FRM Part I · FRM Exam Part I · Sample Moments
Two independent unbiased estimators of the same mean have variances 4 (estimator P) and 12 (estimator Q). A combined estimator is w*P + (1-w)*Q. Which weight w on P gives the minimum-variance unbiased combination, and what is that minimum variance?
The minimum-variance weights are inversely proportional to variances, so the weight on P is 12/16 = 0.75. The combined variance is 4 times 0.5625 plus 12 times 0.0625, which is 3.00. This is lower than the equal-weight variance of 4.
- Aw = 0.50; variance 4.00
- Bw = 0.75; variance 3.00Correct
- Cw = 0.25; variance 3.00
- Dw = 0.75; variance 4.00
Explanation
Variance = 4w^2 + 12(1-w)^2. Minimizing gives w = 12/(4+12) = 0.75. Variance = 4*0.5625 + 12*0.0625 = 2.25 + 0.75 = 3.00. Equal weights give 1 + 3 = 4, which is higher, and any w keeps the estimator unbiased because weights sum to one.
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