変動の分解

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}_{ 1m }$	$3$\hspace{1}{y}_{2}$
3	$3$\hspace{1}{x}_{31}$	$3$\hspace{1}{x}_{32}$	･･･	$3$\hspace{1}{x}_{ 1m }$	$3$\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}$

$3$\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．このデータから求められる回帰式(1次式)は

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

の1次式で表されるとする．

目的変数 $3$Y$ の分散： $3$\displaystyle{{\sigma}_{y}^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{y}_{i}- \bar{y} }\right) }^{2} }}$ ･･････(2)

予測値 $3$\displaystyle{\hat{Y}}$ の分散： $3$\displaystyle{{\sigma}_{\hat{y}}^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{\hat{y}}_{i}- \hat{\hat{y}} }\right) }^{2} }}$ ･･････(3)

残差 $3${\varepsilon}$ の分散： $3$\displaystyle{{\sigma}_{{\varepsilon}}^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} { \left({ \hspace{1}{{\varepsilon}}_{i}- \bar{{\varepsilon}} }\right) }^{2} }}$ ･･････(4)

ただし

$3$\displaystyle{\bar{y}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{y}_{i} }}$ ･･････(5)

$3$\displaystyle{\hat{\hat{y}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{\hat{y}}_{i} }}$ ･･････(6)

$3$\displaystyle{\bar{{\varepsilon}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{{\varepsilon}}_{i} }}$ ･･････(7)

$3$\displaystyle{\hspace{1}{\hat{y}}_{i}=\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 }+b}$ ･･････(8)

$3$\displaystyle{\hspace{1}{{\varepsilon}}_{i}=\hspace{1}{y}_{i}- \hspace{1}{\hat{y}}_{i}}$ ･･････(9)

とおくと

$3$\displaystyle{ {\sigma}_{y}^{2}={\sigma}_{\hat{y}}^{2}+{\sigma}_{{\varepsilon}}^{2} }$ ･･････(10)

が成り立つ．(10)の両辺を $3$n$ で割ると，(2) ，(3)，(4)より

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{y}_{i}- \bar{y} }\right)}^{2}}$ $3$\displaystyle{={\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{\hat{y}}_{i}- \hat{\hat{y}} }\right)}^{2}+{\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{{\varepsilon}}_{i}- \bar{{\varepsilon}} }\right)}^{2}}$ ･･････(11)

備考： $3$\displaystyle{ \bar{y}=\hat{\hat{y}} }$ ， $3$\displaystyle{\bar{{\varepsilon}}=0}$ (証明の内容を参照のこと)

の関係が得られる．

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{y}_{i}- \bar{y} }\right)}^{2}}$ ：全変動，あるいは，全平方和

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{\hat{y}}_{i}- \hat{\hat{y}} }\right)}^{2}}$ ：回帰変動，あるいは，回帰平方和

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}{\left({ \hspace{1}{{\varepsilon}}_{i}- \bar{{\varepsilon}} }\right)}^{2}}$ ：残差変動，あるいは，残差平方和

(11)を書き替えると

全変動(TSS)＝回帰変動(SSR)+残差変動(SSE)　･･････(12)

のような表現もある．(11)，(12)の関係を変動の分解という．

■証明

(8)より

$3$\displaystyle{\begin{array}{lll}\hspace{1}{\hat{y}}_{i}=b+\hspace{1}{a}_{1}\hspace{1}{x}_{ i1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ i2 }& \vspace{6}\\ \hspace{1}{\hat{y}}_{1}=b+\hspace{1}{a}_{1}\hspace{1}{x}_{ 11 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ 12 }& \vspace{6}\\ \hspace{1}{\hat{y}}_{2}=b+\hspace{1}{a}_{1}\hspace{1}{x}_{ 21 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ 22 }& \vspace{6}\\ \text{\hspace{5}}\text{ }\vdots & \vspace{6}\\ \hspace{1}{\hat{y}}_{n}=b+\hspace{1}{a}_{1}\hspace{1}{x}_{ n1 }+\hspace{1}{a}_{2}\hspace{1}{x}_{ n2 }& \vspace{6}\\ \end{array}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。

となる連立方程式が得られる．行列を使って表わすと

$3$\displaystyle{\left({ \begin{array}\hat{y}& \vspace{6}\\ \hspace{1}{\hat{y}}_{2} & \vspace{6}\\ \text{ }\vdots & \vspace{6}\\ \hspace{1}{\hat{y}}_{n} & \vspace{6}\\ \end{array} }\right)=\left({ \begin{array}1& \hspace{1}{x}_{ 11 } & \hspace{1}{x}_{ 12 } & \vspace{6}\\ 1& \hspace{1}{x}_{ 21 } & \hspace{1}{x}_{ 22 } & \vspace{6}\\ \vdots &\vdots &\vdots & \vspace{6}\\ 1& \hspace{1}{x}_{ n1 } & \hspace{1}{x}_{ n2 } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}b& \vspace{6}\\ \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
･･････(13)

となる．

$3$\displaystyle{\left({ \begin{array}\hat{y}& \vspace{6}\\ \hspace{1}{\hat{y}}_{2} & \vspace{6}\\ \text{ }\vdots & \vspace{6}\\ \hspace{1}{\hat{y}}_{n} & \vspace{6}\\ \end{array} }\right)=\displaystyle{\hat{y}}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
， $3$\displaystyle{\left({ \begin{array}1& \hspace{1}{x}_{ 11 } & \hspace{1}{x}_{ 12 } & \vspace{6}\\ 1& \hspace{1}{x}_{ 21 } & \hspace{1}{x}_{ 22 } & \vspace{6}\\ \vdots &\vdots &\vdots & \vspace{6}\\ 1& \hspace{1}{x}_{ n1 } & \hspace{1}{x}_{ n2 } & \vspace{6}\\ \end{array} }\right)=X}$ ， $3$\displaystyle{\left({ \begin{array}b& \vspace{6}\\ \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)=\displaystyle{\alpha }}$

とおくと，(13)は

$3$\displaystyle{\displaystyle{\hat{y}}=X\displaystyle{\alpha }}$ ･･････(14)

となる．

$3$\displaystyle{\left({ \begin{array} \hspace{1}{{\varepsilon}}_{1} & \vspace{6}\\ \hspace{1}{{\varepsilon}}_{2} & \vspace{6}\\ \text{ }\vdots & \vspace{6}\\ \hspace{1}{{\varepsilon}}_{n} & \vspace{6}\\ \end{array} }\right)=\displaystyle{{\varepsilon}}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
， $3$\displaystyle{\left({ \begin{array} \hspace{1}{y}_{1} & \vspace{6}\\ \hspace{1}{y}_{2} & \vspace{6}\\ \text{ }\vdots & \vspace{6}\\ \hspace{1}{y}_{n} & \vspace{6}\\ \end{array} }\right)=\displaystyle{y}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。

とおくと，(9)より

$3$\displaystyle{{\varepsilon}=\displaystyle{y}- \displaystyle{\hat{y}}}$ ･･････(15)

が得られる．

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}_{1}\hspace{1}{x}_{ 1i }+\hspace{1}{a}_{2}\hspace{1}{x}_{ 2i }+b }\right) }\right} }^{2} }}$ ･･････(2)

となる．

$3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ ， $3$b$ を変数としたとき， $3$\displaystyle{\hspace{1}{R}_{ SS }}$ が最小となるとなる $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ ， $3$b$ が回帰係数となる．言い換えると

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

を満たす $3$\hspace{1}{a}_{1}$ ， $3$\hspace{1}{a}_{2}$ ， $3$b$ が回帰係数となる．

(5)と(2)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}- \left({ \hspace{1}{a}_{1}\hspace{1}{x}_{ 1i }+\hspace{1}{a}_{2}\hspace{1}{x}_{ 2i }+b }\right) }\right} }^{2} }=0}$

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

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

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

が得られる．

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

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

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

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

$3$\displaystyle{{\displaystyle{\sum }}\limits_{ i=1 }^{n}1\cdot \hspace{1}{y}_{i}}$ $3$\displaystyle{=\left{{ {\displaystyle{\sum }}\limits_{ i=1 }^{n}1\cdot \left({ \begin{array}1& \hspace{1}{x}_{ 1i } & \hspace{1}{x}_{ 2i } & \vspace{6}\\ \end{array} }\right) }\right}\left({ \begin{array}b& \vspace{6}\\ \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)}$

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

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

$3$\displaystyle{\left({ \begin{array} {\displaystyle{\sum }}\limits_{ i=1 }^{n}1\cdot \hspace{1}{y}_{i} & \vspace{6}\\ {\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ 1i }\hspace{1}{y}_{i} & \vspace{6}\\ {\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ 2i }\hspace{1}{y}_{i} & \vspace{6}\\ \end{array} }\right)}$ $3$\displaystyle{=\left({ \begin{array} {\displaystyle{\sum }}\limits_{ i=1 }^{n}1\cdot \left({ \begin{array}1& \hspace{1}{x}_{ 1i } & \hspace{1}{x}_{ 2i } & \vspace{6}\\ \end{array} }\right) & \vspace{6}\\ {\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ 1i }\left({ \begin{array}1& \hspace{1}{x}_{ 1i } & \hspace{1}{x}_{ 2i } & \vspace{6}\\ \end{array} }\right) & \vspace{6}\\ {\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{ 2i }\left({ \begin{array}1& \hspace{1}{x}_{ 1i } & \hspace{1}{x}_{ 2i } & \vspace{6}\\ \end{array} }\right) & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}b& \vspace{6}\\ \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{\left({ \begin{array}1&1&\cdots &1& \vspace{6}\\ \hspace{1}{x}_{ 11 } & \hspace{1}{x}_{ 12 } &\cdots & \hspace{1}{x}_{ 1n } & \vspace{6}\\ \hspace{1}{x}_{ 21 } & \hspace{1}{x}_{ 22 } &\cdots & \hspace{1}{x}_{ 2n } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array} \hspace{1}{y}_{1} & \vspace{6}\\ \hspace{1}{y}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{y}_{n} & \vspace{6}\\ \end{array} }\right)}$ $3$\displaystyle{=\left({ \begin{array}1&1&\cdots &1& \vspace{6}\\ \hspace{1}{x}_{ 11 } & \hspace{1}{x}_{ 12 } &\cdots & \hspace{1}{x}_{ 1n } & \vspace{6}\\ \hspace{1}{x}_{ 21 } & \hspace{1}{x}_{ 22 } &\cdots & \hspace{1}{x}_{ 2n } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}1& \hspace{1}{x}_{ 11 } & \hspace{1}{x}_{ 21 } & \vspace{6}\\ 1& \hspace{1}{x}_{ 12 } & \hspace{1}{x}_{ 22 } & \vspace{6}\\ \vdots &\vdots &\vdots & \vspace{6}\\ 1& \hspace{1}{x}_{ 1n } & \hspace{1}{x}_{ 2n } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array}b& \vspace{6}\\ \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{\hspace{1}{}_{}^{y}X\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{\alpha }t}\displaystyle{}=\hspace{1}{}_{}^{}X_{\alpha }^{}t$

$3$\displaystyle{\hspace{1}{}_{}^{\displaystyle{y}}X\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{}t}\left({ }\right)=\displaystyle{0}$

$3$\displaystyle{\hspace{1}{}_{}^{{\varepsilon}}X\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{}t}\displaystyle{}=\displaystyle{0}$

$3$\displaystyle{\left({ \begin{array}1& \vspace{6}\\ 1& \vspace{6}\\ \vdots & \vspace{6}\\ 1& \vspace{6}\\ \end{array} }\right)=\hspace{1}{\displaystyle{x}}_{0}}$ ， $3$\displaystyle{\left({ \begin{array} \hspace{1}{x}_{ 11 } & \vspace{6}\\ \hspace{1}{x}_{ 21 } & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{x}_{ n1 } & \vspace{6}\\ \end{array} }\right)=\hspace{1}{\displaystyle{x}}_{1}}$ ， $3$\displaystyle{\left({ \begin{array} \hspace{1}{x}_{ 12 } & \vspace{6}\\ \hspace{1}{x}_{ 22 } & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{x}_{ n2 } & \vspace{6}\\ \end{array} }\right)=\hspace{1}{\displaystyle{x}}_{2}}$

$3$\displaystyle{\hspace{1}{}_{}^{}\displaystyle{x}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{0}t}\hspace{1}{ }_{0}\displaystyle{{\varepsilon}}=0$ ， $3$\displaystyle{\hspace{1}{}_{}^{}\displaystyle{x}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{1}t}\hspace{1}{ }_{1}\displaystyle{{\varepsilon}}=0$ ， $3$\displaystyle{\hspace{1}{}_{}^{}\displaystyle{x}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{2}t}\hspace{1}{ }_{2}\displaystyle{{\varepsilon}}=0$

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

$3$\displaystyle{\hspace{1}{}_{}^{{\varepsilon}}\displaystyle{\hat{y}}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{{\varepsilon}}t}\displaystyle{}=\hspace{1}{}_{}^{}\left({ b\hspace{1}{\displaystyle{x}}_{0}+\hspace{1}{a}_{1}\hspace{1}{\displaystyle{x}}_{1}+\hspace{1}{a}_{2}\hspace{1}{\displaystyle{x}}_{2} }\right)_{{\varepsilon}}^{}t$

$3$\displaystyle{=\left{{ \hspace{1}{}_{}^{\left({ \hspace{1}{a}_{1}\hspace{1}{\displaystyle{x}}_{1} }\right)} \left({ b\hspace{1}{\displaystyle{x}}_{0} }\right) \hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}_{t}^{ \hspace{1}{a}_{2}\hspace{1}{\displaystyle{x}}_{2} }t +\hspace{1}{}_{}^{} _{}^{\left({ \hspace{1}{a}_{2}\hspace{1}{\displaystyle{x}}_{2} }\right)}t}\right}+\hspace{1}{}_{}^{} _{}^{}t}$

$3$\displaystyle{=\left({ b\hspace{1}{}_{}^{}\displaystyle{x}t_{0}^{}t \hspace{1}{ }_{0}+\hspace{1}{a}_{1}\hspace{1}{}_{}^{}\displaystyle{x}_{}^{}t}\right)\hspace{1}{ }_{1}+\hspace{1}{a}_{2}\hspace{1}{}_{}^{\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}0}\displaystyle{x}_{}^{}t}\hspace{1}{ }_{2}$

$3$\displaystyle{=b\hspace{1}{}_{}^{}\displaystyle{x}0_{}^{}t}\hspace{1}{ }_{0}\displaystyle{{\varepsilon}}+\hspace{1}{a}_{1}\hspace{1}{}_{}^{\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}\hspace{5}0}\displaystyle{x}_{}^{}t$

$3$\displaystyle{=0}$

$3$\displaystyle{\displaystyle{\hat{y}}\cdot \displaystyle{{\varepsilon}}=0}$

$3$\displaystyle{{\left|{\displaystyle{y}}\right|}^{2}=\displaystyle{y}\cdot \displaystyle{y}}$

$3$\displaystyle{=\left({ \displaystyle{\hat{y}}+\displaystyle{{\varepsilon}} }\right)\cdot \left({ \displaystyle{\hat{y}}+\displaystyle{{\varepsilon}} }\right)}$

$3$\displaystyle{=\left({ \displaystyle{\hat{y}}+\displaystyle{{\varepsilon}} }\right)\cdot \displaystyle{\hat{y}}+\left({ \displaystyle{\hat{y}}+\displaystyle{{\varepsilon}} }\right)\cdot \displaystyle{{\varepsilon}}}$

$3$\displaystyle{=\displaystyle{\hat{y}}\cdot \displaystyle{\hat{y}}+\displaystyle{{\varepsilon}}\cdot \displaystyle{\hat{y}}++\displaystyle{\hat{y}}\cdot \displaystyle{{\varepsilon}}+\displaystyle{{\varepsilon}}\cdot \displaystyle{{\varepsilon}}}$

$3$\displaystyle{={\left|{\displaystyle{\hat{y}}}\right|}^{2}+{\left|{\displaystyle{{\varepsilon}}}\right|}^{2}}$

$3$\displaystyle{\bar{\displaystyle{y}}=\left({ \begin{array}\bar{y}& \vspace{6}\\ \bar{y}& \vspace{6}\\ \vdots & \vspace{6}\\ \bar{y}& \vspace{6}\\ \end{array} }\right)=\bar{y}\left({ \begin{array}1& \vspace{6}\\ 1& \vspace{6}\\ \vdots & \vspace{6}\\ 1& \vspace{6}\\ \end{array} }\right)=\bar{y}\hspace{1}{\displaystyle{x}}_{0}}$

$3$\displaystyle{\hat{\displaystyle{\hat{y}}}=\left({ \begin{array} \hat{\hat{y}} & \vspace{6}\\ \hat{\hat{y}} & \vspace{6}\\ \vdots & \vspace{6}\\ \hat{\hat{y}} & \vspace{6}\\ \end{array} }\right)=\hat{\hat{y}}\left({ \begin{array}1& \vspace{6}\\ 1& \vspace{6}\\ \vdots & \vspace{6}\\ 1& \vspace{6}\\ \end{array} }\right)=\hat{\hat{y}}\hspace{1}{\displaystyle{x}}_{0}}$

$3$\displaystyle{\left({ \begin{array}1&1&\cdots &1& \vspace{6}\\ \end{array} }\right)\left({ \begin{array} \hspace{1}{{\varepsilon}}_{1} & \vspace{6}\\ \hspace{1}{{\varepsilon}}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{{\varepsilon}}_{n} & \vspace{6}\\ \end{array} }\right)=0}$

$3$\displaystyle{\hspace{1}{{\varepsilon}}_{1}+\hspace{1}{{\varepsilon}}_{2}+\cdots +\hspace{1}{{\varepsilon}}_{n}=0}$

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\left({ \hspace{1}{{\varepsilon}}_{1}+\hspace{1}{{\varepsilon}}_{2}+\cdots +\hspace{1}{{\varepsilon}}_{n} }\right)=0}$

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

$3$\displaystyle{\bar{{\varepsilon}}=0}$

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

$3$\displaystyle{\bar{{\varepsilon}}=\bar{y}- \hat{\hat{y}}}$

$3$\displaystyle{\bar{y}=\hat{\hat{y}}}$

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

$3$\displaystyle{=\left({ \displaystyle{y}- \bar{\displaystyle{y}} }\right)\cdot \left({ \displaystyle{y}- \bar{\displaystyle{y}} }\right)}$

$3$\displaystyle{=\left({ \displaystyle{y}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \displaystyle{y}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)}$

$3$\displaystyle{=\left({ \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)}$

$3$\displaystyle{=\left({ \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \hat{\displaystyle{y}}}$ $3$\displaystyle{+\left({ \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \displaystyle{{\varepsilon}}}$ $3$\displaystyle{- \left({ \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)}$

$3$\displaystyle{=\hat{\displaystyle{y}}\cdot \hat{\displaystyle{y}}+\displaystyle{{\varepsilon}}\cdot \hat{\displaystyle{y}}- \bar{y}\left({ \hspace{1}{\displaystyle{x}}_{0}\cdot \hat{\displaystyle{y}} }\right)}$ $3$\displaystyle{+\hat{\displaystyle{y}}\cdot \displaystyle{{\varepsilon}}+\displaystyle{{\varepsilon}}\cdot \displaystyle{{\varepsilon}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0}\cdot \displaystyle{{\varepsilon}}}$ $3$\displaystyle{- \bar{y}\left({ \hat{\displaystyle{y}}\cdot \hspace{1}{\displaystyle{x}}_{0} }\right)- \bar{y}\left({ \displaystyle{{\varepsilon}}\cdot \hat{\displaystyle{y}} }\right)+{\bar{y}}^{2}\left({ \hspace{1}{\displaystyle{x}}_{0}\cdot \hspace{1}{\displaystyle{x}}_{0} }\right)}$

$3$\displaystyle{=\hat{\displaystyle{y}}\cdot \hat{\displaystyle{y}}- 2\bar{y}\left({ \hspace{1}{\displaystyle{x}}_{0}\cdot \hat{\displaystyle{y}} }\right)+{\bar{y}}^{2}\left({ \hspace{1}{\displaystyle{x}}_{0}\cdot \hspace{1}{\displaystyle{x}}_{0} }\right)+\displaystyle{{\varepsilon}}\cdot \displaystyle{{\varepsilon}}}$

$3$\displaystyle{=\left({ \hat{\displaystyle{y}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \hat{\displaystyle{y}}- \bar{y}\hspace{1}{\displaystyle{x}}_{0} }\right)+\displaystyle{{\varepsilon}}\cdot \displaystyle{{\varepsilon}}}$

$3$\displaystyle{=\left({ \hat{\displaystyle{y}}- \hat{\hat{y}}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \hat{\displaystyle{y}}- \hat{\hat{y}}\hspace{1}{\displaystyle{x}}_{0} }\right)}$

$3$\displaystyle{+\left({ \displaystyle{{\varepsilon}}- \bar{{\varepsilon}}\hspace{1}{\displaystyle{x}}_{0} }\right)\cdot \left({ \displaystyle{{\varepsilon}}- \bar{{\varepsilon}}\hspace{1}{\displaystyle{x}}_{0} }\right)}$

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

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　最終更新日： 2026年4月27日