Under what conditions on a and b is there a distribution $f\_{a,b}$ such that the product $XY$ of two independent realizations $X$ and $Y$ from $f\_{a,b}$ has a Beta(a,b) distribution?


A standard result on deriving the distibution of the product of two variables indicates that the p.d.f. $f\_{a,b}$ needs to satisfy:
$\int\_0^1 f\_{a,b}(x) f\_{a,b}(\frac{v}{x}) \frac{1}{x} dx = \frac{v^{a-1} (1-v)^{b-1}}{B(a,b)}$, but I have no idea how to find such an $f\_{a,b}$. (B denotes the beta function).


This article ( <http://www.jstor.org/stable/2045709> ) might be relevant for the case a=b=1, but my background in the relevant maths is not strong enough to understand it at all.



"Square root" of Beta(a,b) distribution
A partial answer is that $f\_{a,b}$ exists for every positive integer $b$. 


To see this, first recall that for every positive $s$ and $a$ the distribution Gamma$(s,a)$ has density proportional to $z^{s-1}e^{-az}$ on $z\ge0$ and that the sum of independent Gamma$(s,a)$ and Gamma$(t,a)$ is Gamma$(s+t,a)$.


If $b=1$, choose $X=e^{-Z}$ and $Y=e^{-T}$ where $Z$ and $T$ are i.i.d. and Gamma$(1/2,a)$. Then $X$ and $Y$ are i.i.d. and $XY=e^{-(Z+T)}$ where $Z+T$ is Gamma$(1,a)$, that is, exponential of parameter $a$, hence $XY$ is Beta$(a,1)$ and you are done if $b=1$.


From there, recall that the product of two independent Beta$(a,c)$ and Beta$(a+c,b-c)$ is Beta$(a,b)$ for every $c$ in $(0,b)$. Assume that $b$ is a positive integer and choose $X=e^{-Z\_1-Z\_2-\cdots-Z\_b}$ and $Y=e^{-T\_1-T\_2-\cdots-T\_b}$ where all the $Z\_k$ and $T\_i$ are independent and, for each $k$, $Z\_k$ and $T\_k$ are both Gamma$(1/2,a+k-1)$. Then, by the $b=1$ case, each $e^{-(Z\_k+T\_k)}$ is Beta$(a+k-1,1)$ hence $XY$ is the product of some independent Beta$(a,1)$, Beta$(a+1,1)$, ..., Beta$(a+b-1,1)$, hence it is Beta$(a,b)$ and you are done for every positive integer $b$.


