Stochastic Integral I've just learned about stochastic integral and only know how to evaluate $\int\limits^{t}_{0} W(s)\mathrm{d}W(s)$. Could anyone give me some instruction on how to evaluate the following integrals?
$$\mathbb{Var}\left(\int_{1}^2 (W_t)^2\mathrm{d}W_t \right)$$
$$\mathbb{Var}\left(\int_{0}^T (W_t)^2\mathrm{d}t \right)$$
Thanks a lot!!
 A: Let $X = \int\limits_1^2 W_t^2 \mathrm{d} W_t$.
Since $\mathbb{E}\left(X\right) = 0$, the variance $\mathbb{Var}(X) = \mathbb{E}(X^2)$. Here we use Ito isometry:
$$
   \mathbb{E}\left( \left(\int\limits_1^2 W_t^2 \mathrm{d} W_t \right)^2 \right) = 
   \int_{1}^{2} \mathbb{E}(W_t^4) \mathrm{d} t = 
   \int_1^2 3 t^2 \mathrm{d} t = 2^3 - 1^3 = 7
$$ 
Let $Y = \int\limits_0^t W_s^2 \mathrm{d} s$. Expectation is $\mathbb{E}(W_s^2) = s$, thus $\mathbb{E}(Y) = \int\limits_0^t s \mathrm{d} s = \frac{t^2}{2}$.
$$\begin{eqnarray}
  \mathbb{E}(Y^2) &=& \int_0^t \int_0^t \mathbb{E}(W_s^2 W_u^2) \mathrm{d} s \mathrm{d} u
   = 2 \int_0^t \int_0^s \mathbb{E}(W_s^2 W_u^2) \mathrm{d} u \mathrm{d} s \\ 
   &=& 2 \int_0^t \int_0^s \mathbb{E}(W_u^2 (W_u  + W_{s-u})^2) \mathrm{d} u \mathrm{d} s \\
   &=& 2 \int_0^t \int_0^s \left( \mathbb{E}(W_u^4) + \mathbb{E}(W_u^2) \mathbb{E}(W_{s-u}^2) \right) \mathrm{d} u \mathrm{d} s \\
   &=& 2 \int_0^t \int_0^s \left( 3 u^2 + u(s-u) \right) \mathrm{d} u \mathrm{d} s = 2 \int_0^t \left( s^3 + \frac{s^3}{6} \right) \mathrm{d} s = \frac{7}{12} t^4
\end{eqnarray}
$$
Thus $\mathbb{Var}(Y) = \mathbb{E}(Y^2) - (\mathbb{E}(Y))^2 = \frac{7}{12} t^4 - \left(\frac{t^2}{2} \right)^2 = \frac{t^4}{3}$.
