Suppose a random variable X follows the hypergeometric distribution with parameters $N$, $K$, $n$, where the pmf is given as

\begin{equation} Pr(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}. \end{equation}

The question is: how closed is the probability that $X$ is even to $\frac12$, i.e. $\vert Pr(X~is~ even) - \frac12 \vert$?

As mentioned in wolfies' answer, it really depends on the parameters $N$, $K$, $n$. My conjecture is as follows:

Let $Pr_r(even)$ be the probability that the RV is even under the parameters $K_r$, $N_r$, $n_r$. If we assume $K_r= c_1 N_r$, $n_r = c_2 N_r$, for some universal constants $c_1$, $c_2$, then \begin{equation} Pr_r(even) \rightarrow 0.5,~as~N_r\rightarrow \infty. \end{equation} The reason I believe that this is true is that the hypergeometric distribution will converge to binomial distribution as $N \rightarrow \infty$, and in a binomial distribution $Pr(even)=0.5$. If this conjecture is true, the question then becomes: what is the rate of convergence?


1 Answer 1


The answer must be ---- it depends!

To make this clear, the Hypergeometric can have forms where there is a sizeable mass at 0 (which is even). And if you have say 70% of the density mass at $X = 0$, then it is clear that $P(X \text{ is even})$ is not going to be close to $\frac12$. To illustrate, here is a plot of the Hypergeometric pmf when $N = 200$, $n = 10$, and $r = 5$ (whatever notation one uses):

(source: tri.org.au)

If desired, one can formally derive, in say Mathematica, an expression for the sum of the pmf over even values: Sum[pmf, 0, n, 2] which returns a complicated mess involving HypergeometricPFQ functions ... but it comes back to the point made above: it depends on your parameter values.

  • $\begingroup$ Do you mean PDF instead of HypergeometricPFQ? $\endgroup$
    – Sycorax
    Commented Apr 20, 2016 at 20:06
  • $\begingroup$ Thank you so much for your answer! I did not realize that the probability can be far away from 0.5. I modified the question to make it more concrete. $\endgroup$ Commented Apr 20, 2016 at 20:11
  • $\begingroup$ Mathematica yields a "complicated mess" because it doesn't know any better. The pmf for the even terms in any discrete distribution supported on the integers having pmf $p$ is simply $t\to (p(t)+p(-t))/2.$ In this case, $p$ can be expressed as a Riemann hypergeometric function $_2F_1.$ $\endgroup$
    – whuber
    Commented Jun 1, 2019 at 16:34

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