14.5 Practice problems
Problem 14.5.1. With \(\alpha = 0.2\), what weights does simple exponential smoothing give to the last four observations, and what total weight do all observations older than four periods receive?
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Solution. \(c_i = 0.2(0.8)^i\): \(0.200\), \(0.160\), \(0.128\), \(0.102\), adding to \(0.590\). The older observations receive \(1 - 0.590 = 0.8^4 = 0.410\). With small \(\alpha \) the forecast has a long memory.
Problem 14.5.2. For the AR(1) process \(X_t - 50 = 0.7(X_{t-1} - 50) + Z_t\) with \(\sigma _Z^2 = 4\), and \(x_n = 56\), find \(\hat x(n, k)\) for \(k = 1, 2, 3\) and the variance of the \(k\)-step forecast error. What happens as \(k \to \infty \)?
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Solution. \(\hat x(n, k) = 50 + 0.7^k(6)\): \(54.2\), \(52.94\), \(52.06\). The \(\psi \) weights are \(0.7^j\), so the error variances are \(4\), \(4(1.49) = 5.96\), \(4(1.49 + 0.2401) = 6.92\). As \(k \to \infty \) the forecast tends to the mean \(50\) and the variance to \(4/(1 - 0.49) = 7.84\), the variance of the process.
Problem 14.5.3. For the ARIMA(0,1,1) model \(X_t = X_{t-1} + Z_t - 0.4Z_{t-1}\), find the \(\psi \) weights and the variance of the \(k\)-step forecast error.
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Solution. \(\psi (B) = (1 - 0.4B)/(1 - B) = (1 - 0.4B)(1 + B + B^2 + \dots ) = 1 + 0.6B + 0.6B^2 + \dots \), so \(\psi _j = 0.6\) for \(j \ge 1\). Then \(\Var (e_n(k)) = \sigma _Z^2(1 + 0.36(k - 1))\), which grows linearly in \(k\): the interval widens like \(\sqrt k\), as in Figure 12, panel 2.
Problem 14.5.4. For the AR(2) model of Example 14.4.3, the next observation turned out to be \(x_{n+1} = 0.50\). Use the updating formula to revise the forecast of \(X_{n+2}\), and check it directly.
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Solution. The one-step error is \(0.50 - (-0.184) = 0.684\), so \(\hat x(n + 1, 1) = \hat x(n, 2) + \psi _1(0.684) = -0.222 + 0.554(0.684) = 0.157\). Directly from the model, \(\hat x(n + 1, 1) = 0.554(0.50) - 0.402(0.30) = 0.277 - 0.121 = 0.157\).
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