Question

Estimator properties:

6 Estimators properties 6.1 Exercise 1 In order to estimate the average number of hours that children spend watching tv, a Bernoulli sample of size n = 5 children was selected from a primary school. Let X be the variable that represents the hours spent watching tv, let E(X)-μ the parameter to estimate and var(X-σ2 the variance. Compare the following two proposed estimators Τι 1. Compare the two estimators for u on the basis of their bias 2. Compare the two esimators for u on the basis of their efficiency 3. Which estimator is preferable for estimating u?

Answer 1

Solution

Exercise 1

Back-up Theory

A statistic, T (from sample) is an unbiased estimator of a parameter (of population) θ, if

E(T) = θ………………………………………………………………………………… (1)

Bias of an estimator, B = E(T) - θ………………………………………………………(2)

E{∑(i = 1 to n)(a_iX_i)} = ∑(i = 1 to n){a_iE(X_i)}………………………………………….. (3)

V{∑(i = 1 to n)(a_iX_i)} = ∑(i = 1 to n){a_i²V(X_i)}, if X_i are independent………………. (4)

Now to work out the solution,

Given X₁, X₂, X₃, X₄, X₅, are iid with mean µ and variance σ²,

Part (1)

Vide (3),

E(T₁) = E{(1/5)∑(i = 1 to 5)(X_i)}

=(1/5) ∑(i = 1 to 5){E(X_i)}

= (1/5)∑(i = 1 to 5)µ

= µ

So, vide (1), T₁ is an unbiased estimator of µ and hence vide (2), bias (T₁) = 0.

Vide (3),

E(T₂) = E{(1/5)(2X₁ – X₂ + 2X₃ + X₄ + X₅}

= (1/5){2E(X₁) – E(X₂) + 2E(X₃) + E(X₄) + E(X₅)}

= (1/5)(2µ - µ + 2µ + µ + µ)

= µ

So, vide (1), T₂ is an unbiased estimator of µ and hence vide (2), bias (T₂) = 0.

Thus, both estimators have equal bias of zero.Answer

Part (2)

Variance is a measure of efficiency of an estimator.

Vide (4),

V(T₁) = V{(1/5)∑(i = 1 to 5)(X_i)}

= (1/25)∑(i = 1 to 5)V(X_i)}

= (1/25)∑(i = 1 to 5)σ²

= (1/25)5σ²

= (1/5)σ²

V(T₂) = V{(1/5)(2X₁ – X₂ + 2X₃ + X₄ + X₅}

= (1/25){4V(X₁) + V(X₂) + 4V(X₃) + V(X₄) + V(X₅)}

= (1/25)(4σ² + σ² + 4σ² + σ² + σ²)

= (11/25)σ²

Thus, V(T₂) > V(T₁) Answer

Part (3)

Since V(T₁) < V(T₂), T₁ is more efficient that T₂. Answer

DONE