The autocorrelation of an AR(1) process

Question

Consider the stationary AR(1) process Xt = \phi X{t-1} + \varepsilont with \phi = 0.7 and white-noise innovations \varepsilont of variance \sigma^2. Derive the autocorrelation function hok, and find the "half-life" of a shock, the number of steps for the ACF to fall to 	frac12.

Accepted Answer

Multiply Xt = \phi X{t-1} + \varepsilont by X{t-k} (for k \geq 1) and take expectations. Since \varepsilont is uncorrelated with past values, the innovation term drops: \gammak = \phi\,\gamma{k-1} \implies \gammak = \phi^k \gamma0. Dividing by \gamma0 gives the autocorrelation function hok = \phi^k = 0.7^k. So ho1 = 0.7, ho2 = 0.49, ho3 = 0.343, decaying geometrically toward zero. Shocks fade but never vanish exactly. Half-life. Solve \phi^h = 	frac12: h = \frac{\ln 0.5}{\ln 0.7} = \frac{-0.6931}{-0.3567} \approx 1.94 	ext{ steps}. A shock loses half its correlation influence in a l…

The autocorrelation of an AR(1) process

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