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)=∫t0∂f∂t(s,Ws)ds+∫t0∂f∂x(s,Ws)dWs+12∫t0∂2f∂x2(s,Ws)ds=I+∫t02sWsdWs+t22
By rearranging I get
I=tW2t−∫t02sWsdWs−t22
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[tW2t−∫t02sWsdWs−t22]=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=4∫t0W3tdWs+6∫t0W2tds 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(4∫t0W3tdWs)=16∫t0E[W6s]ds=240∫t0s3ds=60t4 on the other hand 2Cov(4∫t0W3tdWs,6∫t0W2tds)=24t4(Why?) Moreover Var(W4t)=Var(4∫t0W3tdWs+6∫t0W2tds) thus 96t4=60t4+36Var(∫t0W2tds)+24t4 i.e Var(∫t0W2tds)=13t4
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