The Mathematics of Arbitrage

(Tina Meador) #1

94 6 The Dalang-Morton-Willinger Theorem


(ii)IfSsatisfies (NA) then the cone


C=K−L^0 +(Ω,F 1 ,P)

is closed inL^0 (Ω,F 1 ,P)too.

Proof.(i): Let (fn)=(Hn,∆S)∞n=1be a sequence inKconverging tof 0 ∈
L^0 (Ω,F 1 ,P) with respect to convergence in measure. We may suppose that
eachHnis in canonical form. Moreover, by passing to a subsequence we
may suppose that (fn)∞n=1converges a.s. tof 0. Proposition 6.4.1 implies that
(Hn)∞n=1converges a.s. to someH^0 ∈L^0 (Ω,F 1 ,P;Rd)sothatf 0 =(H^0 ,∆S),
whencef 0 ∈K.
(ii): Letfn=gn−hnbe a sequence inCconverging in probability tof 0 ∈
L^0 (Ω,F 1 ,P), wheregn=(Hn,∆S), whereHnis an integrand in canonical
form andhn∈L^0 +(Ω,F 1 ,P). We have to show thatf 0 belongs toC. Again
we may suppose that (fn)∞n=1converges a.s. tof 0 .Asgn≥fnwe deduce
that ((Hn,∆S)−)∞n=1is a.s. bounded, so that we may conclude from(NA)
and Proposition 6.4.1 (ii) that (Hn)∞n=1is a.s. bounded too. By passing to a
measurably parameterised subsequence (τk)∞k=1we may suppose thatgτk=
(Hτk,∆S) converges a.s. to (H^0 ,∆S), for someH^0 ∈E.Notethat(fτk)∞k=1
still converges a.s. tof 0 so thathτk=fτk−gτkconverges a.s. to someh 0 ≥0.
Hencef 0 =(H^0 ,∆S)−h 0 ∈K−L^0 +(Ω,F 1 ,P)=C. 


In assertion (ii) of the preceding theorem, the no-arbitrage assumption
cannot be dropped. Indeed, consider the following simple example ([S 92]).
Let Ω = [0,1],F 0 trivial,F 1 the Borelσ-algebra andPLebesgue measure.
LetS 0 ≡0andS 1 (ω)=ω,forω∈[0,1], and


fn(ω)=

{


nω for 0≤ω≤n−^1
1forn−^1 ≤ω≤ 1.

Asfn≤gn:=n∆S,wehavefn∈C,foreachn. Clearly (fn)∞n=1converges
a.s. to the constant functionf 0 =1.Butf 0 is not inC,asforeachf∈Cwe
may find a constantM>0 s.t. almost surelyf(ω)≤Mωso thatf(ω)≤^12
a.s. for 0≤ω≤ 21 M.
Summing up, we have an example of a processS=(St)^1 t=0such thatC
is not closed inL^0 (Ω,F 1 ,P). The crux is, of course, thatSdoes not satisfy
(NA).


6.5 Proof of the Dalang-Morton-Willinger Theorem forT=1


forT=1


As we have proved in Theorem 6.4.2, the coneC∈L^0 (Ω,F 1 ,P)isclosedif
(S 0 ,S 1 ) satisfies the(NA)condition. The Kreps-Yan Theorem can now be
applied in such a way that it yields the existence of an equivalent martingale
measure.

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