The Mathematics of Arbitrage

5.2 No Free Lunch 81

reasoning. However, both arguments amount to the same, and the difference
is only superficial.
Fix 1≤p≤∞,and1≤q≤∞such that^1 p+^1 q =1,andletE =
Lp(Ω,F,P),E′=Lq(Ω,F,P). We denote byE+={f∈Lp|f≥0a.s.}
(resp.E− ={f∈Lp|f ≤0a.s.}) the cone of non-negative (resp. non-
positive) random variables inLp.

Theorem 5.2.3.With the above notation letC ⊂Ebe a σ(E, E′)-closed
convex cone containingE−and suppose thatC∩E+={ 0 }. Then there is
aprobabilitymeasureQonF, which is equivalent toP, satisfying ddQP∈E′,
and so that, for allf∈C, we haveEQ[f]≤ 0.
Conversely, given a probability measureQonF, equivalent toPand sat-
isfyingddQP∈E′,theconeC={f∈E|EQ[f]≤ 0 }isσ(E, E′)-closed and
satisfiesC∩E+={ 0 }.

Proof.For 1≥δ>0, letBδbe the convex set inE

Bδ={f∈E| 0 ≤f≤ 1 ,E[f]≥δ}.

ClearlyBδisσ(E, E′)-compact andC∩Bδ=∅. The Hahn-Banach The-
orem in its version as separating hyperplane (see, e.g., [Sch 99]) gives the
existence of an elementgδ∈E′so that

sup

f∈C

E[gδf]<min

h∈Bδ

E[gδh]. (5.2)

BecauseCis a cone this means that for allf∈Cwe must haveE[gδf]≤0.
This implies that the sup on the left hand side of (5.2) equals 0, so that the
min on the right hand side must be strictly positive. SinceCcontainsE−
we must havegδ≥0 almost surely. Nowgδcannot be identically zero since
I1∈Bδand henceE[gδ1]I>0. We may normalisegδto haveE[gδ]=1.
For eachn≥1weconsiderδn =2−nand letQnbe the probability
measure defined asddQPn=g 2 −n. For later use let us denote byαnthe number
αn=‖g 2 −n‖E′∈[1,∞[. Also remark that forA∈FwithP[A]> 2 −n,we

must haveQn[A]=E[ (^1) Ag 2 −n]>0.
The elementQis now defined as

Q=

∑∞

n=1

βnQn,

wherewechoosetheweightsβn>0 such that

∑∞

n=1βn=1and

∑∞

n=1βnαn<
∞. For instance we could takeβn=^2

−n

cαn wherec=

∑∞

n=1

2 −n

αn <∞.

The probability measureQsatisfies

(i) for allA,P[A]>0wehaveQ[A]=

∑∞

n=1βnQn[A]>0.
(ii) ddQP∈E′since

∥

∥

∥ddQP

∥

∥

∥

E′

≤

∑∞

n=1βnαn<∞.