The Mathematics of Arbitrage

9.4 Proof of the Main Theorem 167

Remark 9.4.5.Let us motivate why we introduced the maximal elementf 0
in the above lemma. As already observed the sequencegnintroduced before
Lemma 9.4.4 is not of immediate use. Our goal is, of course, to find a 1-
admissible integrandH 0 which is, in some sense, a limit of the sequenceHn
of the 1-admissible integrands used to construct the sequencegn.Butthe
convergence of (gn)n≥ 1 (which we may assume by Lemma 9.8.1) does not
imply the convergence of the sequence (Hn)nin any reasonable sense. We
illustrate this with the following example in discrete time. Let (rm)m≥ 1 be
a sequence of Rademacher functions i.e. a sequence of independent identically
distributed variables withP[rm=+1]=P[rm=−1] =^12 .LetSm=

∑m
k=1rk
andS 0 =0.Foreachn, an odd natural number, we take for the strategyHn
the so called doubling strategy. This strategy is defined as

Htn=

{

2 t−^1 if r 1 =...=rt− 1 =1

0elsewhere.

Clearly (Hn·S)t=H 1 nr 1 +···+Htn− 1 rthence we obtain

(Hn·S)t=

{

2 t−1 with probability 2−t

− 1 with probability 1− 2 −t

For o ddnthe final outcomegnsatisfiesgn= limt→∞(Hn·S)t=−1almost
surely.
For eachn, an even natural number, we introduce a “doubling strategy”
Hnstarting at timen. More precisely

Htn=

⎧

⎨

⎩

0fort≤n

2 m−^1 if t=n+mandrn+1=...=rn+m− 1 =1

0elsewhere.

Clearly fort≤n:(Hn·S)t= 0 and fort>n:(Hn·S)t=Hnn(rn+1)+
···+Htn− 1 (rt) hence fort>n:

(Hn·S)t=

{

2 (t−n)−1 with probability 2−(t−n)

− 1 with probability 1− 2 −(t−n).

Again, for each even number n, the final outcomegn satisfies gn =
limt→∞(Hn·S)t =−1 almost surely. Hence all the variablesgn,forodd
as well as for evenn,areequalto−1 almost surely and hence trivially
g= limgn=−1 a.s.. On the other hand the sequenceHn,alongtheeven
numbers, tends to zero onR+×Ω. Along the odd numbers the sequenceHnis
constant and equal to the same doubling strategy. The sequenceHnis there-
fore not converging. Note, however, that the limit functiongis not maximal
in the sense of Lemma 9.4.4. If we take limits along the even numbers then
the pointwise limitHofHnis zero and hence (H·S)∞= 0. The example
suggests that the outcome 0, which is larger thang, can be obtained by look-
ing at limits of the strategiesHn. So the remedy is to replace the functiong