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Suppose we have pair $(X,Y)\sim Normal([\mu_x,\mu_y],{{\sigma_x^2\atop\rho \sigma_x\sigma_y } {\rho \sigma_x\sigma_y \atop \sigma_y^2} }] $ How is $U=X\cdot Y$ distributed? I've tried to compute this by substituting y=u/x in the bivariate normal pdf and taking integral(from $-\infty$ to $\infty$ with respect to x. I find the pdf of U as the sum of two exponential distributions(one for U<0 and one for U>0) that are weighted unequally. Is my method valid, or do I have to deal with cumulative distribution functions instead?

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Arkadiusz gives the answer in the case of two independent Gaussians. A simple technique to reduce the correlated case to the uncorrelated is to diagonalize the system. The intuition which I use is that for two random variables, we need two "independent streams of randomness," which we then mix to get the right correlation structure.

Let $X \sim N(0,\sigma_x)$ and let $Z \sim N(0,1)$ be two independent normals. Define

$Y = \tfrac{\rho \sigma_y}{\sigma_x}X + \sqrt{1-\rho^2}\sigma_y Z$.

Check that $\mathbb E Y^2 = \sigma_y^2$ and $\mathbb E XY = \rho \sigma_x \sigma_y$; this completely determines the bivariate Gaussian case you're interested in.

Now, $XY = \tfrac{\rho \sigma_y}{\sigma_x} X^2 + \sqrt{1-\rho^2}\sigma_y XZ$. The $X^2$ part has a $\chi^2$-distribution, familiar to statistics students; the $XZ$ part is comprised of two independent Gaussians, hence Arkadiusz's answer gives the distribution of that random variable.

Edit: As Robert Israel points out in the comments, I made a mistake in my final conclusion: the random variables $X^2$ and $XZ$ are uncorrelated, though certainly not independent. Nonetheless, the problem is essentially resolved at this point, since we have reduced the problem of understanding the product $XY$ to a sum of uncorrelated random variables $X^2$ and $XZ$ with known distributions.

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    $\begingroup$ No, the density of the sum of two independent random variables is the convolution of their density functions. Unfortunately, $X^2$ and $XZ$ are certainly not independent (although they are uncorrelated). $\endgroup$
    – Robert Israel
    Commented Jun 16, 2011 at 19:07
  • $\begingroup$ @Robert Israel, thank you for your comment. Of course, you are absolutely right. I have edited my response to fix this error. $\endgroup$
    – Tom LaGatta
    Commented Jun 16, 2011 at 20:20
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    $\begingroup$ How does it solve the problem to represent XY as a sum of uncorrelated random variables with known distributions? $\endgroup$
    – Douglas Zare
    Commented Oct 1, 2015 at 10:10
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It seems that no closed-form expression is known for the correlated case.

For the uncorrelated case $\rho=0$ the distribution of XY is $\frac{1}{\pi \sigma_x \sigma_y}K_0(\frac{|u|}{\sigma_x \sigma_y})$, where $K_0(x)$ is the modified Bessel function of the second kind. This distribution differs from the distribution you gave (is it Laplace distribution?). A notable difference: it has kurtosis of 6 (sharper peak), comparing to 3 for Laplace distribution.

Normal Product Distribution

On the Frequency Function of xy. C.Craig, 1936

Edit: answering the second question, the method is valid, but my guess is that the Jacobian determinant was skipped while doing the substitution.

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LaGatta's answer nails it, and may be useful for drawing simulations, etc.

This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity $\operatorname{E}XY= \operatorname{Cov}(X,Y) + (\operatorname{E}X)(\operatorname{E}Y)$.

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  • $\begingroup$ That is a nice observation, pedro. If one is furthermore interested in any higher moments $\mathbb E (XY)^k$, then I suggest using the decomposition $Y = \tfrac{\rho \sigma_y}{\sigma_x}X + \sqrt{1-\rho^2}\sigma_y Z$ and using the independence of $X$ and $Z$. As Arkadiusz says, there is likely no known closed-form expression for the product, though in practice we can estimate probabilities using Chebyshev's inequality and arbitrary moments. $\endgroup$
    – Tom LaGatta
    Commented Jun 16, 2011 at 20:26

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