二項分布の期待値

二項分布 $B(n,p)$

$$B(n,p)=f_X(x)=\{\begin{array}{ll}{\displaystyle {}_n{C}_x{p}^x(1-p{)}^{n-x}}& x\in \{0,1,2,\dots ,n\}\\ {\displaystyle 0}& x\notin \{0,1,2,\dots ,n\}\end{array}$$

二項分布の期待値

$$\begin{array}{rcl}E[X]& =& {\displaystyle \sum _{x=0}^n\left(x\right)({}_n{C}_x{p}^x(1-p{)}^{n-x})}\\ & =& {\displaystyle \sum _{x=0}^n\left(x\right)(\left(\frac{n!}{x!(n-x)!}\right)p^x(1-p{)}^{n-x})}\\ & =& {\displaystyle \sum _{x=0}^n\frac{n!}{(x-1)!(n-x)!}{p}^x(1-p{)}^{n-x}\dots \frac{x}{x!}=\frac{1}{(x-1)!}}\\ & =& {\displaystyle \sum _{x=0}^n\frac{n(n-1)!}{(x-1)!(n-x)!}{p}^x(1-p{)}^{n-x}\dots n!=n(n-1)!}\\ & =& {\displaystyle \sum _{x=0}^n\frac{n(n-1)!}{(x-1)!(n-x)!}p{p}^{x-1}(1-p{)}^{n-x}\dots p^x=p{p}^{x-1}}\\ & =& {\displaystyle np\sum _{x=0}^n\frac{(n-1)!}{(x-1)!(n-x)!}{p}^{x-1}(1-p{)}^{n-x}}\\ & =& {\displaystyle np\sum _{x=0}^n\frac{(n-1)!}{(x-1)!(n-1+1-x)!}{p}^{x-1}(1-p{)}^{n-1+1-x}}\\ & =& {\displaystyle np\sum _{x=0}^n\frac{(n-1)!}{(x-1)!(n-1-(x-1))!}{p}^{x-1}(1-p{)}^{n-1-(x-1)}\dots 1-x=-(x-1)}\\ & =& {\displaystyle np\sum _{x^{\prime }=0}^{n-1}\frac{(n-1)!}{x^{\prime }!(n-1-x^{\prime })!}{p}^{x^{\prime }}(1-p{)}^{n-1-x^{\prime }}\dots x^{\prime }=x-1}\\ & =& {\displaystyle np1\dots B(n-1,p)\mathrm{の}\mathrm{総}\mathrm{和}\mathrm{は}1\mathrm{に}\mathrm{等}\mathrm{し}\mathrm{い}．}\\ & =& {\displaystyle np}\\ & =& \mu \dots \mathrm{母}\mathrm{集}\mathrm{団}\mathrm{の}\mathrm{期}\mathrm{待}\mathrm{値}\end{array}$$