AR(1) variance with a larger shock variance

Question

A stationary process follows Xt = 0.5\,X{t-1} + \varepsilont, where \varepsilont is white noise with variance \sigma^2 = 3. Compute the unconditional variance of Xt.

Accepted Answer

For a stationary AR(1), taking the variance of both sides and using that the shock is uncorrelated with the lagged value gives \gamma0 = \frac{\sigma^2}{1 - \phi^2}. Here \phi = 0.5, so \phi^2 = 0.25, and \sigma^2 = 3: \gamma0 = \frac{3}{1 - 0.25} = \frac{3}{0.75} = 4. The series variance (4) is larger than the shock variance (3) because moderate persistence stacks successive shocks on top of each other. Takeaway. The AR(1) variance scales the shock variance by 1/(1-\phi^2); with \phi = 0.5 that inflation factor is 4/3, lifting a shock variance of 3 up to 4.

AR(1) variance with a larger shock variance

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