Half-life of mean reversion in an AR(1)

Question

A mean-reverting series follows the AR(1) form $X_t = \mu + \phi\,(X_{t-1} - \mu) + \varepsilon_t$ with long-run mean $\mu = 100$ and persistence $\phi = 0.9$ . The current value is $X_t = 130$ .

Find the half-life of mean reversion. (As a check, also forecast 5 steps ahead.)

Accepted Answer

Half-life. A deviation from the mean shrinks by a factor \phi each period, so after h periods a deviation is scaled by \phi^h. The half-life is the number of periods for the deviation to fall to half its size: \phi^{h} = 0.5 \quad\Longrightarrow\quad h = \frac{\ln 0.5}{\ln \phi} = \frac{\ln 0.5}{\ln 0.9} = \frac{-0.6931}{-0.10536} \approx 6.58 	ext{ periods}. Forecast check. The k-step-ahead forecast reverts geometrically toward the mean: E[X{t+k} \mid Xt] = \mu + \phi^{k}\,(Xt - \mu). With the current deviation Xt - \mu = 130 - 100 = 30 and k = 5: E[X{t+5}] = 100 + 0.9^{5}	imes 30 = 100 + …

Half-life of mean reversion in an AR(1)

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