# Distribution of the product of a gamma random variable and a beta random variable

When you multiply a gamma random variable with a beta random variable, you should get a gamma random variable. I'm having a little trouble showing this, though. I figure I'm forgetting some clever integration trick.

Let the densities of $X$ and $Y$ be proportional to $x^{\alpha - 1}e^{-x\beta}$ and $y^{\delta -1}(1-y)^{\gamma - 1}$, respectively. We want the distribution of $Z = XY$. We need an auxilary random variable $U$ to do this. We have two choices for the auxiliary random variable $U$. With choice 1, we use the transformation

$\left[\begin{array}{c}Z \\ U \end{array}\right] = \left[\begin{array}{c}XY \\ X \end{array}\right]$, and we get $f_Z(z) = \int f_x(u)f_Y(z/u)\frac{1}{|z|}du = \int_z^{\infty} u^{\alpha-1}e^{-u\beta}(\frac{z}{u})^{\delta - 1}(1 - \frac{z}{u})^{\gamma - 1}z^{-1}du$.

Here's the other one: $\left[\begin{array}{c}Z \\ U \end{array}\right] = \left[\begin{array}{c}XY \\ Y \end{array}\right]$. Then we get $f_Z(z) = \int f_X(z/u)f_Y(u)\frac{1}{|u|}du = \int_0^1(\frac{z}{u})^{\alpha - 1} e^{-(\frac{z}{u}) \beta} u ^{\delta -1} (1-u)^{\gamma - 1}u^{-1}du$

This is about as far as I can get though. Any tips/tricks I should know?

• What makes you confident that this is true in the general case? Feb 12, 2015 at 1:43
• OP wrote: "When you multiply a gamma random variable with a beta random variable, you should get a gamma random variable." \\\\\\ This is not correct; it is easy to come up with a bimodal counter-example that is plainly not Gamma. Feb 12, 2015 at 2:58
• Can I see the example? It might give me a clue on how to integrate this Feb 12, 2015 at 16:02
• @Taylor Please see exact symbolic solution below, and Monte Carlo check. Feb 14, 2015 at 16:17
• Aug 13, 2019 at 14:16

• Let random variable $X \sim \text{Gamma}(a,b)$ with pdf $f(x)$:

• Let $Y \sim \text{Beta}(c, d)$ with pdf $g(y)$:

We seek the pdf of the product $Z = X*Y$, say $h(z)$, which is given by:

where I am using mathStatica's TransformProduct function to automate the nitty-gritties, and where Hypergeometric1F1 denotes the Kummer confluent hypergeometric function. All done. Note that this does not have the functional form of a Gamma rv.

PDF Plot

The pdf can take a range of possible shapes. Here is a plot to illustrate that it is plainly not Gamma:

• PDF Plot: $a= 3$, $b = 1$, $c = 0.7$ ... and ... $d = 0.1$

Monte Carlo check

Here is a quick Monte Carlo check of the exact symbolic solution derived above, just to make sure no errors have crept in, for the same parameter values:

The blue line is the empirical Monte Carlo pdf, and the red dashed line is the theoretical pdf $h(z)$ above. Looks fine :)

• Nice answer. The result only holds when the parameters are tied together in certain ways. For example, here's one situation: if $X \sim Gamma(a,b)$ and $Y \sim Beta(c(a), (1-c)a)$ with $0<c<1$. Nice looking software, btw. I'm definitely going to play around with this. Feb 16, 2015 at 2:10

Let $X$ and $Y$ be absolutely continuous, independent, and non-negative random variables such that $X$ has bounded support. Then any two of the following 3 conditions imply the third:

(i) $X\sim{}\text{Beta}(a,b)$

(ii) $Y\sim{}\text{Gamma}(a+b,c)$

(iii) $XY\sim{}\text{Gamma}(a,c)$

see Yeo and Milne (1991) at https://www.sciencedirect.com/science/article/pii/016771529190149L

This seems to imply that this is the most general such result.

Yeo, G.F. & Milne, R.K. (1991),
"On characterizations of beta and gamma distributions",
Statistics & Probability Letters, Vol 11, No 3, (March), pp239–242