10.3 Incomplete Markets 2151
n∑nk=1(
logXk−ER[
logXk∣
∣
∣Fνuk− 1])
→ 0.
On the set{νuk− 1 < ∞}we have thatER[
logXk∣
∣Fνu
k− 1]
≤ (^23 −
εk)log^12 +(^13 +εk)log2≤− 31 log 2 + 2εklog 2≤−^16 log 2, at least fork
large enough. It follows that on the set
⋂
∑n n≥^1 {νun < ∞},wehavethat
k=1logXk →−∞, and hence (E(K·N))∞ = 0 on this set. On the
complement, i.e. on
⋃
n≥ 1 {νun =∞}, we find that the maximal function
(E(K·N))∗∞ is bounded by 2nwherenis the first natural number such
thatνun =∞. The probability of this event is bounded byεnand hence
ER[(E(K·N))∞]≤η=
∑
nεn^2
n≤^1
8. Remark 10.3.6.By adjusting theεnwe can actually obtain a predictable pro-
cessKsuch that (E(K·N))∞= 0 on a set with measure arbitrarily close
to 1.
Lemma 10.3.7.IfLis a continuous positive strict local martingale, starting
at 1 , then forα> 0 small enough the processLstopped when it hits the level
αis still a strict local martingale.
Proof of Lemma 10.3.7.Simply letτ=inf{t|Lt<α}. ClearlyER[Lτ]<
α+ER[L∞]<1forα< 1 −ER[L∞].
If we apply the previous lemma on the exponential martingaleL=E(K·N)
and toα =η, we obtain a stopping timeτ and a strict local martingale
E(K·N)τthat is bounded away from zero.
We now use the same integrandK to constructZ=E(K·U)andwe
defineσ=inf{t|Zt=2}.
We will show thatER[Lτ∧σ]<1andthatER[Zτ∧σLτ∧σ] = 1. This will
complete the proof of the theorem since the measureQdefined bydQ=
Zτ∧σdRis an equivalent martingale measure and the elementf=Lτ∧σ− 1
is therefore maximal. On the other handER[f]<0.
Both statements will be shown using a time transform argument. The fact
that the processesK·NandK·Uboth have the same bracket will now turn
out to be useful. The time transform can be used to transform both these
processes into Brownian motions at the same time.
Following [RY 91, Chap. V, Sect. 1], we define
Tt=inf{
u∣
∣∣
∣〈K·N, K·N〉u=∫u0K^2 sd〈N, N〉s>t}
.
As well-known [RY 91], there are
(1) a probability space (Ω ̃,F ̃,R ̃),
(2) a mapπ: ̃Ω→Ω,
(3) a filtration (F ̃t)t≥ 0 onΩ, ̃