# What can we say about hypergeometric distribution with unknown $N$?

Hypergeometric distribution describes outcome of $n$ draws without replacement from the urn containing $K$ white balls and $N-K$ black balls. Binomial distribution describes outcome of $n$ draws with replacement with probability of drawing white ball equal to $p$. As total number of balls in the urn increases, hypergeometric distribution converges to binomial.

What if we are drawing $n$ balls from the urn of unknown size with probability $p=\tfrac{K}{N}$ ? Expected value is obvious no matter of $N$, since it is simply $np$. Can we say anything more about the distribution of such draws besides noticing that the lower boundary for it is hypergeometric distribution with $N = n$ and upper boundary is a binomial distribution?

The question is not about simply estimating $N$ since it can be estimated from the data like any other parameter, but it is rather about what can we say about the distribution nonetheless $N$ is unknown?

• Could you elaborate on the difference between "estimating" and "making educated guesses"? – whuber Nov 1 '16 at 15:33
• @whuber I was thinking of situation where you'd need to make some guesses before seeing the data or when seeing only a very small sample, so your estimates would not be very reliable. – Tim Nov 1 '16 at 15:52
• "Guesses before seeing the data" are usually called a prior distribution, while "guesses ... when seeing only a very small sample" are still called estimates. – whuber Nov 1 '16 at 16:27
• @whuber right, I removed this notion not to make it ambiguous. – Tim Nov 1 '16 at 16:38
• Are you interested in somehow estimating $N$? – Alex R. Nov 1 '16 at 18:41

The simplest way to answer this question is just to state the probability mass function that occurs in the general case, and look at its properties with respect to $N$ (i.e., look at the corresponding likelihood function for the parameter $N$). The probability mass function for the hypergeometric distribution is:

$$\text{HG}(k|N,K,n) = \frac{{K \choose k} {N-K \choose n-k}}{{N \choose n}} \overset{N}{\propto} \frac{(N-K)!(N-n)!}{N!(N-K -n + k)!}.$$

It is possible to investigate all sorts of properties of the above function, and there are all sorts of things we could potentially say about it. One interesting property of this probability mass function is that it is quasi-concave in $N$ (see below). From this result the MLE for that parameter is:

$$\left \lceil{\tfrac{nK}{k}-1}\right \rceil \leqslant \hat{N} \leqslant \left \lfloor{\tfrac{nK}{k}}\right \rfloor.$$

(This gives a unique solution if $\tfrac{nK}{k} \notin \mathbb{N}$, and it gives two solutions if $\tfrac{nK}{k} \in \mathbb{N}$.) Hence, we see that the MLE estimates $N$ in a way that roughly equalises $\tfrac{n}{N} \approx \tfrac{k}{K}$, with convergence as $n \rightarrow \infty$. There is probably plently more we could say about this distribution. It would be interesting to look at what happens once you combine it with various Bayesian priors on $N$. I will leave it to others to give other insights, but the above seems like an interesting aspect of the distribution to me.

Proof of quasi-concavity: For simplicity, we denote $\phi(N) = \text{HG}(k|N,K,n)$ and consider the other variables to be implicit (fixed) values. We can then write:

\begin{aligned} \frac{\phi(N+1)}{\phi(N)} - 1 &= \frac{(N-K+1)(N-n+1)}{(N+1)(N-K-n+k+1)} - 1 \\[6pt] &= \frac{(N-K+1)(N-n+1) - (N+1)(N-K-n+k+1)}{(N+1)(N-K-n+k+1)} \\[6pt] &= \frac{(n-k)(N+1)-n(N-K+1)}{(N+1)(N-K-n+k+1)} \\[6pt] &= \frac{(nN+n-kN-k)-(nN-nK+n)}{(N+1)(N-K-n+k+1)} \\[6pt] &= \frac{nK-k(N+1)}{(N+1)(N-K-n+k+1)} \\[6pt] &= \frac{\tfrac{n}{k/K}-(N+1)}{k(N+1)(N-K-n+k+1)}. \\[6pt] \end{aligned}

Hence, we have:

$$\text{sgn} \Big( \phi(N+1)-\phi(N) \Big) = \text{sgn} \Big( \frac{\phi(N+1)}{\phi(N)} - 1 \Big) = \text{sgn} \Big( \tfrac{n}{k/K} - (N+1) \Big).$$

Hence, we see that the mass function is increasing for $N+1 < \tfrac{n}{k/K}$ and then decreasing for $N+1 > \tfrac{n}{k/K}$. This means that the probability mass function is quasi-concave in $N$.