最小二乗法

ある物理量 $3$\displaystyle{y}$ がある物理量 $3$\displaystyle{x}$ の関数で

$3$\displaystyle{y=ax+b}$

と表されるとする．例えば $3$\displaystyle{x}$ の値を決めて $3$\displaystyle{y}$ の値を測定する作業を $3$\displaystyle{n}$ 回繰り返し，表のように $3$\displaystyle{n}$ 個の $3$\displaystyle{x}$ と $3$\displaystyle{y}$ の対が得られたとする．

測定回数	$3$1$	$3$2$	$3$3$	……	$3$\displaystyle{n- 1}$	$3$\displaystyle{n}$
$3$\displaystyle{x}$ の値	$3$\displaystyle{\hspace{1}{x}_{1}}$	$3$\displaystyle{\hspace{1}{x}_{2}}$	$3$\displaystyle{\hspace{1}{x}_{3}}$	……	$3$\displaystyle{\hspace{1}{x}_{ n- 1 }}$	$3$\displaystyle{\hspace{1}{x}_{n}}$
$3$\displaystyle{y}$ の値	$3$\displaystyle{\hspace{1}{y}_{1}}$	$3$\displaystyle{\hspace{1}{y}_{2}}$	$3$\displaystyle{\hspace{1}{y}_{3}}$	……	$3$\displaystyle{\hspace{1}{y}_{ n- 1 }}$	$3$\displaystyle{\hspace{1}{y}_{n}}$

この $3$\displaystyle{x}$ と $3$\displaystyle{y}$ の対を $3$\displaystyle{xy}$ 座標上にプロットすると

となり，実験条件のバラツキや測定誤差などの要因で，プロットした点は直線上に乗らない．(本来なら $3$\displaystyle{y=ax+b}$ の関係があるので，すべての点は直線上に乗るはずである．)

プロットした点と $3$\displaystyle{y=ax+b}$ との $3$\displaystyle{y}$ 軸方向の値の差 $3$\displaystyle{ \hspace{1}{{\Delta}}_{i} }$ は

$3$\displaystyle{ \hspace{1}{{\Delta}}_{i}=\hspace{1}{y}_{i}- $ a\hspace{1}{x}_{i}+b $ }$

となる． $3$\displaystyle{a}$ ， $3$\displaystyle{b}$ の値によって $3$\displaystyle{ \hspace{1}{{\Delta}}_{i} }$ の値は変化する．すべての点で $3$\displaystyle{ \hspace{1}{{\Delta}}_{i} }$ の値を小さくする $3$\displaystyle{a}$ ， $3$\displaystyle{b}$ の値が， $3$\displaystyle{x}$ と $3$\displaystyle{y}$ の関係を表す最も確からしい値だと考えることができる． $3$\displaystyle{a}$ ， $3$\displaystyle{b}$ の値を決める方法として

$3$\displaystyle{{{\Delta}}^{2}=\displaystyle{{\sum }\limits_{ i=1 }^{n} {\Delta}{i}^{2} }=\displaystyle{{\sum }\limits_{ i=1 }^{n} { \{ \hspace{1}{y}_{i}- $ a\hspace{1}{x}_{i}+b $ \} }^{2} }}$

の値を最小とする $3$\displaystyle{a}$ ， $3$\displaystyle{b}$ を用いる方法がある．この方法のことを最小二乗法という．

● $3$\displaystyle{{{\Delta}}^{2}}$ は最小となる $3$a$ と $3$b$ の値

$3$\displaystyle{a=\frac{\hspace{2} \hspace{1}{{\sigma}}_{ xy } \hspace{2}}{\hspace{2} {\sigma}_{x}^{2} \hspace{2}}}$ ， $3$\displaystyle{b=\bar{y}- a\bar{x}}$

ただし

$3$\displaystyle{\bar{x}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{i}}$ ， $3$\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}}_{ xy }=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{i}\hspace{1}{y}_{i}- \bar{x}\bar{y}}$ (参照：共分散)

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

■導出

偏導関数を用いた計算 ⇒ ここ

$3$\displaystyle{{{\Delta}}^{2}}$ の最小となる $3$\displaystyle{a}$ ， $3$\displaystyle{b}$ の値を平方完成を利用して求める．

$3$\displaystyle{ {{\Delta}}^{2}= \displaystyle{{\sum }\limits_{ i=1 }^{n} { \{ \hspace{1}{y}_{i}- $ a\hspace{1}{x}_{i}+b $ \} }^{2} } }$

$3$\displaystyle{ =\displaystyle{{\sum }\limits_{ i=1 }^{n} \{ y_{i}^{2}- 2 $ a\hspace{1}{x}_{i}+b $ \hspace{1}{y}_{i}+{ $ a\hspace{1}{x}_{i}+b $ }^{2} \} } }$

$3$\displaystyle{ =\displaystyle{{\sum }\limits_{ i=1 }^{n} \{ y_{i}^{2}- 2a\hspace{1}{x}_{i}\hspace{1}{y}_{i}- 2b\hspace{1}{y}_{i}+{a}^{2}x_{i}^{2}+2ab\hspace{1}{x}_{i}+{b}^{2} \} } }$

$3$\displaystyle{ =\displaystyle{{\sum }\limits_{ i=1 }^{n}y_{i}^{2}}+\displaystyle{{\sum }\limits_{ i=1 }^{n} $ - 2a\hspace{1}{x}_{i}\hspace{1}{y}_{i} $ }+\displaystyle{{\sum }\limits_{ i=1 }^{n} $ - 2b\hspace{1}{y}_{i} $ } }$ $3$\displaystyle{ +\displaystyle{{\sum }\limits_{ i=1 }^{n} {a}^{2}x_{i}^{2} } }$ $3$\displaystyle{ +\displaystyle{{\sum }\limits_{ i=1 }^{n} 2ab\hspace{1}{x}_{i} } }$ $3$\displaystyle{ +\displaystyle{{\sum }\limits_{ i=1 }^{n}{b}^{2}} }$

$3$\displaystyle{ =\displaystyle{{\sum }\limits_{ i=1 }^{n}y_{i}^{2}}- 2a\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{i}\hspace{1}{y}_{i} }- 2b\displaystyle{{\sum }\limits_{ i=1 }^{n}\hspace{1}{y}_{i}} }$ $3$\displaystyle{ +{a}^{2}\displaystyle{{\sum }\limits_{ i=1 }^{n}x_{i}^{2}} }$ $3$\displaystyle{ +2ab\displaystyle{{\sum }\limits_{ i=1 }^{n}\hspace{1}{x}_{i}} }$ $3$\displaystyle{ +n{b}^{2} }$

$3$$ ∑ i = 1 n y i 2 = A ， $3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{i}\hspace{1}{y}_{i} }=B}$ ， $3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{y}_{i} }=C}$ ， $3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} x_{i}^{2} }=D}$ ， $3$\displaystyle{\displaystyle{{\sum }\limits_{ i=1 }^{n} \hspace{1}{x}_{i} }=E}$ とおくと

$3$\displaystyle{=A- 2aB- 2bC+{a}^{2}D+2abE+n{b}^{2}}$

$3$\displaystyle{=n{b}^{2}+2 $ Ea- C $ b+D{a}^{2}- 2Ba+A}$

$3$\displaystyle{ =n{ $ b+\frac{\hspace{2} Ea- C \hspace{2}}{\hspace{2}n\hspace{2}} $ }^{2}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{ $ Ea- C $ }^{2} }$ $3$\displaystyle{ +D{a}^{2} }$ $3$\displaystyle{ - 2Ba+A }$

$3$\displaystyle{ =n{ \left({ b+\frac{\hspace{2} Ea- C \hspace{2}}{\hspace{2}n\hspace{2}} }\right) }^{2} }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\left({ - {E}^{2}{a}^{2}+2CEa- {C}^{2}+nD{a}^{2}- 2nBa+nA }\right) }$

$3$\displaystyle{=n{ $ b+\frac{\hspace{2} Ea- C \hspace{2}}{\hspace{2}n\hspace{2}} $ }^{2}}$ $3$\displaystyle{+\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}} $ nD- {E}^{2} $ { $ a+\frac{\hspace{2} CE- nB \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}} $ }^{2}}$ $3$\displaystyle{- \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\frac{\hspace{2} { \left({ CE- nB }\right) }^{2} \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}+A- \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{C}^{2}}$

これより， $3${{\Delta}}^{2}$ が最小になるのは

$3$\displaystyle{b+\frac{\hspace{2} Ea- C \hspace{2}}{\hspace{2}n\hspace{2}}=0}$ ， $3$\displaystyle{a+\frac{\hspace{2} CE- nB \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}=0}$

の時である．よって

$3$\displaystyle{a}$ $3$\displaystyle{=- \frac{\hspace{2} CE- nB \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} nB- CE \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}}$

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

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

ここで

$3$\displaystyle{{\sigma}_{x}^{2}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}x_{i}^{2}- {\bar{x}}^{2}}$ (参照：分散)　･･････(3)

とおくと

$3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{{\sigma}}_{ xy } \hspace{2}}{\hspace{2} {\sigma}_{x}^{2} \hspace{2}} }$ ･･････(4)

$3$\displaystyle{b}$ $3$\displaystyle{=- \frac{\hspace{2} Ea- C \hspace{2}}{\hspace{2}n\hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} C- Ea \hspace{2}}{\hspace{2}n\hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}} $ C- E\frac{\hspace{2} nB- CE \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}\frac{\hspace{2} nCD- C{E}^{2}- nBE+C{E}^{2} \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} CD- BE \hspace{2}}{\hspace{2} nD- {E}^{2} \hspace{2}}}$

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

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

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

(1)より

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

$3$\displaystyle{=\frac{\hspace{2} \displaystyle{\left{{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}x_{i}^{2} - { \bar{x} }^{2} }\right}\bar{y}- \bar{x}\left{{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}\hspace{1}{x}_{i}\hspace{1}{y}_{i} - \bar{x}\bar{y} }\right}} \hspace{2}}{\hspace{2} \displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}n\hspace{2}}{\displaystyle{\sum }}\limits_{ i=1 }^{n}x_{i}^{2}- { \bar{x} }^{2}} \hspace{2}}}$

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

(2)，(3)より

$3$\displaystyle{=\bar{y}- \frac{\hspace{2} \hspace{1}{{\sigma}}_{ xy } \hspace{2}}{\hspace{2} \hspace{1}{{\sigma}}_{x}{ }^{2} \hspace{2}}\bar{x}}$

(4)より

$3$\displaystyle{=\bar{y}- a\bar{x}}$

以上をまとめると

$3$\displaystyle{a=\frac{\hspace{2} \hspace{1}{{\sigma}}_{ xy } \hspace{2}}{\hspace{2} {\sigma}_{x}^{2} \hspace{2}}}$ ， $3$\displaystyle{b=\bar{y}- a\bar{x}}$

とき $3$\displaystyle{{{\Delta}}^{2}}$ は最小となる．

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最終更新日 2025年2月8日

最小二乗法

● は最小となる と の値

■導出

● $3$\displaystyle{{{\Delta}}^{2}}$ は最小となる $3$a$ と $3$b$ の値