# Covariance of squared bivariate normal random vector

$\mathbf{Y}=(Y_1, Y_2)^T$ is a bivariate Gaussian random vector with mean zero.

I want to prove that $Cov[Y_1^2, Y_2^2]=2\times(Cov[Y_1, Y_2])^2$.

I know that

$$Cov[Y_1^2, Y_2^2]=E[Y_1^2Y_2^2]-E[Y_1^2]E[Y_2^2]=E[Y_1^2Y_2^2]-Var[Y_1]Var[Y_2]$$

and

$$2(Cov[Y_1, Y_2])^2=2(E[Y_1Y_2]-E[Y_1]E[Y_2])^2=2(E[Y_1Y_2])^2$$ as $\mu_{Y_1}$,$\mu_{Y_2}$ are 0.

What is the next step I should proceed?

## 1 Answer

Some facts:

1. If $(Y_1,Y_2)$ is jointly normal (with mean 0) with parameters, $\rho, \sigma_1$ and $\sigma_2$, then $$Y_1 | Y_2 \sim N\left(\frac{\rho\sigma_1}{\sigma_2}Y_2, (1-\rho^2)\sigma_1^2\right)$$

2. The fourth central moment of a normal random variable is $\mathbb{E}[Y] = 3\sigma^4$.

Using these facts, we can write,

$$\begin{eqnarray} \mathbb{E}[Y_1^2Y_2^2] &=& \mathbb{E}[\mathbb{E}[Y_1^2Y_2^2 | Y_2]]\\ &=& \mathbb{E}[Y_2^2\mathbb{E}[Y_1^2|Y_2]]\\ &=& (1-\rho^2)\sigma_1^2\mathbb{E}[Y_2^2] + \frac{\rho^2\sigma_1^2}{\sigma_2^2}\mathbb{E}[Y_2^4]\\ &=& (1-\rho^2)\sigma_1^2\sigma_2^2 + 3\rho^2\sigma_1^2\sigma_2^2 \end{eqnarray}$$

Substitute this into your expression for $Cov(Y_1^2,Y_2^2)$ to get $2\rho^2\sigma_1^2\sigma_2^2$ which is identically $2(Cov(Y_1,Y_2))^2$.