I'm interested in the distribution of the logit transform of a $Beta(\alpha, \beta)$ random variable; that is, the distribution of $X$ where $Y\sim Beta(\alpha, \beta)$ and $X = {\rm logit}(Y)$. I've derived the PDF of this distribution as

$$p_X(x;\alpha,\beta) = \frac{1}{B(\alpha,\beta)} \left(\frac{1}{1 + e^{-x}}\right)^\alpha \left(1 - \frac{1}{1 + e^{-x}}\right)^\beta,$$ where $B(\alpha, \beta)$ is the complete beta function. To get the expected value of this distribution, I use the Wikipedia derivation of the geometric mean of the beta distribution, recognizing that

$${\rm logit}(x) = \log(x) - \log(1-x).$$

This gives me $E[X;\alpha,\beta] = \psi(\alpha) - \psi(\beta)$, where $\psi$ is the digamma function. In reviewing the Wikipedia derivation, I notice a peculiar substitution: $$\int_0^1 \log x\ x^{\alpha-1} (1-x)^{\beta-1} dx = \frac{\partial}{\partial\alpha}\int_0^1 x^{\alpha-1} (1-x)^{\beta-1} dx.$$ This step has no citation in the article. Can anyone point me to a resource which shows the derivation of this identity?

It should go without saying that when the identity is established, computing the mean and variance becomes straightforward.


1 Answer 1


I just realized this shortly after posting: the missing step outlined in the question depends on the identity $$\frac{\partial}{\partial\alpha} x^{\alpha-1} = \log(x) x^{\alpha-1},$$ which is easily proven.


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