ネイマン・ピアソンの基本定理の証明

$\mathit{x}=\{x_1,x_2,\cdots ,x_n\}$ の同時確率密度関数を $f(\mathit{x};\theta )$ とし、検定量の空間 $S$ を $$\begin{array}{r}\{\begin{array}{c}C=\{\mathit{x}\in A|L({\theta }_0;\mathit{x})<kL({\theta }_1;\mathit{x})\}\\ D=\{\mathit{x}\in A|L({\theta }_0;\mathit{x})=kL({\theta }_1;\mathit{x})\}\\ E=\{\mathit{x}\in A|L({\theta }_0;\mathit{x})>kL({\theta }_1;\mathit{x})\}\end{array}\end{array}$$ と分割すると、 これらは互いに素であるから $$\begin{array}{r}S=C\cup D\cup E\end{array}$$ と表される。

（i）帰無仮説についての条件
ここで、${\phi }^{\ast }(\theta ;\mathit{x})$ を有意水準 $\alpha$ の他の検定関数とすると、検定関数の定義から $$\begin{array}{c}E\left\{{\phi }^{\ast }(\theta ;\mathit{x})\right|\theta ={\theta }_0\}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\phi }^{\ast }(\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_0)d\mathit{x}\le \alpha \\ E\left\{\phi (\theta ;\mathit{x})\right|\theta ={\theta }_0\}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\phi }^{\ast }(\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_0)d\mathit{x}\le \alpha \end{array}$$ が成り立つ。

（ii）対立仮説についての条件
それぞれの検定の検出力は、 $$\begin{array}{c}E\left\{{\phi }^{\ast }(\theta ;\mathit{x})\right|\theta ={\theta }_1\}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\phi }^{\ast }(\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_1)d\mathit{x}\\ E\left\{\phi (\theta ;\mathit{x})\right|\theta ={\theta }_1\}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_1)d\mathit{x}\end{array}$$ これらの差を取ると、 $$\begin{array}{rl}\mathrm{\Delta }& =E\left\{\phi (\theta ;\mathit{x})\right|\theta ={\theta }_1\}-E\left\{{\phi }^{\ast }(\theta ;\mathit{x})\right|\theta ={\theta }_1\}\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_1)d\mathit{x}-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\phi }^{\ast }(\theta ;\mathit{x})\cdot f(\mathit{x};{\theta }_1)d\mathit{x}\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\{\phi (\theta ;\mathit{x})-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_1)d\mathit{x}\end{array}$$ 積分区間を分割すると、 $$\begin{array}{c}\mathrm{\Delta }={\int }_C\{1-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_1)d\mathit{x}+{\int }_D\{\gamma -{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_1)d\mathit{x}+{\int }_E\{0-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_1)d\mathit{x}\\ \mathrm{\Delta }\ge {\int }_C\{1-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot kf(\mathit{x};{\theta }_0)d\mathit{x}+{\int }_D\{\gamma -{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot kf(\mathit{x};{\theta }_0)d\mathit{x}+{\int }_E\{0-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot kf(\mathit{x};{\theta }_0)d\mathit{x}\\ \mathrm{\Delta }\ge k{\int }_C\{1-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_0)d\mathit{x}+k{\int }_D\{\gamma -{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_0)d\mathit{x}+k{\int }_E\{0-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_0)d\mathit{x}\\ \mathrm{\Delta }\ge k{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\{\phi (\theta ;\mathit{x})-{\phi }^{\ast }(\theta ;\mathit{x})\}\cdot f(\mathit{x};{\theta }_0)d\mathit{x}\\ \mathrm{\Delta }\ge k[E\left\{\phi (\theta ;\mathit{x})\right|\theta ={\theta }_0\}-E\left\{{\phi }^{\ast }(\theta ;\mathit{x})\right|\theta ={\theta }_0\}]\\ \mathrm{\Delta }\ge k(\alpha -\alpha )=0\end{array}$$ したがって、 $$\begin{array}{r}E\left\{\phi (\theta ;\mathit{x})\right|\theta ={\theta }_1\}\ge E\left\{{\phi }^{\ast }(\theta ;\mathit{x})\right|\theta ={\theta }_1\}\end{array}$$ となり、検出力が最大となる。 $◼$