Statistics/Distributions/Hypergeometric

Hypergeometric Distribution

Hypergeometric

Notation	$h(k)={{{m \choose k}{{N-m} \choose {n-k}}} \over {N \choose n}}$
Parameters	???
Support	???
	???
CDF	???
Mean	${nm \over N}$
Median	???
Mode	???
	${nm \over N}\left(1-{n \over N}\right)\left(1-{m-1 \over N-1}\right)$
Skewness	???
Ex. kurtosis	???
Entropy	???
MGF	???
CF	???
PGF	???
Fisher information	???

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:

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

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 its 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:

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

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

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

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

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

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

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

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.

$\operatorname {E} [X]={mn \over N}\sum _{x'=0}^{n'}{{{m' \choose x'}{{N'-m'} \choose {n'-x'}}} \over {{N'} \choose {n'}}}$

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

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

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

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

$\operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}$

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

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

$\operatorname {Var} (X)={nm(N-n)(N-m) \over N^{2}(N-1)}$

or, equivalently,

$\operatorname {Var} (X)={nm \over N}\left(1-{n \over N}\right)\left(1-{m-1 \over N-1}\right)$