Skip to content

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

Using 100 residuals, an analyst computes sample autocorrelations of 0.20 at lag 1, -0.10 at lag 2, and 0.10 at lag 3. The Box-Pierce statistic Q = T Σ ρ̂_k² for m = 3 lags is compared with a chi-square critical value of 7.81 (3 degrees of freedom, 5%). What is Q and the conclusion?

Q equals 6.0 and white noise is not rejected. The squared autocorrelations sum to 0.04 + 0.01 + 0.01 = 0.06, and multiplying by 100 observations gives 6.0, which is below the 5% chi-square critical value of 7.81 for three degrees of freedom.

  1. AQ = 6.0; do not reject white noiseCorrect
  2. BQ = 6.0; reject white noise
  3. CQ = 0.06; do not reject white noise
  4. DQ = 9.0; reject white noise

Explanation

Sum of squares = 0.04 + 0.01 + 0.01 = 0.06. Q = 100 × 0.06 = 6.0. Since 6.0 < 7.81, we fail to reject the null of no autocorrelation. Option C forgets to multiply by T. Option D sums the autocorrelations incorrectly.

Did you get it right without looking?

One question tells you little. A timed set on Stationary Time Series shows your real accuracy, how long you take and where you lose marks.

Practise Stationary Time SeriesFree diagnostic mock

More Stationary Time Series questions