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STAT 410 Final Exam
Wednesday, December 17, 2003 Closed book & Notes. The last page has some useful information.
20 points 1. Suppose X 1 , X 2 , and X 3 are independent, with X 1 ∼ P oisson(λ), X 2 ∼ P oisson(2λ), and X 3 ∼ P oisson(3λ), where λ ∈ (0, ∞). (a) Show that (X 1 , X 2 , X 3 ) has a one-dimensional exponential family distribution. Give T , the natural sufficient statistic, and the natural parameter. Answer: The pmf is
f (x 1 , x 2 , x 3 | λ) = e−λ^ λ
x 1 x 1! e
− 2 λ (2λ)x^2 x 2! e
− 3 λ (3λ)x^3 x 3! = h(x)e(x^1 +x^2 +x^3 ) log(λ)−^6 λ so that h(x) = 2x^2 3 x^3 /x 1 !x 2 !x 3 !. Thus the natural sufficient statistic is T = X 1 + X 2 + X 3 , and the natural parameter is θ = log(λ), with space R. (Also, ψ(θ) = 6eθ.) (b) What is the distribution of T? Answer: T ∼ P oisson(6λ). (c) Find an unbiased estimator that depends on only T. Is it UMVUE? Why or why not? Answer: E[T ] = 6λ, so that T /6 is unbiased. It is UMVUE, because it is a function of the sufficient statistic T , and the model for T is complete because it is a minimal one-dimensional family with natural parameter space R.
20 points 2. Let X 1 ,... , Xn be iid Double Exponential(θ, 1). (a) Are the order statistics X(1),... , X(n) sufficient? Answer: Yes. (b) Let δ(x 1 ,... , xn) = x 1. Is δ an unbiased estimator of θ? Answer: Yes. (c) Find δ∗(t 1 ,... , tn) = E[δ(X 1 ,... , Xn) | X(1) = t 1 ,... , X(n) = tn].
Answer: We know that given the order statistics, X 1 is equally likely to be any one of them, so that E[X 1 | X(1) = t 1 ,... , X(n) = tn] =^1 n (t 1 + · · · + tn). But the average of the order statistics is the same as the average of the xi’s, which means that δ∗^ = x. (d) Is δ∗^ unbiased? What is its variance? Answer: Yes, since δ is (or because E[X] = θ, the mean of each Xi). The variance is V ar[δ∗] = V ar[X] = σ^2 /n = 2/n. (e) The Fisher Information in one observation is 1. (You don’t have to show that.) What is the Cram´er-Rao Lower Bound (CRLB) for unbiased estimates of θ based on all n observations? Does δ∗^ attain this bound? What is the efficiency of δ∗? Answer: The information in n observations is then In(θ) = n, and the CRLB = 1/n. But V ar[δ∗] = 2/n > 1 /n = CRLB, so δ∗^ does not attain the CRLB, and the efficiency is 1/2. (f) Is δ∗^2 an unbiased estimator of θ^2? If so, why? If not, find an unbiased estimator of θ^2. Answer: No, E[δ∗^2 ] = E[X^2 ] = V ar[X] + E[X]^2 = n^2 + θ^2. So X^2 − (^) n^2 is unbiased.
20 points 3. Suppose X 1 ,... , Xn are iid Unif orm(θ, θ + 1), where θ ∈ R. (Assume n > 2.) (a) Find the likelihood function L(θ; x 1 ,... , xn). Answer: The pdf for Xi is
f (xi | θ) =
{ 1 if θ < xi < θ + 1 0 otherwise , so L(θ; x 1 ,... , xn) = ∏^ n i=
f (xi | θ) =
{ 1 if θ < xi < θ + 1 for all i = 1,... , n 0 otherwise.
(b) Suppose the data are 1. 1 , 1. 3 , 1. 2 , 1 .5.
(i) Sketch the likelihood based on these data. Answer: The likelihood is 1 if θ < 1. 1 , 1. 3 , 1. 2 , 1. 5 < θ + 1, and 0 otherwise. Which means it is 1 only if 0. 5 < θ < 1 .1 (because θ < 1 .1 and 1. 5 < θ + 1).
but P [X(n) − X(1) − n n−+1^1 = 0] = 0.
20 points 4. Now suppose X ∼ P oisson(λ) and Y ∼ P oisson(λ+1), where X and Y are independent, and λ ∈ (0∞). (a) Find the score function S(λ; x, y). Answer: The likelihood this time is L(λ; x, y) = e−λλx^ e−λ−^1 (λ + 1)y, so S(λ; x, y) = (^) ∂λ∂ log(L(λ; x, y)) = (^) ∂λ∂ (− 2 λ − 1 + x log(λ) + y log(λ + 1) = −2 + x λ + (^) λ + 1y.
(b) Find E[S(λ; X, Y )] and V ar[S(λ; X, Y )]. (The mean should be 0.) Answer: E[S(λ; X, Y )] = −2 + λ λ + λ λ^ + 1+ 1 = 0, and V ar[S(λ; X, Y )] = (^) λλ 2 + (^) (λλ + 1)+ 1 2 = λ^1 + (^) λ + 1^1.
(c) X is an unbiased estimator for λ. Find an unbiased estimator that is based on X +Y. Answer: E[X + Y ] = λ + λ + 1, so 12 (X + Y − 1) is unbiased. (d) Is the family complete for (X, Y )? Why or why not? Answer: No, because E[Y − X − 1] = 0, but P [Y − X − 1 = 0] < 1. (e) What is the Fisher Information? What is the CRLB for unbiased estimators of λ? Answer: The Fisher Information is V ar[S(λ; X, Y )] = 1/λ + 1/(λ + 1), hence
CRLB = (^) I(^1 λ) = 1 1 λ +^ λ+1^1
= λ 2 (1 +λ + 1^ λ).
(f) Does either of the estimators in part (c) achieve the CRLB?
Answer: No. V ar[X] = λ, so its efficiency is (1 + λ)/(2λ + 1), which is less than 1; and V ar[^12 (X + Y − 1)] = 14 (2λ + 1), so its efficiency is λ(1 + λ) 2 λ + 1 /(
4 (2λ^ + 1)) =^
4 λ + 4λ^2 4 λ + 4λ^2 + 1 <^1.
20 points 5. Let Y 1 , Y 2 , Y 3 be independent, with
Y 1 ∼ N(α − β, 1). Y 2 ∼ N(α, 1). Y 3 ∼ N(α + β, 1), and (α, β) ∈ R^2. (a) Show that the density of (Y 1 , Y 2 , Y 3 ) can be written as a two-dimensional exponential family. What are the natural sufficient statistics and parameters? Answer: The part in the exponent of the pdf is y 1 (α − β) + y 2 (α) + y 3 (α + β) = α(y 1 + y 2 + y 3 ) + β(y 3 − y 1 ), so the natural sufficient statistic is (y 1 + y 2 + y 3 , y 3 − y 1 ), and parameter is (α, β). (b) Find the expected values of the two sufficient statistics in part (a). Answer: E[Y 1 + Y 2 + Y 3 ] = 3α and E[Y 3 − Y 1 ] = 2β.
(c) Write down the equations needed to solve for the maximum likelihood estimates of α and β, and find the MLE’s. Answer: Equate the expected values of the sufficient statistics with their respective observed values: y 1 + y 2 + y 3 = 3 α̂ and y 3 − y 1 = 2 β.̂ So, α̂ = y^1 +^ y 32 +^ y^3 and β̂ = y^3 − 2 y^1.
(d) Are the MLE’s of α and β unbiased? Are they UMVUE’s? Why or why not? Answer: Yes, they are unbiased (seen easily). They are UMVUE’s, because they are functions of the sufficient statistic, and the model is complete for the sufficient statistic because it is a minimal two-dimensional exponential family, and the parameter space is R^2 , which is open.
