The Mathematics of Arbitrage

216 10 Counter-Example

(4) two processes β^1 and β^2 that are Brownian motions with respect to
( ̃Ft)t≥ 0 and such that〈β^1 ,β^2 〉=0,
(5) the variableγ=

∫∞

0 K

2
sd〈N, N〉s◦πis a stopping time with respect to
( ̃Ft)t≥ 0 ,
(6)βt^1 ∧γ=(K·N)Tt◦π,
(7)βt^2 ∧γ=(K·U)Tt◦π,

(8)L ̃=E(β^1 )satisfiesLTt◦π=L ̃t∧γ,

(9)Z ̃=E(β^2 )satisfiesZTt◦π=Z ̃t∧γ,
(10) ̃τ=inf{t|(E(β^1 ))t<^12 }satisfiesτ◦π=T ̃τ,
(11) ̃σ=inf{t|(E(β^2 ))t> 2 }satisfiesσ◦π=T ̃σ.

In this setting we have to show thatER[Lτ∧σ]=E ̃R ̃[ ̃Lτ ̃∧ ̃σ∧γ]<1. But on
the set

{∫∞

0 K

2

sd〈N, N〉s<∞

}

we have, as shown above,

ER

[

(^1) {∫∞
0 K^2 sd〈N,N〉s<∞}
Lτ∧σ

]

≤ER

[

(^1) {∫∞
0 K^2 sd〈N,N〉s<∞}

L∗

]

≤η.

In other words,E ̃R ̃[ (^1) {γ<∞}L ̃∗γ]≤η. So it remains to be shown that

E ̃ ̃

R

[

(^1) {γ=∞}L ̃ ̃τ∧ ̃σ

]

< 1 −η.

Actually, we will show that

E ̃ ̃

R

[

L ̃τ ̃∧ ̃σ

]

< 1 −η.

This is easy and follows from the independence ofβ^1 andβ^2 , a consequence
of〈β^1 ,β^2 〉= 0! As in Sect. 10.2 we have

E ̃ ̃

R

[

L ̃ ̃τ∧ ̃σ

]

=ηR ̃[ ̃σ=∞]+R ̃[σ<∞]=

1

2

η+

1

2

< 1 −η

sinceη≤^18. To show that

ER[Lτ∧σZτ∧σ]=1

we again use the extension and time transform. But ( ̃LZ ̃) ̃τ∧ ̃σis a uniformly
integrable martingale, as follows from the easy calculation in Sect. 10.2, and
hence we obtain

ER[Lτ∧σZτ∧σ]=E ̃R ̃[L ̃τ ̃∧ ̃σ∧γZ ̃τ∧ ̃σ∧γ]=1.

The proof of the theorem is complete now.