# (not asympotic) relation between hypergeometric and binomial distribution

Let $H$ be a random variable with hypergeometric distribution of parameters $n,h,r$ (that is $n$ is the total number of elements, $h$ elements are white and I choose $r$ elements).

Let $B$ be a random variable with binomial distribution of parameters $r$, $h/n$ (that is $r$ independent trials with success probability $h/n$)

Is it true that for any $k$ $$Pr(H\ge k)\ge Pr(B\ge k)?$$

Thanks!

Since $H$ and $B$ have the same expected value, $$\sum_{k} Pr(H\ge k) - Pr(B \ge k) = E[H] - E[B] = 0$$ In any case where the distributions are not the same, your inequality can't be true for all $k$.
• Is the statement valid for $k$ beyond the mean? Since the Chernoff tail bounds for binomial apply to hypergeometric (negative correlation), it seems like binomial should have a heavier tail than hypergeometric. – elexhobby Nov 21 '14 at 17:32