Back out the shock variance of an AR(1)

Question

A stationary AR(1) process Xt = 0.7\,X{t-1} + \varepsilont has unconditional variance \gamma0 = 2. Find the variance \sigma^2 of the white-noise shocks.

Accepted Answer

The AR(1) variance relation is \gamma0 = \frac{\sigma^2}{1 - \phi^2}. Rearranging to solve for the shock variance: \sigma^2 = \gamma0\,(1 - \phi^2) = 2\,\bigl(1 - 0.7^2\bigr) = 2\,(1 - 0.49) = 2 	imes 0.51 = 1.02. The shock variance (1.02) is smaller than the series variance (2), which makes sense: persistence amplifies the raw shocks, so to end up with a spread of 2 you only need to feed in shocks of variance 1.02. Takeaway. The AR(1) variance formula runs both ways: multiply the series variance by (1-\phi^2) to recover the shock variance that generated it.

Back out the shock variance of an AR(1)

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