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I need some help and this is certainly not a home work question, as it appears in my research. If $X$ is a beta distributed random variable, is there a known distribution for \begin{equation} \begin{split} Y & = r.(ln(X)-X) \end{split} \end{equation}

where $r$ is a constant? If not, is there at least an analytic form for the distribution of Y? I tried to derive one, but got stuck when I had to solve for $e^{y/r} = xe^{-x}$. Any help will be appreciated.

Thanks!

user2167741
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    How are you dealing with the case where $X<0$, which is unavoidable since $X$ is normal? – measure_theory Aug 10 '17 at 15:44
  • Thanks for pointing the error! Sorry, in that case $Y = r(ln(-X)+X)$. I will change it. – user2167741 Aug 10 '17 at 15:58
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    Are you trying to solve $e^{y/r} = xe^{-x}$ for $x$? You can use the Lambert W function $W$: if I didn't make a mistake, the solution to $\exp(cx)x=a$ is given by $x=W(ca)/c$, where $c=-1$ and $ a=e^{y/r} $. – user3658307 Aug 11 '17 at 17:06
  • I think this is super useful! Thank you so much. I think this will help in writing an analytic distribution for $Y$. You rock! – user2167741 Aug 11 '17 at 19:09

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From user3658307 insightful reply above, we have $x = -W(-e^{y/r})$, where $W$ is the Lambert $W$ function. Since the pdf of beta distribution is described by $$f(x;\alpha,\beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1}$$ the pdf of $Y$ should technically follow: $$f(y;\alpha,\beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \left(-W(-e^{y/r})\right)^{\alpha-1}\left(1+W(-e^{y/r})\right)^{\beta-1}$$

user2167741
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