Let
$$H(t)=W(1)I_{[1,2)}(t)+W(2)I_{[3,4)}(t), t\ge 0$$
($I$ describes the indicator function and $W(t)$ is Brownian motion)
How can I calculate $I_T(H)=\int_{0}^{T}H(t)dW(t)$ for $T>0$ and $\mathbb E[I(H)^3]$?
$I(H):=\int_{0}^{\infty}H(t)dW(t)$.
I know how to calculate $\mathbb E[I(H)^2]$ if that helps.
$T\ge 4$:
$I_T(H)=I(H)=W_1(W_2-W_1)+W_2(W_4-W_3)$
$T \in [3,4)$:
$I_T(H)=W_1(W_2-W_1)+W_2(W_T-W_3)$
$T \in [2,3)$:
$I_T(H)=W_1(W_2-W_1)$
$T \in [1,2)$:
$I_T(H)=W_1(W_T-W_1)$
$T < 1$:
$I_T(H)=0$