Misalkan terdapat sebanyak \(n\) buah data berpasangan, yaitu \((x_i , y_i)\) di mana \(i=1,2,...,n\). Asumsikan data \(y_i\) dapat didekati dengan persamaan \begin{equation} Y_i = \beta_0 + \beta_1 x_i + \varepsilon \label{1.1} \end{equation} di mana \(\varepsilon\) adalah galat.
Maka jumlah selisih kuadrat \(y_i\) dan \(Y_i\) (dengan mengabaikan galat) dapat ditulis sebagai \begin{equation} \sum_{i=1}^{n}(y_{i}-(\beta_{0}+\beta_{1}x_{i}))^{2} \label{1.2}. \end{equation} Agar nilai \(\beta_0 + \beta_1 x_i\) mendekati \(y_i\), maka dikondisikan agar \begin{equation} \frac{\partial}{\partial \beta_0} \left( \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i)^2 \right) = 0 \label{1.3} \end{equation} \begin{equation} \frac{\partial}{\partial \beta_1} \left( \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i)^2 \right) = 0 \label{1.4}. \end{equation} Kemudian dilakukan evaluasi terhadap persamaan \ref{1.3} seperti berikut. \begin{equation} \begin{aligned} \frac{\partial}{\partial \beta_0} \left( \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i)^2 \right) &= 0 \\ \sum_{i=1}^{n} \frac{\partial}{\partial \beta_0} (y_i - \beta_0 - \beta_1 x_i)^2 &= 0 \\ \sum_{i=1}^{n} -2(y_i - \beta_0 - \beta_1 x_i) &= 0 \\ -2 \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i) &= 0 \\ \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i) &= 0 \\ \sum_{i=1}^{n} y_i - \beta_0 \sum_{i=1}^{n} 1 - \beta_1 \sum_{i=1}^{n} x_i &= 0 \\ \sum_{i=1}^{n} y_i - n \beta_0 - \beta_1 \sum_{i=1}^{n} x_i &= 0 \\ n \beta_0 &= \sum_{i=1}^{n} y_i - \beta_1 \sum_{i=1}^{n} x_i \\ \beta_0 &= \frac{\sum_{i=1}^{n} y_i}{n} - \beta_1 \frac{\sum_{i=1}^{n} x_i}{n} \\ &= \bar{y} - \beta_1 \bar{x}. \end{aligned} \label{1.5} \end{equation} Nilai \(\beta_0\) pada persamaan \ref{1.5} disubstitusikan ke persamaan \ref{1.4} lalu dilakukan evaluasi seperti berikut. \begin{equation} \begin{aligned} \frac{\partial}{\partial \beta_1} \left( \sum_{i=1}^{n} (y_i - \left(\bar{y} - \beta_1 \bar{x} \right) - \beta_1 x_i)^2 \right) &= 0 \\ \sum_{i=1}^{n} \frac{\partial}{\partial \beta_1} (y_i - \bar{y} + \beta_1 \bar{x} - \beta_1 x_i)^2 &= 0 \\ \sum_{i=1}^{n} -2(\bar{x} - x_i) (y_i - \left(y_i - \bar{y} + \beta_1 \bar{x} - \beta_1 x_i\right) &= 0 \\ \sum_{i=1}^{n} 2(x_i - \bar{x}) (y_i - \bar{y} + \beta_1 \bar{x} - \beta_1 x_i) &= 0 \\ 2 \sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y} + \beta_1 \bar{x} - \beta_1 x_i) &= 0 \\ \sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y} + \beta_1 \bar{x} - \beta_1 x_i) &= 0. \end{aligned} \label{1.6} \end{equation} Persamaan \ref{1.6} dapat ditulis sebagai \begin{equation} \begin{aligned} \sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y}) - \sum_{i=1}^{n} \beta_1 (x_i - \bar{x}) (x_i - \bar{x}) &= 0 \\ \sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y}) - \beta_1 \sum_{i=1}^{n} (x_i - \bar{x})^2 &= 0. \end{aligned} \label{1.7} \end{equation} Berikutnya untuk mendapatkan nilai \(\beta_1\) dilakukan manipulasi aljabar pada persamaan \ref{1.7} sehingga diperoleh \begin{equation} \begin{aligned} \beta_1 &= \frac{\sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y})}{\sum_{i=1}^{n} (x_i - \bar{x})^2}. \end{aligned} \label{1.8} \end{equation} Sehingga diperoleh regresi linier \begin{equation} Y = \beta_0 + \beta_1 x + \varepsilon \end{equation} di mana \begin{equation} \begin{aligned} \beta_1 &= \frac{\sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y})}{\sum_{i=1}^{n} (x_i - \bar{x})^2} \\ \beta_0 &= \frac{\sum_{i=1}^{n} y_i}{n} - \beta_1 \frac{\sum_{i=1}^{n} x_i}{n}. \nonumber \end{aligned} \end{equation}