Suppose $Y \sim \text{Inverse Gamma}(\alpha, \beta)$ with scale parameter $\beta$ known, and $\alpha$ unknown, and the pdf is given by

$$f(y) = \frac{\exp(-1/\beta y)}{\Gamma(\alpha) \beta^\alpha y^{\alpha+1}}$$

What is the conjugate prior for the unknown shape parameter $\alpha$?

Here's what I'm thinking:

'Since the form $\frac{1}{\Gamma(\alpha)(\beta y)^\alpha}$ does not belong to any closed-formed kernel, there is no closed-formed conjugate prior for the unknown shape parameter here'.

Any thoughts on this, or link to references? Thanks!


1 Answer 1


The conjugate prior for the shape parameter for the gamma and inverse gamma are essentially of the same form, so you may have better luck looking for information on priors for the gamma distribution. (Alternatively you could take advantage of the gamma priors more directly by writing the model in terms of the inverse of the $y$'s. It's easy to flip back and forth.)

See also the discussion of the gamma with fixed shape near the bottom of this table, and also see Fink, D (1997) A Compendium of Conjugate Priors (p16-17) which give the form $\frac{1}{K} {\frac {a^{\alpha -1}\beta ^{\alpha c}}{\Gamma (\alpha )^{b}}}$ (which generalizes the form you have somewhat).

This is not a "standard closed form density", but there's no obvious reason why you can't deal with something of this form; you'll end up with a posterior of the same form so an integral of something like ${\frac {a^{\alpha -1}\beta ^{\alpha c}}{\Gamma (\alpha )^{b}}}$ will still come up.

If you are avoiding the need to integrate it in the posterior (e.g. by some sampling that avoids it) you could perhaps consider whether something of simpler form is sufficient for your needs. Consider, for example, noting that $\psi^{-\alpha}$ is an exponential (in slightly modified form). If an exponential prior does what you need in some instance, that might suffice in some particular situations.

  • $\begingroup$ Great to know this, thanks for the link. I'll look into and come back later. Appreciate your time! $\endgroup$
    – Demo
    Commented Sep 19, 2017 at 0:48

Your Answer

By clicking “Post 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.