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FRM Part I · FRM Exam Part I · Stationary Time Series

An analyst fits an AR(2) process Y_t = 0.5·Y_(t-1) + 0.2·Y_(t-2) + ε_t (zero mean, stationary). Using the Yule-Walker relationships, what is the lag-1 autocorrelation ρ1 and the lag-2 autocorrelation ρ2?

Solving the Yule-Walker equations gives ρ1 = φ1/(1-φ2) = 0.5/0.8 = 0.625. Then ρ2 = φ1ρ1 + φ2 = 0.5×0.625 + 0.2 = 0.5125. So the autocorrelations are 0.625 at lag 1 and 0.5125 at lag 2.

  1. Aρ1 = 0.625; ρ2 = 0.5125Correct
  2. Bρ1 = 0.50; ρ2 = 0.45
  3. Cρ1 = 0.625; ρ2 = 0.45
  4. Dρ1 = 0.714; ρ2 = 0.5571

Explanation

Yule-Walker: ρ1 = φ1 + φ2ρ1, so ρ1 = 0.5/(1-0.2) = 0.625. Then ρ2 = φ1ρ1 + φ2 = 0.5(0.625) + 0.2 = 0.3125 + 0.2 = 0.5125. Using ρ1 = 0.5 ignores the feedback through φ2; the 0.714 option divides by 0.7 wrongly.

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