Tuesday, February 5, 2019

stochastic calculus - Prove EmathbbQ[Xt|mathscrFu]=Xu given Yt is a martingale


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=t0θsdWs12t0θ2sds


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[(XtXu)Lt|Fu]=0


EQ[(XtXu)LtLT|Fu]=0


EQ[(XtXu)LtLT|Fu]=0


EQ[(XtXu)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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