AR(1) variance with a negative coefficient

Question

A stationary process follows Xt = -0.4\,X{t-1} + \varepsilont, where \varepsilont is white noise with variance \sigma^2 = 0.84. Compute the unconditional variance of Xt.

Accepted Answer

The variance formula uses \phi^2, so the sign of \phi does not matter for the variance: \gamma0 = \frac{\sigma^2}{1 - \phi^2} = \frac{0.84}{1 - (-0.4)^2} = \frac{0.84}{1 - 0.16} = \frac{0.84}{0.84} = 1. Although the variance ignores the sign, the autocorrelations do not: \rhok = \phi^k = (-0.4)^k alternates, giving \rho1 = -0.4, \rho2 = +0.16, \rho3 = -0.064. A negative \phi produces a zigzag where a high value tends to be followed by a low one. Takeaway. The AR(1) variance depends only on \phi^2, so positive and negative coefficients of equal size give the same spread, but a negative coeffic…

AR(1) variance with a negative coefficient

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