Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Simple linear regression ▷ Variance 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 variances of the estimated parameters are

$\label{eq:slr-ols-var} \begin{split} \mathrm{Var}(\hat{\beta}_0) &= \frac{x^\mathrm{T} x}{n} \cdot \frac{\sigma^2}{(n-1) s_x^2} \\ \mathrm{Var}(\hat{\beta}_1) &= \frac{\sigma^2}{(n-1) s_x^2} \end{split}$

where $s_x^2$ is the sample variance of $x$ and $x^\mathrm{T} x$ is the sum of squared values of the covariate.

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

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

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

$\label{eq:sum-ci2} \begin{split} \sum_{i=1}^n c_i^2 &= \sum_{i=1}^n \left( \frac{x_i - \bar{x}}{\sum_{i=1}^n (x_i - \bar{x})^2} \right)^2 \\ &= \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{\left[ \sum_{i=1}^n (x_i - \bar{x})^2 \right]^2} \\ &= \frac{1}{\sum_{i=1}^n (x_i - \bar{x})^2} \; . \end{split}$

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:Var-yi} and \eqref{eq:sum-ci2} as well as invariance, scaling and additivity of the variance, the variance of $\hat{\beta}_1$ is:

$\label{eq:Var-b1} \begin{split} \mathrm{Var}(\hat{\beta}_1) &= \mathrm{Var}\left( \sum_{i=1}^n c_i y_i - \bar{y} \sum_{i=1}^n c_i \right) \\ &= \mathrm{Var}\left( \sum_{i=1}^n c_i y_i \right) \\ &= \sum_{i=1}^n c_i^2 \mathrm{Var}(y_i) \\ &= \sigma^2 \sum_{i=1}^n c_i^2 \\ &= \sigma^2 \frac{1}{\sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \frac{\sigma^2}{(n-1) \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2} \\ &= \frac{\sigma^2}{(n-1) s_x^2} \; . \end{split}$

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

$\label{eq:Var-b0-s1} \begin{split} \mathrm{Var}(\hat{\beta}_0) &= \mathrm{Var}\left( \frac{1}{n} \sum_{i=1}^n y_i - \hat{\beta}_1 \cdot \frac{1}{n} \sum_{i=1}^n x_i \right) \\ &= \mathrm{Var}\left( \frac{1}{n} \sum_{i=1}^n y_i \right) + \mathrm{Var}\left( \hat{\beta}_1 \cdot \bar{x} \right) \\ &= \left( \frac{1}{n} \right)^2 \sum_{i=1}^n \mathrm{Var}(y_i) + \bar{x}^2 \cdot \mathrm{Var}(\hat{\beta}_1) \\ &= \frac{1}{n^2} \sum_{i=1}^n \sigma^2 + \bar{x}^2 \frac{\sigma^2}{(n-1) s_x^2} \\ &= \frac{\sigma^2}{n} + \frac{\sigma^2 \bar{x}^2 }{(n-1) s_x^2} \; . \end{split}$

Applying the formula for the sample variance $s_x^2$, we finally get:

$\label{eq:Var-b0-s2} \begin{split} \mathrm{Var}(\hat{\beta}_0) &= \sigma^2 \left( \frac{1}{n} + \frac{\bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \sigma^2 \left( \frac{\frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2}{\sum_{i=1}^n (x_i - \bar{x})^2} + \frac{\bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \sigma^2 \left( \frac{\frac{1}{n}\sum_{i=1}^n \left( x_i^2 - 2 \bar{x} x_i + \bar{x}^2 \right) + \bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \sigma^2 \left( \frac{\left( \frac{1}{n}\sum_{i=1}^n x_i^2 - 2 \bar{x} \frac{1}{n}\sum_{i=1}^n x_i + \bar{x}^2 \right) + \bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \sigma^2 \left( \frac{\frac{1}{n}\sum_{i=1}^n x_i^2 - 2 \bar{x}^2 + 2 \bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \sigma^2 \left( \frac{\frac{1}{n}\sum_{i=1}^n x_i^2}{(n-1) \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2} \right) \\ &= \frac{x^\mathrm{T} x}{n} \cdot \frac{\sigma^2}{(n-1) s_x^2} \; . \end{split}$
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Metadata: ID: P273 | shortcut: slr-olsvar | author: JoramSoch | date: 2021-10-27, 11:53.