# Is the product of two Gaussian random variables also a Gaussian?

Say I have $X \sim \mathcal N(a, b)$ and $Y\sim \mathcal N(c, d)$. Is $XY$ also normally distributed?

Is the answer any different if we know that $X$ and $Y$ are independent?

• For a negative answer to your first question, take X=Y, then XY=X^2 cannot be Gaussian since it only takes positive values
– user16299
Jan 21, 2012 at 20:42
• Jun 17, 2013 at 20:35
• how does this question related to this paper: tina-vision.net/docs/memos/2003-003.pdf which says that "product and the convolution of Gauss ian probability density functions (PDFs) are also Gaussian functions". Dec 28, 2014 at 5:06
• @AsadIqbal They are related by a confusion between the product of some independent random variables and the product of their PDFs. A random variable product of two independent gaussian random variables is not gaussian except in some degenerate cases such as one random variable in the product being constant. A product of two gaussian PDFs is proportional to a gaussian PDF, always, trivially. Idem for the convolution of PDFs.
– Did
Jun 23, 2015 at 20:13
• @Did suppose $x\sim N(0,1), y|x\sim N(e^x, cos^2(x))$ - the product of their pdf is not a gaussian... Jun 20, 2022 at 10:12

The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables:

$$XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$

Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom.

If $X$ and $Y$ are both zero-mean, then

$$XY \sim c_1 Q - c_2 R$$

where $c_1=\frac{Var(X+Y)}{4}$, $c_2 = \frac{Var(X-Y)}{4}$ and $Q, R \sim \chi^2_1$ are central.

The variables $Q$ and $R$ are independent if and only if $Var(X) = Var(Y)$.

In general, $Q$ and $R$ are noncentral and dependent.

• I am also working on the distribution of the inner-product of two random variables having a normal distribution. The different topics on the subject in this forum helped me a lot. Could you just give some references/proofs about your last sentence that the variables Q and R are independent if and only if Var(X)=Var(Y), cause I exactly faced this problem in my simulations (in signal processing applications). I could not understand empirically why my code was only working when Var(X)=Var(Y). Now it make sense with your last post. Thanks
– user112319
Nov 29, 2013 at 17:19
• Thanks, but I am stuck with finding the expectation of XY when both are gaussian distributed RV Mar 27, 2014 at 5:07
• thanks... what about the var(XY) ? Nov 19, 2014 at 15:28
• Random variables Q and R are independent if and only if (X+Y) and (X-Y) are. Now, these are linear functions of the vector (X,Y). It is a theorem in Multivariate Probability that two linear functions AU and BU of a Gaussian vector U are independent if an only if A.var(U).B^T = 0. This implies that (X+Y) and (X-Y) are independent if and only if Var(X) = Var(Y), as can be easily verified. Nov 22, 2014 at 22:26
• @UlissesBraga-Neto may I ask for a reference in multivariate theory where I can find this theorem? Jul 16, 2017 at 16:47

As @Yemon Choi showed in the first question, without any hypothesis the answer is negative since $P(X^2<0)=0$ whereas $P(U<0)\neq 0$ if $U$ is Gaussian.

For the second question the answer is also no. Take $X$ and $Y$ two Gaussian random variables with mean $0$ and variance $1$. Since they have the same variance, $X-Y$ and $X+Y$ are independent Gaussian random variables. Put $Z:=\frac{X^2-Y^2}2=\frac{X-Y}{\sqrt 2}\frac{X+Y}{\sqrt 2}$. Then $Z$ is the product of two independent Gaussian, but the characteristic function of $Z$ is $\varphi_Z(t)=\frac 1{\sqrt{1+t^2}}$, which is not the characteristic function of a Gaussian.

• This proves that there are cases where X and Y are independent and XY isn't Gaussian, and cases where X and Y are non-independent and XY isn't Gaussian. However, there is also the question of whether cases exist in which XY is Gaussian (which would mean that X and Y would have to be non-independent). It wouldn't surprise me if such cases did exist, especially if the joint density of X and Y is allowed to be a generalized function. (One also needs to rule out the pathological case where a Dirac delta function could be considered to be a Gaussian with zero variance.)
– user13618
Jan 21, 2012 at 23:11
• @jamaicanworm in an answer below I proved that this cannot happen when $X$ and $Y$ follow a bivariate normal distribution unless the variances are zeros.
– Amir
Dec 31, 2023 at 10:25

As pointed out by Davide Giraudo, the characteristic function of Z is $\varphi_Z (t) = \frac{1}{\sqrt{1+t^2}}$. So, the corresponding probability density is given by: \eqalignno{f_Z(z)&={1\over2\pi}\int_{-\infty}^{+\infty}e^{-itz}\varphi_Z(t)\,dt={1\over2\pi}\int_{-\infty}^{+\infty}{\cos zt-i\,{\rm sen}\ zt\over\sqrt{1+t^2}}\,dt\cr &={1\over2\pi}\int_{-\infty}^{+\infty}{\cos zt\over\sqrt{1+t^2}}\,dt-{i\over2\pi}\int_{-\infty}^{+\infty}{{\rm sen}\ zt\over\sqrt{1+t^2}}\,dt.} The last of these integrals is null because the integrand is an odd function. Therefore, ${\Im[f_Z(z)]=0}$ and we have $$f_Z(z)={1\over\pi}\int_0^{+\infty}{\cos zt\over\sqrt{1+t^2}}\,dt.$$ For $z > 0$ this integral converges and represents the zero order modified Bessel function of second species, usually denoted as $K_{0}(z)$. Thus, $$f_Z(z)={1\over\pi}K_0(|\,z\,|), \qquad |\,z\,|>0$$ This shows that the random variable $Z$ has a K-form Bessel distribution. For more information see, for instance, Johnson and Kotz. Distributions in Statistics --- Continuous Univariate Distributions. Boston, Houghton-Mifflin, 1970. See also https://en.wikipedia.org/wiki/K-distribution.

You can use moments to see that the product $XY$ of independent normals cannot be normal except in trivial cases. By trivial, I mean $\mathbb{V}(X)\mathbb{V}(Y)=0.$

Suppose that $X,Y$ are independent normals so that $XY$ is normal.

Case 1: Suppose that $\mathbb{E}(X)=0$. By independence, $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)=0$, so $XY$ is mean zero normal, and hence $$\mathbb{E}((XY)^4)=3\mathbb{E}((XY)^2)^2.$$ By independence we get
$$\mathbb{E}(X^4)\mathbb{E}(Y^4)=3\mathbb{E}(X^2)^2\mathbb{E}(Y^2)^2.$$ Either $\mathbb{V}(X)=0$, or dividing by $\mathbb{E}(X^4)$ gives $\mathbb{E}(Y^4)= \mathbb{E}(Y^2)^2.$ This shows that $Y^2$, and hence $Y$, has zero variance.

Case 2: Suppose that $\mathbb{E}(X^2)>0$ and $\mathbb{E}(Y^2)>0$. Then, without loss of generality, $\mathbb{E}(X^2)=1$ and $\mathbb{E}(Y^2)=1$. In this case, we also have $\mathbb{E}((XY)^2)=1$, so $$\begin{eqnarray} \mathbb{E}(X^3)&=&\mathbb{E}(X)(3-2\mathbb{E}(X)^2)\\ \mathbb{E}(Y^3)&=&\mathbb{E}(Y)(3-2\mathbb{E}(Y)^2)\\ \mathbb{E}((XY)^3)&=&\mathbb{E}(XY)(3-2\mathbb{E}(XY)^2) \end{eqnarray}$$ Subtracting the product of the first two lines from the third line gives $$0=6\mathbb{E}(X)\,\mathbb{E}(Y)\,\mathbb{V}(X)\,\mathbb{V}(Y).$$ Either we are in a trivial case, or back to Case 1.

Thus, the product cannot be normal except in trivial cases.

• I don't think so. Wishart is the product distribution of Gamma distributions. However the Normal distribution is not Gamma (Chi-Squared is), and the type of product taken here of Gamma distributions is a different one to the one meant in the question. Dec 10, 2018 at 23:49

The answer about using a Chi Squared doesn't seem helpful here since 𝑄 and 𝑅 are almost certainly noncentral and dependent. There is a note about using a Normal Product Distribution which seems highly relevant, though I haven't seen it in practice and is less flexible than what I propose. What I have seen performed is the Delta Method or bootstrap to estimate this distribution.

The Delta Method says that an arbitrary function of normals is also approximately normal and in your case the $$Var(XY)\approx Var(X)\hat Y^2 + Var(Y)\hat X^2 + 2Cov(X,Y)\hat X \hat Y$$ where $$E(X) = \hat X$$.

I have seen this used to compute the CI for a ratio of normals X/Y after taking a log transform.

Hope that helps others that stumble upon this as I have, but feel free to correct me if I am wrong.

In an answer above by @Davide Giraudo, it is shown that the product of two normally distributed RVs does not necessarily follow a normal distribution based on a counterexample. In another answer by @user940, for the independent case, it is proved that the product of any two independent normally distributed RVs does not follow a normal distribution. Here I extend this negative result for a more general case and show that the product of any two RVs with a bivariate normal distribution, which can be dependent, does not follow a normal distribution.

Here, using the method detailed here, I derived the characteristic function (CF) of $$XY$$ for a general case that $$X$$ and $$Y$$ jointly follow a bivariate normal distribution (can be dependent), which may be also useful for other purposes.

Let $$X\sim N(\mu_1,\sigma_1^2), Y \sim N(\mu_2,\sigma_2^2)$$, and $$\sigma_{12}$$ denote their covariance. Then, the CF is given by

$$\varphi_{XY}(t)=\frac{\exp\left(\frac{iA^2 C_1 t}{1-2iC_1 t}\right)}{(1-2iC_1 t)^{1/2}} \times \frac{\exp\left(\frac{-iB^2 C_2 t}{1+2iC_2 t}\right)}{(1+2iC_2 t)^{1/2}}$$

where

$$A=\frac{\mu_1+\frac{\sigma_1}{\sigma_2}\mu_2}{\sqrt{2\left(\sigma_1^2+\frac{\sigma_1}{\sigma_2}\sigma_{12} \right)}}, B=\frac{\mu_1-\frac{\sigma_1}{\sigma_2}\mu_2}{\sqrt{2\left(\sigma_1^2-\frac{\sigma_1}{\sigma_2}\sigma_{12} \right)}}, C_1= \frac{\sigma_2}{\sigma_1} \left(\frac{\sigma_1^2+\frac{\sigma_1}{\sigma_2}\sigma_{12}}{2} \right), C_2= \frac{\sigma_2}{\sigma_1} \left( \frac{\sigma_1^2-\frac{\sigma_1}{\sigma_2}\sigma_{12}}{2} \right).$$

By comparing $$\varphi_{XY}$$ with the CF of a normal distribution, we realize that these two become identical only for the degenerate case of $$\sigma_2^2=\sigma_1^2 \rightarrow 0$$ (implying $$\sigma_{12}\rightarrow 0$$).