The Mathematics of Arbitrage

268 13 The Banach Space of Workable Contingent Claims in Arbitrage Theory

Proof.For eacht≤∞we clearly have thatft=Xt− 1 ∈G. Suppose now
that the contingent claimf∞is in the closure of the spaceL∞for the norm

‖g‖=^12 sup{‖g‖L (^1) (Q)|Q∈Me}.Forε=−E[f 4 ∞]>0 we can findgbounded
such that for allQ∈Mewe have‖g−f∞‖≤ε.ForthemeasurePwe find
EP[|f∞−g|]≤εand hence for eacht≤∞we have, by taking conditional
expectations,
EP

[∣

∣ft−EP[g|Ft]

∣

∣

]

≤ε.

In particular, sinceE[ft]=0foreacht<∞,wehaveEP[g]=E

[

EP[g|
Ft]

]

≥−ε. This in turn implies thatEP[f∞]≥− 2 εa contradiction to the
choice ofε.

Theorem 13.4.6.In the setting of the above example, the Banach spaceG
contains a subspace isometric to∞. In other words there is an isometry
u:∞→G. Moreoverucan be chosen such thatu(∞)⊂G∞.

Proof.We start with a partition of Ω into a sequence of pairwise disjoint sets,
defined by the processW. More precisely we putA 1 ={W 1 ∈]−∞,1]}and
forn≥2 we putAn={W 1 ∈]n− 1 ,n]}.LetMbe the stochastic exponential
M=E(B−B^1 ) and let the stopping timeTbe defined as

T=inf{t|Mt≥ 2 }.

The sequence that we will use to construct the subspace isometric to∞is
defined as

fn=2(MT−1) (^1) An.
For each nand eachε>0thereisarealnumberα(n, ε) depending only
onεandnsuch that the random variableφ(n, ε)=α(n, ε) (^1) Ak+ε 1 ⋃m=kAm
is strictly positive and defines a density for a measuredQn,ε=φ(n, ε)dP
which is necessarily inMe, since the random variableφ(n, ε) can be written
as a stochastic integral with respect toW. It is clear thatQn,ε[An]≥ 1 −ε.
This shows that for eachn,sup{Q[An]|Q∈Me}=1.
Clearly eachfnis a 2-admissible maximal contingent claim. Since for each
measureQ∈Mewe haveQ[fn=2|F 1 ]=Q[fn=− 2 |F 1 ]=^121 Anwe
obtain that‖fn‖L (^1) (Q)=2Q(An), hence for theG-norm we find‖fn‖=1.
We now show that for eachx∈∞we can define a contingent claim
u(x)=

∑

k≥ 1

xkfk∈G.

Ifx=(xk)k≥ 1 is an element of∞and ifmis a natural number, we denote
byxmthe element defined asxmk =xkifk≤mandxmk =0otherwise.Letus
already putu(xm)=

∑m

k=1x

m

kfk.Nowifxis a positive element in

∞then

the sequence (u(xm))m≥ 1 is a sequence converging inL^1 (Q) to a contingent
claimu(x)=

∑∞

k=1xkfkand this for eachQ∈M

e. By Corollary 13.2.19 and