Proof. (i) \(U\) is \(N(0, \sigma ^2I)\); the multivariate normal density with \(|\sigma ^2I| = \sigma ^{2n}\) and \((\sigma ^2I)^{-1} = \sigma ^{-2}I\) gives \(f(u)\). The transformation \(u = y - X\beta \) has Jacobian matrix \(I\), with determinant \(1\), so \(f(y \mid X) = f(y - X\beta )\).

(ii) is the logarithm of (i).

(iii) For any fixed \(\sigma ^2\), \(\ell \) is maximised over \(\beta \) by minimising \((y - X\beta )'(y - X\beta )\), which is least squares. Then \(\partial \ell /\partial \sigma ^2 = -\frac {n}{2\sigma ^2} + \frac {e'e}{2\sigma ^4} = 0\) gives \(\hat \sigma ^2 = e'e/n\); this is a maximum since \(\ell \to -\infty \) as \(\sigma ^2 \to 0\) or \(\infty \). Substituting, the last term of \(\ell \) is \(-\frac {e'e}{2e'e/n} = -\frac n2\).

(iv) The first derivatives are \[ \frac {\partial \ell }{\partial \beta } = \frac {1}{\sigma ^2}X'(y - X\beta ), \qquad \frac {\partial \ell }{\partial \sigma ^2} = -\frac {n}{2\sigma ^2} + \frac {u'u}{2\sigma ^4}, \] and the second derivatives are \[ \frac {\partial ^2\ell }{\partial \beta \,\partial \beta '} = -\frac {X'X}{\sigma ^2}, \quad \frac {\partial ^2\ell }{\partial \beta \,\partial \sigma ^2} = -\frac {X'u}{\sigma ^4}, \quad \frac {\partial ^2\ell }{\partial (\sigma ^2)^2} = \frac {n}{2\sigma ^4} - \frac {u'u}{\sigma ^6} . \] Taking expectations with \(E(u) = 0\) and \(E(u'u) = n\sigma ^2\), minus the expected Hessian has blocks \(X'X/\sigma ^2\), \(0\) and \(-\frac {n}{2\sigma ^4} + \frac {n\sigma ^2}{\sigma ^6} = \frac {n}{2\sigma ^4}\). The inverse of a block-diagonal matrix is block-diagonal with the inverted blocks. □