標本平均と標本分散

まず，標本\[X_{1},X_{2},\cdots,X_{n}\]は母平均 $\mu$ をもつ母集団から無作為に抽出されたものなので，\[E[X_{1}]=E[X_{2}]=E[X_{3}]=\cdots=E[X_{n}]\]となります．

$\bar{X}$ の期待値は，\[\begin{eqnarray}E[\bar{X}]&=&E[\frac{1}{n}(X_{1}+X_{2}+X_{3}+\cdots+X_{n})]\\&=&\frac{1}{n}E[X_{1}+X_{2}+X_{3}+\cdots+X_{n}]\\&=&\frac{1}{n}(E[X_{1}]+E[X_{2}]+E[X_{3}]+\cdots+E[X_{n}])\\&=&\frac{1}{n}(\mu+\mu+\mu+\cdots+\mu)\\&=&\frac{1}{n}n\mu=\mu \end{eqnarray}\]となり，$E[\bar{X}=\mu]$ となり，$\bar{X}$ は母平均 $\mu$ の不偏推定量となります．

母平均 $\mu$ ，母分散 $\sigma^{2}$ をもつ母集団から無作為に抽出した標本\[X_{1},X_{2},\cdots,X_{n}\]は，各々，平均 $\mu$，分散 $\sigma^{2}$ をもつ確率分布に従う互いに独立な確率変数であるとみなせます．\[E[X_{i}]=\mu,E[(X_{i}-\mu)^{2}]=\sigma^{2},i=1,2,\cdots,n\]ここで，母分散 $\sigma^{2}$ の不偏推定量 $S^{2}$ の期待値を考えると，\[\begin{eqnarray}E[S^{2}]&=&E[\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}]\\&=&E[\frac{1}{n-1}\sum_{i=1}^{n}\{(X_{i}-\mu)-(\bar{X}-\mu)\}^{2}]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}\{(X_{i}-\mu)-(\bar{X}-\mu)\}^{2}]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}\{(X_{i}-\mu)^{2}-2(X_{i}-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^{2}\}]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X_{i}-\mu)^{2}-\sum_{i=1}^{n} 2(X_{i}-\mu)(\bar{X}-\mu)+\sum_{i=1}^{n}(\bar{X}-\mu)^{2}\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X_{i}-\mu)^{2}-2(\bar{X}-\mu)\sum_{i=1}^{n} (X_{i}-\mu)+(\bar{X}-\mu)^{2}\sum_{i=1}^{n}1]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X_{i}-\mu)^{2}-2(\bar{X}-\mu)(\sum_{i=1}^{n}X_{i}-\mu\sum_{i=1}^{n}1)+(\bar{X}-\mu)^{2}\sum_{i=1}^{n}1]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X_{i}-\mu)^{2}-2(\bar{X}-\mu)(n\bar{X}-n\mu)+(\bar{X}-\mu)^{2}n]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X-\mu)^{2}-2n(\bar{X}-\mu)^{2}+n(\bar{X}-\mu)^{2}]\\&=&\frac{1}{n-1}E[\sum_{i=1}^{n}(X-\mu)^{2}-n(\bar{X}-\mu)^{2}]\\&=&\frac{1}{n-1}{E[\sum_{i=1}^{n}(X-\mu)^{2}]-nE[(\bar{X}-\mu)^{2}]} \end{eqnarray}\]となります．

ここで，\[E[\sum_{i=1}^{n}(X_{i}-\mu)^{2}]=\sum_{i=1}^{n}E[(X_{i}-\mu)^{2}]=\sum_{i=1}^{n}\sigma^{2}=n\sigma^{2}\]であること，および，\[\begin{eqnarray}E[(\bar{X}-\mu)^{2}]&=&Var[\bar{X}]\\&=&Var[\frac{1}{n}(X_{1}+X_{2}+X_{3}+\cdots+X_{n})]\\&=&\frac{1}{n^{2}}(Var[X_{1}]+Var[X_{2}]+Var[X_{3}])+\cdots+Var[X_{n}])\\&=&\frac{1}{n^{2}}(\sigma^{2}+\sigma^{2}+\sigma^{2}+\cdots+\sigma^{2})\\&=&\frac{1}{n^{2}}n\sigma^{2}\\&=&\frac{1}{n}\sigma \end{eqnarray}\]となることから，\[E[S^{2}]=\frac{1}{n-1}(n\sigma^{2}-n\frac{1}{n}\sigma^{2})=\frac{n-1}{n-1}\sigma^{2}=\sigma^{2}\]となり，$E[S^{2}]=\sigma^{2}$ となるため，$S^{2}$ が母分散 $\sigma^{2}$ の不偏推定量となります．□

Mathematics is the language with which God has written the universe.