We are given a filtered probability space (Ω,F,{Ft}t∈[0,T],P), where {Ft}t∈[0,T] is the filtration generated by standard P-Brownian motion.
Let dXt=θtdt+dWt be an Ito process where (θt)t∈[0,T] is Ft-adapated and E[∫T0θ2sds]<∞ and
Yt:=XtLt, Lt=expZt, Zt=−∫t0θsdWs−12∫t0θ2sds
It can be shown that Yt is a (Ft,P)-martingale.
Oh also, Novikov's condition holds. Not sure if that's relevant?
If dQdP=LT, prove EQ[Xt|Fu]=Xu.
What I tried:
By Novikov's Lt is a (Ft,P)-martingale. Then we have that
E[XtLt|Fu]=XuE[Lt|Fu]
→E[(Xt−Xu)Lt|Fu]=0
→EQ[(Xt−Xu)LtLT|Fu]=0
→EQ[(Xt−Xu)LtLT|Fu]=0
→EQ[(Xt−Xu)exp(−ZT+Zt)|Fu]=0
Now what? I don't suppose exp(−ZT+Zt)=1...or is it?
Another thing:
E[Yt|Fu]=Yu
→E[XtLt|Fu]=XuLu
→EP[XtLt|Fu]=XuLu
→EQ[XtLtLT|Fu]=XuLu
→?EQ[Xt|Fu]E[LtLT|Fu]=XuLu
If so, I think we have E[LtLT|Fu]=Lu× some integral that will turn out to be 1 probably by mgf, but I don't think mgf applies as θt is not necessarily deterministic.
What to do?
Something else I tried:
EQ[Xt|Fu]=EQ[YtLt|Fu]
=EP[YtLtLT|Fu]
=E[YtLtLT|Fu]
=E[YtexpZtexpZT|Fu]
=1L2uE[Ytexp(−ZT+Zt)|Fu]
It looks like exp(−ZT+Zt) is independent of Fu, but I don't think
E[Ytexp(−ZT+Zt)|Fu]=E[Yt|Fu]E[exp(−ZT+Zt)|Fu]
Or is it? If so, why? If not, what to do?
Answer
By Bayes' rule for conditional expectation (or here),
EQ[Xt|Fu]E[LT|Fu]=E[XtLT|Fu]
→EQ[Xt|Fu]Lu=E[XtLT|Fu]
→EQ[Xt|Fu]=E[XtLtLu|Fu]
=1LuE[XtLt1|Fu]
=1LuE[Yt|Fu]
=1LuYu
=1LuXuLu
=Xu
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