6.5 The DMW Theorem forT=1 95Theorem 6.5.1.LetS=(S 0 ,S 1 )be an(F 0 ,F 1 )-adaptedRd-valued process
satisfying the (NA) condition. Then we have the existence of an equivalent
probability measureQso that
(i) S 0 ,S 1 ∈L^1 (Q),
(ii) S 0 =EQ[S 1 |F 0 ],
(iii)ddQPis bounded.
Proof.First we take an equivalent probability measureP 1 so that ddPP^1 is
bounded andS 0 ,S 1 ∈L^1 (P 1 ). To do so we could take, for example,ddPP^1 =
cexp(−‖S 0 ‖Rd−‖S 1 ‖Rd), wherecis a suitable normalisation constant.
The next step consists of considering the set
C 1 =C∩L^1 (Ω,F 1 ,P 1 )whereCis defined as in Theorem 6.4.2. BecauseCis closed inL^0 (P), the set
C 1 is closed inL^1 (P 1 ). ObviouslyC 1 is a convex cone (sinceCis a convex
cone). The(NA)condition implies thatC 1 ∩L^1 +(Ω,F 1 ,P 1 )={ 0 }.Theorem
5.2.3 now gives the existence of an equivalent probability measureQso that
dQ
dP 1 is bounded and so thatEQ[f]≤0 for allf∈C^1. ObviouslyS^0 ,S^1 ∈
L^1 (Q)sinceddPQ 1 is bounded. Since for each coordinatej=1,...,dand each
A∈F 0 we have (^1) A(S 1 j−Sj 0 )∈C 1 and− (^1) A(Sj 1 −Sj 0 )∈C 1 ,wemusthave
EQ[ (^1) ASj 1 ]=EQ[ (^1) ASj 0 ]. This showsS 0 =EQ[S 1 |F 0 ]. SinceddQP=ddPQ 1 ddPP^1
we also have thatddQPis bounded.
Remark 6.5.2.One may ask whether it is possible to replace in Theorem 6.5.1
assertion (iii) by the assertion thatddPQis bounded, i.e., by
(iii’) there is a constantc>0 such thatddQP>calmost surely.
A moment’s reflexion reveals that, in general, this is not possible. Indeed,
if it happens that‖St‖ 1 =E[|St|]=∞, then for each probability measure
QwithddQP≥c>0wealsohaveEQ[|St|]=∞,sothatScannot be aQ-
martingale. But even ifSis uniformly bounded, we cannot replace (iii) by
(iii’) as the subsequent example shows.
Let Ω =N,theσ-algebraF 0 be generated by the partition of Ω into the
sets ({ 2 n− 1 , 2 n})∞n=1andF 1 be the power set of Ω. Define the probability
measurePonF 1 byP[2n−1] =P[2n]=2−(n+1).LetS 0 ≡0,S 1 =1
on{ 2 n− 1 }andS 1 =− 2 −non{ 2 n}to obtain anR-valued adapted process
S=(St)^1 t=0satisfying the(NA)condition. IfQis a probability measure onF 1
withddQP≥c>0, we haveQ[2n−1]≥c 2 −(n+1). If in additionEQ[S 1 |F 0 ]=0,
we must haveQ[2n]=2nQ[2n−1]≥ c 2 , which is a contradiction to the
finiteness ofQ. Hence, there cannot exist a measureQsatisfying (i) and (ii)
of Theorem 6.5.1 and (iii’) above. We refer to [R 04] and [RS 05] for a more
thorough analysis of condition (iii’).