Proof. (i) is Proposition 3.1.2. (ii) Let the \(X_t\) be independent, with \(X_t \sim N(0, 1)\) for odd \(t\) and \(X_t\) uniform on \((-\sqrt 3, \sqrt 3)\) for even \(t\). Both distributions have mean \(0\) and variance \(1\), and different variables are uncorrelated, so \(\mu = 0\) and \(\gamma (k) = 0\) for \(k \ne 0\), \(\gamma (0) = 1\): the process is second-order stationary. But \(X_1\) and \(X_2\) have different distributions, so it is not first-order stationary. (iii) A multivariate normal distribution is determined by its mean vector and covariance matrix. For a weakly stationary process, the mean vector of \((X_{t_1+k}, \dots , X_{t_n+k})\) is \((\mu , \dots , \mu )\) and its covariance matrix has entries \(\gamma (t_j - t_i)\), neither depending on \(k\), so the joint distribution does not depend on \(k\). □

Solution. \(E(X_t) = \cos (\lambda t)E(A) + \sin (\lambda t)E(B) = 0\). Since \(\Cov (A, B) = 0\), \begin {align*} \Cov (X_t, X_{t+k}) &= \cos (\lambda t)\cos (\lambda (t + k))\Var (A) \\ &\qquad + \sin (\lambda t)\sin (\lambda (t + k))\Var (B) \\ &= \sigma ^2\bigl [\cos (\lambda t)\cos (\lambda t + \lambda k) + \sin (\lambda t)\sin (\lambda t + \lambda k)\bigr ] \\ &= \sigma ^2\cos (\lambda k) , \end {align*}

by the identity \(\cos a\cos b + \sin a\sin b = \cos (b - a)\). The mean is constant and the covariance depends only on \(k\), so the process is stationary, with \(\gamma (0) = \sigma ^2\) and \(\rho (k) = \cos (\lambda k)\). Each realisation is a sinusoid with a random amplitude and phase, and its correlations never die away: values a whole period apart are perfectly correlated. Processes built from such components are the basis of the frequency-domain view in Section 16.