0
$\begingroup$

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 $k$.

$\endgroup$
2
  • $\begingroup$ 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

2
$\begingroup$

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 $\alpha\rightarrow\infty$.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.