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FRM Part I · FRM Exam Part I · Nonstationary Time Series

A quadratic trend model for annual output is y_t = 100 + 6t - 0.1t^2 + e_t. At which value of t does the fitted trend reach its maximum, and what is the maximum fitted value?

The trend peaks at t = 30 with a fitted value of 190. Setting the derivative 6 minus 0.2t to zero gives t = 30, and substituting gives 100 plus 180 minus 90, which equals 190. The negative quadratic coefficient makes this a maximum.

  1. At = 30; fitted value 190Correct
  2. Bt = 60; fitted value 280
  3. Ct = 30; fitted value 100
  4. Dt = 15; fitted value 167.5

Explanation

The derivative is 6 - 0.2t = 0, so t = 30. The fitted value is 100 + 180 - 0.1(900) = 100 + 180 - 90 = 190. The t = 60 option ignores the factor of 2 when differentiating.

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