# Covariance of bi-variate normal distribution

$\left( \begin{array}{c} X_1 \\ X_2 \end{array} \right) \sim N\left( \left( \begin{array}{c} 0 \\ 0 \end{array} \right) , \left( \begin{array}{cc} 1 & r \\ r & 1 \end{array} \right) \right)$

How do you to calculate Cov$(X_1^2,X_2^2)$?

I know Cov$(X_1^2,X_2^2)=E(X_1^2X_2^2)-E(X_1^2)E(X_2^2)$ and I could calculate $E(X_1^2)$ and $E(X_2^2)$. However, I got stuck at the $E(X_1^2X_2^2)$.

Any thought on how to do that part? Thanks!

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Given the value of $X_2$, $X_1$ is normal with known mean and variance, and so you can get $E[X_1^2X_2^2 \mid X_2] = X_2^2 E[X_1^2 \mid X_2]$ which should be a quartic $X_2$, and then get the expected value of the result? –  Dilip Sarwate Oct 18 '11 at 16:34

As mentioned by @Robert Israel, Isserlis' theorem is a nice way to get the result. A more ad hoc method is to note that this covariance matrix means that $X_2=rX_1+\sqrt{1-r^2}X_3$ where $X_1$ and $X_3$ are i.i.d. standard gaussian random variables. Hence $$X_1^2X_2^2=r^2X_1^4+2rX_1^3X_3+(1-r^2)X_1^2X_3^2.$$ Since $\mathrm E(X_i^2)=1$, $\mathrm E(X_i^3)=0$, $\mathrm E(X_i^4)=3$, and $X_1$ and $X_3$ are independent, one gets $$\mathrm E(X_1^2X_2^2)=r^2\cdot3+2r\cdot0+(1-r^2)\cdot1=1+2r^2,$$ from which the value of $\mathrm{Cov}(X_1^2,X_2^2)$ should be clear.