How can I compute $\mathbb{E}_{q}\Big[\log (1-x^a)\Big]$ when the distribution of $q$ is given as $q(x)\sim\mathrm{Beta}(\alpha,\beta)$?
1 Answer
Mathematica can do nothing with this expectation in general:
So, it is highly unlikely that this expectation can be expressed in terms of elementary, or even special, functions.
However, we have this:
and this:
Also, writing $\ln(1-x^a)=-\sum_{k=1}^\infty\dfrac{x^{ak}}k$, we see that the expectation in question is $$-\frac{\Gamma (\alpha+\beta )} {\Gamma (\alpha)}\sum_{k=1}^\infty\frac{\Gamma (a k+\alpha )}{k\,\Gamma (a k+\alpha +\beta )}.$$ It is highly unlikely as well that the latter sum can be expressed in terms of elementary, or even special, functions:
Of course, one can compute this expectation numerically with any degree of accuracy. E.g., we have this:
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$\begingroup$ @losifPinelis What about some sort of Taylor expansion? $\endgroup$– DalekCommented Oct 20, 2021 at 16:42
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$\begingroup$ @Dalek : I have added the use of a Taylor expansion. $\endgroup$ Commented Oct 20, 2021 at 17:15
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$\begingroup$ @Dalek : Do you have a further response to this answer? $\endgroup$ Commented Oct 24, 2021 at 1:56
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$\begingroup$ When $a = 1$, how does $\mathbb{E}[log (1 - x)]$ become that difference of polygamma functions? $\endgroup$ Commented Feb 15, 2022 at 23:44
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$\begingroup$ @RylanSchaeffer : This was done by Mathematica, not me. However, this can be done by hand too, by differentiating (with respect to the second argument) both the defining expression for the beta function (en.wikipedia.org/wiki/Beta_function) and its expression in terms of the gamma function. $\endgroup$ Commented Feb 16, 2022 at 3:11