University of Ljubljana Doctoral Programme in Statistics Methodology of Statistical Research Written examination February 16

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(1)University of Ljubljana Doctoral Programme in Statistics Methodology of Statistical Research Written examination February 16th , 2018. ID number:. io n. Instructions. s. Name and surname:. Read carefully the wording of the problem before you start. There are four problems altogeher. You may use a A4 sheet of paper and a mathematical handbook. Please write all the answers on the sheets provided. You have two hours.. a.. b.. c.. So. lu t. Problem 1. 2. 3. 4. Total. •. d. • • • •.

(2) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. 1. (25) Suppose a stratified sample is taken from a population of size N . The strata are of size N1 , N2 , . . . , NK , and the simple random samples are of size n1 , n2 , . . . , nK . Denote by µ the population mean and by σ 2 the population variance for the entire population, and by µk and σk2 the population means and the population variances for the strata. a. (5) Show that 2. σ =. K X. wk σk2. +. k=1. where wk =. Nk N. K X. wk (µk − µ)2. k=1. for k = 1, 2, . . . , K.. Solution: By definition we have Nk K X X. 1 σ2 = N. ! (yki − µ)2. k=1 i=1. where yki is the value for the i-th unit in the k-th stratum. Note that Nk X. (yki − µ)2 =. i=1. =. =. =. Nk X i=1 Nk X i=1 Nk X. (yki − µk + µk − µ)2 Nk Nk X X 2 (yki − µk ) + (µk − µ) + 2(µk − µ) (yki − µ) 2. (yki − µk )2 +. i=1. =. Nk σk2. i=1 Nk X. i=1. (µk − µ)2. i=1 2. + Nk (µk − µ) .. Using this in the above summation gives the result. b. (10) P Let Ȳk be the sample average in the k-th stratum for k = 1, 2, . . . , K and Ȳ = K k=1 wk Ȳk the unbiased estomator of the population mean. The estimators Ȳ1 , . . . , Ȳn are assumed to be independent. To estimate σ 2 we need to estimate the quantity K K X X 2 2 wk (µk − µ) = wk µ2k − µ2 . σb = k=1. k=1. The estimator σ̂b2. =. K X k=1 2. wk Ȳk2 − Ȳ 2.

(3) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. is suggested. Show that E(σ̂b2 ). =. K X. wk (1 − wk )var(Ȳk ) +. k=1. K X. wk µ2k − µ2 .. k=1. Solution: We know that E(Ȳk2 ) = var(Ȳk2 ) + µ2k and E(Ȳ 2 ) = var(Ȳ ) + µ2 . We have E(σ̂b2 ) =. K X. wk var(Ȳk2 ) + µ2k − var(Ȳ ) − µ2 .. k=1. Taking into account that var(Ȳ ) =. K X. wk2 var(Ȳk ). k=1. the result follows. c. (10) Is there an unbiased estimator of σ 2 ? Explain your answer. Solution: We know that 2. σ =. K X. wk σk2. k=1. +. K X. wk (µk − µ)2. k=1. We have unbiased estimators for σk2 . The second term can be estimated by K X k=1. wk Ȳk2 − Ȳ 2 −. K X k=1. wk (1 − wk ). σ̂k2 Nk − nk · . nk Nk − 1. This last term is an unbiased estimator of the second term.. 3.

(4) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. 2. (25) Let X1 , X2 , . . . , Xn be an i.i.d. sample from the inverse Gaussian distribution I(µ, τ ) with density r. τ τ 2 exp − (x − µ) , x > 0, τ > 0, µ > 0. 2πx3 2xµ2 The expectation of the inverse Gaussian distribution is E(X1 ) = µ. a. (10) Find the MLE for (µ, τ ) based on observations x1 , . . . , xn . Solution: The log-likelihood function is `=. n n n n 3X τ X (xk − µ)2 log τ − log 2π − . log xk − 2 2 2 2 k=1 2µ k=1 xk. The partial derivatives are n n 1 X (xk − µ)2 ∂` = − 2 ∂τ 2τ 2µ k=1 xk n. X (xk − µ) ∂` =τ ∂µ µ3 k=1 Equating the partial derivatives to 0 we get n. µ̂ =. 1X xk n k=1. and τ̂ =. µ̂2 1 n. (xk −µ̂)2 k=1 xk. Pn. .. b. (10) Compute the Fisher information matrix I(µ, τ ). Solution: The second derivatives of the log-likelihood function for n = 1 are ∂ 2` 1 = − ∂τ 2 2τ 2 2 ∂ ` µ−x =− 3 ∂τ µ µ 2 ∂ ` τ (−2µ + 3x) =− 2 ∂µ µ4 Replacing x by X and taking expectations we have 1 0 2 2τ I(τ, µ) = 0 µτ3. 4.

(5) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. c. (5) Give a formula for the approximate 95% confidence interval for µ based on x1 , x2 , . . . , xn . Solution: The interval is. √ 2τ µ̂ ± 1.96 · √ . n. 5.

(6) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. 4. (25) Assume the following regression model Yi1 = α + βxi1 + i1 Yi2 = α + βxi2 + i2 for i = 1, 2, . . . , n. Assume that: • E(i1 ) = E(i2 ) = 0, var(i1 ) = var(i2 ) = σ 2 and corr(i1 , i2 ) = ρ ∈ (−1, 1); • the Pn pairs (i1 , i2 ) are uncorrelated. • i=1 xi1 xi2 = 0. Define the following vectors and matrices:       Y11 1 x11 11  Y12  1 x12   12         Y21  1 x21   21        α       β= , Y =  Y22  , X = 1 x22  , η =  22  β  ..   ..   ..   .  .   .        Yn1  1 xn1  n1  Yn2 1 xn2 n2 With the above notation we can write Y = Xβ + η. a. (10) Assume that ρ is known. Let α̂ and β̂ be ordinary least squares estimators of α and β. In other words −1 T α̂ = XT X X Y. β̂ Show that the estimators are unbiased and express their standard errors with quantities Pokažite, da sta cenilki nepristranski, ter izrazite njuni standardni napaki σ 2 , ρ and covariates xij . Solution: Since −1 T −1 T E β̂ = XT X X E (Y) = XT X X Xβ = β , the estimator is unbiased. For the standard error compute −1 T T var β̂ = var X X X Y −1 −1 = XT X XT var (Y) X XT X The variance of Y is of the form σ 2 Σ where Σ is the 2n × 2n matrix having 2 × 2 block matrices of the form 1 ρ , ρ 1 6.

(7) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. on the diagonal and zeros else. Denote n X S= (xk1 + xk2 ) ,. T =. k=1. n X. (x2k1 + x2k2 ) .. k=1. We have 2n S X X= , S T. 2n(1 + ρ) (1 + ρ)S X ΣX = . (1 + ρ)S T P For the second product we took into account that nk=1 xk1 xk2 = 0. Invert the matrix XT X to get T. . T. (XT X)−1 XT ΣX(XT X)−1 1 2n(1 + ρ)T 2 − (1 + 2ρ)S 2 T (1 + ρ)S 3 − 2nST . = (1 + ρ)S 3 − 2nST 4n2 T − 2n(1 + ρ)S 2 (2nT − S 2 )2 Collecting all the terms we get 2n(1 + ρ)T 2 − (1 + 2ρ)S 2 T 2 var α̂ = σ , (2nT − S 2 )2 4n2 T − 2n(1 + ρ)S 2 2 σ . var β̂ = (2nT − S 2 )2 b. (10) Let η̂ be the residuals thet we get for the ordinary least squares. Express the quantity " n # X E ˆ2i1 + ˆ2i2 i=1. with ρ, σ 2 and the elements of the matrix H = X XT X. −1. Solution: We have " E. # ˆ2i1 + ˆ2i2 =. n X i=1. = E η T (I − H) η =. 2n X. (δij − hij )E (ηi ηj ). i,j=1. = σ2. 2n X. (1 − hii ) − 2σ 2 ρ. i=1. n X. h2i−1,2i. i=1. = (2n − 2)σ 2 − σ 2 ρ. n X i=1. Here δij = 1, if i = j and 0 else. 7. h2i−1,2i .. XT ..

(8) Methodology of Statistical Research, 2017/2018, M. Perman. M. Pohar-Perme. c. (5) Suggest estimates for σ 2 and γ = ρσ 2 . Solution: In b. we replace the expected values with random variables. We get a system of linear equations for σ 2 and ρσ 2 . The solutions are the estimators.. 8.

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