09_26_11 - STAT 409 Examples for 09/26/2011 Fall 2011 Gamma...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
STAT 409 Examples for 09/26/2011 Fall 2011 Gamma Distribution : ( ) ( ) x e x x f λ 1 α α α λ - - Γ = , 0 x < OR ( ) ( ) θ 1 α 1 α θ α x e x x f - - Γ = , 0 x < If T has a Gamma ( α , θ = 1 / λ ) distribution, where α is an integer, then F T ( t ) = P ( T t ) = P ( X t α ) , P ( T > t ) = P ( X t α – 1 ) , where X t has a Poisson ( λ 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 α and θ ? M X ( t ) = M χ 2 ( r ) ( t ) = ( ) 2 2 1 1 r t - , t < 2 1 . M Gamma ( α , θ ) ( t ) = ( ) α 1 1 θ t - , t < θ 1 . If X has a chi-square distribution with r degrees of freedom, then X has a Gamma distribution with α = r / 2 and θ = 2. b) Let Y be a random variable with a Gamma distribution with parameters α and θ = 1 / λ . Assume α is an integer. Show that 2 Y / θ has a chi-square distribution. What is the number of degrees of freedom? M Y ( t ) = M Gamma ( α , θ ) ( t ) = ( ) α 1 1 θ t - , t < θ 1 . If W = a Y + b , then M W ( t ) = e b t M Y ( a t ). M 2 Y / θ ( t ) = M Y ( 2 t / θ ) = ( ) α 2 1 1 t - , t < 2 1 . 2 Y / θ has a chi-square distribution with r = 2 α degrees of freedom.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. Let Y be a random variable with a Gamma distribution with α = 5 and θ = 3. Find the probability P ( Y > 18 ) … a) … by integrating the p.d.f. of the Gamma distribution; P ( Y > 18 ) = ( ) Γ - - 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 / θ = 6 ). EXCEL: = POISSON( x , λ , 0 ) gives P( X = x ) = POISSON( x , λ , 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. P ( Y > 18 ) = P > 18 3 2 Y 3 2 = P ( X > 12 ) where X is χ 2 ( 5 2 = 10 ). = CHIINV ( α , v ) gives ( ) v 2 α χ for 2 χ distribution with v degrees of freedom, x s.t. P ( ( ) v 2 χ > x ) = α . = CHIDIST ( x , v ) gives the upper tail probability for 2 χ distribution with v degrees of freedom, P ( ( ) v 2 χ > x ) . A B A B 1 =CHIDIST(12,10) 1 0.285057 2
Background image of page 2
3. Let X be a random variable with a Gamma distribution with α = 3 and θ = 5 ( i.e., λ = 0.2 ). Find the probability P ( X > 31.48 ) … a) … by integrating the p.d.f. of the Gamma distribution; P ( X > 31.48 ) = - 48 . 31 5 2 3 5
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

09_26_11 - STAT 409 Examples for 09/26/2011 Fall 2011 Gamma...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online