FRM Part I · FRM Exam Part I · Sample Moments
A sample of 100 monthly returns has skewness of 0.4 and kurtosis of 4.0. Using the Jarque-Bera statistic, JB = (n/6)[S^2 + (K-3)^2/4], which is distributed chi-square with 2 degrees of freedom (5% critical value 5.991), what is the conclusion?
JB equals 6.83, which exceeds the 5% chi-square critical value of 5.991 with two degrees of freedom, so normality is rejected. The statistic is 16.667 times the sum of squared skewness, 0.16, and one quarter of squared excess kurtosis, 0.25.
- AJB = 6.83; reject normality at the 5% levelCorrect
- BJB = 6.83; do not reject normality at the 5% level
- CJB = 2.67; do not reject normality at the 5% level
- DJB = 19.33; reject normality at the 5% level
Explanation
JB = (100/6)(0.16 + 1/4) = 16.667 x 0.41 = 6.83. This exceeds the critical value of 5.991, so normality is rejected at 5%. Using (K-3)^2 without dividing by 4 gives 19.33, and counting only skewness gives 2.67.
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