Proof: The sum of residuals is zero in simple linear regression
Index:
The Book of Statistical Proofs ▷
Statistical Models ▷
Univariate normal data ▷
Simple linear regression ▷
Sum of residuals is zero
Metadata: ID: P276 | shortcut: slr-ressum | author: JoramSoch | date: 2021-10-27, 13:07.
Theorem: In simple linear regression, the sum of the residuals is zero when estimated using ordinary least squares.
Proof: The residuals are defined as the estimated error terms
\[\label{eq:slr-res} \hat{\varepsilon}_i = y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i\]where $\hat{\beta}_0$ and $\hat{\beta}_1$ are parameter estimates obtained using ordinary least squares:
\[\label{eq:slr-ols} \hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} \quad \text{and} \quad \hat{\beta}_1 = \frac{s_{xy}}{s_x^2} \; .\]With that, we can calculate the sum of the residuals:
\[\label{eq:slr-ressum} \begin{split} \sum_{i=1}^n \hat{\varepsilon}_i &= \sum_{i=1}^n (y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i) \\ &= \sum_{i=1}^n (y_i - \bar{y} + \hat{\beta}_1 \bar{x} - \hat{\beta}_1 x_i) \\ &= \sum_{i=1}^n y_i - n \bar{y} + \hat{\beta}_1 n \bar{x} - \hat{\beta}_1 \sum_{i=1}^n x_i \\ &= n \bar{y} - n \bar{y} + \hat{\beta}_1 n \bar{x} - \hat{\beta}_1 n \bar{x} \\ &= 0 \; . \end{split}\]Thus, the sum of the residuals is zero under ordinary least squares, if the model includes an intercept term $\beta_0$.
∎
Sources: - Wikipedia (2021): "Simple linear regression"; in: Wikipedia, the free encyclopedia, retrieved on 2021-10-27; URL: https://en.wikipedia.org/wiki/Simple_linear_regression#Numerical_properties.
Metadata: ID: P276 | shortcut: slr-ressum | author: JoramSoch | date: 2021-10-27, 13:07.