$\mathbb E[W_s^2W_t^2]$ for Brownian motion

Question

Consider a Brownian motion $(W_t)_{t\ge0}$ on $(\Omega ,\mathcal F, \mathbb P)$.

How can I calculate $\mathbb E[W_s^2W_t^2]$?

I know that $\mathbb E[W_sW_t^2]=0$ but I don't know if that helps here.

Answer 1

Suppose $s\geq t$. $$E(W_s^2W_t^2) = E((W_s-W_t)^2W_t^2+2W_t^3W_s-W_t^4) \\ = E(W_{s-t}^2)E(W_t^2)+2E(W_t^3W_s)-E(W_t^4) =\\ (s-t)t+2E(W_t^3(W_s-W_t))+E(W_t^4) = \\ (s-t)t+2E(W_t^3)E(W_{s-t})+3t^2 = (s-t)t+3t^2 = 2t^2+st $$