Proof. Under A1–A4 the \(y_i\) are independent with \(y_i \sim N(\beta _0 + \beta _1 x_i, \sigma ^2)\), so the likelihood is the product of normal densities, \begin {align*} L(\beta _0, \beta _1, \sigma ^2) &= \prod _{i=1}^n \frac {1}{\sqrt {2\pi \sigma ^2}} \exp \left \{ -\frac {(y_i - \beta _0 - \beta _1 x_i)^2}{2\sigma ^2} \right \} \\ &= (2\pi \sigma ^2)^{-n/2} \exp \left \{ -\frac {S(\beta _0, \beta _1)}{2\sigma ^2} \right \}, \end {align*}

where \(S\) is the least-squares criterion. The log-likelihood is \[ \ell = -\frac n2 \log (2\pi ) - \frac n2 \log \sigma ^2 - \frac {S(\beta _0, \beta _1)}{2\sigma ^2} . \] For any fixed \(\sigma ^2\), \(\ell \) is largest when \(S(\beta _0, \beta _1)\) is smallest, that is, at \(\hat \beta _0, \hat \beta _1\), whatever the value of \(\sigma ^2\). Putting these in, \(S\) becomes \(SS_{\mathrm {Res}}\), and \[ \frac {\partial \ell }{\partial \sigma ^2} = -\frac {n}{2\sigma ^2} + \frac {SS_{\mathrm {Res}}}{2\sigma ^4} = 0 \quad \text {gives}\quad \tilde \sigma ^2 = \frac {SS_{\mathrm {Res}}}{n} . \] (The derivative is positive below this value and negative above it, so it is a maximum.) Finally, by Theorem 2.5.3, \(E(\tilde \sigma ^2) = E(SS_{\mathrm {Res}})/n = (n-2)\sigma ^2/n\). □