How do I calculate the probability density function for a joint beta + uniform distribution? 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.
 A: 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 $(a,b):$

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.
