超幾何分布
\(N\)個の要素からなる母集団の内,\(M\)個(\(M \leq N\))が\(1\)で,他が\(0\)だとする. これから,\(K\)個(\(K \leq N\))を非復元抽出(sampling without replacement)した時, その合計が\(X\)(\(K\)個取り出した内\(1\)が\(X\)個である)の確率を考える.
まず,\(N\)個の要素から\(K\)個取り出したときの組み合わせは,\(_N C _K\). \(1\)の\(M\)個内\(X\)を取り出す組み合わせは,\( _M C _X\). \(0\)の\(N-M\)個内\(K-X\)を取り出す組み合わせは,\( _{N-M} C _{K-X}\). から, \[ P(X=x:N,M,K) = \frac{\left(\begin{array}{c} M \\ x \end{array}\right)\left(\begin{array}{c} N-M \\ K-x \end{array}\right)}{\left( \begin{array}{c} N \\ K \end{array} \right)}, x=0,1,…,K \] となりこれを超幾何分布(hypergeometric distribution)と言う.
まず,\(\sum^{K} _{x=0} P(X=x:N,M,K) = 1\)を確かめる.
\[ (a+b)^N = (a+b)^{N-M}(a+b)^M \]
から
\[ \begin{align} \sum_{x=0}^N \left(\begin{array}{c} N \\ x \end{array}\right) a^xb^{x-N} &= \sum_{y=0}^{N-M} \left(\begin{array}{c} N-M-y \\ y \end{array}\right) a^y b^{N-M-y} \sum_{z=0}^{M} \left(\begin{array}{c} M \\ z \end{array}\right) a^z b^{M-z} \\ \sum_{y=0}^{N-M} \left(\begin{array}{c} N-M-y \\ y \end{array}\right) a^y b^{N-M-y} \sum_{z=0}^{M} \left(\begin{array}{c} M \\ z \end{array}\right) a^z b^{M-z} &= \left(\left(\begin{array}{c} N-M \\ 0 \end{array}\right)a^0 b^{N-M-0} + \left(\begin{array}{c} N-M \\ 1 \end{array}\right)a^1 b^{N-M-1} + \left(\begin{array}{c} N-M \\ 2 \end{array}\right)a^2 b^{N-M-2} + … + \left(\begin{array}{c} N-M \\ N-(M-2) \end{array}\right)a^{N-(M-2)} b^{2} + \left(\begin{array}{c} N-M \\ N-(M-1) \end{array}\right)a^{N-(M-1)} b^{1} + \left(\begin{array}{c} N-M \\ N-M \end{array}\right)a^{N-M} b^{0} \right) \left( \left(\begin{array}{c} M \\ 0 \end{array}\right)a^0 b^{M-0} + \left(\begin{array}{c} M \\ 1 \end{array}\right)a^1 b^{M-1} + \left(\begin{array}{c} M \\ 2 \end{array}\right)a^2 b^{M-2} + … + \left(\begin{array}{c} M \\ M-2 \end{array}\right)a^{M-2} b^{2} + \left(\begin{array}{c} M \\ M-1 \end{array}\right)a^{M-1} b^{1} + \left(\begin{array}{c} M \\ M \end{array}\right)a^{M} b^{0} \right) \\ &=\left(\left(\begin{array}{c} N-M \\ 0 \end{array}\right)a^0 b^{N-M-0}\right)\left( \left(\begin{array}{c} M \\ 0 \end{array}\right)a^0 b^{M-0} \right) +\left(\left(\begin{array}{c} N-M \\ 1 \end{array}\right)a^1 b^{N-M-1} \right)\left( \left(\begin{array}{c} M \\ 0 \end{array}\right)a^0 b^{M-0} \right) +\left(\left(\begin{array}{c} N-M \\ 0 \end{array}\right)a^0 b^{N-M-0}\right)\left(\left(\begin{array}{c} M \\ 1 \end{array}\right)a^1 b^{M-1} \right) +\left(\left(\begin{array}{c} N-M \\ 0 \end{array}\right)a^0 b^{N-M-0}\right)\left(\left(\begin{array}{c} M \\ 2 \end{array}\right)a^2 b^{M-2}\right) +\left(\left(\begin{array}{c} N-M \\ 1 \end{array}\right)a^1 b^{N-M-1}\right)\left(\left(\begin{array}{c} M \\ 1 \end{array}\right)a^1 b^{M-1}\right) +\left(\left(\begin{array}{c} N-M \\ 2 \end{array}\right)a^2 b^{N-M-2}\right)\left(\left(\begin{array}{c} M \\ 0 \end{array}\right)a^0 b^{M-0}\right)+…\\ &=\left(\left(\begin{array}{c} N-M \\ 0 \end{array}\right)\left(\begin{array}{c} M \\ 0 \end{array}\right)\right)a^0 b^{N-0}+ \left(\left(\begin{array}{c} N-M \\ 1 \end{array}\right)\left(\begin{array}{c} M \\ 0 \end{array}\right)+\left(\begin{array}{c} N-M \\ 0 \end{array}\right)\left(\begin{array}{c} M \\ 1 \end{array}\right)\right)a^1 b^{N-1}+ \left(\left(\begin{array}{c} N-M \\ 2 \end{array}\right)\left(\begin{array}{c} M \\ 0 \end{array}\right)+\left(\begin{array}{c} N-M \\ 1 \end{array}\right)\left(\begin{array}{c} M \\ 1 \end{array}\right)+\left(\begin{array}{c} N-M \\ 0 \end{array}\right)\left(\begin{array}{c} M \\ 1 \end{array}\right)\right)a^2 b^{N-2} + …+ \left(\left(\begin{array}{c} N-M \\ N-(M-1) \end{array}\right)\left(\begin{array}{c} M \\ M \end{array}\right)+\left(\begin{array}{c} N-M \\ N-M \end{array}\right)\left(\begin{array}{c} M \\ M-1 \end{array}\right)\right)a^{N-1} b^1+ \left(\left(\begin{array}{c} N-M \\ N-M \end{array}\right)\left(\begin{array}{c} M \\ M \end{array}\right)\right)a^N b^0 \\ &= \left(\sum^0_{x=0}\left(\begin{array}{c} N-M \\ 0-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right)\right)a^0b^{N-0}+ \left(\sum^1_{x=0}\left(\begin{array}{c} N-M \\ 1-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right)\right)a^1b^{N-1}+ \left(\sum^2_{x=0}\left(\begin{array}{c} N-M \\ 2-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right)\right)a^2b^{N-2}+ …+ \left(\sum^M_{x=0}\left(\begin{array}{c} N-M \\ M-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right)\right)a^Nb^{N-N} \end{align} \]
項の係数を比較して,
\[ \left(\begin{array}{c} N \\ x \end{array}\right) = \sum^x_{k=0}\left(\begin{array}{c} N-M \\ x-k \end{array}\right)\left(\begin{array}{c} M \\ k \end{array}\right) \]
改めて\(x=K,k=x\)とおいて
\[ \left(\begin{array}{c} N \\ K \end{array}\right) = \sum^K_{k=0}\left(\begin{array}{c} N-M \\ K-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right) \\ 1 = \frac{\sum^K_{k=0}\left(\begin{array}{c} N-M \\ K-x \end{array}\right)\left(\begin{array}{c} M \\ x \end{array}\right)}{ \left(\begin{array}{c} N \\ K \end{array}\right)} \] より,\(\sum^{K} _{x=0} P(X=x:N,M,K) = 1\)