13.4 Forecasting
Autocorrelation is a nuisance for estimation but an asset for forecasting: the last residual carries information about the next error.
Theorem 13.4.1. Under the AR(1) model with known parameters, the best predictor of \(y_{n+h}\) given \(y_1, \dots , y_n\) is \[ \hat y_{n+h} = \beta _0 + \beta _1x_{n+h} + \rho ^h\varepsilon _n , \] and its prediction error has variance \(\sigma _a^2(1 + \rho ^2 + \dots + \rho ^{2(h-1)}) = \sigma _a^2(1 - \rho ^{2h})/(1 - \rho ^2)\).
Proof. Applying the AR(1) recursion \(h\) times, \[ \varepsilon _{n+h} = \rho ^h\varepsilon _n + a_{n+h} + \rho a_{n+h-1} + \dots + \rho ^{h-1}a_{n+1} . \] The \(a\)’s after time \(n\) are independent of everything observed up to time \(n\) and have mean zero, so the best predictor of \(\varepsilon _{n+h}\) is \(\rho ^h\varepsilon _n\), and the error is the sum of the \(a\) terms, with variance \(\sigma _a^2(1 + \rho ^2 + \dots + \rho ^{2(h-1)})\). The geometric series sums to the stated form. □
For one step ahead the error variance is \(\sigma _a^2\), smaller than the \(\sigma ^2 = \sigma _a^2/(1 - \rho ^2)\) that would apply if the last residual were ignored. As \(h\) grows the adjustment \(\rho ^h\varepsilon _n\) fades and the variance rises to \(\sigma ^2\). In practice the parameters are replaced by estimates and \(\varepsilon _n\) by the last residual, \(e_n = y_n - \hat \beta _0 - \hat \beta _1x_n\), which makes the intervals somewhat too narrow.
Example 13.4.2. For the sales data, advertising next year will be \(x_{16} = 15\). Forecast sales, using the Cochrane–Orcutt fit.
Solution. The regression part gives \(9.29 + 2.079(15) = 40.48\). The last residual from this fit is \(e_{15} = 40.3 - (9.29 + 2.079 \times 14) = 1.90\), and with \(\hat \rho = 0.792\), \[ \hat y_{16} = 40.48 + 0.792(1.90) = 41.98 . \] Because sales have been running above the line, the forecast is raised by \(1.5\). Ordinary least squares, which ignores the autocorrelation, would forecast \(10.20 + 1.964(15) = 39.67\). With \(\hat \sigma _a^2 = 1.563\) and \(t_{0.025,12} = 2.179\), an approximate \(95\%\) prediction interval is \(41.98 \pm 2.179\sqrt {1.563} = 41.98 \pm 2.72\), from \(39.3\) to \(44.7\).
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