Question

Consider the situation of this chapter, where we have to estimate the parameter N from a sample x1,...,xn drawn without replacement from the numbers {1,...,N}. To keep it simple, we consider n = 2. Let M = M2 be the maximum of X1 and X2. We have found that T2 = 3M/2 − 1 is a good unbiased estimator for N. We want to construct a new unbiased estimator T3 based on the minimum L of X1 and X2. In the following you may use that the random variable L has the same distribution as the random variable N + 1 − M (this follows from symmetry considerations). a. Show that T3 = 3L − 1 is an unbiased estimator for N. b. Compute Var(T3) using that Var(M)=(N + 1)(N − 2)/18. (The latter has been computed in Remark 20.1.) c. What is the relative efficiency of T2 with respect to T3?

Answer 1

(a)

Solution:- T3=3L-1

E[T3]=3E[L]-1

E[T3]=3E[N+1-M]-1 where we substitute L=N+1-M because both L and N+1-M has same distribution.

E[T3]=3E[N]+3-3E[M]-1

E[T3]= 3N-3E[M]+2 ........................(1)

E[T3] = 3N-2N-2+2 .......................(2)

As -3E[M] can be calculated from given estimator T2. We know that T2 is an unbiased estimator that means E[T2]=N and we know that T2=3M/2-1 that means E[T2]=3/2 E[M]-1 implies N=3/2E[M]-1 implies 3/2E[M]=N+1 implies E[M]=2/3 N + 2/3 implies -3E[M]=-2/3 N + 2/3 ...........................(3)

When we substitute equation (3) in equation (1) we get equation (2).

FInally equation (2) results in E[T3]=N. Hence we prove that T3 is an unbiased estimator.

(B)

Solution:- Var[T3]=9Var[L]

But we substitute L=N+1-M

implies Var[T3]=9Var[N+1-M]

implies Var[T3]=9Var[M]

implies Var[T3]=9 * (N + 1)(N − 2)/18

implies Var[T3]=(N+1)(N-2)/2

(c)

Solution :- relative efficiency = Var[T3]/Var[T2]

=[(N+1)(N-2)/2]/[(N + 1)(N − 2)/18]

= 9

Hence estimator T3 is 9 times more efficient than T2.