. Problem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . . Solution . . . 5 / 21 . Example from Stigler (1972) Am. Stat. Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N . Is T ( X ) = X a complete statistic? Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ ∈ N . Note that f X ( x ) = 1 θ I ( x ∈ { 1 , · · · , θ } ) = 1 θ I N θ ( x ) . θ θ 1 θ g ( x ) = 1 ∑ ∑ E [ g ( T ) | θ ] = E [ g ( X ) | θ ] = g ( x ) = 0 θ x =1 x =1 θ ∑ g ( x ) = 0 x =1
• if • . • if g k . . k , x g x g . . . Therefore, g x for all x , and T X X is a complete statistic for . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 g k g . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary g Solution (cont’d) , x g x g 6 / 21 . . . . . . . . . for all θ ∈ N , which implies • if θ = 1 , ∑ θ x =1 g ( x ) = g (1) = 0
• . • if . k , x g x g g k g k . . for all x , and T X X is a complete statistic for . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 Therefore, g x . . Complete Statistics . . . . . . . . . . . . . . Basu’s Theorem . Summary Solution (cont’d) 6 / 21 . . . . . . . . . for all θ ∈ N , which implies • if θ = 1 , ∑ θ x =1 g ( x ) = g (1) = 0 • if θ = 2 , ∑ θ x =1 g ( x ) = g (1) + g (2) = g (2) = 0 .
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . X is a complete statistic for , and T X for all x Therefore, g x . . 6 / 21 . Solution (cont’d) . . Summary . . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . for all θ ∈ N , which implies • if θ = 1 , ∑ θ x =1 g ( x ) = g (1) = 0 • if θ = 2 , ∑ θ x =1 g ( x ) = g (1) + g (2) = g (2) = 0 . • . • if θ = k , ∑ θ x =1 g ( x ) = g (1) + · · · + g ( k − 1) = g ( k ) = 0 .
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . . Solution (cont’d) 6 / 21 . Basu’s Theorem . . . . . . . . . . . . . . Complete Statistics . . . . . . . . . for all θ ∈ N , which implies • if θ = 1 , ∑ θ x =1 g ( x ) = g (1) = 0 • if θ = 2 , ∑ θ x =1 g ( x ) = g (1) + g (2) = g (2) = 0 . • . • if θ = k , ∑ θ x =1 g ( x ) = g (1) + · · · + g ( k − 1) = g ( k ) = 0 . Therefore, g ( x ) = 0 for all x ∈ N , and T ( X ) = X is a complete statistic for θ ∈ Ω = N .
. x . . Define a nonzero g x as follows g x x n x n otherwise E g T g x . n n Because does not include n , g x for all n , and T X X is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Is the previous example barely complete? . Modified Problem . . Is T X X a complete statistic? . Solution . . 7 / 21 . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N − { n } .
. g x . Define a nonzero g x as follows g x x n x n otherwise E g T x n . n Because does not include n , g x for all n , and T X X is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . Is the previous example barely complete? . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary . . Modified Problem . . . Solution . . . 7 / 21 . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N − { n } . Is T ( X ) = X a complete statistic?
. Because . otherwise E g T x g x n n does not include n , g x . for all n , and T X X is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . 7 / 21 Solution . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Is the previous example barely complete? . Modified Problem . . . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N − { n } . Is T ( X ) = X a complete statistic? Define a nonzero g ( x ) as follows 1 x = n g ( x ) = − 1 x = n + 1 0
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang X is not a complete statistic. T X n , and for all does not include n , g x Because otherwise . . . Solution 7 / 21 . . . Modified Problem . Summary . . Basu’s Theorem . . Is the previous example barely complete? . . . . . . . . . . Complete Statistics . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N − { n } . Is T ( X ) = X a complete statistic? Define a nonzero g ( x ) as follows 1 x = n g ( x ) = − 1 x = n + 1 0 { 0 θ 1 θ ̸ = n ∑ E [ g ( T ) | θ ] = g ( x ) = 1 θ = n θ θ x =1
. Modified Problem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang otherwise . . . Solution . . . 7 / 21 . Complete Statistics . . . Is the previous example barely complete? . . . . . . . . . . . . Summary Basu’s Theorem . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N − { n } . Is T ( X ) = X a complete statistic? Define a nonzero g ( x ) as follows 1 x = n g ( x ) = − 1 x = n + 1 0 { 0 θ 1 θ ̸ = n ∑ E [ g ( T ) | θ ] = g ( x ) = 1 θ = n θ θ x =1 Because Ω does not include n , g ( x ) = 0 for all θ ∈ Ω = N − { n } , and T ( X ) = X is not a complete statistic.
• We know that R X n • Define g T X n • Then E g T . . . X is an ancillary statistic, which do not depend on . . . X E R . Note that E R is constant to . E R E R , so T is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . Last Lecture : Ancillary and Complete Statistics . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary . . Problem . . i.i.d. . A Simple Proof . 8 / 21 . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • Is T ( X ) = ( X (1) , X ( n ) ) a complete statistic?
• Define g T X n • Then E g T . E R . Note that E R is constant to A Simple Proof . . X . . E R E R , so T is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . 8 / 21 Summary Last Lecture : Ancillary and Complete Statistics . . . . . . . . . . Complete Statistics Basu’s Theorem . i.i.d. . . . . Problem . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • Is T ( X ) = ( X (1) , X ( n ) ) a complete statistic? • We know that R = X ( n ) − X (1) is an ancillary statistic, which do not depend on θ .
• Then E g T . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang , so T is not a complete statistic. E R E R . . A Simple Proof . . i.i.d. . Problem . . . . . . . . . . . . . . Complete Statistics 8 / 21 Basu’s Theorem . Summary Last Lecture : Ancillary and Complete Statistics . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • Is T ( X ) = ( X (1) , X ( n ) ) a complete statistic? • We know that R = X ( n ) − X (1) is an ancillary statistic, which do not depend on θ . • Define g ( T ) = X ( n ) − X (1) − E ( R ) . Note that E ( R ) is constant to θ .
. Problem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . A Simple Proof . i.i.d. . . . . Complete Statistics . . . . Last Lecture : Ancillary and Complete Statistics . . . . . . . . . . Basu’s Theorem . Summary 8 / 21 . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • Is T ( X ) = ( X (1) , X ( n ) ) a complete statistic? • We know that R = X ( n ) − X (1) is an ancillary statistic, which do not depend on θ . • Define g ( T ) = X ( n ) − X (1) − E ( R ) . Note that E ( R ) is constant to θ . • Then E [ g ( T ) | θ ] = E ( R ) − E ( R ) = 0 , so T is not a complete statistic.
. Define g T . . . . . . . r T Proof E r T , which does not depend on the parameter because r T is ancillary. Then E g T for a non-zero function g T , and T X is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . Basu’s Theorem . . . . . . . . . . . . . . Complete Statistics . Summary Useful Fact 1 : Ancillary and Complete Statistics . Fact . . 9 / 21 . . . . . . . . . For a statistic T ( X ) , If a non-constant function of T , say r ( T ) is ancillary, then T ( X ) cannot be complete
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang g T , and T X is not a complete statistic. for a non-zero function Then E g T . . Proof . . . Fact . . . . . . . . . . . . . . . Complete Statistics 9 / 21 Basu’s Theorem . Summary Useful Fact 1 : Ancillary and Complete Statistics . . . . . . . . . For a statistic T ( X ) , If a non-constant function of T , say r ( T ) is ancillary, then T ( X ) cannot be complete Define g ( T ) = r ( T ) − E [ r ( T )] , which does not depend on the parameter θ because r ( T ) is ancillary.
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . Proof . . . Fact . Useful Fact 1 : Ancillary and Complete Statistics Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . 9 / 21 . . . . . . . . . For a statistic T ( X ) , If a non-constant function of T , say r ( T ) is ancillary, then T ( X ) cannot be complete Define g ( T ) = r ( T ) − E [ r ( T )] , which does not depend on the parameter θ because r ( T ) is ancillary. Then E [ g ( T ) | θ ] = 0 for a non-zero function g ( T ) , and T ( X ) is not a complete statistic.
. , then E g . . . E g T E g r T Assume that E g T for all r T . holds for all too. Because T X is a complete statistic, Pr g r T . Therefore Pr g T , and T is a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Useful Fact 2 : Arbitrary Function of Complete Statistics . . Fact . . complete. . Proof . 10 / 21 . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also
. holds for all . . Assume that E g T for all , then E g r T too. Because T X is a complete statistic, Pr g . r T . Therefore Pr g T , and T is a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . Proof complete. . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary Useful Fact 2 : Arbitrary Function of Complete Statistics . Fact . . 10 / 21 . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also E [ g ( T ∗ ) | θ ] = E [ g ◦ r ( T ) | θ ]
. too. Because T X is a complete statistic, Pr g Proof . . then E g r T holds for all r T complete. . Therefore Pr g T , and T is a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . 10 / 21 . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary . Useful Fact 2 : Arbitrary Function of Complete Statistics . Fact . . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also E [ g ( T ∗ ) | θ ] = E [ g ◦ r ( T ) | θ ] Assume that E [ g ( T ∗ ) | θ ] = 0 for all θ ,
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang , and T is a complete statistic. Therefore Pr g T . r T Because T X is a complete statistic, Pr g too. . . Proof . complete. . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem 10 / 21 Useful Fact 2 : Arbitrary Function of Complete Statistics Summary . Fact . . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also E [ g ( T ∗ ) | θ ] = E [ g ◦ r ( T ) | θ ] Assume that E [ g ( T ∗ ) | θ ] = 0 for all θ , then E [ g ◦ r ( T ) | θ ] = 0 holds for all θ
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang , and T is a complete statistic. Therefore Pr g T . r T Pr g . . Proof . . complete. . Basu’s Theorem . . . . . . . . . . . . . . Complete Statistics 10 / 21 . Summary Useful Fact 2 : Arbitrary Function of Complete Statistics . Fact . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also E [ g ( T ∗ ) | θ ] = E [ g ◦ r ( T ) | θ ] Assume that E [ g ( T ∗ ) | θ ] = 0 for all θ , then E [ g ◦ r ( T ) | θ ] = 0 holds for all θ too. Because T ( X ) is a complete statistic,
. Fact January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . Proof . complete. . . . . . . . . . . . . . . . . . Useful Fact 2 : Arbitrary Function of Complete Statistics . 10 / 21 Complete Statistics Basu’s Theorem . Summary . . . . . . . . . If T ( X ) is a complete statistic, then a function of T , say T ∗ = r ( T ) is also E [ g ( T ∗ ) | θ ] = E [ g ◦ r ( T ) | θ ] Assume that E [ g ( T ∗ ) | θ ] = 0 for all θ , then E [ g ◦ r ( T ) | θ ] = 0 holds for all θ too. Because T ( X ) is a complete statistic, Pr [ g ◦ r ( T ) = 0] = 1 , ∀ θ ∈ Ω . Therefore Pr [ g ( T ∗ ) = 0] = 1 , and T ∗ is a complete statistic.
. . . . . Any complete, and sufficient statistic is also a minimal sufficient statistic . The converse is NOT true . . . . . . . . A minimal sufficient statistic is not necessarily complete. (Recall the example in the last lecture). Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Theorem 6.2.28 - Lehman and Schefle (1950) . . The textbook version . . If a minimal sufficient statistic exists, then any complete statistic is also a minimal sufficient statistic. . Paraphrased version . 11 / 21 . . . . . . . . .
. . Any complete, and sufficient statistic is also a minimal sufficient statistic . The converse is NOT true . . . . . . . . A minimal sufficient statistic is not necessarily complete. (Recall the example in the last lecture). Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . Paraphrased version . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Theorem 6.2.28 - Lehman and Schefle (1950) . The textbook version . . If a minimal sufficient statistic exists, then any complete statistic is also a minimal sufficient statistic. 11 / 21 . . . . . . . . .
. minimal sufficient statistic. January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang example in the last lecture). A minimal sufficient statistic is not necessarily complete. (Recall the . . The converse is NOT true . Any complete, and sufficient statistic is also a minimal sufficient statistic . . Paraphrased version . If a minimal sufficient statistic exists, then any complete statistic is also a . Complete Statistics . . . . . . . . . . . . . . Basu’s Theorem . . Summary Theorem 6.2.28 - Lehman and Schefle (1950) . The textbook version . 11 / 21 . . . . . . . . .
. s Suppose that S X is an ancillary statistic. We want to show that Pr S X s T X t Pr S X s t Alternatively, we can show that Pr T X t S X Pr T X . t Pr T X t S X s Pr T X t Pr S X s Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Basu’s Theorem . Theorem 6.2.24 . . every ancillary statistic. . Proof strategy - for discrete case . . . . 12 / 21 . . . . . . . . . If T ( X ) is a complete sufficient statistic, then T ( X ) is independent of
. t Alternatively, we can show that Pr T X t S X s Pr T X t Pr T X S X . s Pr T X t Pr S X s Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . . Proof strategy - for discrete case . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Basu’s Theorem . Theorem 6.2.24 . . every ancillary statistic. 12 / 21 . . . . . . . . . If T ( X ) is a complete sufficient statistic, then T ( X ) is independent of Suppose that S ( X ) is an ancillary statistic. We want to show that Pr ( S ( X ) = s | T ( X ) = t ) = Pr ( S ( X ) = s ) , ∀ t ∈ T
. S X . . . Alternatively, we can show that Pr T X t s every ancillary statistic. Pr T X t Pr S X s Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . Proof strategy - for discrete case 12 / 21 . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Basu’s Theorem . Theorem 6.2.24 . . . . . . . . . . If T ( X ) is a complete sufficient statistic, then T ( X ) is independent of Suppose that S ( X ) is an ancillary statistic. We want to show that Pr ( S ( X ) = s | T ( X ) = t ) = Pr ( S ( X ) = s ) , ∀ t ∈ T Pr ( T ( X ) = t | S ( X ) = s ) = Pr ( T ( X ) = t )
. Theorem 6.2.24 January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang Alternatively, we can show that . . Proof strategy - for discrete case . . . . every ancillary statistic. . . . . . . . . . . . . . . Basu’s Theorem . Complete Statistics Basu’s Theorem . Summary 12 / 21 . . . . . . . . . If T ( X ) is a complete sufficient statistic, then T ( X ) is independent of Suppose that S ( X ) is an ancillary statistic. We want to show that Pr ( S ( X ) = s | T ( X ) = t ) = Pr ( S ( X ) = s ) , ∀ t ∈ T Pr ( T ( X ) = t | S ( X ) = s ) = Pr ( T ( X ) = t ) Pr ( T ( X ) = t ∧ S ( X ) = s ) = Pr ( T ( X ) = t ) Pr ( S ( X ) = s )
• As T X is sufficient, by definition, f X X T X • Because S X is a function of X , Pr S X T X • We need to show that . of . Pr S X s T X Pr S X t s t . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 is also independent . . is independent of Proof of Basu’s Theorem Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . 13 / 21 . . . . . . . . . • As S ( X ) is ancillary, by definition, it does not depend on θ .
• Because S X is a function of X , Pr S X T X • We need to show that is also independent January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . t s Pr S X t s T X Pr S X . of . . Proof of Basu’s Theorem Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . 13 / 21 . . . . . . . . . • As S ( X ) is ancillary, by definition, it does not depend on θ . • As T ( X ) is sufficient, by definition, f X ( X | T ( X )) is independent of θ .
• We need to show that . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . t s Pr S X t s T X Pr S X 13 / 21 Proof of Basu’s Theorem Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . . . . . • As S ( X ) is ancillary, by definition, it does not depend on θ . • As T ( X ) is sufficient, by definition, f X ( X | T ( X )) is independent of θ . • Because S ( X ) is a function of X , Pr ( S ( X ) | T ( X )) is also independent of θ .
. Basu’s Theorem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang Proof of Basu’s Theorem . . Summary Complete Statistics . . . . . . . . . . . . . . 13 / 21 . . . . . . . . . • As S ( X ) is ancillary, by definition, it does not depend on θ . • As T ( X ) is sufficient, by definition, f X ( X | T ( X )) is independent of θ . • Because S ( X ) is a function of X , Pr ( S ( X ) | T ( X )) is also independent of θ . • We need to show that Pr ( S ( X ) = s | T ( X ) = t ) = Pr ( S ( X ) = s ) , ∀ t ∈ T .
. s Define g t Pr S X s T X t Pr S X s . Taking (1)-(3), t Pr S X s T X t Pr S X Pr T X t t t g t Pr T X t E g T X T X is complete, so g t almost surely for all possible t . Therefore, S X is independent of T X . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 (3) s Pr T X . Pr S X . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Proof of Basu’s Theorem (cont’d) 14 / 21 Pr S X s Pr S X s t Pr T X t (2) t . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T
. Pr T X Pr S X s T X t Pr S X s . Taking (1)-(3), t Pr S X s T X t Pr S X s t (3) t g t Pr T X t E g T X T X is complete, so g t almost surely for all possible t . Therefore, S X is independent of T X . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 Define g t t . s Pr T X . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Proof of Basu’s Theorem (cont’d) 14 / 21 t (2) Pr S X . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T
. Pr T X Pr S X s T X t Pr S X s . Taking (1)-(3), t Pr S X s T X t Pr S X s t (3) t g t Pr T X t E g T X T X is complete, so g t almost surely for all possible t . Therefore, S X is independent of T X . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 Define g t 14 / 21 . . . (2) . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Proof of Basu’s Theorem (cont’d) . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T
. t Taking (1)-(3), t Pr S X s T X t Pr S X s Pr T X t g t Pr T X . t E g T X T X is complete, so g t almost surely for all possible t . Therefore, S X is independent of T X . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 (3) 14 / 21 (2) . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Proof of Basu’s Theorem (cont’d) . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T Define g ( t ) = Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s ) .
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang Therefore, S X is independent of T X . . almost surely for all possible t T X is complete, so g t E g T X t g t Pr T X t (3) (2) 14 / 21 Summary . Proof of Basu’s Theorem (cont’d) Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T Define g ( t ) = Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s ) . Taking (1)-(3), ∑ [ Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s )] Pr ( T ( X ) = t | θ ) = 0 t ∈T
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang Therefore, S X is independent of T X . . almost surely for all possible t T X is complete, so g t (3) (2) 14 / 21 Proof of Basu’s Theorem (cont’d) . . . . . . . . . . . . Basu’s Theorem . . . Summary Complete Statistics . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T Define g ( t ) = Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s ) . Taking (1)-(3), ∑ [ Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s )] Pr ( T ( X ) = t | θ ) = 0 t ∈T ∑ g ( t ) Pr ( T ( X ) = t | θ ) = E [ g ( T ( X )) | θ ] = 0 t ∈T
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang Therefore, S X is independent of T X . (3) (2) . Proof of Basu’s Theorem (cont’d) 14 / 21 . Basu’s Theorem . . . . . . . . . . . . . . Complete Statistics . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T Define g ( t ) = Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s ) . Taking (1)-(3), ∑ [ Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s )] Pr ( T ( X ) = t | θ ) = 0 t ∈T ∑ g ( t ) Pr ( T ( X ) = t | θ ) = E [ g ( T ( X )) | θ ] = 0 t ∈T T ( X ) is complete, so g ( t ) = 0 almost surely for all possible t ∈ T .
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang (3) (2) . Proof of Basu’s Theorem (cont’d) 14 / 21 . . . . . . . . . . . . . . Complete Statistics . Basu’s Theorem . . . . . . . . . ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s | T ( X ) = t ) Pr ( T ( X ) = t | θ ) (1) t ∈T ∑ Pr ( S ( X ) = s | θ ) = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T ∑ = Pr ( S ( X ) = s ) Pr ( T ( X ) = t | θ ) t ∈T Define g ( t ) = Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s ) . Taking (1)-(3), ∑ [ Pr ( S ( X ) = s | T ( X ) = t ) − Pr ( S ( X ) = s )] Pr ( T ( X ) = t | θ ) = 0 t ∈T ∑ g ( t ) Pr ( T ( X ) = t | θ ) = E [ g ( T ( X )) | θ ] = 0 t ∈T T ( X ) is complete, so g ( t ) = 0 almost surely for all possible t ∈ T . Therefore, S ( X ) is independent of T ( X ) .
• Calculate E X n X n • We know that X n is sufficient statistic. • We know that X n is complete, too. • We can easily show that X X n is an ancillary statistic. • Then we can leverage Basu’s Theorem for the calculation. . . A strategy for the solution . . . . . . X . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . and E X . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Application of Basu’s Theorem . Problem . . i.i.d. X 15 / 21 . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) .
• We know that X n is sufficient statistic. • We know that X n is complete, too. • We can easily show that X X n is an ancillary statistic. • Then we can leverage Basu’s Theorem for the calculation. . . . A strategy for the solution . . . . . . . . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 and E 15 / 21 . . . . . . . . . . . Basu’s Theorem . Summary Application of Basu’s Theorem Problem . . i.i.d. . . . . Complete Statistics . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) . [ X (1) ] [ X (1) + X (2) ] • Calculate E X ( n ) X ( n )
• We know that X n is complete, too. • We can easily show that X X n is an ancillary statistic. • Then we can leverage Basu’s Theorem for the calculation. . i.i.d. January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . A strategy for the solution . and E . 15 / 21 . Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Application of Basu’s Theorem Problem . . . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) . [ X (1) ] [ X (1) + X (2) ] • Calculate E X ( n ) X ( n ) • We know that X ( n ) is sufficient statistic.
• We can easily show that X X n is an ancillary statistic. • Then we can leverage Basu’s Theorem for the calculation. . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . A strategy for the solution . and E . i.i.d. 15 / 21 . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary Application of Basu’s Theorem . Problem . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) . [ X (1) ] [ X (1) + X (2) ] • Calculate E X ( n ) X ( n ) • We know that X ( n ) is sufficient statistic. • We know that X ( n ) is complete, too.
• Then we can leverage Basu’s Theorem for the calculation. . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . A strategy for the solution . and E . i.i.d. . 15 / 21 Problem . . . . . . . . . . . Complete Statistics Basu’s Theorem . . . . Summary Application of Basu’s Theorem . . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) . [ X (1) ] [ X (1) + X (2) ] • Calculate E X ( n ) X ( n ) • We know that X ( n ) is sufficient statistic. • We know that X ( n ) is complete, too. • We can easily show that X (1) / X ( n ) is an ancillary statistic.
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . . A strategy for the solution . and E . i.i.d. . 15 / 21 Problem . . . . . . . . . . . Complete Statistics . . Basu’s Theorem . . . Summary Application of Basu’s Theorem . . . . . . . . . • X 1 , · · · , X n ∼ Uniform (0 , θ ) . [ X (1) ] [ X (1) + X (2) ] • Calculate E X ( n ) X ( n ) • We know that X ( n ) is sufficient statistic. • We know that X ( n ) is complete, too. • We can easily show that X (1) / X ( n ) is an ancillary statistic. • Then we can leverage Basu’s Theorem for the calculation.
f Y y X n Y n Y n does not depend on X n is I y X Y . Because the distribution of Y Uniform , X an ancillary statistic for . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 . , and Y . , then dx dy . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Let y x 16 / 21 . . . . . . . . . Showing that X (1) / X ( n ) is Ancillary 1 f X ( x | θ ) = θ I (0 < x < θ )
f Y y X n Y n Y n does not depend on X n is . y X Y , X Because the distribution of Y an ancillary statistic for . Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 I 16 / 21 . Complete Statistics . . . . . . . . . . . . . . Basu’s Theorem . Summary . . . . . . . . . Showing that X (1) / X ( n ) is Ancillary 1 f X ( x | θ ) = θ I (0 < x < θ ) Let y = x / θ , then | dx / dy | = θ , and Y ∼ Uniform (0 , 1) .
X n Y n Y n does not depend on X n is . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . an ancillary statistic for , X Because the distribution of Y Y X 16 / 21 . . . . . Summary Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . Showing that X (1) / X ( n ) is Ancillary 1 f X ( x | θ ) = θ I (0 < x < θ ) Let y = x / θ , then | dx / dy | = θ , and Y ∼ Uniform (0 , 1) . f Y ( y | θ ) = I (0 < y < 1)
Y n does not depend on X n is . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . an ancillary statistic for , X Because the distribution of Y 16 / 21 Summary . . . . . . . . . . . . . . Complete Statistics . Basu’s Theorem . . . . . . . . . Showing that X (1) / X ( n ) is Ancillary 1 f X ( x | θ ) = θ I (0 < x < θ ) Let y = x / θ , then | dx / dy | = θ , and Y ∼ Uniform (0 , 1) . f Y ( y | θ ) = I (0 < y < 1) X (1) Y (1) = X ( n ) Y ( n )
. Basu’s Theorem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . Summary . 16 / 21 Complete Statistics . . . . . . . . . . . . . . . . . . . . . . . Showing that X (1) / X ( n ) is Ancillary 1 f X ( x | θ ) = θ I (0 < x < θ ) Let y = x / θ , then | dx / dy | = θ , and Y ∼ Uniform (0 , 1) . f Y ( y | θ ) = I (0 < y < 1) X (1) Y (1) = X ( n ) Y ( n ) Because the distribution of Y 1 , · · · , Y n does not depend on θ , X (1) / X ( n ) is an ancillary statistic for θ .
• If X and Y are independent, E XY X n E X n X n X n X n E X n E Y n E Y n E Y E Y E X X E Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 X E . . X . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Applying Basu’s Theorem E X E Y . E X E 17 / 21 . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) .
X n E X n X n X n X n E X n E Y n E Y n E Y E Y E X Hyun Min Kang X E Biostatistics 602 - Lecture 07 January 29th, 2013 X E . . X . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Applying Basu’s Theorem E X E 17 / 21 . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) .
E X n X n X n E X n E Y n E Y n X E X . E X . E Y E Y Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 E 17 / 21 . . . . . . . . . . . . E . . Applying Basu’s Theorem Summary . Basu’s Theorem Complete Statistics . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] E [ X (1) ] = X ( n ) X ( n )
X n E X n E Y n E Y n . E January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang E Y E Y E X X E E . 17 / 21 Applying Basu’s Theorem . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] [ X (1) ] [ ] E [ X (1) ] = = E X ( n ) X ( n ) X ( n ) X ( n )
E X n E Y n E Y n . E January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang E Y E Y E X E E . 17 / 21 Basu’s Theorem . Complete Statistics Summary . . . . . . . . . . . Applying Basu’s Theorem . . . . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] [ X (1) ] [ ] E [ X (1) ] = = E X ( n ) X ( n ) X ( n ) X ( n ) [ X (1) ] X ( n )
E Y n E Y n . E January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang E Y E Y E E . 17 / 21 Basu’s Theorem . Complete Statistics . . . . . . . . . . . Summary . Applying Basu’s Theorem . . . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] [ X (1) ] [ ] E [ X (1) ] = = E X ( n ) X ( n ) X ( n ) X ( n ) E [ X (1) ] [ X (1) ] = E [ X ( n ) ] X ( n )
E Y n . Applying Basu’s Theorem January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang E Y E E . E 17 / 21 Summary . . . . . . . . . . . . . . . Basu’s Theorem Complete Statistics . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] [ X (1) ] [ ] E [ X (1) ] = = E X ( n ) X ( n ) X ( n ) X ( n ) E [ X (1) ] [ X (1) ] = E [ X ( n ) ] X ( n ) E [ θ Y (1) ] = E [ θ Y ( n ) ]
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang E E . E Applying Basu’s Theorem 17 / 21 . . . . . . . . . . . . . . . Basu’s Theorem Complete Statistics . . . . . . . . . • By Basu’s Theorem, X (1) / X ( n ) is independent of X ( n ) . • If X and Y are independent, E ( XY ) = E ( X ) E ( Y ) . [ X (1) ] [ X (1) ] [ ] E [ X (1) ] = = E X ( n ) X ( n ) X ( n ) X ( n ) E [ X (1) ] [ X (1) ] = E [ X ( n ) ] X ( n ) E [ θ Y (1) ] = E [ θ Y ( n ) ] E [ Y (1) ] = E [ Y ( n ) ]
f Y y F Y y f Y y F Y y y n . n n I y n I y y Y Beta n E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 n I y f Y . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary . Y I y yI y 18 / 21 . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1)
F Y y f Y y F Y y y n . n n I y n I y y Y Beta n E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 n I y f Y . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary yI Y y 18 / 21 . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1)
f Y y F Y y y n . I n n I y n y y Y Beta n E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 n f Y . . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Summary Y 18 / 21 . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1)
y n . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y n Beta Y y I n 18 / 21 Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem Y . . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [1 − F Y ( y )] n − 1 I (0 < y < 1) f Y (1) ( y ) =
. Y January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y n Beta Y . 18 / 21 . Basu’s Theorem . . . . . . . . . . . . . . Summary Complete Statistics . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [1 − F Y ( y )] n − 1 I (0 < y < 1) f Y (1) ( y ) = n (1 − y ) n − 1 I (0 < y < 1) =
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y . Y 18 / 21 . . . . . . . . . . . Complete Statistics . . . . Basu’s Theorem . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [1 − F Y ( y )] n − 1 I (0 < y < 1) f Y (1) ( y ) = n (1 − y ) n − 1 I (0 < y < 1) = ∼ Beta (1 , n ) Y (1)
. . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang . Y Summary 18 / 21 Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . . . . . Obtaining E [ Y (1) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [1 − F Y ( y )] n − 1 I (0 < y < 1) f Y (1) ( y ) = n (1 − y ) n − 1 I (0 < y < 1) = ∼ Beta (1 , n ) Y (1) 1 E [ Y (1) ] = n + 1
f Y y F Y y f Y n y f Y y F Y y Y n E Y n X n E Y n E Y I n January 29th, 2013 I y ny n Biostatistics 602 - Lecture 07 Hyun Min Kang y n Beta n n n Therefore, E X . n n . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Y I y yI y I y 19 / 21 . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1)
F Y y f Y n y f Y y F Y y Y n E Y n X n E Y n n I y ny n I y . Beta n n Therefore, E X E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 n n n Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . 19 / 21 . Y yI y I y . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1)
f Y n y f Y y F Y y Y n E Y n X n E Y n n I y ny n I y . Beta n n Therefore, E X E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 n n . Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . 19 / 21 n Y . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1)
Y n E Y n X n E Y n . n I y Beta n n X Therefore, E . E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 ny n 19 / 21 Complete Statistics Y . . . Basu’s Theorem . Summary . . . . . . . . . . . . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [ F Y ( y )] n − 1 I (0 < y < 1) f Y ( n ) ( y ) =
Y n E Y n X n E Y n . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y X Therefore, E n n Beta n 19 / 21 Y Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [ F Y ( y )] n − 1 I (0 < y < 1) f Y ( n ) ( y ) = ny n − 1 I (0 < y < 1) =
E Y n X n E Y n . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y X Therefore, E n n 19 / 21 . Complete Statistics Basu’s Theorem . Summary Y . . . . . . . . . . . . . . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [ F Y ( y )] n − 1 I (0 < y < 1) f Y ( n ) ( y ) = ny n − 1 I (0 < y < 1) = ∼ Beta ( n , 1) Y ( n )
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n Therefore, E n . Y 19 / 21 . Complete Statistics . . . . . . . . . . . . . . Basu’s Theorem . . . . . . . . . Obtaining E [ Y ( n ) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 1)! f Y ( y ) [ F Y ( y )] n − 1 I (0 < y < 1) f Y ( n ) ( y ) = ny n − 1 I (0 < y < 1) = ∼ Beta ( n , 1) Y ( n ) E [ Y ( n ) ] = n + 1 [ X (1) ] E [ Y (1) ] E [ Y ( n ) ] = 1 = X ( n )
f Y y F Y y F Y y f Y y F Y y y n X n E Y n E Y n y I E Y Y Beta n n n y Therefore, E n X X E Y Y E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 y . I . . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Y 20 / 21 I y yI y I y f Y y n n n . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1)
F Y y F Y y f Y y F Y y y n X n E Y n E Y n n y n n y I y Y Beta Therefore, E E Y n X X E Y Y E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 I . . n . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Y 20 / 21 yI y I y f Y y n n . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1)
F Y y f Y y F Y y y n X n E Y n E Y n E Y y n n y I y Y Beta n X n Therefore, E X E Y Y E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 I . . n . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Summary Y 20 / 21 y f Y n n . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1)
y n X n E Y n E Y n X I y Y Beta n E Y n Therefore, E . X n n E Y Y E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 y 20 / 21 . Summary . . . . . . . . . . . . . . Complete Statistics Basu’s Theorem . Y . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 2)! [1 − F Y ( y )] n − 2 f Y ( y ) [ F Y ( y )] I (0 < y < 1) f Y (2) ( y ) =
X n E Y n E Y n . Beta n E Y n Therefore, E X X Y E Y . E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 Y 20 / 21 Y . . . . . . . . . . Basu’s Theorem . Summary . Complete Statistics . . . . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 2)! [1 − F Y ( y )] n − 2 f Y ( y ) [ F Y ( y )] I (0 < y < 1) f Y (2) ( y ) = n ( n − 1) y (1 − y ) n − 2 I (0 < y < 1) =
X n E Y n E Y n Y n Therefore, E X X E Y . . E Y E Y n Hyun Min Kang Biostatistics 602 - Lecture 07 January 29th, 2013 E Y 20 / 21 Y . . . . . . . . . . Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 2)! [1 − F Y ( y )] n − 2 f Y ( y ) [ F Y ( y )] I (0 < y < 1) f Y (2) ( y ) = n ( n − 1) y (1 − y ) n − 2 I (0 < y < 1) = ∼ Beta (2 , n − 1) Y (2)
X n E Y n E Y n . . January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n E Y E Y Y E Y X X Therefore, E 20 / 21 Y . Summary . Basu’s Theorem Complete Statistics . . . . . . . . . . . . . . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 2)! [1 − F Y ( y )] n − 2 f Y ( y ) [ F Y ( y )] I (0 < y < 1) f Y (2) ( y ) = n ( n − 1) y (1 − y ) n − 2 I (0 < y < 1) = ∼ Beta (2 , n − 1) Y (2) 2 E [ Y (2) ] = n + 1
. Summary January 29th, 2013 Biostatistics 602 - Lecture 07 Hyun Min Kang n Therefore, E . Y 20 / 21 . Basu’s Theorem . . Complete Statistics . . . . . . . . . . . . . . . . . . . . . Obtaining E [ Y (2) ] ∼ Uniform (0 , 1) f Y ( y ) = I (0 < y < 1) F Y ( y ) = yI (0 < y < 1) + I ( y ≥ 1) n ! ( n − 2)! [1 − F Y ( y )] n − 2 f Y ( y ) [ F Y ( y )] I (0 < y < 1) f Y (2) ( y ) = n ( n − 1) y (1 − y ) n − 2 I (0 < y < 1) = ∼ Beta (2 , n − 1) Y (2) 2 E [ Y (2) ] = n + 1 [ X (1) + X (2) ] E [ Y (1) + Y (2) ] E [ Y (1) ]+ E [ Y (2) ] = 3 = = E [ Y ( n ) ] E [ Y ( n ) ] X ( n )
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