13.2: Regression analysis

According to Fisher's maximum likelihood method (Theorem5.6 in $\S$5.2) and the existence theorem of the realized causal observable, we have the following theorem:

Theorem 13.4 [Regression analysis]

Let $(T{{=}} \{ t_0,t_1,$ $\ldots,$ $t_N\} , \pi: T \setminus \{ t_0 \} \to T)$ be a tree. Let $\widehat{\mathsf O}_{T{}}$ ${{=}} ( \times_{t \in T} X_t,$ ${\boxtimes_{t \in T} {\cal F}_t},$ ${\widehat F}_{t_0})$ be the realized causal observable of a sequential causal observable $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{\pi(t), t }{}:$ $L^\infty ({\Omega}_{t})\to L^\infty ({\Omega }_{\pi(t)}) \}_{ t \in T\setminus \{t_0\} }$ $]$. Consider a measurement

\begin{align} {\mathsf M}_{L^\infty (\Omega_{t_0})} (\widehat{\mathsf O}_{T{}} {{=}} ( \times_{t \in T} X_t, {\boxtimes_{t \in T} {\cal F}_t}, {\widehat F}_{t_0}), S_{[\ast]}) \qquad \end{align}

Assume that a measured value obtained by the measurement belongs to ${\widehat \Xi} \;(\in {\boxtimes_{t \in T} {\cal F}_t} )$. Then, there is a reason to infer that

\begin{align} [{}\ast{}] = {\omega_{t_0}} \end{align} where ${\omega_{t_0}} \;(\in \Omega_{t_0})$ is defined by \begin{align} [\widehat F_{t_0} ( {\widehat \Xi} )](\omega_{t_0}) = \max_{\omega \in \Omega_{t_0}} [\widehat F_{t_0} ( {\widehat \Xi} ){}](\omega) \end{align}

The poof is a direct consequence of Axiom 2 (causality; $\S$10.3) and Fisher maximum likelihood method (Theorem 5.6Theorem in $\S$5.2). Thus, we omit it.

It should be noted that
 $(\sharp):$ regression analysis is related to both Axiom 1 (measurement; $\S$2.7) and Axiom 2 (causality; $\S$10.3)
That is, \begin{align} & \fbox{regression analysis} \\ \\ = & \color{blue}{ \color{magenta}{ \underset{\mbox{or, some inference method}}{\fbox{Fisher's maximum likelihho method}} } + \fbox{realized causal observable} } \end{align}

Now we will answer Problem 13.1 in terms of quantum language, that is, in terms of regression analysis (Theorem 13.4).

Answer 13.5 [(Continued from Problem 13.1 (Inference problem))Regression analysis]

Let $(T{{=}}$ $\{ 0,1,2\} , \pi: T \setminus \{ 0 \} \to T)$ be the parent map representation of a tree, where it is assumed that

\begin{align} \pi(1)=\pi(2)=0 \end{align}

Put $\Omega_0 =\{\omega_1, \omega_2,\ldots,\omega_5\}$, $\Omega_1 = \mbox{interval} [100,200]$, $\Omega_2 = \mbox{interval}[30,110]$. Here, we consider that

\begin{align} \Omega_0 \ni \omega_n \cdots \cdots \mbox{a state such that "the boy is helped by a student $\omega_n$" } \quad (n=1,2,...,5) \end{align}

For each $t \;(\in \{ 1,2 \})$, the deterministic map $\phi_{0,t}{}: \Omega_0 \to \Omega_t$ is defined by $\phi_{0,1}=h$(height function), $\phi_{0,2}=w$(weight function). Thus, for each $t \;(\in \{ 1,2\})$, the deterministic causal operator $\Phi_{0,t}{}: L^\infty (\Omega_t) \to L^\infty (\Omega_0)$ is defined by

\begin{align} [\Phi_{0,t} f_t] (\omega) = f_t (\phi_{0,t} (\omega)) \qquad (\forall \omega \in \Omega_0, \forall f_t \in L^\infty (\Omega_t)) \end{align}

For each $t=1,2$, let ${\mathsf O}_{G_{\sigma_t}} {{=}} ({\mathbb R}, {\cal B}_{\mathbb R}, G_{\sigma_t})$ be the normal observable with a standard deviation $\sigma_t >0$ in $L^\infty (\Omega_t)$. That is,

\begin{align} [G_{\sigma_t}(\Xi)] (\omega) = \frac{1}{\sqrt{2 \pi \sigma_t^2}} \int_{\Xi} e^{- \frac{(x - \omega)^2}{2 \sigma_t^2}} dx \quad (\forall \Xi \in {\cal B}_{\mathbb R}, \forall \omega \in \Omega_t ) \end{align}

Thus, we have a deterministic sequence observable $[ \{{\mathsf O}_{G_{\sigma_t}}\}_{t=1,2} , \{ \Phi_{0, t} : L^\infty (\Omega_t) \to L^\infty (\Omega_0)\}_{t=1,2} ]$. Its realization $\widehat{\mathbb O}_{T{}}$ ${{=}}$ $({\mathbb R}^2, {\cal F}_{{\mathbb R}^2}, {\widehat F}_0)$ is defined by

\begin{align} & [{\widehat F}_0(\Xi_1 \times \Xi_2 )] (\omega) = [\Phi_{0,1} G_{\sigma_1} ] (\omega) \cdot [\Phi_{0,2} G_{\sigma_2}] (\omega) = [G_{\sigma_1} ({\Xi_1})] (\phi_{0,1}(\omega)) \cdot [G_{\sigma_2} ({\Xi_2})] (\phi_{0,2}(\omega)) \\ \\ & \qquad \qquad \qquad \qquad (\forall \Xi_1, \Xi_2 \in {\cal B}_{\mathbb R} , \; \forall \omega \in \Omega_0 = \{\omega_1, \omega_2,\ldots,\omega_5\} ) \end{align}

Let $N$ be sufficiently large. Define intervals $\Xi_1, \Xi_2 \subset {\mathbb R}$ by

\begin{align} \Xi_1 =\left[{}165 - \frac1{N}, 165 + \frac1{N}\right], \qquad \Xi_2 =\left[{}65 - \frac1{N}, 65+ \frac1{N} \right] \end{align} The measured data obtained by a measurement ${\mathsf M}_{L^\infty (\Omega_0)} (\widehat{\mathbb O}_{T{}}, S_{[\ast]})$ is \begin{align} (165,65) \; (\in {\mathbb R}^2) \end{align}

Thus, measured value belongs to $\Xi_1 \times \Xi_2$. Using regression analysis ( Theorem 13.4 ) is characterized as follows:

 $(\sharp):$ Find $\omega_0$ $(\in \Omega_0)$ such as \begin{align} [{\widehat F}_0(\{\Xi_1 \times \Xi_2)] (\omega_0 ) = \max_{\omega \in \Omega } [{\widehat F}_0(\{\Xi_1 \times \Xi_2)] (\omega ) \end{align}
Since $N$ is sufficiently large, \begin{align} ({{\sharp}} ) \Longrightarrow & \max_{ \omega \in \Omega_0 } \frac1{{\sqrt{(2 \pi)^2 \sigma_1^2 \sigma_2^2}{}}} \mathop{ \int\int}_{ \Xi_1 \times \Xi_2 } \exp { [ {}- \frac{ ({}{x_1} - h(\omega))^2 }{2 \sigma_1^2} - \frac{ ({}{x_2} - {}w(\omega))^2 }{2 \sigma_2^2} {}] } d {}{x_1} d {}{x_2} \\ \Longrightarrow & \max_{ \omega \in \Omega_0 } \exp { [ {}- \frac{ ({}{165} - h(\omega))^2 }{2 \sigma_1^2} - \frac{ ({}{65} - {}w(\omega))^2 }{2 \sigma_2^2} {}] } \\ \Longrightarrow & \min_{ \omega \in \Omega_0 } { [ \frac{ ({}{165} - h(\omega))^2 }{2 \sigma_1^2} + \frac{ ({}{65} - {}w(\omega))^2 }{2 \sigma_2^2} {}] } \qquad \mbox{( for simplicity, assume that $\sigma_1=\sigma_2$)} \\ \Longrightarrow & \mbox{When $\omega_4$, minimum value }\frac{ ({}{165} - 170)^2+({}{65} - 60)^2 }{2 \sigma_1^2} \mbox{is obtained} \\ \Longrightarrow & \mbox{The student is $\omega_4$} \end{align} Therefore, we can infer that the student who helps the boy is $\omega_4$.
$\square \quad$

Now, let us answer Problem 13.2 in terms of quantum language (or, by using regression analysis (Theorem 13.4).

Answer 13.6 [(Continued from Problem 13.2 (Control problem))Regression analysis]

In Problem 13.2, it is natural to consider that the tree $T=\{0,1,2,3\}$ is discrete time, that is, the linear ordered set with the parent map $\pi :T\setminus\{0\} \to T$ such that $\pi ( t ) = t-1$ $\;(t=1,2,3)$. For example, put

\begin{align} \Omega_0 = [0,\; 1] \times [0,\; 2], \; \Omega_1 =[0,\; 4] \times [0,\; 2], \;\; \Omega_2 = [0,,\; 6] \times [0,\; 2], \;\; \Omega_3 = [0, \; 8] \times [0,\; 2] \end{align}

For each $t=1,2,3$, define the deterministic causal map $\phi_{\pi(t),t}{}: \Omega_{\pi(t)} \to \Omega_t$ by (13.3), that is,

\begin{align} & \phi_{0,1}(\omega_0 ) = (\alpha + \beta, \beta) &\quad& (\forall \omega_0 =(\alpha, \beta ) \in \Omega_0 = [0,\; 1] \times [0,\; 2]) \\ & \phi_{1,2 }(\omega_1 ) = (\alpha + \beta, \beta) &\quad& (\forall \omega_1 =(\alpha, \beta ) \in \Omega_1=[0,\; 4] \times [0,\; 2] ) \\ & \phi_{2,3 }(\omega_2 ) = (\alpha + \beta, \beta) &\quad& (\forall \omega_2 =(\alpha, \beta ) \in \Omega_2 =[0,\; 6] \times [0,\; 2] ) \end{align}

Thus, we get the deterministic sequence causal map $\{\phi_{\pi(t),t}{}: \Omega_{\pi(t)} \to \Omega_t\}_{t \in \{1,2,3\}}$, and the deterministic sequence causal operator $\{\Phi_{\pi(t),t}{}: L^\infty (\Omega_{t}) \to L^\infty (\Omega_{\pi(t)} ) \}_{t \in \{1,2,3\}}$. That is,

\begin{align} (\Phi_{0,1}f_1)(\omega_0) \!=\! f_1( \phi_{0,1}(\omega_0)) & \quad \; (\forall f_1 \in L^\infty (\Omega_1), \forall \omega_0 \in \Omega_0) \\ (\Phi_{1,2}f_2)(\omega_1) \!=\! f_2( \phi_{1,2}(\omega_1)) & \quad \; ( \forall f_2 \in L^\infty (\Omega_2), \forall \omega_1 \in \Omega_1) \\ (\Phi_{2,3}f_3)(\omega_2) \!=\! f_3( \phi_{2,3}(\omega_2)) & \quad \; ( \forall f_3 \in L^\infty (\Omega_3), \forall \omega_1 \in \Omega_2). \end{align}

Illustrating by the diagram, we see

\begin{align} { {\mbox{${L^\infty (\Omega_0)}$}} } \mathop{\longleftarrow}^{\Phi_{0, 1 } } {\mbox{${L^\infty (\Omega_{1})}$}} \mathop{\longleftarrow}^{\Phi_{1,2 } } {\mbox{${L^\infty (\Omega_{2})}$}} \mathop{\longleftarrow}^{\Phi_{2,3 } } {\mbox{$L^\infty (\Omega_{3})$}} \end{align}

And thus, $\phi_{0,2}(\omega_0)=\phi_{1,2}(\phi_{0,1}(\omega_0))$, $\phi_{0,3}(\omega_0)=\phi_{2,3}(\phi_{1,2}(\phi_{0,1}(\omega_0)))$, Therefore, note that $\Phi_{0,2}=\Phi_{0,1}\cdot \Phi_{1,2}$, $\;\;$ $\Phi_{0,3}=\Phi_{0,1}\cdot \Phi_{1,2} \cdot \Phi_{2,3}$.

Let ${\mathbb R}$ be the set of real numbers. Fix $\sigma>0$. For each $t=0,1,2$, define the normal observable ${\mathsf O}_{t} {{\equiv}} ({\mathbb R}, {\cal B}_{\mathbb R}, G_{\sigma})$ in $L^\infty (\Omega_t)$ such that

\begin{align} & [G_{\sigma}(\Xi)] (\omega_t ) = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{\Xi} \exp({- \frac{(x - \alpha)^2}{2 \sigma^2}}) dx \\ & (\forall \Xi \in {\cal B}_{\mathbb R}, \forall \omega_t =(\alpha, \beta) \in \Omega_t {{=}} [{}0, \; 2t+2{}]\times [0, \;2]). \end{align}

Thus, we have the deterministic sequential causal observable $[ \{{\mathsf O}_{t}\}_{t=1,2,3} , \{\Phi_{\pi(t),t}{}: L^\infty (\Omega_{t}) \to L^\infty (\Omega_{\pi(t)} ) \}_{t \in \{1,2,3\}} ]$. And thus, we have the realized causal observable $\widehat{\mathsf O}_{T{}}$ ${{=}}$ $({\mathbb R}^3, {\cal F}_{{\mathbb R}^3}, {\widehat F}_0)$ in ${L^\infty (\Omega_0)}$ such that ( using Theorem 12.8 )

\begin{align} & [{\widehat F}_0(\Xi_1 \times \Xi_2 \times \Xi_3)] (\omega_0) = \big[ \Phi_{0,1} \big(G_{\sigma} ({\Xi_1}) \Phi_{1,2} (G_{\sigma} ({\Xi_2}) \Phi_{2,3} (G_{\sigma} ({\Xi_3}) )) \big) \big] (\omega_0) \\ = & [\Phi_{0,1} G_{\sigma} ({\Xi_1})] (\omega_0) \cdot [\Phi_{0,2} G_{\sigma}({\Xi_2})] (\omega_0) \cdot [\Phi_{0,3} G_{\sigma}({\Xi_3})] (\omega_0) \\ = & [G_{\sigma} ({\Xi_1})] (\phi_{0,1}(\omega_0)) \cdot [G_{\sigma} ({\Xi_2})] (\phi_{0,2}(\omega_0)) \cdot [G_{\sigma}({\Xi_3})] (\phi_{0,3}(\omega_0)) \\ & \qquad \qquad (\forall \Xi_1, \Xi_2, \Xi_3 \in {\cal B}_{\mathbb R}, \; \forall \omega_0 =(\alpha, \beta) \in \Omega_0 = [0, \; 1] \times [0, \; 2]) \end{align}

Our problem (i.e., Problem 13.2Problem}) is as follows,

 $(\sharp_1):$ Determine the parameter $(\alpha, \beta )$ such that the measured value of ${\mathsf M}_{L^\infty (\Omega_0)} ($ $\widehat{\mathsf O}_{T{}},$ $S_{[\ast]})$ is equal to $(1.9, 3.0, 4.7)$

For a sufficiently large natural number $N$, put

\begin{align} \Xi_1 =\left[{}1.9 - \frac1{N}, 1.9 + \frac1{N}\right], \Xi_2 =\left[{}3.0 - \frac1{N}, 3.0 + \frac1{N}\right], \Xi_3 =\left[{}4.7 - \frac1{N}, 4.7 + \frac1{N}\right] \end{align}

Fisher's maximum likelihood method (Theorem 5.6) says that the above ($\sharp_1$) is equivalent to the following problem

 $(\sharp_2):$ Find $(\alpha, \beta)$ $(=\omega_0 \in \Omega_0)$ such that \begin{align} [{\widehat F}_0(\Xi_1 \times \Xi_2 \times \Xi_3 )] (\alpha, \beta) = \max_{ (\alpha, \beta )} [{\widehat F}_0(\Xi_1 \times \Xi_2 \times \Xi_3 )] \end{align}
Since $N$ is assumed to be sufficiently large, we see

\begin{align} ({{\sharp_2}} ) \Longrightarrow & \max_{ (\alpha, \beta) \in \Omega_0 } [{\widehat F}_0(\Xi_1 \times \Xi_2 \times \Xi_3 )] (\alpha, \beta) \\ \Longrightarrow & \max_{ (\alpha, \beta) \in \Omega_0 } \frac1{{\sqrt{2 \pi \sigma^2}{}}^3} \mathop{ \int\int\int}_{ \Xi_1 \times \Xi_2 \times \Xi_3 } e^{ [{}- \frac{ ({x_1} - (\alpha + \beta))^2 + ({x_2} - (\alpha + 2 \beta))^2 + ({x_3} - (\alpha + 3 \beta))^2 }{2 \sigma^2}{}] } \\ & \hspace{4cm} \times d {}{x_1} d {}{x_2} d {}{x_3} \\ \Longrightarrow & \max_{ (\alpha, \beta) \in \Omega_0 } \exp(-J/(2 \sigma^2)) \\ \Longrightarrow & \min_{ (\alpha, \beta) \in \Omega_0 } J \end{align} where $J = (1.9 -(\alpha + \beta))^2 + (3.0 - (\alpha + 2 \beta))^2 + (4.7 - (\alpha + 3 \beta))^2$
( $\frac{\partial}{\partial \alpha}\{\cdots \}=0, \frac{\partial}{\partial \beta}\{ \cdots \}=0 \mbox{ and thus,}$ )
\begin{align} \Longrightarrow & \left\{\begin{array}{ll} (1.9 -(\alpha + \beta)) + (3.0 - (\alpha + 2 \beta)) + (4.7 - (\alpha + 3 \beta)) = 0 \\ (1.9 -(\alpha + \beta)) + 2 (3.0 - (\alpha + 2 \beta)) + 3 (4.7 - (\alpha + 3 \beta)) = 0 \end{array}\right. \\ \Longrightarrow & \quad (\alpha, \beta) = (0.4, 1.4) \end{align} Therefore, in order to obtain a measured value $(1.9, \; 3.0, \; 4.7)$, it suffices to put \begin{align} (\alpha, \beta) =(0.4, \; 1.4) \end{align} Remark 13.7 For completeness, note that,
 $\bullet$ From the theoretical point of view, "inference" = "control"
Thus, we conclude that statistics and dynamical system theory are essentially the same.