最小二乗法(偏微分を用いた計算)

${\displaystyle {\Delta }^2={\displaystyle \sum _{i=1}^n\Delta i^2}={\displaystyle \sum _{i=1}^n{\left\{y_i-\left(a{x}_i+b\right)\right\}}^2}}$

${\displaystyle {\sigma }_{xy}=\frac{1}{n}{\displaystyle \sum }_{i=1}^n{x}_i{y}_i-\overline{x}\overline{y}}$ (参照：共分散)

${\displaystyle {\sigma }_x^2=\frac{1}{n}{\displaystyle \sum }_{i=1}^n{x}_i^2-{\overline{x}}^2}$ (参照：分散)

${\displaystyle ={\displaystyle \sum }_{i=1}^n\left\{y_i{}^2-2\left(a{x}_i+b\right)y_i+{\left(a{x}_i+b\right)}^2\right\}}$

${\displaystyle ={\displaystyle \sum }_{i=1}^n\left(y_i{}^2-2a{x}_i{y}_i-2b{y}_i+a{x}_i{}^2+2ab{x}_i+b^2{x}_i{}^2\right)}$

${\displaystyle ={\displaystyle \sum }_{i=1}^n\left(x_i{}^2{a}^2+x_i{}^2{b}^2+2x_{i}ab-2x_i{y}_{i}a-2y_{i}b+y_i{}^2\right)}$

${\displaystyle \frac{\partial }{\partial a}{\Delta }^2=\frac{\partial }{\partial a}{\displaystyle \sum _{i=1}^n{\left\{y_i-\left(a{x}_i+b\right)\right\}}^2}}$

${\displaystyle ={\displaystyle \sum _{i=1}^{n}2\left\{y_i-\left(a{x}_i+b\right)\right\}\frac{\partial }{\partial a}\left\{y_i-\left(a{x}_i+b\right)\right\}}}$

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

${\displaystyle {\displaystyle \sum }_{i=1}^n{y}_i{x}_i-a{\displaystyle \sum }_{i=1}^n{x}_i{}^2-b{\displaystyle \sum }_{i=1}^n{x}_i=0}$　･･････(4)

${\displaystyle \frac{\partial }{\partial b}{\Delta }^2=\frac{\partial }{\partial b}{\displaystyle \sum _{i=1}^n{\left\{y_i-\left(a{x}_i+b\right)\right\}}^2}}$

${\displaystyle ={\displaystyle \sum _{i=1}^{n}2\left\{y_i-\left(a{x}_i+b\right)\right\}\frac{\partial }{\partial b}\left\{y_i-\left(a{x}_i+b\right)\right\}}}$

${\displaystyle {\displaystyle \sum _{i=1}^n{y}_i{x}_i}-a{\displaystyle \sum _{i=1}^n{x}_i{}^2}-\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}-a\cdot \frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}\right){\displaystyle \sum _{i=1}^n{x}_i}=0}$

${\displaystyle -a\left\{{\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2\right\}={\displaystyle \sum _{i=1}^n{y}_i{x}_i}-\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}\right){\displaystyle \sum _{i=1}^n{x}_i}}$

${\displaystyle a=\frac{{\displaystyle \sum _{i=1}^n{y}_i{x}_i}-\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}\right){\displaystyle \sum _{i=1}^n{x}_i}}{{\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}}$　･･････(8)

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

${\displaystyle {\sigma }_{xy}=\frac{1}{n}{\displaystyle \sum }_{i=1}^n{x}_i{y}_i-\overline{x}\overline{y}}$ (参照：共分散)　･･････(12)

${\displaystyle {\sigma }_x^2=\frac{1}{n}{\displaystyle \sum }_{i=1}^n{x}_i^2-{\overline{x}}^2}$ (参照：分散)　･･････(13)