6.5:二つの正規測定の母平均の差の信頼区間法と統計的仮説検定

二つの正規測定の並行測定${\mathsf M}_{L^\infty (({\mathbb R} \times {\mathbb R}_+) \times ({\mathbb R} \times {\mathbb R}_+))}$ $({\mathsf O}_G^n \otimes {\mathsf O}_G^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G}}^n} \otimes {{{G}}^m}) ,$ $S_{[(\mu_1, \sigma_1, \mu_2 , \sigma_2)]})$ (in $L^\infty(({\mathbb R} \times {\mathbb R}_+) \times ({\mathbb R} \times {\mathbb R}_+))$) を考えよう. $\sigma_1$と$\sigma_2$は固定されているとする.したがって, この並行測定 を,$L^\infty({\mathbb R} \times {\mathbb R} )$内の ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$と見なして議論を進める. ここで,${\mathsf O}_{\sigma} =({\mathbb R}, {\mathcal B}_{{\mathbb R}}, G_\sigma)$は次で定まる.

\begin{align} & \small{ [{{{G_\sigma}}}({\Xi})] ({} {}{\mu} {}) = \frac{1}{{\sqrt{2 \pi }\sigma{}}} \int_{{\Xi}} \exp[{}- \frac{({}{}{x} - {}{\mu} {})^2 }{2 \sigma^2} {}] d {}{x} \quad ({}\forall {\Xi} \in {\cal B}_{{\mathbb R}{}}\mbox{(=Borel field in ${\mathbb R}$))}, \quad \forall \mu \in {\mathbb R}). } \\ & \tag{6.65} \end{align} したがって,,状態空間を $\Omega ={\mathbb R}^2 = \{ \omega=(\mu_1, \mu_2) \;:\; \mu_1,\mu_2 \in {\mathbb R} \}$とする. さて,第二状態空間$\Theta$を$\Theta={\mathbb R}$として, 距離 $d_\Theta^{(1)} ( \theta_1, \theta_2 )= |\theta_1-\theta_2|$を持つとする. システム量 $\pi:{\mathbb R}^2(=\Omega) \to {\mathbb R}(=\Theta) $ を \begin{align} \pi (\mu_1, \mu_2)= \mu_1-\mu_2 \tag{6.66} \end{align} として, 推定量 $E: \widehat{X}(=X \times Y = {{\mathbb R}^n \times {\mathbb R}^m}) \to \Theta(={\mathbb R})$を を次のように定める. \begin{align} E(x_1, \ldots, x_n,y_1, \ldots, y_m) = \frac{\sum_{k=1}^n x_k}{n} - \frac{\sum_{k=1}^m y_k}{m} \tag{6.67} \end{align} 任意の状態 $ \omega=(\mu_1, \mu_2 ) ({}\in \Omega= {\mathbb R} \times {\mathbb R} )$に対して, 正数 $\eta^\alpha_{\omega}$ $({}> 0)$ を次のように定める. \begin{align} \eta^\alpha_{\omega} = \inf \{ \eta > 0: [F ({}E^{-1} ({} {{\rm Ball}^c_{d_\Theta^{(1)}}}(\pi(\omega) ; \eta{}))](\omega ) \ge \alpha \} \nonumber \end{align} ここに, ${{\rm Ball}^c_{d_\Theta^{(1)} }}(\pi(\omega) ; \eta)$ $= (-\infty, \mu_1 - \mu_2 - \eta] \cup [ \mu_1 - \mu_2 + \eta , \infty)$であった. さて,　この$\eta^\alpha_{\omega}$を以下のように計算しよう. \begin{align} & E^{-1}({{\rm Ball}^c_{d_\Theta^{(1)} }}(\pi(\omega) ; \eta )) = E^{-1}( (-\infty, \mu_1 - \mu_2 - \eta] \cup [ \mu_1 - \mu_2 + \eta , \infty) ) \nonumber \\ = & \{ (x_1, \ldots , x_n, y_1, \ldots, y_m ) \in {\mathbb R}^n \times {\mathbb R}^m \;: \; | \frac{\sum_{k=1}^n x_k}{n} - \frac{\sum_{k=1}^m y_k}{m} -(\mu_1 - \mu_2)| \ge \eta \} \nonumber \\ = & \{ (x_1, \ldots , x_n, y_1, \ldots, y_m ) \in {\mathbb R}^n \times {\mathbb R}^m \;: \; | \frac{\sum_{k=1}^n (x_k - \mu_1)}{n} - \frac{\sum_{k=1}^m (y_k- \mu_2)}{m} | \ge \eta \} \\ & \tag{6.68} \end{align} したがって, \begin{align} & [ ({{{G_{\sigma_1}}}}^n \otimes {{{G_{\sigma_2}}}}^m ) (E^{-1}({{\rm Ball}^c_{d_\Theta^{(1)} }}(\pi(\omega) ; \eta ))] ({}\omega{}) \nonumber \\ = & \frac{1}{({{\sqrt{2 \pi }\sigma_1{}}})^n({{\sqrt{2 \pi }\sigma_2{}}})^m} \nonumber \\ & \small{ {{{{\times}}}} \!\!\!\!\!\! \underset{{ | \frac{\sum_{k=1}^n( x_k - \mu_1)}{n} - \frac{\sum_{k=1}^m (y_k- \mu_2)}{m} | \ge \eta }}{\int \cdots \int} \exp[ {}- \frac{\sum_{k=1}^n ({}{}{x_k} - {}{\mu_1} {})^2 } {2 \sigma_1^2} {}- \frac{\sum_{k=1}^m ({}{}{y_k} - {}{\mu_2} {})^2 } {2 \sigma_2^2} ] d {}{x_1} d {}{x_2}\cdots dx_nd {}{y_1} d {}{y_2}\cdots dy_m } \nonumber \\ = & \small{ \frac{1}{({{\sqrt{2 \pi }\sigma_1{}}})^n({{\sqrt{2 \pi }\sigma_2{}}})^m} \underset{{ | \frac{\sum_{k=1}^n x_k }{n} - \frac{\sum_{k=1}^m y_k}{m} | \ge \eta }}{\int \cdots \int} \exp[ - \frac{ \sum_{k=1}^n {x_k}^2 } {2 \sigma_1^2} - \frac{ \sum_{k=1}^m {y_k}^2 } {2 \sigma_2^2} ] d {}{x_1} d {}{x_2}\cdots dx_nd {}{y_1} d {}{y_2}\cdots dy_m } \nonumber \\ = & 1- \frac{1}{{\sqrt{2 \pi }(\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2}{}}} \int_{{- \eta}}^{\eta} \exp[{}- \frac{{x}^2 }{2 (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})}] d {x} \\ & \tag{6.69} \end{align} ここで,$z(\alpha/2)$を使って, 次を得る. \begin{align} \eta^\alpha_{\omega} = (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z(\frac{\alpha}{2}) \tag{6.70} \end{align} ここで, \begin{align*} \overline{\mu}(x)=\frac{\sum_{k=1}^n x_k }{n}, \quad \overline{\mu}(y) = \frac{\sum_{k=1}^m y_k}{m} \end{align*} としておく。

このとき, 次の信頼区間 $D_{\widehat{x}}^{1- \alpha, \Theta}$ ( $({}1- \alpha {})$-信頼区間 of ${\widehat x}$ ) を得る.

6.5.3 仮説検定 [棄却域: 帰無仮説$H_N=\{\mu_0\} \subseteq \Theta = {\mathbb R}$]

したがって, 棄却域${\widehat R}_{\widehat{x}}^{\alpha}$ ( $({}\alpha{})$-棄却域 of $H_N =\{\theta_0\}( \subseteq \Theta)$ ) は,次のようになる \begin{align} {\widehat R}_{H_N}^{\alpha,\Theta} & = \bigcap_{\omega =(\mu_1, \mu_2 ) \in \Omega (={\mathbb R}^2 ) \mbox{ such that } \pi(\omega)= \mu_1-\mu_2 \in {H_N}(=\{\theta_0 \} )} \{ E(\widehat{x}) (\in \Theta) : d_\Theta^{(1)} ({}E(\widehat{x}), \pi(\omega)) \ge \eta^\alpha_{\omega } \} \nonumber \\ & = \{ \overline{\mu}(x)-\overline{\mu}(y) \in \Theta (={\mathbb R}) \;:\; | \overline{\mu}(x)-\overline{\mu}(y) -\theta_0| \ge (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z(\frac{\alpha}{2}) \} \tag{6.72} \end{align} または, \begin{align} {\widehat R}_{H_N}^{\alpha, X} & = \bigcap_{\omega =(\mu_1, \mu_2 ) \in \Omega (={\mathbb R}^2 ) \mbox{ such that } \pi(\omega)= \mu_1-\mu_2 \in {H_N}(=\{\theta_0 \} )} \{ \widehat{x} (\in {\mathbb R}^n \times {\mathbb R}^m) : d_\Theta^{(1)} ({}E(\widehat{x}), \pi(\omega)) \ge \eta^\alpha_{\omega } \} \nonumber \\ & = \{ \widehat{x} (\in {\mathbb R}^n \times {\mathbb R}^m) \;:\; | \overline{\mu}(x)-\overline{\mu}(y) -\theta_0| \ge (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z(\frac{\alpha}{2}) \} \tag{6.73} \end{align}