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Unformatted text preview: Solution for 36217 Wanjie Wang Teacher: Jiashun Jin October 19, 2009 If there is any question, please contact me at wwang@stat.cmu.edu 1. (10 Points) What is the value of ( n + 1 2 ) , ( n ) ? Calculate (6 . 5) and (11) ... ( 1 2 )( 1 2 ) = 1 / 2 = . So, ( 1 2 ) = Then, we can calculate ( n + 1 2 ) with it, ( n + 1 2 ) = ( n 1 2 )( n 1 + 1 2 ) = ( n 1 2 )( n 3 2 ) 1 2 ( 1 2 ) = (2 n 1)!! 2 n ( n ) = ( n 1)( n 1) = ( n 1)( n 2) (1) = ( n 1)! With these results, we have (6 . 5) = (11)!! 2 6 = 10395 64 287 . 8853 (11) = 10! = 3628800 2. (10 Points) Find E [ X 3 ] , E [ X 4 ] , E [ X 5 ] with X distributed as Expo nential(2). E [ X 3 ] = R x 3 f ( x ) dx = R x 3 e x dx = R  x 3 d ( e x ) = x 3 e x   R e x d ( x 3 ) = R 3 x 2 e x dx = = 6 / 3 1 Similarly, E [ X 4 ] = Z x 4 f ( x ) dx = Z x 4 e x dx = 24 / 4 E [ X 5 ] = Z x 5 f ( x ) dx = Z x 5 e x dx = 120 / 5 If the probability density function is given right, there is one point. It is the same with Problem 4. 3. (10 Points) Find E [ X 4 ] , E [ X 5 ] , E [ X 6 ] with X distributed as N (0 , 4) . The probability density function of X is f ( x ) = 1 8 exp( x 2 8 ) Let y = x/ 2, so Y N (0 , 1). Then we have E [ X 4 ] = R x 4 f ( x ) dx = R  x 4 1 8 exp( x 2 8 ) dx = 1 2 R  16 y 4 exp(...
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This note was uploaded on 09/30/2010 for the course STATISTICS 36217 taught by Professor Jin during the Fall '09 term at Carnegie Mellon.
 Fall '09
 Jin

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