Suppose we have the sample x1 x2 and x3 drawn randomly from

Suppose we have the sample x1, x2 and x3, drawn randomly from a distribution with mean 4 and variance 9.

a) Does mu*=(x1+x2+x3)/3 or mu**=(x1+x2+x3)/4 have a smaller MSE as an estimator of beta=4?

b) Is mu* a better estimator that mu**?

c) relative efficiency,

E(mu*) = 1/3(12) =4

Var(mu*) = 1/9(9+9+9) = 9

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E(MU**) = 1/4 (12) =3

Var (mu**) = 1/16 (9) = 9/16

Obviously mu* has a smaller mse for beta =4

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mu* is a better estimator than mu.

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c) Relative efficiency is mse will be the least for mu*

The mse for mu* = sum of |(x-3)|

mse for mu** = sum of |(x-4)|

The difference will be |n| where n is the number of entries in x1,x2,x3 together.