Can a weighted average be an unbiased estimator of the population mean?
Say we have 3 variables from a sample of known mu: X1, X2, X3
Would the expected value of Z = (1/2)X1 + (1/4)X2 + (1/4)X3 evaluate to mu?
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yes, as long as .5 +.25+.25 equals one, you have an unbiased estimator.
HOWEVER the variance of this linear combination will be larger than the sample mean's variance.
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