Solution. \(E(X_t) = 0\). \(\gamma (0) = \Var (Z_t) + 0.25\Var (Z_{t-1}) = 1.25\sigma _Z^2\). For \(k = 1\), \(\Cov (Z_t + 0.5Z_{t-1}, Z_{t+1} + 0.5Z_t) = 0.5\Var (Z_t) = 0.5\sigma _Z^2\). For \(|k| \ge 2\) the two variables involve no common \(Z\), so \(\gamma (k) = 0\). None of this depends on \(t\), so the process is stationary, with \(\rho (1) = 0.5/1.25 = 0.4\) and \(\rho (k) = 0\) for \(|k| \ge 2\).

Solution. \(E(X_t) = 0\) and \(\Cov (X_t, X_{t+k}) = (-1)^{2t+k}\Var (Z) = (-1)^k\sigma ^2\), which depends only on \(k\). So it is stationary, with \(\rho (k) = (-1)^k\). Each realisation simply alternates between \(Z\) and \(-Z\): one random number decides the whole path. Stationarity does not mean the series is random-looking, only that its moments do not change with time.

Solution. \(X_t - X_{t-k} = Z_{t-k+1} + \dots + Z_t\), a sum of \(k\) independent terms, so its mean is \(k\mu \) and its variance \(k\sigma _Z^2\), whatever \(t\). As a moving sum of white noise it is stationary: \(\Cov (X_t - X_{t-k}, X_{t+j} - X_{t+j-k})\) is \(\sigma _Z^2\) times the number of \(Z\)’s the two sums share, which is \(\max (k - |j|, 0)\), a function of \(j\) alone.

Solution. Expanding \(\det \mathbf {R}_n\) along its first row gives the recursion, since \(\mathbf {R}_n\) is tridiagonal. Then \(D_3 = 0.64 - 0.36 = 0.28\), \(D_4 = 0.28 - 0.36(0.64) = 0.0496\) and \(D_5 = 0.0496 - 0.36(0.28) = -0.0512 < 0\). So \(\mathbf {R}_5\) is not non-negative definite, and the function is not an ac.f., even though the \(3 \times 3\) and \(4 \times 4\) checks pass. Checking a few small matrices is not enough in general; for this shape of function, the exact condition is \(|\rho (1)| \le \frac 12\), and \(0.6\) exceeds it.