226 11 Change of Num ́eraire
Theorem 11.4.2.LetW be a semi-martingale, taking values inRd.LetV
be a strictly positive semi-martingale such thatV∞= limt→∞Vtexists and
is strictly positive a.s.. The semi-martingaleX is the(d+2)-dimensional
processX=(W, 1 ,V). The processZdefined asZ=(WV,V^1 ,1)is a(d+2)-
dimensional semi-martingale. It satisfies the (NA) property with respect to
general admissible integrands if and only ifV∞− 1 is maximal in the set of
outcomes of 1 -admissible integrands forX.
Proof.Using the symmetry between the processesXandZwe first reformulate
the statement of the theorem. We can regard the processXas obtained from
Zby dividing it by the processV^1. The processV^1 is also strictly positive and
at infinity its limit exists a.s. and is still strictly positive. If we change the role
ofXandZ,resp.VandV^1 , we see that the proof of the theorem is equivalent
to the proof of the following two statements
(1) IfXsatisfies the(NA)property with respect to general admissible in-
tegrands thenV^1 ∞−1 is maximal in the set of outcomes of 1-admissible
integrands forZ.
(2) IfZpermits arbitrage with respect to general admissible integrands then
V∞−1 is not maximal in the set of outcomes of 1-admissible integrands
forX.
The proof depends on the following calculation from vector stochastic calculus.
FromX=VZwe deduce that
dXt=dVtZt−+Vt−dZt+d[V, Z]t.IfKis a (d+ 2)-dimensional predictable process that is a 1-admissible inte-
grand for the systemZ=(WV,V^1 ,1) then we letY=(1+K·Z)V.Remark
thatYis a process that describes a portfolio obtained by using an investment
described by the systemZthat afterwards is converted, through the change
of num ́eraireV, into values that fit in the systemX.Wehavethat
dYt=dVt(1 + (K·Z)t−)+Vt−KtdZt+Ktd[V, Z]t.Using the expression fordXwe may convert this into
dYt=dVt(1 + (K·Z)t−)+KtdXt−dVtKtZt−which is of the form
dYt=LtdXt
for some (d+ 2)-dimensional predictable andX-integrable processL.Since
Kwas 1-admissible forZ,wehavethatY is positive and thereforeLis 1-
admissible forX. We now apply the above equality in two different cases. To
prove (1) we suppose thatV^1 ∞−1 is not maximal. TakeKa 1-admissible
integrand forZsuch that the limit at infinity exists and such that 1 + (K·
Z)∞≥V^1 ∞, with strict inequality on a non-negligible set. In that case we have