Monday, April 3, 2017

stochastic processes - Variance of time integral of squared Brownian motion


I want to calculate the variance of


I=t0W2sds


I was thinking I could define the function f(t,Wt)=tW2t and then apply Ito's lemma so I get



f(t,Wt)f(0,0)=t0ft(s,Ws)ds+t0fx(s,Ws)dWs+12t02fx2(s,Ws)ds=I+t02sWsdWs+t22


By rearranging I get


I=tW2tt02sWsdWst22


We then get that (I'm not sure here but i think the expectation is zero of any integral w.r.t BM?)


E[I]=t22


And variance


V[I]=V[tW2tt02sWsdWst22]=t2V[W2t]+E[(t02sWsdWs)2]=2t4+E[t04s2W2sds](Isometry property)


Not sure if it is OK to change order of integration and expectation here, but if I do that, I get


V[I]=2t4+t04s2E[W2s]ds=2t4+t04s2E[W2s]ds=2t4+t04s3ds=3t4


However, the answer says the variance should be t43, so I guess I do something wrong?




Answer



Other Way


By application of Ito's lemma , we have W4t=4t0W3tdWs+6t0W2tds We know



{E[W2n+1(t)]=0E[W2n(t)]=(2n)!2nn!tn



therefore Var(W4t)=E[W8t]E[W4t]2=105t4(3t2)2=96t4 By application of Ito's Isometry, we have Var(4t0W3tdWs)=16t0E[W6s]ds=240t0s3ds=60t4 on the other hand 2Cov(4t0W3tdWs,6t0W2tds)=24t4(Why?) Moreover Var(W4t)=Var(4t0W3tdWs+6t0W2tds) thus 96t4=60t4+36Var(t0W2tds)+24t4 i.e Var(t0W2tds)=13t4


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