{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dr. Hackney STA Solutions pg 188

Dr. Hackney STA Solutions pg 188 - 11-16 Solutions Manual...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
11-16 Solutions Manual for Statistical Inference 11.39 Let A i be the set A i = ˆ α, ˆ β : α + ˆ βx 0 i ) - ( α + βx 0 i ) S 1 n + ( x 0 i - ¯ x ) 2 S xx t n - 2 ,α/ 2 m . Then P ( m i =1 A i ) is the probability of simultaneous coverage, and using the Bonferroni In- equality (1.2.10) we have P ( m i =1 A i ) m i =1 P ( A i ) - ( m - 1) = m i =1 1 - α m - ( m - 1) = 1 - α. 11.41 Assume that we have observed data ( y 1 , x 1 ) , ( y 2 , x 2 ) , . . . , ( y n - 1 , x n - 1 ) and we have x n but not y n . Let φ ( y i | x i ) denote the density of Y i , a n( a + bx i , σ 2 ). a. The expected complete-data log likelihood is E n i =1 log φ ( Y i | x i ) = n - 1 i =1 log φ ( y i | x i ) + E log φ ( Y | x n ) , where the expectation is respect to the distribution φ ( y | x n ) with the current values of the parameter estimates. Thus we need to evaluate E log φ ( Y | x n ) = E - 1 2 log(2 πσ 2 1 ) - 1 2 σ 2 1 ( Y - μ 1 ) 2 , where Y n( μ 0 , σ 2 0 ). We have E( Y - μ 1 ) 2 = E([ Y - μ 0 ] + [ μ 0 - μ 1 ]) 2 = σ 2 0 + [ μ 0 - μ 1 ] 2 , since the cross term is zero. Putting this all together, the expected complete-data log likelihood is - n 2 log(2 πσ 2 1 ) - 1 2 σ 2 1 n - 1 i =1 [ y i - ( a 1 + b 1 x i )] 2 - σ 2 0 + [( a 0 + b 0 x n ) - ( a 1 + b 1 x n )] 2 2 σ 2 1 = - n 2 log(2 πσ 2 1 ) - 1 2 σ 2 1 n i =1 [ y i - ( a 1
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern