13.5 Practice problems
Problem 13.5.1. The first six residual autocorrelations from an ARMA(1,1) model fitted to \(n = 100\) observations are \(0.12\), \(-0.15\), \(0.08\), \(0.21\), \(-0.05\), \(0.10\). Compute the Box–Pierce statistic and test the adequacy of the model at the \(5\%\) level.
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Solution. \(\sum r_k^2 = 0.0144 + 0.0225 + 0.0064 + 0.0441 + 0.0025 + 0.0100 = 0.0999\), so \(Q = 100 \times 0.0999 = 9.99\) on \(6 - 2 = 4\) degrees of freedom. Since \(\chi ^2_{0.05,4} = 9.49\), the model is (just) rejected. Notice \(r_4 = 0.21\), outside \(\pm 0.2\): with quarterly data this would suggest a seasonal term at lag \(4\).
Problem 13.5.2. Two models fitted to \(n = 200\) observations gave residual variances \(\hat \sigma ^2 = 2.40\) for AR(1) and \(\hat \sigma ^2 = 2.34\) for AR(3). Which does each criterion prefer?
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Solution. \(n\log \hat \sigma ^2\) is \(200\log 2.40 = 175.09\) and \(200\log 2.34 = 170.03\). AIC: \(175.09 + 2 = 177.09\) against \(170.03 + 6 = 176.03\), so AIC prefers AR(3). BIC: \(175.09 + \log 200 = 180.39\) against \(170.03 + 3\log 200 = 185.93\), so BIC prefers AR(1). The criteria disagree because the gain in fit, \(5.06\), lies between AIC’s extra penalty of \(4\) and BIC’s of \(2\log 200 = 10.6\).
Problem 13.5.3. Explain why the Ljung–Box statistic is always larger than the Box–Pierce statistic computed from the same residual autocorrelations, and why this is a correction in the right direction.
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Solution. Each term of \(Q^*\) is \(r_k^2\) multiplied by \(n(n + 2)/(n - k)\), which exceeds \(n\) because \(n + 2 > n - k\). The true variance of \(r_k\) for white noise is \((n - k)/(n(n + 2))\), smaller than \(1/n\), so \(Q = n\sum r_k^2\) understates the standardised sum and the Box–Pierce test rejects too rarely in small samples. Dividing by the correct variance makes the \(\chi ^2\) approximation accurate.
Problem 13.5.4. After fitting an MA(1) model to monthly data, the residual ac.f. is small everywhere except \(r_{12} = 0.35\), with \(n = 120\). What does this suggest?
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Solution. The band is \(\pm 2/\sqrt {120} = \pm 0.18\), and \(r_{12}\) lies well outside it, at the seasonal lag of monthly data. The model has missed a seasonal dependence: values twelve months apart are related. A seasonal term is needed, for example a seasonal MA term at lag \(12\), giving a SARIMA model (Section 15).
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