粗率と加重最小2乗和の関係

二重同次ポアソンモデルの仮定より、 $$\begin{array}{r}E\left(d_i\right)=V\left(d_i\right)=\lambda t_i\end{array}$$ 各被験者の率の定義より、 $$\begin{array}{r}{r}_i=\frac{d_i}{t_i}\end{array}$$ 両辺の期待値を取ると、 $$\begin{array}{rl}E\left(r_i\right)& =E\left(\frac{d_i}{t_i}\right)\\ & =\frac{1}{t_i}E\left(d_i\right)\\ & =\frac{1}{t_i}\cdot \lambda t_i\\ & =\lambda \end{array}$$ 両辺の分散を取ると、 $$\begin{array}{rl}V\left(r_i\right)& =V\left(\frac{d_i}{t_i}\right)\\ & =\frac{1}{t_i^2}V\left(d_i\right)\\ & =\frac{\lambda t_i}{t_i^2}\\ & =\frac{\lambda }{t_i}\end{array}$$ 率の分散の最尤推定量は、 $$\begin{array}{r}\hat{V}\left(r_i\right)=\frac{\hat{\lambda }}{t_i}\end{array}$$ 定義より、 $$\begin{array}{r}{\hat{\tau }}_i=\frac{1}{\hat{V}\left(r_i\right)}=\frac{t_i}{\hat{\lambda }}\end{array}$$ 同様に、加重最小2乗和の定義より、 $$\begin{array}{rl}\overline{r}& =\frac{\sum _{i=1}^N{\hat{\tau }}_i{r}_i}{\sum _{i=1}^N{\hat{\tau }}_i}\\ & =\frac{\sum _{i=1}^N\frac{t_i}{\hat{\lambda }}{r}_i}{\sum _{i=1}^N\frac{t_i}{\hat{\lambda }}}\\ & =\frac{\sum _{i=1}^N{t}_i{r}_i}{T}\\ & =\frac{\sum _{i=1}^N{t}_i\frac{d_i}{t_i}}{T}\\ & =\frac{\sum _{i=1}^N{d}_i}{T}\\ & =\frac{D}{T}\\ & =\hat{\lambda }\end{array}$$ $◼$

まず、 $$\begin{array}{rl}\sum _{i=1}^N{\hat{\tau }}_i& =\sum _{i=1}^N\frac{t_i}{\hat{\lambda }}\\ & =\frac{1}{\hat{\lambda }}\sum _{i=1}^N{t}_i\\ & =\frac{T}{\hat{\lambda }}\\ \frac{1}{\sum _{i=1}^N{\hat{\tau }}_i}& =\frac{\hat{\lambda }}{T}\\ & =\hat{V}\left(\hat{\lambda }\right)\end{array}$$ 加重最小2乗和の分散を取ると、 $$\begin{array}{rl}\hat{V}\left(\overline{r}\right)& =V\left(\frac{\sum _{i=1}^N{\hat{\tau }}_i{r}_i}{\sum _{i=1}^N{\hat{\tau }}_i}\right)\\ & ={\left(\frac{1}{\sum _{i=1}^N{\hat{\tau }}_i}\right)}^{2}V\left(\sum _{i=1}^N{\hat{\tau }}_i{r}_i\right)\\ & =\frac{{\hat{\lambda }}^2}{T^2}\cdot V\left(\sum _{i=1}^N\frac{t_i}{\hat{\lambda }}\cdot \frac{d_i}{t_i}\right)\\ & =\frac{{\hat{\lambda }}^2}{T^2}\cdot V\left(\sum _{i=1}^N\frac{d_i}{\hat{\lambda }}\right)\\ & =\frac{{\hat{\lambda }}^2}{T^2}\cdot \frac{1}{{\hat{\lambda }}^2}\sum _{i=1}^N\hat{V}\left(d_i\right)\\ & =\frac{1}{T^2}\sum _{i=1}^N\hat{\lambda }{t}_i\\ & =\frac{\hat{\lambda }\cdot T}{T^2}\\ & =\frac{\hat{\lambda }}{T}\\ & =\frac{1}{\sum _{i=1}^N{\hat{\tau }}_i}\end{array}$$ $◼$