How to Compute $E[(W_t)^2(W_s)^2] $ , with $W_t$, $W_s$ are standard Wiener Process

Question

I can compute $E(W_tW_s)$, but I cannot compute $E[(W_t)^2(W_s)^2] $. please help me! Thanks a lot!

Answer 1

Hints:

1. For $t>s$, $f(W_t-W_s)$ and $f(W_s)$ are independent random variables with some measurable function $f$ since $W_t-W_s$ and $W_s$ are independent (see here).
2. $W_t-W_s\sim\mathcal N(0,t-s)$ for $0\le s< t$.
3. The fourth moment of the Gaussian random variable $\xi\sim\mathcal N(0,\sigma^2)$ is equal to $3\sigma^4$ (see here).

Edit:

Without loss of generality, let us assume that $t>s$. Then \begin{align*} \operatorname E[W_t^2W_s^2] &=\operatorname E[(W_t-W_s+W_s)^2W_s^2]\\ &=\operatorname E[((W_t-W_s)^2+2(W_t-W_s)W_s+W_s^2)W_s^2]\\ &=\operatorname E[(W_t-W_s)^2W_s^2]+2\operatorname E[(W_t-W_s)W_s^3]+\operatorname EW_s^4\\ &=\operatorname E(W_t-W_s)^2\operatorname EW_s^2+2\operatorname E(W_t-W_s)\operatorname EW_s^3+\operatorname EW_s^4. \end{align*} Can you take it from here?