$E[(\int_{0}^{\infty}f(t)dW_t)^2]$ for $f(t)=(W_2-W_1)1_{[2,3)}(t)+(W_3-W_1)1_{[3,5)}(t)$

Question

Let $f(t)=(W_2-W_1)1_{[2,3)}(t)+(W_3-W_1)1_{[3,5)}(t), t \ge 0$.

$(W_t)_{t\ge0}$ is Brownian motion.

What's the 'best' method to calculate $\mathbb E[(\int_{0}^{\infty}f(t)dW_t)^2]$?

I would try $\mathbb E[(\int_{0}^{\infty}f(t)dW_t)^2]=\mathbb E[\int_{0}^{\infty}f(t)^2dt]=\int_{0}^{\infty}\mathbb E[f(t)^2]dt$ and then using the properties of Brownian motion.

Answer 1

If $I$ and $J$ are two disjoint intervals, then $1_I 1_J =0$ and therefore

$$\big(x 1_{I}(t) + y 1_{J}(t) \big)^2 = x^2 1_I(t) + y^2 1_J(t)$$

for any real numbers $x,y \in \mathbb{R}$. This implies

$$f(t)^2 = (W_2-W_1)^2 1_{[2,3)}(t) + (W_3-W_1)^2 1_{[3,5)}.$$

Hence,

$$\int_0^{\infty} \mathbb{E}(f(t)^2) \, dt = \int_2^3 \mathbb{E}((W_2-W_1)^2) \, dt + \int_3^5 \mathbb{E}((W_3-W_1)^2) \, dt.$$

Use $W_t-W_s \sim N(0,t-s)$ to compute the remaining expectations.