Show that if t" = ½(t + t') where t and t' are both most efficient estimators with variance v, then var(t'') = ½v(1 + p).
Answers
Answer:
X1 , X2 and X3 is a random sample of size 3 from a population with mean mu and
variance sigma^2. T1, T2 and T3 are the estimators to estimate mu, and are given by
T1 = X1 + X2 - X3; T2 = 2X1 + 3X3 - 4X2 and T3 = 1/3(lambda X1 + X2 + X3).
(i) Are T 1 and T2 unbiased ? Give reason.
(ii) Find the value of A, such that T3 is
unbiased.
(iii) Which is the best estimator ? State
giving reasons.
Expert's answer
(i) \text{We will show that } T_1 \text{ and } T_2 \text{ are unbiased estimators}\\ \text{of the parameter } \mu \text{ using the definition of an unbiased estimator}.\\ E(T_1)=E(X_1+X_2-X_3)=\mu+\mu-\mu=\mu.\\ E(T_2)=E(2X_1+3X_3-4X_2)=2\mu+3\mu-4\mu=\mu.\\ \text{So } T_1 \text{ and } T_2 \text{ are unbiased estimators of } \mu.\\ (ii) E(T_3)=\mu.\\ \frac{1}{3}(\lambda\mu+\mu+\mu)=\mu.\\ \text{Then }\lambda=1.\\ (iii) D(T_1)=D(X_1+X_2-X_3)=\sigma^2+\sigma^2-\sigma^2=\sigma^2.\\ D(T_2)=D(2X_1+3X_3-4X_2)=2\sigma^2+3\sigma^2-4\sigma^2=\sigma^2.\\ D(T_3)=D\big(\frac{1}{3}(X_1+X_2+X_3)\big)=\frac{1}{9}(\sigma^2+\sigma^2+\sigma^2)=\frac{1}{3}\sigma^2.\\ \text{Since } T_3 \text{ has the lowest variance then } T_3\\ \text{is the best estimator of } \mu.(i)We will show that T
1
and T
2
are unbiased estimators
of the parameter μ using the definition of an unbiased estimator.
E(T
1
)=E(X
1
+X
2
−X
3
)=μ+μ−μ=μ.
E(T
2
)=E(2X
1
+3X
3
−4X
2
)=2μ+3μ−4μ=μ.
So T
1
and T
2
are unbiased estimators of μ.
(ii)E(T
3
)=μ.
3
1
(λμ+μ+μ)=μ.
Then λ=1.
(iii)D(T
1
)=D(X
1
+X
2
−X
3
)=σ
2
+σ
2
−σ
2
=σ
2
.
D(T
2
)=D(2X
1
+3X
3
−4X
2
)=2σ
2
+3σ
2
−4σ
2
=σ
2
.
D(T
3
)=D(
3
1
(X
1
+X
2
+X
3
))=
9
1
(σ
2
+σ
2
+σ
2
)=
3
1
σ
2
.
Since T
3
has the lowest variance then T
3
is the best estimator of μ.