Question

Suppose that X₁,……,X_n form a random sample from a normal distribution for which the mean is known and the variance is unknown. Construct an efficient estimator that is not identically equal to a constant, and determine the expectation and the variance of this estimator.

Short answer

The expectations and variance of the estimator is E (x̄_n) = µ, Var (x̄_n) = σ² /n

Step-by-step solution

Given the information

It is given that X_1,….,X_n is ii d variable from a normal distribution with known mean µ and unknown variance. Therefore X_1,….,X_n are ii d Normal (µ , σ)

Define the pdf

f(x| µ, σ ) = 1/√ 2πσ²) exp(- 1/2 ((x - µ)/σ)²)

Define an efficient estimator

The most efficient estimator among a group of unbiased estimators is the one with the minor variance.

An efficient estimator also fetches a slight variance or mean square error. Therefore, there is a slight deviation between the estimated and parameter values.

Define fisher information

So, to establish efficiency, they have to compare the estimator's variance with the Cramer-Rao bound.

Assume X~ f (x| θ) (pdf or pmf) with θ ∈ ʘ ⊂ R

Then the fisher information is defined by

I_x(θ) = E_θ[(∂ / ∂θ logf(X|θ))²]

= E_θ[(-∂² / ∂θ logf(X|θ)]

And

I_x(θ) = nl_x1 (θ)

Calculating fisher information for normal distribution

Let X =X₁

From the definition

I_x(σ) = E_θ[(∂² / ∂ σ² logf(X|σ)]

= -3(x-µ)²/ σ⁴ + 1/σ²

= 2/ σ²

I_x(σ²) = nI_X1(σ²)

= 2n/ σ²

Therefore, the fisher information is 2n/ σ²

Therefore 1/l (σ) = σ²/ 2n

Applying CRLB bound

Now, by the Cramer Rao bound

V (σ)≥ I_x(σ²)^-1= n/2σ²

Since the lowest bound of variance is attained by CRLB equality, the M.L.E. X̄_n is the most efficient estimator of µ.

Calculating the expectation and variance

The expectations and variance of the estimator is E (x̄_n) = µ, Var (x̄_n) = σ² /n