)Z(0)- J α (0) Z′ (0))}} then [(H- H )+(T)+ (J- J )+(T)] Z′ (T)-[ (H- H )- (T)+(J- J )- (T)]Z(T)≥x+y then ((H+J)-( H + J ))+(T) Z′ (T)- ((H+J)-( H + J ))-(T)Z(T)≥x+y then (x+y)= inf{ Lim {((H+J)(0)Z(0)-( H α + J α )(0) Z′ (0))}} = inf{ Lim {H(0)Z(0)+J(0)Z(0)- H α (0) Z′ (0)- J α (0) Z′ (0)}} = inf{ Lim {H(0)Z(0)- H α (0) Z′ (0)}}+ inf{ Lim {J(0)Z(0)- J α (0) Z′ (0)}} =(x) +(y). Let 0≤≤1, let xM. x (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) for some (H, H ) + (1-)y (1-)(J- J ) (T) Z′ (T)-(1-) (J- J ) (T)Z(T) for some (J, J ) x+(1-)y(H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)+(1-)(J- J )+(T) Z′ (T)-(1-) (J- J )(T)Z(T) [ (H- H )++(J- J )+](T) Z′ (T)-[(1-)((H- H )--(J- J )-](T)Z(T) ((H+J)-( H + J ))+(T) Z′ (T)-(1-)(( H+J)-( H + J ))-(T)Z(T). Then (x+(1-)y)= inf{ Lim { (H+J)(0)Z(0)- (1-) ( H J ) α (0) Z′ (0)}} inf{ Lim {((H +J)(0)Z(0)- (1-) ( H + J )(0) Z′ (0))}} 846 Lim inf{ {(H(0)Z(0) )( H (0)Z (0) (1 ) J (0) Z (0)}} inf{ Lim {(H(0)Z(0)- H +J(0)Z(0)- (0)Z(0))}+inf{ Lim (1- {(1-)(J(0)Z(0)- J α (0) Z (0))}} (x)+ (1- ) (y) . 3. Main Results In this section we start by the following theorem: Theorem (3.1): Let (,F,P) be a probability space , then (M,) is a non-multi free lunch if there exist a Probability measure Q equivalent to P and a process Z* satisfying Z′ Z* Z such that Z′ is super- martingale and Z is sub-martingale. Proof: Let Q be a probability measure equivalent to P and let Z* with Z′ Z* Z, such that Z is super- martingale and Z′ sub- martingale with respect to Q and { F t}. To prove (M,) is non-multi free lunch, let mM there exist (H, H ) such that: (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) m, since Z′ Z* Z and (H, H ) is self financing and non-decreasing, then EQ[(H(tn)-H(tn-1))Z*(tn)-( H (tn)- H (tn-1)) Z*(tn)| F t n 1 ] EQ[(H(tn)-H(tn-1))Z(tn)-( H (tn)- H (tn-1)) Z (tn)| F t n 1 ] 0 n=1,2,…,N. Moreover EQ[(H(tn)-H(tn-1))Z (tn)-( H (tn)- H (tn-1) Z (tn))] 0 and EQ[(H(tn)-H(tn-1))Z*(tn)| F t n 1 ))] EQ[( H (tn)- H (tn-1) )Z*(tn)| F * then EQ[(H- H )(tn) Z*(tn)| F * t n 1 )] EQ[(H- H )(tn-1) Z*(tn)| F t n 1 t n 1 )] )] (H- H )(tn-1) Z*(tn-1). By iteration EQ[(H- H )(T) Z*(T)] (H- H )(0) Z*(0) H(0) Z*(0)- H (0) Z*(0) Since (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) {(H- H )+(T) Z*(T)- (H- H )- (T)}Z*(T). We have EQ[(H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)] H(0)Z(0)- H (0) Z (0). Since (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)X+ , then EQ[H- H )+(T) Z′ (T)- (H- H )- (T)Z(T)] > 0 Thus H(0)Z(0)- H (0) Z′ (0) > 0 This implies that, there is no-multi free-lunch. The converse of above proposition is not true, see [ W] 847 Journal of Babylon University/Pure and Applied Sciences/ No.(3)/ Vol.(19): 2011 Theorem (3.2): Let (,F,P) be a probability space. Let EQ be denote the expectation operator associated to the (super/ sub) martingale measure Q. Then there exists one-to-one correspondence between the set of such expectation operators and the set of convex functional such that M=. Proof: Let be a convex functional such that M=. Let Q: F R defined by Q (B) = (IB) BF [B] By positively of , Q is positive. Suppose Z0= Z′ 0 =1 Q () = (I) =1 [B] Since is continuous, then by Riesz representation theorem, there exist random variables X such that: (x)=E(x) , for each xX. dQ Q(B)=E(IB) for each BF and = is square integrable [ D] dp It remains to show that, there exists a process Z* with Z′ Z* Z such that Z is a sub martingale and Z′ is super martingale with respect to Q and {Ft}. Suppose Z*k (t,w)=Sup{EQ(H0(T)- H 0 (T))| Ft} for K=1,…,k , for each w. Z*k (t,w)=inf{EQ(-H0(T)- H 0 (T))| Ft} for K=k+1,…,s , for each w. then Z*k (t) Zk(t) for K=1,…,k t [0,T] Assume there exist t[0,T] and BFt such that P(B) > 0 and Z (t,w) > Z*(t,w) , for each wB. then (H, H ) such that EQ( Z*k (t)IB)>EQ(-[H0(T)- H 0 (T)](IB)) Suppose ((, θ ) with 0(T) =H0 (T). Z′ (T) , (k- θ k )(T)=0 in all other cases. EQ((0- θ 0 )(T)IB) > 0 EQ((- θ )+(T) Z′ (T)-(- θ )-(T) Z(T) IB)> 0 Which contradict the fact that EQ((- θ )+(T) Z′ (T)-(- θ )-(T) Z(T)) [(- θ )+(T) Z′ (T)-(- θ )-(T) Z(T) IB] 0 Hence Z′ (t) > Z*(t) for each t. Hence Z* lies between Z and Z′. Moreover, we must have: EQ( Z*k (s) Ft) EQ( Z k (s) Ft) for each K=k+1,…,s , For each s > t Since EQ( Z k (s) Ft) Z k (t) [R.Ash] and Z k (t) < Z*k (t) So that EQ( Z*k (s) Ft) Z*k (t) K=k+1,…,s. 848 Also, EQ( Z*k (s) Ft) EQ(Z′ k(t) Ft) Z′ k(s), This implies EQ( Z*k (s) Ft) Z′ k(s) K=1,…,k. Zk (t) EQ( Z*k (s) Ft)) Z k (t), since Z′ k(t) Z*k (t) Z k (t) then EQ( Z*k (s) Ft)= Z*k (t) So Z* is martingale with respect to Q and { Ft} Conversely, Let Q be a probability measure equivalent to P and Z′ Z* Z such that Z* is a martingale with respect o Q. Define : XR be (x)=EQ(x) for each xX =EP(x) for each xX. Since is square integrable ,then is continuous. Since is positive, and Q equivalent measure to P then is positive. Since E is linear functional then is linear functional so that is convex functional. It remains to prove M=. Let mM such that Q(m)=EQ(m) so there exist (H, H ) such that (H- H )+(T) Z′ (T)- (H- H )- (T)Z(T) m EQ[(H- H )(tn-1) Z(tn-1)-(H- H )(tn-1) Z′ (tn-1) F t n 1 )]=EQ[(H- H )(tn) Z(tn)-(H- H )(tn) Z′ (tn) F t n 1 )] Since EQ[(H- H )(tn) Z(tn)-(H- H )(tn) Z′ (tn) F t n 1 ] =( H- H )(tn-1) Z(tn-1)-(H- H )(tn-1) Z′ (tn-1). Iterating this equality yields: EQ [(H- H )(T) Z(T)-(H- H )(T) Z′ (T)]= (H- H )(0) Z(0)-(H- H )(0) Z′ (0) EQ(m)= (H- H )(0) Z(0)-(H- H )(0) Z′ (0) Since (m)= (H- H )(0) Z(0)-(H- H )(0) Z′ (0) Then EQ(m)=(m) T hen Q(m)=(m) for each mM. Proposition (3.3): Let (,F,P) be a probability space and let Z′ be super-martingale, Z is submartingale with respect to Q=P, with E(2) < , then there exist a martingale Z* with respect to { Ft} and Q such that Z′ Z* Z. Proof: Suppose that Z is super martingale and Z is sub martingale with respect to { Ft} and Q. Then EQ(Z′ (tn) F t n 1 ) Z (tn-1) ,and EQ(Z(tn) F t n 1 ) Z(tn-1) Assume Dn={ t 1n ,…, t nn } where t 1n =0 and t nn =T 849 Journal of Babylon University/Pure and Applied Sciences/ No.(3)/ Vol.(19): 2011 We have Dn is increasing sequence of finite subset of [0,T]. Suppose that DnD then D[0,T] {Since [0,T] is complete closed in R} Suppose that Yn:DnR such that Z′ ( t in )Yn( t in )Z( t in ) for each i=1,…,n. For each tD, let Y(t):XR such that Yn(t)Y(t) then Z′ (t)Y(t)Z(t) To prove that Y(t) is martingale. E(Z′ (tn) F t n 1 ) E(Y(tn) F t n 1 )E(Z(tn) F t n 1 ) ……………………(1) E(Z′ (tn-1) F t n 1 ) E(Y(tn-1) F t n 1 )E(Z(tn-1) F t n 1 ) …………………(2) E(Z′ (tn)- Z′ (tn-1) F t n 1 ) E(Y(tn)- Y(tn-1) F t n 1 )E(Z(tn)- Z(tn-1) F t n 1 ) Since Z′ is super martingale then E(Z′ (tn)- Z (tn-1) F t n 1 ) 0 0 E(Y(tn)- Y(tn-1) F t n 1 ) E(Z(tn)- Z(tn-1) F t n 1 ) Since Z is sub-martingale So that E(Z(tn)- Z(tn-1) F t n 1 )0 Then 0 E(Y(tn)- Y(tn-1) F t n 1 )0 E(Y(tn)- Y(tn-1) F t n 1 )=0 Y ( t in ) is martingale as t in t Hence, Z′ (t)Y(t)Z(t). If tD then Z*(t)=Y(t) If tD , we can take Z*(t)=Lim inf {Y(s)}. st,s>t,sD * Thus, Z is martingale with respect to Q and -field { Ft} lying between Z and Z`. References Andrew, W. (2008) " Efficient Markets hypothesis and Arbitrage in Financial Market, Journal of finance and Economics,vol.19,p.p.1450-2887. Ash, R. (1972). 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