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Unformatted text preview: Chapter 6 Principles of Data Reduction 6.1 By the Factorization Theorem,  X  is sufficient because the pdf of X is f ( x  2 ) = 1 2 e x 2 / 2 2 = 1 2 e x  2 / 2 2 = g (  x  2 ) 1  {z } h ( x ) . 6.2 By the Factorization Theorem, T ( X ) = min i ( X i /i ) is sufficient because the joint pdf is f ( x 1 ,...,x n  ) = n Y i =1 e i x i I ( i, + ) ( x i ) = e in I ( , + ) ( T ( x ))  {z } g ( T ( x )  ) e i x i  {z } h ( x ) . Notice, we use the fact that i &gt; 0, and the fact that all x i s &gt; i if and only if min i ( x i /i ) &gt; . 6.3 Let x (1) = min i x i . Then the joint pdf is f ( x 1 ,...,x n  , ) = n Y i =1 1 e ( x i ) / I ( , ) ( x i ) = e / n e i x i / I ( , ) ( x (1) )  {z } g ( x (1) , i x i  , ) 1  {z } h ( x ) . Thus, by the Factorization Theorem, ( X (1) , i X i ) is a sufficient statistic for ( , ). 6.4 The joint pdf is n Y j =1 h ( x j ) c ( ) exp k X i =1 w i ( ) t i ( x j ) ! = c ( ) n exp k X i =1 w i ( ) n X j =1 t i ( x j )  {z } g ( T ( x )  ) n Y j =1 h ( x j )  {z } h ( x ) . By the Factorization Theorem, n j =1 t 1 ( X j ) ,..., n j =1 t k ( X j ) is a sufficient statistic for . 6.5 The sample density is given by n Y i =1 f ( x i  ) = n Y i =1 1 2 i I ( i (  1) x i i ( + 1)) = 1 2 n n Y i =1 1 i ! I min x i i  (  1) I max x i i + 1 . Thus (min X i /i, max X i /i ) is sufficient for . 62 Solutions Manual for Statistical Inference 6.6 The joint pdf is given by f ( x 1 ,...,x n  , ) = n Y i =1 1 ( ) x i  1 e x i / = 1 ( ) n n Y i =1 x i !  1 e i x i / . By the Factorization Theorem, ( Q n i =1 X i , n i =1 X i ) is sufficient for ( , ). 6.7 Let x (1) = min i { x 1 ,...,x n } , x ( n ) = max i { x 1 ,...,x n } , y (1) = min i { y 1 ,...,y n } and y ( n ) = max i { y 1 ,...,y n } . Then the joint pdf is f ( x , y  ) = n Y i =1 1 ( 3 1 )( 4 2 ) I ( 1 , 3 ) ( x i ) I ( 2 , 4 ) ( y i ) = 1 ( 3 1 )( 4 2 ) n I ( 1 , ) ( x (1) ) I ( , 3 ) ( x ( n ) ) I ( 2 , ) ( y (1) ) I ( , 4 ) ( y ( n ) )  {z } g ( T ( x )  ) 1 {z} h ( x ) . By the Factorization Theorem, ( X (1) ,X ( n ) ,Y (1) ,Y ( n ) ) is sufficient for ( 1 , 2 , 3 , 4 ). 6.9 Use Theorem 6.2.13. a. f ( x  ) f ( y  ) = (2 ) n/ 2 e i ( x i ) 2 / 2 (2 ) n/ 2 e i ( y i ) 2 / 2 = exp 1 2 &quot; n X i =1 x 2 i n X i =1 y 2 i ! +2 n ( y x ) # . This is constant as a function of if and only if y = x ; therefore X is a minimal sufficient statistic for ....
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This note was uploaded on 04/18/2010 for the course STAT 622 taught by Professor Peruggia,m during the Spring '08 term at Ohio State.
 Spring '08
 Peruggia,M

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