What is the expectation of $B_s^2B_t^2$ where $B_t$ is a standard Brownian motion?

Question

I am tring to find the variance of the integral $I=\int_0^T B_t^2 dt$.

I have found that $\mathbb{E}[I] = \int_0^T \mathbb{E}[B_t^2] dt = \frac{T^2}{2}$.

Since $\mathbb{E}[I^2] = \mathbb{E}[\int_0^T \int_0^T B_t^2 B_s^2 \ dt\ ds] = \int_0^T \int_0^T \mathbb{E}[B_t^2 B_s^2] \ dt \ ds$, I want to know the value of $\mathbb{E}\left[B_t^2 B_s^2\right]$.

Answer 1

For $s<t$ we have $EB_t^{2}B_s^{2}=E((B_t-B_s)+B_s)^{2}B_s^{2}=E((B_t-B_s))^{2}B_s^{2}+EB_s^{4}$ since $E((B_t-B_s)B_s)B_s^{2}E((B_t-B_s)) EB_s^{3}=0$. So $EB_t^{2}B_s^{2}=E((B_t-B_s))^{2}EB_s^{2}+EB_s^{4}=(t-s) s+EB_s^{4}$. $EB_s^{4}=s^{2}m_4$ where $m_4$ is the 4th moment of the standard normal distribution.