Lag-3 autocorrelation of an AR(1)

Question

A stationary process follows Xt = 0.8\,X{t-1} + \varepsilont with white-noise shocks. Compute the lag-3 autocorrelation \rho3.

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For a stationary AR(1) with coefficient \phi, the autocorrelation at lag k decays geometrically: \rhok = \phi^k. This comes from multiplying the AR(1) equation by X{t-k} and taking expectations, which gives \gammak = \phi\,\gamma{k-1}, so each step down the lags multiplies by \phi. With \phi = 0.8 and k = 3: \rho3 = 0.8^3 = 0.512. For comparison, \rho1 = 0.8 and \rho2 = 0.64, so the correlation fades smoothly but slowly because \phi is large. Takeaway. An AR(1) has autocorrelations \rhok = \phi^k, a clean geometric decay; the larger \phi is, the more slowly the memory of the series dies out.

Lag-3 autocorrelation of an AR(1)

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