Given prob space (Ω,F,P) and a Wiener process (Wt)t≥0, define filtration Ft=σ(Wu:u≤t)
Let (Bt)t≥0 where Bt=W3t−3tWt. 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(W2t−3t)|Fs]
=E[Wt(W2t−t−2t)|Fs]
=E[Wt(W2t−t)−2tWt)|Fs]
=E[Wt(W2t−t)|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(W3t−3tWt∣Fs)=E((Wt−Ws+Ws)3−3t(Wt−Ws+Ws)∣Fs)=E((Wt−Ws)3+W3s+3(Wt−Ws)2Ws+3(Wt−Ws)W2s−3t(Wt−Ws)−3tWs∣Fs)=E((Wt−Ws)3)+W3s+3WsE((Wt−Ws)2)+3W2sE(Wt−Ws)−3tE(Wt−Ws)−3tWs=W3s+3Ws(t−s)−3tWs=W3s−3sWs, by noting that E((Wt−Ws)3)=E(Wt−Ws)=0, and E((Wt−Ws)2)=t−s.
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