Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Simple linear regression ▷ Expectation of estimates

Theorem: Assume a simple linear regression model with independent observations

$\label{eq:slr} y = \beta_0 + \beta_1 x + \varepsilon, \; \varepsilon_i \sim \mathcal{N}(0, \sigma^2), \; i = 1,\ldots,n$

and consider estimation using ordinary least squares. Then, the expected values of the estimated parameters are

$\label{eq:slr-ols-mean} \begin{split} \mathrm{E}(\hat{\beta}_0) &= \beta_0 \\ \mathrm{E}(\hat{\beta}_1) &= \beta_1 \end{split}$

which means that the ordinary least squares solution produces unbiased estimators.

Proof: According to the simple linear regression model in \eqref{eq:slr}, the expectation of a single data point is

$\label{eq:E-yi} \mathrm{E}(y_i) = \beta_0 + \beta_1 x_i \; .$ $\label{eq:slr-ols} \begin{split} \hat{\beta}_0 &= \frac{1}{n} \sum_{i=1}^n y_i - \hat{\beta}_1 \cdot \frac{1}{n} \sum_{i=1}^n x_i \\ \hat{\beta}_1 &= \frac{\sum_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2} \; . \end{split}$

If we define the following quantity

$\label{eq:ci} c_i = \frac{x_i - \bar{x}}{\sum_{i=1}^n (x_i - \bar{x})^2} \; ,$

we note that

$\label{eq:sum-ci} \begin{split} \sum_{i=1}^n c_i &= \frac{\sum_{i=1}^n (x_i - \bar{x})}{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{\sum_{i=1}^n x_i - n \bar{x}}{\sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \frac{n \bar{x} - n \bar{x}}{\sum_{i=1}^n (x_i - \bar{x})^2} = 0 \; , \\ \end{split}$

and

$\label{eq:sum-ci-xi} \begin{split} \sum_{i=1}^n c_i x_i &= \frac{\sum_{i=1}^n (x_i - \bar{x}) x_i}{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{\sum_{i=1}^n \left( x_i^2 - \bar{x} x_i \right)}{\sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \frac{\sum_{i=1}^n x_i^2 - 2 n \bar{x} \bar{x} + n \bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{\sum_{i=1}^n \left( x_i^2 - 2 \bar{x} x_i + \bar{x}^2 \right)}{\sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{\sum_{i=1}^n (x_i - \bar{x})^2} = 1 \; . \end{split}$

With \eqref{eq:ci}, the estimate for the slope from \eqref{eq:slr-ols} becomes

$\label{eq:slr-ols-sl} \begin{split} \hat{\beta}_1 &= \frac{\sum_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \sum_{i=1}^n c_i (y_i - \bar{y}) \\ &= \sum_{i=1}^n c_i y_i - \bar{y} \sum_{i=1}^n c_i \end{split}$

and with \eqref{eq:E-yi}, \eqref{eq:sum-ci} and \eqref{eq:sum-ci-xi}, its expectation becomes:

$\label{eq:E-b1} \begin{split} \mathrm{E}(\hat{\beta}_1) &= \mathrm{E}\left( \sum_{i=1}^n c_i y_i - \bar{y} \sum_{i=1}^n c_i \right) \\ &= \sum_{i=1}^n c_i \mathrm{E}(y_i) - \bar{y} \sum_{i=1}^n c_i \\ &= \beta_1 \sum_{i=1}^n c_i x_i + \beta_0 \sum_{i=1}^n c_i - \bar{y} \sum_{i=1}^n c_i \\ &= \beta_1 \; . \end{split}$

Finally, with \eqref{eq:E-yi} and \eqref{eq:E-b1}, the expectation of the intercept estimate from \eqref{eq:slr-ols} becomes

$\label{eq:E-b0} \begin{split} \mathrm{E}(\hat{\beta}_0) &= \mathrm{E}\left( \frac{1}{n} \sum_{i=1}^n y_i - \hat{\beta}_1 \cdot \frac{1}{n} \sum_{i=1}^n x_i \right) \\ &= \frac{1}{n} \sum_{i=1}^n \mathrm{E}(y_i) - \mathrm{E}(\hat{\beta}_1) \cdot \bar{x} \\ &= \frac{1}{n} \sum_{i=1}^n (\beta_0 + \beta_1 x_i) - \beta_1 \cdot \bar{x} \\ &= \beta_0 + \beta_1 \bar{x} - \beta_1 \bar{x} \\ &= \beta_0 \; . \end{split}$
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Metadata: ID: P272 | shortcut: slr-olsmean | author: JoramSoch | date: 2021-10-27, 09:54.