SOLUTIONS%20FINAL%202008

# SOLUTIONS%20FINAL%202008 - 1 2 3 4 5 7 8 9 10 total/70...

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1 2 3 4 5 7 8 9 10 total/70 UNIVERSITY OF TORONTO Faculty of Arts & Science APRIL-MAY EXAMINATIONS 2008 STA 261H1 S Prof. D. Brenner Duration - 3 hours Examination Aids: Non-programmable Calculators Instructions Please show all your work clearly in the space provided to obtain partial credit; you may use the backs of the pages for rough work. There are three parts to the exam: PARTS A, B and C You are to answer: 4 questions in part A, 1 in part B &2 in part C for a total of 7 questions altogether. All complete questions will be valued equally, but partial grades are shown to the left of each part. Tables for the N (0 , 1) , χ 2 ( m ) ,t ( m ) , & F ( m,n ) are appended. Name SOLUTIONS Student number TA (1)

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PART A [Do any 4 questions of the 5 provided in this part] 1. Suppose that T G (3 - 1 ) and we observe t =3 / 2. a) Let ± λ = aT - 1 and Fnd the value of the constant a for which E ± λ = λ . λT d = G (3) E ( λT ) - 1 = Γ(2) / Γ(3) = 1 / 2 (3) E 2 T - 1 = λ so a =2 ± b) Obtain var ± λ . var ± λ = E ± λ 2 - λ 2 = a 2 ET - 2 - λ 2 (3) =4 λ 2 E ( λT ) - 2 - λ 2 = λ 2 ² 4 Γ(1) Γ(3) - 1 ³ = λ 2 ± c) Obtain a 95% conFdence upper bound for the rate parameter λ . 2 λT d = χ 2 (6) (4) 95 / 100 = P ( χ 2 (6) 12 . 6) = P (2 λT 12 . 6) = P ( λ 12 . 6 / 2 T ) UB ( λ ; . 95) = 12 . 6 / 2 t = 12 . 6 / 3 = 4 . 2 ±
2. Suppose that X 1 ,X 2 IID X U (0 ) with observed values x 1 =3 ,x 2 =4 and consider the, so-called, maximum likelilood estimator of θ : ± θ = X (2) = max ( X 1 2 ) . a) Obtain both the exact distribution function, F (2) ( u ), as well as the probability density function, f (2) ( u ), for U (2) = X (2) = ± θ/θ . F (2) ( u )= P ( U (2) u u 2 , 0 <u< 1 (2) Thus f (2) ( u F ± (2) ( u ) = 2 u, 0 1 b) Using the density f (2) ( u ) for U (2) obtained above, derive both the mean and variance of ± θ as these each depend on θ . E ± θ/θ = EU (2) = ² 1 0 2 u 2 du =2 / 3 so E ± θ θ/ 3 ± (4) On the other hand E ( ± θ/θ ) 2 = 2 (2) = ² 1 0 2 u 3 du / 4 = 1 / 2 and so var ± θ = θ 2 ³ 1 2 - 4 9 ´ = θ 2 / 18 ± c) Use the distribution function, F (2) ( u ), to obtain a 99% conFdence upper bound for θ .

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## This note was uploaded on 12/21/2010 for the course STA sta261 taught by Professor Brenner during the Spring '10 term at University of Toronto- Toronto.

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SOLUTIONS%20FINAL%202008 - 1 2 3 4 5 7 8 9 10 total/70...

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