The Mathematics of Arbitrage

292 14 The FTAP for Unbounded Stochastic Processes

we obtain anF 1 -measurable density of a probability measure. Assertion (i)
of Lemma 14.3.5 implies thatQ̂∼Pand‖Q̂−P‖<ε. Assertion (ii) implies
that
EQ̂[‖∆S 1 ‖Rd|F 0 ]<∞, a.s.

and
EQ̂[∆S 1 |F 0 ]=0, a.s..

We are not quite finished yet as this only shows that (St)^1 t=0is a Q̂-
sigma-martingale but not necessarily aQ̂-martingale as it may happen that
EQ̂[‖∆S 1 ‖Rd]=∞. But it is easy to overcome this difficulty: find a strictly
positiveF 0 -measurable functionw(ω), normalised so thatEQ̂[w]=1and

such thatEQ̂[w(ω)E[‖∆S 1 ‖Rd|F 0 ]]<∞. We can constructwis such a way
that the probability measureQdefined by

dQ(ω)

dQ̂(ω)

=w(ω),

still satisfies‖Q−P‖<ε.Then

EQ[‖∆S 1 ‖Rd]<∞

and
EQ[∆S 1 |F 0 ]=0, a.s.,

i.e.,Sis aQ-martingale.
To extend the above argument fromT= 1 to arbitraryT∈Nwe need yet
another small refinement: an inspection of the proof of Lemma 14.3.5 above
reveals that in addition to assertions (i) and (ii) of Lemma 14.3.5, and given
M>1, we may chooseGηsuch that

(iii)

∥

∥

∥

∥

dGη

dFη

∥

∥

∥

∥

L∞(Rd,Fη)

≤M, π-a.s..

We have not mentioned this additional assertion in order not to overload
Lemma 14.3.5 and as we shall only need (iii) in the present proof.
Using (iii), withM= 2 say, and, choosingwabove also uniformly bounded
by 2, the argument in the first part of the proof yields a probabilityQ∼P,
‖Q−P‖<ε, such that‖ddQP‖L∞(P)≤4.
Now letT ∈Nand (St)Tt=0, based on (Ω,(Ft)Tt=0,F,P), be given. By
backward induction ont=T,...,1 apply the first part of the proof to find
Ft-measurable densitiesZtsuch that, defining the probability measureQ(t)by

dQ(t)

dP

=Zt,

we have that the two-step process (Su

∏T

v=t+1Zv)

t

u=t− 1 is aQ

(t)-martingale

with respect to the filtration (Fu)tu=t− 1 ,Q(t)∼P,‖Q(t)−P‖ 1 <ε 4 −TT−^1 ,
and such that‖Zt‖L∞(P)≤4.