Proof. Both are linear combinations of the independent normal \(y_i\) (Lemma 2.4.1, and \(\hat \beta _0 = \bar y - \bar x \hat \beta _1\)), so both are normal by (F1). Their means and variances are those of Theorems 2.4.2 and 2.4.4. □

Proof. Every \(e_i\), together with \(\bar y\) and \(\hat \beta _1\), is a linear combination of the independent normal \(y_j\), so by (F1) it is enough to show that the covariances are zero. Lemma 2.4.3 gives \(\operatorname {Cov}(\bar y, \hat \beta _1) = 0\). Next, \(e_i = y_i - \bar y - \hat \beta _1 (x_i - \bar x)\). From \(\hat \beta _1 = \sum c_j y_j\), \(\operatorname {Cov}(y_i, \hat \beta _1) = c_i \sigma ^2\), and from \(\bar y = \sum y_j/n\), \(\operatorname {Cov}(y_i, \bar y) = \sigma ^2/n\). So \begin {align*} \operatorname {Cov}(e_i, \hat \beta _1) &= c_i\sigma ^2 - 0 - (x_i - \bar x)\frac {\sigma ^2}{S_{xx}} = \frac {(x_i - \bar x)\sigma ^2}{S_{xx}} - \frac {(x_i - \bar x)\sigma ^2}{S_{xx}} = 0, \\ \operatorname {Cov}(e_i, \bar y) &= \frac {\sigma ^2}{n} - \frac {\sigma ^2}{n} - (x_i - \bar x) \cdot 0 = 0 . \qedhere \end {align*} □

Proof. \(SS_{\mathrm {Res}} = \sum e_i^2\) is a function of the residuals alone, and \(\hat \beta _0\) and \(\hat \beta _1\) are functions of \((\bar y, \hat \beta _1)\), so the independence follows from Lemma 2.7.2.

For the distribution, split each error into three parts. Since \(\hat \beta _0 = \bar y - \hat \beta _1 \bar x\), \[ \varepsilon _i = y_i - \beta _0 - \beta _1 x_i = e_i + (\bar y - \beta _0 - \beta _1 \bar x) + (\hat \beta _1 - \beta _1)(x_i - \bar x) , \] as can be checked by expanding \(e_i = y_i - \bar y - \hat \beta _1(x_i - \bar x)\). Square and sum over \(i\). All three cross terms vanish, because \(\sum e_i = 0\), \(\sum (x_i - \bar x) e_i = \sum x_i e_i - \bar x \sum e_i = 0\) (Theorem 2.3.1) and \(\sum (x_i - \bar x) = 0\). So \[ \sum \varepsilon _i^2 = SS_{\mathrm {Res}} + n(\bar y - \beta _0 - \beta _1 \bar x)^2 + (\hat \beta _1 - \beta _1)^2 S_{xx} . \] Divide by \(\sigma ^2\) and look at each term.

• Left side: \(\varepsilon _i/\sigma \sim N(0,1)\) independently, so \(\sum \varepsilon _i^2/\sigma ^2 \sim \chi ^2_n\) by (F2).
• \(\bar y \sim N(\beta _0 + \beta _1 \bar x, \sigma ^2/n)\), so \(n(\bar y - \beta _0 - \beta _1 \bar x)^2/\sigma ^2 = Z_1^2 \sim \chi ^2_1\).
• By Theorem 2.7.1, \((\hat \beta _1 - \beta _1)^2 S_{xx}/\sigma ^2 = Z_2^2 \sim \chi ^2_1\).

By Lemma 2.7.2 the three terms on the right are independent. Taking moment generating functions, with \(M(t)\) that of \(SS_{\mathrm {Res}}/\sigma ^2\), \[ (1 - 2t)^{-n/2} = M(t)\,(1 - 2t)^{-1/2}(1 - 2t)^{-1/2}, \] so that \[ M(t) = (1 - 2t)^{-(n-2)/2}, \] which is the moment generating function of \(\chi ^2_{n-2}\). □