1)


Two random samples of sizes n and 3n are taken from normal populations with means µ and 3µ and variances σ2 and 3σ2 respectively.





If  X 1 and  X 2 are the sample means, show that the estimator a X 1 + b X 2 is an unbiased estimator for µ if a + 3b = 1.



  Also, find the values of a and b if this estimator is to have minimum variance.


(所有X上面都有一劃, 即係mean)

Denote M1 and M2 be the sample mean for X1 and X2 respectively
Then E[aM1 + bM2]
= aE[M1] + bE[M2]
= aµ + b(3µ )
= (a + 3b)µ
Hence if a + 3b = 1, E[aM1 + bM2] = µ, that is,
aM1 + bM2 is an unbiased estimator for µ

Var[aM1 + bM2] = a^2Var[M1] + b^2Var[M2]
= a^2(σ^2/n) + b^2(3σ^2/3n)
= (a^2 + b^2)σ^2/n
Given this estimator is unbiased, a = 1 - 3b and hence
a^2 + b^2 = (1 - 3b)^2 + b^2
= 9b^2 - 6b + 1 + b^2
= 10(b^2 - 2*3b/10 + 9/100) + 1 - 9*10/100
= 10(b - 3/10)^2 + 1/10
Hence the minimum of a^2 + b^2 occurs at b = 3/10, a = 1/10
Var[aM1 + bM2] attains minimum
⇔ a^2 + b^2 attains minimum
⇔ a = 1/10 and b = 3/10