0

I am working on a problem using a multivariate hypergeometric likelihood. The multivariate hypergeometric distribution does not belong to the exponential family of distributions, so (to my knowledge) we cannot guarantee that a conjugate prior exists.

However, Wikipedia claims on its page about conjugate priors that the univariate hypergeometric distribution is conjugate with a beta binomial. This post shows that the posterior in the univariate case is for M-x, the "number of target individuals in the population shifted by the number observed in the sample".

Does the multivariate hypergeometic distribution have a conjugate prior? My hunch is that it would be the multivariate form of the beta-binomial, the dirichlet-multinomial. However, I don't know how to show that analytically. If it does have a conjugate prior, what would the posterior's hyper-parameters be?

Xi'an
  • 90,397
  • 9
  • 157
  • 575
Ryan Folks
  • 1
  • 2
    What are the parameters of the multivariate hypergeometric on which you want to use a conjugate prior? Using [the standard definition](https://en.wikipedia.org/wiki/Hypergeometric_distribution#Multivariate_hypergeometric_distribution), with the $K_i$'s being the parameters there is no conjugate prior. – Xi'an Nov 01 '21 at 17:26
  • The $K_i$'s would be the parameters. Is there a simple explanation as to why? Or has one just not been found yet? – Ryan Folks Nov 01 '21 at 19:41
  • As you wrote, this distribution is not from an exponential family, hence cannot enjoy a conjugate prior. – Xi'an Nov 01 '21 at 19:42

0 Answers0