SOLUTIONS%20FINAL%202007

# SOLUTIONS%20FINAL%202007 - SOLUTIONS 2007 STA261 ( c David...

This preview shows pages 1–5. Sign up to view the full content.

SOLUTIONS 2007 STA261 ( c ± David Brenner, 2007, 2010) revised Mar. 8, 2010

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1 2 3 4 5 7 8 9 10 total/70 UNIVERSITY OF TORONTO Faculty of Arts & Science APRIL-MAY EXAMINATIONS 2007 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)
PART A [Do any 4 questions of the 5 provided in this part] 1. Suppose that we have four competing probability functions { f θ | θ =1 , 2 , 3 , 4 } for a certain random variable X distributed on X = { 1 , 2 , 3 , 4 , 5 } f 1 ± 12345 1 / 15 2 / 15 5 / 15 4 / 15 3 / 15 ² ,f 2 ± 1 / 15 4 / 15 3 / 15 2 / 15 5 / 15 ² f 3 ± 6 / 15 2 / 15 2 / 15 2 / 15 3 / 15 ² 4 ± 1 / 15 9 / 15 2 / 15 2 / 15 1 / 15 ² a) For θ = 2, what is the variance of X ?( i.e. What is var θ X ?) (3) NA b) Determine the maximum likelihood estimator as a function ³ θ : X -→ Θ. (2) c) For θ = 2, determine the distance of ³ θ from θ : i.e. d ( ³ θ, θ )= ´ E θ ( ³ θ - θ ) 2 (5)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. Suppose that X N ( μ,σ 2 ) and that P ( X> 2 | 1) = P ( 1) = 2 P ( 2). a) Obtain μ , σ 2 and thus fnd P ( 3) P ( 2 | 1) = P ( 2) /P ( 1) = P ( 1) = 2 P ( 2) (6) P ( 2) = P ( 1) 2 = P ( 1) / 2 P ( 1) = 1 / 2 and P ( 2) = 1 / 4 P ± Z> 1 - μ σ ² = 1 2 and P ± 2 - μ σ ² = 1 4 1 - μ σ = 0 and 2 - μ σ = . 67 μ = 1 and σ =1 /. 67 P ( 3) = P ± 3 - μ σ ² = P ( 1 . 34) . 09 ± b) For what value o± b is P ( X 2 / 2 b ) 1 / 100?
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 12

SOLUTIONS%20FINAL%202007 - SOLUTIONS 2007 STA261 ( c David...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online