$E[X_1^2X_2^2]$ for a gaussian vector

Question

I would like to calculate the following expectation $E[X_1^2X_2^2]$, where $X=(X_1,X_2)$ ha a gaussian distribution with mean vector $(0,0)$ and covariance matrix Q (non diadonal and non invertible). Since Q is not invertible I can't write the density of X (otherwise it's easy). Can anyone give me any idea? Thank you in advance.

Answer 1

As the covariance matrix is not inversible, there is a relation (assuming $X_1\neq 0$ and using the fact that the expectation of both random variables is $0$) $$ X_1 = aX_2 $$

and hence $$ E[X_1^2 X_2^2] = a^2E[X_1^4] = 3a^2\sigma_1^4 $$ where $\sigma_1^2$ is the variance of $X_1$, that is $Q(1,1)$.

To get rid $a^2$ write the same thing with $X_2$:

$$ E[X_1^2 X_2^2] = \frac3{a^2}\sigma_2^4 \\\implies E[X_1^2 X_2^2] = \sqrt{\frac3{a^2}Q(2,2)^2 \times 3a^2 Q(1,1)^2} = 3Q(1,1)Q(2,2) $$