AR(2) that fails the first inequality

Question

Consider the AR(2) process Xt = 0.6\,X{t-1} + 0.5\,X{t-2} + \varepsilont with white-noise shocks. Is it stationary? Answer yes or no.

Accepted Answer

Test the three stationarity conditions with \phi1 = 0.6, \phi2 = 0.5: 1. \phi1 + \phi2 = 0.6 + 0.5 = 1.1 < 1? No, this fails (1.1 exceeds 1). 2. \phi2 - \phi1 = 0.5 - 0.6 = -0.1 < 1. OK. 3. 0.5 < 1. OK. Because the first condition fails, the process is not stationary. Intuitively, \phi1 + \phi2 \ge 1 means the series has (at least) a unit root: shocks do not fully die out and the series behaves like a random walk or worse. You can also see it in the characteristic equation, z = 1 solves 1 - 0.6z - 0.5z^2 = 0 only if \phi1 + \phi2 = 1; here the sum exceeds 1, pushing a root inside the unit cir…

AR(2) that fails the first inequality

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