Is square of Wiener process an orthogonal process?

Question

I'm trying to prove: Let $t_1 < t_2 \leq t_3 < t_4$ and $(X)_t$ is the square of Wiener process. Then $E(X_{t_2} - X_{t_1})(X_{t_4}- X_{t_3}) \neq 0.$

Progress

Maybe the fact $E(X_{t_2} - X_{t_1}) = t_2 - t_1$ for $t_2 > t_1$ can help, but i don't see how.

Answer 1

Hint: It suffices to calculate expectations of the form

$$\mathbb{E}(X_t \cdot X_s)$$

where $s \leq t$. Setting $X_t = W_t^2$, we have

$$\begin{align*} \mathbb{E}(X_t X_s) &= \mathbb{E}\big( ((W_t-W_s)+W_s)^2 W_s^2 \big) \\ &= \mathbb{E}((W_t-W_s)^2 W_s^2) + 2 \mathbb{E}((W_t-W_s) W_s^3) + \mathbb{E}(W_s^4). \end{align*}$$

Now use that $W_s \sim N(0,s)$ and $W_t-W_s \sim N(0,t-s)$ and the fact that $W_t-W_s$ is independent from $W_s$.