Is this AR(2) process stationary?

Question

Consider the AR(2) process Xt = 0.5\,X{t-1} + 0.3\,X{t-2} + \varepsilont with white-noise shocks. Is it stationary? Answer yes or no.

Accepted Answer

An AR(2) with coefficients \phi1 and \phi2 is stationary when all three of these inequalities hold (the "stationarity triangle"): 1. \phi1 + \phi2 < 1 2. \phi2 - \phi1 < 1 3. -1 < \phi2 < 1 Check each with \phi1 = 0.5, \phi2 = 0.3: 1. 0.5 + 0.3 = 0.8 < 1. OK. 2. 0.3 - 0.5 = -0.2 < 1. OK. 3. 0.3 < 1. OK. All three hold, so the process is stationary. Equivalently, the roots of the characteristic equation 1 - 0.5z - 0.3z^2 = 0 lie outside the unit circle, which is what these three inequalities encode in a form you can check by hand. Takeaway. An AR(2) is stationary exactly when \phi1+\phi2<1, \p…

Is this AR(2) process stationary?

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