スコア凾数の期待値と分散

$$ \href{https://shikitenkai.blogspot.com/2020/04/blog-post.html}{\frac{\partial \log{f(x;\theta)}}{\partial \theta}:スコア凾数}\\ $$

スコア凾数の期待値

$$ \begin{eqnarray} \mathrm{E}\left[ \frac{\partial \log{f(x;\theta)}}{\partial \theta} \right]\ &=&\int{ \frac{\partial \log{f(x;\theta)}}{\partial \theta} f(x;\theta)\mathrm{d}x}\\ &=&\int{ \frac{1}{f(x;\theta)} \frac{\partial f(x;\theta)}{\partial \theta} f(x;\theta)\mathrm{d}x} \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}\log{f(x)}=\frac{1}{f(x)}\frac{\mathrm{d}f(x)}{\mathrm{d}x}\\ &=&\int{ \frac{\partial f(x;\theta)}{\partial \theta} \mathrm{d}x}\\ &=&\frac{\partial }{\partial \theta} \int{f(x;\theta) \mathrm{d}x}\\ &=&\frac{\partial }{\partial \theta} 1\;\cdots\;\int{f(x;\theta) \mathrm{d}x}=1\\ &=&0\;\cdots\;定数の微分は0\\ \end{eqnarray} $$

スコア凾数の分散

$$ \mathcal{I}=\mathrm{V}\left[\frac{ \partial \log{ f(x;\theta) }}{ \partial \theta }\right]:フィッシャー情報量\\ $$