Question

(1 point) A normal distribution with mean u and variance o2 is independently sampled three times, yielding values X1, X2, and X3 . Consider the three estimators în1 = x1 + 4x2, Û2 = x1 – x2 + x3 , and из şx2 + 3x2 + zxz Find the expected value of each estimator (type mu for u and sigma for o): ECÂ1) = E@2) = ECÂ3) = Which estimator(s) are biased and which are unbiased? Estimator în1: ? Estimator Û2: ? Estimator ûz: ? Find the variance of each estimator (type mu for u and sigma for o): = var(în) var(Û2) = var(û3) = Which estimator is the most efficient? ?

Answer 1

Mean=μ and Variance=ơ²

X1,X2 and X3 are independent of one another

So, E(X1)=E(X2)=E(X3)=μ

and Var(X1)=Var(X2)=Var(X3)=ơ²

The covariances are 0 since X1,X2 and X3 are independent.

Now given:

μ1^=X1+4X2

μ2^=X1-X2+X3

μ3^=2/5*X1+1/5*X2+2/5*X3

Solution to Part 1:

E(μ1^) =E(X1+4X2) =E(X1)+E(4X2) =E(X1)+4*E(X2)=μ+4*μ=5μ

Therefore, E(μ1^)=5μ

E(μ2^)=E(X1-X2+X3)=E(X1)-E(X2)+E(X3)=μ-μ+μ=μ

Therefore, E(μ2^)=μ

E(μ3^)=E(2/5*X1+1/5*X2+2/5*X3) =2/5*E(X1) +1/5*E(X2) +2/5*E(X3) =2/5*μ+1/5*μ+2/5*μ=μ

Therefore, E(μ3^)=μ

[Since E(aX+bY) =a*E(X)+b*E(Y), where a and b are constants and E(X1)=E(X2)=E(X3)=μ]

Ans: E(μ1^)=5μ,E(μ2^)=μ and E(μ3^)=μ

Solution to Part 2:

The estimator X is said to be unbiased of the parameter μ if:

E(X)=μ

Otherwise it is biased.

From part (1) we see,

E(μ1^)=5μ ≠ μ,

Hence μ1^ is biased.

E(μ2^)=μ and E(μ3^)=μ ,

Hence both μ2^ and μ3^ are unbiased.

Ans: Estimator μ1^ is biased, Estimator μ2^ is unbiased and Estimator μ3^ is unbiased .

Solution to part 3:

We know, Var(aX+bY) =a^2*Var(X)+b^2*Var(Y), where X and Y are independent and a,b are constants. And, here, Var(X1)=Var(X2)=Var(X3)=ơ²

So,

Var(μ1^) =Var(X1+4X2) =Var(X1)+Var(4X2) =(1)^2*Var(X1)+(4)^2*Var(X2)=1.ơ²+16*ơ²=17ơ²

Therefore, Var(μ1^)=17ơ²

Var(μ2^)=Var(X1-X2+X3)=(1)^2*Var(X1)+(-1)^2*Var(X2)+(1)^2*(X3)=1.ơ²+1.ơ²+1.ơ²=3ơ²

Therefore, Var(μ2^)=3ơ²

Var(μ3^)=Var(2/5*X1+1/5*X2+2/5*X3) =(2/5)^2*Var(X1) +(1/5)^2*Var(X2) +(2/5)^2*Var(X3) =4/25*ơ²+1/25*ơ²+4/25*ơ²=9/25*ơ²=0.36ơ²

Therefore, Var(μ3^)=0.36ơ²

Ans: Var(μ1^)=17ơ², Var(μ2^)=3ơ² and Var(μ3^)=0.36ơ²

Solution to part 4:

The unbiased estimator with the minimum variance is the most efficient estimator.

From part 2, μ1^ is biased. So it can't be considered.

μ2^ and μ3^ are unbiased.

Hence, from μ2^ and μ3^ we need to see which estimator has lower variance.

From part 3, Var(μ2^)=3ơ² and Var(μ3^)=0.36ơ²

Now,

Var(μ3^)=0.36ơ² < 3ơ²=Var(μ2^) [ Since ơ²>0]

So μ3^ is the most efficient estimator.

Ans: Most efficiet estimator = μ3^.