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The PDF of a Beta distribution is $$ f_X(x) = {{x^{a-1}(1-x)^{b-1}}\over {B(a,b)}} $$ and CDF $$ F_X(x) = I_x(a,b) $$

The PDF of a uniform distribution is $$ f_Y(y) = {1 \over {b-a}} \space for \space x \in [a,b] $$ and CDF $$ { x - a} \over {b - a} $$

I've know that if X and Y are two independent, continuous random variables, described by probability density functions above of $f_X(x)$ and $f_Y(x)$ then probability density function of $Z = XY$ is $$ f_Z(z) = \int_\infty^\infty f_X(x)f_Y({z\over x}){1\over abs(x)}dx $$ but I'm struggling with the algebra to get from this to the PDF and the CDF of the joint distribution. I'm using the uniform distribution between 0 and 1.

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  • $\begingroup$ Is the mod(x) in your equation supposed to be abs(x)? $\endgroup$ – R Carnell Feb 4 at 13:43
  • $\begingroup$ yes - sorry I've amended $\endgroup$ – andrew Feb 4 at 14:58
  • $\begingroup$ Although your question is unclear--you never specify what the distributions of $X$ and $Y$ are--the title suggests they are both Beta distributions, indicating the duplicate thread provides an answer. $\endgroup$ – whuber Feb 4 at 15:31
  • $\begingroup$ sorry - amended to make clear that i'm looking for the joint PDF and CDF of a uniform distribution and a beta distribution $\endgroup$ – andrew Feb 4 at 15:48
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    $\begingroup$ Because the duplicate at stats.stackexchange.com/questions/3359/… has no explicit answers and because the uniform distribution (which is a Beta(1,1) distribution) is special, conceivably a simple, effective answer could be provided in this special case, so I have voted to reopen this question. $\endgroup$ – whuber Feb 5 at 15:59
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The density of a $B(a,b)$ variable $X$ is $$f(x,a,b) = \frac{1}{B(a,b)}x^{a-1}(1-x)^{b-1}$$

for $0\le x\le 1.$ Let $F(x,a,b)$ be its distribution function

$$F(x,a,b) = \int_{-\infty}^x f(x,a,b)\,\mathrm{d}x = \int_0^x f(x,a,b)\,\mathrm{d}x,$$

so that (as usual) $\frac{d}{dx} F(x,a,b) = f(x,a,b).$

When $U$ is a uniform variable on $[0,1]$ and $0\le z\le 1,$ $\Pr(U\le z) = z.$ Otherwise, when $z\gt 1,$ $\Pr(U\le z)=1.$ These (defining) formulas can be expressed as

$$\Pr(U\le z) = z\wedge 1$$

(the smaller of $z$ and $1$), provided $z\ge 0.$

To evaluate the distribution of $UX$ (where $U$ and $X$ are independent), note it's certain that $0\le UX \le 1.$ So let $z$ be in this range and, assuming $a\gt 1,$ compute

$$\begin{aligned} F_{UX}(z) &= \Pr(UX\le z) = \Pr\left(U\le \frac{z}{X}\right) = E\left[\frac{z}{X}\wedge 1\right]\\ &= \frac{1}{B(a,b)}\int_z^1 \frac{z}{x} x^{a-1}(1-x)^{b-1}\,\mathrm{d}x + \int_0^z (1)f(x,a,b)\,\mathrm{d}x\\ &= \frac{z B(a-1,b)}{B(a,b)} (1-F(z,a-1,b)) + F(z,a,b)\\ &= \frac{z (a+b-1)}{a-1} (1-F(z,a-1,b)) + F(z,a,b). \end{aligned}$$

That determines the CDF of $UX$ (by definition). To obtain its PDF, differentiate with respect to $z,$ giving

$$f_{UX}(z) = \frac{d}{dz}F_{UX}(z) = \frac{a+b-1}{a-1}\left(1 - F(z,a-1,b) - z f(z,a-1,b)\right) + f(z,a,b).$$

Here are the results of four separate simulations of one million values of $UX$ for various values of $(a,b):$

Figure

The red curves plot $f_{UX}:$ they agree perfectly with the simulations.


Since this question appears to be for educational purposes, I leave the evaluation of the case $0\lt a \le 1,$ should that be of interest, as an exercise. Note that even in this case $UX$ exists, and there is no problem evaluating integrals from $z$ to $1$ even when the exponent of $x$ (equal to $a-2$ in the formula) is less than $-1;$ however, such integrals do not correspond to Beta distributions and need to be expressed in terms of the Incomplete Beta function.

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