26 2 Models of Financial Markets on Finite Probability Spaces
Remark 2.4.5.Before we prove the theorem let us remark that the “sup” and
the “min” are taken in the spaceL^0 (Ω,F 0 ,P)ofF 0 -measurable functions.
Both sets are lattice ordered. Indeed, ifEQ 1 [f|F 0 ]andEQ 2 [f |F 0 ]are
given, whereQ 1 ,Q 2 ∈Me(S,F 0 ), then there is an elementQ 3 ∈Me(S,F 0 )
so thatEQ 3 [f|F 0 ]=max{EQ 1 [f|F 0 ],EQ 2 [f|F 0 ]}. The construction is
rather straightforward. LetA={EQ 1 [f|F 0 ]>EQ 2 [f|F 0 ]}∈F 0 and let
Q 3 [B]=Q 1 [A∩B]+Q 2 [Ac∩B]. BecauseQ 1 |F 0 =Q 2 |F 0 =Pwe get that
Q 3 is a probability and thatQ 3 ∈Me(S,F 0 ). AlsoEQ 3 [f|F 0 ]=EQ 1 [f|
F 0 ]∨EQ 2 [f|F 0 ].
Similarly, the set on the right is stable for the “min” operation. Indeed,
leth 1 +g 1 ≥fandh 2 +g 2 ≥f.ForA={h 1 <h 2 },anF 0 -measurable
set, we defineh=h 11 A+h 21 Ac andg 11 A+g 21 Ac=g. The functionhis
F 0 -measurable andg∈K(becauseA∈F 0 ). Clearlyh+g≥f.
Proof of Theorem 2.4.4.Iff≤h+g,wherehisF 0 -measurable andg∈K,
then forQ∈Me(S,F 0 )wehaveEQ[f|F 0 ]≤h+0=h. This shows that
a 1 := sup{EQ[f|F 0 ]|Q∈Me(S,F 0 )}
≤inf{h|hF 0 -measurable,h+g≥f,for someg∈K}
=:a 2.To prove the converse inequality, we show that there isg∈Kwitha 1 +g≥
f. If this were not be true then (a 1 +K)∩(f+L∞+)=∅and we could find,
using the separating hyperplane theorem, a linear functionalφandε>0, so
that∀g∈K,∀l≥0wehaveε+φ(a 1 +g)<φ(f+l). This implies that
φ≥0andφ(g) = 0 for allg∈K. Of course we can normaliseφso that
it comes from a probability measureQ.SowegetEQ[a 1 ]+ε′<EQ[f]and
Q∈Ma(S), whereε′>0.
By the density ofMe(S)inMa(S)wemayperturbQa little bit to make
it an element ofMe(S). We still getEQ[a 1 ]+ε<EQ[f], but this time for
ameasureQ∈Me(S). Let nowZt= ddQP
∣
∣
∣Ftand setLt=ZZ 0 t. The process(Lt)∞t=0defines a measureQ^0 ∈Me(S,F 0 )viadQ
0
dP =LT.Furthermore
EQ 0 [f|F 0 ]=EP[fLT|F 0 ]=EP[fZT|F 0 ]
Z 0
=EQ[f|F 0 ]ThereforeEQ[f |F 0 ]≤a 1 and henceEQ[f]≤EQ[a 1 ], contradicting the
choice ofQ.
Corollary 2.4.6.Under the assumptions of Theorem 2.4.4 we have
{EQ[f|F 0 ]|Q∈Me(S)}={EQ[f|F 0 ]|Q∈Me(S,F 0 )}.Hence, forf∈L∞+(Ω,F,P), we havesupQ∈Me(S)EQ[f]=‖a 1 ‖∞where
a 1 =sup{EQ[f|F 0 ]|Q∈Me(S,F 0 )}.