Solved – Distribution of the product of a gamma random variable and a beta random variable

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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?

Best Answer

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

enter image description here

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

enter image description here

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

enter image description here

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 ... $d = 0.1$

enter image description here

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:

enter image description here

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

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