# Statistics/Distributions/Hypergeometric

## Contents

### Hypergeometric Distribution

Notation $h(k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}$ ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???

The hypergeometric distribution describes the number of successes in a sequence of n draws without replacement from a population of N that contained m total successes.

Its probability mass function is:

$f(k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}\text{ for all }x \in[0,n]$

Technically the support for the function is only where x∈[max(0, n+m-N), min(m, n)]. In situations where this range is not [0,n], f(x)=0 since for k>0, ${0\choose k}=0$.

#### Probability Density Function

We first check to see that f(x) is a valid pmf. This requires that it is non-negative everywhere and that it's total sum is equal to 1. The first condition is obvious. For the second condition we will start with Vandermonde's identity

$\sum_{x=0}^n{a \choose x}{b \choose n-x}={a+b \choose n}$
$\sum_{x=0}^n{{a \choose x}{b \choose n-x} \over {a+b \choose n}}=1$

We now see that if a=m and b=N-m that the condition is satisfied.

#### Mean

We derive the mean as follows:

$\operatorname{E}[X] = \sum^n_{x=0} x \cdot f(x;n,m,N) = \sum^n_{x=0} x \cdot {{{m \choose x} {{N-m} \choose {n-x}}}\over {N \choose n}}$
$\operatorname{E}[X] = 0\cdot {{{m \choose 0} {{N-m} \choose {n-0}}}\over {N \choose n}}+\sum^n_{x=1} x \cdot {{{m \choose x} {{N-m} \choose {n-x}}}\over {N \choose n}}$

We use the identity $\binom{a}{b} = \frac{a}{b} \binom{a-1}{b-1}$ in the denominator.

$\operatorname{E}[X] = 0+\sum^n_{x=1} x \cdot {{{m \choose x} {{N-m} \choose {n-x}}}\over {{N \over n}{{N-1} \choose {n-1}}}}$
$\operatorname{E}[X] = {n \over N}\sum^n_{x=1} x \cdot {{{m \choose x} {{N-m} \choose {n-x}}}\over {{N-1} \choose {n-1}}}$

Next we use the identity $b \binom{a}{b} = a \binom{a-1}{b-1}$ in the first binomial of the numerator.

$\operatorname{E}[X] = {n \over N}\sum^n_{x=1} {m {{m-1 \choose x-1} {{N-m} \choose {n-x}}}\over {{N-1} \choose {n-1}}}$

Next, for the variables inside the sum we define corresponding prime variables that are one less. So N′=N−1, m′=m−1, x′=x−1, n′=n-1.

$\operatorname{E}[X] = {m n \over N}\sum^{n'}_{x'=0} {{{m' \choose x '} {{N'-m'} \choose {n'-x'}}}\over {{N'} \choose {n'}}}$
$\operatorname{E}[X] = {m n \over N}\sum^{n'}_{x'=0} f(x';n',m',N')$

Now we see that the sum is the total sum over a Hypergeometric pmf with modified parameters. This is equal to 1. Therefore

$\operatorname{E}[X] = {n m\over N}$

#### Variance

We first determine E(X2).

$\operatorname{E}[X^2] = \sum_{x=0}^n f(x;n,m,N) \cdot x^2 = \sum_{x=0}^n {{{m \choose x} {{N-m} \choose {n-x}}}\over {N \choose n}} \cdot x^2$
$\operatorname{E}[X^2] = {{{m \choose 0} {{N-m} \choose {n-0}}}\over {N \choose n}} \cdot 0^2+\sum_{x=1}^n {{{m \choose x} {{N-m} \choose {n-x}}}\over {N \choose n}} \cdot x^2$
$\operatorname{E}[X^2] = 0+\sum_{x=1}^n {{m {m-1 \choose x-1} {{N-m} \choose {n-x}}}\over {{N \over n}{{N-1} \choose {n-1}}}} \cdot x$
$\operatorname{E}[X^2] = {mn \over N} \sum_{x=1}^n {{{m-1 \choose x-1} {{N-m} \choose {n-x}}}\over {{{N-1} \choose {n-1}}}} \cdot x$

We use the same variable substitution as when deriving the mean.

$\operatorname{E}[X^2] = {mn \over N} \sum_{x'=0}^{n'} {{{m' \choose x'} {{N'-m'} \choose {n'-x'}}}\over {{{N'} \choose {n'}}}} (x'+1)$
$\operatorname{E}[X^2] = {mn \over N} \left[\sum_{x'=0}^{n'} {{{m' \choose x'} {{N'-m'} \choose {n'-x'}}}\over {{{N'} \choose {n'}}}} x'+\sum_{x'=0}^{n'} {{{m' \choose x'} {{N'-m'} \choose {n'-x'}}}\over {{{N'} \choose {n'}}}}\right]$

The first sum is the expected value of a hypergeometric random variable with parameteres (n',m',N'). The second sum is the total sum that random variable's pmf.

$\operatorname{E}[X^2] = {mn \over N} \left[{n'm' \over N'}+1\right]$
$\operatorname{E}[X^2] = {mn \over N} \left[{(n-1)(m-1) \over (N-1)}+1\right]={mn \over N} \left[{{(n-1)(m-1) +(N-1)}\over (N-1)}\right]$

We then solve for the variance

$\operatorname{Var}(X) = \operatorname{E}[X^2]-(\operatorname{E}[X])^2$
$\operatorname{Var}(X) = {mn \over N} \left[{{(n-1)(m-1) +(N-1)}\over (N-1)}\right]-\left({mn \over N}\right)^2$
$\operatorname{Var}(X) = {Nmn \over N^2} \left[{{(n-1)(m-1) +(N-1)}\over (N-1)}\right]-{(N-1)(mn)^2 \over (N-1)N^2}$
$\operatorname{Var}(X) = {nm(N-n)(N-m)\over N^2(N-1)}$