尤度凾数の期待値の平均情報量(離散/標本からの)

尤度凾数の期待値の平均情報量(離散/標本からの)

$$ \begin{eqnarray} T_n &=&-\frac{1}{n}\sum^n_{i=1}\log\left(\; \mathrm{E}_\Theta\left[q(X;\theta)\right] \;\right) \;\cdots\;尤度凾数の期待値の平均情報量\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left(\; \mathrm{E}_\Theta\left[q(X;\theta_0)\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \;\right) \;\cdots\;q(X;\theta)=q(X;\theta_0)\exp\left(-f(X_i,\theta_0,\theta)\right)\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left(\; q(X;\theta_0)\mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \;\right) \;\cdots\;\mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ &=&-\frac{1}{n}\sum^n_{i=1}\left\{ \log\left(q(X;\theta_0)\right) +\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right) \right\} \;\cdots\;\log(AB)=\log(A)+\log(B)\\ &=&-\frac{1}{n}\left\{\sum^n_{i=1}\log\left(q(X;\theta_0)\right) +\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right)\right\} \;\cdots\;\sum(A+B)=\sum A+\sum B\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left(q(X;\theta_0)\right) -\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right)\\ &=&L_n(\theta_0) -\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right) \;\cdots\;-\frac{1}{n}\sum^n_{i=1}\log\left(q(X;\theta_0)\right)=L_n(\theta_0)\\ &=&L_n(\theta_0)+T_n^{(0)} \;\cdots\;-\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right)=T_n^{(0)}\\ \end{eqnarray} $$

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$$ \begin{eqnarray} T_n^{(0)} &=&-\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left(-f(X_i,\theta_0,\theta)\right)\right] \right)\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left( -\log\frac{q(X_i;\theta_0)}{q(X_i;\theta)} \right)\right] \right)\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[\exp\left( \log\frac{q(X_i;\theta)}{q(X_i;\theta_0)} \right)\right] \right)\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left( \mathrm{E}_\Theta\left[ \frac{q(X_i;\theta)}{q(X_i;\theta_0)} \right] \right)\\ &=&-\frac{1}{n}\sum^n_{i=1}\log\left( \frac{\mathrm{E}_\Theta\left[q(X_i;\theta)\right]}{q(X_i;\theta_0)} \right)\\ &=&\frac{1}{n}\sum^n_{i=1}-\log\left( \frac{\mathrm{E}_\Theta\left[q(X_i;\theta)\right]}{q(X_i;\theta_0)} \right)\\ &=&\frac{1}{n}\sum^n_{i=1}\log\left( \frac{q(X_i;\theta_0)}{\mathrm{E}_\Theta\left[q(X_i;\theta)\right]} \right)\\ \end{eqnarray} $$