LØSNINGSFORSLAG EXAM IN TMA4295 STATISTICAL INFERENCE Friday 6 June 2008 Time: 09:00 13:00 - PDF

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Norges teknisk naturvitenskapelige universitet Institutt for matematiske fag Side 1 av 6 LØSNINGSFORSLAG EXAM IN TMA4295 STATISTICAL INFERENCE Friday 6 June 2008 Time: 09:00 13:00 Oppgave 1 Let X 1,...,

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Norges teknisk naturvitenskapelige universitet Institutt for matematiske fag Side 1 av 6 LØSNINGSFORSLAG EXAM IN TMA4295 STATISTICAL INFERENCE Friday 6 June 2008 Time: 09:00 13:00 Oppgave 1 Let X 1,..., X n be iid from a beta distribution with parameters (θ, 4θ, i.e. from a distribution with pdf Γ(5θ Γ(θΓ(4θ xθ 1 (1 x 4θ 1, 0 x 1, θ 0. a List at least three different one-dimensional sufficient statistics. Solution. The likelihood function is where f(x θ = ( ( n Γ(5θ n Γ(θΓ(4θ Therefore (Factorization theorem θ ( n 1 x i (1 x i 4 x i (1 x i = g(t (x; θh(x, T (x = n x i (1 x i 4, ( n Γ(5θ g(t; θ = t θ, Γ(θΓ(4θ ( n 1 h(x = x i (1 x i. T (X = n X i (1 X i 4 Side 2 av 6 is a (one-dimensional sufficient statistic. Any one-to-one transformation of a sufficient statistic is also a sufficient statistic, therefore, for example, each is a sufficient statistic. T k (X = [T (X] k, k = 2, 3,..., b Can the method of moments estimator (MME of θ be found using the first moment (expectation? Find MME using the second moment. Solution. The first moment is 1/5 independently of θ, therefore it cannot be used. Let µ 2 be the second moment. Then or Thus MME is where µ 2 = θ = ˆθ = 4 25(5θ (25µ (25m , m 2 = 1 n Xi 2. Oppgave 2 Let X 1,..., X n be iid from a geometric distribution with parameter θ, i.e. from a distribution with pmf θ(1 θ x 1, x = 1, 2,...; 0 θ 1. a Find the maximum likelihood estimator ˆτ 1 of τ(θ = 1/θ. Solution. MLE of θ is solution of the equation that is ln f(x θ θ = n θ n X i n 1 θ ˆθ MLE = 1 X. = 0 Therefore (the invariance property of MLE ˆτ 1 = X. Side 3 av 6 b Find the asymptotic variance of the estimator ˆτ 1. Solution. The asymptotic variance is (see theory v(θ = [τ (θ] 2 I 0 (θ, where I 0 (θ is the Fisher information of one observation. I 0 (θ = E 2 ln f(x i θ θ 2 = 1 θ 2 (1 θ. Therefore, since τ (θ = 1/θ 2, the asymptotic variance is v(θ = 1 θ θ 2. c Find asymptotic (1 α maximum likelihood confidence interval for θ. Solution. Since the variance is estimated by Then 1 α P ( z α/2 ˆθ θ V arˆθ n(ˆθ θ ( D 1 N 0,, I 0 (θ V arˆθ = 1 ni 0 (ˆθ = ˆθ 2 (1 ˆθ n = X 1 n X 3. ( 1 X 1 z α/2 = P X z α/2 n X θ 1 X X 1 + z 3 α/2 n X. 3 Thus [ 1 X 1 X z α/2 n X, 1 X ] X 1 + z 3 α/2 n X 3 is an asymptotic (1 α maximum likelihood confidence interval for θ. d Suppose that the first ten observations and each even observation are lost, and τ(θ = 1/θ is estimated by n/2 2 ˆτ 2 = X 2i 1 n 10 i=6 Side 4 av 6 (assume for simplicity that the sample size n is always even. Find the asymptotic efficiency of ˆτ 2 (that is asymptotic relative efficiency of ˆτ 2 with respect to asymptotically efficient estimator ˆτ 1, for which all observations are used. Solution. Since n 10 D (ˆτ 2 τ(θ N 2 and since we have n(ˆτ2 τ(θ n n 10 2 D N Therefore the asymptotic efficiency of ˆτ 2 is 1/2. 2, (0, [τ (θ] 2 I 0 (θ (0, 2[τ (θ] 2 I 0 (θ., Oppgave 3 Let X 1,..., X n be a random sample drawn from a Poisson distribution with parameter θ. a Show that for testing H 0 : θ θ 0 versus H 1 : θ θ 0, the rejection region of a uniformly most powerful test has form { } R = x : x i c. Let α be the significance level. Find (approximately c if n is large enough so that the Central Limit Theorem can be used. Solution. The likelihood function is therefore, if θ θ, then the ratio L(θ; X = e nθ θ P X i ( Xi! 1, ( L(θ ; X L(θ ; X = en(θ θ θ P X i θ is a monotone (decreasing function of T (X = X i. Therefore the rejection region of UMP test has form { } R = x : x i c. Side 5 av 6 To find c let us use CLT. We have EX i = θ, V ar(x i = θ therefore and α = P θ0 ( X i c = P θ0 ( Xi nθ 0 nθ0 c nθ 0 nθ0 1 Φ Thus the hypothesis is accepted if c = nθ 0 + nθ 0 z α. ( c nθ0 nθ0 or X i nθ 0 + nθ 0 z α X θ 0 + θ0 n z α. b Prove that the test of part (a is unbiased. Solution. Prove the following. Let Y 1 Poisson(λ 1, Y 2 Poisson(λ 2, and λ 1 λ 2. Then P (Y 2 c P (Y 1 c for any c 0. Indeed, consider Y 3 independent on Y 1 and such that Y 3 Poisson(λ 2 λ 1. Then Y 1 + Y 3 Poisson(λ 2, and Now part (b follows from the fact that P (Y 2 c = P (Y 1 + Y 3 c P (Y 1 c. X i Poisson(nθ. Less strong but also valid solution is based on the normal approximation (see solution of part (c. c Find (approximately and plot the power function π(θ of the test of part (a. Find, in particular, lim θ 0 π(θ, π(θ 0 and lim θ π(θ Solution. π(θ = P θ ( X i c = P θ ( Xi nθ nθ ( n(θ0 θ + nθ 0 z α 1 Φ. nθ n(θ 0 θ + nθ 0 z α nθ Side 6 av 6 Simple analysis shows that lim π(θ = 0, θ 0 π(θ 0 = α and lim π(θ = 1. θ d Find the (1 α one-sided confidence interval that results from inverting the test of part (a. Solution. Inverting the test of part (a, i.e. solving the inequality θ X θ + n z α with respect to θ, we obtain the following (1 α one-sided confidence interval: [ ( 1 z 2 α 4 n + 4 X z 2 α,. n
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