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This is a follow-up to the question Square root of a Beta(1,1) random variable, which received two great answers.

If $XY \sim \text{Beta}(\alpha, \beta)$, and $X$ and $Y$ are two independent identically distributed random variables, do we know their distribution? If there is no simple general solution, can we figure out the distribution of $X$ and $Y$ for the case $XY \sim \text{Beta}(1, 1)$?

From my search so far I guess this is a deconvolution problem for which no readily worked-out solution, such as the Kumaraswamy distribution in the previous question, exists. However, I would appreciate to be proven wrong in this regard.

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In the case of $Z=XY\sim \text{Beta}(\alpha,1)$, the moment generating function (mgf) of $-\ln(XY)=-\ln X-\ln Y$ is \begin{align} M_{-\ln(XY)}(t) &= E(e^{-t\ln Z}) \\ &=E(Z^{-t}) \\ &=\int_0^1 z^{-t}\alpha z^{\alpha-1}dz \\ &= \frac{1}{1-t/\alpha} \end{align} which is the mgf of an exponentially distributed random variable with rate parameter $\alpha$.

Thus, if $X$ and $Y$ are iid, this means that $-\ln X$ and $-\ln Y$ must be Gamma distributed with shape parameter $1/2$ and rate parameter $\alpha$ (see wikipedia). Backtransforming, the pdf of $X$ and $Y$ is $$ f(x)=\sqrt{-\frac\alpha{\pi\ln x}}x^{\alpha-1} $$ for $0\le x\le 1$.

A similar calculation in the general Beta$(\alpha,\beta)$-case leads to $$ M_{-\ln(XY)}(t)=\frac{\Gamma(\alpha+t)\Gamma(\alpha+\beta)}{\Gamma(\alpha+\beta+t)\Gamma(\alpha)} $$ and the mgf of $-\ln X$ and $-\ln Y$ would need to be the square root of this. But that is not the mgf of any known distribution and such a distribution (that when convolved with itself produces the target density) may not even exist.

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  • $\begingroup$ This looks extremely useful. Can you maybe add a hint, how you derived the relationships in the first paragraph? Or how they are called? $\endgroup$ – LuckyPal May 4 at 9:08
  • $\begingroup$ @LuckyPal I added some more explanation and made the result slightly more general. $\endgroup$ – Jarle Tufto May 4 at 9:37

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