• E g T • In other words, g T • Equivalently, T X is called a complete statistic • T X • g • g . . . . . T X = Pr g T X . . T X I T X = Pr g T X Pr T X . In this case, g T X is almost surely true. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Complete Statistics . Definition . . for all implies Pr g T for all . almost surely. . Example 7 / 26 . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if
• In other words, g T • Equivalently, T X is called a complete statistic • T X • g • g . . . T X = Pr g T X . T X . I T X = Pr g T X Pr T X . In this case, g T X is almost surely true. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Complete Statistics Definition . . . almost surely. . Example . . 7 / 26 . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if • E [ g ( T ) | θ ] = 0 for all θ implies Pr [ g ( T ) = 0 | θ ] = 1 for all θ .
• Equivalently, T X is called a complete statistic • T X • g • g . T X . . T X = Pr g T X . I T X . = Pr g T X Pr T X . In this case, g T X is almost surely true. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Complete Statistics Definition . . . . Example . . 7 / 26 . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if • E [ g ( T ) | θ ] = 0 for all θ implies Pr [ g ( T ) = 0 | θ ] = 1 for all θ . • In other words, g ( T ) = 0 almost surely.
• T X • g • g . T X . . . T X = Pr g T X . I T X . = Pr g T X Pr T X . In this case, g T X is almost surely true. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Complete Statistics Definition . . . . Example . 7 / 26 . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if • E [ g ( T ) | θ ] = 0 for all θ implies Pr [ g ( T ) = 0 | θ ] = 1 for all θ . • In other words, g ( T ) = 0 almost surely. • Equivalently, T ( X ) is called a complete statistic
• g . Pr g Example . . T X I T X = T X . Pr T X . In this case, g T X is almost surely true. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . 7 / 26 Complete Statistics . Ancillary Statistics Complete Statistics . Summary . Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if • E [ g ( T ) | θ ] = 0 for all θ implies Pr [ g ( T ) = 0 | θ ] = 1 for all θ . • In other words, g ( T ) = 0 almost surely. • Equivalently, T ( X ) is called a complete statistic • T ( X ) ∼ N (0 , 1) • g 1 ( T ( X )) = 0 = ⇒ Pr [ g 1 ( T ( X )) = 0] = 1 .
. Definition January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . . Example . . . . 7 / 26 . Ancillary Statistics . . . . Complete Statistics . . . . . Complete Statistics . Summary . . . . . . . . . . . . . . . . . . . • Let T = { f T ( t | θ ) , θ ∈ Ω } be a family of pdfs or pmfs for a statistic T ( X ) . • The family of probability distributions is called complete if • E [ g ( T ) | θ ] = 0 for all θ implies Pr [ g ( T ) = 0 | θ ] = 1 for all θ . • In other words, g ( T ) = 0 almost surely. • Equivalently, T ( X ) is called a complete statistic • T ( X ) ∼ N (0 , 1) • g 1 ( T ( X )) = 0 = ⇒ Pr [ g 1 ( T ( X )) = 0] = 1 . • g 2 ( T ( X )) = I ( T ( X ) = 0) = ⇒ Pr [ g 2 ( T ( X )) = 0] = 1 − Pr [ T ( X ) = 0)] . In this case, g 2 ( T ( X )) = 0 is almost surely true.
• For example, X • The above example is only for a particular distribution, not a family • If X • Therefore, the family of . satisfies E g X of distributions. , then no function of X except for g X . for all but Pr g X distributions, , is complete. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 instead of 1. , . E X . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Notes on Complete Statistics distributions, not of a particular distribution. and g x x makes E g X 8 / 26 . . . . . . . . . . . . . . . . . . . • Notice that completeness is a property of a family of probability
• The above example is only for a particular distribution, not a family • If X • Therefore, the family of . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete. , is distributions, . for all satisfies E g X g X , then no function of X except for of distributions. 8 / 26 . distributions, not of a particular distribution. Notes on Complete Statistics Summary . Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Notice that completeness is a property of a family of probability • For example, X ∼ N (0 , 1) and g ( x ) = x makes E [ g ( X )] = E X = 0 , but Pr ( g ( X ) = 0) = 0 instead of 1.
• If X • Therefore, the family of . of distributions. January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete. , is distributions, . for all satisfies E g X g X , then no function of X except for 8 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . Summary Notes on Complete Statistics distributions, not of a particular distribution. . . . . . . . . . . . . . . . . . . . • Notice that completeness is a property of a family of probability • For example, X ∼ N (0 , 1) and g ( x ) = x makes E [ g ( X )] = E X = 0 , but Pr ( g ( X ) = 0) = 0 instead of 1. • The above example is only for a particular distribution, not a family
• Therefore, the family of . Notes on Complete Statistics January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete. , is distributions, of distributions. . distributions, not of a particular distribution. Summary Ancillary Statistics . . . . . . . . . 8 / 26 Complete Statistics . . . . . . . . . . . . . . . . . . . . • Notice that completeness is a property of a family of probability • For example, X ∼ N (0 , 1) and g ( x ) = x makes E [ g ( X )] = E X = 0 , but Pr ( g ( X ) = 0) = 0 instead of 1. • The above example is only for a particular distribution, not a family • If X ∼ N ( θ, 1) , −∞ < θ < ∞ , then no function of X except for g ( X ) = 0 satisfies E [ g ( X ) | θ ] for all θ .
. Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete. of distributions. distributions, not of a particular distribution. . Notes on Complete Statistics 8 / 26 . Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Notice that completeness is a property of a family of probability • For example, X ∼ N (0 , 1) and g ( x ) = x makes E [ g ( X )] = E X = 0 , but Pr ( g ( X ) = 0) = 0 instead of 1. • The above example is only for a particular distribution, not a family • If X ∼ N ( θ, 1) , −∞ < θ < ∞ , then no function of X except for g ( X ) = 0 satisfies E [ g ( X ) | θ ] for all θ . • Therefore, the family of N ( θ, 1) distributions, −∞ < θ < ∞ , is
• Requiring g T to satisfy the definition puts a restriction on g . The . likely to appreciate its usefulness, January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the he is just as likely to be puzzled by 9 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very
• Requiring g T to satisfy the definition puts a restriction on g . The . likely to appreciate its usefulness, he is just as likely to be puzzled by January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: and wonder what connection (if any) there is between the its name, 9 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very
• Requiring g T to satisfy the definition puts a restriction on g . The . likely to appreciate its usefulness, he is just as likely to be puzzled by January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. and the dictionary definition: statistical use of the term ”complete”, its name, and wonder what connection (if any) there is between the 9 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very
• Requiring g T to satisfy the definition puts a restriction on g . The . likely to appreciate its usefulness, he is just as likely to be puzzled by January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the . . Ancillary Statistics . . . . . . . . . Complete Statistics . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . 9 / 26 . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T larger the family of pdfs/pmfs, the greater the restriction on g . When The lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the likely to appreciate its usefulness, he is just as likely to be puzzled by 9 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very • Requiring g ( T ) to satisfy the definition puts a restriction on g .
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the , it rules out all g except for the trivial g T for all the family of pdfs/pmfs is augmented to the point that E g T When larger the family of pdfs/pmfs, the greater the restriction on g . lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the likely to appreciate its usefulness, he is just as likely to be puzzled by 9 / 26 . Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Why ”Complete” Statistics? . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very • Requiring g ( T ) to satisfy the definition puts a restriction on g . The
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. , then the it rules out all g except for the trivial g T larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the likely to appreciate its usefulness, he is just as likely to be puzzled by . . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . Ancillary Statistics 9 / 26 Complete Statistics . Summary Why ”Complete” Statistics? . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very • Requiring g ( T ) to satisfy the definition puts a restriction on g . The the family of pdfs/pmfs is augmented to the point that E [ g ( T )] = 0 for all θ ,
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. then the larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the likely to appreciate its usefulness, he is just as likely to be puzzled by . . Stigler (1972) Am. Stat. 26(2):28-9 . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Why ”Complete” Statistics? 9 / 26 . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very • Requiring g ( T ) to satisfy the definition puts a restriction on g . The the family of pdfs/pmfs is augmented to the point that E [ g ( T )] = 0 for all θ , it rules out all g except for the trivial g ( T ) = 0 ,
. Stigler (1972) Am. Stat. 26(2):28-9 January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang family is said to be complete. larger the family of pdfs/pmfs, the greater the restriction on g . When lacking none of the parts, whole, entire. statistical use of the term ”complete”, and the dictionary definition: its name, and wonder what connection (if any) there is between the likely to appreciate its usefulness, he is just as likely to be puzzled by . . . . Why ”Complete” Statistics? . . . . . . . . . 9 / 26 Ancillary Statistics Complete Statistics . Summary . . . . . . . . . . . . . . . . . . . • While a student encountering completeness for the first time is very • Requiring g ( T ) to satisfy the definition puts a restriction on g . The the family of pdfs/pmfs is augmented to the point that E [ g ( T )] = 0 for all θ , it rules out all g except for the trivial g ( T ) = 0 , then the
• We need to find a counter example, • which is a function g such that E g T . Proof strategy . . . . . . . . . for but g T . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . 10 / 26 Summary . . . . . . . . . . Problem . Example - Poisson distribution . . Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . { } f T : f T ( t | λ ) = λ t e − λ • Suppose T = for t ∈ { 0 , 1 , 2 , · · · } . Let t ! λ ∈ Ω = { 1 , 2 } . Show that this family is NOT complete
• which is a function g such that E g T . . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . g T but for . . Proof strategy . . 10 / 26 . Summary . . . . . . . . . Ancillary Statistics Complete Statistics . Example - Poisson distribution . Problem . . . . . . . . . . . . . . . . . . . { } f T : f T ( t | λ ) = λ t e − λ • Suppose T = for t ∈ { 0 , 1 , 2 , · · · } . Let t ! λ ∈ Ω = { 1 , 2 } . Show that this family is NOT complete • We need to find a counter example,
. Problem January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . . Proof strategy . . . . 10 / 26 . Example - Poisson distribution Summary . Complete Statistics . . . . . . . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . { } f T : f T ( t | λ ) = λ t e − λ • Suppose T = for t ∈ { 0 , 1 , 2 , · · · } . Let t ! λ ∈ Ω = { 1 , 2 } . Show that this family is NOT complete • We need to find a counter example, • which is a function g such that E [ g ( T ) | λ ] = 0 for λ = 1 , 2 but g ( T ) ̸ = 0 .
. The above equation can be rewritten as g t t e t E g T t g t t e t t E g T g t t t t g t t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 t . Thus, . for . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Poisson distribution example : Proof The function g must satisfy 11 / 26 . . . . . . . . . . . . . . . . . . . ∞ g ( t ) λ t e − λ ∑ E [ g ( T ) | λ ] = = 0 t ! t =0
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang t t g t t t g t t The above equation can be rewritten as 11 / 26 The function g must satisfy . Complete Statistics Ancillary Statistics . . . . . . . . Poisson distribution example : Proof . Summary . . . . . . . . . . . . . . . . . . . ∞ g ( t ) λ t e − λ ∑ E [ g ( T ) | λ ] = = 0 t ! t =0 for λ ∈ { 1 , 2 } . Thus, t =0 g ( t ) 1 t e − 1 E [ g ( T ) | λ = 1] = ∑ ∞ = 0 t ! t =0 g ( t ) 2 t e − 2 E [ g ( T ) | λ = 2] ∑ ∞ = = 0 t !
. Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang The above equation can be rewritten as . The function g must satisfy Poisson distribution example : Proof 11 / 26 . . . . . . . . . . Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . ∞ g ( t ) λ t e − λ ∑ E [ g ( T ) | λ ] = = 0 t ! t =0 for λ ∈ { 1 , 2 } . Thus, t =0 g ( t ) 1 t e − 1 E [ g ( T ) | λ = 1] = ∑ ∞ = 0 t ! t =0 g ( t ) 2 t e − 2 E [ g ( T ) | λ = 2] ∑ ∞ = = 0 t ! ∑ ∞ t =0 g ( t )/ t ! = 0 ∑ ∞ t =0 2 t g ( t )/ t ! = 0
. Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang Then otherwise . Poisson distribution example : Proof (cont’d) 12 / 26 . Complete Statistics . . . . . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . Define g ( t ) as t = 0 ∨ t = 2 2 g ( t ) = − 3 t = 1 0 ∞ ∑ g ( t )/ t ! = g (0)/0! + g (1)/1! + g (2)/2! = 2 − 3 + 2/2 = 0 t =0 ∞ ∑ g (0)/0! + 2 g (1)/1! + 2 2 g (2)/2! = 2 − 6 + 8/2 = 0 2 t g ( t )/ t ! = t =0 There exists a non-zero function g that satisfies E [ g ( T ) λ ] = 0 for all λ ∈ Ω . Therefore this family is NOT complete.
• Need to find the distribution of T X • Show that there is no non-zero function g such that E g T . . . . . . . . . . for all . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 Proof strategy 13 / 26 . . . . . . . . . . . Ancillary Statistics Complete Statistics Summary Another example with Poisson distribution . Problem . . i.i.d. . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n ∼ Poisson ( λ ) , λ > 0 . • Show that T ( X ) = ∑ n i =1 X i is a complete statistic.
. Problem January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . . Proof strategy . i.i.d. . . . . Another example with Poisson distribution . . . . . . . . . Ancillary Statistics Complete Statistics . Summary 13 / 26 . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n ∼ Poisson ( λ ) , λ > 0 . • Show that T ( X ) = ∑ n i =1 X i is a complete statistic. • Need to find the distribution of T ( X ) • Show that there is no non-zero function g such that E [ g ( T ) | λ ] = 0 for all λ .
e s e e s f Poisson x e s . x e x e x e e e s e s x e s e s e e e s x e e s Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 x e . x . . . . . . . . . Ancillary Statistics Complete Statistics . Summary 14 / 26 Proof : Finding the moment-generating function of X . . . . . . . . . . . . . . . . . . . ∞ e sx e − λ λ x ∑ M X ( s ) = E [ e sX ] = x ! x =0
f Poisson x e s . x e e e s e s x e e s x e e e s . e e s Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 x 14 / 26 Complete Statistics . . Summary Proof : Finding the moment-generating function of X . . . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . ∞ e sx e − λ λ x ∑ M X ( s ) = E [ e sX ] = x ! x =0 ∞ e − λ ( e s λ ) x ∑ e − e s λ e e s λ = x ! x =0
f Poisson x e s . Proof : Finding the moment-generating function of X January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang e s e x e e e s . 14 / 26 Summary . Ancillary Statistics . . . . . . . . . Complete Statistics . . . . . . . . . . . . . . . . . . . ∞ e sx e − λ λ x ∑ M X ( s ) = E [ e sX ] = x ! x =0 ∞ e − λ ( e s λ ) x ∑ e − e s λ e e s λ = x ! x =0 ∞ e − λ e e s λ ( e s λ ) x e − e s λ ∑ = x ! x =0
. Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang e s e . Proof : Finding the moment-generating function of X 14 / 26 . . . . . . . . . Ancillary Statistics . Complete Statistics . . . . . . . . . . . . . . . . . . . ∞ e sx e − λ λ x ∑ M X ( s ) = E [ e sX ] = x ! x =0 ∞ e − λ ( e s λ ) x ∑ e − e s λ e e s λ = x ! x =0 ∞ e − λ e e s λ ( e s λ ) x e − e s λ ∑ = x ! x =0 ∞ ∑ e λ e e s λ = f Poisson ( x | e s λ ) x =0
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . Proof : Finding the moment-generating function of X Summary 14 / 26 Complete Statistics . . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . ∞ e sx e − λ λ x ∑ M X ( s ) = E [ e sX ] = x ! x =0 ∞ e − λ ( e s λ ) x ∑ e − e s λ e e s λ = x ! x =0 ∞ e − λ e e s λ ( e s λ ) x e − e s λ ∑ = x ! x =0 ∞ ∑ e λ e e s λ = f Poisson ( x | e s λ ) x =0 e λ ( e s − 1) =
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . . e n e s . Theorem 2.3.11 (b) . . . . . . . e s By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n e . n . . . . . . . . . Ancillary Statistics Complete Statistics . Summary 15 / 26 E n i e sX i n i E e sX i E e sX i . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i E ( e s ∑ X i ) = M T ( s ) =
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . . e n e s . Theorem 2.3.11 (b) . . . . . . . e s By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n e . n . . . . . . . . . Ancillary Statistics Complete Statistics . Summary 15 / 26 n e sX i i E e sX i E e sX i . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . . e e s n e n e s . Theorem 2.3.11 (b) . . . . . . E e sX i . By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n 15 / 26 . . . Summary Ancillary Statistics e sX i . . . . . n . . E . Complete Statistics . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) = = i =1
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . . e e s n e n e s . Theorem 2.3.11 (b) . . . . . . . . By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 E 15 / 26 Ancillary Statistics n Summary . . . . . . e sX i . Complete Statistics . E . . . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) [ ( e sX i )] n = = i =1
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . e s . Theorem 2.3.11 (b) . . . . . . . . . By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 e n E E n . . . . . . . . . Ancillary Statistics Complete Statistics . Summary e sX i 15 / 26 . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) [ ( e sX i )] n = = i =1 [ e − λ ( e s − 1) ] n = =
Let F X x and F Y y be two cdfs all of whose moments exists. If the moment generating functions exists and M X t M Y t for all t in some neighborhood of 0, then F X u F Y u for all u . . E . Theorem 2.3.11 (b) . . . . . . . . E By Theorem 2.3.11, T X Poisson n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . 15 / 26 Summary . e sX i . Ancillary Statistics . . . . . Complete Statistics . . n . . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) [ ( e sX i )] n = = i =1 [ e − λ ( e s − 1) ] n = e n λ ( e s − 1) =
. e sX i January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . Poisson n By Theorem 2.3.11, T X . . Theorem 2.3.11 (b) . E . n E 15 / 26 . . . . . . . . . Ancillary Statistics Complete Statistics . Summary . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) [ ( e sX i )] n = = i =1 [ e − λ ( e s − 1) ] n = e n λ ( e s − 1) = Let F X ( x ) and F Y ( y ) be two cdfs all of whose moments exists. If the moment generating functions exists and M X ( t ) = M Y ( t ) for all t in some neighborhood of 0, then F X ( u ) = F Y ( u ) for all u .
. e sX i January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . . Theorem 2.3.11 (b) . E E . n 15 / 26 Ancillary Statistics . . . . . . . . Complete Statistics . Summary . . . . . . . . . . . . . . . . . . . . Proof : Finding the MGF of T ( X ) = ∑ n i =1 X i ( n ) ∏ E ( e s ∑ X i ) = E M T ( s ) = i =1 ∏ ( e sX i ) [ ( e sX i )] n = = i =1 [ e − λ ( e s − 1) ] n = e n λ ( e s − 1) = Let F X ( x ) and F Y ( y ) be two cdfs all of whose moments exists. If the moment generating functions exists and M X ( t ) = M Y ( t ) for all t in some neighborhood of 0, then F X ( u ) = F Y ( u ) for all u . By Theorem 2.3.11, T ( X ) ∼ Poisson ( n λ ) .
g t n t t X i is . Because the function above is a power series expansion of t t Which is equivalent to t g t n t t t for all . , t for all t . and g t for all t . Therefore T X n i a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 g t n n . Summary . . . . . . . . . Ancillary Statistics Complete Statistics . 16 / 26 e E g T t e n n t t . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 .
g t n t t X i is . Because the function above is a power series expansion of t t Which is equivalent to t g t n t t t for all . , t for all t . and g t for all t . Therefore T X n i a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 g t n n . Summary . . . . . . . . . Ancillary Statistics Complete Statistics . 16 / 26 e . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0
g t n t t X i is . t g t n t t t for all . Because the function above is a power series expansion of , for all t . and g t . for all t . Therefore T X n i a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 Which is equivalent to 16 / 26 Summary . . Complete Statistics . . . . . . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0 ∞ g ( t )( n λ ) t ∑ e − n λ = = 0 t ! t =0
g t n t t X i is . . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang a complete statistic. i n for all t . Therefore T X for all t . and g t , . Because the function above is a power series expansion of for all Which is equivalent to 16 / 26 . Complete Statistics Summary Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0 ∞ g ( t )( n λ ) t ∑ e − n λ = = 0 t ! t =0 ∞ g ( t ) n t ∑ λ t = 0 t ! t =0
g t n t t X i is . . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang a complete statistic. i n for all t . Therefore T X for all t . and g t , Because the function above is a power series expansion of Which is equivalent to 16 / 26 Summary . . . . . . . . . Ancillary Statistics Complete Statistics . . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0 ∞ g ( t )( n λ ) t ∑ e − n λ = = 0 t ! t =0 ∞ g ( t ) n t ∑ λ t = 0 t ! t =0 for all λ > 0 .
X i is . . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang a complete statistic. i n Therefore T X Which is equivalent to 16 / 26 . . Ancillary Statistics . . . . . . . . Summary Complete Statistics . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0 ∞ g ( t )( n λ ) t ∑ e − n λ = = 0 t ! t =0 ∞ g ( t ) n t ∑ λ t = 0 t ! t =0 for all λ > 0 . Because the function above is a power series expansion of λ , g ( t ) n t / t ! = 0 for all t . and g ( t ) = 0 for all t .
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang a complete statistic. Which is equivalent to . Summary 16 / 26 Complete Statistics . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . Proof : Showing E [ g ( T ) | λ ] = 0 ⇐ ⇒ Pr [ g ( T ) = 0] = 1 . Suppose that there exists a g ( T ) such that E [ g ( T ) | λ ] = 0 for all λ > 0 . ∞ e − n λ ( n λ ) t ∑ E [ g ( T ) | λ ] = t ! t =0 ∞ g ( t )( n λ ) t ∑ e − n λ = = 0 t ! t =0 ∞ g ( t ) n t ∑ λ t = 0 t ! t =0 for all λ > 0 . Because the function above is a power series expansion of λ , g ( t ) n t / t ! = 0 for all t . and g ( t ) = 0 for all t . Therefore T ( X ) = ∑ n i =1 X i is
X n . Let f X x , then its cdf is F X x f X t F X t n f T t n I x x I x I x . n . . n t n I t n n t n I t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 We need to obtain the distribution of T X . . Problem . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Example : Uniform Distribution . . . . i.i.d. . Proof . . . . . 17 / 26 . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete.
f X x , then its cdf is F X x f X t F X t n f T t . x x I x I x . n n t n n I t n n t n I t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 I Let . Example : Uniform Distribution . . . . . . . . . Ancillary Statistics Complete Statistics . Summary . Problem . . i.i.d. . Proof . . 17 / 26 . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) .
F X x f X t F X t n f T t t x I x I x . n n n . . I t n n t n I t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n 17 / 26 . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Example : Uniform Distribution . Problem . . i.i.d. . Proof . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) . Let f X ( x ) = 1 θ I (0 < x < θ ) , then its cdf is
f X t F X t n f T t . n . n n n t I Proof t n n t n I t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . Example : Uniform Distribution . . . . . . . . . Ancillary Statistics Complete Statistics . Summary . Problem . . i.i.d. 17 / 26 . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) . Let f X ( x ) = 1 θ I (0 < x < θ ) + I ( x ≥ θ ) . θ I (0 < x < θ ) , then its cdf is F X ( x ) = x
. t . . n t n I n . n t n I t Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 Proof . i.i.d. . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Example : Uniform Distribution 17 / 26 . Problem . . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) . Let f X ( x ) = 1 θ I (0 < x < θ ) + I ( x ≥ θ ) . θ I (0 < x < θ ) , then its cdf is F X ( x ) = x n ! ( n − 1)! f X ( t ) F X ( t ) n − 1 f T ( t | θ ) =
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang t I n t n n n . . Proof . i.i.d. . . Complete Statistics . . . . . . . . . Ancillary Statistics 17 / 26 Problem . . Example : Uniform Distribution Summary . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) . Let f X ( x ) = 1 θ I (0 < x < θ ) + I ( x ≥ θ ) . θ I (0 < x < θ ) , then its cdf is F X ( x ) = x n ! ( n − 1)! f X ( t ) F X ( t ) n − 1 f T ( t | θ ) = ) n − 1 ( t = I (0 < t < θ ) θ θ
. Problem January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang n . . . . i.i.d. . . Proof . Ancillary Statistics . . . . Example : Uniform Distribution . . . . . 17 / 26 . Summary Complete Statistics . . . . . . . . . . . . . . . . . . . Let X 1 , · · · , X n ∼ Uniform (0 , θ ) , θ > 0 , Ω = (0 , ∞ ) . Show that X ( n ) is complete. We need to obtain the distribution of T ( X ) = X ( n ) . Let f X ( x ) = 1 θ I (0 < x < θ ) + I ( x ≥ θ ) . θ I (0 < x < θ ) , then its cdf is F X ( x ) = x n ! ( n − 1)! f X ( t ) F X ( t ) n − 1 f T ( t | θ ) = ) n − 1 ( t I (0 < t < θ ) = n θ − n t n − 1 I (0 < t < θ ) = θ θ
n g X n is a complete statistic. . n n n n g t t n dt ng n n g t t n Taking derivative of both sides, dt nE g T Because g T holds for all , T X Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n dt . Summary . . . . . . . . . Ancillary Statistics Complete Statistics . Proof : Uniform Distribution (cont’d) g t t n E g T g t n n t n I t dt n n 18 / 26 . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0
n g X n is a complete statistic. . n n n n g t t n dt ng n n g t t n Taking derivative of both sides, dt nE g T Because g T holds for all , T X Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n dt . g t t n . . . . . . . . . Ancillary Statistics Complete Statistics . Summary n Proof : Uniform Distribution (cont’d) n 18 / 26 . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0
n g X n is a complete statistic. . n n n n g t t n dt ng n n g t t n Taking derivative of both sides, dt nE g T Because g T holds for all , T X Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 n 18 / 26 . Summary . . . . . . . . . Ancillary Statistics Complete Statistics . Proof : Uniform Distribution (cont’d) n . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0
X n is a complete statistic. . dt Taking derivative of both sides, n ng n n n g t t n nE g T n Because g T holds for all , T X Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . 18 / 26 . Complete Statistics Summary Proof : Uniform Distribution (cont’d) Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ n 2 θ n g ( θ ) θ n − 1 − g ( t ) t n − 1 dt = 0 θ n +1 0
X n is a complete statistic. . n January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang , T X holds for all Because g T n n . Taking derivative of both sides, 18 / 26 Summary . . . . . . . . . Proof : Uniform Distribution (cont’d) Ancillary Statistics Complete Statistics . . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ n 2 θ n g ( θ ) θ n − 1 − g ( t ) t n − 1 dt = 0 θ n +1 0 ∫ θ ng ( θ ) g ( t ) t n − 1 dt = n = n θ E [ g ( T ) | θ ] = 0 θ θ θ n 0
X n is a complete statistic. . n January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang , T X holds for all Because g T n n . Taking derivative of both sides, 18 / 26 Summary . . . . . . . . . Proof : Uniform Distribution (cont’d) Ancillary Statistics Complete Statistics . . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ n 2 θ n g ( θ ) θ n − 1 − g ( t ) t n − 1 dt = 0 θ n +1 0 ∫ θ ng ( θ ) g ( t ) t n − 1 dt = n = n θ E [ g ( T ) | θ ] = 0 θ θ θ n 0
. Proof : Uniform Distribution (cont’d) January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang n n Taking derivative of both sides, . n 18 / 26 Summary . . . . . . Ancillary Statistics . . Complete Statistics . . . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ n 2 θ n g ( θ ) θ n − 1 − g ( t ) t n − 1 dt = 0 θ n +1 0 ∫ θ ng ( θ ) g ( t ) t n − 1 dt = n = n θ E [ g ( T ) | θ ] = 0 θ θ θ n 0 Because g ( T ) = 0 holds for all θ > 0 , T ( X ) = X ( n ) is a complete statistic.
X n is a . g dt g t t n dt Taking derivative of both sides, g n . Because g T for all n holds for all , T X complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 g t t n n . Complete Statistics . . . . . . . . . Ancillary Statistics . Summary A simpler proof (how it was solved in the class) 19 / 26 . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0
X n is a . for all dt Taking derivative of both sides, g n g holds for all . Because g T . , T X complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 g t t n n 19 / 26 Summary . . . . . . . . . Ancillary Statistics Complete Statistics . A simpler proof (how it was solved in the class) . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0
X n is a . n January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete statistic. , T X holds for all . Because g T for all g n g Taking derivative of both sides, . 19 / 26 Summary . . . . . . . . . Ancillary Statistics Complete Statistics . A simpler proof (how it was solved in the class) . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ g ( t ) t n − 1 dt = 0 0
X n is a . n January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete statistic. , T X holds for all . Because g T for all Taking derivative of both sides, . 19 / 26 Summary . . . . . A simpler proof (how it was solved in the class) . . . . . Ancillary Statistics Complete Statistics . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ g ( t ) t n − 1 dt = 0 0 g ( θ ) θ n − 1 = 0 g ( θ ) = 0
. Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang complete statistic. Taking derivative of both sides, . n A simpler proof (how it was solved in the class) 19 / 26 . . . . . . Complete Statistics . . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . Consider a function g ( T ) such that E [ g ( T ) | θ ] = 0 for all θ > 0 ∫ θ g ( t ) n θ − n t n − 1 I (0 < t < θ ) dt E [ g ( T ) | θ ] = 0 ∫ θ g ( t ) t n − 1 dt = 0 = θ n 0 ∫ θ g ( t ) t n − 1 dt = 0 0 g ( θ ) θ n − 1 = 0 g ( θ ) = 0 for all θ > 0 . Because g ( T ) = 0 holds for all θ > 0 , T ( X ) = X ( n ) is a
• We have previously shown that T X X n • Show that T X is not a complete statistic. X n r n f R r . . . . . . Define R X . We have previously shown that n n . r r Then R Beta n , and E R n n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . Proof - Using a range statistic . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Another Example of Uniform Distribution . Problem . . i.i.d. X is a minimal sufficient statistic for . . 20 / 26 . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R .
• Show that T X is not a complete statistic. X n r n f R r n n . . . . Define R X . We have previously shown that . . r r Then R Beta n , and E R n n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . Another Example of Uniform Distribution . . . . . . . . . Ancillary Statistics Complete Statistics . Summary . . Problem . . i.i.d. . Proof - Using a range statistic 20 / 26 . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • We have previously shown that T ( X ) = ( X (1) , X ( n ) ) is a minimal sufficient statistic for θ .
X n r n f R r . . . . . Define R X . We have previously shown that n n r . r Then R Beta n , and E R n n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Another Example of Uniform Distribution . Problem . . i.i.d. . Proof - Using a range statistic 20 / 26 . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • We have previously shown that T ( X ) = ( X (1) , X ( n ) ) is a minimal sufficient statistic for θ . • Show that T ( X ) is not a complete statistic.
r n f R r . r Proof - Using a range statistic . . We have previously shown that n n r Then R . Beta n , and E R n n . Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . 20 / 26 Summary . . . . . . . Ancillary Statistics Complete Statistics . Another Example of Uniform Distribution . Problem . . . i.i.d. . . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • We have previously shown that T ( X ) = ( X (1) , X ( n ) ) is a minimal sufficient statistic for θ . • Show that T ( X ) is not a complete statistic. Define R = X ( n ) − X (1) .
. i.i.d. January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . n n , and E R Beta n Then R . . Proof - Using a range statistic . . 20 / 26 Summary . . . . . Complete Statistics . . . Problem . Another Example of Uniform Distribution Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • We have previously shown that T ( X ) = ( X (1) , X ( n ) ) is a minimal sufficient statistic for θ . • Show that T ( X ) is not a complete statistic. Define R = X ( n ) − X (1) . We have previously shown that n ( n − 1) r ( n − 2) (1 − r ) f R ( r | θ ) = , 0 < r < 1
. Problem January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . . Proof - Using a range statistic . . i.i.d. . . 20 / 26 . Ancillary Statistics . . . Another Example of Uniform Distribution . . . . . . . Complete Statistics Summary . . . . . . . . . . . . . . . . . . . • Let X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , θ ∈ R . • We have previously shown that T ( X ) = ( X (1) , X ( n ) ) is a minimal sufficient statistic for θ . • Show that T ( X ) is not a complete statistic. Define R = X ( n ) − X (1) . We have previously shown that n ( n − 1) r ( n − 2) (1 − r ) f R ( r | θ ) = , 0 < r < 1 Then R ∼ Beta ( n − 1 , 2) , and E [ R | θ ] = n − 1 n +1 .
E X n X n for all n n n n Therefore, there exist a g T such that Pr g T . , so n X is not a complete statistic. Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 T X X n Ancillary Statistics . . . . . . . . . Complete Statistics . . Summary Proof E g T 21 / 26 . . . . . . . . . . . . . . . . . . . Define g ( T ( X )) = X ( n ) − X (1) − n − 1 n +1
X n . . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang is not a complete statistic. X T X , so for all Therefore, there exist a g T such that Pr g T n n n n 21 / 26 Complete Statistics . . Proof Summary . . Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . Define g ( T ( X )) = X ( n ) − X (1) − n − 1 n +1 E [ X ( n ) − X (1) | θ ] − n − 1 E [ g ( T ) | θ ] = n + 1
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang . Proof Summary 21 / 26 Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . Define g ( T ( X )) = X ( n ) − X (1) − n − 1 n +1 E [ X ( n ) − X (1) | θ ] − n − 1 E [ g ( T ) | θ ] = n + 1 n − 1 n + 1 − n − 1 = n + 1 = 0 Therefore, there exist a g ( T ) such that Pr [ g ( T ) | θ ] < 1 for all θ , so T ( X ) = ( X (1) , X ( n ) ) is not a complete statistic.
• Define g T X n • Then E g T . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang , so T is not a complete statistic. E R E R . E R . Note that E R is constant to X 22 / 26 . Even Simpler Proof Summary . Complete Statistics Ancillary Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . • We know that R = X ( n ) − X (1) is an ancillary statistic, which do not depend on θ .
• Then E g T . Summary January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang , so T is not a complete statistic. E R E R . Even Simpler Proof . Complete Statistics Ancillary Statistics . . . . . . . . . 22 / 26 . . . . . . . . . . . . . . . . . . . • 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 .
. Complete Statistics January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang Even Simpler Proof Summary . . Ancillary Statistics . . . . . . . . . 22 / 26 . . . . . . . . . . . . . . . . . . . • 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.
Note that f X x . x . . . . Consider a function g T such that E g T for all . I x I E g T . E g X x g x x g x x g x Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . Example from Stigler (1972) Am. Stat. . . . . . . . . . Ancillary Statistics Complete Statistics . Summary . . Problem . . Is T X X a complete statistic? . Solution . 23 / 26 . . . . . . . . . . . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N .
Note that f X x . E g T . . Consider a function g T such that E g T for all . I x I x . E g X . x g x x g x x g x Hyun Min Kang Biostatistics 602 - Lecture 05 January 24th, 2013 . . . . . . . . . . . . . Ancillary Statistics Complete Statistics . Summary Example from Stigler (1972) Am. Stat. . Problem . . . Solution . . 23 / 26 . . . . . . . . . . . . . . . . . . . Let X is a uniform random sample from { 1 , · · · , θ } where θ ∈ Ω = N . Is T ( X ) = X a complete statistic?
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang g x x g x x g x x E g X E g T . . Solution . 23 / 26 . . . . . . . . . . Ancillary Statistics Complete Statistics Summary Example from Stigler (1972) Am. Stat. . Problem . . . . . . . . . . . . . . . . . . . . . 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 ) .
. . January 24th, 2013 Biostatistics 602 - Lecture 05 Hyun Min Kang g x x . . . Solution . . 23 / 26 Problem . Example from Stigler (1972) Am. Stat. Summary . Complete Statistics Ancillary 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
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