09_23 - STAT 410 Examples for 09/23/2011 Fall 2011 Gamma...

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Unformatted text preview: STAT 410 Examples for 09/23/2011 Fall 2011 Gamma Distribution : ( ) ( ) x e x x f 1 -- = , 0 x < OR ( ) ( ) 1 1 x e x x f-- = , 0 x < If T has a Gamma ( , = 1 / ) distribution, where is an integer, then F T ( t ) = P ( T t ) = P ( X t ) , P ( T > t ) = P ( X t 1 ) , where X t has a Poisson ( t ) distribution. 1. a) Let X be a random variable with a chi-square distribution with r degrees of freedom. Show that X has a Gamma distribution. What are and ? M X ( t ) = M 2 ( r ) ( t ) = ( ) 2 2 1 1 r t- , t < 2 1 . M Gamma ( , ) ( t ) = ( ) 1 1 t- , t < 1 . If X has a chi-square distribution with r degrees of freedom, then X has a Gamma distribution with = r / 2 and = 2. b) Let Y be a random variable with a Gamma distribution with parameters and = 1 / . Assume is an integer. Show that 2 Y / has a chi-square distribution. What is the number of degrees of freedom? M Y ( t ) = M Gamma ( , ) ( t ) = ( ) 1 1 t- , t < 1 . If W = a Y + b , then M W ( t ) = e b t M Y ( a t ). M 2 Y / ( t ) = M Y ( 2 t / ) = ( ) 2 1 1 t- , t < 2 1 . 2 Y / has a chi-square distribution with r = 2 degrees of freedom. 2. Let Z be a N ( 0, 1 ) standard normal random variable. Show that X = Z 2 has a chi-square distribution with 1 degree of freedom. M X ( t ) = E ( e t Z 2 ) = -- z e e d z z t 2 2 2 2 1 = ( ) ( ) --- z e d t z 2 2 1 2 2 1 = ( ) 2 1 2 1 1 t- , t < 1 / 2 , since ( ) ( ) ( ) 2 2 1 2 2 1 2 2 1 t z e t- -- is the p.d.f. of a N ( 0, t 2 1 1- ) random variable. X has a 2 ( 1 ) distribution. OR F X ( x ) = P ( X x ) = P ( Z 2 x ) = P ( x- Z x ) = -- x x d z z e 2 2 2 1 = F Z ( x ) F Z ( x- ). f X ( x ) = F ' X ( x ) = x 2 1 f Z ( x ) - x 2 1 f Z ( x- ) = - -- - 2 2 2 1 2 1 2 1 2 1 x x e x e x = 2 2 1 2 1 2 1 x e x-- = ( ) ( ) 2 1 2 1 2 1 2 2 1 1 x e x-- , x > 0. X ~ 2 ( 1 ) 3. Let Y be a random variable with a Gamma distribution with = 5 and = 3. Find the probability P ( Y > 18 ) a) by integrating the p.d.f. of the Gamma distribution; P ( Y > 18 ) = ( ) -- 18 3 1 5 5 3 5 1 dx e x x = - 18 3 4 5,832 1 dx e x x = b) by using the relationship between Gamma and Poisson distributions; P ( Y > 18 ) = P ( X 18 4 ) = 0.285 where X 18 is Poisson ( 18 / = 6 ). EXCEL: = POISSON( x , , ) gives P( X = x ) = POISSON( x , , 1 ) gives P( X x ) A B A B 1 =POISSON(4,18/3,1) 1 0.285057 2 2 c) by using the relationship between Gamma and Chi-square distribution....
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This note was uploaded on 10/31/2011 for the course MATH 464 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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09_23 - STAT 410 Examples for 09/23/2011 Fall 2011 Gamma...

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