Half-life of a very persistent AR(1)

Question

A mean-reverting series follows an AR(1) with persistence \phi = 0.95. Find the half-life of mean reversion.

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A deviation from the mean is multiplied by \phi each period, so the half-life h solves \phi^h = 0.5: h = \frac{\ln 0.5}{\ln \phi} = \frac{\ln 0.5}{\ln 0.95} = \frac{-0.6931}{-0.051293} \approx 13.51 	ext{ periods}. Compare with \phi = 0.9, which gives a half-life of about 6.58: raising \phi from 0.9 to 0.95 roughly doubles the half-life. As \phi approaches 1 the half-life shoots toward infinity, which is the random-walk limit where shocks never decay and there is no mean reversion at all. Takeaway. Half-life = \ln(0.5)/\ln(\phi) grows steeply as \phi nears 1; at \phi = 0.95 a shock takes abo…

Half-life of a very persistent AR(1)

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