Question

Suppose that Y1, . . . , Yn is a sample of size n from a Exp(θ), with θ > 0, E[Y1] = θ and V(Y1) = θ^2

.
a) Find the distribution of Y¯Bar and Y(1) = min(Y1, . . . , Yn).
b) Find the constants c1 and c2 such that T1 = c1Y(1) and T2 = c2Y¯Bar are unbiased the estimators of θ. Compare the MSE of T1 and T2
c) Show that Q(Y1, . . . , Yn; θ) = 2nY(1)/θ ∼ Chi-sqaure(2).
d) Construct a two-sided 1 − α confidence interval for θ using the pivot in c).

Answer 1

See below.

Explanation:

a)

Since Y1, ..., Yn are independent and identically distributed, we have Y¯Bar ~ Exp(θ/n) and Y(1) ~ Exp(nθ).

b)

We want to find c1 and c2 such that E[T1] = θ and E[T2] = θ. We have E[Y(1)] = 1/θ and E[Y¯Bar] = θ/n, so setting T1 = c1Y(1) gives E[T1] = c1/θ = θ, and setting T2 = c2Y¯Bar gives E[T2] = c2θ/n = θ. Thus, c1 = θ^2 and c2 = n. To compare the MSE of T1 and T2, we compute,

MSE(T1) = V(T1) + [E(T1) - θ]^2

= V(θY(1)) + [θ - θ]^2

= θ^2V(Y(1))

= θ^2/θ^2

= 1

MSE(T2) = V(T2) + [E(T2) - θ]^2

= V(nY¯Bar) + [θ - θ]^2

= n^2V(Y¯Bar)/n^2

= V(Y¯Bar)/n

= (θ^2/n^2)(n/θ)

= θ^2/n

Since MSE(T2) < MSE(T1) for n > 1, T2 is the preferred estimator.

c)

Let Q(Y1, ..., Yn; θ) = 2nY(1)/θ. We have E[Q] = 2n/θ and V(Q) = 4n^2V(Y(1))/θ^2 = 4n^2/θ^2. To show that Q ~ Chi-squared(2), we need to show that Q has a gamma distribution with parameters k = 2 and θ = 1/2. We have,

fQ(q) = (1/θ^k)(q^(k-1) exp(-q/θ))/Γ(k) = (1/2^2)(q/2)exp(-q/2) = (1/4)q/2 exp(-q/2)

which is the pdf of a gamma distribution with k = 2 and θ = 1/2.

d)

To construct a two-sided 1 - α confidence interval for θ, we use the fact that 2nY(1)/Q(Y1, ..., Yn; θ) ~ F(2, 2n). Let Fα/2 be the (1 - α/2) quantile of the F distribution with 2 and 2n degrees of freedom. Then we have,

P(2nY(1)/(Fα/2) < θ < 2nY(1)/(F1-α/2)) = 1 - α

So the confidence interval is [2nY(1)/(Fα/2), 2nY(1)/(F1-α/2)].