286 14 The FTAP for Unbounded Stochastic Processes
Proof.ClearlyBis convex. Let us also remark that it is non-empty. To see
this let us takeδ>0 and let us defineddQQ 1 =EQexp(−δ‖S^1 ‖)
1 [exp(−δ‖S^1 ‖)]
. ClearlyS 1 is
Q-integrable and from Lebesgue’s dominated convergence theorem we deduce
that forδsmall enough‖Q−Q 1 ‖<ε.
If 0 were not in the relative interior ofBwe could find by the Minkowski
separation theorem, an elementx∈Rd, such thatBis contained in the
halfspaceHx={y∈Rd|(x, y)≥ 0 }and such that (x, y)>0forsomey∈B.
In order to obtain a contradiction we distinguish two cases:
Case 1:xfails to be admissible, i.e., (x, S 1 ) fails to be (essentially) uniformly
bounded from below.
First find, as above, a probability measureQ 2 ∼P,‖Q 1 −Q 2 ‖<ε 2 and
such thatEQ 2 [S 1 ] is well-defined.
By assumption the random variable (x, S 1 ) is not (essentially) uniformly
bounded from below, i.e., forM∈R+,theset
ΩM={ω|(x, S 1 (ω))<−M}has strictly positiveQ 2 -measure. ForM∈R+define the measureQMby
dQM
dQ 2=
{
1 −ε 4 on Ω\ΩM
1 −ε 4 + 4 Q 2 ε(ΩM) on ΩM.It is straightforward to verify thatQMis a probability measure,QM∼P,‖QM−Q 2 ‖<ε 2 ,dQ
M
dQ 2 ∈L∞and that(x,EQM[S 1 ]) =EQM[
(x, S^1 )]
≤
(
1 −
ε
4)
EQ 2 [(x, S 1 )]−
εM
4.
ForM>0 big enough the right hand side becomes negative which gives the
desired contradiction.
Case 2:xis admissible, i.e., (x, S 1 ) is (essentially) uniformly bounded from
below.
In this case we know from the Beppo-Levi theorem that the random vari-
able (x, S 1 )isQ 1 -integrable and thatEQ 1 [(x, S 1 )]≤0; (note that, for each
M∈R+,wehavethat(x, S 1 )∧Mis inCand thereforeEQ 1 [(x, S 1 )∧M]≤
0).
Also note that (x, S 1 ) cannot be equal to 0 a.s., because as we saw above
there is ay∈Bsuch that (x, y)>0 and hence (x, S 1 )cannotequalzeroa.s.
either. The no-arbitrage property then tells us thatQ 1 [(x, S 1 )>0] as well as
Q 1 [(x, S)<0] are both strictly positive.
We next observe that for allη>0thevariableexp(η(x, S 1 )−) is bounded.
The measureQ 2 ,givenbyddQQ^21 =
exp(η(x,S 1 )−)
EQ 1 [exp(η(x,S 1 )−)]is therefore well-defined.
Forηsmall enough we also have that‖Q 2 −Q 1 ‖<ε.ButQ 2 also satisfies:
EQ 2 [(x, S 1 )]< 0.