線形重回帰分析の回帰係数の導出(説明変数が2個の場合)

■説明変数が2つの場合

データNo.	データ $3$\displaystyle{\hspace{1}{X}_{1}}$ （説明変数1）	データ $3$\displaystyle{\hspace{1}{X}_{2}}$ （説明変数2）	データ $3$Y$ （目的変数）
1	$3$\hspace{1}{x}_{11}$	$3$\hspace{1}{x}_{12}$	$3$\hspace{1}{y}_{1}$
2	$3$\hspace{1}{x}_{21}$	$3$\hspace{1}{x}_{22}$	$3$\hspace{1}{y}_{2}$
3	$3$\hspace{1}{x}_{31}$	$3$\hspace{1}{x}_{32}$	$3$\hspace{1}{y}_{3}$
$3$\vdots$	$3$\vdots$	$3$\vdots$	$3$\vdots$
$3$n$	$3$\displaystyle{\hspace{1}{x}_{ n1 }}$	$3$\displaystyle{\hspace{1}{x}_{ n2 }}$	$3$\hspace{1}{y}_{n}$

線形重回帰式

$3$\displaystyle{\hat{y}=\hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{1}+\hspace{1}{a}_{2}\hspace{1}{x}_{2}}$ ･･････(1)

の定数項 $3$\hspace{1}{a}_{0}$ ，および，偏回帰係数 $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ を求める．

残差平方和 $3$\displaystyle{\hspace{1}{R}_{ SS }}$ は

$3$\displaystyle{\hspace{1}{R}_{ SS }=\displaystyle{{\sum }\limits_{ i=1 }^{n} {\varepsilon}_{i}^{2} }}$

$3$\displaystyle{=\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{y}_{i}- \hat{ \hspace{1}{y}_{i} } }\right) }^{2} }}$

$3$\displaystyle{={\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right}}^{2}}$ ･･････(2)

となる．

回帰式では， $3$\displaystyle{\hspace{1}{R}_{ SS }}$ が最小(極小)となるとなるように $3$\hspace{1}{a}_{0}$ ， $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ が求められている．よって， $3$\displaystyle{\hspace{1}{R}_{ SS }}$ が極小となるときに成り立つ式

$3$\displaystyle{\frac{\hspace{2} {\partial}\hspace{1}{R}_{ SS } \hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{0} \hspace{2}}=0}$ ･･････(3)
$3$\displaystyle{\frac{\hspace{2} {\partial}\hspace{1}{R}_{ SS } \hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{1} \hspace{2}}=0}$ ･･････(4)
$3$\displaystyle{\frac{\hspace{2} {\partial}\hspace{1}{R}_{ SS } \hspace{2}}{\hspace{2} {\partial} \hspace{1}{a}_{2} \hspace{2}}=0}$ ･･････(5)

を満たす $3$\hspace{1}{a}_{0}$ ， $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ を求めればよい．

備考：残差平方和の特徴として，極小は存在するが，極大は存在しない．よって，(3)，(4)，(5)を満たすのは極小のときのみである．

(3)と(2)より(偏微分の計算はここを参照)

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{0} \hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right} }^{2} }=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{0} \hspace{2}}{ \left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right} }^{2} }=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} 2\left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right}\left({ - 1 }\right) }=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right} }=0}$ ･･････(6)

が得られる．

(6)の両辺を $3$n$ で割る．

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left{{ \hspace{1}{y}_{i}- \left({ \hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right} }=0}$

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{y}_{i}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{a}_{0}- \hspace{1}{a}_{1}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ i1 }- \hspace{1}{a}_{2}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ i2 }=0}$

ここで

$3$\displaystyle{\bar{y}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{y}_{i} }}$ ･･････(7)
$3$\displaystyle{\bar{ \hspace{1}{x}_{1} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{ i1 } }}$ ･･････(8)
$3$\displaystyle{\bar{ \hspace{1}{x}_{2} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{ i2 } }}$ ･･････(9)

とおくと

$3$\displaystyle{\bar{y}- \hspace{1}{a}_{0}- \hspace{1}{a}_{1}\bar{ \hspace{1}{x}_{1} }- \hspace{1}{a}_{2}\bar{ \hspace{1}{x}_{2} }=0}$

$3$\displaystyle{\hspace{1}{a}_{0}=\bar{y}- \hspace{1}{a}_{1}\bar{ \hspace{1}{x}_{1} }- \hspace{1}{a}_{2}\bar{ \hspace{1}{x}_{2} }}$ ･･････(10)

(10)を(2)に代入する．

$3$\displaystyle{\hspace{1}{R}_{ SS }={\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left{{ \hspace{1}{y}_{i}- \left({ \bar{y}- \hspace{1}{a}_{1}\bar{ \hspace{1}{x}_{1} }- \hspace{1}{a}_{2}\bar{ \hspace{1}{x}_{2} }+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 } }\right) }\right}}^{2}}$

$3$\displaystyle{=\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right} }^{2} }}$ ･･････(11)

(4)と(11)より

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{1} \hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right} }^{2} }=0}$

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{1} \hspace{2}}{\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right}}^{2}=0}$

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}2\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right}\left{{ - \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right) }\right}=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right} }=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{1}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right) }^{2} }}$ $3$\displaystyle{- \hspace{1}{a}_{2}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }=0}$ ･･････(12)

(12)の両辺を $3$n$ で割る．

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{1}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right) }^{2} }}$ $3$\displaystyle{- \hspace{1}{a}_{2}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }=0}$

ここで

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ ･･･(13)
$3$\displaystyle{{\sigma}_{ \hspace{1}{x}_{1} }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right) }^{2} }}$ ･･･(14)
$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }}$ ･･･(15)

おくと

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y }- \hspace{1}{a}_{1}{\sigma}_{ \hspace{1}{x}_{1} }^{2}- \hspace{1}{a}_{2}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }=0}$

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y }=\hspace{1}{a}_{1}{\sigma}_{ \hspace{1}{x}_{1} }^{2}+\hspace{1}{a}_{2}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }}$ ･･････(16)

(5)と(11)より

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{2} \hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right} }^{2} }=0}$

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}\hspace{1}{a}_{2} \hspace{2}}{\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right}}^{2}=0}$

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}2\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right}\left{{ - \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right}=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right)\left{{ \hspace{1}{y}_{i}- \bar{y}- \hspace{1}{a}_{1}\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)- \hspace{1}{a}_{2}\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }\right} }=0}$

$3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{1}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{2}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }^{2} }=0}$ ･･････(17)

(13)の両辺を $3$n$ で割る．

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{1}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }}$ $3$\displaystyle{- \hspace{1}{a}_{2}\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }^{2} }=0}$

ここで

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right)\left({ \hspace{1}{y}_{i}- \bar{y} }\right) }}$ ･･･(18)
$3$\displaystyle{{\sigma}_{ \hspace{1}{x}_{2} }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right) }^{2} }}$ ･･･(19)

とおく．（18)，(15)，(19)より

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y }- \hspace{1}{a}_{1}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }- \hspace{1}{a}_{2}{\sigma}_{ \hspace{1}{x}_{2} }^{2}=0}$

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y }=\hspace{1}{a}_{1}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }+\hspace{1}{a}_{2}{\sigma}_{ \hspace{1}{x}_{2} }^{2}}$ ･･････(20)

(16)と(20)の連立方程式

$3$\displaystyle{\left{{\begin{array}{lll}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y }=\hspace{1}{a}_{1}{\sigma}_{ \hspace{1}{x}_{1} }^{2}+\hspace{1}{a}_{2}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }& \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y }=\hspace{1}{a}_{1}\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }+\hspace{1}{a}_{2}{\sigma}_{ \hspace{1}{x}_{2} }^{2}& \vspace{6}\\ \end{array}}$

を行列を使って表わすと

$3$\displaystyle{\left({ \begin{array} \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y } & \vspace{6}\\ \end{array} }\right)=\left({ \begin{array} {\sigma}_{ \hspace{1}{x}_{1} }^{2} & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & {\sigma}_{ \hspace{1}{x}_{2} }^{2} & \vspace{6}\\ \end{array} }\right)\left({ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)}$

となる．クラメルの公式を用いると

$3$\displaystyle{\hspace{1}{a}_{1}=\frac{\hspace{2} \left|{ \begin{array} \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y } & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y } & {\sigma}_{ \hspace{1}{x}_{2} }^{2} & \vspace{6}\\ \end{array} }\right| \hspace{2}}{\hspace{2} \left|{ \begin{array} {\sigma}_{ \hspace{1}{x}_{1} }^{2} & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & {\sigma}_{ \hspace{1}{x}_{2} }^{2} & \vspace{6}\\ \end{array} }\right| \hspace{2}}}$ ， $3$\displaystyle{\hspace{1}{a}_{2}=\frac{\hspace{2} \left|{ \begin{array} \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y } & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y } & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{2}y } & \vspace{6}\\ \end{array} }\right| \hspace{2}}{\hspace{2} \left|{ \begin{array} {\sigma}_{ \hspace{1}{x}_{1} }^{2} & \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & \vspace{6}\\ \hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} } & {\sigma}_{ \hspace{1}{x}_{2} }^{2} & \vspace{6}\\ \end{array} }\right| \hspace{2}}}$

となる．

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　最終更新日： 2026年5月12日