The Mathematics of Arbitrage

5.2 No Free Lunch 77

Definition 5.2.1.(compare [K 81])Ssatisfies the condition ofno free lunch
(NFL) if the closureCofCsimple, taken with respect to the weak-star topology
ofL∞(Ω,F,P),satisfies

C∩L∞+(Ω,F,P)={ 0 }.

This strengthening of the condition of no-arbitrage is tailor-made so that
the subsequent version of the fundamental theorem of asset pricing holds.

Theorem 5.2.2 (Kreps-Yan).A locally bounded stochastic processSsatis-
fies the condition of no free lunch (NFL), iff condition (EMM) of the existence
of an equivalent local martingale measure is satisfied:

(NFL) ⇐⇒ (EMM).

Proof. (EMM)⇒(NFL): This is still the easy part. By Lemma 5.1.3 we have
EQ[f]≤0, for eachQ∈Me(S)andf∈Csimple, and this inequality also
extends to the weak-star-closureCasf→EQ[f] is a weak-star continuous
functional. On the other hand, if(EMM)were true and(NFL)were violated,
there would exist aQ∈Me(S)andf∈C,f≥0notvanishingalmostsurely,
whenceEQ[f]>0, which yields a contradiction.
(NFL)⇒(EMM): We follow the strategy of the proof for the case of finite
Ω, but have to refine the argument:
Step 1 (Hahn-Banach argument):We claim that, for fixedf∈L∞+,f≡0,
there isg∈L^1 +which, viewed as a linear functional onL∞,islessthanor
equal to zero onC,andsuchthat(f, g)>0. To see this, apply the separation
theorem (e.g., [Sch 99, Theorem II, 9.2]) to theσ∗-closed convex setCand the
compact set{f}to findg∈L^1 andα<βsuch thatg|C≤αand (f, g)>β.
Since 0∈Cwe haveα≥0. AsCis a cone, we have thatgis zero or negative
onCand, in particular, non-negative onL∞+, i.e.g∈L^1 +. Noting thatβ> 0
we have proved step 1.
Step2(Exhaustionargument):Denote byGthe set of allg∈L^1 +,g≤ 0
onC.Since0∈G(or by step 1),Gis non-empty.
LetSbe the family of (equivalence classes of) subsets of Ω formed by the
supports{g> 0 }of the elementsg∈G.NotethatSis closed under countable
unions, as for a sequence (gn)∞n=1∈G, we may find strictly positive scalars
(αn)∞n=1, such that

∑∞

n=1αngn∈G. Hence there isg^0 ∈Gsuch that, for
{g 0 > 0 },wehave

P[{g 0 > 0 }]=sup{P[{g> 0 }]|g∈G}.

We now claim thatP[{g 0 > 0 }] = 1, which readily shows thatg 0 is strictly
positive almost surely. Indeed, ifP[{g 0 > 0 }]<1, then we could apply step 1
tof=χ{g 0 =0}to findg 1 ∈Gwith

E[fg 1 ]=〈f, g 1 〉=

∫

{g 0 =0}

g 1 (ω)dP(ω)> 0