推定したパラメタ(不偏推定量)の分散(平均二乗誤差)
$$
\begin{eqnarray}
\mathrm{E}\left[\left(\hat{\theta}-\theta\right)^2\right]
&=&\mathrm{E}\left[
\hat{\theta}^2-2\hat{\theta}\theta+\theta^2
\right]\\
&=&\mathrm{E}\left[\hat{\theta}^2\right]
-2\theta\mathrm{E}\left[\hat{\theta}\right]
+\theta^2\mathrm{E}\left[1\right]\\
&=&\mathrm{E}\left[\hat{\theta}^2\right]
-2\theta\mathrm{E}\left[\hat{\theta}\right]
+\theta^2
+\mathrm{E}\left[\hat{\theta}\right]^2
-\mathrm{E}\left[\hat{\theta}\right]^2
\;\cdots\;\mathrm{E}\left[\hat{\theta}\right]^2
-\mathrm{E}\left[\hat{\theta}\right]^2=0\\
&=&\mathrm{E}\left[\hat{\theta}\right]^2
-2\theta\mathrm{E}\left[\hat{\theta}\right]
+\theta^2
+\mathrm{E}\left[\hat{\theta}^2\right]
-\mathrm{E}\left[\hat{\theta}\right]^2\\
&=&\left(\mathrm{E}\left[\hat{\theta}\right]-\theta\right)^2
+\mathrm{V}\left[\hat{\theta}\right]
\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{\mathrm{V}\left[X\right]=\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]^2}\\
&=&\left(\theta-\theta\right)^2
+\mathrm{V}\left[\hat{\theta}\right]
\;\cdots\;\hat{\theta}は不偏推定量なので\mathrm{E}\left[\hat{\theta}\right]=\theta\\
&=&\mathrm{V}\left[\hat{\theta}\right]\\
\end{eqnarray}
$$
0 件のコメント:
コメントを投稿