ch6sol - Chapter 6 Principles of Data Reduction 6.1 By the...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 Principles of Data Reduction 6.1 By the Factorization Theorem, | X | is sufficient because the pdf of X is f ( x | σ 2 ) = 1 2 πσ e - x 2 / 2 σ 2 = 1 2 πσ e -| x | 2 / 2 σ 2 = g ( | x || σ 2 ) · 1 h ( x ) . 6.2 By the Factorization Theorem, T ( X ) = min i ( X i /i ) is sufficient because the joint pdf is f ( x 1 , . . . , x n | θ ) = n i =1 e - x i I ( iθ, + ) ( x i ) = e inθ I ( θ, + ) ( T ( x )) g ( T ( x ) | θ ) · e - Σ i x i h ( x ) . Notice, we use the fact that i > 0, and the fact that all x i s > iθ if and only if min i ( x i /i ) > θ . 6.3 Let x (1) = min i x i . Then the joint pdf is f ( x 1 , . . . , x n | μ, σ ) = n i =1 1 σ e - ( x i - μ ) I ( μ, ) ( x i ) = e μ/σ σ n e - Σ i x i I ( μ, ) ( x (1) ) g ( x (1) , Σ i x i | μ,σ ) · 1 h ( x ) . Thus, by the Factorization Theorem, ( X (1) , i X i ) is a sufficient statistic for ( μ, σ ). 6.4 The joint pdf is n j =1 h ( x j ) c ( θ ) exp k i =1 w i ( θ ) t i ( x j ) = c ( θ ) n exp k i =1 w i ( θ ) n j =1 t i ( x j ) g ( T ( x ) | θ ) · n j =1 h ( x j ) h ( x ) . By the Factorization Theorem, n j =1 t 1 ( X j ) , . . . , n j =1 t k ( X j ) is a sufficient statistic for θ . 6.5 The sample density is given by n i =1 f ( x i | θ ) = n i =1 1 2 I ( - i ( θ - 1) x i i ( θ + 1)) = 1 2 θ n n i =1 1 i I min x i i ≥ - ( θ - 1) I max x i i θ + 1 . Thus (min X i /i, max X i /i ) is sufficient for θ .
Image of page 1

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

View Full Document Right Arrow Icon
6-2 Solutions Manual for Statistical Inference 6.6 The joint pdf is given by f ( x 1 , . . . , x n | α, β ) = n i =1 1 Γ( α ) β α x i α - 1 e - x i = 1 Γ( α ) β α n n i =1 x i α - 1 e - Σ i x i . By the Factorization Theorem, ( n i =1 X i , n i =1 X i ) is sufficient for ( α, β ). 6.7 Let x (1) = min i { x 1 , . . . , x n } , x ( n ) = max i { x 1 , . . . , x n } , y (1) = min i { y 1 , . . . , y n } and y ( n ) = max i { y 1 , . . . , y n } . Then the joint pdf is f ( x , y | θ ) = n i =1 1 ( θ 3 - θ 1 )( θ 4 - θ 2 ) I ( θ 1 3 ) ( x i ) I ( θ 2 4 ) ( y i ) = 1 ( θ 3 - θ 1 )( θ 4 - θ 2 ) n I ( θ 1 , ) ( x (1) ) I ( -∞ 3 ) ( x ( n ) ) I ( θ 2 , ) ( y (1) ) I ( -∞ 4 ) ( y ( n ) ) g ( T ( x ) | θ ) · 1 h ( x ) . By the Factorization Theorem, ( X (1) , X ( n ) , Y (1) , Y ( n ) ) is sufficient for ( θ 1 , θ 2 , θ 3 , θ 4 ). 6.9 Use Theorem 6.2.13. a. f ( x | θ ) f ( y | θ ) = (2 π ) - n/ 2 e - Σ i ( x i - θ ) 2 / 2 (2 π ) - n/ 2 e - Σ i ( y i - θ ) 2 / 2 = exp - 1 2 n i =1 x 2 i - n i =1 y 2 i +2 θn y - ¯ x ) . This is constant as a function of θ if and only if ¯ y = ¯ x ; therefore ¯ X is a minimal sufficient statistic for θ . b. Note, for X location exponential( θ ), the range depends on the parameter. Now f ( x | θ ) f ( y | θ ) = n i =1 ( e - ( x i - θ ) I ( θ, ) ( x i ) ) n i =1 ( e - ( y i - θ ) I ( θ, ) ( y i ) ) = e e - Σ i x i n i =1 I ( θ, ) ( x i ) e e - Σ i y i n i =1 I ( θ, ) ( y i ) = e - Σ i x i I ( θ, ) (min x i ) e - Σ i y i I ( θ, ) (min y i ) . To make the ratio independent of θ we need the ratio of indicator functions independent of θ . This will be the case if and only if min { x 1 , . . . , x n } = min { y 1 , . . . , y n } . So T ( X ) = min { X 1 , . . . , X n } is a minimal sufficient statistic. c. f ( x | θ ) f ( y | θ ) = e - Σ i ( x i - θ ) n i =1 ( 1 + e - ( x i - θ ) ) 2 n i =1 ( 1 + e - ( y i - θ ) ) 2 e - Σ i ( y i - θ ) = e - Σ i ( y i - x i ) n i =1 ( 1 + e - ( y i - θ ) ) n i =1 ( 1 + e - ( x i - θ ) ) 2 . This is constant as a function of θ if and only if x and y have the same order statistics.
Image of page 2
Image of page 3
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