Variance of sum of two ito integrals I don't really understand how to solve the following problem:
Var(X) where X = $\int_0^2 2t dW(t) + \int_4^6 W(t) dW(t)$ 
If I use $E [(A+B)^2] = E(A^2) + E(B^2) + 2E(AB)$ I get to the point where I have the cross product of the two integrals, but I dont know how to handle that.. But maybe there is an easier way?
 A: Hint Note that
$$\mathbb{E} \left[ \left( \int_0^t f(s) \, dW_s \right) \cdot \left( \int_0^t g(s) \, dW_s \right) \right] = \mathbb{E} \left[ \left( \int_0^t f(s) \cdot g(s) \, ds \right) \right]$$
This follows by polarization from Itô's isometry:
$$\begin{align*} & \mathbb{E} \left[ \left( \int_0^t f(s) \, dW_s \right) \cdot \left( \int_0^t g(s) \, dW_s \right) \right]\\ &= \frac{1}{4} \left( \mathbb{E} \left[ \left( \int_0^t f(s)+g(s) dW_s \right)^2 \right] - \mathbb{E} \left[ \left( \int_0^t f(s)-g(s) \, dW_s \right)^2 \right] \right) \\ &= \frac{1}{4} \left( \mathbb{E} \left( \int_0^t (f(s)+g(s))^2  \, ds \right) - \mathbb{E}  \left( \int_0^t (f(s)-g(s))^2 \, ds \right)\right) \\ &= \mathbb{E} \left[ \left( \int_0^t f(s) \cdot g(s) \, ds \right) \right] \end{align*}$$

This is not the most direct way to compute the variance and the following argumentation is easier: First of all, $\mathbb{E}X=0$ follows from the fact that any stochastic integral of the form
$$M_t := \int_0^t f(s) \, dW_s$$
is a martingale (...if $f$ is nice - and they are nice; in the given example) and therefore $\mathbb{E}M_t= \mathbb{E}M_0=0$. It remains to calculate $\mathbb{E}(X^2)$. To this end, we rewrite $X$ in the following way:
$$X = \int_0^6 \underbrace{(2s 1_{[0,2]}(s) + W(s) 1_{[2,4]}(s))}_{=:f(s)} \, dW(s).$$
It follows from Itô's isometry that
$$\begin{align*} \mathbb{E}(X^2) &= \mathbb{E} \left( \int_0^6 f(s)^2 \, ds \right) \\ &= \int_0^6 (4s^2 1_{[0,2]}(s)  + \mathbb{E}(W_s^2) 1_{[4,6]}(s)) \, ds \\ &= \int_0^2 4s^2 \, ds + \int_4^6 s \, ds \end{align*}$$ (Here we used that $\mathbb{E}(W_s^2)=s$ since $W_s \sim N(0,s)$.)
A: Write the second integral as 
$$
\int_4^6[W(t) - W(4)]\,dW(t)+W(4)[W(6) - W(4)]=I_1+I_2,
$$
and note that $I_1$ is independent of the first integral defining $X$. (Why?)  The expectation of the product of the first integral $\int_0^2 2t\,dW(t)$ (call it $Y$) and $I_2$ is zero because $Y\cdot I_2 = [Y\cdot W(4)]\cdot [W(6)-W(4)]$ and the increment $W(6) -W(4)$ is independent of $\mathcal{F}_4:=\sigma\{W(s):0\le s\le 4\}$ while $Y\cdot W(4)$ is $\mathcal{F}_4$ measurable.
