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

A monthly series follows the seasonal AR model Y_t = 0.6*Y_{t-12} + e_t, with e_t white noise of variance 1. What is the autocorrelation of Y at lag 24, and what is the variance of Y?

The series is an AR(1) at the seasonal lag of 12, so autocorrelation at lag 24 is 0.6 squared, or 0.36. Its variance is 1/(1 - 0.36) = 1.5625. Autocorrelation at non-multiples of 12 is zero.

  1. AAutocorrelation 0.36; variance 1.5625Correct
  2. BAutocorrelation 0.60; variance 1.5625
  3. CAutocorrelation 0.36; variance 1.0
  4. DAutocorrelation 0.36; variance 2.5

Explanation

This is an AR(1) in the 12-month lag with phi = 0.6, so autocorrelation at lag 12k is 0.6^k. At lag 24 it is 0.36. Variance = sigma^2/(1 - phi^2) = 1/(1 - 0.36) = 1/0.64 = 1.5625. Variance 2.5 wrongly uses 1/(1 - 0.6).

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