Multi-step forecast of a mean-reverting series

Question

A mean-reverting series follows Xt = \mu + \phi\,(X{t-1} - \mu) + \varepsilont with long-run mean \mu = 50 and persistence \phi = 0.8. The current value is Xt = 70. Forecast the value 3 steps ahead, E[X{t+3} \mid Xt].

Accepted Answer

The k-step-ahead forecast of an AR(1) reverts toward the mean geometrically: E[X{t+k} \mid Xt] = \mu + \phi^{k}\,(Xt - \mu). The current deviation from the mean is Xt - \mu = 70 - 50 = 20. With \phi = 0.8 and k = 3: \phi^3 = 0.8^3 = 0.512, E[X{t+3}] = 50 + 0.512 	imes 20 = 50 + 10.24 = 60.24. The original deviation of 20 has decayed to about 10.24 after three steps, a little more than halfway back to the mean, consistent with the half-life \ln(0.5)/\ln(0.8) \approx 3.1 periods for this \phi. Takeaway. AR(1) forecasts pull toward the mean as \mu + \phi^k(Xt-\mu); three steps out at \phi = 0.8…

Multi-step forecast of a mean-reverting series

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