FRM Part I · FRM Exam Part I · Sample Moments
Four equally likely scenarios give the following deviations from the mean. X: -2, 0, 0, 2. Y: 2, -1, -1, 0. Using population moments, what is the coskewness S(X,X,Y) = E[(X-μx)²(Y-μy)] / (σx²·σy)?
The coskewness is about 0.82. The cross moment E[x²y] is 2, and it is divided by σx² = 2 and σy = 1.2247. Dividing by σy² instead of σy would give 0.67, which is the wrong normalization.
- A0.67
- B0.82Correct
- C1.00
- D1.63
Explanation
E[x²y] = (4·2 + 0 + 0 + 4·0)/4 = 2. σx² = (4+0+0+4)/4 = 2. σy² = (4+1+1+0)/4 = 1.5, so σy = 1.2247. S = 2/(2×1.2247) = 0.816. Using σy² instead of σy gives 0.67, and omitting σx² gives 1.63.
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