In class, I mentioned that the sample has the lowest variance (and MSE) of an unbiased estimator of the population. Cons
Posted: Sun Oct 03, 2021 12:52 pm
In class, I mentioned that the sample has the lowest variance
(and MSE) of an unbiased estimator of the population. Consider once
again the example of car ownership and consider a sample of two
observations from the population: X1 and X2. The sample mean, X,
places equal weight on each observation. Therefore, assuming
observations are independent, the variance of the sample mean is: V
ar(X) =V ar( 1 n Xn i Xi) =V ar( 1 2 X 2 i Xi) =V ar( 1 2 X1 + 1 2
X2) =V ar( 1 2 X1) + V ar( 1 2 X2) X1 and X2 are just random draws
from the population so you can plug in your answer to 2(b) or 2(c)
to obtain the equivalent variances of X1 and X2.
(a) Complete the calculation for the variance of X.
(b) Confirm that X is an unbiased estimator of µX.
(and MSE) of an unbiased estimator of the population. Consider once
again the example of car ownership and consider a sample of two
observations from the population: X1 and X2. The sample mean, X,
places equal weight on each observation. Therefore, assuming
observations are independent, the variance of the sample mean is: V
ar(X) =V ar( 1 n Xn i Xi) =V ar( 1 2 X 2 i Xi) =V ar( 1 2 X1 + 1 2
X2) =V ar( 1 2 X1) + V ar( 1 2 X2) X1 and X2 are just random draws
from the population so you can plug in your answer to 2(b) or 2(c)
to obtain the equivalent variances of X1 and X2.
(a) Complete the calculation for the variance of X.
(b) Confirm that X is an unbiased estimator of µX.