The Hypergeometric Distribution and Its Properties

Definition

$X\sim{H(n,K,N)}$, X is said to obey the hypergeometric distribution.

$$\boxed{P(X=k)=\frac{C_{K}^{k}C_{N-K}^{n-k}}{C_{N}^{n}}(k=0,1,2,...,n)}$$

Preliminary knowledge

Commonly used combined identities

1. $C_{n}^{k}=\frac{n}{k}C_{n-1}^{k-1}=\frac{n(n-1)}{k(k-1)}C_{n-2}^{k-2}=...$
2. $\sum_{k=0}^{r}C_{m}^{k}C_{n}^{r-k}=C_{m+n}^{r}$

Proof 1:

$$\begin{gathered} C_{n}^{k}=\frac{n!}{k!(n-k)!}\\=\frac{n}{k}\frac{(n-1)!}{(k-1)![(n-1)-(k-1)]!}\\=\frac{n}{k}C_{n-1}^{k-1}=\frac{n(n-1)}{k(k-1)}C_{n-2}^{k-2} \end{gathered}$$

Proof 2:

Consider this scenario: select r people $(r\le \min{m,n})$among m boys and n girls. It is not difficult to find that 2 is the result of counting twice.

To add, 2 is exactly the famous Vandermonde Identity

Properties of random variables

$$\boxed{D(X)=E(X^2)-[E(X)]^2}$$

Starting from the definition of variance: $$\begin{gathered} D(X)=\sum_{i=0}^{n}p_i[X_i-E(X)]^2\\ =\sum_{i=0}^n(p_iX_i^2)-2E(X)\sum_{i=0}^n(p_iX_i)+[E(X)]^2\\ =E(X^2)-[E(X)]^2 \end{gathered}$$

Properties

1. $E(X)=n\frac{K}{N}$
2. $D(X)=n\frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1}$

$$\begin{gathered} E(X)=\sum_{i=0}^n[P(X=i)i]\\ =\sum_{i=0}^n[\frac{C_{K}^{i}C_{N-K}^{n-i}}{C_{N}^{n}}i]\\ \text{apply combinatorial identity 1 to eliminate }i\\ =\sum_{i=1}^n[\frac{K}{i}\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}i]\\ =K\sum_{i=1}^n[\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}]\\ \text{apply combinatorial identity 2}\\ =\frac{K}{C_{N}^n}C_{N-1}^{n-1}\\ \text{apply combinatorial identity 1}\\ =\frac{nK}{N} \end{gathered}$$

$$\begin{gathered} E(X^2)=\sum_{i=0}^n[P(X=i)i^2]\\ =\sum_{i=0}^n[\frac{C_{K}^{i}C_{N-K}^{n-i}}{C_{N}^{n}}i^2]\\ \text{apply combinatorial identity 1 to eliminate }i\\ =K\sum_{i=1}^n[\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}i]\\ =K\sum_{i=1}^n[(i-1+1)\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}]\\ =K\sum_{i=1}^n[(i-1)\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}+\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}]\\ \text{apply combinatorial identity 2}\\ =K\sum_{i=1}^n[(i-1)\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}]+K\frac{C_{N-1}^{n-1}}{C_{N}^{n}}\\ =K\sum_{i=1}^n[(i-1)\frac{C_{K-1}^{i-1}C_{N-K}^{n-i}}{C_{N}^{n}}]+\frac{Kn}{N}\\ \text{apply combinatorial identity 1 to eliminate }i-1\\ =K\sum_{i=2}^n[(K-1)\frac{C_{K-2}^{i-2}C_{N-K}^{n-i}}{C_{N}^{n}}]+\frac{Kn}{N}\\ \text{apply combinatorial identity 2}\\ =K[(K-1)\frac{C_{N-2}^{n-2}}{C_{N}^{n}}]+\frac{Kn}{N}\\ =K(K-1)\frac{n(n-1)}{N(N-1)}+\frac{Kn}{N}\\ =\frac{Kn}{N}[\frac{(K-1)(n-1)}{N-1}+1]\\ D(X)=E(X^2)-[E(X)]^2\\ =\frac{Kn}{N}[\frac{(K-1)(n-1)}{N-1}+1]-(\frac{Kn}{N})^2\\ =\frac{Kn}{N}[\frac{N^2-(K+n)N+Kn}{N(N-1)}]\\ =n\frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1} \end{gathered}$$

Summary

At this point, we have completely derived the distribution law based on only two combination identities.

$$E(X)=n\frac{K}{N},\qquad D(X)=n\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}.$$

There are two key points in the whole process: one is to repeatedly use identity 1 to reduce the combination number and eliminate $i$in the summation term with $i$; the other is to use the Vandermonde identity to merge the summation into a single combination number. The reason why $i^2$is split into $(i-1)+1$when calculating the variance is precisely to create $i(i-1)$, a structure that can be reduced to two orders again. In essence, it is to find the factorial moment $E[X(X-1)]$.

Note $p=K/N$, and the result can be written as $E(X)=np$, $D(X)=np(1-p)\cdot\dfrac{N-n}{N-1}$. Compared with $np(1-p)$of the binomial distribution, the extra factor $\dfrac{N-n}{N-1}$is called the finite population correction factor - it is less than 1, which depicts the fact that sampling without replacement makes the variance smaller; when $N\gg n$, the factor tends to 1, and the hypergeometric distribution is approximated as a binomial distribution.