ロジスティック・モデルによる対数リスク比の最尤推定

積二項尤度の基本式より、 $$\begin{array}{r}L({\pi }_1,{\pi }_2)={}_{n_1}{C}_a\cdot {\pi }_1^a{(1-{\pi }_1)}^{n_1-a}\cdot {}_{n_0}{C}_b\cdot {\pi }_0^b{(1-{\pi }_0)}^{n_0-b}\end{array}$$ 対数尤度関数の定義 $l(\theta ,\mathit{x})=\log L(\theta ,\mathit{x})$ より、 $$\begin{array}{rl}l({\pi }_1,{\pi }_2)& =\log [{}_{n_1}{C}_a\cdot {\pi }_1^a{(1-{\pi }_1)}^{n_1-a}\cdot {}_{n_0}{C}_b\cdot {\pi }_0^b{(1-{\pi }_0)}^{n_0-b}]\\ & =\log {}_{n_1}{C}_a+a\log {\pi }_1+(n_1-a)\log (1-{\pi }_1)+\log {}_{n_0}{C}_b+b\log {\pi }_0+(n_0-b)\log (1-{\pi }_0)\end{array}$$ $\log {}_{n_1}{C}_a+\log {}_{n_0}{C}_b=C$ とすると、各セルどうしの関係、対数リスクモデルの仮定より、 $$\begin{array}{c}{n}_1-a=c{n}_0-b=da+b=m_1\log {\pi }_1=\alpha +\beta \log {\pi }_0=\alpha \end{array}$$ よって、 $$\begin{array}{rl}l({\pi }_1,{\pi }_2)& =a\log {\pi }_1+c\log (1-{\pi }_1)+b\log {\pi }_0+d\log (1-{\pi }_0)+C\\ & =a\log \left(e^{\alpha +\beta }\right)+c\log (1-e^{\alpha +\beta })+b\log \left(e^{\alpha }\right)+d\log (1-e^{\alpha })+C\\ & =a(\alpha +\beta )+c\log (1-e^{\alpha +\beta })+b\alpha +d\log (1-e^{\alpha })+C\\ & =(a+b)\alpha +a\beta +c\log (1-e^{\alpha +\beta })+d\log (1-e^{\alpha })+C\\ & =m_1\alpha +a\beta +c\log (1-e^{\alpha +\beta })+d\log (1-e^{\alpha })+C\end{array}$$ パラメータ $\alpha$ のスコア関数は、この対数尤度関数を $\alpha$ で偏微分して、 $$\begin{array}{rl}{U}_{\alpha }\left(\mathit{\theta }\right)& =m_1-c\cdot \frac{e^{\alpha +\beta }}{1-e^{\alpha +\beta }}-d\cdot \frac{e^{\alpha }}{1-e^{\alpha }}\\ & =m_1-\frac{c{\pi }_1}{1-{\pi }_1}-\frac{d{\pi }_0}{1-{\pi }_0}\end{array}$$ 同様に、パラメータ $\beta$ のスコア関数は、この対数尤度関数を $\beta$ で偏微分して、 $$\begin{array}{rl}{U}_{\beta }\left(\mathit{\theta }\right)& =a-c\cdot \frac{e^{\alpha +\beta }}{1-e^{\alpha +\beta }}\\ & =a-\frac{c{\pi }_1}{1-{\pi }_1}\end{array}$$

最尤推定量の定義 $U\left(\hat{\theta }\right)=0$ より、 $$\begin{array}{}\text{(1)}& U_{\alpha }(\hat{\alpha },\hat{\beta })=m_1-\frac{c{\pi }_1}{1-{\pi }_1}-\frac{d{\pi }_0}{1-{\pi }_0}=0\text{(2)}& U_{\beta }(\hat{\alpha },\hat{\beta })=a-\frac{c{\pi }_1}{1-{\pi }_1}=0\end{array}$$ この連立方程式を解くと、 $$\begin{array}{c}{m}_1-a-\frac{d{e}^{\hat{\alpha }}}{1-e^{\hat{\alpha }}}=0\\ \frac{e^{\hat{\alpha }}}{1-e^{\hat{\alpha }}}=\frac{b}{d}\\ e^{\hat{\alpha }}=\frac{b}{d}-\frac{b}{d}{e}^{\hat{\alpha }}\\ (1+\frac{b}{d})e^{\hat{\alpha }}=\frac{b}{d}\\ \left(\frac{d+b}{d}\right)e^{\hat{\alpha }}=\frac{b}{d}\\ e^{\hat{\alpha }}=\frac{b}{d}\cdot \frac{d}{d+b}\\ e^{\hat{\alpha }}=\frac{d}{d+b}=\frac{b}{n_0}\\ \hat{\alpha }=\log \frac{b}{n_0}\end{array}$$ $$\begin{array}{c}\frac{e^{\hat{\alpha }+\hat{\beta }}}{1-e^{\hat{\alpha }+\hat{\beta }}}=\frac{a}{c}\\ e^{\hat{\alpha }+\hat{\beta }}=\frac{a}{c}-\frac{a}{c}{e}^{\hat{\alpha }+\hat{\beta }}\\ (1+\frac{a}{c})e^{\hat{\alpha }+\hat{\beta }}=\frac{a}{c}\\ e^{\hat{\alpha }+\hat{\beta }}=\frac{a}{c}\cdot \frac{c}{a+c}=\frac{a}{n_1}\\ e^{\hat{\beta }}\frac{b}{n_0}=\frac{a}{n_1}\\ e^{\hat{\beta }}=\frac{a{n}_0}{b{n}_1}\\ \hat{\beta }=\log \frac{a{n}_0}{b{n}_1}\end{array}$$ $◼$

ヘッセ行列の成分の定義と $c=n_1(1-{\pi }_1),d=n_0(1-{\pi }_0)$ より、 $$\begin{array}{rl}{H}_{\alpha }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial {\alpha }^2}l\left(\mathit{\theta }\right)\\ & =-c\cdot \frac{e^{\alpha +\beta }}{{(1-e^{\alpha +\beta })}^2}-d\cdot \frac{e^{\alpha }}{{(1-e^{\alpha })}^2}\\ & =-c\frac{{\pi }_1}{{(1-{\pi }_1)}^2}-d\frac{{\pi }_0}{{(1-{\pi }_0)}^2}\\ & =-n_1\cdot \frac{{\pi }_1}{1-{\pi }_1}-n_0\cdot \frac{{\pi }_0}{1-{\pi }_0}\end{array}$$ $$\begin{array}{rl}{H}_{\beta }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial {\beta }^2}l\left(\mathit{\theta }\right)\\ & =-c\cdot \frac{e^{\alpha +\beta }}{{(1-e^{\alpha +\beta })}^2}\\ & =-c\frac{{\pi }_1}{{(1-{\pi }_1)}^2}\\ & =-n_1\cdot \frac{{\pi }_1}{1-{\pi }_1}\end{array}$$ $$\begin{array}{rl}{H}_{\alpha \beta }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial \alpha \partial \beta }l\left(\mathit{\theta }\right)\\ & =c\cdot \frac{e^{\alpha +\beta }}{{(1-e^{\alpha +\beta })}^2}\\ & =-c\frac{{\pi }_1}{{(1-{\pi }_1)}^2}\\ & =-n_1\cdot \frac{{\pi }_1}{1-{\pi }_1}\end{array}$$ 観測情報行列の定義 $\mathit{i}\left(\mathit{\theta }\right)=-\mathit{H}\left(\mathit{\theta }\right)$ と二項分布における期待情報行列との関係 $\mathit{i}\left(\mathit{\theta }\right)=\mathit{I}\left(\mathit{\theta }\right)$ から、${\psi }_i=\frac{{\pi }_i}{1-{\pi }_i}$ とすると、 $$\begin{array}{rl}\mathit{I}\left(\mathit{\theta }\right)& =\mathit{i}\left(\mathit{\theta }\right)\\ & =-\left[\begin{array}{cc}{H}_{\alpha }\left(\mathit{\theta }\right)& H_{\alpha \beta }\left(\mathit{\theta }\right)\\ H_{\alpha \beta }\left(\mathit{\theta }\right)& H_{\beta }\left(\mathit{\theta }\right)\end{array}\right]\\ & =\left[\begin{array}{cc}{n}_1{\psi }_1+n_0{\psi }_0& n_1{\psi }_1\\ n_1{\psi }_1& n_1{\psi }_1\end{array}\right]\end{array}$$

期待情報行列の成分は、 $$\begin{array}{r}\mathit{I}\left(\mathit{\theta }\right)=\left[\begin{array}{cc}{\mathit{I}}_{\alpha }\left(\mathit{\theta }\right)& {\mathit{I}}_{\alpha \beta }\left(\mathit{\theta }\right)\\ {\mathit{I}}_{\alpha \beta }{\left(\mathit{\theta }\right)}^T& {\mathit{I}}_{\beta }\left(\mathit{\theta }\right)\end{array}\right]\end{array}$$ このとき、推定量の共分散行列は、$\mathit{V}\left(\hat{\mathit{\theta }}\right)={\mathit{I}\left(\mathit{\theta }\right)}^{-1}$ となる。期待情報行列の逆行列は、 $$\begin{array}{r}{\mathit{I}\left(\mathit{\theta }\right)}^{-1}=\frac{\mathbf{1}}{\left|\mathit{I}\left(\mathit{\theta }\right)\right|}\left[\begin{array}{cc}{\mathit{I}}_{\beta }\left(\mathit{\theta }\right)& -{\mathit{I}}_{\alpha \beta }{\left(\mathit{\theta }\right)}^T\\ -{\mathit{I}}_{\alpha \beta }\left(\mathit{\theta }\right)& {\mathit{I}}_{\alpha }\left(\mathit{\theta }\right)\end{array}\right]\end{array}$$ 2×2行列の行列式の公式 $\left|\mathit{A}\right|=a_{11}{a}_{22}-a_{12}{a}_{21}$ より、 $$\begin{array}{rl}\left|\mathit{I}\left(\mathit{\theta }\right)\right|& =(n_1{\psi }_1+n_0{\psi }_0)n_1{\psi }_1-{\left\{n_1{\psi }_1\right\}}^2\\ & =(n_1{\psi }_1+n_0{\psi }_0-n_1{\psi }_1)n_1{\psi }_1\\ & =n_1{\psi }_1{n}_0{\psi }_0\end{array}$$ したがって、 $$\begin{array}{rl}\mathit{V}(\hat{\alpha },\hat{\beta })& ={\mathit{I}\left(\mathit{\theta }\right)}^{-1}\\ & =\frac{1}{n_1{\psi }_1{n}_0{\psi }_0}\left[\begin{array}{cc}{n}_1{\psi }_1& -n_1{\psi }_1\\ -n_1{\psi }_1& n_1{\psi }_1+n_0{\psi }_0\end{array}\right]\\ & =\left[\begin{array}{cc}\frac{1}{n_0{\psi }_0}& -\frac{1}{n_0{\psi }_0}\\ -\frac{1}{n_0{\psi }_0}& \frac{n_1{\psi }_1+n_0{\psi }_0}{n_1{\psi }_1{n}_0{\psi }_0}\end{array}\right]\end{array}$$ したがって、推定量の共分散行列の成分は、 $$\begin{array}{rl}V\left(\hat{\alpha }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha }}\\ & =\frac{1-{\pi }_0}{n_0{\pi }_0}\end{array}$$ $$\begin{array}{rl}V\left(\hat{\beta }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\beta }}\\ & =\frac{1}{n_1{\psi }_1}+\frac{1}{n_0{\psi }_0}\\ & =\frac{1-{\pi }_1}{n_1{\pi }_1}+\frac{1-{\pi }_0}{n_0{\pi }_0}\end{array}$$ $$\begin{array}{rl}\mathrm{Cov}(\hat{\alpha },\hat{\beta })& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha },\mathit{\beta }}\\ & =-\frac{1-{\pi }_0}{n_0{\pi }_0}\end{array}$$

この一致推定量は、${\hat{\pi }}_1=p_1=e^{\hat{\alpha }+\hat{\beta }}=\frac{a{n}_2}{b{n}_1},{\hat{\pi }}_0=p_0=e^{\hat{\alpha }}=\frac{b}{n_0}$ を代入して、 $$\begin{array}{rl}\hat{V}\left(\hat{\beta }\right)& =\hat{V}(\log \widehat{\mathrm{RR}})\\ & =\frac{1-p_1}{n_1{p}_1}+\frac{1-p_0}{n_0{p}_0}\\ & =\frac{1-p_1}{a}+\frac{1-p_0}{b}\\ & =\frac{1}{a}-\frac{p_1}{a}+\frac{1}{b}-\frac{p_0}{b}\end{array}$$ $a=n_1{p}_1,b=n_0{p}_0$ なので、 $$\begin{array}{r}\hat{V}(\log \widehat{\mathrm{RR}})=\frac{1}{a}-\frac{1}{n_1}+\frac{1}{b}-\frac{1}{n_0}\end{array}$$ $◼$

対数尤度の式より、帰無仮説 $\beta =0$ における対数尤度は、 $$\begin{array}{rl}l({\pi }_1,{\pi }_0)& =m_1\alpha +a\beta +c\log (1-e^{\alpha +\beta })+d\log (1-e^{\alpha })\\ l(\hat{\alpha },0)& =\log L(\hat{\alpha },0)\\ & =m_1\alpha +a\cdot 0+c\log (1-e^{\alpha +0})+d\log (1-e^{\alpha })\\ & =m_1\alpha +c\log (1-e^{\alpha })+d\log (1-e^{\alpha })=m_1\alpha +m_0\log (1-e^{\alpha })\end{array}$$ パラメータ $\alpha$ のスコア関数は、この対数尤度関数を $\alpha$ で偏微分すると、 $$\begin{array}{r}{U}_{\alpha }\left(\mathit{\theta }\right)=m_1-m_0\frac{e^{\alpha }}{1-e^{\alpha }}\end{array}$$ 尤度方程式 $U_{\alpha }\left(\mathit{\theta }\right)=0$ を解くことによって得られる $\alpha$ の最尤推定量は、 $$\begin{array}{c}{m}_1-m_0\frac{e^{{\hat{\alpha }}_0}}{1-e^{{\hat{\alpha }}_0}}=0\\ \frac{e^{{\hat{\alpha }}_0}}{1-e^{{\hat{\alpha }}_0}}=\frac{m_1}{m_0}\\ (1+\frac{m_1}{m_0})e^{{\hat{\alpha }}_0}=\frac{m_1}{m_0}\\ e^{{\hat{\alpha }}_0}=\frac{m_1}{m_0}\cdot \frac{m_0}{m_1+m_0}\\ e^{{\hat{\alpha }}_0}=\frac{m_1}{N}\\ {\hat{\alpha }}_0=\log \frac{m_1}{N}\end{array}$$ したがって、共通の発症確率 ${\pi }_1={\pi }_0=\pi$ の最尤推定量は、 $$\begin{array}{r}\hat{\pi }=\frac{m_1}{N}\end{array}$$

$\beta$ に関するスコア関数 $U_{\beta }\left(\mathit{\theta }\right)=a-c\cdot \frac{e^{\alpha +\beta }}{1-e^{\alpha +\beta }}$ は、帰無仮説 $\beta =0$ においては、 $$\begin{array}{rl}{U}_{\beta }\left(\mathit{\theta }\right)& =a-n_1\cdot \frac{e^{{\hat{\alpha }}_0}}{1+e^{{\hat{\alpha }}_0}}\\ & =a-c\cdot \frac{\frac{m_1}{N}}{1-\frac{m_1}{N}}\\ & =a-c\cdot \frac{m_1}{m_0}\end{array}$$

また、帰無仮説のもとで $\beta$ に対応する推定情報量の逆行列の成分は、 $$\begin{array}{rl}V\left(\hat{\beta }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\beta }}\\ & =\frac{1-\hat{\pi }}{n_1\hat{\pi }}+\frac{1-\hat{\pi }}{n_0\hat{\pi }}\\ & =\frac{1-\hat{\pi }}{\hat{\pi }}(\frac{1}{n_1}+\frac{1}{n_0})\\ & =\frac{1-\frac{m_1}{N}}{\frac{m_1}{N}}(\frac{1}{n_1}+\frac{1}{n_0})\\ & =\frac{m_0}{m_1}(\frac{1}{n_1}+\frac{1}{n_0})\\ & =\frac{m_0}{m_1}\cdot \frac{n_1+n_0}{n_1{n}_0}\\ & =\frac{m_0}{m_1}\cdot \frac{N}{n_1{n}_0}\end{array}$$

結果として有効スコア検定 ${\chi }^2=\frac{{\left\{U_{\beta }\left(\mathit{\theta }\right)\right\}}^2}{\hat{V}\left(\hat{\beta }\right)}$ は、 $$\begin{array}{r}{\chi }^2={(a-c\cdot \frac{m_1}{m_0})}^2(\frac{m_1}{m_0}\cdot \frac{n_1{n}_0}{N})\end{array}$$

スコアにもとづく推定値の公式 ${\hat{\theta }}_0=\frac{U\left({\theta }_0\right)}{I\left({\theta }_0\right)}$ より、 $$\begin{array}{r}\hat{\beta }=(a-c\cdot \frac{m_1}{m_0})(\frac{m_1}{m_0}\cdot \frac{n_1{n}_0}{N})\end{array}$$ また、スコアにもとづく推定値の分散の公式 $V\left({\hat{\theta }}_0\right)=\frac{1}{I\left({\theta }_0\right)}$ より、 $$\begin{array}{r}V\left(\hat{\beta }\right)=\frac{m_1}{m_0}\cdot \frac{n_1{n}_0}{N}\end{array}$$ $◼$