13.5 Practice problems
Problem 13.5.1. For an AR(1) process with \(\rho = 0.6\) and \(\sigma _a^2 = 4\), find \(\operatorname {Var}(\varepsilon _t)\), \(\operatorname {Corr}(\varepsilon _t, \varepsilon _{t-3})\), and the variance of the error in forecasting \(\varepsilon _{n+2}\) from \(\varepsilon _n\).
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Solution. \(\operatorname {Var}(\varepsilon _t) = 4/(1 - 0.36) = 6.25\); \(\operatorname {Corr}(\varepsilon _t, \varepsilon _{t-3}) = 0.6^3 = 0.216\); the two-step forecast error variance is \(4(1 + 0.36) = 5.44\), between the one-step value \(4\) and the unconditional \(6.25\).
Problem 13.5.2. The residuals from a regression on \(n = 6\) quarterly observations are \(2\), \(1\), \(-1\), \(-2\), \(-1\), \(1\). Compute the Durbin–Watson statistic and \(r_1\), and verify the identity of Theorem 13.2.1.
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Solution. \(\sum e_t^2 = 4 + 1 + 1 + 4 + 1 + 1 = 12\). The differences are \(-1\), \(-2\), \(-1\), \(1\), \(2\), with squares adding to \(11\), so \(d = 11/12 = 0.917\). The lagged products are \(2\), \(-1\), \(2\), \(2\), \(-1\), adding to \(4\), so \(r_1 = 4/12 = 0.333\). Then \(2(1 - 0.333) - (4 + 1)/12 = 1.333 - 0.417 = 0.917\), as the theorem says. (With so few observations the end term is large, and \(d\) differs considerably from \(2(1 - r_1)\).)
Problem 13.5.3. A regression with \(n = 15\) and one regressor gives \(d = 3.1\). What does this indicate, using \(d_L = 1.08\) and \(d_U = 1.36\)?
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Solution. For negative autocorrelation the rule is applied to \(4 - d = 0.9\), which is below \(d_L = 1.08\). So there is significant negative autocorrelation: the residuals tend to alternate in sign. This is less common in practice, and can be produced by over-correcting, for example by differencing data that did not need it.
Problem 13.5.4. Show that if the errors follow a random walk, \(\varepsilon _t = \varepsilon _{t-1} + a_t\), regressing \(y_t - y_{t-1}\) on \(x_t - x_{t-1}\) through the origin gives an estimator of \(\beta _1\) that satisfies the standard assumptions. What happens to \(\beta _0\)?
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Solution. Differencing the model, \(y_t - y_{t-1} = \beta _1(x_t - x_{t-1}) + (\varepsilon _t - \varepsilon _{t-1}) = \beta _1(x_t - x_{t-1}) + a_t\), with independent errors of constant variance and no intercept. This is Theorem 13.3.1 with \(\rho = 1\). The intercept \(\beta _0\) cancels and cannot be estimated from the differenced data. Under a random walk the errors have no fixed level to return to, so \(\beta _0\) is not well determined in any case; the differenced model is about how changes in \(x\) produce changes in \(y\).
Problem 13.5.5. Leaving out a regressor that trends over time can produce autocorrelated residuals, even when the true errors are independent. Explain why, and say what should be tried before transforming the data.
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Solution. If the true model is \(y_t = \beta _0 + \beta _1x_t + \beta _2z_t + \varepsilon _t\) and \(z_t\) is omitted, the error of the fitted model is \(\beta _2z_t + \varepsilon _t\). A regressor that moves smoothly over time has neighbouring values that are alike, so \(\beta _2z_t\), and with it the residuals, are positively autocorrelated. Before transforming, look for the missing variable (a trend term, a price, population, season indicators) and add it; the autocorrelation may disappear, and the model will be better for reasons beyond the error structure. Transformations such as Cochrane–Orcutt treat the symptom, not the cause.
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