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Unformatted text preview: Exam 1 Solutions (1) (a) The joint density of W 1 ,W 2 ,...,W n is f a ( w 1 ,w 2 ,...,w n ) = 1 2 a n n Y i =1 1 { a w i a } = 1 2 a n 1 { a min w i max w i a } = 1 2 a n 1 { max { w (1)  ,  w ( n ) } a } , where w (1) = min i w i , and w ( n ) = max i w i . Thus, by the factorization theorem, max(  W (1)  ,  W ( n )  ) = max 1 i n { W i } is sufficient. Clearly (min W i , max W i ) cannot be minimal, since it cannot be written as a function of max 1 i n { W i } . (b) ( W (1) ,W ( n ) ) is not complete. To see this, let U i = W i . Consider U 1 ,U 2 ,...,U n . Then U i s are i.i.d. Unif( a,a ) and note that U (1) = W ( n ) . So, E U (1) = E W ( n ) . But U (1) d = W (1) so E U (1) = E W (1) . Thus, E( W (1) + W ( n ) ) = 0 but W (1) + W ( n ) 6 = 0, showing that W (1) + W ( n ) is not complete. (2) (a) By using the tower property of conditional expectation, E Y = E [E [ Y  X ]] = E [ X ] = E [ X ] = 2 ....
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This note was uploaded on 04/14/2010 for the course STATS 610 taught by Professor Moulib during the Fall '09 term at University of Michigan.
 Fall '09
 moulib

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