The Mathematics of Arbitrage

98 6 The Dalang-Morton-Willinger Theorem

oddnandg′′n:=gn′ for evenn, we may also conclude that (gn′)∞n=1converges
in probability tog 0.

To sketch the idea of Rogers’ proof let us first assume thatF 0 is trivial
so thatL^0 (Ω,F 0 ,P,Rd) may be identified withRd. Then the problem (6.5)
reduces to find the optimal vectorĤ∈Rd.
So, let (Hn)∞n=1be a sequence inRdsuch that

lim

n→∞

E[U((Hn,∆S))] = sup

H∈Rd

E[U((H,∆S))].

From the previous lemma we know that the sequencefn:= (Hn,∆S)
converges in measure to a functiong 0 ∈L^0 (Ω,F 1 ,P,]−∞,∞]); we also know
that every sequencegn∈conv{fn,fn+1,...}converges tog 0 in measure.
We want to conclude that (Hn)∞n=1converges; for this we still need two
ingredients: first we have to suppose that eachHnis in the predictable range
ofS(see 6.4.1 above); in the present case of a trivialσ-algebraF 0 this just
means thatHnlies in the smallest subspaceRofRdsuch that ∆Stakes a.s.
its values inR. We may always pass from an arbitraryHn∈RdtoP(Hn)
wherePdenotes the orthogonal projection ontoR,as(Hn,∆S)=(PHn,∆S)
almost surely.
The second ingredient is the assumption of no-arbitrage. Using this as-
sumption we claim that (Hn)∞n=1is bounded inRdand thereforeg 0 cannot
assume the value +∞.
First we have to isolate the trivial case when ∆S= 0 a.s. so thatR={ 0 }
andP= 0. In this case we may define the martingale measureQviaddQP=1,
which — for trivial reasons — coincides with (6.7).
Hence we may suppose thatRis a subspace ofRdof dimension dim(R)≥ 1
and define

γ=inf{E[(H,∆S)−∧1]|H∈R,‖H‖Rd=1}. (6.9)

AsH→E[(H,∆S)−∧1] is continuous on the unit sphere ofR(by Lebesgue’s
theorem) and strictly positive (by(NA)and the construction ofR) we deduce
from the compactness of the unit sphere ofRdthatγis a strictly positive
number.
Next we show that (Hn)∞n=1is bounded. Indeed, otherwise there is an
increasing sequence (nk)∞k=1such that‖Hnk‖Rd≥kso that

E[U((Hnk,∆S))]≤−‖Hnk‖RdE

[(

Hnk

‖Hnk‖Rd

,∆S

)

−

]

≤−kγ,

in contradiction to the assumption that (Hn)∞n=1is a maximising sequence.
Hence we obtain thatg 0 is a.s. finite.
Finally we show that (Hn)∞n=1converges inRd. Indeed, otherwise there is
α>0 and sequences (nk)∞k=1,(mk)∞k=1tending to infinity such that