Sunday, March 20, 2016

stochastic processes - Show that E[Bt|mathscrFs]=Bs for Bt=W3t3tWt


Given prob space (Ω,F,P) and a Wiener process (Wt)t0, define filtration Ft=σ(Wu:ut)


Let (Bt)t0 where Bt=W3t3tWt. Show that E[Bt|Fs]=Bs whenever s<t.


I think this all comes down to manipulation since there are martingales somewhere


My attempt:


Splitting up into E[W3t|Fs]3E[tWt|Fs] doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:



E[Wt(W2t3t)|Fs]


=E[Wt(W2tt2t)|Fs]


=E[Wt(W2tt)2tWt)|Fs]


=E[Wt(W2tt)|Fs]2E[tWt|Fs]


Wt is not Fs-measurable, so we can't take that out...


tW1/t is Brownian and thus a martingale, but I don't know about tWt...


cWt/c2 is Brownian and thus a martingale, but I don't think we can set c = t...


Help please?



Answer



E(W3t3tWtFs)=E((WtWs+Ws)33t(WtWs+Ws)Fs)=E((WtWs)3+W3s+3(WtWs)2Ws+3(WtWs)W2s3t(WtWs)3tWsFs)=E((WtWs)3)+W3s+3WsE((WtWs)2)+3W2sE(WtWs)3tE(WtWs)3tWs=W3s+3Ws(ts)3tWs=W3s3sWs,

by noting that E((WtWs)3)=E(WtWs)=0,
and E((WtWs)2)=ts.



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