Covariance function of the square of a Gaussian process

Question

Let $X_t$, $t\in\mathbb{R}$, be a Gaussian process with mean $0$. Prove that $$ Cov(X_s^2,X_t^2)=2Cov(X_s,X_t)^2 $$

I don't know how to handle the squares here.

Answer 1

Let $\sigma_s^2$ and $\sigma_t^2$ be the respective variances of $X_s$ and $X_t$, and let $\rho_{s,t}$ be their correlation. Then if $Z$ is standard normal and independent of $X_s$, we can redefine $X_t$ as $$X_t/\sigma_t \overset{d}{=} \rho_{s,t} X_s/\sigma_s + \sqrt{1-\rho_{s,t}^2} Z$$ without changing the joint bivariate normal distribution $(X_s, X_t)$.

We have $E[X_s^2] = \sigma_s^4$ and $E[X_t^2] = \sigma_t^4$. Also, \begin{align} E[X_s^2 X_t^2] &= E[X_s^2 \sigma_t^2 (\rho_{s,t} X_s/\sigma_s + \sqrt{1-\rho_{s,t}^2} Z)^2] \\ &= \sigma_t^2 \rho_{s,t}^2 E[X_s^4]/ \sigma_s^2 + 0 + \sigma_t^2 (1-\rho^2_{s,t}) E[X_s^2]E[Z^2] \\ &= 3\sigma_s^2 \sigma_t^2 \rho^2_{s,t} + \sigma_s^2 \sigma_t^2 (1-\rho^2_{s,t}) \\ &= \sigma_s^2 \sigma_t^2 (2\rho_{s,t}^2 + 1). \end{align} So $\text{Cov}(X_s^2, X_t^2) = E[X_s^2 X_t^2] - E[X_s^2]E[X_t^2] = 2\sigma_s^2 \sigma_t^2 \rho^2_{s,t} = 2 \text{Cov}(X_s, X_t)^2$.