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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? This

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.

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? 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.

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.

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"Square root" of Beta(a,b) distribution

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? 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.