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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.
'
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‫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),
ts.
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+={xX: p(x  0) =1 and p(x0)=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}tT, and
E (Zk2(t)  , E(Z′k2(t)  tT and K=1,…..,k .
We assume that Z, Z′ and the filtration {Ft}tT are right continuous .
2.Multi-Free lunch:we start this section by the following definition:-
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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}tT , and
(2) E((Hk Z′k)2(t))< ∞ , E(( Hk Zk)2(t))< ∞  tT and K=1,2,…,k
(3) Z1(t)= Z′1 (t)=1 ,  tT.
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={xX: (H- H  )+(T). Z  (T)- (H- H  )-(T).Z (T) ≥ x for some (H, H  )}
And :MR 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 xαx
Definition (2.3),[J,W]:
Let (,F, P) be a probability space. A multi-free lunch is a net (x, m) in XM
such that:
(1) xx*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,yM
 (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
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Journal of Babylon University/Pure and Applied Sciences/ No.(3)/ Vol.(19): 2011
where =H+J , θ = H  + J 
Thus x+yM.
Let >0, xM
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 xM
Thus M is convex cone.
Now, to prove  is convex functional.
Let x, yM
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 xM.
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 mM
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]
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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)  BF
[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 xX.
dQ
Q(B)=E(IB) for each BF 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 BFt such that P(B) > 0 and Z  (t,w) > Z*(t,w) , for
each wB.
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.
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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.
Zk (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  : XR be (x)=EQ(x) for each xX
=EP(x) for each xX.
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 mM 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 mM.
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
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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 DnD
then D[0,T] {Since [0,T] is complete closed in R}
Suppose that Yn:DnR such that Z′ ( t in )Yn( t in )Z( t in ) for each i=1,…,n.
For each tD, let Y(t):XR 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 tD then Z*(t)=Y(t)
If tD , we can take Z*(t)=Lim inf {Y(s)}.
st,s>t,sD
*
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). " Real analysis and Probability", Academic Press, New York .
Dalang, R.C. ; Morton A. and Willinger, W. (1992). ”Equivalent martingale measure and
no-arbitrage in stochastic securities markets models”, stoch.Rep(29).
Delbaen, F. ; Schacharmayer, W. (1994). ”A General version of the fundamental theorem
of Asset pricing", Math. Annal, Vol.300,p.p. 463-520.
Gaussel, N. (2002). ”The martingale property of priced when arbitrage opportunities exist",
university of Paris, Ph.D.
Harrison, J.M. ; Kreps, D.M. (1979),”Martingales and arbitrage in multiperiod securities
markets", economic theory,Vol.901,pp.381-408.
Kannan, D. (1979). ”An introduction to stochastic Processes”, New york-Oxford.
Youssif Hussein., B. (2006). " On equivalent martingale measures", Thesis Ph.D, ALMustasiriya university.
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