Distribution of the product of two random variables

{{main|Algebra of random variables}}

The product is one type of algebra for random variables: Related to the product distribution are the ratio distribution, sum distribution (see List of convolutions of probability distributions) and difference distribution. More generally, one may talk of combinations of sums, differences, products and ratios.

Many of these distributions are described in Melvin D. Springer's book from 1979 The Algebra of Random Variables.{{Cite book

| last = Springer

| first = Melvin Dale

| title = The Algebra of Random Variables

| url = https://archive.org/details/algebraofrandomv0000spri

| url-access = registration

| access-date = 24 September 2012

| publisher = Wiley

| year = 1979

| isbn = 978-0-471-01406-5

}}

If $X$ and $Y$ are two independent, continuous random variables, described by probability density functions $f\_X$ and $f\_Y$ then the probability density function of $Z\; =\; XY$ is{{cite book|last=Rohatgi|first=V. K.|title=An Introduction to Probability Theory and Mathematical Statistics|year=1976|publisher=Wiley|location=New York|isbn=978-0-19-853185-2|doi=10.1002/9781118165676|series=Wiley Series in Probability and Statistics}}

:$f\_Z(z)\; =\; \backslash int^\backslash infty\_\{-\backslash infty\}\; f\_X(x)\; f\_Y(\; z/x)\; \backslash frac\{1\}$

We first write the cumulative distribution function of $Z$ starting with its definition

:$\backslash begin\{align\}$

F_Z(z) & \, \stackrel{\text{def}}{=}\ \mathbb{P}(Z\leq z) \\

& = \mathbb{P}(XY\leq z) \\

& = \mathbb{P}(XY\leq z , X \geq 0) + \mathbb{P}(XY\leq z , X \leq 0)\\

& = \mathbb{P}(Y\leq z/X , X \geq 0) + \mathbb{P}(Y\geq z/X , X \leq 0)\\

& = \int^\infty_0 f_X(x) \int^{z/x}_{-\infty} f_Y(y)\, dy \,dx

+\int^0_{-\infty} f_X(x) \int^\infty_{z/x} f_Y(y)\, dy \,dx

\end{align}

We find the desired probability density function by taking the derivative of both sides with respect to $z$. Since on the right hand side, $z$ appears only in the integration limits, the derivative is easily performed using the fundamental theorem of calculus and the chain rule. (Note the negative sign that is needed when the variable occurs in the lower limit of the integration.)

: $\backslash begin\{align\}$

f_Z(z) & = \int^\infty_0 f_X(x) f_Y(z/x) \frac{1}{x}\,dx

-\int^0_{-\infty} f_X(x) f_Y(z/x) \frac{1}{x} \,dx \\

& = \int^\infty_0 f_X(x) f_Y(z/x) \frac{1}

+ \int_{-\infty}^0 f_X(x) f_Y(z/x) \frac{1}

& = \int^\infty_{-\infty} f_X(x) f_Y(z/x) \frac{1}

\end{align}

where the absolute value is used to conveniently combine the two terms.{{cite book|last1=Grimmett|first1=G. R. |last2=Stirzaker |first2=D.R.|title=Probability and Random Processes |year=2001|publisher=Oxford University Press|location=Oxford|isbn=978-0-19-857222-0|url=http://ukcatalogue.oup.com/product/9780198572220.do?keyword=grimmett+stirzaker&sortby=bestMatches|access-date=4 October 2015}}

A faster more compact proof begins with the same step of writing the cumulative distribution of $Z$ starting with its definition:

: $\backslash begin\{align\}$

F_Z(z) & \overset{\underset{\mathrm{def}}{}}{=} \ \ \mathbb{P}(Z\leq z) \\

& = \mathbb{P}(XY\leq z) \\

& = \int^\infty_{-\infty} \int^\infty_{-\infty} f_X(x) f_Y(y) u(z-xy) \, dy \,dx

\end{align}

where $u(\backslash cdot)$ is the Heaviside step function and serves to limit the region of integration to values of $x$ and $y$ satisfying $xy\backslash leq\; z$.

We find the desired probability density function by taking the derivative of both sides with respect to $z$.

: $\backslash begin\{align\}$

f_Z(z) & = \int^\infty_{-\infty} \int^\infty_{-\infty} f_X(x) f_Y(y) \delta(z-xy) \, dy \,dx\\

& = \int^\infty_{-\infty} f_X(x) \left[\int^\infty_{-\infty} f_Y(y) \delta(z-xy) \, dy \right]\,dx\\

& = \int^\infty_{-\infty} f_X(x) f_Y(z/x) \frac{1}

\end{align}

where we utilize the translation and scaling properties of the Dirac delta function $\backslash delta$.

A more intuitive description of the procedure is illustrated in the figure below. The joint pdf $f\_X(x)\; f\_Y(y)$ exists in the $x$-$y$ plane and an arc of constant $z$ value is shown as the shaded line. To find the marginal probability $f\_Z(z)$ on this arc, integrate over increments of area $dx\backslash ,dy\; \backslash ;f(x,y)$ on this contour.

File:WikiPicProd.jpg

Starting with $y=\; \backslash frac\{z\}\{x\}$, we have $dy\; =\; -\backslash frac\{z\}\{x^2\}\; \backslash ,\; dx\; =\; -\backslash frac\{y\}\{x\}\; \backslash ,\; dx$. So the probability increment is $\backslash delta\; p\; =\; f(x,y)\; \backslash ,dx\backslash ,|dy|\; =\; f\_X(x)f\_Y(z/x)\; \backslash frac\; \{y\}$

Let $X\; \backslash sim\; f(x)$ be a random sample drawn from probability distribution $f\_x(x)$. Scaling $X$ by $\backslash theta$ generates a sample from scaled distribution $\backslash theta\; X\; \backslash sim\; \backslash frac\; \{1\}$

Letting $\backslash theta$ be a random variable with pdf $f\_\backslash theta\; (\; \backslash theta\; )$, the distribution of the scaled sample becomes $f\_X\; (\; \backslash theta\; x\; )\; =\; g\_X(x\backslash mid\; \backslash theta)\; f\_\backslash theta\; (\backslash theta)$ and integrating out $\backslash theta$ we get $h\_x(x)\; =\; \backslash int\_\{-\backslash infty\}^\backslash infty\; g\_X(x|\; \backslash theta)\; f\_\backslash theta\; (\backslash theta)\; d\backslash theta$ so $\backslash theta\; X$ is drawn from this distribution $\backslash theta\; X\; \backslash sim\; h\_X(x)$. However, substituting the definition of $g$ we also have

$h\_X(x)\; =\; \backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash frac\{1\}$

For the case of one variable being discrete, let $\backslash theta$ have probability $P\_i$ at levels $\backslash theta\_i$ with $\backslash sum\_i\; P\_i\; =\; 1$. The conditional density is $f\_X\; (\; x\; \backslash mid\; \backslash theta\_i\; )\; =\; \backslash frac\; \{1\}$

When two random variables are statistically independent, the expectation of their product is the product of their expectations. This can be proved from the law of total expectation:

:$\backslash operatorname\{E\}(X\; Y)\; =\; \backslash operatorname\{E\}\; (\; \backslash operatorname\{E\}\; (X\; Y\; \backslash mid\; Y))$

In the inner expression, {{mvar|Y}} is a constant. Hence:

:$\backslash operatorname\{E\}\; (X\; Y\; \backslash mid\; Y)\; =\; Y\backslash cdot\; \backslash operatorname\{E\}[X\backslash mid\; Y]$

:$\backslash operatorname\{E\}(X\; Y)\; =\; \backslash operatorname\{E\}\; (\; Y\backslash cdot\; \backslash operatorname\{E\}[X\backslash mid\; Y])$

This is true even if {{mvar|X}} and {{mvar|Y}} are statistically dependent in which case $\backslash operatorname\{E\}[X\backslash mid\; Y]$ is a function of {{mvar|Y}}. In the special case in which {{mvar|X}} and {{mvar|Y}} are statistically

independent, it is a constant independent of {{mvar|Y}}. Hence:

:$\backslash operatorname\{E\}(X\; Y)\; =\; \backslash operatorname\{E\}(\; Y\backslash cdot\; \backslash operatorname\{E\}[X])$

:$\backslash operatorname\{E\}(X\; Y)\; =\; \backslash operatorname\{E\}(X)\; \backslash cdot\; \backslash operatorname\{E\}(Y)$

Let $X,\; Y$ be uncorrelated random variables with means $\backslash mu\_X,\; \backslash mu\_Y,$ and variances $\backslash sigma\_X^2,\; \backslash sigma\_Y^2$.

If, additionally, the random variables $X^2$ and $Y^2$ are uncorrelated, then the variance of the product XY is{{cite journal |last1=Goodman |first1=Leo A. |author-link1=Leo Goodman |date=1960 |title=On the Exact Variance of Products |jstor=2281592 |journal=Journal of the American Statistical Association |volume=55 |issue=292 |pages=708–713 |doi=10.2307/2281592}}

:$\backslash operatorname\{Var\}(XY)\; =\; (\backslash sigma\_X^2\; +\; \backslash mu\_X^2\; )(\backslash sigma\_Y^2\; +\; \backslash mu\_Y^2\; )\; -\backslash mu\_X^2\; \backslash mu\_Y^2$

In the case of the product of more than two variables, if $X\_1\; \backslash cdots\; X\_n,\; \backslash ;\backslash ;\; n>2$ are statistically independent then{{Cite web |url=https://stats.stackexchange.com/q/52699 |title=Variance of product of multiple random variables |last=Sarwate |first=Dilip |date=March 9, 2013 |work=Stack Exchange }} the variance of their product is

:$\backslash operatorname\{Var\}(X\_1X\_2\backslash cdots\; X\_n)\; =\; \backslash prod\_\{i=1\}^n\; (\backslash sigma\_i^2\; +\; \backslash mu\_i^2\; )\; -\backslash prod\_\{i=1\}^n\; \backslash mu\_i^2$

Assume X, Y are independent random variables. The characteristic function of X is $\backslash varphi\_X(t)$, and the distribution of Y is known. Then from the law of total expectation, we have{{cite web |title=How to find characteristic function of product of random variables |date=January 3, 2013 |work=Stack Exchange |url=https://math.stackexchange.com/q/269579 }}

:

\\ & = \operatorname{E} ( \operatorname{E} (e^{itX Y} \mid Y))

\\ & = \operatorname{E} ( \varphi_X(tY))

\end{align}

If the characteristic functions and distributions of both X and Y are known, then alternatively, $\backslash varphi\_Z(t)\; =\; \backslash operatorname\{E\}(\backslash varphi\_Y(tX))$

also holds.

The Mellin transform of a distribution $f(x)$ with support only on $x\; \backslash ge\; 0$ and having a random sample $X$ is

:$\backslash mathcal\{M\}f(x)\; =\; \backslash varphi(s)=\backslash int\_0^\backslash infty\; x^\{s-1\}\; f(x)\; \backslash ,\; dx\; =\; \backslash operatorname\{E\}[\; X^\{s-1\}].$

The inverse transform is

:$\backslash mathcal\{M\}^\{-1\}\backslash varphi\; (s)\; =\; f(x)=\backslash frac\{1\}\{2\; \backslash pi\; i\}\; \backslash int\_\{c-i\; \backslash infty\}^\{c+i\; \backslash infty\}\; x^\{-s\}\; \backslash varphi(s)\backslash ,\; ds.$

if $X\; \backslash text\{\; and\; \}\; Y$ are two independent random samples from different distributions, then the Mellin transform of their product is equal to the product of their Mellin transforms:

: $$

\mathcal{M}_{XY}(s) = \mathcal{M}_X(s)\mathcal{M}_Y(s)

If s is restricted to integer values, a simpler result is

:$\backslash operatorname\{E\}[(XY)^n]\; =\; \backslash operatorname\{E\}[X^n]\; \backslash ;\; \backslash operatorname\{E\}[Y^n]$

Thus the moments of the random product $XY$ are the product of the corresponding moments of $X\; \backslash text\{\; and\; \}\; Y$ and this extends to non-integer moments, for example

:$\backslash operatorname\{E\}[\; \{\; (XY)^\{1/p\}\}]\; =\; \backslash operatorname\{E\}[X^\{1/p\}]\; \backslash ;\; \backslash operatorname\{E\}[\; Y^\{1/p\}].$

The pdf of a function can be reconstructed from its moments using the saddlepoint approximation method.

A further result is that for independent X, Y

:$\backslash operatorname\{E\}[X^pY^q]\; =\; \backslash operatorname\{E\}[X^p]\; \backslash operatorname\{E\}[Y^q]$

Gamma distribution example To illustrate how the product of moments yields a much simpler result than finding the moments of the distribution of the product, let $X,Y$ be sampled from two Gamma distributions, $f\_\{Gamma\}(x;\backslash theta,1)\; =\; \backslash Gamma(\backslash theta)^\{-1\}\; x^\{\; \backslash theta\; -1\}\; e^\{-x\}$ with parameters $\backslash theta\; =\; \backslash alpha,\; \backslash beta$

whose moments are

:$\backslash operatorname\{E\}[X^p]\; =\; \backslash int\_0^\backslash infty\; x^p\; \backslash Gamma\; (x,\; \backslash theta)\backslash ,dx\; =\; \backslash frac\; \{\backslash Gamma(\backslash theta+p)\}\{\backslash Gamma(\backslash theta)\; \}.$

Multiplying the corresponding moments gives the Mellin transform result

:$\backslash operatorname\{E\}[(XY)^p]\; =\; \backslash operatorname\{E\}[X^p]\; \backslash ;\; \backslash operatorname\{E\}[Y^p]\; =\; \backslash frac\{\; \backslash Gamma(\backslash alpha+p)\}\{\backslash Gamma(\backslash alpha)\}\; \backslash ;\; \backslash frac\{\; \backslash Gamma(\backslash beta+p)\}\{\backslash Gamma(\backslash beta)\}$

Independently, it is known that the product of two independent Gamma-distributed samples (~Gamma(α,1) and Gamma(β,1)) has a K-distribution:

:$f(z,\backslash alpha,\backslash beta)\; =\; 2\; \backslash Gamma(\backslash alpha)^\{-1\}\; \backslash Gamma(\backslash beta)^\{-1\}z^\{\backslash frac\{\backslash alpha+\backslash beta\}\{2\}-1\}K\_\{\backslash alpha-\backslash beta\}(2\backslash sqrt\; z)\; =\; \backslash frac\{1\}\{\backslash alpha\; \backslash beta\}f\_\{K\}\backslash left(\backslash frac\{z\}\{\backslash alpha\; \backslash beta\};1,\backslash alpha,\backslash beta\backslash right),\; \backslash ;\; z\backslash ge\; 0$

To find the moments of this, make the change of variable $y\; =\; 2\; \backslash sqrt\; z$, simplifying similar integrals to:

:$\backslash int\_0^\backslash infty\; z^p\; K\_\backslash nu\; (2\; \backslash sqrt\; z)\; \backslash ,\; dz\; =\; 2^\{-2p-1\}\; \backslash int\_0^\backslash infty\; y^\{2p+1\}\; K\_\backslash nu\; (y)\; \backslash ,\; dy$

thus

:$2\; \backslash int\_0^\backslash infty\; z^\{\backslash frac\{\backslash alpha+\backslash beta\}\{2\}-1\}\; K\_\{\backslash alpha-\backslash beta\}\; (2\; \backslash sqrt\; z)\; \backslash ,\; dz\; =$

2^{-(\alpha+\beta) - 2p+1} \int_0^\infty y^{(\alpha+\beta) + 2p -1} K_{\alpha-\beta}(y) \, dy

The definite integral

:$\backslash int\_0^\backslash infty\; y^\backslash mu\; K\_\backslash nu\; (y)\; \backslash ,dy\; =\; 2^\{\backslash mu\; -1\}\; \backslash Gamma\; \backslash left\; (\; \backslash frac\{1+\; \backslash mu\; +\; \backslash nu\}\{2\}\; \backslash right\; )\; \backslash Gamma\; \backslash left\; (\backslash frac\{1+\; \backslash mu\; -\; \backslash nu\}\{2\}\; \backslash right\; )$ is well documented and we have finally

: $\backslash begin\{align\}$

E[Z^p] & = \frac {2^{-(\alpha+\beta) - 2p+1} \; 2^{(\alpha+\beta) + 2p -1} }{\Gamma(\alpha) \; \Gamma(\beta)} \Gamma \left ( \frac{(\alpha+\beta+2p)+(\alpha-\beta)}{2} \right ) \Gamma \left( \frac{(\alpha+\beta+2p)-(\alpha-\beta)}{2} \right ) \\ \\

& = \frac{\Gamma( \alpha+ p) \, \Gamma(\beta + p)} {\Gamma(\alpha) \, \Gamma(\beta)}

\end{align}

which, after some difficulty, has agreed with the moment product result above.

If X, Y are drawn independently from Gamma distributions with shape parameters $\backslash alpha,\; \backslash ;\; \backslash beta$ then

:$\backslash operatorname\{E\}[X^pY^q]\; =\; \backslash operatorname\{E\}[X^p]\; \backslash ;\; \backslash operatorname\{E\}[Y^q]\; =\; \backslash frac\{\; \backslash Gamma(\backslash alpha+p)\}\{\backslash Gamma(\backslash alpha)\}\; \backslash ;\; \backslash frac\{\; \backslash Gamma(\backslash beta+q)\}\{\backslash Gamma(\backslash beta)\}$

This type of result is universally true, since for bivariate independent variables $f\_\{X,Y\}(x,y)\; =\; f\_X(x)\; f\_Y(y)$ thus

:

\\ & = \int_{x=-\infty}^\infty x^p \Big [ \int_{y=-\infty}^\infty y^q f_Y(y)\, dy \Big ] f_X(x) \, dx

\\ & = \int_{x=-\infty}^\infty x^p f_X(x) \, dx \int_{y=-\infty}^\infty y^q f_Y(y) \, dy

\\ & = \operatorname{E}[X^p] \; \operatorname{E}[Y^q]

\end{align}

or equivalently it is clear that $X^p\; \backslash text\{\; and\; \}\; Y^q$ are independent variables.

The distribution of the product of two random variables which have lognormal distributions is again lognormal. This is itself a special case of a more general set of results where the logarithm of the product can be written as the sum of the logarithms. Thus, in cases where a simple result can be found in the list of convolutions of probability distributions, where the distributions to be convolved are those of the logarithms of the components of the product, the result might be transformed to provide the distribution of the product. However this approach is only useful where the logarithms of the components of the product are in some standard families of distributions.

Let $Z$ be the product of two independent variables $Z=\; X\_1\; X\_2$ each uniformly distributed on the interval [0,1], possibly the outcome of a copula transformation. As noted in "Lognormal Distributions" above, PDF convolution operations in the Log domain correspond to the product of sample values in the original domain. Thus, making the transformation $u=\backslash ln\; (x)$, such that $p\_U(u)\; \backslash ,\; |du|\; =\; p\_X(x)\; \backslash ,\; |dx|$, each variate is distributed independently on u as

:.

and the convolution of the two distributions is the autoconvolution

:

Next retransform the variable to $z=e^y$ yielding the distribution

: $c\_2(z)\; =\; c\_Y(y)/|dz/dy|\; =\; \backslash frac\; \{-y\; e^y\; \}\{e^y\}\; =\; -y\; =\; \backslash ln\; (1/z)$ on the interval [0,1]

For the product of multiple (> 2) independent samples the characteristic function route is favorable. If we define $\backslash tilde\{y\}\; =\; -y$ then $c(\backslash tilde\{y\})$ above is a Gamma distribution of shape 1 and scale factor 1, $c(\backslash tilde\{y\})\; =\; \backslash tilde\{y\}\; e^\{-\backslash tilde\{y\}\}$ , and its known CF is $(1\; -\; i\; t)^\{-1\}$. Note that $|d\backslash tilde\{y\}|\; =\; |dy|$ so the Jacobian of the transformation is unity.

The convolution of $n$ independent samples from $\backslash tilde\{Y\}$ therefore has CF $(1\; -\; i\; t)^\{-n\}$ which is known to be the CF of a Gamma distribution of shape $n$:

:$c\_n\; (\backslash tilde\{y\})\; =\; \backslash Gamma(n)^\{-1\}\; \backslash tilde\{y\}^\{(n-1)\}\; e^\{-\backslash tilde\{y\}\}\; =\; \backslash Gamma(n)^\{-1\}\; (-y)^\{(n-1)\}\; e^y$.

Make the inverse transformation $z=e^y$ to extract the PDF of the product of the n samples:

:$f\_n(z)\; =\; \backslash frac\; \{c\_n(y)\}$

The following, more conventional, derivation from Stackexchange{{Cite web |url=https://math.stackexchange.com/q/659278 |title=product distribution of two uniform distribution, what about 3 or more |author=heropup |date=1 February 2014 |work=Stack Exchange }} is consistent with this result.

First of all, letting $Z\_2\; =\; X\_1\; X\_2$ its CDF is

:

\\ & = \int_{x=0}^z 1 dx + \int_{x=z}^1 \frac{z}{x} \, dx

\\ & = z - z\log z, \;\; 0 < z \le 1

\end{align}

The density of $z\_2\; \backslash text\{\; is\; then\; \}\; f(z\_2)\; =\; -\backslash log(z\_2)$

Multiplying by a third independent sample gives distribution function

: $$

\begin{align} F_{Z_3}(z) = \Pr \Big [Z_3 \le z \Big ] & = \int_{x=0}^1 \Pr \Big [ X_3 \le \frac{z}{x} \Big] f_{Z_2}(x) \, dx

\\ & = -\int_{x=0}^z \log(x) \, dx - \int_{x=z}^1 \frac{z}{x} \log(x) \,dx

\\ & = -z \Big (\log(z) - 1 \Big ) + \frac{1}{2}z \log^2 (z)

\end{align}

Taking the derivative yields

f_{Z_3}(z) = \frac{1}{2} \log^2 (z), \;\; 0 < z \le 1.

The author of the note conjectures that, in general,

File:UnifProd4.jpg

The figure illustrates the nature of the integrals above. The area of the selection within the unit square and below the line z = xy, represents the CDF of z. This divides into two parts. The first is for 0 < x < z where the increment of area in the vertical slot is just equal to dx. The second part lies below the xy line, has y-height z/x, and incremental area dx z/x.

The product of two independent Normal samples follows a modified Bessel function. Let $x,\; y$ be independent samples from a Normal(0,1) distribution and $z=xy$.

Then

:

The variance of this distribution could be determined, in principle, by a definite integral from Gradsheyn and Ryzhik,{{Cite book|title=Tables of Integrals, Series and Products|last1=Gradsheyn|first1=I S|last2=Ryzhik|first2=I M|publisher=Academic Press|year=1980|pages=section 6.561}}

:$$

\int_0^\infty x^\mu K_\nu (ax) \, dx = 2^{\mu - 1} a^{-\mu - 1} \Gamma \Big ( \frac{1 + \mu + \nu}{2} \Big ) \Gamma \Big ( \frac{1 + \mu - \nu}{2} \Big), \;\; a>0, \;\nu + 1 \pm \mu >0

thus

\operatorname{E}[Z^2] = \int_{-\infty}^\infty \frac{z^2 K_0 (|z|)}{\pi} \, dz = \frac{4}{\pi} \; \Gamma^2 \Big (\frac{3}{2} \Big ) = 1

A much simpler result, stated in a section above, is that the variance of the product of zero-mean independent samples is equal to the product of their variances. Since the variance of each Normal sample is one, the variance of the product is also one.

The product of two Gaussian samples is often confused with the product of two Gaussian PDFs. The latter simply results in a bivariate Gaussian distribution.

The product of correlated Normal samples case was recently addressed by Nadarajaha and Pogány.{{Cite journal|last1=Nadarajah|first1=Saralees|last2=Pogány|first2=Tibor|date=2015|title=On the distribution of the product of correlated normal random variables|journal=Comptes Rendus de l'Académie des Sciences, Série I |volume=354 |issue=2 |pages=201–204|doi=10.1016/j.crma.2015.10.019|doi-access=free}}

Let $X\; \backslash text\{,\; \}\; Y$ be zero mean, unit variance, normally distributed variates with correlation coefficient $\backslash rho\; \backslash text\; \{\; and\; let\; \}\; Z\; =\; XY$

Then

:$f\_Z(z)\; =\; \backslash frac\{1\}\{\backslash pi\; \backslash sqrt\{1-\backslash rho^2\}\; \}\backslash exp\; \backslash left\; (\; \backslash frac\{\; \backslash rho\; z\}\{\; 1\; -\; \backslash rho^2\; \}\; \backslash right\; )\; K\_0\; \backslash left\; (\; \backslash frac$

and

: $\backslash begin\{align\}$

p(z) & \rightarrow \frac{1}{\pi \sqrt{1-\rho^2} } \exp \left (\frac{\rho z}{1-\rho^2 } \right)

\sqrt{\frac{\pi (1-\rho^2)}{2z} } \exp\left (-\frac

& = \frac{1}{\sqrt{2\pi z } } \exp \Bigg (\frac{ -|z| +\rho z}{(1-\rho)(1+\rho) } \Bigg) \\

& = \frac{1}{\sqrt{2\pi z } } \exp \Bigg (\frac{ -z }{1+\rho } \Bigg), \;\; z>0 \\

& \rightarrow \frac{1}{\Gamma(\tfrac{1}{2})\sqrt{2 z } } e^{-\tfrac{ z }{2 } }, \;\; \text { as }\rho \rightarrow 1 \\

\end{align}

which is a Chi-squared distribution with one degree of freedom.

Multiple correlated samples. Nadarajaha et al. further show that if $Z\_1,\; Z\_2,..Z\_n\; \backslash text\{\; are\; \}\; n$ iid random variables sampled from $f\_Z(z)$ and $\backslash bar\{Z\}\; =\; \backslash tfrac\{1\}\{n\}\; \backslash sum\; Z\_i$ is their mean then

:$f\_\backslash bar\; Z(\; z)=\; \backslash frac\; \{n^\{n/2\}\; 2^\{-n/2\}\; \}\; \{\; \backslash Gamma\; (\backslash frac\; \{n\}\{2\}\; )\; \}\; |z|^\{n/2-1\; \}$

\exp \left( \frac{\beta-\gamma}{2}z \right) {W}_{0,\frac{1-n}{2}}(| z |), \;\; -\infty < z < \infty.

where W is the Whittaker function while $\backslash beta\; =\; \backslash frac\{n\}\{1\; -\; \backslash rho\},\; \backslash ;\backslash ;\; \backslash gamma\; =\; \backslash frac\{n\}\{1\; +\; \backslash rho\}$

.

Using the identity $W\_\{0,\backslash nu\}(x)=\; \backslash sqrt\; \{\backslash frac\{x\}\{\backslash pi\}\}K\_\backslash nu(x/2),\; \backslash ;\backslash ;\; x\; \backslash ge\; 0$, see for example the DLMF compilation. eqn(13.13.9),{{Cite web|url=https://dlmf.nist.gov|title=Digital Library of Mathematical Functions|last=Equ(13.18.9)|website=NIST: National Institute of Standards and Technology}} this expression can be somewhat simplified to

:$f\_\backslash bar\; z(\; z)=\; \backslash frac\; \{n^\{n/2\}\; 2^\{-n/2\}\; \}\; \{\; \backslash Gamma\; (\backslash frac\; \{n\}\{2\}\; )\; \}\; |z|^\{n/2-1\; \}\; \backslash exp\; \backslash left\; (\backslash frac\{\backslash beta-\backslash gamma\}\{2\}z\; \backslash right\; )\; \backslash sqrt\{\; \backslash frac\{\; \backslash beta+\backslash gamma\; \}\{\backslash pi\}\; |z|\; \}\; \backslash ;$

K_{ \frac {1-n}{2}} \left ( \frac{\beta+\gamma}{2} |z| \right ), \;\; -\infty < z < \infty.

The pdf gives the marginal distribution of a sample bivariate normal covariance, a result also shown in the Wishart Distribution article. The approximate distribution of a correlation coefficient can be found via the Fisher transformation.

Multiple non-central correlated samples. The distribution of the product of correlated non-central normal samples was derived by Cui et al.{{Cite journal|last=Cui|first=Guolong|date=2016|title=Exact Distribution for the Product of Two Correlated Gaussian Random Variables|doi=10.1109/LSP.2016.2614539|journal=IEEE Signal Processing Letters|volume=23 |issue=11|pages=1662–1666|bibcode=2016ISPL...23.1662C|s2cid=15721509 }} and takes the form of an infinite series of modified Bessel functions of the first kind.

Moments of product of correlated central normal samples

For a central normal distribution N(0,1) the moments are

: $$

\operatorname{E}[X^p] =

\frac{1}{\sigma \sqrt{2\pi}}\int_{-\infty}^\infty x^p \exp (-\tfrac {x^2}{2\sigma^2}) \, dx=

\begin{cases}

0 & \text{if }p\text{ is odd,} \\

\sigma^p (p-1)!! & \text{if }p\text{ is even.}

\end{cases}

where $n!!$ denotes the double factorial.

If $X,Y\; \backslash sim\; \backslash text\{Norm\}(0,1)$ are central correlated variables, the simplest bivariate case of the multivariate normal moment problem described by Kan,{{Cite journal |last=Kan |first=Raymond |date=2008 |title=From moments of sum to moments of product |journal=Journal of Multivariate Analysis |volume=99 |issue=3 |pages=542–554 |doi=10.1016/j.jmva.2007.01.013 |doi-access=free }} then

: $$

\operatorname{E}[X^pY^q] =

\begin{cases}

0 & \text{if }p+q \text{ is odd,} \\

\frac {p!q!}{2^{\tfrac{p+q}{2} } }\sum_{k=0}^t \frac { (2\rho)^{2k} }{\Big(\frac{p}{2}- k \Big)! \;

\Big(\frac{q}{2}- k \Big)! \;(2k)! } & \text{if } p \text{ and } q \text{ are even} \\

\frac {p!q!}{2^{\tfrac{p+q}{2} } }\sum_{k=0}^t \frac { (2\rho)^{2k+1} }{\Big(\frac{p-1}{2}- k \Big)! \;

\Big(\frac{q-1}{2}- k \Big)! \; (2k+1)! } & \text{if } p \text{ and } q \text{ are odd}

\end{cases}

where

:$\backslash rho$ is the correlation coefficient and $t\; =\; \backslash min([p,q]/2)$

[needs checking]

The distribution of the product of non-central correlated normal samples was derived by Cui et al. and takes the form of an infinite series.

These product distributions are somewhat comparable to the Wishart distribution. The latter is the joint distribution of the four elements (actually only three independent elements) of a sample covariance matrix. If $x\_t,\; y\_t$ are samples from a bivariate time series then the $W\; =\; \backslash sum\_\{t=1\}^K\; \backslash dbinom\; \{x\_t\}\{y\_t\}\; \{\backslash dbinom\; \{x\_t\}\{y\_t\}\; \}^T$ is a Wishart matrix with K degrees of freedom. The product distributions above are the unconditional distribution of the aggregate of K > 1 samples of $W\_\{2,1\}$.

Let $u\_1,\; v\_1,\; u\_2,\; v\_2$ be independent samples from a normal(0,1) distribution.
Setting

$z\_1\; =\; u\_1\; +\; i\; v\_1\; \backslash text\{\; and\; \}\; z\_2\; =\; u\_2\; +\; i\; v\_2\; \backslash text\{\; then\; \}\; z\_1,\; z\_2$ are independent zero-mean complex normal samples with circular symmetry. Their complex variances are $\backslash operatorname\{Var\}\; |z\_i|\; =\; 2.$

The density functions of

:$r\_i\; \backslash equiv\; |z\_i|\; =\; (u\_i^2\; +\; v\_i^2)^\{\backslash frac\{1\}\{2\}\},\; \backslash ;\backslash ;i=1,2$ are Rayleigh distributions defined as:

:$f\_r(r\_i)\; =\; r\_i\; e^\{-r\_i^2\; /\; 2\}\; \backslash text\{\; of\; mean\; \}\; \backslash sqrt\{\backslash tfrac\{\backslash pi\}\{2\}\}\; \backslash text\{\; and\; variance\}$

\frac{4-\pi}{2}

The variable $y\_i\; \backslash equiv\; r\_i^2$ is clearly Chi-squared with two degrees of freedom and has PDF

:$f\_\{y\_i\}(y\_i)\; =\; \backslash tfrac\{1\}\{2\}\; e^\{-y\_i/2\}\; \backslash text\{\; of\; mean\; value\; \}\; 2$

Wells et al.{{Cite journal|last1=Wells|first1=R T|last2=Anderson|first2=R L|last3=Cell|first3=J W|date=1962|title=The Distribution of the Product of Two Central or Non-Central Chi-Square Variates|journal=The Annals of Mathematical Statistics|volume=33| issue = 3|pages=1016–1020|doi=10.1214/aoms/1177704469|doi-access=free}} show that the density function of $s\; \backslash equiv\; |z\_1\; z\_2|$ is

:$f\_s\; (s)\; =\; s\; K\_0(s),\; \backslash ;\backslash ;\; s\; \backslash ge\; 0$

and the cumulative distribution function of $s$ is

:$P(a)\; =\; \backslash Pr\; [s\; \backslash le\; a\; ]\; =\; \backslash int\_\{s=0\}^a\; s\; K\_0(s)\; ds\; =\; 1\; -\; a\; K\_1(a)$

Thus the polar representation of the product of two uncorrelated complex Gaussian samples is

:$f\_\{s,\backslash theta\}(s,\backslash theta)\; =\; f\_s(s)\; p\_\backslash theta(\backslash theta)\; \backslash text\{\; where\; \}\; p(\backslash theta)\; \backslash text\{\; is\; uniform\; on\; \}[0,2\backslash pi]$.

The first and second moments of this distribution can be found from the integral in Normal Distributions above

:$m\_1\; =\; \backslash int\_0^\backslash infty\; s^2\; K\_0\; (s)\; \backslash ,\; dx\; =\; 2\; \backslash Gamma^2\; (\; \backslash tfrac\{3\}\{2\}\; )\; =\; 2\; (\backslash tfrac\; \{\backslash sqrt\; \backslash pi\}\{2\})^2\; =\; \backslash frac\{\backslash pi\}\{2\}$

:$m\_2\; =\; \backslash int\_0^\backslash infty\; s^3\; K\_0\; (s)\; \backslash ,\; dx\; =\; 2^2\; \backslash Gamma^2\; (\; \backslash tfrac\{4\}\{2\}\; )\; =\; 4$

Thus its variance is $\backslash operatorname\{Var\}\; (s)\; =\; m\_2\; -\; m\_1^2\; =\; 4\; -\; \backslash frac\{\backslash pi^2\}\{4\}$.

Further, the density of $z\; \backslash equiv\; s^2\; =$

:$f\_Z(z)\; =\; \backslash tfrac\{1\}\{2\}\; K\_0(\backslash sqrt\{z\})\; \backslash text\{\; with\; expectation\; \}\; \backslash operatorname\{E\}(z)\; =\; 4$

Let $u\_1,\; v\_1,\; u\_2,\; v\_2,\; \backslash ldots,\; u\_\{2N\},\; v\_\{2N\},$ be $4N$ independent samples from a normal(0,1) distribution.Setting $z\_1\; =\; u\_1\; +\; i\; v\_1,\; z\_2\; =\; u\_2\; +\; i\; v\_2,\; \backslash ldots,\; \backslash text\{\; and\; \}\; z\_\{2N\}\; =\; u\_\{2N\}\; +\; i\; v\_\{2N\},$ then $z\_1,\; z\_2,\; \backslash ldots,\; z\_\{2N\}$are independent zero-mean complex normal samples with circular symmetry.

Let of $s\; \backslash equiv\; \backslash sum\_\{i=1\}^\{N\}z\_\{2i-1\}z\_\{2i\}$ , Heliot et al.{{Cite journal |last=Héliot |first=Fabien |last2=Tafazolli |first2=Rahim |date=2024 |title=On the Sum of Products of Independent Complex Normal Variables: Understanding the Fundamental SNR Gain Limit of MIMO-RIS |url=https://ieeexplore.ieee.org/document/10533263/ |journal=IEEE Transactions on Signal Processing |volume=72 |pages=2622–2636 |doi=10.1109/TSP.2024.3402345 |issn=1053-587X}} show that the joint density function of the real and imaginary parts of $s$ , denoted $s\_\backslash textrm\{R\}$ and $s\_\backslash textrm\{I\}$, respectively, is given by

$p\_\{s\_\{\backslash textrm\{R\}\},s\_\{\backslash textrm\{I\}\}\}(s\_\{\backslash textrm\{R\}\},s\_\{\backslash textrm\{I\}\})=\backslash frac\{2\backslash left(s\_\{\backslash textrm\{R\}\}^2+s\_\{\backslash textrm\{I\}\}^2\backslash right)^\{\backslash frac\{N-1\}\{2\}\}\}\{\backslash pi\backslash Gamma(n)\backslash sigma\_s^\{N+1\}\}K\_\{n-1\}\backslash !\backslash left(\backslash !2\backslash frac\{\backslash sqrt\{s\_\{\backslash textrm\{R\}\}^2+s\_\{\backslash textrm\{I\}\}^2\}\}\{\backslash sigma\_s\}\backslash right),$ where $\backslash sigma\_s$is the standard deviation of $s$. Note that $\backslash sigma\_s=1$ if all the $u\_i,\; v\_i$variables are normal(0,1).

Besides, they also prove that the density function of the magnitude of $s$, $|s|$, is

$p\_$

The first moment of this distribution, i.e. the mean of $|s|$, can be expressed as

$E\backslash \{|s|\backslash \}=\backslash sqrt\{\backslash pi\}\backslash sigma\_s\backslash frac\{\backslash Gamma(N+\backslash frac\{1\}\{2\})\}\{2\backslash Gamma(N)\},$ which further simplifies as $E\backslash \{|s|\backslash \}\backslash sim\backslash frac\{\backslash sigma\_s\backslash sqrt\{\backslash pi\; N\}\}\{2\},$when $N$is asymptotically large (i.e., $N\backslash rightarrow\backslash infty$) .

The product of non-central independent complex Gaussians is described by O’Donoughue and Moura{{Cite journal|last1=O’Donoughue|first1=N|last2=Moura|first2=J M F|date=March 2012|title=On the Product of Independent Complex Gaussians|journal=IEEE Transactions on Signal Processing|volume=60|issue=3|pages=1050–1063|doi=10.1109/TSP.2011.2177264|bibcode=2012ITSP...60.1050O|s2cid=1069298}} and forms a double infinite series of modified Bessel functions of the first and second types.

The product of two independent Gamma samples, $z\; =\; x\_1\; x\_2$, defining $\backslash Gamma(x;k\_i,\backslash theta\_i)\; =\; \backslash frac\; \{x^\{k\_i-1\}e^\{-x/\backslash theta\_i\}\}\{\backslash Gamma(k\_i)\; \backslash theta\_i^\{k\_i\}\; \}$, follows{{Cite web|url=https://math.stackexchange.com/q/2396324 |title=PDF of the product of two independent Gamma random variables|last=Wolfies|date=August 2017|website=stackexchange}}

:

K_{k_1 -k_2 } \left( 2 \sqrt{ \frac {z}{\theta_1 \theta_2} } \right) \\ \\

& = \frac {2}{\Gamma(k_1) \Gamma(k_2) } \frac {y^{\frac {k_1 + k_2}{2 }-1}}

{\theta_1 \theta_2 } K_{k_1 -k_2 } \left(2 \sqrt y \right) \text { where } y = \frac {z}{\theta_1 \theta_2}\\

\end{align}

Nagar et al.{{Cite journal|last1=Nagar|first1=D K|last2=Orozco-Castañeda|first2=J M|last3=Gupta|first3=A K|date=2009|title=Product and quotient of correlated beta variables|journal=Applied Mathematics Letters|volume=22|pages=105–109|doi=10.1016/j.aml.2008.02.014|doi-access=free}} define a correlated bivariate beta distribution

:

where

:$B(a,b,c)\; =\; \backslash frac\; \{\backslash Gamma(a)\; \backslash Gamma(b)\; \backslash Gamma(c)\}\{\; \backslash Gamma(a+b+c)\; \}$

Then the pdf of Z = XY is given by

:$f\_Z(z)\; =\; \backslash frac\; \{B(a+c,b+c)\; z^\{a-1\}\; (1-z)^\{c-1\}\; \}\; \{B(a,b,c)\; \}$

{_2F_1} (a+c,a+c; a+b+2c; 1-z), \;\;\; 0< z <1

where $\{\_2F\_1\}$ is the Gauss hypergeometric function defined by the Euler integral

:$\{\_2F\_1\}(a,b,c,z)\; =\; \backslash frac\; \{\backslash Gamma(c)\; \}\{\backslash Gamma(a)\; \backslash Gamma\; (c-a)\; \}$

\int_0^1 v^{a-1} (1-v)^{c-a-1} (1-vz)^{-b} \, dv

Note that multivariate distributions are not generally unique, apart from the Gaussian case, and there may be alternatives.

The distribution of the product of a random variable having a uniform distribution on (0,1) with a random variable having a gamma distribution with shape parameter equal to 2, is an exponential distribution.

{{cite book

|last1=Johnson |first1=Norman L.

|last2=Kotz |first2=Samuel

|last3=Balakrishnan |first3= N.

|title=Continuous Univariate Distributions Volume 2, Second edition

|url=http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471584940.html

|access-date=24 September 2012

|page=306

|year=1995

|publisher=Wiley

|isbn=978-0-471-58494-0

}} A more general case of this concerns the distribution of the product of a random variable having a beta distribution with a random variable having a gamma distribution: for some cases where the parameters of the two component distributions are related in a certain way, the result is again a gamma distribution but with a changed shape parameter.

The K-distribution is an example of a non-standard distribution that can be defined as a product distribution (where both components have a gamma distribution).

The product of n Gamma and m Pareto independent samples was derived by Nadarajah.{{Cite journal|last=Nadarajah|first=Saralees|date=June 2011|title=Exact distribution of the product of n gamma and m Pareto random variables|journal=Journal of Computational and Applied Mathematics|volume=235|issue=15|pages=4496–4512|doi=10.1016/j.cam.2011.04.018|doi-access=free}}

• Algebra of random variables
• Sum of independent random variables

{{Reflist}}

• {{cite journal

| doi = 10.1137/0118065

| title = The distribution of products of beta, gamma and Gaussian random variables

| last1 = Springer

| first1 = Melvin Dale

| last2 = Thompson

| first2 = W. E.

| journal = SIAM Journal on Applied Mathematics

| volume = 18

| issue = 4

| pages = 721–737

| year = 1970

| jstor = 2099424

}}

• {{cite journal

| title = The distribution of products of independent random variables

| last1 = Springer

| first1 = Melvin Dale

| last2 = Thompson

| first2 = W. E.

| journal = SIAM Journal on Applied Mathematics

| pages = 511–526

| year = 1966

| jstor = 2946226

| volume = 14

| issue = 3

| doi = 10.1137/0114046

}}

{{DEFAULTSORT:Product Distribution}}

Category:Types of probability distributions

Category:Algebra of random variables