二項分布の期待値

二項分布 \(B(n, p)\)

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases} $$

二項分布の期待値

$$\begin{array}{rcl} E[X]&=&\displaystyle \sum_{x=0}^{n}\left(x\right)\left( _nC_x\,p^x(1-p)^{n-x} \right)\\ &=& \displaystyle \sum_{x=0}^{n}\left(x\right)\left(\left( \frac{n!}{x!(n-x)!}\right)p^x(1-p)^{n-x}\right)\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\dotso \frac{x}{x!}=\frac{1}{(x-1)!}\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\dotso n!=n(n-1)!\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)!}{(x-1)!(n-x)!}pp^{x-1}(1-p)^{n-x}\dotso p^x=pp^{x-1}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1}(1-p)^{n-x}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-1+1-x)!}p^{x-1}(1-p)^{n-1+1-x}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-1-(x-1))!}p^{x-1}(1-p)^{n-1-(x-1)}\dotso 1-x=-(x-1)\\ &=& \displaystyle np\sum_{x'=0}^{n-1} \frac{(n-1)!}{x'!(n-1-x')!}p^{x'}(1-p)^{n-1-x'}\dotso x'=x-1\\ &=& \displaystyle np\,1 \dotso B(n-1, p)の総和は1に等しい．\\ &=& \displaystyle np\\ &=& \mu \dotso 母集団の期待値\\ \end{array}$$