The Mathematics of Arbitrage

(Tina Meador) #1

12.4 The Existence of an Absolutely Continuous Local Martingale Measure 247


This settles the problem of the usual hypotheses. Each time we need anF ̃-
predictable process, we can without danger replace it by a predictable process
forF. Without further notice we will do this.
The processSis a semi-martingale with respect to the system (F ̃,PF).
This is well-known, see [P 90].
Note also that forPFwe have that


∫∞


0 h


ud〈M, M〉uhu<∞a.s.. We will
need this later on.
As a first step we will decomposeSintoasumofaPF-local martingale
and a predictable process of finite variation. BecausePFis only absolutely
continuous with respect toPwe need an extension of the Girsanov-Maruyama
formula for this case. The generalisation was given by [L 77]. We need the
cadlag martingaleUdefined as


Ut=E

[


(^1) F
P[F]





∣Ft

]


.


Note thatUis not necessarily continuous, as we only assumed thatSis
continuous and not that eachFt-martingale is continuous.
Together with the processUwe need the stopping time


ν=inf{t|Ut=0}=inf{t> 0 |Ut−=0}

(see [DM 80] for this equality).


Lemma 12.4.5.We haveν=TP-almost surely.


ProofofLemma.We first show that for an arbitrary stopping timeσwe
have thatLσ>0ontheset{Uσ > 0 }.LetAbe a set inFσsuch that
P[A∩{Uσ> 0 }]>0. This already implies thatP[A∩F]>0. Indeed we have
that


E[ (^1) A (^1) F|Fσ]=P[F] (^1) AUσ
and hence we necessarily have thatP[A∩F]>0. The following chain of
equalities is almost trivial

A∩{Uσ> 0 }
Lσ=



A

Lσ (^1) {Uσ> 0 }≥P[F]



A

LσUσ=


A

Lσ (^1) F=



A∩F

Lσ.

The last term is strictly positive sinceLσ>0onF. This proves that for each
setAsuch thatP[A∩{Uσ> 0 }]>0wemusthave



A∩{Uσ> 0 }Lσ>0. This
implies thatLσ>0ontheset{Uσ> 0 }, henceν≤T.
The converse inequality is less trivial and requires the use of the(NA)
property ofS. We proceed in the same way. TakeG∈Fσsuch thatG⊂
{Lσ> 0 }andP[G]>0. Suppose thatUσ=0onG. We will show that this
leads to a contradiction. IfUσ=0onGthen clearlyG∩F=∅. However, on
Fcwe have thatLttends to zero and henceL^1 t tends to∞. We know that
1
Lt−1 can be obtained as a stochastic integral with respect toS.Wetakethe

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