Let us explain the moment method, which as well as Fisher's maximum likelihood method are frequently used.

Consider the measurement ${\mathsf M}_{\cal A} \bigl({\mathsf O} \equiv (X,{\mathcal F}, F) ,$ $S_{[\rho] } \bigl)$, and its parallel measurement $\otimes_{k=1}^n {\mathsf M}_{\cal A} \bigl({\mathsf O}$ $\equiv (X,{\mathcal F}, F) ,$ $S_{[\rho] } \bigl)$ (= ${\mathsf M}_{\otimes {\cal A}} \big( \bigotimes_{k=1}^n {\mathsf O} := ( X^n,{\mathcal F}^n, \bigotimes_{k=1}^n F ) , S_{ [{}\otimes_{k=1}^n \rho{}]}{}\big)$. Assume that the measured value $(x_1,$ $x_2 ,$ $...$, $x_n ) (\in X^n)$ is obtained by the parallel measurement. Assume that $n$ is sufficiently large. By the law of large numbers (Theorem 4.5), we can assure that

\begin{align} {\mathcal M_{+1}(X)} \ni \nu_n \Big( \equiv \frac{\delta_{x_1}+\delta_{x_2}+\cdots + \delta_{x_n}}{n} \Big) \doteqdot \rho ( F (\cdot )) \in {\mathcal M_{+1}(X)} \tag{5.18} \end{align} Thus,
 $(A):$ in order to infer the unknown state $\rho (\in {\frak S}^p({\mathcal A}^*))$, it suffices to solve the equation (5.18)
For example, we have several methods to solve the equation (5.18)as follows.
 $(B_1):$ Solve the following equation: \begin{align} \| \nu_n(\cdot ) - \rho(F(\cdot ))\|_{{\mathcal M}(X)} = \min \{ \| \nu_n(\cdot ) - \rho_1(F(\cdot ))\|_{{\mathcal M}(X)} \;| \; \rho_1 (\in {\frak S}^p({\mathcal A}^*)) \} \tag{5.19} \end{align} $(B_2):$ For some $f_1, f_2, \cdots , f_n$ $\in C(X)$ $(= \mbox{the set of all continuous functions on$X$} )$, it suffices to find $\rho (\in {\frak S}^p({\mathcal A}^*))$ such that $\Delta (\rho)=\min_{\rho_1 (\in {\frak S}^p({\mathcal A}^*))} \Delta(\rho_1)$, where \begin{align} \Delta (\rho)= & \sum_{k=1}^n \Big| \int_X f_k(\xi) \nu_n (d \xi ) - \int_X f_k(\xi) \rho( F(d \xi )) \Big| \\ = & \sum_{k=1}^n \Big| \frac{f_k({x_1})+f_k({x_2})+ \cdots + f_k({x_n})}{n} - \int_X f_k(\xi) \rho( F(d \xi )) \Big| \end{align} $(B_3):$ In the cases of the classical measurement ${\mathsf M}_{L^\infty (\Omega)} \bigl({\mathsf O} \equiv (X,{\mathcal F}, F) ,$ $S_{[\rho] } \bigl)$ (putting $\rho=\delta_{\omega}$), it suffices to solve \begin{align} 0= \sum_{k=1}^n \Big| \frac{f_k({x_1})+f_k({x_2})+ \cdots + f_k({x_n})}{n} - \int_X f_k(\xi) [F(d \xi )](\omega) \Big| \tag{5.20} \end{align} or, it suffices to solve \begin{align} \left\{\begin{array}{ll} \frac{f_1({x_1})+f_1({x_2})+ \cdots + f_1({x_n})}{n} - \int_X f_1(\xi) [F(d \xi )](\omega)=0 \\ \\ \frac{f_2({x_1})+f_2({x_2})+ \cdots + f_2({x_n})}{n} - \int_X f_2(\xi) [F(d \xi )](\omega)=0 \\ \qquad \dots \dots \\ \qquad \dots \dots \\ \frac{f_m({x_1})+f_m({x_2})+ \cdots + f_m({x_n})}{n} - \int_X f_m(\xi) [F(d \xi )](\omega)=0 \end{array}\right. \end{align} $(B_4)$: Particularly, in the case that $X=\{\xi_1, \xi_2, \cdots , \xi_m \}$ is finite, define $f_1, f_2, \cdots , f_m$ $\in C(X)$ by \begin{align} f_k(\xi) = \chi_{{}_{\{\xi_k \}}}(\xi)=\left\{\begin{array}{ll} 1 \quad & (\xi= \xi_k) \\ 0 & ( \xi \not=\xi_k ) \end{array}\right. \end{align} and, it suffices to find the $\rho (= \delta_\omega )$ such that \begin{align} & \sum_{k=1}^n \Big| \frac{ \chi_{{}_{\{\xi_k \}}}({x_1})+ \chi_{{}_{\{\xi_k \}}}({x_2})+ \cdots +\chi_{{}_{\{\xi_k \}}} ({x_n})}{n} - \int_X \chi_{{}_{\{\xi_k \}}}( \xi) \rho( F(d \xi )) \Big| \\ = & \sum_{k=1}^n \Big| \frac{\sharp[\{x_m \;:\; \xi_k=x_m \}]}{n} - [F( \{ \xi_k\} ](\omega ) ) \Big|=0 \end{align}
The above methods are all { the moment method}. Note that
 $(C_1):$ It is desirable that $n$ is sufficiently large, but the moment method may be valid even when $n=1$. $(C_2)$: The choice of $f_k$ is artificial ( on the other hand, Fisher' maximum likelihood method is natural).
Problem 5.11 [=Problem5.2: Urn problem: by the moment method] You do not know which the urn behind the curtain is, $U_1$ or $U_2$. Assume that you pick up a white ball from the urn. The urn is $U_1$ or $U_2$? $\quad$ Which do you think?

Consider the measurement ${\mathsf M}_{{L^\infty (\Omega) }} ({\mathsf O} {{=}}$ $( \{ w,$ $b \},$ $2^{\{ w, b \} } ,$ $F) , S_{ [{}{\ast}]})$. Here, recall that the observable ${\mathsf O}_{wb} = ( \{ w, b \}, 2^{\{ w, b \} } , F_{wb})$ in $L^\infty(\Omega)$ is defined by

\begin{align} & [F_{wb}(\{ w \})](\omega_1)= 0.8, & \quad & [ F_{wb}(\{ b \})](\omega_1)= 0.2 \nonumber \\ & [F_{wb}(\{ w \})](\omega_2)= 0.4, & \quad & [F_{wb}(\{ b \})] (\omega_2)= 0.6 \nonumber \end{align}

Since a measured value "w" is obtained, the approximate sample space $( \{ w, b \}, 2^{\{ w, b \} } , \nu_1)$ is obtained as

\begin{align} \nu_1(\{ w \})=1, \quad \nu_1(\{b \})=0 \end{align}

[(i): when the unknown state $[\ast]$ is $\omega_1$]

\begin{align} (5.19)=|1-0.8|+|0-0.2| \end{align}

[(ii):when the unknown state $[\ast]$ is $\omega_2$]

\begin{align} (5.19)=|1-0.4|+|0-0.6| \end{align}

Thus, by the moment method, we can infer that $[\ast]=\omega_1$, that is, the urn behind the curtain is $U_1$.

The above may be too easy. Thus, we add the following problem.

Problem 5.12 [Sampling with replacement]: As mentioned in the above, assume that "white ball" is picked. and the ball is returned to the urn. And further, we pick "black ball", and it is returned to the urn. Repeat this, after all, assume that we get \begin{align} "w", "b", "b", "w", "b", "w", "b", \end{align} Then, we have the following problem:
 $(a):$ Which the urn behind the curtain is $U_1$ or $U_2$?

Consider the simultaneous measurement ${\mathsf M}_{{L^\infty (\Omega) }} (\times_{k=1}^{7}{\mathsf O} {{=}}$ $( \{ w, b \}^7,$ $2^{{\{ w, b \} }^7} ,$ $\times_{k=1}^{7} F) ,$ $S_{ [{}{\ast}]})$. And assume that the measured value is $(w, b, b, w, b, w, b)$. Then, [(i): when $[\ast]$ is $\omega_1$]

\begin{align} (5.19) =|3/7-0.8|+|4/7-0.2|=52/70 \end{align}

[(ii):when $[\ast]$ is $\omega_2$]

\begin{align} (5.19) =|3/7-0.4|+|4/7-0.6|=10/70 \end{align}

Thus, by the moment method, we can infer that $[\ast]=\omega_2$, that is, the urn behind the curtain is $U_2$.

Example 5.13 [The most important example of moment method]

Putting $\Omega={\mathbb R} \times {\mathbb R}_+$ $=\{ \omega =(\mu , \sigma ) \; |\; \mu \in {\mathbb R}, \sigma >0 \}$ with Lebesgue measure $\nu$, Consider the classical basic structure

\begin{align} [ C_0(\Omega ) \subseteq L^\infty(\Omega,\nu) \subseteq B(L^2(\Omega,\nu))] \end{align}

Assume that the observable ${\mathsf O}_{G}$ ${{=}}$ $(X(={}{\mathbb R}) , {\cal B}_{{\mathbb R}}^{} , G)$ in $L^\infty ({\Omega}{}, \nu)$ satisfies that

\begin{align} & \int_{\mathbb R} \xi [G( d \xi )](\mu, \sigma )=\mu, \quad \int_{\mathbb R} (\xi -\mu)^2 [G( d \xi )](\mu, \sigma )=\sigma^2 \\ & \quad \qquad ( \forall {\omega}=(\mu, \sigma) \in \Omega (={\mathbb R} \times {\mathbb R}_+ )) \end{align}

Here, assume that a measured value $(x_1, x_2, x_3 ) (\in {\mathbb R}^3)$ is obtained by the simultaneous measurement $\times_{k=1}^3{\mathsf M}_{L^\infty(\Omega )} ({\mathsf O}_{G}, S_{[{}\ast{}] })$. That is, we have the $3$-sample distribution $\nu_3$ such that

\begin{align} \nu_3 = \frac{ \delta_{x_1}+\delta_{x_2}+\delta_{x_3}}{3} \in {\mathcal M}_{+1}({\mathbb R}) \end{align} Put $f_1(\xi)=\xi, f_2(\xi)=\xi^2$. Then, by the moment method (5.20), we see: \begin{align} 0=& \sum_{k=1}^2 \Big| \int_{\mathbb R} \xi^k \nu_3 (d \xi ) - \int_{\mathbb R} \xi^k [G(d \xi )](\omega) \Big| \\ = & \sum_{k=1}^2 \Big| \frac{({x_1})^k+({x_2})^k+ ({x_n})^k}{3} - \int_{\mathbb R} \xi^k [G(d \xi )](\mu, \sigma) \Big| \\ = & \Big| \frac{{x_1}+{x_2}+ {x_3}}{3} - \mu \Big| + \Big| \frac{({x_1})^2+({x_2})^2+ ({x_3})^2}{3} - (\sigma^2 + \mu^2 ) \Big| \end{align} Thus, we get: \begin{align} \mu &=\frac{{x_1}+{x_2}+ {x_n}}{3} \\ \sigma^2 & = \frac{({x_1})^2+({x_2})^2+ ({x_3})^2}{3} -\mu^2 \\ & = \frac{({x_1}- \frac{{x_1}+{x_2}+ {x_n}}{3} )^2+({x_2}- \frac{{x_1}+{x_2}+ {x_n}}{3} )^2+ ({x_3} - \frac{{x_1}+{x_2}+ {x_n}}{3} )^2}{3} \end{align} which is the same as the (5.13) concerning the normal measurement.
$\fbox{Note 5.3}$ Consider the measurement ${\mathsf M}_{L^\infty(\Omega)}({\mathsf O} {{=}} (X,2^X,$ $F),$ $S_{[\ast]} )$, where $X= \{x_1,x_2,...,x_n\}$ is finite. Then, we see that \begin{align} \mbox{"Fisher's maximum likelihood method"="moment method"} \end{align} [Answer] Assume that a measured value$x_m (\in X)$ is obtained by the measurement ${\mathsf M}_{\overline{\mathcal A}}({\mathsf O} {{=}} (X,2^X,$ $F),$ $S_{[\ast]} )$
[Fisher's maximum likelihood method]:
 $(a):$ Find $\omega_0 (\in \Omega )$ such that \begin{align} [ F(\{ x_m\})](\omega_0 ) = \max_{\omega \in \Omega } [ F(\{ x_m\})](\omega ) \end{align}
[Moment method]:
 $(b):$ Since we get the approximate sample probability space $(X, 2^X, \delta_{x_m} )$, we see \begin{align} & |0- [ F(\{ x_1 \})](\omega ) |+ \cdots +|0- [ F(\{ x_{m-1} \})](\omega )| + |1- [ F(\{ x_{m} \})](\omega )| \\ & \qquad +|0- [ F(\{ x_{m+1} \})](\omega )|+\cdots +|0- [ F(\{ x_{n} \})](\omega )| \\ = & [ F(\{ x_1 \})](\omega ) + \cdots +[ F(\{ x_{m-1} \})](\omega ) + [ F(\{ x_{m} \})](\omega ) \\ & \qquad + [ F(\{ x_{m+1} \})](\omega )+\cdots +[ F(\{ x_{n} \})](\omega ) \\ = & 1- 2[ F(\{ x_{m} \})](\omega ) \end{align} Thus, it suffice to find $\omega_0 (\in \Omega )$ such that \begin{align} 1- 2[ F(\{ x_{m} \})](\omega_0 ) = \min_{\omega} (1- 2[ F(\{ x_{m} \})](\omega )) \end{align}
Thus, Fisher's maximum likelihood method and the moment method are the same in this case.