Given two estimators U and V of parameter , we say that U is more effficient tha

Question

Given two estimators U and V of parameter , we say that U is more effficient than V if MSE(U)<MSE(V).

(a) Suppose that we sample X1, X2 independently from a population. Which of U= (X1+ 2X2)/3 and V= (2X1+ 3X2)/5 is the more efficient estimator of the mean of the population?

(b) We say that is aconvex combination of X1and X2 if there exists [0,1] such that =X1+ (1)X2. For example, the estimators Uand V in Part A are both convex combinations ofthe samples X1and X2; indeed, U corresponds to = 1/3 and V corresponds to = 2/5. Show that the average of X1and X2, i.e., the sample mean Y= (X1+X2)/2 is a more efficient estimator of the population mean than any other convex combination of X1andX2.

Hint: You can express the MSE of as a function of the variance of the population 2 and the convex combination coefficient . You can use techniques from calculus, e.g., the second derivative test, to establish that this function is minimized when = 1/2.

Explanation / Answer

a) when estimator is unbiased MSE is variance of estimator

here both U and V areunbiased

Var(U) = 1/9 (1 + 2^2) Var(X) = 5/9 Var(X)

Var(V) = 1/25* (2^2 + 3^2 ) *Var(X) = 13/25 Var(X)

13/25 < 5/9

hence V is more efficient

b)

Var(Ybar) = 1/4*(1 + 1) Var(X) = 1/2 Var(X)

Var(X1+ (1)X2.) = var(X) (^2 + (1-)^2)

^2 + (1-)^2 is minimum when = 1/2

Z = ^2 + (1-)^2

dZ/d = 2 -2 (1- ) = 0

= (1- )

= 1/2

second derivative = 2 + 2 =4   > 0 ,hence minima is attended at = 1/2

hence

the sample mean Y= (X1+X2)/2 is a more efficient estimator of the population mean than any other convex combination of X1andX2.

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