Contents:

6.5.1: Preparation (simultaneous normal measurement)

Consider the parallel measurement${\mathsf M}_{L^\infty (({\mathbb R} \times {\mathbb R}_+) \times ({\mathbb R} \times {\mathbb R}_+))}$ $({\mathsf O}_G^n \otimes {\mathsf O}_G^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G}}^n} \otimes {{{G}}^m}) ,$ $S_{[(\mu_1, \sigma_1, \mu_2 , \sigma_2)]})$ (in $L^\infty(({\mathbb R} \times {\mathbb R}_+) \times ({\mathbb R} \times {\mathbb R}_+))$ ) of two normal measurements. Assume that $\sigma_1$ and $\sigma_2$ are fixed and known. Thus, this parallel measurement is represented by ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$ in $L^\infty ({\mathbb R} \times {\mathbb R} )$. Here, recall the normal observable (6.1), i.e.,

\begin{align} & [{{{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}

Therefore, we have the state space $\Omega ={\mathbb R}^2 = \{ \omega=(\mu_1, \mu_2) \;:\; \mu_1,\mu_2 \in {\mathbb R} \}$. Put $\Theta={\mathbb R}$ with the distance $d_\Theta^{(1)} ( \theta_1, \theta_2 )= |\theta_1-\theta_2|$ and consider the quantity $\pi:{\mathbb R}^2 \to {\mathbb R}$ by

\begin{align} \pi (\mu_1, \mu_2)= \mu_1-\mu_2 \tag{6.66} \end{align}

The estimator $E: \widehat{X}(=X \times Y = {{\mathbb R}^n \times {\mathbb R}^m}) \to \Theta(={\mathbb R})$ is defined by

\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}

For any $ \omega=(\mu_1, \mu_2 ) (\in \Omega= {\mathbb R} \times {\mathbb R} )$, define the positive number $\eta^\alpha_{\omega} (= \delta_\omega^{1-\alpha} )$ $(> 0)$ such that:

\begin{align} \eta^\alpha_{\omega} (= \delta_\omega^{1-\alpha} ) = \inf \{ \eta > 0: [F (E^{-1} ( {{ Ball}^C_{d_\Theta^{(1)}}}(\pi(\omega) ; \eta))](\omega ) \ge \alpha \} \nonumber \end{align}

where ${{ Ball}^C_{d_\Theta^{(1)} }}(\pi(\omega) ; \eta)$ $= (-\infty, \mu_1 - \mu_2 - \eta] \cup [ \mu_1 - \mu_2 + \eta , \infty)$. Define the null hypothesis $H_N$ $(\subseteq \Theta = {\mathbb R})$ such that

\begin{align} H_N =\{ \theta_0 \} \end{align}

Now let us calculate the $\eta^\alpha_{\omega}$ as follows:

\begin{align} & E^{-1}({{ 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} Thus, \begin{align} & [ ({{{N_{\sigma_1}}}}^n \otimes {{{N_{\sigma_2}}}}^m ) (E^{-1}({{ 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 \\ & \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 \\ = & \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}

Using the $z(\alpha/2)$ in (6.33), we get that

\begin{align} \eta^\alpha_{\omega} = \delta_\omega^{1- \alpha } = (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z(\frac{\alpha}{2}) \tag{6.70} \end{align}

6.5.2: Confidence interval

Our present problem is as follows

Problem 6.15 [ Confidence interval for the difference of population means].

Let $\sigma_1$ and $\sigma_2$ be positive numbers which are assumed to be fixed. Consider the parallel measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$. Assume that a measured value $\widehat{x}$ $=$ $(x,y)$ $=(x_1,\ldots, x_n,y_1,\ldots, y_m)$ $(\in {\mathbb R}^n\times {\mathbb R}^m)$ is obtained by the measurement. Let $0 < \alpha \ll 1$.


Then, find the confidence interval ${D}_{(x,y)}^{1- \alpha; \Theta}( \subseteq \Theta)$ (which may depend on $\sigma_1$ and $\sigma_2$) such that

$\bullet$ the probability that $\mu_1 - \mu_2 \in {D}_{(x,y)}^{1- \alpha; \Theta}$ is more than $1-\alpha$.

Here, the more the confidence interval ${D}_{(x,y)}^{1- \alpha; \Theta}$ is small, the more it is desirable.

Therefore, for any $\widehat{x}$ $=$ $(x,y)$ $=(x_1,\ldots, x_n,y_1,\ldots, y_m)$ $(\in {\mathbb R}^n\times {\mathbb R}^m)$, we get $D_{\widehat{x}}^{{1 - \alpha }}$ ( the $({1 - \alpha })$-confidence interval of ${\widehat x}$ ) as follows:

\begin{align} D_{\widehat{x}}^{{1 - \alpha, \Omega }} & = \{ {\omega} (\in \Omega) : d_\Theta (E(\widehat{x}), \pi(\omega)) \le \delta^{1 - \alpha }_{\omega } \} \nonumber \\ & = \{ (\mu_1, \mu_2 ) \in {\mathbb R} \times {\mathbb R} \;:\; | \frac{\sum_{k=1}^n x_k }{n} - \frac{\sum_{k=1}^m y_k}{m} -(\mu_1 - \mu_2 )| \le (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z(\frac{\alpha}{2}) \} \\ & \tag{6.71} \end{align}

6.5.3 Statistical hypothesis testing [rejection region: null hypothesis$H_N=\{\mu_0\} \subseteq \Theta = {\mathbb R}$]

Our present problem is as follows

Problem 6.16 [Statistical hypothesis testing for the difference of population means].

Consider the parallel measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$. Assume that

\begin{align} \pi(\mu_1, \mu_2 ) =\mu_1 - \mu_2 =\theta_0 \in \Theta ={\mathbb R} \end{align}

that is, assume the null hypothesis$H_N$ such that

\begin{align} H_N=\{ \theta_0 \} (\subseteq \Theta= {\mathbb R} ) ) \end{align}

Let $0 < \alpha \ll 1$.


Then, find the rejection region ${\widehat R}_{{H_N}}^{\alpha; \Theta}( \subseteq \Theta)$ (which may depend on $\mu$) such that

$\bullet$ the probability that a measured value$(x,y) (\in{\mathbb R}^n\times {\mathbb R}^m )$ obtained by ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$ sa tisfies \begin{align} E(x,y)=\frac{x_1+x_2+ \cdots + x_n }{n}-\frac{y_1+y_2+ \cdots + y_m }{m} \in {\widehat R}_{{H_N}}^{\alpha; \Theta} \end{align} is less than $\alpha$.

Here, the more the rejection region ${\widehat R}_{{H_N}}^{\alpha; \Theta}$ is large, the more it is desirable.

By the formula (6.70), we see that the rejection region${\widehat R}_{\widehat{x}}^{\alpha}$ ( $(\alpha)$-rejection region of $H_N =\{\theta_0\}( \subseteq \Theta)$ ) is defined by

\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} or \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}

Here,

\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}

6.5.4 Statistical hypothesis testing [rejection region: null hypothesis$H_N=(- \infty , \theta_0] \subseteq \Theta = {\mathbb R}$]

Our present problem is as follows

Problem 6.17 [Statistical hypothesis testing for the difference of population means].

Consider the parallel measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$　$({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal　B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$　$S_{[(\mu_1, \mu_2 )]})$. Assume that

\begin{align} \pi(\mu_1, \mu_2 ) =\mu_1 - \mu_2 =(-\infty , \theta_0] \subseteq \Theta ={\mathbb R} \end{align}

that is,　assume the　null hypothesis$H_N$　such that

\begin{align} H_N=(-\infty , \theta_0] (\subseteq \Theta= {\mathbb R} ) ) \end{align} Let　$0 < \alpha \ll 1$.

Then, find the rejection region ${\widehat R}_{{H_N}}^{\alpha; \Theta}( \subseteq \Theta)$　(which may depend on　$\mu$)　such that

$\bullet$　the probability that a measured value$(x,y) (\in{\mathbb R}^n\times {\mathbb R}^m )$ obtained by ${\mathsf M}_{L^\infty ({\mathbb R} \times{\mathbb R} )}$ $({\mathsf O}_{G_{\sigma_1}}^n \otimes {\mathsf O}_{G_{\sigma_1}}^m= ({\mathbb R}^n \times {\mathbb R}^m \ , {\mathcal B}_{\mathbb R}^n \boxtimes {\mathcal B}_{\mathbb R}^m, {{{G_{\sigma_1}}}^n} \otimes {{{G_{\sigma_2}}}^m}) ,$ $S_{[(\mu_1, \mu_2 )]})$ satisfies \begin{align} E(x,y)=\frac{x_1+x_2+ \cdots + x_n }{n}-\frac{y_1+y_2+ \cdots + y_m }{m} \in {\widehat R}_{{H_N}}^{\alpha; \Theta} \end{align}

is less than $\alpha$.

Here, the more the rejection region ${\widehat R}_{{H_N}}^{\alpha; \Theta}$ is large, the more it is desirable.

Since the null hypothesis $H_N$ is assumed as follows:

\begin{align} H_N = (- \infty , \theta_0], \end{align}

it suffices to define the semi-distance $d_\Theta^{(1)} $ in $ \Theta(= {\mathbb R}) $ such that

\begin{align} d_\Theta^{(1)} (\theta_1, \theta_2) = \left\{\begin{array}{ll} |\theta_1-\theta_2| \quad & ( \forall \theta_1, \theta_2 \in \Theta={\mathbb R} \mbox{ such that } \theta_0 \le \theta_1, \theta_2 ) \\ \max \{ \theta_1, \theta_2 \} - \theta_0 \quad & ( \forall \theta_1, \theta_2 \in \Theta={\mathbb R} \mbox{ such that } \min \{ \theta_1, \theta_2 \} \le \theta_0 \le \max \{ \theta_1, \theta_2 \} ) \\ 0 & ( \forall \theta_1, \theta_2 \in \Theta={\mathbb R} \mbox{ such that } \theta_1, \theta_2 \le \theta_0 ) \end{array}\right. \\ & \tag{6.74} \end{align}

Then, we can easily see that

\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}(=(- \infty, \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 {\mathbb R} \;:\; \overline{\mu}(x)-\overline{\mu}(y) -\theta_0 \ge (\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{m})^{1/2} z({\alpha}) \} \tag{6.75} \end{align}