# Expected value of product of square of random variable

Suppose we have two Gaussian distributed random variable $X$~$N(0,\sigma^2)$ and $Y$~$N(0,\sigma^2)$. These variables are not independent. What will be the expected value of product of square of this random variables

$E[X^2Y^2]$ = ??

Edit 1: They are jointly Gaussian distributed with correlation coefficient $\rho$

Edit 2: $X$~$N(0,\sigma^2)$, $Y$~$N(0,\sigma^2)$

• Please show any work that you have done, and your thoughts on the problem. – Clarinetist Sep 5 '16 at 14:10
• What is the joint distribution of $(X,Y)$? This will be needed as the variables are stated to be dependent. – wolfies Sep 5 '16 at 14:19
• Sorry: I should have mentioned that they are jointly Gaussian distributed with correlation coefficient $\rho$ – Marcus Sep 5 '16 at 15:08
• You still have not defined the variances, nor what you have tried. In any event, the mgf is your friend. – wolfies Sep 5 '16 at 16:23

Try using the Law of total expectation - set $Z = X^2Y^2$ and use:

$$\mathbb{E}[X^2Y^2]=\mathbb{E}[Z]=\int_{y}\mathbb{E}[Z\mid Y=y]\cdot\Pr[Y=y]$$

• Since $Y$ is a normally-distributed random variable, the sum should be an integral. – Clarinetist Sep 5 '16 at 14:22
• Youre right, I missed that part – Snufsan Sep 5 '16 at 14:23
• Now I agree that this is a clever method - but how do you propose dealing with $\mathbb{E}[X^2 \mid Y = y]$? – Clarinetist Sep 5 '16 at 14:24
• I guess you should know somthing about the relation between those variables. It seems like OP didnt wrote all the details that matter :) – Snufsan Sep 5 '16 at 14:25
• I was just proposing a way to try and approach that problem – Snufsan Sep 5 '16 at 14:26

See Variance of product of dependent variables at CrossValidated, specifically the second answer, for the outline of a derivation. In your case, because $X$ and $Y$ have zero mean and the same variance, the formula is simpler.

As explained there, the conditional density of $Y$ given $X = x$ is normal, with mean $\rho x$ and variance $\sigma^2(1-\rho^2)$. It follows that \begin{align*} E[X^2 Y^2 \mid X]&= X^2 E[Y^2 \mid X] \\ &= X^2 \left[ \sigma^2 (1-\rho^2) + \rho^2 X^2 \right]. \end{align*} By the law of iterated expectation, $$E[X^2Y^2] = E\left[ E[X^2Y^2 \mid X] \right] = \sigma^2(1-\rho^2) E[X^2] + \rho^2 E[X^4]$$ and you can can look up the moments $E[X^2]$ and $E[X^4]$ at the Wikipedia article on the normal distribution. You'll end up with $\sigma^4(1+2\rho^2)$.