19.1: Belief, probability and odds

For instance, we want to formulate the following "probability":

$(A):$ the " probability" that Japan will win the victory in the next FIFA World Cup

This is possible (cf. ref. [8] in $\S$0.0), if " parimutuel betting ( or, odds in bookmaker )" is formulated by Axiom$^{(m)}$ 1 ( mixed measurement ). The purpose of this chapter is to show it, and further, to propose the principle of equal weight, that is,
$(B):$ the principle that, in the absence of any reason to expect one event rather than another, all the possible events should be assigned the same probability}.
whose justice is not solved yet. That is, it is one of the most important unsolved problems in statistics.

In Chapter 9, we studied the mixed measurement: that is,

\[ \underset{\mbox{ (=quantum language)}}{\fbox{mixed measurement theory (A)}} := \underbrace{ \color{red}{ \underset{\mbox{ (\(\S\)9.1)}}{ \overset{ [\mbox{ (mixed) Axiom 1}] }{\fbox{mixed measurement}} } } + \underset{\mbox{ ( \(\S \)10.3)}}{ \overset{ [{\mbox{ Axiom 2}}] }{\fbox{Causality}} } }_{\mbox{ a kind of incantation (a priori judgment)}} + \underbrace{ \underset{\mbox{ (\(\S\)3.1) }} { \overset{ {}}{\fbox{Linguistic interpretation}} } }_{\mbox{ the manual on how to use spells}} \tag{9.2} \]

The purpose of this chapter is to characterize "belief" as a kind of mixed measurement.

19.1.1: A simple example; how to describe "belief" in quantum language

We begin with a simplest example (cf. Problem 9.5) as follows.

Problem 19.1 [$=$Problem 9.5Problem; Bayes' method] Assume the following situation:
$(C):$ You do not know which the urn behind the curtain is, $U_1$ or $U_2$, but the "probability": $p$ and $1-p$.
Here, consider the following problem:

Answer 19.2 (=Answer 9.13)

Put $\Omega=\{ \omega_1, \omega_2 \}$ with the discrete metric and the counting measure $\nu_c$, thus, note that $C_0(\Omega)$ $=C(\Omega )$ $= L^\infty (\Omega, \nu )$. Thus, in this chapter, we devote ourselves to the $C^*$-algebraic formulation: Define the observables ${\mathsf O} = ( \{ {{{W}},} {{B}} \},$ $ 2^{\{ {{W}}, {{B}} \} } ,$ $ F)$ and ${\mathsf O}_U = ( \{ {{{U_1}},} {{U_2}} \},$ $ 2^{\{ {{U_1}}, {{U_2}} \} } ,$ $ G_U)$ in $C(\Omega)$ by \begin{align} & F(\{ {{W}} \})(\omega_1)= 0.8, F(\{ {{B}} \})(\omega_1)= 0.2, F(\{ {{W}} \})(\omega_2)= 0.4, F(\{ {{B}} \})(\omega_2)= 0.6 \\ & G_U(\{ {{U_1}} \})(\omega_1)= 1, G_U(\{ {{U_2}} \})(\omega_1)= 0, G_U(\{ {{U_1}} \})(\omega_2)= 0, G_U(\{ {{U_2}} \})(\omega_2)= 1 \end{align}

Here "$W$" and "$B$" means "white" and "black" respectively. Under the identification: $U_1 \approx \omega_1$ and $U_2 \approx \omega_2$, the above situation is represented by the mixed state $\rho^{(p)}_{\mbox{ prior}} ( \in {\mathcal M}_{+1} (\Omega))$ such that \begin{align} \rho^{(p)}_{\mbox{ prior}} = p \delta_{\omega_1} + (1-p) \delta_{\omega_2} \end{align} where $\delta_{\omega}$ is the point measure at $\omega$. Thus, we have the mixed measurement: \begin{align} {\mathsf M}_{C (\Omega)}({\mathsf O} \times {\mathsf O}_{U} := ( \{ {{{W}},} {{B}} \} \times \{U_1, U_2\} ,2^{ \{ {{{W}},} {{B}} \} \times \{U_1, U_2\} } , F \times G_U), S_{[*]}( \rho^{(p)}_{\mbox{ prior}} )) \tag{19.2} \end{align}

Axiom${}^{(m)}$ 1 gives the answer to the (i) in Problem 19.1 as follows.

$(D):$ the probability that a measured value $(x,y) $ obtained by the mixed measurement ${\mathsf M}_{C (\Omega)}({\mathsf O} \times {\mathsf O}_{U}, S_{[*]}( \rho^{(p)}_{\mbox{ prior}} ))$ belongs to $\{W \} \times \{U_1, U_2 \}$ is given by

$$ {}_{{\mathcal M}(\Omega)}(\rho^{(p)}_{\mbox{ prior}}, F(\{W\}) )_{C_{}(\Omega )} = 0.8 p + 0.4 (1-p). $$

Since a white ball is obtained, Answer 9.13Answer} (=Bayes' theorem ) says that a new {mixed state} $\rho^{(p)}_{\mbox{ post}} (\in {\mathcal M}_{+1}(\Omega ))$ is given by

\begin{align} \rho^{(p)}_{\mbox{ post}} & = \frac{ F(\{ {{W}}\} ) \rho^{(p)}_{\mbox{ prior}} }{\int_\Omega [F( \{ {{W}}\} )](\omega) \rho^{(p)}_{\mbox{ prior}}(d \omega )} = \frac{\displaystyle 0.8 p }{\displaystyle 0.8p+0.4(1-p) } \delta_{\omega_1} + \frac{\displaystyle 0.4(1- p) }{\displaystyle 0.8p+0.4(1-p) } \delta_{\omega_2} \tag{19.3} \end{align} Hence, the answer of the (ii) is given by $$ {}_{{\mathcal M}(\Omega)}(\rho^{(p)}_{\mbox{ post}}, G_U(\{ U_1 \}) )_{C_{}(\Omega )} = \frac{\displaystyle 0.8 p }{\displaystyle 0.8p+0.4(1-p) } $$
$\square \quad$

By an analogy of the above Problem 19.1Problem} ( for simplicity, we put: $ p=1/4, \;\; 1-p = 3/4 $ ), we consider as follows.
Assume that there are 100 people. And moreover assume the following sitiation (E) such that, for some reasons,

\begin{align} \mbox{(E)} \; \left\{\begin{array}{ll} \quad \mbox{25 people believe ( or vote) that $[\ast]= U_1$ (i.e., $U_1$ is behind the curtain) } \\ \quad \mbox{75 people believe ( or vote) that $[\ast]= U_2$ (i.e., $U_2$ is behind the curtain) } \end{array}\right. \end{align} That is, we have the following picture (instead of Figure 19.1):

Now, we have the following problem:

Problem 19.3 Consider Situation (E) and Situation (C) ( $ p=1/4, \;\; 1-p = 3/4 $ ). Then,

$(F_1):$ Can Situation (E) be understood like Situation (C) ?
or, in the same sense,
$(F_2):$ Can Situation (E) be formulated in mixed measurement (i.e., Axiom${}^{(m)}$ 1)? $\;\;\;$ That is, can Situation (E) be described in quantum language?

19.1.2: The affirmative answer to Problem 19.3.

Since 100 people know the situation of the urn (i.e., Figure 19.2, the assumption (E) ) implies (G)(=Figure 19.3), that is,
$(G):$ $ \left\{\begin{array}{ll} \mbox{25 people (in 100 people) believe that $[\ast]= U_1$} \\ \quad \Longrightarrow \left\{\begin{array}{ll} \mbox{(G$_1$): 20 people guess (or bet) that a white ball will be picked} \\ \mbox{(G$_2$): 5 people guess (or bet) that a black ball will be picked} \end{array}\right. \\ \mbox{75 people (in 100 people) believe that $[\ast]= U_2$} \\ \quad \Longrightarrow \left\{\begin{array}{ll} \mbox{(G$_3$): 30 people guess (or bet) that a white ball will be picked} \\ \mbox{(G$_4$): 45 people guess (or bet) that a black ball will be picked} \end{array}\right. \end{array}\right. $

Assume that a white ball is picked in the above figure. Then, the above (G$_2$) and (G$_4$) are vanished as follows.

After all, we get the following figure:

Thus we see that \begin{align} \overset{{\mbox{ (prior state)}}}{\underset{\frac{1}{4} \delta_{\omega_1} + \frac{3}{4} \delta_{\omega_2}}{\fbox{Fig. 19.3}}} \xrightarrow[{\mbox{$\qquad \qquad$}}]{} \overset{{\mbox{ (a white ball is picked)}}}{\fbox{Fig. 19.4}} \xrightarrow[{\mbox{$\qquad \qquad$}}]{} \overset{{\mbox{ (post state)}}}{\underset{\frac{2}{5} \delta_{\omega_1} + \frac{3}{5} \delta_{\omega_2}}{\fbox{Fig.19.5}}} \tag{19.4} \end{align}

Considering the mixed measurement (i.e., the (19.2) in the case that $p=1/4$, $1-p=3/4$ ):

\begin{align} {\mathsf M}_{C_{} (\Omega)}({\mathsf O} \times{\mathsf O}_U = ( \{ {{{W}},} {{B}} \} \times \{U_1, U_2\},2^{ \{ {{{W}},} {{B}} \} \times \{U_1, U_2\} } , F\times G_U) , S_{[*]}( \rho^{(1/4)}_{\mbox{ prior}} )) \tag{19.5} \end{align} we see that the above (19.4) is the same as the Bayesian result (19.3).
Note that the measurement (8.5) is interpreted as
$(H):$ choose one person from the 100 people at random, and ask him/her "Do you guess that a white ball (or, a black ball) will be picked from the urn behind the curtain, and its urn is $U_1$ or $U_2$?"

In what follows, let us explain it. Consider the product observable $ \widehat{\mathsf O} \times \widehat{\mathsf O}_{U} $ of $\widehat{\mathsf O}= ( \{W,B \}, 2^{\{W,B \} },$ $ \widehat{F})$ and $\widehat{\mathsf O}_{U}= ( \{U_1, U_2 \}, 2^{\{U_1, U_2 \} }, \widehat{G}_U)$ in $C(\Theta)$ (where $\Theta=\{ \theta_1, \theta_2, ..., \theta_{100} \}$) such that

\begin{align} & [\widehat{F} (\{W\})](\theta_k)=4/5,\;\; [\widehat{F}(\{B\})](\theta_k)=1/5, \;\; (k=1,2,...,25) \nonumber \\ & [\widehat{F}(\{W\})](\theta_k)=2/5,\;\; [\widehat{F}(\{B\})](\theta_k)=3/5, \;\; (k=26,27,...,100) \tag{19.6} \\ & [\widehat{G}_{U}(\{ U_1 \})](\theta_k)=1,\;\; [\widehat{G}_{U}(\{ U_2 \})](\theta_k)=0, \;\; (k=1,2,...,25) \nonumber \\ & [\widehat{G}_{U}(\{ U_1 \})](\theta_k)=0,\;\; [\widehat{G}_{U}(\{ U_2 \})](\theta_k)=1, \;\; (k=26,27,...,100) \tag{19.7} \end{align}

And put $\nu_0= (1/100) \sum_{k=1}^{100} \delta_{\theta_k} (\in {\mathcal M}_{+1}(\Theta))$. Then, the above measurement (H) is formulated by

\begin{align} {\mathsf M}_{C_{} (\Theta)}(\widehat{\mathsf O} \times \widehat{\mathsf O}_U = ( \{ {{{W}},} {{B}} \} \times \{U_1, U_2\},2^{ \{ {{{W}},} {{B}} \} \times \{U_1, U_2\} } , \widehat{F} \times \widehat{G}_U) , S_{[*]}( \nu_0 )) \tag{19.8} \end{align}

which is identified with the measurement (19.5) under the deterministic causal operator $\Phi:C(\Omega) \to C(\Theta)$ such that $\Phi^* (\delta_{\theta_k} ) =\delta_{\omega_1}$ $(k=1,2,...,25)$, $ =\delta_{\omega_2}\; (k=26,27,...,100)$. That is, we see, symbolically,

$$ \fbox{(19.8): the Heisenberg picture} \xleftarrow[\mbox{ identification}]{\Phi} \fbox{(19.5): the Schrödinger picture} $$ Thus, we can answer Problem 19.3 such that
$(I_1):$ Situation (E) can be understood like Situation (C).
That is,
$(I_2):$ Situation (E) can be formulated in mixed measurement (i.e., Axiom${}^{(m)}$ 1). $\;\;\;$ In the same sense, Situation (E) can be described in quantum language.