線形重回帰分析

■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}$

$3$\displaystyle{\hspace{1}{X}_{1}}$ ， $3$\displaystyle{\hspace{1}{X}_{2}}$ ， $3$Y$ のデータの組が $3$n$ 個あるとする．このデータを用いて $3$\displaystyle{Y}$ の予測値 $3$\displaystyle{\hat{y}}$ を求める線形重回帰式は

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

ただし

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

$3$\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}}}$

$3$\displaystyle{\bar{ \hspace{1}{x}_{1} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\sum }\limits_{ i=1 }^{n}\hspace{1}{x}_{ i1 }}$ ， $3$\displaystyle{\bar{ \hspace{1}{x}_{2} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\sum }\limits_{ i=1 }^{n}\hspace{1}{x}_{ i2 }}$ ， $3$\displaystyle{\bar{y}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{y}_{i}}$ (参照：平均)

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{1}y }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\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}{{\sigma}}_{ \hspace{1}{x}_{2}y }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\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}{{\sigma}}_{ \hspace{1}{x}_{1}\hspace{1}{x}_{2} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\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)}$ (参照：共分散)

$3$\displaystyle{{\sigma}_{ \hspace{1}{x}_{1} }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\sum }\limits_{ i=1 }^{n}{\left({ \hspace{1}{x}_{ i1 }- \bar{ \hspace{1}{x}_{1} } }\right)}^{2}}$ ， $3$\displaystyle{{\sigma}_{ \hspace{1}{x}_{2} }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\sum }\limits_{ i=1 }^{n}{\left({ \hspace{1}{x}_{ i2 }- \bar{ \hspace{1}{x}_{2} } }\right)}^{2}}$ (参照：分散)

である．

定数項 $3$\hspace{1}{a}_{0}$ ，および，回帰係数 $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ の求め方はここを見てください．

■説明変数が $3$m$ 個の場合

$3$\displaystyle{\hspace{1}{X}_{1}}$ ， $3$\displaystyle{\hspace{1}{X}_{2}}$ ，･･･， $3$\displaystyle{\hspace{1}{X}_{m}}$ ， $3$Y$ のデータの組が $3$m$ 個あるとする．

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 $3$\displaystyle{\hspace{1}{X}_{m}}$ （説明変数m）	データ $3$Y$ （目的変数）
1	$3$\hspace{1}{x}_{11}$	$3$\hspace{1}{x}_{12}$	･･･	$3$\hspace{1}{x}_{ 1m }$	$3$\hspace{1}{y}_{1}$
2	$3$\hspace{1}{x}_{21}$	$3$\hspace{1}{x}_{22}$	･･･	$3$\hspace{1}{x}_{ 2m }$	$3$\hspace{1}{y}_{2}$
3	$3$\hspace{1}{x}_{31}$	$3$\hspace{1}{x}_{32}$	･･･	$3$\hspace{1}{x}_{ 3m }$	$3$\hspace{1}{y}_{3}$
$3$\vdots$	$3$\vdots$	$3$\vdots$	$3$\vdots$	$3$\vdots$	$3$\vdots$
$3$n$	$3$\displaystyle{\hspace{1}{x}_{ n1 }}$	$3$\displaystyle{\hspace{1}{x}_{ n2 }}$	･･･	$3$\hspace{1}{x}_{ nm }$	$3$\hspace{1}{y}_{n}$

このデータから求められる線形重回帰式は

$3$\displaystyle{\hat{y}=\hspace{1}{a}_{0}+\hspace{1}{a}_{1}\hspace{1}{x}_{1}+\hspace{1}{a}_{2}\hspace{1}{x}_{2}+\cdots +\hspace{1}{a}_{m}\hspace{1}{x}_{m}}$

ただし

$3$\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} }- \cdots - \hspace{1}{a}_{m}\bar{ \hspace{1}{x}_{m} }}$

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

$3$\displaystyle{{\sigma}_{ \hspace{1}{x}_{j} }^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{x}_{ ij }- \bar{ \hspace{1}{x}_{j} } }\right) }^{2} }}$

$3$\displaystyle{\hspace{1}{{\sigma}}_{ \hspace{1}{x}_{j}\hspace{1}{x}_{k} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \left({ \hspace{1}{x}_{ ij }- \bar{ \hspace{1}{x}_{j} } }\right)\left({ \hspace{1}{x}_{ ik }- \bar{ \hspace{1}{x}_{k} } }\right) }}$

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

$3$\displaystyle{\bar{ \hspace{1}{x}_{j} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{ ij } }}$

$3$\displaystyle{\bar{y}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{y}_{i} }}$

である．

定数項および回帰係数の求め方は，(説明変数が2つの場合の証明)を参考にして考えてみてください．

ホーム>>カテゴリー分類>>>>確率統計>>線形回帰

　最終更新日： 2026年5月20日

線形重回帰分析

■説明変数が2つの場合

■説明変数が 個の場合

■説明変数が $3$m$ 個の場合