11.5 Forecasting with autocorrelated disturbances
When the disturbances are autocorrelated, the last residual contains information about the next disturbance, and a good forecast uses it.
Theorem 11.5.1. Let \(Y_t = x_t'\beta + u_t\) with \(u_t = \rho u_{t-1} + \varepsilon _t\), and suppose \(\beta \), \(\rho \), \(u_n\) and the future regressor values \(x_{n+s}\) are known. The minimum mean square error forecast of \(Y_{n+s}\) is \[ \hat Y_{n+s} = x_{n+s}'\beta + \rho ^su_n , \] and its forecast error has variance \[ \sigma _\varepsilon ^2(1 + \rho ^2 + \dots + \rho ^{2(s-1)}) = \sigma _\varepsilon ^2\,\frac {1 - \rho ^{2s}}{1 - \rho ^2} , \] which increases with \(s\) towards \(\sigma _u^2 = \sigma _\varepsilon ^2/(1 - \rho ^2)\).
Proof. Iterating the AR(1) recursion \(s\) times, \[ u_{n+s} = \rho ^su_n + \varepsilon _{n+s} + \rho \varepsilon _{n+s-1} + \dots + \rho ^{s-1}\varepsilon _{n+1} . \] The future \(\varepsilon \) are independent of everything known at time \(n\), so the conditional mean of \(Y_{n+s}\) is \(x_{n+s}'\beta + \rho ^su_n\), which is the minimum mean square error predictor by Theorem 1.4.3. The error is the sum of the future \(\varepsilon \) terms, with variance \(\sigma _\varepsilon ^2\sum _{j=0}^{s-1}\rho ^{2j}\). □
In practice \(\beta \) and \(\rho \) are replaced by their feasible GLS or maximum likelihood estimates and \(u_n\) by the last residual \(e_n = Y_n - x_n'\hat \beta \), which adds estimation error, of order \(1/n\), to the forecast variance. The correction \(\rho ^se_n\) fades geometrically, so for distant horizons the forecast returns to the regression line \(x_{n+s}'\hat \beta \).
Example 11.5.2. Quarterly sales have been modelled as \(\hat Y_t = 20 + 3x_t\), with AR(1) disturbances, \(\hat \rho = 0.6\) and \(\hat \sigma _\varepsilon = 1.2\). The last residual is \(e_n = 2.5\), and the regressor will be \(x_{n+1} = 10\) and \(x_{n+2} = 11\). Then \[ \hat Y_{n+1} = 20 + 30 + 0.6(2.5) = 51.5, \qquad \hat Y_{n+2} = 20 + 33 + 0.36(2.5) = 53.9 , \] with standard errors \(1.2\) and \(1.2\sqrt {1 + 0.36} = 1.40\), ignoring estimation error. A forecaster who ignored the autocorrelation would predict \(50\) and \(53\), missing the information that sales were running above the line, and would quote the standard error \(\sigma _u = 1.2/\sqrt {1 - 0.36} = 1.5\) for both horizons, too large at short horizons.
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