# An inequality for a bi-modal hypergeometric distribution

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:

$$P(X=d+1)>P(X=d-2)$$

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

So far I have found:

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

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

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

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 $$k\lt \frac12m or $$n\lt \frac12m .

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

• Exactly what I need! Subtracting works much better than dividing here indeed... – Michel de Ruiter Apr 7 '20 at 11:03
• – Michel de Ruiter Apr 29 '20 at 16:03