The Mathematics of Arbitrage

106 6 The Dalang-Morton-Willinger Theorem

proved as follows. Let (H 1 ,∆S 1 )+f 2 = 0 whereH 1 is in canonical form and
f 2 ∈K 2 .OnthesetA={(H 1 ,∆S 1 )< 0 }we have thatf 2 >0 and since

(^1) Af 2 ∈K 2 we have thatP[A]>0 gives an arbitrage opportunity. On the set
{(H 1 ,∆S 1 )> 0 }we replacef 2 by−f 2 and get the same result. This means
that (H 1 ,∆S 1 ) = 0 and sinceH 1 is in canonical form we getH 1 =0.
(iii): This assertion is immediate from the closedness ofK (Proposition
6.8.1) and the measurable sub-sequence principle (Proposition 6.3.3). The
reason why we stated it explicitly is to avoid confusion: we only claim that
the random subsequence (σn)∞n=1is measurably parameterised w.r. toFT;we
do not claim thatHσn=(H 1 σn,...,HTσn)ispredictable.
(ii): To prove (ii) we again use induction onT.ForT =1wereferto
Proposition 6.4.1. So let us suppose that assertions (ii) and (iii) hold true
forT−1 and fix the horizonT. We now show that the assumption of (ii)
implies that (H 1 n)∞n=1is a.s. bounded. We proceed in the same way as in the
proof of 6.8.1. LetA={lim sup|Hn 1 |=+∞}and letτnbe anF 0 -measurably
parameterised subsequence selected in such a way that|H 1 τn|→∞onA
and
H 1 τn
|H 1 τn| →ψ^1 onA.OnthesetAwe must have|ψ^1 |=1andwemay
putψ 1 = 0 on the complement ofA. We remark that this is possible by
the selection principle and thatψ 1 ∈H 1 i.e.ψ 1 is in canonical form. Now
(Hτn·S)T=(Hτ 1 n,∆S 1 )+f 2 τn,wheref 2 τn∈K 2 .OnthesetAwe get
(
Hτn
|H 1 τn|

·S

)

T

=

(

H 1 τn

|H 1 τn|

,∆S 1

)

+

1

|H 1 τn|

f 2 τn.

Since the first term on the right hand side is bounded by|∆S 1 |we get

lim sup

n→∞

(

1

|H 1 τn|

f 2 τn

)−

≤|∆S 1 |+ lim sup

n→∞

1

|Hτ 1 n|

(Hτn·S)−T≤|∆S 1 |.

By the induction hypothesis this means that the sequence

H ̃n:=

(

0 , (^1) A
H 2 τn
|H 1 τn|
,..., (^1) A
HτTn
|Hτ 1 n|

)

=H ̃n

is a.s. bounded. It follows that the functions (H ̃n·S)T=f 2 τn (^1) A|H^1 τn
1 |
are also
a.s. bounded. By (iii) there is aFT-measurably parameterised subsequence
σnofτnso that

(

H ̃σn·S

)

T

→fandf∈K 2. Since (σn)nis a subsequence

ofτnwe still have H

σ 1 n

|Hσ 1 n|→ψ^1 and hence

(ψ 1 ,∆S 1 )+f= lim

n→∞

(

H 1 τn

|H 1 τn|

,∆S 1

)

+(H ̃σn·S)T

= lim

n→∞

(^1) A((H 1 σn,∆S 1 )+f 2 σn)

1

|Hτ 1 n|

= lim

n→∞

(^1) A(Hσn·S)T

1

|H 1 τn|

.