4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ What makes you confident that this is true in the general case? $\endgroup$ – Glen_b Feb 12 '15 at 1:43
  • 1
    $\begingroup$ 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. $\endgroup$ – wolfies Feb 12 '15 at 2:58
  • 1
    $\begingroup$ Can I see the example? It might give me a clue on how to integrate this $\endgroup$ – Taylor Feb 12 '15 at 16:02
  • 1
    $\begingroup$ @Taylor Please see exact symbolic solution below, and Monte Carlo check. $\endgroup$ – wolfies Feb 14 '15 at 16:17
  • $\begingroup$ related: stats.stackexchange.com/q/104390/8402 $\endgroup$ – Stéphane Laurent Aug 13 at 14:16
7
$\begingroup$
  • 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 :)

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Taylor Feb 16 '15 at 2:10
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.