間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

正規化された分配凾数

正規化された分配凾数

$$ \begin{eqnarray} \prod^n_{i=1}q(X_i;\theta)^{\beta} &=&\exp\left(\log\left(\displaystyle\prod^n_{i=1}q(X_i;\theta)^{\beta}\right)\right) \;\cdots\;A=\exp\left(\log\left(A\right)\right)\\ &=&\exp\left(\sum^n_{i=1}\log\left(q(X_i;\theta)^{\beta}\right)\right) \;\cdots\;\log\left(\prod_{i=1}^n A_i\right) = \sum_{i=1}^n \log\left(A_i\right)\\ &=&\exp\left(\beta\sum^n_{i=1}\log\left(q(X_i;\theta)\right)\right)\\ &=&\exp\left(-n\beta \left(-\frac{1}{n}\sum^n_{i=1}\log\left(q\left(X_i;\theta\right)\right)\right)\right) \;\cdots\;(-n)(-\frac{1}{n})=1\\ &=&\exp\left(-n\beta L_n(\theta)\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_1.html}{L_n(\theta)=-\frac{1}{n}\displaystyle\sum^n_{i=1}\log\left(q(X_i;\theta)\right)}\\ \end{eqnarray} $$
$$ \begin{eqnarray} Z_n(\beta) &=&\int_\Theta \pi(\theta) \prod^n_{i=1}q(X_i;\theta)^{\beta} \mathrm{d}\theta\\ &=&\int_\Theta \pi(\theta) \exp\left(-n\beta L_n(\theta)\right) \mathrm{d}\theta \;\cdots\;\prod^n_{i=1}q(X_i;\theta)^{\beta}=\exp\left(-n\beta L_n(\theta)\right)\\ &=&\int_\Theta \pi(\theta) \exp\left(-n\beta \left( L_n(\theta_0) + K_n(\theta) \right)\right) \mathrm{d}\theta \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_1.html}{L_n(\theta) = L_n(\theta_0) + K_n(\theta)}\\ &=&\int_\Theta \pi(\theta) \exp(-n\beta L_n(\theta_0) -n\beta K_n(\theta)) \mathrm{d}\theta\\ &=&\int_\Theta \pi(\theta) \exp(-n\beta L_n(\theta_0)) \exp(-n\beta K_n(\theta)) \mathrm{d}\theta\\ &=&\exp(-n\beta L_n(\theta_0))\int_\Theta \pi(\theta)\exp(-n\beta K_n(\theta)) \mathrm{d}\theta\\ &=&\exp(-n\beta L_n(\theta_0))Z_n^{(0)}(\beta) \;\cdots\;Z_n^{(0)}(\beta)=\int_\Theta \pi(\theta)\exp(-n\beta K_n(\theta)) \mathrm{d}\theta:正規化された分配凾数\\ &=& Z_n^{(0)}(\beta)\exp(-n\beta L_n(\theta_0))\\ \end{eqnarray} $$

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