Statistics GIDP Ph.D. Qualifying Exam Theory Jan 11, 2016, 9:00am-1:00pm

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1 Statistics GIDP Ph.D. Qualifying Exam Theory Jan, 06, 9:00am-:00pm Instructions: Provide answers on the supplied pads of paper; write on only one side of each sheet. Complete exactly 5 of the 6 problems. Turn in only those sheets you wish to have graded. Stay calm and do your best; good luck.. Hidden inside each box of Primate brand breakfast cereal is a small plastic figure of an animal: an ape or a baboon. Suppose a fraction p of the very large population of cereal boxes contain apes, a fraction q = p contain baboons. Let X be the minimal number of boxes you need to buy to collect both figures. (a) Find P (X = 3). (b) Find E(X).. Two random variables X and Y are jointly normal and marginally standard normal with correlation ρ. (a) Show that X Y = max{x, Y } + max{ X, Y }. (b) Find E(max{X, Y }). 3. Let X,...,X n be independent and identically distribution random variables with mean and finite variance. Show that n(n+) j= jx j in probability.. Let X,...,X n be independent and identically distributed observations from a normal distribution N(θ, θ ), with θ > 0 unknown. (a) Show that T = ( X i, X i ) is sufficient for θ. (b) Find two unbiased estimators based on T that are unbiased for θ. (c) Is T complete? Justify your answer. (d) Suppose that we test H 0 : θ = vs H : θ > by using the rule Reject H 0 if X i > c. Find the value of c to give a test of size α. Assume α > 0 is given. 5. Let X,, X n be independent and identically distributed random variables with the probability density function (pdf) { x exp{ } if x 0 πθ πθ f(x; θ) = 0 if x < 0,

2 where θ > 0 is an unknown parameter. Note that E(X ) = θ, E(X ) = ( π ) θ, E(X ) = ( 3π ) θ. (a) Derive the Cramér-Rao lower bound for variance of an unbiased estimator of θ. (b) Find the maximum likelihood estimator (MLE) for θ. (c) Let Y = πθ X i. Specify the distribution of Y. (d) Construct a 95% confidence interval for θ. 6. Let X,...,X n are independent and identically distributed samples with the pdf f(x θ) = e (x θ), x θ where θ is an unknown parameter. Let X () = min i n X i and X (n) = max i n X i. You may use the fact that X () is a complete statistic in the following questions. (a) Provide with justification a minimum sufficient statistic for θ. (b) Derive the method-of-moment estimator (MME) of θ. Call the estimator ˆθ MME. (c) Derive the maximum likehood estimator (MLE) of θ. Call the estimator ˆθ MLE. (d) Between ˆθ MME and ˆθ MLE, which is a better estimator for θ in terms of the mean squared error (MSE)? Justify your answer. (e) Find an unbiased estimator of θ which is better than both ˆθ MME and ˆθ MLE.

3 Solutions. (a) P (X = 3) = P (aab) + P (bba) = p q + q p = pq. (b) It is easy to see P (X = x) = pq(p x + q x ) for x. So E(X) =.... (a) X Y = max{x, Y } min{x, Y } = max{x, Y } + max{ X, Y }. (b) By symmetry, max{x, Y } and max{ X, Y } have the same distribution. So E(max{X, Y }) = E( X Y ). X Y is normal with mean 0 and variance ρ. E(max{X, Y }) = E( X Y ) = ( ρ)/π. 3. Simply check that the mean goes to and variance goes to 0.. (a) The joint density of X,..., X n is f(x θ) = (θ π) n exp [ θ x i + θ ] x i exp{ n }. By factorization theorem, T = ( X i, X i ) is sufficient for θ. Define T = X i and T = X i. (b) The sample variance is always unbiased for variance. So g (T ) = (X i X) n Secondly, consider X = (T /n). Note that = X i n X n = n [T T n ]. E( X ) = E( X) + V ar( X) = θ + θ /n = n + n θ. So the second unbiased estimator is (c) T is not complete since g (T ) = but g g is not a zero function. (d) Suppose that I were to test ans use a test of the form: n n + X = T n(n + ) E[g (T ) g (T )] = 0, H 0 : θ = vs H : θ > Reject if X i c. Note that X i N(nθ, nθ ). The size of the test P ( X i c) = P ( X i c θ = ) = P ( sup θ= which leads to c = n + z α n. 3 X i n n c n n ) = α,

4 5. (a) Note that s (x, θ) = θ log f(x; θ) = θ + πθ 3 x. Then I (θ) = Var(s ) = π θ 6 Var(X ). ( ) Since V ar(x ) = E(X ) [E(X )] 3π = θ ( ) π θ = π θ, we have I (θ) = π θ 6 π θ = θ. So the Cramér lower variance bound for θ is (b) It is easy to show that ˆθ MLE = (θ) I n (θ) = (θ) n/θ = θ n. nπ x i and ˆθ MLE = nπ x i. (c) Let W = X πθ. Then the pdf of W is f W (w) = πw exp{ w }I(w > 0) = w Γ( I(w > 0), )e w/ which is χ. Then Y = W i χ n. (d) Using Y as a pivotal, then we look for a and b such that P (a Y b) = P (a πθ Xi b) = 95%. Setting a = χ n,0.975 and b = χ n,0.05, where χ n,α is ( α) quantile of χ n distribution. Then 95% CI for θ is [ X i, πχ X i ]. n,0.05 πχ n, (a) Let X = (X,..., X n ). Its joint pdf is f(x θ) = e nθ I(θ X () )e x i, where X () = min i n X i. By factorization theorem, X () is a sufficient statistic. Since it is also complete, it is a minimum sufficient statistic. (b) Let X = E(X ) = θ +, X is the sample mean. Therefore ˆθMME = X. (c) The likelihood function is L(θ) = e nθ I(θ X () )e x i is increasing in θ (, X () ]; and is equal to zero for θ (X (), ). Therefore ˆθ MLE = X ().

5 (d) ˆθ MME is unbiased. V ar(x i ) =, so V ar(ˆθ MME ) = V ar( X) =.So the MSE n of ˆθ MME =. The density of X n () is f X() (x θ) = ne n(θ x) I(x θ), we have E(X () ) = θ + n and V ar(x ()) = n. Therefore, the MSE of ˆθ MLE is n. If n =, then ˆθ MME is better; if n =, two estimators are the same; if n >, the MLE is better. (e) Consider ˆθ = X () n. Its MSE is n, smaller than both ˆθ MME and ˆθ MLE for n >. When n =, ˆθ is better than MLE and is the same as MME. 5