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      Question

      An AR(2) process: Yₜ = 1.2Yₜ₋₁ −

      0.32Yₜ₋₂ + εₜ, εₜ ~ WN(0,1). Which is CORRECT?
      A Stationary because sum of AR coefficients (1.2 − 0.32 = 0.88) is less than 1 Correct Answer Incorrect Answer
      B Non-stationary because characteristic equation roots are inside the unit circle Correct Answer Incorrect Answer
      C Stationary because both characteristic equation roots lie outside the unit circle: z₁ = 2.5, z₂ = 1.25 Correct Answer Incorrect Answer
      D Non-stationary; mean undefined and variance grows without bound Correct Answer Incorrect Answer

      Solution

      Characteristic equation: 1 − 1.2z + 0.32z² = 0, or 0.32z² − 1.2z + 1 = 0 Solving: z = [1.2 ± √(1.44 − 1.28)] / 0.64 = [1.2 ± 0.4] / 0.64 z₁ = 1.6/0.64 = 2.5 and z₂ = 0.8/0.64 = 1.25. Both |z| > 1 ⇒ STATIONARY. ✔ Note: Trap — Option (A): Sum of coefficients < 1 is necessary but NOT sufficient for AR(2) stationarity. Always check characteristic roots, not just the sum.

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