6.2: The reverse relation between confidence interval method and statistical hypothesis testing

Contents:
In what follows, we shall mention the reverse relation (such as "the two sides of a coin") between confidence interval method and statistical hypothesis testing. We devote ourselves to the classical systems, i.e., the classical basic structure: \begin{align} [ C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))] \end{align} 6.2.1: The confidence interval method

Consider an observable ${\mathsf O} = (X, {\cal F} , F){}$ in ${L^\infty (\Omega)}$. Let $\Theta$ be a locally compact space (called the second state space) which has the semi-metric $d^x_{\Theta}$ $(\forall x \in X)$ such that,

 $(\sharp):$ for each $x\in X$, the map $d^x_{\Theta}: \Theta^2 \to [0,\infty)$ satisfies (i):$d^x_\Theta (\theta, \theta )=0$, (ii):$d^x_\Theta (\theta_1, \theta_2 )$ $=d^x_\Theta (\theta_2, \theta_1 )$, (ii):$d^x_\Theta (\theta_1, \theta_3 )$ $\le d^x_\Theta (\theta_1, \theta_2 ) + d^x_\Theta (\theta_2, \theta_3 )$.

Furthermore, consider two maps $E:X \to \Theta$ and $\pi: \Omega \to \Theta$. Here, $E:X \to \Theta$ and $\pi: \Omega \to \Theta$ is respectively called an estimator and a system quantity

Theorem 6.3 [Confidence interval method].
Let a positive number $\alpha$ be $0 < \alpha \ll 1$, for example, $\alpha = 0.05$. For any state $\omega (\in \Omega)$, define the positive number $\delta^{1-\alpha}_{\omega}$ $(> 0)$ such that: \begin{align} \delta^{1-\alpha}_{\omega} = \inf \{ \delta > 0: [F(\{ x \in X \;:\; d^x_\Theta ( E(x) , \pi( \omega ) ) < \delta \} )](\omega ) \ge {1 -\alpha } \} \\ & \tag{6.9} \end{align} Then we say that:
 $(A):$ the probability, that the measured value $x$ obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$ $S_{[\omega_0 {}] } \big)$ satisfies the following condition (6.10), is more than or equal to ${1 -\alpha }$ (e.g., ${1 -\alpha }= 0.95$).
\begin{align} d^x_\Theta (E(x), \pi(\omega_0)) \le {\delta }^{1 -\alpha }_{\omega_0} \tag{6.10} \end{align} And further, put \begin{align} D_{x}^{{1 -\alpha, \Theta }} = \{ \pi(\omega) (\in \Theta) : d^x_\Theta (E(x), \pi(\omega ) ) \le \delta^{1-\alpha}_{\omega } \}. \tag{6.11} \end{align} which is called $({1 -\alpha })$-confidence interval. Here, we see the following equivalence: \begin{align} (6.10) \; \Longleftrightarrow \; \; D_{x}^{1 -\alpha, \Theta } \ni \pi (\omega_0). \tag{6.12} \end{align}

Remark 6.4 [(B$_1$):The meaning of confidence interval]. Consider the parallel measurement $\bigotimes_{j=1}^J {\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$ $S_{[\omega_0 {}] } \big)$, and assume that a measured value $x=(x_1,x_2, \ldots , x_J)( \in X^J)$ is obtained by the parallel measurement. Recall the formula (6.12). Then, it surely holds that

\begin{align} \lim_{J \to \infty } \frac{\mbox{Num} [\{ j \;|\; D_{x_j}^{{1 -\alpha, \Theta }} \ni \pi( \omega_0)]}{J} \ge {1 -\alpha } (= 0.95) \tag{6.13} \end{align}

where $\mbox{Num} [A]$ is the number of the elements of the set $A$. Hence Theorem 6.3 can be tested by numerical analysis (with random number). Similarly, Theorem 6.5 ( mentioned later ) can be tested.

[(B$_2$)] Also, note that \begin{align} (6.9) & = \delta_\omega^{1-\alpha} = \inf \{ \delta > 0: [F(\{ x \in X \;:\; d^x_\Theta ( E(x) , \pi( \omega ) ) < \delta \} )](\omega ) \ge {1-\alpha} \} \nonumber \\ &= \inf \{ \eta > 0: [F(\{ x \in X \;:\; d^x_\Theta ( E(x) , \pi( \omega ) ) \ge \eta \} )](\omega ) \le \alpha \} \tag{6.14} \end{align}

6.2.2 Statistical hypothesis testing

Next, we will explain the statistical hypothesis testing, which is characterized as the reverse of the confident interval method.

Theorem 6.5 [Statistical hypothesis testing]
Let $\alpha$ be a real number such that $0 < \alpha \ll 1$, for example, $\alpha = 0.05$. For any state $\omega (\in \Omega)$, define the positive number $\eta^\alpha_{\omega}$ $(> 0)$ such that: \begin{align} \eta^\alpha_{\omega} & = \inf \{ \eta > 0: [F(\{ x \in X \;:\; d^x_\Theta ( E(x) , \pi( \omega ) ) \ge \eta \} )](\omega ) \le \alpha \} \tag{6.15} \\ & \mbox{ ( by the (6.14), note that $\delta_\omega^{1 - \alpha}=\eta_\omega^\alpha$) } \nonumber \end{align} Then we say that:
 $(C):$ the probability, that the measured value $x$ obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$ $S_{[\omega_0 {}] } \big)$ satisfies the following condition (6.16), is less than or equal to $\alpha$ (e.g., $\alpha= 0.05$). \begin{align} d^x_\Theta (E(x), \pi(\omega_0)) \ge {\eta }^\alpha_{\omega_0} . \tag{6.16} \end{align}
Furthermore, consider a subset $H_N$ of $\Theta$, which is called a "null hypothesis". Put \begin{align} & {\widehat R}_{H_N}^{\alpha, \Theta} = \bigcap_{\omega \in \Omega \mbox{ such that } \pi(\omega) \in {H_N}} \{ E({x}) (\in \Theta) : d^x_\Theta (E(x), \pi(\omega ) ) \ge \eta^\alpha_{\omega } \}. \\ & \tag{6.17} \end{align} which is called the $(\alpha)$-rejection region of the null hypothesis ${H_N}$. Then we say that:
 $(D):$ the probability, that@the measured value $x$@obtained@by the measurement@${\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$@$S_{[\omega_0 {}] } \big)$@$($where $\pi(\omega_0) \in H_N )$@satisfies the following@condition (6.18),@is less than or equal to@$\alpha$@(e.g., $\alpha= 0.05$).
\begin{align} {\widehat R}_{H_N}^\alpha \ni E(x). \tag{6.18} \end{align}
6.2.3: The reverse relation between Confidence interval and statistical hypothesis testing
Corollary6.6[The reverse relation between Confidence intervaland statistical hypothesis testing]

Let $0 < \alpha \ll 1$. Consider an observable ${\mathsf O} = (X, {\cal F} , F){}$ in ${L^\infty (\Omega)}$, and the second state space $\Theta$ (i.e., locally compact space with a semi-metric $d_\Theta^x (x \in X)$ ). And consider the estimator $E:X \to \Theta$ and the system quantity $\pi: \Omega \to \Theta$. Define $\delta_\omega^{1-\alpha}$ by (6.9), and define $\eta_\omega^{\alpha}$ by (6.15) ( and thus, $\delta_\omega^{1-\alpha}= \eta_\omega^{\alpha}$).

$(E):$[Confidence interval method]. for each $x \in X$,define $(1- \alpha)$-confidence interval by \begin{align} & D_{x}^{1- \alpha, \Theta } = \{ \pi(\omega) (\in \Theta) : d^x_\Theta (E(x), \pi(\omega ) ) < \delta^{1- \alpha}_{\omega } \} \tag{6.19} \end{align} \begin{align} & D_{x}^{1- \alpha, \Omega} = \{ \omega (\in \Omega) : d^x_\Theta (E(x), \pi(\omega ) ) < \delta^{1- \alpha}_{\omega } \} \tag{6.20} \end{align} Here, assume that a measured value $x (\in X)$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$ $S_{[\omega_0 {}] } \big)$. Then, we see that
 $(E_1):$ the probability that \begin{align} D_x^{1-\alpha, \Theta} \ni \pi(\omega_0) \quad \mbox{ or,in the same sense } \quad D_x^{1-\alpha, \Omega} \ni \omega_0 \end{align} is more than $1- \alpha$.
And,
$(F):$ [statistical hypothesis testing]. Consider the null hypothesis $H_N ( \subseteq \Theta )$. Assume that the state $\omega_0(\in \Omega )$ satisfies: \begin{align} \pi(\omega_0) \in H_N ( \subseteq \Theta ) \end{align} Here, put, \begin{align} & {\widehat R}_{{H_N}}^{\alpha; \Theta} = \bigcap_{\omega \in \Omega \mbox{ such that } \pi(\omega) \in {H_N}} \{ E({x}) (\in \Theta) : d^x_\Theta (E(x), \pi(\omega ) ) \ge \eta^\alpha_{\omega } \}. \\ & \tag{6.21} \end{align} \begin{align} & {\widehat R}_{{H_N}}^{\alpha; X} = E^{-1}( {\widehat R}_{{H_N}}^{\alpha; \Theta}) = \bigcap_{\omega \in \Omega \mbox{ such that } \pi(\omega) \in {H_N}} \{ x (\in X) : d^x_\Theta (E(x), \pi(\omega ) ) \ge \eta^\alpha_{\omega } \}. \\ & \tag{6.22} \end{align} which is called the $(\alpha)$-rejection region of the null hypothesis ${H_N}$. Assume that a measured value $x (\in X)$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} \big({\mathsf O}:= (X, {\cal F} , F) ,$ $S_{[\omega_0 {}] } \big)$. Then, we see that
 $(F_1):$ the probability that \begin{align} & "E(x) \in {\widehat R}_{{H_N}}^{\alpha; \Theta}" \quad \mbox{ or,in the same sense, } \quad "x \in {\widehat R}_{{H_N}}^{\alpha; X}" \\ & \tag{6.23} \end{align} is less than $\alpha$.