What is the distribution of a random variable that is the product of two normal random variables?

Let $X\sim N(\mu_1,\sigma_1), Y\sim N(\mu_2,\sigma_2)$ and $Z=XY$

That is, what is its probability density function, its expected value, and its variance?

  • $\begingroup$ Please check if this post answers your questions. $\endgroup$
    – Sasha
    Commented Jun 22, 2012 at 18:04
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    $\begingroup$ Saying that each is normally distributed falls short of saying what the joint distribution is. Often it's intended that they are independent but that doesn't get mentioned. But it should be. $\endgroup$ Commented Jun 22, 2012 at 18:06
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    $\begingroup$ The distribution is fairly messy. For independent normals with mean $0$, we are dealing with the product normal, which has been studied. For general independent normals, mean and variance of the product are not hard to compute from general properties of expectation. $\endgroup$ Commented Jun 22, 2012 at 18:13
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    $\begingroup$ @Clara, I attempted to edit your post to make it more clear. Please let me know if anything was incorrect. $\endgroup$
    – Justin
    Commented Jun 22, 2012 at 19:56

5 Answers 5


I will assume $X$ and $Y$ are independent. By scaling, we may assume for simplicity that $\sigma_1 = \sigma_2 = 1$. You might then note that $XY = (X+Y)^2/4 - (X-Y)^2/4$ where $X+Y$ and $X-Y$ are independent normal random variables; $(X+Y)^2/2$ and $(X-Y)^2/2$ have noncentral chi-squared distributions with $1$ degree of freedom. If $f_1$ and $f_2$ are the densities for those, the PDF for $XY$ is $$ f_{XY}(z) = 2 \int_0^\infty f_1(t) f_2(2z+t)\ dt$$

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    $\begingroup$ If $X$ and $Y$ are independent standard gaussians, the characteristic function should be $$\phi_{XY}(t)=\frac{1}{\sqrt{1+4t^2}}$$ as done here math.stackexchange.com/questions/101062/… $\endgroup$
    – qwertyuio
    Commented Jan 18, 2013 at 22:54
  • $\begingroup$ @Robert Israel: Why are $X+Y$ and $X-Y$ are independent? $\endgroup$
    – May
    Commented Sep 13, 2013 at 18:05
  • $\begingroup$ Two normal random variables are independent iff they are uncorrelated. Calculate the covariance of $X+Y$ and $X-Y$. $\endgroup$ Commented Sep 14, 2013 at 1:40
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    $\begingroup$ Yes, but I said "By scaling, we may assume for simplicity that $\sigma_1 = \sigma_2 = 1$. $\endgroup$ Commented Sep 15, 2013 at 6:57
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    $\begingroup$ @DylanDijk Yes if $X$ or $Y$ has mean $0$ (because then $XY$ and $-XY$ have the same distribution), but in general no. $\endgroup$ Commented Apr 27, 2023 at 2:59

For the special case that both Gaussian random variables $X$ and $Y$ have zero mean and unit variance, and are independent, the answer is that $Z=XY$ has the probability density $p_Z(z)={\rm K}_0(|z|)/\pi$. The brute force way to do this is via the transformation theorem: \begin{align} p_Z(z)&=\frac{1}{2\pi}\int_{-\infty}^\infty{\rm d}x\int_{-\infty}^\infty{\rm d}y\;{\rm e}^{-(x^2+y^2)/2}\delta(z-xy) \\ &= \frac{1}{\pi}\int_0^\infty\frac{{\rm d}x}{x}{\rm e}^{-(x^2+z^2/x^2)/2}\\ &= \frac{1}{\pi}{\rm K}_0(|z|) \ . \end{align}


Given the densities $\varphi$ and $\psi$ of two independent random variables, the probability that their product is less than $z$ is $$ \iint_{xy< z}\varphi(x)\psi(y)\,\mathrm{d}x\,\mathrm{d}y\tag{1} $$ Letting $w=xy$ so that $x=w/y$ yields $$ \iint_{w< z}\varphi\left(\frac{w}{y}\right)\psi(y)\,\mathrm{d}\frac{w}{y}\,\mathrm{d}y=\iint_{w< z}\varphi\left(\frac{w}{y}\right)\psi(y)\,\mathrm{d}w\,\frac{\mathrm{d}y}{y}\tag{2} $$ Taking the derivative of $(2)$ with respect to $z$ gives the density of the product of the random variables to be $$ \phi(z)=\int\varphi\left(\frac{z}{y}\right)\psi(y)\,\frac{\mathrm{d}y}{y}\tag{3} $$ We can compute the expected value using this distribution as $$ \begin{align} \mathrm{E}(Z) &=\int z\phi(z)\,\mathrm{d}z\\ &=\iint z\,\varphi\left(\frac{z}{y}\right)\psi(y)\,\frac{\mathrm{d}y}{y}\,\mathrm{d}z\\ &=\iint xy\,\varphi(x)\psi(y)\,\mathrm{d}y\,\mathrm{d}x\tag{4} \end{align} $$ which is exactly what one would expect when computing the expected value of the product directly.

In the same way, we can also compute $$ \begin{align} \mathrm{E}(Z^2) &=\int z^2\phi(z)\,\mathrm{d}z\\ &=\iint z^2\,\varphi\left(\frac{z}{y}\right)\psi(y)\,\frac{\mathrm{d}y}{y}\,\mathrm{d}z\\ &=\iint x^2y^2\,\varphi(x)\psi(y)\,\mathrm{d}y\,\mathrm{d}x\tag{5} \end{align} $$ again getting the same result as when computing this directly.

The variance is then, as usual, $\mathrm{E}(Z^2)-\mathrm{E}(Z)^2$.

  • $\begingroup$ Note however that in this case $X$ and $Y$ are not positive. $\endgroup$ Commented Jun 22, 2012 at 21:33
  • $\begingroup$ @RobertIsrael: actually, I don't think the positivity is needed. $\endgroup$
    – robjohn
    Commented Jun 22, 2012 at 21:36
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    $\begingroup$ There is a book that deals strictly with products of random variables by an author named Springer. It was published many year ago by Wiley (late 1970s or early 1980s). I will look for a link maybe from amazon. $\endgroup$ Commented Jun 22, 2012 at 21:50
  • $\begingroup$ @Michael Chernick: The lowest price for a used copy at amazon is more than 400\$'s, and for a new copy the double of that! $\endgroup$ Commented Apr 19, 2017 at 10:53

To your first point, let's look at how we calculate the expectation of a product. The idea is easy enough but Gaussian distributions can look a little messier than they really are.

Specifically, to look at the distribution function, I would start here: http://en.wikipedia.org/wiki/Product_distribution

For expectation though, recall (or note):

For independent random variables, the joint probability distribution function, $h(x,y)$ can be found simply as the product of the marginal distributions, say $f(x)$ and $g(y)$.

That is $h(x,y)=f(x)*g(y)$. You find the expectation in the same way you would find it for a single variable with single pmf. Namely,

$E(XY)=E(Z)=\int\int xy*h(x,y)dydx$

=$\int\int xy (f(x)g(y)dydx)=[\int xf(x)dx][\int yg(y)dy]=E(X)E(Y)$

For standard normal RVs, this is simple to compute. If, in fact, your variables are not independent, then you need to incorporate a covariance term into your calculations.

Hope that helps.

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    $\begingroup$ More in general, if $X,Y$ are independent, then $E(Z^m)=E(X^m)E(Y^m)$. In this way, we get $\sigma^2_Z=\sigma^2_X \sigma^2_Y + \sigma^2_Y \mu_X^2 +\sigma^2_X \mu_Y^2 $ $\endgroup$
    – leonbloy
    Commented Jun 22, 2012 at 20:12
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    $\begingroup$ ... and more generally, $E[X^m] = \sum_{k=0}^{\lfloor m/2 \rfloor} \dfrac{\mu_1^{m-2k} \sigma_1^{2k} m!}{2^k k! (m-2k)!}$ and similarly for $E[Y^m]$, so you can get all the moments. $\endgroup$ Commented Jun 22, 2012 at 20:33

It is called The Algebra of Random Variables by Melvin D. Spinger (Wiley, 1979) and includes a lot on products: http://www.amazon.com/Algebra-Variables-Probability-Mathematical-Statistics/dp/0471014060/ref=sr_1_1?s=books&ie=UTF8&qid=1340403029&sr=1-1&keywords=the+algebra+of+random+variables

In searching I also found this book by Galambos and Simonelli: http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Dstripbooks&field-keywords=product+of+random+variables


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