18.1.3: Reliability coefficient

When we suppose the group test, we can consider the reliability coefficient which can be represented by a proportion of variance of mathematical intelligences to obtained variance.

Definition 18.7[Reliability coefficient] Let ${\mathsf O}_\tau := (X_{\mathbb R},{\cal F}_{X_{\mathbb R}},F_\tau)$ [resp. ${\mathsf O}_E := (X_{\mathbb R},{\cal F}_{X_{\mathbb R}},E)$] be a test observable [resp.~an exact observable] in $L^\infty(\Omega_{\mathbb R},d\omega)$. And, let $${\mathsf M}_{{\mathsf O}_\tau}^\otimes := \otimes_{\theta_i \in \Theta} {\mathsf M}_{L^\infty(\Omega_{\mathbb R},d\omega)}({\mathsf O}_\tau, S_{[\ast]}(\Phi_\ast(1_{\theta_i})))$$ be a group test. The reliability coefficient ${ RC} [{\mathsf M}_{{\mathsf O}_\tau}^\otimes]$ of the group test ${\mathsf M}_{{\mathsf O}_\tau}^\otimes$ is defined by \begin{align} { RC} [{\mathsf M}_{{\mathsf O}_\tau}^\otimes] = \frac{{ Var} [{\mathsf M}_{{\mathsf O}_E}^\otimes]}{{ Var} [{\mathsf M}_{{\mathsf O}_\tau}^\otimes]}. \end{align}

Now let us consider the measurement error. First, when the intelligence (true value) is $\omega \;(\in \Omega)$, the measurement error $\Delta_\omega$ is as follows:

\begin{align} \Delta_\omega := \Big( \int_{X_{\mathbb R}} (x-\omega)^2 \, [F_\tau(dx)](\omega) \Big)^{1/2} \quad (\forall \omega \in \Omega). \tag{18.9} \end{align}

Note that the error $\Delta_\omega$ ($\forall \omega \in \Omega$) depends on $\omega \;(\in \Omega)$ in general , that is, we do not assume that $\Delta_\omega = \Delta_{\omega'}$ ($\forall \omega, \forall\omega' \in \Omega$).

Next, for each $\theta_i \;(\in \Theta)$, the error $\Delta_i$ for the student $\theta_i \;(\in \Theta)$ is as follows:

\begin{align} \Delta_i &:= \Big( \int_{X_{\mathbb R}} \Delta_\omega \, [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \Big)^{1/2} \nonumber \\ &= \Big( \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} (x-\omega)^2 \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \Big)^{1/2} \quad (i=1,2,\dots,n). \tag{18.10} \end{align}

Finally, the group average of the student $\theta_i$'s error $\Delta_i$ ($i=1,2,\dots,n$) is as follows:

\begin{align} \Delta_g &:= \Big( \frac1n \sum_{i=1}^n \Delta_i^2 \Big)^{1/2}. \tag{18.11} \end{align} From what we have seen, we can get the following theorem.

Theorem 18.8 (i: The variance ${ Var}[{\mathsf M}_{{\mathsf O}_\tau}^{(i)}]$) Let ${\mathsf M}_{{\mathsf O}_\tau}^{(i)} := {\mathsf M}_{L^\infty(\Omega_{\mathbb R},d\omega)}({\mathsf O}_\tau, S_{[\ast]}(\Phi_\ast(1_{\theta_i})))$ be the measurement of test observable ${\mathsf O}_\tau$ for the statistical state $\Phi_\ast(1_{\theta_i})$. Then, we see

\begin{align} { Var}[{\mathsf M}_{{\mathsf O}_\tau}^{(i)}] = { Var}[{\mathsf M}_{{\mathsf O}_E}^{(i)}] + \Delta_i^2. \tag{18.12} \end{align}

(ii: The variance ${ Var}[{\mathsf M}_{{\mathsf O}_\tau}^\otimes]$) We consider the group test ${\mathsf M}_{{\mathsf O}_\tau}^\otimes := \otimes_{\theta_i \in \Theta}{\mathsf M}_{{\mathsf O}_\tau}^{(i)}$ $= \otimes_{\theta_i \in \Theta} {\mathsf M}_{L^\infty(\Omega_{\mathbb R},d\omega)}({\mathsf O}_\tau,$ $S_{[\ast]}(\Phi_\ast(1_{\theta_i})))$. And, we obtain the following:

\begin{align} { Var}[{\mathsf M}_{{\mathsf O}_\tau}^\otimes] = { Var}[{\mathsf M}_{{\mathsf O}_E}^\otimes] + \Delta_g^2 \tag{18.13} \end{align}

Proof Let $\mu_i$ be an expectation of $\Phi_\ast(1_{\theta_i})$. Then, we see

\begin{align} &{ Var}[{\mathsf M}_{{\mathsf O}_\tau}^{(i)}] = \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} (x-\mu_i)^2 \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ &= \int_{\Omega_{\mathbb R}} (\omega-\mu_i)^2 \, [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega + \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} (x-\omega)^2 \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ & \quad + \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} 2(x-\omega)(\omega-\mu_i) \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ &= { Var}[{\mathsf M}_{{\mathsf O}_E}^{(i)}] + \Delta_i^2. \end{align} From the above formula, the group average of ${ Var}[{\mathsf M}_{{\mathsf O}_\tau}^{(i)}]$ follows that \begin{align} { Var}[{\mathsf M}_{{\mathsf O}_\tau}^\otimes] &= \int_{\Omega_{\mathbb R}} \cdots \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} \cdots \int_{X_{\mathbb R}} \frac1n \sum_{i=1}^n (x_i-\overline\mu)^2 \times_{i=1}^n [F_\tau(dx_i)](\omega_i) \Big) \times_{i=1}^n [\Phi_\ast(1_{\theta_i})](\omega_i) \, d\omega_i \\ &= \frac1n \sum_{i=1}^n \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} (\omega-\overline\mu+x-\omega)^2 \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ &= \frac1n \sum_{i=1}^n \int_{\Omega_{\mathbb R}} (\omega-\overline\mu)^2 \, [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ & \quad + \frac1n \sum_{i=1}^n \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} (x-\omega)^2 \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ & \quad + \frac1n \sum_{i=1}^n \int_{\Omega_{\mathbb R}} \Big( \int_{X_{\mathbb R}} 2(x-\omega)(\omega-\overline\mu) \, [F_\tau(dx)](\omega) \Big) [\Phi_\ast(1_{\theta_i})](\omega) \, d\omega \\ &= \int_{\Omega_{\mathbb R}} \cdots \int_{\Omega_{\mathbb R}} \frac1n \sum_{i=1}^n (\omega_i-\overline\mu)^2 \times_{i=1}^n [\Phi_\ast(1_{\theta_i})](\omega_i) \, d\omega_i + \frac1n \sum_{i=1}^n \Delta_i^2 \\ &= { Var}[{\mathsf M}_{{\mathsf O}_E}^\otimes] + \Delta_g^2. \end{align}
$\square \quad$