10.3 Practice problems
Problem 10.3.1. For \(X_t = 0.8X_{t-1} + Z_t + 0.4Z_{t-1}\), check stationarity and invertibility, find the \(\psi \) weights, and find \(\rho (1)\) and \(\rho (2)\).
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Solution. The roots are \(1/0.8 = 1.25\) for \(\phi \) and \(-1/0.4 = -2.5\) for \(\theta \), both outside the unit circle. \(\psi _j = (0.8 + 0.4)0.8^{j-1} = 1.2(0.8)^{j-1}\): \(1.2, 0.96, 0.768, \dots \). By Theorem 9.3.1, \(\rho (1) = (1 + 0.32)(1.2)/(1 + 0.64 + 0.16) = 1.584/1.8 = 0.88\) and \(\rho (2) = 0.8(0.88) = 0.704\).
Problem 10.3.2. Show that \(X_t = 1.3X_{t-1} - 0.4X_{t-2} + Z_t - 0.5Z_{t-1}\) is really an AR(1) process.
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Solution. \(\phi (B) = 1 - 1.3B + 0.4B^2 = (1 - 0.5B)(1 - 0.8B)\) and \(\theta (B) = 1 - 0.5B\). The common factor \((1 - 0.5B)\) cancels, leaving \((1 - 0.8B)X_t = Z_t\), an AR(1) with \(\alpha = 0.8\).
Problem 10.3.3. Write the ARIMA(1,1,1) model with \(\alpha = 0.5\), \(\beta = 0.3\) as a difference equation in \(X_t\).
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Solution. \((1 - 0.5B)(1 - B)X_t = (1 + 0.3B)Z_t\), and \((1 - 0.5B)(1 - B) = 1 - 1.5B + 0.5B^2\), so \(X_t = 1.5X_{t-1} - 0.5X_{t-2} + Z_t + 0.3Z_{t-1}\).
Problem 10.3.4. The series \(3, 6, 11, 18, 27\) follows a quadratic trend exactly. Find its first and second differences.
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Solution. First differences: \(3\), \(5\), \(7\), \(9\), still trending. Second differences: \(2\), \(2\), \(2\), constant. A quadratic trend needs \(d = 2\) (Proposition 2.4.4).
Problem 10.3.5. After differencing a series once, the \(n = 150\) differenced values have sample autocorrelations \(r_1 = -0.48\), \(r_2 = 0.03\) and \(r_3 = -0.05\). What does this suggest?
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Solution. The correlogram cuts off after lag \(1\), with \(r_1\) close to \(-0.5\). This is the signature described in Theorem 10.2.1(ii): the original series was probably already stationary (perhaps white noise around a constant), and the differencing was unnecessary. Fit a model to the undifferenced series instead.
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