Suppose that $X$ has the beta distribution Beta$(1,K-1)$ and $Y$ follows a chi-squared with $2K$ degrees. In addition, we assume that $X$ and $Y$ are independent.

What is the distribution of the product $Z=XY$ .

My attempt: \begin{align} f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{0}^{+\infty} \frac{1}{B(1,K-1)2^K \Gamma(K)} \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{0}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} [-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})]_0^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1,-z/2) \end{align}

Is it correct? if yes, how we call this distribution?

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    $\begingroup$ If this is homework or self-study, please add the appropriate tag. We don't (usually) solve such problems for you, but rather help guide you to a solution yourself, which in general will give you a better understanding of how to solve such problems in the future. $\endgroup$ – jbowman Nov 25 '15 at 16:38
  • $\begingroup$ Not sure but maybe this is of some help: en.wikipedia.org/wiki/Noncentral_beta_distribution $\endgroup$ – user95564 Nov 25 '15 at 16:40
  • $\begingroup$ Have you tried creating a second variable? Say $W=X+Y$? Then you could get the joint distribution of $W,Z$ and integrate out $W$ to get teh distribution of $Z$. $\endgroup$ – user95564 Nov 25 '15 at 16:42
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    $\begingroup$ I do not see where you are using the fact that the Beta density function is zero on the complement of the interval $[0,1]$. $\endgroup$ – whuber Nov 25 '15 at 19:44
  • $\begingroup$ @whuber I think I found the error. Would you like to provide a full answer or I do it by myself? $\endgroup$ – tam Nov 25 '15 at 20:00

After some valuable remarks, I was able to find the solution:

We have $f_X(x)=\frac{1}{B(1,K-1)} (1-x)^{K-2}$ and $f_Y(y)=\frac{1}{2^K \Gamma(K)} y^{K-1} e^{-y/2}$.

Also, we have $0\le x\le 1$. Thus, if $x=\frac{z}{y}$, we get $0 \le \frac{z}{y} \le 1$ which implies that $z\le y \le \infty$.

Hence: \begin{align} f_Z &= \int_{y=-\infty}^{y=+\infty}\frac{1}{|y|}f_Y(y) f_X \left (\frac{z}{y} \right ) dy \\ &= \int_{z}^{+\infty} \frac{1}{B(1,K-1)2^K \Gamma(K)} \frac{1}{y} y^{K-1} e^{-y/2} (1-z/y)^{K-2} dy \\ &= \frac{1}{B(1,K-1)2^K \Gamma(K)}\int_{z}^{+\infty} e^{-y/2} (y-z)^{K-2} dy \\ &=\frac{1}{B(1,K-1)2^K \Gamma(K)} \left[-2^{K-1}e^{-z/2}\Gamma(K-1,\frac{y-z}{2})\right]_z^\infty \\ &= \frac{2^{K-1}}{B(1,K-1)2^K \Gamma(K)} e^{-z/2} \Gamma(K-1) \\ &= \frac{1}{2} e^{-z/2} \end{align} where the last equality holds since $B(1,K-1)=\frac{\Gamma(1)\Gamma(K-1)}{\Gamma(K)}$.

So $Z$ follows an exponential distribution of parameter $\frac{1}{2}$; or equivalently, $Z \sim\chi_2^2$.


There is a pleasant, natural statistical solution to this problem for integral values of $K$, showing that the product has a $\chi^2(2)$ distribution. It relies only on well-known, easily established relationships among functions of standard normal variables.

When $K$ is integral, a Beta$(1,K-1)$ distribution arises as the ratio $$\frac{X}{X+Z}$$ where $X$ and $Z$ are independent, $X$ has a $\chi^2(2)$ distribution, and $Z$ has a $\chi^2(2K-2)$ distribution. (See the Wikipedia article on the Beta distribution for instance.)

Any $\chi^2(n)$ distribution is that of the sum of squares of $n$ independent standard Normal variates. Consequently, $X+Z$ is distributed as the squared length of a $2 + 2K-2 = 2K$ vector with a standard multinormal distribution in $\mathbb{R}^{2K}$ and $X/(X+Z)$ is the squared length of the first two components when that vector is radially projected to the unit sphere $S^{2K-1}$.

The projection of a standard multinormal $n$-vector onto the unit sphere has a uniform distribution because the multinormal distribution is spherically symmetric. (That is, it is invariant under the orthogonal group, a result that follows immediately from two simple facts: (a), the orthogonal group fixes the origin and by definition does not change covariances; and (b) the mean and covariance completely determine the multivariate normal distribution. I illustrated this for the case $n=3$ at https://stats.stackexchange.com/a/7984). In fact, the spherical symmetry immediately shows this distribution is uniform conditional on the length of the original vector. The ratio $X/(X+Z)$ therefore is independent of the length.

What all this implies is that multiplying $X/(X+Z)$ by an independent $\chi^2(2K)$ variable $Y$ creates a variable with the same distribution as $X/(X+Z)$ multiplied by $X+Z$; to wit, the distribution of $X$, which has a $\chi^2(2)$ distribution.

  • $\begingroup$ Very nice analogy! I feel a bit uncertain about the final paragraph though as the simplification only occur because $X+Z$ is on both sides of the multiplication, which cannot work for an independent $\chi^2(2K)$. $\endgroup$ – Xi'an Nov 26 '15 at 14:55
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    $\begingroup$ But after some further musing in the Paris métro, I realised that because $X/(X+Z)$ and $(X+Z)$ are independent, using $(X+Z)\times X/(X+Z)$ or using $Y \times X/(X+Z)$ lead to the same distribution. Congrats! $\endgroup$ – Xi'an Nov 26 '15 at 16:43
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    $\begingroup$ addendum: the reasoning goes for non-integer K's as well, if one defines a $\chi^2_q$ as a Gamma $\text{Ga}(q/2,1/2)$. $\endgroup$ – Xi'an Nov 26 '15 at 16:52
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    $\begingroup$ @Xi'an Thank you for those revealing comments. Indeed, one way to exploit the recognition that $X/(X+Z)$ and $X+Z$ are independent is to pursue the implication that their density functions will be separable: and that idea applies without modification to the general case of non-integral $K$. Even for those who prefer to compute the convolution $XY$ directly, these statistical insights suggest a simple and effective way of proceeding with the integration by means of an appropriate change of variables. $\endgroup$ – whuber Nov 27 '15 at 14:59

I greatly deprecate the commonly used tactic of finding the density of $Z = g(X,Y)$ by computing first computing the joint density of $Z$ and $X$ (or $Y$) because it is "easy" to use Jacobians, and then getting $f_Z$ as a marginal density (cf. Rusty Statistician's answer). It is much easier to find the CDF of $Z$ directly and then differentiate to find the pdf. This is the approach used below.

$X$ and $Y$ are independent random variables with densities $f_X(x) = (K-1)(1-x)^{K-2}\mathbf 1_{(0,1)}(x)$ and $f_Y(y) = \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2}\mathbf 1_{(0,\infty)}(y)$. Then, with $Z = XY$, we have for $z > 0$, \begin{align} P\{Z > z\} &= P\{XY > z\}\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2} \left[\int_{x=\frac{z}{y}}^1 (K-1)(1-x)^{K-2}\,\mathrm dx \right] \,\mathrm dy\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} y^{K-1} e^{-y/2} \left(1-\frac{z}{y}\right)^{K-1}\,\mathrm dy\\ &= \int_{y=z}^\infty \frac{1}{2^K (K-1)!} (y-z)^{K-1} e^{-y/2} \,\mathrm dy\\ &= e^{-z/2}\int_0^\infty \frac{1}{2^K (K-1)!}t^{K-1} e^{-t/2} \,\mathrm dy~~~\scriptstyle{\text{on setting}~y-z = t}\\ &= e^{-z/2}\qquad\scriptstyle{\text{on noting that the integral is that of a Gamma pdf}}\\ \end{align}

It is well-known that if $V \sim \mathsf{Exponential}(\lambda)$, then $P\{V > v\} = e^{-\lambda v}$. It follows that $Z = XY$ has an exponential density with parameter $\lambda = \frac 12$, which is also the $\chi^2(2)$ distribution.


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