The Mathematics of Arbitrage

2.5 Change of Num ́eraire 29

V. It suffices to show that, for a one-dimensional predictable processL,the
quantitiesLt∆Vtare inK(S). This is easy, since

Lt∆Vt=Lt

(

Ht^0 ,∆St

)

=

(

LtH^0 t,∆St

)

∈K(S)

by definition ofK(S). This shows thatK(Sext)=K(S).

Lemma 2.5.2.Fix 0 ≤t≤T,andletf∈K(S)=K(Sext)beFt-measurable.

Then the random variableVftis of the form f

′

VT wheref

′∈K(S).

Proof.Clearly

f

Vt

−

f

VT

=

1

VT

(

f

VT−Vt

Vt

)

=

1

VT

∑T

s=t+1

f

Vt

(Vs−Vs− 1 ).

We see thatf′′=

∑T

s=t+1

f

Vt(Vs−Vs−^1 ) belongs toK(S

ext) because f
Vt is
Ft-measurable and the summation is ons>t. Hencef′=f′′+fdoes the
job.

Proposition 2.5.3.Assume thatXis defined as in (2.10). Then

K(X)=

{

f

VT

∣

∣

∣

∣f∈K(S)

}

.

Proof.We have thatg∈K(X) if and only if there is a (d+1)-dimensional pre-
dictable processH,withg=

∑T

t=1(Ht,∆Xt)=

∑T

t=1

∑d+1

j=1H

j

t∆X

j
t. Clearly,
forj=1,...,dandt=1,...,T,

∆Xjt=

(

Stj

Vt

−

Stj− 1

Vt− 1

)

=

∆Stj

Vt

+Stj− 1

(

1

Vt

−

1

Vt− 1

)

=

∆Stj

Vt

−

Stj− 1

Vt− 1

∆Vt

Vt

=

1

Vt

(

∆Stj−Xtj− 1 ∆Vt

)

.

So we get thatHtj∆Xtj = V^1 t

(

Htj∆Stj−

(

HtjXtj− 1

)

∆Vt

)

,whichisofthe

formVftfor somef∈K(Sext)=K(S). Forj=d+1 andt=1,...,Tthe

same argument applies by replacingStjandStj− 1 by 1.

By the previous lemma we haveVft=f

′

VT for somef

′∈K(S). This shows

thatK(X)⊂V^1 TK(S).