According to the table of conjugate distributions on Wikipedia, the hypergeometric distribution has as conjugate prior a beta-binomial distribution, where the parameter of interest is "$M$, the number of target members." I interpret "target members" to mean, I am modeling as hypergeometric the number of blue balls in a sample from an urn containing $N$ total balls of which $M$ are blue, $N-M$ non-blue.

But then I cannot make sense of the claim of conjugacy. After the data is observed, say $b$ blue balls in the sample, then it is known that $M>b$. But a beta-binomial distribution is supported on $0,...,F$ (for some parameter $F$). So how can the posterior for $M$ also be beta-binomial?

  • $\begingroup$ I may be wrong but it looks to me like that there is a slight misunderstanding. Hypergeometric distribution describes the probability of drawing $k$ blue balls out of $n$ draws (without replacement) from an urn containing $N$ balls out of which $M$ are blue. It is not the probability distribution over $M$, the number of blue balls in the urn. $\endgroup$ Oct 31, 2017 at 23:04
  • $\begingroup$ @MossMurderer No, this is a bayesian setup. $\endgroup$
    – Hasse1987
    Nov 1, 2017 at 18:59

2 Answers 2


The problem with the Wikipedia article and the reference therein (Fink D., 1997) is that there is some key information missing.

Specifically, the given posterior is for $M-x$ (i.e. the number of target individuals in the population shifted by the number observed in the sample), not for $M$. Furthermore, the posterior parameter corresponding to the number of observations is missing and should be $N-n$ (i.e. the population size minus the sample size). These two corrections fixes the support problem that you correctly noticed, as shown below.

Suppose that $0 \leq X \leq n$ is the number of target individuals in a sample of size $n$ from a population of size $N$ with $0 \leq M \leq N$ total target individuals.

Then, $X \sim \text{HG}(n, M, N)$ with support in $[\max(0, n-N+M), \min(n, M)]$.

If $M \sim \text{BB}(N, \alpha, \beta)$ is the prior distribution of $M$, the posterior distribution for $M - x$ is also Beta-Binomial-distributed: $$M - x\,|\,x,\alpha,\beta \sim \text{BB}(N-n, \alpha + x, \beta + n - x)$$

If you write the probability mass function for $M$ you will find @Tim's answer above.

As an illustration, for $N = 20$ and $n = 10$, let's assume a non-informative prior distribution for $M$ with $M \sim \text{BB}(N, .5, .5)$. Suppose that we observe $x = 9$.

N = 20
n = 10
a0 <- b0 <- .5
x <- 9
  m = 0:N
) %>% 
    prior = dbbinom(m, size = N, alpha = a0, beta = b0),
    post = dbbinom(m-x, size = N-n, a0+x, b0+n-x)
  ) %>% 
  gather(key, dens, -m) %>% 
  ggplot(aes(m, dens, col = key)) +
  geom_line() +

Created on 2018-10-10 by the reprex package (v0.2.1)

Note that the posterior support is correctly [x, N − n + x].

Dyer, D. and Pierce, R.L. (1993). On the Choice of the Prior Distribution in Hypergeometric Sampling. Communications in Statistics - Theory and Methods, 22(8), 2125-2146.


Hypergeometric distribution describes sampling without replacement from the urn containing $N$ balls, out of which $M \le N$ are target balls, let's say blue. The conjugate beta-binomial prior distribution leads to posterior distribution for unknown $M \in \{x,x+1,\dots,N-n+x\}$ in form

$$ f(M\mid x,N,\alpha,\beta) = {N-n \choose M-x} \frac{\Gamma(\alpha+M)\,\Gamma(\beta+N-x)\,\Gamma(\alpha+\beta+n)}{\Gamma(\alpha+x)\,\Gamma(\beta+n-x)\,\Gamma(\alpha+\beta+N)} $$

as described in

Dyer, D. and Pierce, R.L. (1993). On the Choice of the Prior Distribution in Hypergeometric Sampling. Communications in Statistics - Theory and Methods, 22(8), 2125-2146.

  • 1
    $\begingroup$ I agree it makes sense to have the prior distribution on $M$ be beta-binomial. What doesn't make sense to me is that the posterior is also beta-binomial (because of the different supports issue mentioned in the question). $\endgroup$
    – Hasse1987
    Oct 31, 2017 at 22:50
  • $\begingroup$ @Hasse1987 I don't understand what do you mean, the support is $0,1,\dots,N$, where BB distribution is parametrized by $\alpha,\beta,N$, where $N$ is known population size. $\endgroup$
    – Tim
    Oct 31, 2017 at 23:22
  • $\begingroup$ But the posterior distribution of $M$ cannot have that support. After the data is observed you know there are at least $b$ target members. So the support of the posterior is contained on $b,\ldots,N,$ right? $\endgroup$
    – Hasse1987
    Oct 31, 2017 at 23:27
  • $\begingroup$ @Hasse1987 a priori it has $0,\dots,N$ support, after you observe some data it gets centered on more realistic values. $\endgroup$
    – Tim
    Nov 1, 2017 at 11:46
  • $\begingroup$ Of course. The point is that those "more realistic values" preclude a beta-binomial distribution. $\endgroup$
    – Hasse1987
    Nov 1, 2017 at 18:59

Your Answer

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

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