Given that $\boldsymbol{\theta} \sim \mathrm{Dir}(\boldsymbol{\alpha})$, then $E_{p(\boldsymbol{\theta} \mid \boldsymbol{\alpha})}[\log{\theta_k}] = \Psi(\alpha_k) - \Psi(\sum_{k'=1}^K \alpha_{k'})$, How is $E_{p(\boldsymbol{\theta}\mid \boldsymbol{\alpha})}[\log{(a\cdot\theta_k + b)}]$ calculated, where a and b are constants?
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$\begingroup$ First $a$ is superfluous, since it can be exfiltrated from the logarithm, replacing $b$ by $b/a$. Second, $\theta_k$ is a Beta variate with parameters $\alpha_k$ and $\sum_{j\ne k}\alpha_j$, hence the question is truly about a Beta integral. Third, there is no reason that the expectation is closed form (if one considers the function $\Psi$ to be closed form. $\endgroup$– Xi'anCommented Apr 10 at 14:00
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3$\begingroup$ As @Xi'an mentions: a closed form is in the eye of the beholder. Mathematica gives $\frac{a}{b} \frac{\alpha_k}{\sum_{j=1}^K \alpha_j} \, _3F_2\left(1,1,\alpha_k +1;2,\left(\sum_{j=1}^K \alpha_j\right) +1;-\frac{a}{b}\right)+\log(b)$ for the expectation when $b\neq 0$. $\endgroup$– JimBCommented Apr 10 at 16:05
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