A GENERAL inequality for a bi-modal hypergeometric distribution

Question

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.

See my previous question. I got a great answer proving $P(X=d+1) > P(X=d-2)$. That got me wondering: can we make a more general statement? More specifically (for natural $c \leq d-2$):

$P(X=d+c) > P(X=d-1-c)$

This is true for $c = 1$, but also in many cases when $c \geq 2$. I have not found any counterexamples yet. Can this be proven? Or where to start?

Answer 1

You can turn the answer from the other question into an inductive proof for this question*.

$$\tfrac{P(X=d+c+1)}{P(X=d+c)}-\tfrac{P(X=d-c-2)}{P(X=d-c-1)} = \tfrac{(k-d-c)(n-d-c)}{(d+1+c) (m-k-n+d+1+c)} -\tfrac{ (d-c-1) (m-k-n+d-c-1)}{(k-d+c+2)(n-d+c+2)} \\= \tfrac{(k-d-c)(n-d-c)(k-d+c+2)(n-d+c+2)-(d-c-1) (m-k-n+d-c-1)(d+1+c) (m-k-n+d+1+c)}{(d+1+c) (m-k-n+d+1+c)(k-d+c+2)(n-d+c+2)}$$

again the denominator is positive, and we only need to show that the numerator is positive.

We can do the same steps, substituting $d=(k+1)(n+1)/(m+2)$ gives for the numerator:

$$(c+1)^2(m-2k)(m-2n)$$

which is positive when both $k< \frac{1}{2}m$ and $n < \frac{1}{2}m$.

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Some other interesting points

• For $c = 0$ you get the previous answer.

• For $c=-1$ you get $\frac{P(X=d)}{P(X=d-1)}-\frac{P(X=d-1)}{P(X=d)} = 0$, which is true by the assumption $P(X=d) = P(X=d-1)$.

• Also for $n=\frac{1}{2}m$ you get that the term $(m-2n)$ equals zero and you get the symmetry $P(X=d+c) = P(X=d-c-1)$

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*If $\tfrac{P(X=d+c+1)}{P(X=d+c)}-\tfrac{P(X=d-c-2)}{P(X=d-c-1)}> 0$ and $$P(X=d+c) \geq P(X=d-1-c)$$ then $P(X=d+(c+1)) > P(X=d-1-(c+1))$