07_08ans - STAT 410 Examples for 07/08/09 Summer 2009 Gamma...

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STAT 410 Examples for 07/08/09 Summer 2009 Gamma Distribution : ( 29 ( 29 x e x x f λ 1 α α α λ - - Γ = , 0 x < OR ( 29 ( 29 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 ) = ( 29 2 2 1 1 r t - , t < 2. M Gamma ( α , θ ) ( t ) = ( 29 α 1 1 θ t - , t < θ . 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 ) = ( 29 α 1 1 θ t - , t < θ . 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 ) = ( 29 α 2 1 1 t - . 2 Y / θ has a chi-square distribution with r = 2 α degrees of freedom.
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2. Let Y be a random variable with a Gamma distribution with α = 5 and θ = 4 ( i.e., λ = 0.25 ). Find the probability P ( Y 15 ) … a) … by integrating the p.d.f. of the Gamma distribution; P ( Y 15 ) = ( 29 - - Γ 15 0 4 1 5 5 4 5 1 dx e x x = b) … by using the relationship between Gamma and Poisson distributions; P ( Y 15 ) = P ( X 15 5 ) = 1 – P ( X 15 4 ) where X 15 is Poisson ( 15 λ = 3.75 ). EXCEL: = POISSON( x , λ , 0 ) gives P( X = x ) = POISSON( x , λ , 1 ) gives P( X x ) A B A B 1 =1-POISSON(4,15/4,1) 1 0.322452 2 2 c) … by using the relationship between Gamma and Chi-square distribution. P ( Y 15 ) = P 15 4 2 Y 4 2 = P ( X 7.5 ) where X is χ 2 ( 5 2 = 10 ). = CHIINV ( α , v ) gives ( 29 v 2 α χ for 2 χ distribution with v degrees of freedom, x s.t. P ( ( 29 v 2 χ > x ) = α . = CHIDIST ( x , v ) gives the upper tail probability for 2 χ distribution with v degrees of freedom, P ( ( 29 v 2 χ > x ) . A B A B 1 =1-CHIDIST(7.5,10) 1 0.322452 2
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3. Traffic accidents at a particular intersection follow Poisson distribution with an average rate of 1.4 per week. a) What is the probability that the next week is accident-free? 1 week λ = 1.4. P ( X = 0 ) = ! 0 4 . 1 4 . 1 0 e - 0.2466 . b) What is the probability that there will be exactly 3 accidents next week? 1 week λ = 1.4. P ( X = 3 ) = ! 3 4 . 1 4 . 1 3 e - 0.1128 . c)
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07_08ans - STAT 410 Examples for 07/08/09 Summer 2009 Gamma...

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