I want to compute $$E[\int_0^t W_r dr \int_0^s W_r^2 dW_r].$$ Here $t,s$ are arbitrary. I have thought about this a lot but not sure how to proceed. I tried to apply Ito's formula to one of the factors in the product, but that did not seem to help. Any idea will be appreciated!
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$$\int_0^tW_rdr=\int_0^t(t-r)dW_r\implies E\left(\int_0^tW_rdr\int_0^sW_r^2dW_r\right)=\int_0^{\min(s,t)}(t-r)E(W_r^2)dr={}{}{}\ldots$$ – Did Mar 21 '16 at 00:32
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I'd start by using Ito's formula to write $$ \int_0^s W^2_r\,dW_r={1\over 3}W^3_s-\int_0^s W_u\,du. $$
John Dawkins
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I know how to compute $E[\int_0^t W_r dr \int_0^s W_u du]$, but not sure how to deal with $E[W_s^3/3 \int_0^t W_r dr]$... – S in NT Mar 21 '16 at 01:18
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Swap expectation and integration: $\int_0^t E[W_s^3W_r],dr$. Now split the integral into $r\le s$ and $r>s$. There might be a couple of cases to worry about, depending on which of $s$ and $t$ is bigger. – John Dawkins Mar 21 '16 at 01:22