What is the covariance function of the following process? $Y(t) = X^2(t)$, $t \ge 0$, where $X(t)$ is a Brownian motion with variance parameter $\sigma^2$.
I note that $$E(Y(t)) = E(X^2(t)) = \text{Var}(X) + E(X(t))^2 = t + 0 = t$$
and
$$\text{Cov} (Y(t), Y(s)) = E(X^2(t)X^2(s)) - E(X^2(t))E(X^2(s)) = E(X^2(t)X^2(s)) - ts$$
If it were $E(X(t)X(s))$, I could have solved it using $X(t) = X(t) - X(s) + X(s)$, but this case it confuses me because $X_s$ are squared.
Any help would be appreciated.
Actually writing $X_t = X_t-X_s+X_s$ is a good idea - there are some more terms than "usual", but this is not too bad. Note that
$$X_t^2 X_s^2 = ((X_t-X_s)+X_s)^2 X_s^2 = (X_t-X_s)^2 X_s^2 + 2 (X_t-X_s) X_s^3 + X_s^4.$$
Use the fact that $X_t-X_s$ and $X_s$ are independent to conclude that
$$\mathbb{E}(X_t^2 X_s^2) = \mathbb{E}((X_t-X_s)^2) \mathbb{E}(X_s^2)+2 \mathbb{E}(X_t-X_s) \mathbb{E}(X_s^3) +\mathbb{E}(X_s^4).$$
Calculate the expectations on the right-hand side using the stationarity of the increments and the fact that $X_u$ is Gaussian with mean $0$ and variance $u$ for any $u \geq 0$.