Expectation of the square of integral of BM

Question

I got stuck with the following integral:

$$ E[\Big(\int_{0}^{T}\frac{W_{t}^{2}}{2}dt\Big)^{2}], $$ where $W_{t}$ is a standard Brownian Motion.

Answer 1

$$\begin{align} \mathbb{E}\left[\left(\int_0^T \frac{W_t^2}{2} \,\mathrm{d}t \right)^2\right] &= \mathbb{E}\left[\int_0^T \int_0^T \frac{W_t^2W_s^2}{4} \,\mathrm{d}s\,\mathrm{d}t\right] \\ &=\frac{1}{4}\int_0^T\int_0^T \mathbb{E}(W_t^2W_s^2) \,\mathrm{d}s\,\mathrm{d}t \\ &=\frac{1}{2}\int_0^T\left(\int_0^t \mathbb{E}(W_t^2W_s^2) \,\mathrm{d}s\right)\,\mathrm{d}t. \end{align} $$

Assuming $0\leq s \leq t$, we have

$$\begin{align} \mathbb{E}(W_t^2W_s^2) &=\mathbb{E}\bigl((W_t-W_s+W_s)^2W_s^2\bigr) \\ &=\mathbb{E}\bigl((W_t-W_s)^2W_s^2\bigr)+2\mathbb{E}\bigl((W_t-W_s)W_s^3\bigr)+\mathbb{E}\bigl(W_s^4\bigr) \\ &=(t-s)s+0+3s^2 \\ &=ts+2s^2. \end{align} $$

Hence

$$\begin{align} \mathbb{E}\left[\left(\int_0^T \frac{W_t^2}{2} \,\mathrm{d}t \right)^2\right] &=\frac{1}{2}\int_0^T \frac{t^3}{2}+\frac{2t^3}{3} \,\mathrm{d}t \\ &=\frac{7}{48}T^4. \end{align} $$