ロジスティック・モデルによる対数オッズ比の最尤推定

積二項尤度の基本式より、 $$\begin{array}{r}L({\pi }_1,{\pi }_0)={}_{n_1}{C}_a{\pi }_1^a{(1-{\pi }_1)}^{n_1-a}\cdot {}_{n_0}{C}_b{\pi }_0^b{(1-{\pi }_0)}^{n_0-b}\end{array}$$ 対数尤度関数の定義式 $l(\theta ,\mathit{x})=\log L(\theta ,\mathit{x})$ より、 $$\begin{array}{rl}l({\pi }_1,{\pi }_0)& =\log [{}_{n_1}{C}_a{\pi }_1^a{(1-{\pi }_1)}^{n_1-a}\cdot {}_{n_0}{C}_b{\pi }_0^b{(1-{\pi }_0)}^{n_0-b}]\\ & =\log {}_{n_1}{C}_a+a\log {\pi }_1+(n_1-a)\log (1-{\pi }_1)+\log {}_{n_0}{C}_b+b\log {\pi }_0+(n_0-b)\log (1-{\pi }_0)\end{array}$$ $\log {}_{n_1}{C}_a+\log {}_{n_0}{C}_b=C$ とすると、$n_1-a=c,n_0-b=d$ より、 $$\begin{array}{r}l({\pi }_1,{\pi }_0)=a\log {\pi }_1+c\log (1-{\pi }_1)+b\log {\pi }_0+d\log (1-{\pi }_0)+C\end{array}$$

対数尤度をパラメータ $\alpha ,\beta$ およびの周辺度数 $n_1,n_0,m_1,m_0$ で表すと、 $$\begin{array}{rl}l({\pi }_1,{\pi }_0)& =a\log \left(\frac{e^{\alpha +\beta }}{1+e^{\alpha +\beta }}\right)+c\log \left(\frac{1}{1+e^{\alpha +\beta }}\right)+b\log \left(\frac{e^{\alpha }}{1+e^{\alpha }}\right)+d\log \left(\frac{1}{1+e^{\alpha }}\right)+C\\ & =a\log \left(e^{\alpha +\beta }\right)-a\log (1+e^{\alpha +\beta })-c\log (1+e^{\alpha +\beta })+b\log \left(e^{\alpha }\right)-b\log (1+e^{\alpha })-d\log (1+e^{\alpha })+C\\ & =a\log \left(e^{\alpha +\beta }\right)-(a+c)\log (1+e^{\alpha +\beta })+b\log \left(e^{\alpha }\right)-(b+d)\log (1+e^{\alpha })+C\end{array}$$ 各セル度数どうしの関係より、 $$\begin{array}{c}a+c=n_1\\ b+d=n_0\\ a+b=m_1\end{array}$$ よって、 $$\begin{array}{rl}l({\pi }_1,{\pi }_0)& =a(\alpha +\beta )-n_1\log (1+e^{\alpha +\beta })+b\alpha -n_0\log (1+e^{\alpha })+C\\ & =(a+b)\alpha +a\beta -n_1\log (1+e^{\alpha +\beta })-n_0\log (1+e^{\alpha })+C\\ & =m_1\alpha +a\beta -n_1\log (1+e^{\alpha +\beta })-n_0\log (1+e^{\alpha })+C\end{array}$$ この対数尤度関数を $\alpha$ で偏微分すると、 $$\begin{array}{r}{U}_{\alpha }\left(\mathit{\theta }\right)=m_1-n_1\cdot \frac{e^{\alpha +\beta }}{1+e^{\alpha +\beta }}-n_0\cdot \frac{e^{\alpha }}{1+e^{\alpha }}\end{array}$$ ロジスティック・モデルの仮定より、 $$\begin{array}{r}{\pi }_1=\frac{e^{\alpha +\beta }}{1+e^{\alpha +\beta }}{\pi }_0=\frac{e^{\alpha }}{1+e^{\alpha }}\end{array}$$ よって、 $$\begin{array}{r}{U}_{\alpha }\left(\mathit{\theta }\right)=m_1-n_1{\pi }_1-n_0{\pi }_0\end{array}$$ 同様に、対数尤度関数を $\beta$ で偏微分すると、 $$\begin{array}{rl}{U}_{\beta }\left(\mathit{\theta }\right)& =a-n_1\cdot \frac{e^{\alpha +\beta }}{1+e^{\alpha +\beta }}\\ & =a-n_1{\pi }_1\end{array}$$

ヘッセ行列の成分の定義より、 $$\begin{array}{rl}{H}_{\alpha }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial {\alpha }^2}l\left(\mathit{\theta }\right)\\ & =-n_1\cdot \frac{e^{\alpha +\beta }}{{(1+e^{\alpha +\beta })}^2}-n_0\cdot \frac{e^{\alpha }}{{(1+e^{\alpha })}^2}\\ & =-n_1{\pi }_1(1-{\pi }_1)-n_0{\pi }_0(1-{\pi }_0)\end{array}$$ $$\begin{array}{rl}{H}_{\beta }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial {\beta }^2}l\left(\mathit{\theta }\right)\\ & =-n_1\cdot \frac{e^{\alpha +\beta }}{{(1+e^{\alpha +\beta })}^2}\\ & =-n_1{\pi }_1(1-{\pi }_1)\end{array}$$ $$\begin{array}{rl}{H}_{\alpha \beta }\left(\mathit{\theta }\right)& =\frac{{\partial }^2}{\partial \alpha \partial \beta }l\left(\mathit{\theta }\right)\\ & =-n_1\cdot \frac{e^{\alpha +\beta }}{{(1+e^{\alpha +\beta })}^2}\\ & =-n_1{\pi }_1(1-{\pi }_1)\end{array}$$ 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 $$\begin{array}{rl}\mathit{I}\left(\mathit{\theta }\right)& =\mathit{i}\left(\mathit{\theta }\right)\\ & =-\left[\begin{array}{cc}{H}_{\alpha }\left(\mathit{\theta }\right)& H_{\alpha \beta }\left(\mathit{\theta }\right)\\ H_{\alpha \beta }\left(\mathit{\theta }\right)& H_{\beta }\left(\mathit{\theta }\right)\end{array}\right]\\ & =\left[\begin{array}{cc}{n}_1{\pi }_1(1-{\pi }_1)+n_0{\pi }_0(1-{\pi }_0)& n_1{\pi }_1(1-{\pi }_1)\\ n_1{\pi }_1(1-{\pi }_1)& n_1{\pi }_1(1-{\pi }_1)\end{array}\right]\end{array}$$

最尤推定量 $U\left(\hat{\theta }\right)=0$ を求めると、 $$\begin{array}{}\text{(1)}& U_{\alpha }(\hat{\alpha },\hat{\beta })=m_1-n_1{\pi }_1-n_0{\pi }_0=0\text{(2)}& U_{\beta }(\hat{\alpha },\hat{\beta })=a-n_1{\pi }_1=0\end{array}$$ この連立方程式を解くと、 $$\begin{array}{rl}{\pi }_0& =\frac{m_1-a}{n_0}\\ & =\frac{b}{n_0}\\ \frac{e^{\hat{\alpha }}}{1+e^{\hat{\alpha }}}& =\frac{b}{n_0}\\ e^{\hat{\alpha }}& =\frac{b}{n_0-b}\\ & =\frac{b}{d}\\ \hat{\alpha }& =\log \frac{b}{d}\end{array}$$ この結果と式 $(2)$ より、 $$\begin{array}{c}a-n_1\cdot \frac{e^{\hat{\alpha }+\hat{\beta }}}{1+e^{\hat{\alpha }+\hat{\beta }}}=0\end{array}$$ これを $\hat{\beta }$ について解くと、 $$\begin{array}{c}(1+e^{\hat{\alpha }+\hat{\beta }})a-n_1\cdot e^{\hat{\alpha }+\hat{\beta }}=0\\ a+a\cdot e^{\hat{\alpha }+\hat{\beta }}-n_1\cdot e^{\hat{\alpha }+\hat{\beta }}=0\\ a+(a-n_1)\cdot e^{\hat{\alpha }}\cdot e^{\hat{\beta }}=0\\ a-c\cdot \frac{b}{d}\cdot e^{\hat{\beta }}=0\\ e^{\hat{\beta }}=\frac{ad}{bc}\\ \hat{\beta }=\log \frac{ad}{bc}\end{array}$$ $◼$

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、 $$\begin{array}{rl}\left|\mathit{I}\left(\mathit{\theta }\right)\right|& =\{n_1{\pi }_1(1-{\pi }_1)+n_0{\pi }_0(1-{\pi }_0)\}{n}_1{\pi }_1(1-{\pi }_1)-{\left\{n_1{\pi }_1(1-{\pi }_1)\right\}}^2\\ & =\{n_1{\pi }_1(1-{\pi }_1)+n_0{\pi }_0(1-{\pi }_0)-n_1{\pi }_1(1-{\pi }_1)\}{n}_1{\pi }_1(1-{\pi }_1)\\ & =n_1{\pi }_1(1-{\pi }_1)n_0{\pi }_0(1-{\pi }_2)\end{array}$$ ${\pi }_1(1-{\pi }_1)={\psi }_1,{\pi }_0(1-{\pi }_0)={\psi }_0$ とすると、 $$\begin{array}{r}\left|\mathit{I}\left(\mathit{\theta }\right)\right|=n_1{\psi }_1{n}_0{\psi }_0\end{array}$$ したがって、 $$\begin{array}{rl}{\mathit{I}\left(\mathit{\theta }\right)}^{-1}& =\frac{1}{n_1{\psi }_1{n}_0{\psi }_0}\left[\begin{array}{cc}{n}_1{\psi }_1& -n_1{\psi }_1\\ -n_1{\psi }_1& n_1{\psi }_1+n_0{\psi }_0\end{array}\right]\\ & =\left[\begin{array}{cc}\frac{1}{n_0{\psi }_0}& -\frac{1}{n_0{\psi }_0}\\ -\frac{1}{n_0{\psi }_0}& \frac{n_1{\psi }_1+n_0{\psi }_0}{n_1{\psi }_1{n}_0{\psi }_0}\end{array}\right]\end{array}$$

別な表現として、 $$\begin{array}{r}{\mathit{I}\left(\mathit{\theta }\right)}^{-1}=\left[\begin{array}{cc}{\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha }}& {\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha },\mathit{\beta }}\\ {\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha },\mathit{\beta }}^T& {\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\beta }}\end{array}\right]\end{array}$$ したがって、推定量の共分散行列は、 $$\begin{array}{rl}V\left(\hat{\alpha }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha }}\\ & =\frac{1}{n_0{\pi }_0(1-{\pi }_0)}\end{array}$$ $$\begin{array}{rl}V\left(\hat{\beta }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\beta }}\\ & =\frac{n_1{\pi }_1(1-{\pi }_1)+n_0{\pi }_0(1-{\pi }_0)}{n_1{\pi }_1(1-{\pi }_1)n_0{\pi }_0(1-{\pi }_0)}\\ & =\frac{1}{n_1{\pi }_1(1-{\pi }_1)}+\frac{1}{n_0{\pi }_0(1-{\pi }_0)}\end{array}$$ $$\begin{array}{rl}\mathrm{Cov}(\hat{\alpha },\hat{\beta })& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\alpha },\mathit{\beta }}\\ & =-\frac{1}{n_0{\pi }_0(1-{\pi }_0)}\end{array}$$

推定量の共分散行列に、ロジスティック・モデルの各値を代入すると、 $$\begin{array}{rl}V\left(\hat{\beta }\right)& =\frac{1}{n_1\cdot \frac{e^{\alpha +\beta }}{1+e^{\alpha +\beta }}\cdot \frac{1}{1+e^{\alpha +\beta }}}+\frac{1}{n_0\cdot \frac{e^{\alpha }}{1+e^{\alpha }}\cdot \frac{1}{1+e^{\alpha }}}\\ & =\frac{{(1+e^{\alpha +\beta })}^2}{n_1{e}^{\alpha +\beta }}+\frac{{(1+e^{\alpha })}^2}{n_0{e}^{\alpha }}\end{array}$$ この最尤推定量は、$\alpha =\hat{\alpha },\beta =\hat{\beta }$ を代入して、 $$\begin{array}{rl}\hat{V}\left(\hat{\beta }\right)& =\frac{{(1+e^{\hat{\alpha }+\hat{\beta }})}^2}{n_1{e}^{\hat{\alpha }+\hat{\beta }}}+\frac{{(1+e^{\hat{\alpha }})}^2}{n_0{e}^{\hat{\alpha }}}\\ & =\frac{{(1+\frac{a}{c})}^2}{n_1\frac{a}{c}}+\frac{{(1+\frac{b}{d})}^2}{n_0\frac{b}{d}}\\ & =\frac{{(a+c)}^2}{c^2}\cdot \frac{c}{a{n}_1}+\frac{{(b+d)}^2}{d^2}\cdot \frac{d}{b{n}_0}\\ & =\frac{{(a+c)}^2}{ac{n}_1}+\frac{{(b+d)}^2}{bd{n}_0}\end{array}$$ $a+c=n_1,b+d=n_0$ より、 $$\begin{array}{r}\hat{V}\left(\hat{\beta }\right)=\frac{a+c}{ac}+\frac{b+d}{bd}\end{array}$$ 各項を部分分数分解すると、 $$\begin{array}{r}\hat{V}\left(\hat{\beta }\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\end{array}$$ $◼$

ロジスティック・モデルのパラメータと対数オッズ比の関係から、問題としている帰無仮説と対立仮説を評価すると、 $$\begin{array}{c}{H}_0:\log \mathrm{OR}=0H_1:\log \mathrm{OR}\ne 0\end{array}$$ すなわち、この検定問題は、通常の分割表における曝露と発症の関係に関する検定問題と同値である。

〔1〕ワルド検定の検定統計量
ワルド検定の検定統計量の定義 ${\chi }_W^2=\frac{{\hat{\beta }}^2}{\hat{V}\left(\hat{\beta }\right)}$ より、 $$\begin{array}{r}{\chi }_W^2=\frac{{(\log \frac{ad}{bc})}^2}{[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}]}\end{array}$$ $◼$

〔2〕尤度比検定の検定統計量
対数尤度の式より、帰無仮説 $\beta =0$ における対数尤度は、 $$\begin{array}{rl}l({\pi }_1,{\pi }_2)& =m_1\alpha +a\beta -n_1\log (1+e^{\alpha +\beta })-n_0\log (1+e^{\alpha })\\ l(\hat{\alpha },0)& =m_1\alpha +a\cdot 0-n_1\log (1+e^{\alpha +0})-n_0\log (1+e^{\alpha })\\ & =m_1\alpha -(n_1+n_0)\log (1+e^{\alpha })\\ & =m_1\alpha -N\log (1+e^{\alpha })\end{array}$$ この対数尤度関数を $\alpha$ で偏微分すると、スコア関数は、 $$\begin{array}{r}{U}_{\alpha }\left(\mathit{\theta }\right)=m_1-N\frac{e^{\alpha }}{1+e^{\alpha }}\end{array}$$ 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、 $$\begin{array}{c}{m}_1-N\frac{e^{{\hat{\alpha }}_0}}{1+e^{{\hat{\alpha }}_0}}=0\\ \frac{e^{{\hat{\alpha }}_0}}{1+e^{{\hat{\alpha }}_0}}=\frac{m_1}{N}\\ (1-\frac{m_1}{N})e^{{\hat{\alpha }}_0}=\frac{m_1}{N}\\ e^{{\hat{\alpha }}_0}=\frac{m_1}{N}\cdot \frac{N}{N-m_1}\\ e^{{\hat{\alpha }}_0}=\frac{m_1}{m_0}\\ {\hat{\alpha }}_0=\log \frac{m_1}{m_0}\end{array}$$ 尤度比検定の検定統計量の定義 ${\chi }_{LR}^2=2\log L(\hat{\alpha },\hat{\beta })-2\log L(\hat{\alpha },0)$ より、 $$\begin{array}{rl}{\chi }_{LR}^2& =2\{m_1\hat{\alpha }+a\hat{\beta }-n_1\log (1+e^{\hat{\alpha }+\hat{\beta }})-n_0\log (1+e^{\alpha })\}-2\{m_1{\hat{\alpha }}_0-N\log (1+e^{{\hat{\alpha }}_0})\}\\ & =2\{m_1(\hat{\alpha }-{\hat{\alpha }}_0)+a\hat{\beta }-n_1\log (1+e^{\hat{\alpha }+\hat{\beta }})-n_0\log (1+e^{\alpha })+N\log (1+e^{{\hat{\alpha }}_0})\}\\ & =2\{m_1(\log \frac{b}{d}-\log \frac{m_1}{m_0})+a\log \frac{ad}{bc}-n_1\log (1+\frac{a}{c})-n_0\log (1+\frac{b}{d})+N\log (1+\frac{m_1}{m_0})\}\\ & =2\{m_1\log \frac{b{m}_0}{d{m}_1}+a\log \frac{ad}{bc}-n_1\log \frac{n_1}{c}-n_0\log \frac{n_0}{d}+N\log \frac{N}{m_0}\}\\ & =2(m_1\log \frac{b{m}_0}{d{m}_1}+a\log \frac{ad}{bc}+n_1\log \frac{c}{n_1}+n_0\log \frac{d}{n_0}+N\log \frac{N}{m_0})\\ & =2(\log \frac{b^{m_1}{m}_0^{m_1}}{d^{m_1}{m}_1^{m_1}}+\log \frac{a^a{d}^a}{b^a{c}^a}+\log \frac{c^{n_1}}{n_1^{n_1}}+\log \frac{d^{n_0}}{n_0^{n_0}}+\log \frac{N^N}{m_0^N})\\ & =2\log [\frac{b^{m_1}{m}_0^{m_1}}{d^{m_1}{m}_1^{m_1}}\cdot \frac{a^a{d}^a}{b^a{c}^a}\cdot \frac{c^{n_1}}{n_1^{n_1}}\cdot \frac{d^{n_0}}{n_0^{n_0}}\cdot \frac{N^N}{m_0^N}]\\ & =2\log [\frac{a^a{N}^N}{n_1^{n_1}{n}_0^{n_0}{m}_1^{m_1}}\cdot \frac{b^{m_1}}{b^a}\cdot \frac{c^{n_1}}{c^a}\cdot \frac{d^a{d}^{n_0}}{d^{m_1}}\cdot \frac{m_0^{m_1}}{m_0^N}]\\ & =2\log [\frac{a^a{N}^N}{n_1^{n_1}{n}_0^{n_0}{m}_1^{m_1}}\cdot b^{a+b-a}\cdot c^{a+c-a}\cdot d^{a+b+d-(a+b)}\cdot m_0^{-(m_1+m_0)-m_1}]\\ & =2\log \left[\frac{a^a{b}^b{c}^c{d}^d{N}^N}{n_1^{n_1}{n}_0^{n_0}{m}_1^{m_1}{m}_0^{m_0}}\right]\end{array}$$ $◼$

〔3〕有効スコア検定の検定統計量
$\beta$ に関する、スコア関数は、帰無仮説 $\beta =0$ においては、 $$\begin{array}{rl}{U}_{\beta }\left(\mathit{\theta }\right)& =a-n_1\cdot \frac{e^{{\hat{\alpha }}_0}}{1+e^{{\hat{\alpha }}_0}}\\ & =a-n_1\cdot \frac{\frac{m_1}{m_0}}{1+\frac{m_1}{m_0}}\\ & =a-\frac{n_1{m}_1}{N}\\ & =a-\hat{E}\left(a\right)\end{array}$$ 帰無仮説のもとで ${\pi }_1={\pi }_0=\pi$ であり、 $$\begin{array}{r}\hat{\pi }=\frac{e^{{\hat{\alpha }}_0}}{1+e^{{\hat{\alpha }}_0}}=\frac{m_1}{N}\end{array}$$ これを、期待情報行列 $\mathit{I}\left(\mathit{\theta }\right)$ の式に代入すると、 $$\begin{array}{rl}\mathit{I}\left({\hat{\mathit{\theta }}}_{\mathbf{0}}\right)& =\left[\begin{array}{cc}{n}_1\hat{\pi }(1-\hat{\pi })+n_0\hat{\pi }(1-\hat{\pi })& n_1\hat{\pi }(1-\hat{\pi })\\ n_1\hat{\pi }(1-\hat{\pi })& n_1\hat{\pi }(1-\hat{\pi })\end{array}\right]\\ & =\hat{\pi }(1-\hat{\pi })\left[\begin{array}{cc}N& n_1\\ n_1& n_1\end{array}\right]\end{array}$$ 期待情報行列の逆行列 ${\mathit{I}\left(\mathit{\theta }\right)}^{-1}$ の式に代入すると、 $$\begin{array}{rl}{\mathit{I}\left({\hat{\mathit{\theta }}}_{\mathbf{0}}\right)}^{-1}& =\left[\begin{array}{cc}\frac{1}{n_0\hat{\pi }(1-\hat{\pi })}& -\frac{1}{n_0\hat{\pi }(1-\hat{\pi })}\\ -\frac{1}{n_0\hat{\pi }(1-\hat{\pi })}& \frac{n_1\hat{\pi }(1-\hat{\pi })+n_0\hat{\pi }(1-\hat{\pi })}{n_1\hat{\pi }(1-\hat{\pi })n_0\hat{\pi }(1-\hat{\pi })}\end{array}\right]\\ & =\frac{1}{\hat{\pi }(1-\hat{\pi })}\left[\begin{array}{cc}\frac{1}{n_0}& -\frac{1}{n_0}\\ -\frac{1}{n_0}& \frac{N}{n_1{n}_0}\end{array}\right]\\ & =\frac{1}{\hat{\pi }(1-\hat{\pi })}\cdot \frac{1}{N{n}_1-n_1^2}\left[\begin{array}{cc}{n}_1& -n_1\\ -n_1& N\end{array}\right]\end{array}$$ 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、 $$\begin{array}{rl}V\left(\hat{\beta }\right)& ={\left[{\mathit{I}\left(\mathit{\theta }\right)}^{-1}\right]}_{\mathit{\beta }}\\ & =\frac{N}{n_1{n}_0\hat{\pi }(1-\hat{\pi })}\\ & =\frac{N^3}{n_1{n}_0{m}_1{m}_0}\end{array}$$ 結果として有効スコア検定 ${\chi }^2=\frac{{\hat{\beta }}^2}{\hat{V}\left(\hat{\beta }\right)}$ は、 $$\begin{array}{rl}{\chi }^2& =\frac{{[a-\frac{n_1{m}_1}{N}]}^2}{\frac{n_1{n}_0{m}_1{m}_0}{N^3}}\\ & =\frac{{[a-\hat{E}\left(a\right)]}^2}{\hat{V}[a-\hat{E}\left(a\right)]}\end{array}$$ これは、通常の分割表に対するコクラン検定の検定統計量と等しい。 $◼$