The following expression arises in the study of hierarchical models. I suspect that the sum of the underlined $4$ terms become constant as $\alpha\rightarrow \infty$. Mathematica agrees when prompted with 'toy' versions, but I'm having some difficulty seeing how it generalizes.

I would greatly appreciate any help or observations.

\begin{align} \log \text{P}\propto \underbrace{ \log \Gamma(k \alpha) - \log \Gamma(k\alpha + length(\ell)) - k \log \Gamma(\alpha) + \sum^k_{i=1} \log \Gamma(c_{i} + \alpha)} \ + \ f(\text{other parameters}) \end{align} Where: $\ell$ is a list of items; Each item in $\ell$ must be assigned to exactly one bin, so $c_{i}$ is the count of items in $\ell$ assigned to bin $i$, and there are a total of $k$ bins.

Current thinking: It is well known that as $\alpha\rightarrow \infty$, the argument of the sum in the rightmost term $\log \Gamma(c_{i}+a)$ increases only linearly in its arguments rather than superlinearly. This is because $\displaystyle \lim_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial\alpha^2}\log \Gamma(c_{i} + \alpha)\big) = \lim_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial c_{i}^2}\log \Gamma(c_{i} + \alpha)\big) = \displaystyle \lim_{\alpha\rightarrow\infty} \big(\frac{\partial}{\partial c_{\ell,i}}\Psi(c_{i} + \alpha)\big)=0$, where $\Psi$ is the Digamma function. Therefore any partition of $\ell$ (i.e. any choices of the different $c_{i}$) will cause this term to sum to the same constant.

The other terms are all constant for fixed $\alpha$, $\ell$, and $k$.

  • $\begingroup$ in the third term, is the argument of the Gamma function $k\alpha+\text{length}(\ell)$ or is it $k\alpha+k\,\text{length}(\ell)$ ? $\endgroup$ Feb 23, 2023 at 8:49
  • $\begingroup$ Thank you -- question edited to clarify (that term's argument is (π‘˜π›Ό+π‘™π‘’π‘›π‘”π‘‘β„Ž(β„“)). $\endgroup$ Feb 23, 2023 at 9:31

1 Answer 1


For each $k>0$, $c\in\mathbb{C}$ it holds that $$\lim_{\alpha\rightarrow\infty}\frac{\Gamma(\alpha+c)}{\Gamma(\alpha)\alpha^c}=1\Rightarrow\lim_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha+c)}{\Gamma(k\alpha)}-c\log k\alpha\right)=0.$$ Apply this to $$I= - \log \frac{\Gamma(k\alpha + L)}{\Gamma(k\alpha)} + \sum^k_{i=1} \log\frac{\Gamma(\alpha+c_i )}{\Gamma(\alpha)} $$ and you find (using $\sum_{i=1}^k c_i=L$) that $$\lim_{\alpha\rightarrow\infty} I=-L\log k\alpha+\sum_{i=1}^k c_i\log\alpha=-L\log k.$$ So indeed, the function $\log P$ approaches an $\alpha$-independent limit for $\alpha\rightarrow\infty$.

  • $\begingroup$ Ahh of course it applies a second time -- thank you! May I ask where you found the very useful property $\lim_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha+c)}{\Gamma(k\alpha)}-c\log k\alpha\right)=0.$? $\endgroup$ Feb 23, 2023 at 13:25
  • $\begingroup$ this follows from the asymptotics $\Gamma(x+y)\rightarrow\Gamma(x)x^y$ for $x\rightarrow\infty$ at fixed $y\in\mathbb{C}$, see Wikipedia $\endgroup$ Feb 23, 2023 at 13:37
  • $\begingroup$ This makes sense--Thank you for this very helpful answer. $\endgroup$ Feb 23, 2023 at 13:52

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