Question

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?

Answer 1

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. Let $f_{X^2}$ and $f_{XZ}$ be the two density functions, respectively. The density of a sum of two random variables is the convolution of their density functions: $f_{XY} = f_{X^2} * f_{XZ}$.

Answer 2

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.