Example 5.9 [Normal observable(i): $\Omega={\mathbb R}$] As mentioned before, we again discuss the normal observable in what follows. Consider the classical basic structure: \begin{align} \mbox{ $[ C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]\qquad$ (where, $\Omega={\mathbb R} )$ } \end{align} Fix $\sigma > 0$, and consider the normal observable ${\mathsf O}_{G_\sigma} $ ${{=}}$ $({\mathbb R} , {\cal B}_{{\mathbb R}}^{} , {G_\sigma})$ in $L^\infty ({\mathbb R})$ (where $\Omega={\mathbb R}$) such that \begin{align} &[{G_\sigma}(\Xi)] ( {\mu} ) = \frac{1}{{\sqrt{2 \pi } \sigma}} \int_{\Xi} \exp[{}- \frac{1}{2 \sigma^2 } ({x} - {\mu} )^2 ] d{x} \\ & \qquad (\forall \Xi \in {\cal B}_{{\mathbb R}}^{}, \quad \forall {\mu} \in \Omega= {\mathbb R}) \end{align} Thus, the simultaneous observable $\times_{k=1}^3 {\mathsf O}_{G_\sigma}$ (in short, ${\mathsf O}_{G_\sigma}^3 $) ${{=}}$ $({\mathbb R}^3 , {\cal B}_{{\mathbb R}^3}^{} ,$ $ G_{\sigma}^3)$ in $L^\infty ({\mathbb R})$ is defined by \begin{align} & [G_{\sigma}^3({\Xi_1 \times \Xi_2 \times \Xi_3 })] ( {\mu} ) = [{G_\sigma}({\Xi_1})] (\mu) \cdot [{G_\sigma}({\Xi_2})] (\mu) \cdot [{G_\sigma}({\Xi_3})](\mu) \\ = & \frac{1}{({\sqrt{2 \pi } \sigma)^3}} \iiint_{\Xi_1 \times \Xi_2 \times \Xi_3} \exp[{}- \frac{({x_1} - {\mu} )^2 +({x_2} - {\mu} )^2 + ({x_3} - {\mu} )^2 }{2 \sigma^2 } ] \\ &\hspace{5cm} \times d{x_1} d{x_2} d{x_3} \\ & \qquad \qquad \qquad \qquad (\forall \Xi_k \in {\cal B}_{{\mathbb R}}^{},k=1,2,3, \quad \forall {\mu} \in \Omega= {\mathbb R}) \end{align} Thus, we get the measurement ${\mathsf M}_{L^\infty({\mathbb R})} ({\mathsf O}_{G_\sigma}^3, S_{[{}\ast{}] })$ Now we consider the following problem:

$(a):$

Assume that a measured value $(x^0_1, x^0_2 , x^0_3)$ $(\in {\mathbb R}^3)$ is obtained by the measurement ${\mathsf M}_{L^\infty({\mathbb R})} ({\mathsf O}_{G_\sigma}^3,$ $ S_{[{}\ast{}] })$. Then,　infer the unknown state　$[\ast]　(\in {\mathbb R})$.

Answer(a)：　Put \begin{align} \Xi_i = [ x_i^0 -\frac{1}{N},x_i^0 +\frac{1}{N}] \qquad(i=1,2,3) \end{align} Assume that　$N$　is sufficiently large.　Fisher's maximum likelihood method (Theorem5.6)　says that　the　unknown state　$[{}\ast{}]$　$= \mu_0$　is found in what follows. \begin{align} [G_{\sigma}^3({\Xi_1 \times \Xi_2 \times \Xi_3 })] ( {\mu_0} ) = \max_{\mu \in {\mathbb R}} [G_{\sigma}^3({\Xi_1 \times \Xi_2 \times \Xi_3 })] ( {\mu} ) \end{align} Since $N$ is sufficiently large,　we see \begin{align} & \frac{1}{({\sqrt{2 \pi } \sigma)^3}} \exp[{}- \frac{({x^0_1} - {\mu_0} )^2 +({x^0_2} - {\mu_0} )^2 + ({x^0_3} - {\mu_0} )^2 }{2 \sigma^2 } ] \\ = & \max_{\mu \in {\mathbb R}} \Big[ \frac{1}{({\sqrt{2 \pi } \sigma)^3}} \exp[{}- \frac{({x^0_1} - {\mu} )^2 +({x^0_2} - {\mu} )^2 + ({x^0_3} - {\mu} )^2 }{2 \sigma^2 } ] \Big] \end{align} That is, \begin{align} ({x^0_1} - {\mu_0} )^2 +({x^0_2} - {\mu_0} )^2 + ({x^0_3} - {\mu_0} )^2 = \min_{\mu \in {\mathbb R}} \big\{ ({x^0_1} - {\mu} )^2 +({x^0_2} - {\mu} )^2 + ({x^0_3} - {\mu} )^2 \big\} \end{align} Therefore,　solving　$\frac{d}{d\mu} \{ \cdots \}=0$,　we conclude that \begin{align} \mu_0 = \frac{{x^0_1}+{x^0_2}+{x^0_3}}{3} \end{align} [Normal observable　(ii)] Next consider the classical basic structure: \begin{align} [ C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]\quad (\mbox{where ,} \Omega = {\mathbb R} \times {\mathbb R}_+ ) \end{align} and consider the case:

$\bullet$

we know that the length of the pencil $\mu$ is satisfied that 10cm $\le $ $\mu$ cm $\le $30cm.

And we assume that

$(\sharp):$

the length of the pencil $\mu$ and the roughness $\sigma$ of the ruler are unknown.

That is, assume that the state space $\Omega$ $=$ $[10,30{}] \times {\mathbb R}_+ $ $ \big( {{=}} \{ \mu \in {\mathbb R} \; |\; 10 {{\; \leqq \;}}\mu {{\; \leqq \;}}30 \} \times \{ \sigma \in {\mathbb R} \;|\; \sigma > 0 \} \big) $ Define the observable ${\mathsf O}$ ${{=}}$ $({\mathbb R} , {\cal B}_{{\mathbb R}}^{} , G)$ in $L^\infty ([10,30 {}]\times {\mathbb R}_+)$ such that \begin{align} [G(\Xi)](\mu, \sigma ) = [{G_\sigma}({\Xi})] (\mu) \quad (\forall \Xi \in {\cal B}_{{\mathbb R}}^{}, \;\; \forall ({\mu}, \sigma) \in \Omega= [10,30] \times {\mathbb R}_+ ) \end{align} Therefore, the simultaneous observable ${\mathsf O}^3 $ ${{=}}$ $({\mathbb R}^3 , {\cal B}_{{\mathbb R}^3}^{} ,$ $ G^3)$ in $C ([10,30{}]\times {\mathbb R}_+)$ is defined by \begin{align} & [G^3({\Xi_1 \times \Xi_2 \times \Xi_3 })] ( {\mu},\sigma ) = [G(\Xi_1)](\mu, \sigma ) \cdot [G(\Xi_2)](\mu, \sigma ) \cdot [G(\Xi_3)](\mu, \sigma ) \\ = & \frac{1}{({\sqrt{2 \pi } \sigma)^3}} \int_{\Xi_1 \times \Xi_2 \times \Xi_3} \exp[{}- \frac{({x_1} - {\mu} )^2 +({x_2} - {\mu} )^2 + ({x_3} - {\mu} )^2 }{2 \sigma^2 } ] d{x_1} d{x_2} d{x_3} \\ & \qquad \qquad \qquad (\forall \Xi_k \in {\cal B}_{{\mathbb R}}^{},k=1,2,3, \quad \forall ({\mu},\sigma) \in \Omega= [10,30{}]\times {\mathbb R}_+) \end{align} Thus, we get the simultaneous measurement ${\mathsf M}_{L^\infty([10,30]\times {\mathbb R}_+)} ({\mathsf O}^3, S_{[{}\ast{}] })$. Here, we have the following problem:

$(b):$

When a measured value $(x^0_1, x^0_2 , x^0_3)$ $(\in {\mathbb R}^3)$ is obtained by the measurement ${\mathsf M}_{L^\infty([10,30]\times {\mathbb R}_+)}$ $ ({\mathsf O}^3,$ $ S_{[{}\ast{}] })$, infer the unknown state $[\ast] ( = (\mu_0,\sigma_0) \in [10,30]\times {\mathbb R}_+)$, i.e., the length $\mu_0$ of the pencil and the roughness $\sigma_0$ of the ruler.

Answer (b): By the same way of (a), Fisher's maximum likelihood method (Theorem5.6) says that the unknown state $[{}\ast{}]$ $= (\mu_0,\sigma_0)$ such that \begin{align} & \frac{1}{({\sqrt{2 \pi } \sigma_0)^3}} \exp[{}- \frac{({x^0_1} - {\mu_0} )^2 +({x^0_2} - {\mu_0} )^2 + ({x^0_3} - {\mu_0} )^2 }{2 \sigma_0^2 } ] \nonumber \\ = & \max_{(\mu,\sigma) \in [10, 30] \times {\mathbb R}_+} \Big\{ \frac{1}{({\sqrt{2 \pi } \sigma)^3}} \exp[{}- \frac{({x^0_1} - {\mu} )^2 +({x^0_2} - {\mu} )^2 + ({x^0_3} - {\mu} )^2 }{2 \sigma^2 } ] \Big\} \tag{5.12} \end{align} Thus, solving $\frac{\partial }{\partial \mu}\{\cdots \}=0$, $\frac{\partial }{\partial \sigma}\{\cdots \}=0$ we see \begin{align} & \mu_0 = \left\{\begin{array}{ll} 10 \quad \qquad & (\mbox{when } \mbox{} (x^0_1+x^0_2+x^0_3)/3< 10\; ) \\ \\ ({x^0_1}+{x^0_2}+{x^0_3})/3 \quad \qquad& (\mbox{when } \mbox{} 10 {{\; \leqq \;}} (x^0_1+x^0_2+x^0_3)/3{{\; \leqq \;}}30 \; \mbox) \\ \\ 30 \quad& (\mbox{when } 30 < (x^0_1+x^0_2+x^0_3)/3 \;) \end{array}\right. \tag{5.13} \\ & \sigma_0 = \sqrt{ \{ (x^0_1-{\widetilde \mu})^2 + (x^0_2-{\widetilde \mu})^2 + (x^0_3-{\widetilde \mu})^2 \}/3 } \nonumber \end{align} where \begin{equation*} {\widetilde \mu}=({x^0_1}+{x^0_2}+{x^0_3} )/3 \end{equation*}

Example 5.10 [Fisher's maximum likelihood method for the simultaneous normal measurement] Consider the simultaneous normal observable ${\mathsf O}_G^n = ({\mathbb R}^n, {\mathcal B}_{\mathbb R}^n, {{{G}}^n} )$ in $L^\infty ({\mathbb R} \times {\mathbb R}_+)$ (such as defined in formula (5.3). This is essentially the same as the simultaneous observable ${\mathsf O}^n$ $=$ $({\mathbb R}^n, {\mathcal B}_{{\mathbb R}^n}, \times_{k=1}^n {G_\sigma})$ in $L^\infty ({\mathbb R} \times {\mathbb R}_+)$. That is, \begin{align} & [(\times_{k=1}^n {G_\sigma})(\Xi_1 \times \Xi_2 \times \cdots \times \Xi_n )](\omega) = \times_{k=1}^n [G_{\sigma}(\Xi_k) ]( {\omega} ) \nonumber \\ = & \times_{k=1}^n \frac{1}{{\sqrt{2 \pi } \sigma}} \int_{\Xi_k} \exp \left[ {}- \frac{1}{2 \sigma^2 } ({x_k} - {\mu} )^2 \right] d{x_k} \nonumber \\ & \quad (\forall \Xi_k \in {\cal B}_{{X}}^{}( ={\cal B}_{{\mathbb R}}^), \; \forall {\omega}=(\mu, \sigma ) \in \Omega (={\mathbb R}\times {\mathbb R}_+ )) \nonumber \end{align} Assume that a measured value $x=(x_1, x_2, \ldots, x_n ) (\in {\mathbb R}^n )$ is obtained by the measurement ${\mathsf M}_{L^\infty ({\mathbb R} \times {\mathbb R}_+ )} ({\mathsf O}^n = ({\mathbb R}^n, {\mathcal B}_{\mathbb R}^n, {{{G}}_\sigma^n} )$,$S_{[\ast]})$. The likelihood function $L_x(\mu, \sigma)(=L(x, (\mu,\sigma)) $ is equal to \begin{align} L_x(\mu, \sigma) & = \frac{1}{({{\sqrt{2 \pi }\sigma{}}})^n} \exp[{}- \frac{\sum_{k=1}^n ({}{x_k} - {}{\mu} )^2 } {2 \sigma^2} {}] \nonumber \end{align} in the sense of (5.9), \begin{align} L_x(\mu, \sigma) & = \frac{ \frac{1}{({{\sqrt{2 \pi }\sigma{}}})^n} \exp[{}- \frac{\sum_{k=1}^n ({}{x_k} - {}{\mu} )^2 } {2 \sigma^2} {}] } { \frac{1}{({{\sqrt{2 \pi }\overline{\sigma}(x){}}})^n} \exp[{}- \frac{\sum_{k=1}^n ({}{x_k} - {}{\overline{\mu}(x)} )^2 } {2 \overline{\sigma}(x)^2} {}] } \\ & (\forall x = (x_1, x_2, \ldots , x_n ) \in {\mathbb R}^n, \quad \forall {}{\omega}=(\mu, \sigma ) \in \Omega = {\mathbb R}\times {\mathbb R}_+). \end{align} Therefore,we get the following likelihood equation: \begin{align} \frac{\partial L_x(\mu, \sigma)}{\partial \mu}=0, \quad \frac{\partial L_x(\mu, \sigma)}{\partial \sigma}=0 \tag{5.15} \end{align} which is easily solved. That is, Fisher's maximum likelihood method (Theorem5.6) says that the unknown state $[\ast]=(\mu, \sigma)$ $(\in {\mathbb R} \times {\mathbb R}_+)$ is inferred as follows. \begin{align} & \mu=\overline{\mu}(x) =\frac{x_1 + x_2+ \ldots + x_n }{n}, \quad \tag{5.16} \\ & \sigma= \overline{\sigma}(x) = \sqrt{\frac{\sum_{k=1}^n (x_k - \overline{\mu}(x))^2}{n}} \tag{5.17} \end{align}