Thread: Weighted Average / Pop Mean

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?

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.