Proof. (i) Write \(\Sigma = CC'\) with \(C = \Sigma ^{1/2}\) non-singular, and let \(Z = C^{-1}(X - \mu )\). Changing variables in the density, with Jacobian \(|C|\) and \((x - \mu )'\Sigma ^{-1}(x - \mu ) = z'z\), shows that \(Z\) has density \((2\pi )^{-n/2}e^{-z'z/2}\): its elements are independent \(N(0, 1)\). Each has moment generating function \(e^{s^2/2}\), so \(E(e^{s'Z}) = e^{s's/2}\). Then \(X = \mu + CZ\) and \[ E(e^{t'X}) = e^{t'\mu }E\bigl (e^{(C't)'Z}\bigr ) = e^{t'\mu }e^{t'CC't/2} = \exp \bigl (t'\mu + \tfrac 12t'\Sigma t\bigr ) . \]

(ii) For \(s \in \mathbb R^m\), \(E(e^{s'(AX + b)}) = e^{s'b}E(e^{(A's)'X}) = \exp \bigl (s'(A\mu + b) + \frac 12s'A\Sigma A's\bigr )\), the moment generating function of \(N(A\mu + b, A\Sigma A')\). \(A\Sigma A'\) is pd because \(A\) has full row rank, and the moment generating function determines the distribution. A subvector is \(AX\) for a matrix \(A\) that selects rows of the identity.

(iii) Independence always implies zero covariance. Conversely, if \(\Sigma = \operatorname {diag}(\Sigma _{11}, \Sigma _{22})\), then \(\Sigma ^{-1}\) is block diagonal and \(|\Sigma | = |\Sigma _{11}||\Sigma _{22}|\), so the density factorises into a function of \(x_1\) times a function of \(x_2\). □