6.4(1): Population variance (Confidence interval method)
6.4.1: Preparation (simultaneous normal measurement)

Consider the simultaneous normal measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times {\mathbb R}_+)}$ $({\mathsf O}_G^n = ({\mathbb R}^n, {\mathcal B}_{\mathbb R}^n, {{{G}}^n}) ,$ $S_{[(\mu, \sigma)]})$ in $L^\infty({\mathbb R} \times {\mathbb R}_+)$. Here,recall that the simultaneous normal observable ${\mathsf O}_G^n = ({\mathbb R}^n, {\mathcal B}_{\mathbb R}^n, {{{G}}^n} )$ is defined by

\begin{align} & [{{{G}}}^n ({\mathop{\mbox{$\times$}}}_{k=1}^n \Xi_k)] (\omega) = {\mathop{\mbox{ $\times$}}}_{k=1}^n [{{{G}}}(\Xi_k)](\omega) \nonumber \\ = & \frac{1}{({{\sqrt{2 \pi }\sigma{}}})^n} \underset{{{\mathop{\mbox{$\times$}}}_{k=1}^n \Xi_k }}{\int \cdots \int} \exp[{}- \frac{\sum_{k=1}^n ({}{x_k} - {}{\mu} )^2 } {2 \sigma^2} {}] d {}{x_1} d {}{x_2}\cdots dx_n \tag{6.45} \\ & \qquad (\forall \Xi_k \in {\cal B}_{{\mathbb R}{}}^{} (k=1,2,\ldots, n), \quad \forall {}{\omega}=(\mu, \sigma ) \in \Omega = {\mathbb R}\times {\mathbb R}_+). \nonumber \end{align} where, note that \begin{align} & \Omega = {\mathbb R} \times {\mathbb R}_+, \qquad X={\mathbb R}^n \end{align} The second state space $\Theta$ is \begin{align} & \Theta = {\mathbb R}_+ \end{align} Putting \begin{align} \overline{\mu}(x) = \frac{x_1 + x_2 + \cdots + x_n}{n} \end{align}

we define the estimator $E: {\mathbb R}^n \to \Theta (\equiv {\mathbb R}_+ )$ by

\begin{align} & E(x)=E(x_1, x_2, \ldots , x_n ) = \sqrt{ \frac{(x_1-\overline{\mu}(x))^2 + (x_2-\overline{\mu}(x))^2 + \cdots + ( x_n-\overline{\mu}(x))^2 }{n} } \end{align} and the system quantity $\pi: \Omega \to \Theta$ by \begin{align} & \Omega={\mathbb R} \times {\mathbb R}_+ \ni \omega = (\mu, \sigma ) \mapsto \pi (\omega ) = \sigma \in \Theta={\mathbb R}_+ \end{align} 6.4.2: Confidence interval Our present problem is as follows.
Problem 6.12 [Confidence interval for population variance]. Consider the simultaneous normal measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times {\mathbb R}_+)}$ $({\mathsf O}_G^n = ({\mathbb R}^n, {\mathcal B}_{\mathbb R}^n, {{{G}}^n}) ,$ $S_{[(\mu, \sigma)]})$. Assume that a measured value$x \in X ={\mathbb R}^n$ is obtained by the measurement. Let $0 < \alpha \ll 1$.

Then, find the ${D}_{x}^{1- \alpha; \Theta}( \subseteq \Theta)$ (which may depend on $\mu$) such that

 $\bullet$ the probability that $\sigma \in {D}_{x}^{1- \alpha; \Theta}$ is more than $1-\alpha$

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

Consider the following semi-distance $d_{\Theta}^{(1)}$ in $\Theta (={\mathbb R}_+ )$:

\begin{align} d_{\Theta}^{(1)}(\theta_1, \theta_2) = | \int_{\sigma_1}^{\sigma_2} \frac{1}{\sigma} d \sigma | = |\log{\sigma_1} - \log{\sigma_2} | \tag{6.46} \end{align}

For any $\omega=(\mu, {\sigma} ) (\in \Omega= {\mathbb R} \times {\mathbb R}_+ )$, define the positive number $\delta^{1- \alpha}_{\omega}$ $(> 0)$ such that:

\begin{align} \delta^{1-\alpha}_{\omega} & = \inf \{ \eta > 0: [F (E^{-1} ( {{ Ball}_{d_{\Theta}^{(1)}}}(\omega ; \eta))](\omega ) \ge 1- \alpha \} \nonumber \\ & = \inf \{ \eta > 0: [F (E^{-1} ( {{ Ball}^C_{d_{\Theta}^{(1)}}}(\omega ; \eta))](\omega ) \le \alpha \} \tag{6.47} \end{align} where \begin{align} & {{ Ball}^C_{d_{\Theta}^{(1)}}}(\omega ; \eta ) = {{ Ball}^C_{d_{\Theta}^{(1)}}}((\mu ; {\sigma} ), \eta ) = {\mathbb R} \times \{ \sigma' \;:\; |\log(\sigma'/{\sigma})| \ge \eta \} \\ = & {\mathbb R} \times \big( (0,{\sigma} e^{-\eta}] \cup [{\sigma} e^{\eta} , \infty ) \big) \tag{6.48} \end{align} Then, \begin{align} & E^{-1}( {{ Ball}^C_{d_{\Theta}^{(1)}}}(\omega ; \eta )) = E^{-1} \Big( {\mathbb R} \times \big( (0,{\sigma} e^{-\eta}] \cup [{\sigma} e^{\eta} , \infty ) \big) \Big) \nonumber \\ = & \{ (x_1, \ldots , x_n ) \in {\mathbb R}^n \;: \; \Big( \frac{\sum_{k=1}^n ( x_k - \overline{\mu} (x))^2}{n} \Big)^{1/2} \le {\sigma} e^{ -\eta } \mbox{ or } {\sigma} e^{ \eta } \le \Big( \frac{\sum_{k=1}^n ( x_k - \overline{\mu} (x))^2}{n} \Big)^{1/2} \} \\ & \tag{6.49} \end{align}

Hence we see, by the Gauss integral (6.7), that

\begin{align} & [{{{G}}}^n (E^{-1}({{ Ball}^C_{d_{\Theta}^{(1)}}}(\omega; \eta ))] (\omega) \nonumber \\ = & \frac{1}{({{\sqrt{2 \pi }{\sigma}{}}})^n} \underset{{ E^{-1} \Big( {\mathbb R} \times \big( (0,{\sigma} e^{-\eta}] \cup [{\sigma} e^{\eta} , \infty ) \big) \Big) }}{\int \cdots \int} \exp[{}- \frac{\sum_{k=1}^n ({}{x_k} - {}{\mu} )^2 } {2 {\sigma}^2} {}] d {}{x_1} d {}{x_2}\cdots dx_n \nonumber \\ = & \int_0^{{n} e^{- 2 \eta}} p^{\chi^2}_{n-1} (x ) dx + \int_{{n} e^{ 2 \eta}}^\infty p^{\chi^2}_{n-1} (x ) dx = 1- \int_{{n} e^{- 2 \eta}}^{{n} e^{ 2 \eta}} p^{\chi^2}_{n-1} (x ) dx \tag{6.50} \end{align}

Using the chi-squared distribution $p^{{\chi}^2}_{n-1}({ x} )$ (with $n-1$ degrees of freedom) in (6.8), define the $\delta^{1-\alpha}_{\omega}$ such that

\begin{align} 1- \alpha = \int_{{n} e^{-2 \delta^{1-\alpha}_{\omega}}}^{{n} e^{2 \delta^{1-\alpha}_{\omega}}} p^{\chi^2}_{n-1} (x ) dx \tag{6.51} \end{align}

where it should be noted that the $\delta^{1-\alpha}_{\omega}$ depends on only $\alpha$ and $n$. Thus, put

\begin{align} \delta^{1-\alpha}_{\omega} = \delta^{1-\alpha}_{n} \tag{6.52} \end{align}

Hence we get, for any $x$ $(\in X)$, the $D_x^{{1 - \alpha, \Omega }}$ ( the $({1 - \alpha })$-confidence interval of $x$ ) as follows:

\begin{align} D_x^{{1 - \alpha, \Omega }} & = \{ {\omega} (\in \Omega) : d^{(1)}_\Theta (E(x), \pi(\omega) ) \le \delta^{1 - \alpha}_{n} \} \nonumber \\ & = \{ (\mu, \sigma ) \in {\mathbb R} \times {\mathbb R}_+ \;: \; \sigma e^{- \delta^{1 - \alpha }_{n} } \le \Big( \frac{\sum_{k=1}^n ( x_k - \overline{\mu} (x))^2}{n} \Big)^{1/2} \le \sigma e^{ \delta^{1 - \alpha }_{n} } \} \tag{6.53} \end{align}

Recalling (6.4), i.e., $\overline{\sigma}(x) = \Big( \frac{\sum_{k=1}^n ( x_k - \overline{\mu} (x))^2}{n} \Big)^{1/2} ={(\frac{{\overline{SS}}(x)}{n})}^{1/2}$, we conclude that

\begin{align} D_x^{{1 - \alpha, \Omega }} & = \{ (\mu, \sigma ) \in {\mathbb R} \times {\mathbb R}_+ \;: \; \overline{\sigma}(x) e^{ - \delta^{1 - \alpha }_{n}} \le \sigma \le \overline{\sigma}(x) e^{ \delta^{1 - \alpha }_{n}} \} \nonumber \\ & = \{ (\mu, \sigma ) \in {\mathbb R} \times {\mathbb R}_+ \;: \; \frac{e^{ - 2\delta^{1 - \alpha }_{n}}}{n}{\overline{SS}}(x) \le \sigma^2 \le \frac{e^{ 2\delta^{1 - \alpha }_{n}}}{n}{\overline{SS}}(x) \} \tag{6.54} \end{align} And \begin{align} D_x^{{1 - \alpha, \Theta }} & = \{ \sigma \in {\mathbb R}_+ \;: \; \overline{\sigma}(x) e^{ - \delta^{1 - \alpha }_{n}} \le \sigma \le \overline{\sigma}(x) e^{ \delta^{1 - \alpha }_{n}} \} \nonumber \\ & = \{ (\mu, \sigma ) \in {\mathbb R} \times {\mathbb R}_+ \;: \; \frac{e^{ - 2\delta^{1 - \alpha }_{n}}}{n}{\overline{SS}}(x) \le \sigma^2 \le \frac{e^{ 2\delta^{1 - \alpha }_{n}}}{n}{\overline{SS}}(x) \} \nonumber \end{align}