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I have 2 stochastic integrals :

$ I(T):= \int_0^T W(t)^2 \, d(t)$

$ J(T):= \int_0^T W(t)^2 \, dW(t)$

As $W(t)$ is the standard wiener process.

I need to show $Var(I(T))= \frac{T^4}{3}$ and $Var(J(T))=T^3$

I know I need to use Ito Isometry but I cant get the results I need to prove. Please Help!

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    Where exactly are you stuck? In order to calculate the variance you need the expectation value of the integrals... any ideas on that? – 2017-02-26
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    $\text{var} (J(T)) = \mathbb{E}((J(T))^2) = \mathbb{E} \left[ \left( \int_0^T W(t)^2 \, dW(t) \right)^2 \right]= \left[ \left( \int_0^T\mathbb{E}( W(t)^2) \, d(t) \right)^2 \right]=\left[ \left( \int_0^T\mathbb{E}( W(t)^4) \, d(t) \right) \right]=\left[ \left( \int_0^T3t^2 \, d(t) \right) \right]=T^3$ – 2017-02-26
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    Is it OK? for $Var(I(T))$ I have no clue... – 2017-02-26
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    If you know why $\mathbb{E}(J(T))=0$, then this is correct. – 2017-02-26

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