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. . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang January 15th, 2013 Hyun Min Kang Factorization Theorem Lecture 02 Biostatistics 602 - Statistical Inference . Summary . . Factorization Theorem Recap . . . . . .

  1. • and for all parameter points • f X x . Theorem 6.2.6 - Factorization Theorem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x . g T x , . . . . Recap . . . . . . . . Factorization Theorem . Factorization Theorem Summary 6 / 27 . . . . . . . . . . . . . . . . . . . . . • Let f X ( x | θ ) denote the joint pdf or pmf of a sample X . • A statistic T ( X ) is a sufficient statistic for θ , if and only if • There exists function g ( t | θ ) and h ( x ) such that, • for all sample points x ,

  2. • f X x . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x . g T x . . . Theorem 6.2.6 - Factorization Theorem Factorization Theorem Recap . . . . . . . . 6 / 27 Factorization Theorem Summary . . . . . . . . . . . . . . . . . . . . . . • Let f X ( x | θ ) denote the joint pdf or pmf of a sample X . • A statistic T ( X ) is a sufficient statistic for θ , if and only if • There exists function g ( t | θ ) and h ( x ) such that, • for all sample points x , • and for all parameter points θ ,

  3. . Summary January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . Theorem 6.2.6 - Factorization Theorem . Factorization Theorem . . Factorization Theorem Recap . . . . . . . . 6 / 27 . . . . . . . . . . . . . . . . . . . . . • Let f X ( x | θ ) denote the joint pdf or pmf of a sample X . • A statistic T ( X ) is a sufficient statistic for θ , if and only if • There exists function g ( t | θ ) and h ( x ) such that, • for all sample points x , • and for all parameter points θ , • f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) .

  4. • Choose g t • and h x • Because T X is sufficient, h x does not depend on f X x Pr X x Pr X x T X T x Pr T X T x Pr X . T x Pr T X T x Pr X x T X T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 x T X T x . . . . . . . . . . Recap Factorization Theorem . Summary Factorization Theorem : Proof The proof below is only for discrete distributions. . only if part . . Pr T X t Pr X x T X 7 / 27 . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic

  5. • and h x • Because T X is sufficient, h x does not depend on f X x x Pr X x T X T x Pr T X T x Pr X x T X T x Pr T X T x Pr X x T X T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 Pr X . . . . . . . . . . . Recap Factorization Theorem . Summary Factorization Theorem : Proof The proof below is only for discrete distributions. . only if part . . Pr X x T X T x 7 / 27 . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ )

  6. • Because T X is sufficient, h x does not depend on f X x T x Pr X x T X T x Pr T X T x Pr X x T X Pr T X Pr X T x Pr X x T X T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 x . . . . . . . . . . . Recap Factorization Theorem . Summary Factorization Theorem : Proof The proof below is only for discrete distributions. . only if part . . 7 / 27 . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x ))

  7. f X x T x Pr X x T X T x Pr T X T x Pr X x T X Pr T X Pr X T x Pr X x T X T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 x . . . . . . . . . . . Recap Factorization Theorem Summary Factorization Theorem : Proof The proof below is only for discrete distributions. . only if part . . 7 / 27 . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ .

  8. . Pr T X x T X T x Pr T X T x Pr X x T X T x T x . Pr X x T X T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 Pr X 7 / 27 Factorization Theorem : Proof The proof below is only for discrete distributions. . . . . Recap Factorization Theorem . Summary . . . only if part . . . . . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ . f X ( x | θ ) Pr ( X = x | θ ) =

  9. . Pr X T x Pr X x T X T x Pr T X T x x T X . T x g T x h x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 Pr T X 7 / 27 The proof below is only for discrete distributions. Factorization Theorem . . . . . . . . . Recap . only if part . Factorization Theorem : Proof Summary . . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ . f X ( x | θ ) Pr ( X = x | θ ) = = Pr ( X = x ∧ T ( X ) = T ( x ) | θ )

  10. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x g T x T x x T X Pr X T x Pr T X . . 7 / 27 only if part Factorization Theorem The proof below is only for discrete distributions. Factorization Theorem : Proof . . . . . . . . Recap . Summary . . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ . f X ( x | θ ) Pr ( X = x | θ ) = = Pr ( X = x ∧ T ( X ) = T ( x ) | θ ) = Pr ( T ( X ) = T ( x ) | θ ) Pr ( X = x | T ( X ) = T ( x ) , θ )

  11. . The proof below is only for discrete distributions. January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x g T x . . . only if part . 7 / 27 Factorization Theorem : Proof Summary . Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ . f X ( x | θ ) Pr ( X = x | θ ) = = Pr ( X = x ∧ T ( X ) = T ( x ) | θ ) = Pr ( T ( X ) = T ( x ) | θ ) Pr ( X = x | T ( X ) = T ( x ) , θ ) = Pr ( T ( X ) = T ( x ) | θ ) Pr ( X = x | T ( X ) = T ( x ))

  12. . Factorization Theorem : Proof January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . . only if part . The proof below is only for discrete distributions. 7 / 27 Summary . Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Suppose that T ( X ) is a sufficient statistic • Choose g ( t | θ ) = Pr ( T ( X ) = t | θ ) • and h ( x ) = Pr ( X = x | T ( X ) = T ( x )) • Because T ( X ) is sufficient, h ( x ) does not depend on θ . f X ( x | θ ) Pr ( X = x | θ ) = = Pr ( X = x ∧ T ( X ) = T ( x ) | θ ) = Pr ( T ( X ) = T ( x ) | θ ) Pr ( X = x | T ( X ) = T ( x ) , θ ) = Pr ( T ( X ) = T ( x ) | θ ) Pr ( X = x | T ( X ) = T ( x )) = g ( T ( x ) | θ ) h ( x )

  13. • Let q t • Define A t f X y . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang y A t t Pr T X q t t . T y y be the pmf of T X . . Recap . . . . . . . . Factorization Theorem . Summary Factorization Theorem : Proof (cont’d) . if part . 8 / 27 . . . . . . . . . . . . . . . . . . . . . • Assume that the factorization f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) exists.

  14. • Define A t f X y . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang y A t t Pr T X q t t . T y y . . Recap . . . . . . . . Factorization Theorem if part . Summary Factorization Theorem : Proof (cont’d) . 8 / 27 . . . . . . . . . . . . . . . . . . . . . • Assume that the factorization f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) exists. • Let q ( t | θ ) be the pmf of T ( X )

  15. f X y . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang y A t t Pr T X q t . . if part . Factorization Theorem : Proof (cont’d) Summary . Factorization Theorem Recap . . . . . . . . 8 / 27 . . . . . . . . . . . . . . . . . . . . . • Assume that the factorization f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) exists. • Let q ( t | θ ) be the pmf of T ( X ) • Define A t = { y : T ( y ) = t } .

  16. f X y . Factorization Theorem : Proof (cont’d) January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang y A t . . . . if part Summary . . . . . . . . . 8 / 27 Recap Factorization Theorem . . . . . . . . . . . . . . . . . . . . . • Assume that the factorization f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) exists. • Let q ( t | θ ) be the pmf of T ( X ) • Define A t = { y : T ( y ) = t } . q ( t | θ ) = Pr ( T ( X ) = t | θ )

  17. . Summary January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . if part . Factorization Theorem : Proof (cont’d) . . Factorization Theorem . . . 8 / 27 . . . . . Recap . . . . . . . . . . . . . . . . . . . . . • Assume that the factorization f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) exists. • Let q ( t | θ ) be the pmf of T ( X ) • Define A t = { y : T ( y ) = t } . q ( t | θ ) = Pr ( T ( X ) = t | θ ) ∑ = f X ( y | θ ) y ∈ A t

  18. y A T x g T y T x h y A T x h y f X x h x h y g T x h x g T x A y . h x . Thus, T X is a sufficient statistic for , if and only if g T x h x . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 g T x 9 / 27 Factorization Theorem : Proof (cont’d) . Factorization Theorem . Summary . . . . . if part (cont’d) . Recap . . . . . . . . . . . . . . . . . . . . . . . . . f X ( x | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = = q ( T ( x ) | θ ) q ( T ( x ) | θ ) ∑ y ∈ A T ( x ) f X ( y | θ )

  19. T x h y A T x h y f X x . h x g T x A y h x , if and only if Thus, T X is a sufficient statistic for . g T x h x . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 g T x 9 / 27 if part (cont’d) Recap . . . . Factorization Theorem . . . . Summary . . . . Factorization Theorem : Proof (cont’d) . . . . . . . . . . . . . . . . . . . . . f X ( x | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = = q ( T ( x ) | θ ) q ( T ( x ) | θ ) ∑ y ∈ A T ( x ) f X ( y | θ ) g ( T ( x ) | θ ) h ( x ) = ∑ y ∈ A T ( x ) g ( T ( y ) | θ ) h ( y )

  20. A T x h y f X x . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x . g T x , if and only if Thus, T X is a sufficient statistic for h x . . 9 / 27 if part (cont’d) . Factorization Theorem . Recap . . . . . . Summary Factorization Theorem : Proof (cont’d) . . . . . . . . . . . . . . . . . . . . . . . f X ( x | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = = q ( T ( x ) | θ ) q ( T ( x ) | θ ) ∑ y ∈ A T ( x ) f X ( y | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = y ∈ A T ( x ) g ( T ( y ) | θ ) h ( y ) = ∑ g ( T ( x ) | θ ) ∑ A y ∈ T ( x ) h ( y )

  21. f X x . if part (cont’d) January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang h x . g T x , if and only if Thus, T X is a sufficient statistic for . . . 9 / 27 . Factorization Theorem : Proof (cont’d) Summary . . . . . . . . . Recap Factorization Theorem . . . . . . . . . . . . . . . . . . . . . f X ( x | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = = q ( T ( x ) | θ ) q ( T ( x ) | θ ) ∑ y ∈ A T ( x ) f X ( y | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = y ∈ A T ( x ) g ( T ( y ) | θ ) h ( y ) = ∑ g ( T ( x ) | θ ) ∑ A y ∈ T ( x ) h ( y ) h ( x ) = ∑ A T ( x ) h ( y )

  22. . Factorization Theorem : Proof (cont’d) January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . . if part (cont’d) . 9 / 27 Summary . Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . f X ( x | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = = q ( T ( x ) | θ ) q ( T ( x ) | θ ) ∑ y ∈ A T ( x ) f X ( y | θ ) g ( T ( x ) | θ ) h ( x ) g ( T ( x ) | θ ) h ( x ) = y ∈ A T ( x ) g ( T ( y ) | θ ) h ( y ) = ∑ g ( T ( x ) | θ ) ∑ A y ∈ T ( x ) h ( y ) h ( x ) = ∑ A T ( x ) h ( y ) Thus, T ( X ) is a sufficient statistic for θ , if and only if f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) .

  23. Then f X x . t n i x i x Because T X X n , we have g t Pr T X exp n n t h x g T x holds, and T X X is a sufficient statistic for by the factorization theorem. Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 exp h x . Example 6.2.7 - Factorization of Normal Distribution . . . . . . . . Recap Factorization Theorem . Summary From Example 6.2.4, we know that . We can define h x , so that it does not depend on 10 / 27 . . . . . . . . . . . . . . . . . . . . . i =1 ( x i − x ) 2 + n ( x − µ ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp f X ( x | µ ) = − 2 σ 2

  24. Then f X x . X n , we have g t Pr T X t exp n t h x g T x . holds, and T X X is a sufficient statistic for by the factorization theorem. Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 Because T X 10 / 27 From Example 6.2.4, we know that . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.7 - Factorization of Normal Distribution . . . . . . . . . . . . . . . . . . . . . i =1 ( x i − x ) 2 + n ( x − µ ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp f X ( x | µ ) = − 2 σ 2 We can define h ( x ) , so that it does not depend on µ . i =1 ( x i − x ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp h ( x ) = − 2 σ 2

  25. Then f X x . From Example 6.2.4, we know that January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang by the factorization theorem. statistic for X is a sufficient holds, and T X h x g T x . 10 / 27 Example 6.2.7 - Factorization of Normal Distribution . Recap . . . . . . . . Summary Factorization Theorem . . . . . . . . . . . . . . . . . . . . . i =1 ( x i − x ) 2 + n ( x − µ ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp f X ( x | µ ) = − 2 σ 2 We can define h ( x ) , so that it does not depend on µ . i =1 ( x i − x ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp h ( x ) = − 2 σ 2 Because T ( X ) = X ∼ N ( µ, σ 2 / n ) , we have − n ( t − µ ) 2 ( ) g ( t | µ ) = Pr ( T ( X ) = t | µ ) = exp 2 σ 2

  26. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . From Example 6.2.4, we know that Example 6.2.7 - Factorization of Normal Distribution Summary 10 / 27 Factorization Theorem . . . . Recap . . . . . . . . . . . . . . . . . . . . . . . . . i =1 ( x i − x ) 2 + n ( x − µ ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp f X ( x | µ ) = − 2 σ 2 We can define h ( x ) , so that it does not depend on µ . i =1 ( x i − x ) 2 ( ∑ n ) (2 πσ 2 ) − n /2 exp h ( x ) = − 2 σ 2 Because T ( X ) = X ∼ N ( µ, σ 2 / n ) , we have − n ( t − µ ) 2 ( ) g ( t | µ ) = Pr ( T ( X ) = t | µ ) = exp 2 σ 2 Then f X ( x | µ ) = h ( x ) g ( T ( x ) | µ ) holds, and T ( X ) = X is a sufficient statistic for µ by the factorization theorem.

  27. . Example 6.2.8 - Uniform Sufficient Statistic January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang otherwise . . Problem . . Summary . . . . . 11 / 27 Recap . . . . Factorization Theorem . . . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n are iid observations uniformly drawn from { 1 , · · · , θ } . { 1 x = 1 , 2 , · · · , θ f X ( x | θ ) θ = 0 • Find a sufficient statistic for θ using factorization theorem.

  28. . . Define h x . . . . . . . h x otherwise x n otherwise Note that h x is independent of . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 12 / 27 . . . . . . . . . . Recap Factorization Theorem Summary Example 6.2.8 - Uniform Sufficient Statistic . Joint pmf . . . . . . . . . . . . . . . . . . . . . . . The joint pmf of X 1 , · · · , X n is { θ − n x ∈ { 1 , 2 , · · · , θ } n f X ( x | θ ) = 0

  29. . Joint pmf January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang otherwise . . . otherwise . . . 12 / 27 . Factorization Theorem . . . . . . . . Example 6.2.8 - Uniform Sufficient Statistic Recap Summary . . . . . . . . . . . . . . . . . . . . . . The joint pmf of X 1 , · · · , X n is { θ − n x ∈ { 1 , 2 , · · · , θ } n f X ( x | θ ) = 0 Define h ( x ) { 1 x ∈ { 1 , 2 , · · · } n h ( x ) = 0 Note that h ( x ) is independent of θ .

  30. • f X x • Thus, by the factorization theorem, T X max i X i is a sufficient . Putting things together . . . . . . . . otherwise g T x h x holds. statistic for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 13 / 27 . Summary . . . . . . . . Recap Factorization Theorem . Example 6.2.8 - Uniform Sufficient Statistic . . . i . . . . . . . . . . . . . . . . . . . . . Define T ( X ) and g ( t | θ ) Define T ( X ) = max i x i , then { θ − n t ≤ θ g ( t | θ ) = Pr ( T ( x ) = t | θ ) = Pr ( max x i = t | θ ) = 0

  31. • Thus, by the factorization theorem, T X max i X i is a sufficient . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . statistic for . . Putting things together . otherwise . i . Summary . . . . . . . . Recap Factorization Theorem . Example 6.2.8 - Uniform Sufficient Statistic . 13 / 27 . . . . . . . . . . . . . . . . . . . . . Define T ( X ) and g ( t | θ ) Define T ( X ) = max i x i , then { θ − n t ≤ θ g ( t | θ ) = Pr ( T ( x ) = t | θ ) = Pr ( max x i = t | θ ) = 0 • f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) holds.

  32. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . Putting things together . otherwise i . . 13 / 27 Recap Example 6.2.8 - Uniform Sufficient Statistic . Summary . . . . . . . . Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . Define T ( X ) and g ( t | θ ) Define T ( X ) = max i x i , then { θ − n t ≤ θ g ( t | θ ) = Pr ( T ( x ) = t | θ ) = Pr ( max x i = t | θ ) = 0 • f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) holds. • Thus, by the factorization theorem, T ( X ) = max i X i is a sufficient statistic for θ .

  33. . Factorization Theorem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang Summary . 14 / 27 . Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example of h ( x ) when θ = 5 , n = 1

  34. . Factorization Theorem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang Summary . 15 / 27 . Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example of g ( x ) when θ = 5 , n = 1

  35. . Factorization Theorem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang Summary . 16 / 27 . Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example of f ( x ) when θ = 5 , n = 1

  36. f X x f X x f X x x i , max i x i is a sufficient statistic. I T x n I T x I x i i i n x i i n . I I x i can be factorized into g t n I t and h x n i I and T x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 max I x i . . . . . . . . . Recap Factorization Theorem . Summary Alternative Solution - Using Indicator Functions • , and n i n i n x i I i x i I I i n n x i 17 / 27 . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise.

  37. f X x f X x f X x x i , max i x i is a sufficient statistic. . I n I T x I x i n i n x i i T x x i i I I can be factorized into g t n I t and h x n i I and T x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 max x i . I . . . . . . . . Recap Factorization Theorem . Summary Alternative Solution - Using Indicator Functions n i I n i n x i I i x i x i I i n n 17 / 27 . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise. • N = { 1 , 2 , · · · } , and N θ = { 1 , 2 , · · · , θ }

  38. f X x f X x x i , max i x i is a sufficient statistic. . x i n I T x I i I n n x i i max T x I i x i x i can be factorized into g t n I t and h x n i I and T x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 I I . n . . . . . . . . Recap Factorization Theorem . Summary Alternative Solution - Using Indicator Functions i 17 / 27 n i n x i I n . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise. • N = { 1 , 2 , · · · } , and N θ = { 1 , 2 , · · · , θ } 1 ∏ ∏ f X ( x | θ ) = θ I N θ ( x i ) = θ − n I N θ ( x i ) i =1 i =1

  39. f X x f X x x i , max i x i is a sufficient statistic. . i x i n I T x n i I x i n I can be factorized into g t . t and h x n i I and T x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 max 17 / 27 n Recap Summary . n Factorization Theorem n . . . . . . . . Alternative Solution - Using Indicator Functions . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise. • N = { 1 , 2 , · · · } , and N θ = { 1 , 2 , · · · , θ } 1 ∏ ∏ f X ( x | θ ) = θ I N θ ( x i ) = θ − n I N θ ( x i ) i =1 i =1 ( n ) ( n ) [ ] ∏ ∏ ∏ I N θ ( x i ) = I N ( x i ) = I N ( x i ) I N θ [ T ( x )] I N θ i =1 i =1 i =1

  40. f X x x i , max i x i is a sufficient statistic. . n I . max i x i n can be factorized into g t n t and h x n i I and T x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 n 17 / 27 . Recap . Factorization Theorem Summary . . . n Alternative Solution - Using Indicator Functions . . . . . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise. • N = { 1 , 2 , · · · } , and N θ = { 1 , 2 , · · · , θ } 1 ∏ ∏ f X ( x | θ ) = θ I N θ ( x i ) = θ − n I N θ ( x i ) i =1 i =1 ( n ) ( n ) [ ] ∏ ∏ ∏ I N θ ( x i ) = I N ( x i ) = I N ( x i ) I N θ [ T ( x )] I N θ i =1 i =1 i =1 ∏ θ − n I N θ [ T ( x )] f X ( x | θ ) = I N ( x i ) i =1

  41. . n January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang n x i i max . n n 17 / 27 . . . Alternative Solution - Using Indicator Functions . . . . . . Recap Summary Factorization Theorem . . . . . . . . . . . . . . . . . . . . . • I A ( x ) = 1 if x ∈ A , and I A ( x ) = 0 otherwise. • N = { 1 , 2 , · · · } , and N θ = { 1 , 2 , · · · , θ } 1 ∏ ∏ f X ( x | θ ) = θ I N θ ( x i ) = θ − n I N θ ( x i ) i =1 i =1 ( n ) ( n ) [ ] ∏ ∏ ∏ I N θ ( x i ) = I N ( x i ) = I N ( x i ) I N θ [ T ( x )] I N θ i =1 i =1 i =1 ∏ θ − n I N θ [ T ( x )] f X ( x | θ ) = I N ( x i ) i =1 f X ( x | θ ) can be factorized into g ( t | θ ) = θ − n I N θ ( t ) and h ( x ) = ∏ n i =1 I N ( x i ) , and T ( x ) = max i x i is a sufficient statistic.

  42. • Both • The parameter is a vector : • The problem is to use the Factorization Theorem to find the sufficient • Propose T X • Use Factorization Theorem to decompose f X x . . . . . . . . . T . X T X as sufficient statistic for and . . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 How to solve it statistics for . . . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.9 - Normal Sufficient Statistic . Problem . . i.i.d. and are unknown . 18 / 27 . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n

  43. • The parameter is a vector : • The problem is to use the Factorization Theorem to find the sufficient • Propose T X • Use Factorization Theorem to decompose f X x . . . . . . . . X T How to solve it T X as sufficient statistic for and . . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . . . . . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.9 - Normal Sufficient Statistic . Problem . . i.i.d. . statistics for 18 / 27 . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown

  44. • The problem is to use the Factorization Theorem to find the sufficient • Propose T X • Use Factorization Theorem to decompose f X x . T . . . . . . . X How to solve it T X as sufficient statistic for and . . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . . . . . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.9 - Normal Sufficient Statistic . Problem . . i.i.d. statistics for 18 / 27 . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown • The parameter is a vector : θ = ( µ, σ 2 ) .

  45. • Propose T X • Use Factorization Theorem to decompose f X x . T . . . . . . . T X How to solve it X as sufficient statistic for and . . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . . . Summary . . . . . . . . Recap Factorization Theorem . Example 6.2.9 - Normal Sufficient Statistic . Problem . . i.i.d. 18 / 27 . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown • The parameter is a vector : θ = ( µ, σ 2 ) . • The problem is to use the Factorization Theorem to find the sufficient statistics for θ .

  46. • Use Factorization Theorem to decompose f X x . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . . How to solve it . i.i.d. . . 18 / 27 Problem . . . . . . . . . Recap Factorization Theorem Summary . Example 6.2.9 - Normal Sufficient Statistic . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown • The parameter is a vector : θ = ( µ, σ 2 ) . • The problem is to use the Factorization Theorem to find the sufficient statistics for θ . • Propose T ( X ) = ( T 1 ( X ) , T 2 ( X )) as sufficient statistic for µ and σ 2 .

  47. . Problem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . How to solve it . i.i.d. . . . 18 / 27 . Recap Summary . Factorization Theorem . . . . . . . . Example 6.2.9 - Normal Sufficient Statistic . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown • The parameter is a vector : θ = ( µ, σ 2 ) . • The problem is to use the Factorization Theorem to find the sufficient statistics for θ . • Propose T ( X ) = ( T 1 ( X ) , T 2 ( X )) as sufficient statistic for µ and σ 2 . • Use Factorization Theorem to decompose f X ( x | µ, σ 2 ) .

  48. . Problem January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang . . How to solve it . i.i.d. . . . 18 / 27 . Recap Summary . Factorization Theorem . . . . . . . . Example 6.2.9 - Normal Sufficient Statistic . . . . . . . . . . . . . . . . . . . . . ∼ N ( µ, σ 2 ) • X 1 , · · · , X n • Both µ and σ 2 are unknown • The parameter is a vector : θ = ( µ, σ 2 ) . • The problem is to use the Factorization Theorem to find the sufficient statistics for θ . • Propose T ( X ) = ( T 1 ( X ) , T 2 ( X )) as sufficient statistic for µ and σ 2 . • Use Factorization Theorem to decompose f X ( x | µ, σ 2 ) .

  49. . x exp n i x i n exp n i x i x n . exp n i x i x n x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 n 19 / 27 Summary . . . . . . . . Recap n Factorization Theorem . . . . Example 6.2.9 - Solution . . . . . . . . . . . . . . . . . . . . . Decomposing f X ( x | µ, σ 2 ) - Similarly to Example 6.2.4 − ( x i − µ ) 2 ( ) 1 ∏ f X ( x | µ, σ 2 ) √ = 2 σ 2 2 πσ 2 exp i =1

  50. . exp n exp n i x i x x n n . i x i x n x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 n 19 / 27 . Recap n Example 6.2.9 - Solution Summary . Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Decomposing f X ( x | µ, σ 2 ) - Similarly to Example 6.2.4 − ( x i − µ ) 2 ( ) 1 ∏ f X ( x | µ, σ 2 ) √ = 2 σ 2 2 πσ 2 exp i =1 ( ) ( x i − µ ) 2 (2 πσ 2 ) − n /2 exp ∑ = − 2 σ 2 i =1

  51. . n January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang x n x x i i n exp n n n . 19 / 27 Summary . . . . . . . Example 6.2.9 - Solution . . Recap Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . . . Decomposing f X ( x | µ, σ 2 ) - Similarly to Example 6.2.4 − ( x i − µ ) 2 ( ) 1 ∏ f X ( x | µ, σ 2 ) √ = 2 σ 2 2 πσ 2 exp i =1 ( ) ( x i − µ ) 2 (2 πσ 2 ) − n /2 exp ∑ = − 2 σ 2 i =1 ( ) ( x i − x + x − µ ) 2 (2 πσ 2 ) − n /2 exp ∑ = − 2 σ 2 i =1

  52. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang n n n n . n . 19 / 27 . Recap . . . . Factorization Theorem . . . . . Example 6.2.9 - Solution Summary . . . . . . . . . . . . . . . . . . . . . Decomposing f X ( x | µ, σ 2 ) - Similarly to Example 6.2.4 − ( x i − µ ) 2 ( ) 1 ∏ f X ( x | µ, σ 2 ) √ = 2 σ 2 2 πσ 2 exp i =1 ( ) ( x i − µ ) 2 (2 πσ 2 ) − n /2 exp ∑ = − 2 σ 2 i =1 ( ) ( x i − x + x − µ ) 2 (2 πσ 2 ) − n /2 exp ∑ = − 2 σ 2 i =1 ( ) − 1 ( x i − x ) 2 − (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1

  53. . n T X T X T X T x x n i . x i T x n i x i x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 n 20 / 27 Propose a sufficient statistic . . . . . Summary Example 6.2.9 - Solution . Recap . . . . . . n Factorization Theorem . . . . . . . . . . . . . . . . . . . . . ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1

  54. . x i n T x x n n i T n x n i x i x Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 20 / 27 . Recap Propose a sufficient statistic Summary . . Factorization Theorem . Example 6.2.9 - Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 T ( X ) = ( T 1 ( X ) , T 2 ( X ))

  55. . n January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang x x i i n x T x i n n n . 20 / 27 . . Summary Recap . Example 6.2.9 - Solution . . Propose a sufficient statistic . . . . Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . . . ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 T ( X ) = ( T 1 ( X ) , T 2 ( X )) x = 1 ∑ T 1 ( x ) = i =1

  56. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang n x i n n n . n . 20 / 27 Propose a sufficient statistic Recap . . . . . . . . . Factorization Theorem . Summary Example 6.2.9 - Solution . . . . . . . . . . . . . . . . . . . . . ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 T ( X ) = ( T 1 ( X ) , T 2 ( X )) x = 1 ∑ T 1 ( x ) = i =1 ∑ ( x i − x ) 2 T 2 ( x ) = i =1

  57. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang n x i n n n . n . 20 / 27 Propose a sufficient statistic Recap . . . . . . . . . Factorization Theorem . Summary Example 6.2.9 - Solution . . . . . . . . . . . . . . . . . . . . . ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 T ( X ) = ( T 1 ( X ) , T 2 ( X )) x = 1 ∑ T 1 ( x ) = i =1 ∑ ( x i − x ) 2 T 2 ( x ) = i =1

  58. f X x . T n exp t n t g T x T x h x Thus, T X x g t T x x n i x i x is a sufficient statistic for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 t h x . . . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.9 - Solution . . n n 21 / 27 . . . . . . . . . . . . . . . . . . . . . Factorize f X ( x | µ, σ 2 ) ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1

  59. f X x . T n exp t n t g T x T x h x Thus, T X x g t T x x n i x i x is a sufficient statistic for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 t 21 / 27 . . . . . . . . . . Recap Factorization Theorem . Summary Example 6.2.9 - Solution . . n n . . . . . . . . . . . . . . . . . . . . . Factorize f X ( x | µ, σ 2 ) ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 h ( x ) = 1

  60. f X x . T n g T x T x h x Thus, T X T x x n x n i x i x is a sufficient statistic for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 21 / 27 . . . n . Example 6.2.9 - Solution Summary Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Factorize f X ( x | µ, σ 2 ) ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 h ( x ) = 1 ( − 1 ) g ( t 1 , t 2 | µ, σ 2 ) (2 πσ 2 ) − n /2 exp 2 σ 2 ( t 1 − µ ) 2 = 2 σ 2 t 2 −

  61. . n . n Thus, T X T x T x x i n x i x is a sufficient statistic for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 n 21 / 27 . . . . . . . Example 6.2.9 - Solution Summary . Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . Factorize f X ( x | µ, σ 2 ) ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 h ( x ) = 1 ( − 1 ) g ( t 1 , t 2 | µ, σ 2 ) (2 πσ 2 ) − n /2 exp 2 σ 2 ( t 1 − µ ) 2 = 2 σ 2 t 2 − f X ( x | µ, σ 2 ) g ( T 1 ( x ) , T 2 ( x ) | µ, σ 2 ) h ( x ) =

  62. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang is a sufficient statistic n . n . . n Example 6.2.9 - Solution Recap . . . . . . . . Summary 21 / 27 . Factorization Theorem . . . . . . . . . . . . . . . . . . . . . Factorize f X ( x | µ, σ 2 ) ( ) − 1 ( x i − x ) 2 − f X ( x | µ, σ 2 ) (2 πσ 2 ) − n /2 exp ∑ 2 σ 2 ( x − µ ) 2 = 2 σ 2 i =1 h ( x ) = 1 ( − 1 ) g ( t 1 , t 2 | µ, σ 2 ) (2 πσ 2 ) − n /2 exp 2 σ 2 ( t 1 − µ ) 2 = 2 σ 2 t 2 − f X ( x | µ, σ 2 ) g ( T 1 ( x ) , T 2 ( x ) | µ, σ 2 ) h ( x ) = i =1 ( x i − x ) 2 ) ( x , ∑ n Thus, T ( X ) = ( T 1 ( x ) , T 2 ( x )) = for θ = ( µ, σ 2 ) .

  63. Rewriting f X x f X x f X x . x i if x otherwise I x n i I I . x x n I min i x i max i x i Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . . . . . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic Problem . . . i.i.d. . . . . . 22 / 27 . . . . . . . . . . . . . . . . . . . . . Assume X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , −∞ < θ < ∞ . Find a sufficient statistic for θ .

  64. f X x . x n otherwise n i I x i I x I . min i x i max i x i Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 22 / 27 . . . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic Problem . . i.i.d. . . . . . . . . . . . . . . . . . . . . . . Assume X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , −∞ < θ < ∞ . Find a sufficient statistic for θ . Rewriting f X ( x | θ ) { 1 if θ < x < θ + 1 f X ( x | θ ) = = I ( θ < x < θ + 1) 0

  65. . I . . otherwise n I x x n min . i x i max i x i Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 22 / 27 Factorization Theorem . . Summary One parameter, two-dimensional sufficient statistic . Problem Recap . . . . . . . . . i.i.d. . . . . . . . . . . . . . . . . . . . . . Assume X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , −∞ < θ < ∞ . Find a sufficient statistic for θ . Rewriting f X ( x | θ ) { 1 if θ < x < θ + 1 f X ( x | θ ) = = I ( θ < x < θ + 1) 0 ∏ f X ( x | θ ) = I ( θ < x i < θ + 1) i =1

  66. . i . . otherwise n I min x i . max i x i Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 22 / 27 i.i.d. . . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic . . Problem . . . . . . . . . . . . . . . . . . . . . Assume X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , −∞ < θ < ∞ . Find a sufficient statistic for θ . Rewriting f X ( x | θ ) { 1 if θ < x < θ + 1 f X ( x | θ ) = = I ( θ < x < θ + 1) 0 ∏ f X ( x | θ ) = I ( θ < x i < θ + 1) i =1 I ( θ < x 1 < θ + 1 , · · · , θ < x n < θ + 1) =

  67. . i.i.d. January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang i i min I n otherwise . . . . 22 / 27 . Summary . . . . . . . . Recap . Factorization Theorem . One parameter, two-dimensional sufficient statistic Problem . . . . . . . . . . . . . . . . . . . . . . Assume X 1 , · · · , X n ∼ Uniform ( θ, θ + 1) , −∞ < θ < ∞ . Find a sufficient statistic for θ . Rewriting f X ( x | θ ) { 1 if θ < x < θ + 1 f X ( x | θ ) = = I ( θ < x < θ + 1) 0 ∏ f X ( x | θ ) = I ( θ < x i < θ + 1) i =1 I ( θ < x 1 < θ + 1 , · · · , θ < x n < θ + 1) = ( ) = x i > θ ∧ max x i < θ + 1

  68. f X x min i x i max i x i is a sufficient statistic . T I t t I min i x i max i g T x x g t h x Thus, T x T x T x for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 t x i . Factorization . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic . . i . T x min i x i T x max 23 / 27 . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1

  69. f X x min i x i max i x i is a sufficient statistic . T I t t I min i x i max i g T x x g t h x Thus, T x T x T x for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 t x i . Factorization . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic . . i . min i x i max 23 / 27 . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 T 1 ( x ) = T 2 ( x ) =

  70. f X x min i x i max i x i is a sufficient statistic . x I min i x i max i g T x T h x . Thus, T x T x T x for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 x i i max Factorization . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic . . . min i x i 23 / 27 . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 T 1 ( x ) = T 2 ( x ) = g ( t 1 , t 2 | θ ) = I ( t 1 > θ ∧ t 2 < θ + 1)

  71. min i x i max i x i is a sufficient statistic . Thus, T x max . x i I min i i x T i T x for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 x i i min . . . . . . . . . Recap Factorization Theorem . Summary One parameter, two-dimensional sufficient statistic . Factorization 23 / 27 . . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 T 1 ( x ) = T 2 ( x ) = g ( t 1 , t 2 | θ ) = I ( t 1 > θ ∧ t 2 < θ + 1) ( ) f X ( x | θ ) = x i > θ ∧ max < θ + 1 = g ( T 1 ( x ) , T 2 ( x ) | θ ) h ( x )

  72. . min January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang i i min I x i i max . x i i 23 / 27 . Recap . . . . . . . Factorization Theorem . Summary . One parameter, two-dimensional sufficient statistic . Factorization . . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 T 1 ( x ) = T 2 ( x ) = g ( t 1 , t 2 | θ ) = I ( t 1 > θ ∧ t 2 < θ + 1) ( ) f X ( x | θ ) = x i > θ ∧ max < θ + 1 = g ( T 1 ( x ) , T 2 ( x ) | θ ) h ( x ) Thus, T ( x ) = ( T 1 ( x ) , T 2 ( x )) = ( min i x i , max i x i ) is a sufficient statistic for θ .

  73. • Define order statistics x x n as an ordered permutation of • Is the order statistic a sufficient statistic for T n x x n . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang x x T T x ? x i.i.d. 24 / 27 . . . . . . . . . Recap Factorization Theorem . Summary Sufficient Order Statistics . Problem . . . . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n ∼ f X ( x | θ ) . • f X ( x | θ ) = ∏ n i =1 f X ( x i | θ )

  74. • Is the order statistic a sufficient statistic for T n x x n . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang x x T T x ? x . i.i.d. . Factorization Theorem . . . . . . . . Recap . Summary Sufficient Order Statistics . Problem 24 / 27 . . . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n ∼ f X ( x | θ ) . • f X ( x | θ ) = ∏ n i =1 f X ( x i | θ ) • Define order statistics x (1) ≤ · · · ≤ x ( n ) as an ordered permutation of

  75. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang x i.i.d. . . . Problem 24 / 27 Sufficient Order Statistics Summary . . . . . . . . Recap Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . • X 1 , · · · , X n ∼ f X ( x | θ ) . • f X ( x | θ ) = ∏ n i =1 f X ( x i | θ ) • Define order statistics x (1) ≤ · · · ≤ x ( n ) as an ordered permutation of • Is the order statistic a sufficient statistic for θ ? T ( x ) = ( T 1 ( x ) , · · · , T n ( x )) = ( x (1) , · · · , x ( n ) )

  76. f X t i f X x T n x T n x x n ) x n x h x (Note that T x Therefore, T x is a permutation of x x is a sufficient statistics for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 g T . . i . . . . . . . . Recap Factorization Theorem . Summary Factorization of Order Statistics g t t n n 25 / 27 . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1

  77. f X x T n x T n x x n ) x n . x h x (Note that T x Therefore, T x is a permutation of x x is a sufficient statistics for . Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 g T 25 / 27 . . . . . . . . . . Recap Factorization Theorem Summary n Factorization of Order Statistics . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 ∏ g ( t 1 , · · · , t n | θ ) = f X ( t i | θ ) i =1

  78. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang n . Factorization of Order Statistics Summary 25 / 27 Factorization Theorem Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . h ( x ) = 1 ∏ g ( t 1 , · · · , t n | θ ) = f X ( t i | θ ) i =1 f X ( x | θ ) = g ( T 1 ( x ) , · · · , T n ( x ) | θ ) h ( x ) (Note that ( T 1 ( x ) , · · · , T n ( x )) is a permutation of ( x 1 , · · · , x n ) ) Therefore, T ( x ) = ( x (1) , · · · , x ( n ) ) is a sufficient statistics for θ .

  79. f X x Then f X x . x . . exp x Define h x T x exp g t . t g T x h x holds, and T X X is a sufficient statistic by the Factorization Theorem. Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . . . Summary . . . . . . . . Recap Factorization Theorem . Exercise 6.1 . . Problem . . . Solution . . 26 / 27 . . . . . . . . . . . . . . . . . . . . . X is one observation from a N (0 , σ 2 ) . Is | X | a sufficient statistic for σ 2 ?

  80. Then f X x t . Define T x x g t exp . . g T x h x holds, and T X X is a sufficient statistic by the Factorization Theorem. Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 26 / 27 Solution Recap . Problem . . Factorization Theorem . Exercise 6.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . X is one observation from a N (0 , σ 2 ) . Is | X | a sufficient statistic for σ 2 ? − x 2 1 ( ) f X ( x | σ 2 ) = √ 2 σ 2 2 πσ 2 exp h ( x ) = 1

  81. Then f X x . . . Define g t exp t g T x Solution h x holds, and T X X is a sufficient statistic by the Factorization Theorem. Hyun Min Kang Biostatistics 602 - Lecture 02 January 15th, 2013 . 26 / 27 . Factorization Theorem Summary Exercise 6.1 Recap . . . . . . . . . Problem . . . . . . . . . . . . . . . . . . . . . . . . X is one observation from a N (0 , σ 2 ) . Is | X | a sufficient statistic for σ 2 ? − x 2 1 ( ) f X ( x | σ 2 ) = √ 2 σ 2 2 πσ 2 exp h ( x ) = 1 T ( x ) = | x |

  82. Then f X x . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang statistic by the Factorization Theorem. X is a sufficient h x holds, and T X g T x Define . . . Solution 26 / 27 . . Problem . Exercise 6.1 . . . . . . . . Recap Summary Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . X is one observation from a N (0 , σ 2 ) . Is | X | a sufficient statistic for σ 2 ? − x 2 1 ( ) f X ( x | σ 2 ) = √ 2 σ 2 2 πσ 2 exp h ( x ) = 1 T ( x ) = | x | − t 2 1 ( ) g ( t | θ ) √ = 2 σ 2 2 πσ 2 exp

  83. . . January 15th, 2013 Biostatistics 602 - Lecture 02 Hyun Min Kang statistic by the Factorization Theorem. Define . . . Solution . . 26 / 27 Problem . . . . . . . . Recap . Factorization Theorem . Summary Exercise 6.1 . . . . . . . . . . . . . . . . . . . . . X is one observation from a N (0 , σ 2 ) . Is | X | a sufficient statistic for σ 2 ? − x 2 1 ( ) f X ( x | σ 2 ) = √ 2 σ 2 2 πσ 2 exp h ( x ) = 1 T ( x ) = | x | − t 2 1 ( ) g ( t | θ ) √ = 2 σ 2 2 πσ 2 exp Then f X ( x | θ ) = g ( T ( x ) | θ ) h ( x ) holds, and T ( X ) = | X | is a sufficient

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