Question

In estimating the average price of a gallon of gasoline in a region we plan to select a random sample (independent and identically distributed) of size 10. Let X1 , X2 , ... , X10 denote the selected sample. The four estimators for estimating the average price, mu, are:

U1 = ( X1 + X2 + ... + X10 ) /10

U2 = ( X1 + X2 + ... + X8 ) /8 + X10 - X9

U3 = ( X1 + X2 + ... + X4 ) /4

U4 = X10

Which estimator is preferred for estimating mu?

Answer 1

We prefer the estimator with the lowest MSE where

MSE = Var(Ui) +

Bias(Ui) = E[Ui] -

For U1,

E[U1] = E[( X1 + X2 + ... + X10 ) /10] = (E[ X1] + E[X2]+ ... + E[X10] ) /10 = ( + + ... + )/10 = 10 / 10 =

Bias(U1) = E[U1] - = - = 0

Var[U1] = Var[( X1 + X2 + ... + X10 ) /10] = (Var[ X1] + Var[X2]+ ... + Var[X10] ) /10^2 = ( + + ... + )/100 = 10 / 100 = /10

MSE = Var(U1) + = /10 = 0.1

For U2,

E[U1] = E[( X1 + X2 + ... + X8 ) /8 + X10 - X9] = (E[ X1] + E[X2]+ ... + E[X8] ) /8 + E[X10] - E[X9]

= ( + + ... + )/8 + - = 8 / 8 =

Bias(U2) = E[U2] - = - = 0

Var[U2] = Var[( X1 + X2 + ... + X8 ) /8 + X10 - X9] = (Var[ X1] + Var[X2]+ ... + Var[X8] ) /8^2 + Var[X10] + Var[X9]

= ( + + ... + )/64 + + = 8 / 64 + 2 = 17/8 = 2.125

MSE = Var(U2) + = 2.125

For U3,

E[U1] = E[( X1 + X2 + ... + X4 ) /4] = (E[ X1] + E[X2]+ ... + E[X4] ) /4 = ( + + ... + )/4 = 4 / 4 =

Bias(U3) = E[U3] - = - = 0

Var[U3] = Var[( X1 + X2 + ... + X4 ) /4] = (Var[ X1] + Var[X2]+ ... + Var[X4] ) /4^2 = ( + + ... + )/16 = 4 / 16 = /4

MSE = Var(U1) + = /4 = 0.25

For U4,

E[U4] =E[X10] =

Bias(U4) = E[U4] - = - = 0

Var[U4] = Var[X10 ] =  

MSE = Var(U4) + =

The lowest MSE is 0.1for U1. Thus, we prefer U1 for estimating .