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I want to find the variance of the Ito integral:

$X_t=\displaystyle \int_0^t\sqrt{s}W_sdW_s$,

where W is a Brownian motion and s is the variable of integration.

This is what I have done so far:

$$Var(X_t)=E\left[\left(\displaystyle\int_0^t\sqrt{s}W_sdW_s\right)^2\right]=E\left[\displaystyle\int_0^tsW_s^2ds\right]$$

However, I'm not sure where to go from here. I think what's really confusing me is what to do with the fact that s is the variable of integration.

I would appreciate any guidance!

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  • $\begingroup$ why $dW_s$ changed to $ds$ in last part after applying ^2 $\endgroup$ – mahdi Mar 26 at 20:37
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$$E\left(\int_0^tsW_s^2\,\mathrm ds\right)=\int_0^tsE(W_s^2)\,\mathrm ds=\int_0^ts\cdot s\,\mathrm ds=\cdots$$

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