Variance of a stationary AR(1) process

Question

A stationary process follows Xt = 0.6\,X{t-1} + \varepsilont, where \varepsilont is white noise with variance \sigma^2 = 1. Compute the unconditional variance of Xt.

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For a stationary AR(1), take the variance of both sides. Using stationarity, \operatorname{Var}(Xt) = \operatorname{Var}(X{t-1}) = \gamma0, and the shock \varepsilont is uncorrelated with the past value X{t-1}: \gamma0 = \phi^2 \gamma0 + \sigma^2 \quad\Longrightarrow\quad \gamma0 = \frac{\sigma^2}{1 - \phi^2}. Plugging in \phi = 0.6 and \sigma^2 = 1: \gamma0 = \frac{1}{1 - 0.36} = \frac{1}{0.64} = 1.5625. Notice the variance of Xt is larger than the variance of the shock: persistence lets past shocks accumulate. As \phi approaches 1 the denominator shrinks toward 0 and the variance blows up, …

Variance of a stationary AR(1) process

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