The likelihood of an observation $x$ under a gamma distribution is

$$L(x | \alpha, \beta) \propto \beta^\alpha x^{\alpha-1} \frac {\exp(-x\beta)} {\Gamma(\alpha)}$$

Suppose I have some observations d from a gamma distribution with unknown shape and reciprocal scale parameters $\alpha$ and $\beta$

I wish to evaluate $ \int_{\forall \alpha} \int_{\forall \beta} L(x | \alpha, \beta) \times L(\alpha,\beta | \boldsymbol{d}) \ d\alpha \ d\beta $ in order to find the likelihood of a new observation $x$ from this distribution.

Is this integral tractable if I use Miller's prior (http://www.leg.ufpr.br/lib/exe/fetch.php/projetos:mci:tabelasprioris.pdf) for the distribution of the parameters of the gamma distribution, eg:

$$L(\alpha,\beta | \boldsymbol{d}) \propto \frac {p^{\alpha - 1} \exp(-\beta s)} {(\Gamma(\alpha)\beta^{-\alpha})^n} $$

Where $n$ is the number of observations, $s$ is their sum and $p$ is their product?

In other words, is it possible to exactly calculate

$$\int_0^\infty \int_0^\infty \beta^\alpha x^{\alpha-1} \frac {\exp(-x\beta)} {\Gamma(\alpha)} \frac {p^{\alpha - 1} \exp(-\beta s)} {(\Gamma(\alpha)\beta^{-\alpha})^n} \ d\alpha \ d\beta$$

If this is not possible analytically, do I have to resort to maximum likelihood estimation, or is there another approach that will be viable? Such as an alternative prior? Or numerical techniques?

  • 2
    $\begingroup$ The $\beta$ integral is straightforward to evaluate, obviously producing a multiple of $\Gamma(1+(n+1)\alpha)$. For any $n\gt 0$, the ratio of that to $\Gamma(\alpha)$ blows up so fast that the resulting integral over $\alpha$ diverges. $\endgroup$ – whuber Oct 4 '15 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.