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?

  • 2
    $\begingroup$ What are the parameters of the multivariate hypergeometric on which you want to use a conjugate prior? Using the standard definition, with the $K_i$'s being the parameters there is no conjugate prior. $\endgroup$
    – Xi'an
    Commented Nov 1, 2021 at 17:26
  • $\begingroup$ The $K_i$'s would be the parameters. Is there a simple explanation as to why? Or has one just not been found yet? $\endgroup$
    – Ryan Folks
    Commented Nov 1, 2021 at 19:41
  • $\begingroup$ As you wrote, this distribution is not from an exponential family, hence cannot enjoy a conjugate prior. $\endgroup$
    – Xi'an
    Commented Nov 1, 2021 at 19:42

1 Answer 1


According to this compendium of conjugate priors, the prior for the multivariate hypergeometric distribution's parameters is a Dirichlet-Multinomial distribution.

(The fact that the hypergeometric distribution does not belong to the exponential family does not preclude the possibility for conjugate prior. Simply, its existence is not guaranteed.)


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