Say $X$ has a hypergeometric distribution with parameters $m$, $n$ and $k$, with $k\leq n<\frac12m$.

I know that $X$ has a dual mode if and only if $d=\frac{(k+1)(n+1)}{m+2}$ is integer. In that case $P(X=d)=P(X=d-1)$ equals the maximum probability.

I am wondering if I can say anything about $P(X=d+1)$ versus $P(X=d-2)$ then. When is the former higher than the latter? I.e. when is:


Always? I tried many combinations programmatically and did not find any counterexample.

So far I have found:


Because $d=\frac{(k+1)(n+1)}{m+2}$, this can be simplified to:


I have tried further combining this with $d=\frac{(k+1)(n+1)}{m+2}$ being integer, but that gets quite complex and gives me no further clue.

I feel there is something relatively easy to prove here...?

For $n=\frac12m$, $P(X=d+1)=P(X=d-2)$ due to symmetry.


In the case you are considering you have $P(X=d)=P(X=d-1)$

so let's consider the sign of $$\frac{P(X=d+1)}{P(X=d)}-\frac{P(X=d-2)}{P(X=d-1)} = \tfrac{(k-d)(n-d)}{(d+1) (m-k-n+d+1)} -\tfrac{ (d-1) (m-k-n+d-1)}{(k-d+2)(n-d+2)} \\= \tfrac{(k-d)(n-d)(k-d+2)(n-d+2)-(d-1) (m-k-n+d-1)(d+1) (m-k-n+d+1)}{(d+1) (m-k-n+d+1)(k-d+2)(n-d+2)}$$

The denominator is positive so does not affect the sign. In the numerator, we can make the substitution $d=\frac{(k+1)(n+1)}{m+2}$ and then multiply through by the positive $(m+2)^4$. Expanding the result and factorising gives

$$\tfrac{m^6 +(8-2n-2k)m^5 +(24-16n-16k+kn)m^4 +(32-48n-48k+32kn)m^3 +(16-64n-64k+96kn)m^2 +(-32n-32k+128kn)m +64kn}{\text{something positive}} \\ = \frac{(m+2)^4(m-2n)(m-2k)}{\text{something positive}} $$

and this is positive, i.e. $P(X=d+1) > P(X=d-2)$, when $k\leq n<\frac12m$. $\blacksquare$

As a check, the difference is actually $\frac{(m-2n)(m-2k)}{(d+1) (m-k-n+d+1)(k-d+2)(n-d+2)}$.

It is also positive when $n\lt k<\frac12m$, and when both $k > \frac12m$ and $n>\frac12m$.

It is negative , i.e. $P(X=d+1) < P(X=d-2)$ when $k\lt \frac12m <n$ or $n\lt \frac12m <k$.

Finally, it is zero, i.e. $P(X=d+1) = P(X=d-2)$ when $k= \frac12m$ or $n= \frac12m$.


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