# How to derive the bound formula of hypergeometric distribution?

The bound formula is:

$$H(M,N,n,k) \leq (( \frac{p}{p+t} )^{p+t} (\frac{1-p}{1-p-t})^{1-p-t})^n$$

So how to proof it?

Some papers discuss it: this one, that one.

They say the proof is in the following paper: The tail of the hypergeometric distribution. Discrete Mathematics

But I can not find the online version.

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The paper is online here, but you need a university subscription. dx.doi.org/10.1016/0012-365X(79)90084-0 – t.b. Feb 14 '11 at 12:56
Please don't post the same questions (more or less) simultaneously here and on MathOverflow. I suggest the following procedure: Ask here first and if you didn't get any helpful answer in a few days you might consider asking on MathOverflow. mathoverflow.net/questions/55402 – t.b. Feb 14 '11 at 12:58
Also you have the MR entry ams.org/mathscinet-getitem?mr=MR0534946 – Willie Wong Feb 14 '11 at 13:04
That, plus the fact that the inequality is a straightforward consequence of the usual exponential inequalities applied to sums of random variables. That is, coming back to the physical interpretation of the hypergeometric distribution, one considers $S_n=X_1+\cdots+X_n$ where $X_i$ is Bernoulli and $X_i=1$ iff the ball drawn at time $i$ is white. The only twist here is that the $X_i$s are not independent. Fortunately, they are negatively correlated, hence the upper bound one would get by assuming independence is still valid. – Did Feb 14 '11 at 13:36
@Fan: That would be absurd and is not what I wrote. Rather, for every negatively correlated $U$ and $V$, $E(UV)\le E(U)E(V)$. (And if you merely look at the definition, you will see that it is meaningless to say that a random variable is negatively correlated, on its own, all alone...) – Did Feb 14 '11 at 17:57