Download STAT 9220 Test #1 Solutions: Exponential Families, Confidence Intervals, Independence - Pr and more Exams Statistics in PDF only on Docsity! STAT 9220 Test # 1 Mar 3, 2009 Name: Instructions: Please return for credit by 1 PM Wed, Mar 3rd. Show the justification for your calculations for full credit and round the answers to two decimal points when appropriate. Solve any four problems for total of 100 points. 1. (25 points) Let X1, . . . , Xn be a sequence of iid random variables with an unknown mean ν, unit variance, and a finite third moment. Show that the (a) Lindeberg and (b) Liapunov conditions are satisfied. Derive the asymptotic confidence interval for µ. Answer: Lindeberg’s condition was checked in class. As to Liapunov condition:∑n i=1E|Xi|3 n3/2 = nE|X1|3 n3/2 → 0 The confidence interval for µ based on normal limit theorem is X̄ ± zα√ n . 2. (25 points) Let Pθ = {N (θ, θ2) : θ ∈ R}. (a) Is Pθ an exponential family ? what is its rank ? can it we written as natural exponential family of full rank ? Justify you answer. (b) Find the sufficient statistic for θ. Is it complete ? Justify your answer. Be careful, it not as easy as it seems... Answer: (a) Since dPθ dx (x) = (2π)−1/2 exp ( − x 2 2θ2 + x θ − 1/2− log θ ) hence it is an exponential family of rank one since the dimension of the subspace (θ, θ2) is one. Natural exponential family parametrized by η1 = (2θ)−2, η2 = θ−1 will not be of full rank since η2 is a function of η1. (b) By factorization theorem (−x2, x) is sufficient. But it is not complete, take h(x, y) = x+ y2. STAT 9220 Test # 1 3. (25 points) Show that if the distribution of a positive random variable X is in a scale family, then log(X) is in a location family. Answer: Consider families of probability distributions Pσ which is a scale family. Let B be any Borel set on R Pσ(X ∈ B) = Pσ(X/σ ∈ B/σ) since Pσ is a scale family (1) = P (Z ∈ B/σ) where Z = X/σ is a r.v. free of σ (2) = P (logZ ∈ log(B/σ)) (3) = P (logZ ∈ logB − log σ) (4) = Pσ(logX − log σ ∈ logB − log σ) (5) = Pσ(logX ∈ logB). (6) The result follows from (4) and (6). 4. (25 points) Let X1, . . . , Xn be iid with exponential distribution E(a, 1) i.e., with Lebesgue density f(x) = exp(−(x− a))I(a,+∞)(x). Are ∑ (Xi − a) and X(1) independent ? Justify your answer. Answer: There is no way they are independent. Basu’s theorem doesn’t apply since ∑ (Xi−a) is not a statistic. Consider the related homework problem. We have shown that ∑ (Xi −X(1)) and X(1) are independent. Thus∑ (Xi − a) = ∑ (Xi −X(1)) + n(X(1) − a) and ∑ (Xi − a) and X(1) are correlated, and thus can’t be independent. To complete the argument we perhaps need to argue that Basu’s theorem applies to ∑ (Xi − X(1)) and X(1). But the distribution of ∑ (Xi −X(1)) is clearly uneffected by adding or sub- tracting a constant, hence it is ancilliary. Sufficiency of X(1) follows from factorization theorem, completeness can be shown exactly like in the case of arguing the completeness of X(n) in the family U(0, θ). Page 2
