Dr. Hackney STA Solutions pg 32

Dr. Hackney STA Solutions pg 32 - E X 2 =-1 log p ∞ X x...

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Second Edition 3-5 3.13 For any X with support 0 , 1 ,. .. , we have the mean and variance of the 0 - truncated X T are given by E X T = X x =1 xP ( X T = x ) = X x =1 x P ( X = x ) P ( X > 0) = 1 P ( X > 0) X x =1 xP ( X = x ) = 1 P ( X > 0) X x =0 xP ( X = x ) = E X P ( X > 0) . In a similar way we get E X 2 T = E X 2 P ( X> 0) . Thus, Var X T = E X 2 P ( X > 0) - ± E X P ( X > 0) ² 2 . a. For Poisson( λ ), P ( X > 0) = 1 - P ( X = 0) = 1 - e - λ λ 0 0! = 1 - e - λ , therefore P ( X T = x ) = e - λ λ x x !(1 - e - λ ) x = 1 , 2 ,. .. E X T = λ/ (1 - e - λ ) Var X T = ( λ 2 + λ ) / (1 - e - λ ) - ( λ/ (1 - e - λ )) 2 . b. For negative binomial( r, p ), P ( X > 0) = 1 - P ( X = 0) = 1 - ( r - 1 0 ) p r (1 - p ) 0 = 1 - p r . Then P ( X T = x ) = ( r + x - 1 x ) p r (1 - p ) x 1 - p r , x = 1 , 2 ,. .. E X T = r (1 - p ) p (1 - p r ) Var X T = r (1 - p ) + r 2 (1 - p ) 2 p 2 (1 - p r ) - ³ r (1 - p ) p (1 - p r ) 2 ´ . 3.14 a. x =1 - (1 - p ) x x log p = 1 log p x =1 - (1 - p ) x x = 1 , since the sum is the Taylor series for log p . b. E X = - 1 log p " X x =1 (1 - p ) x # = - 1 log p " X x =0 (1 - p ) x - 1 # == - 1 log p ³ 1 p - 1 ´ = - 1 log p ± 1 - p p ² . Since the geometric series converges uniformly,
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Unformatted text preview: E X 2 =-1 log p ∞ X x =1 x (1-p ) x = (1-p ) log p ∞ X x =1 d dp (1-p ) x = (1-p ) log p d dp ∞ X x =1 (1-p ) x = (1-p ) log p d dp ³ 1-p p ´ =-(1-p ) p 2 log p . Thus Var X =-(1-p ) p 2 log p ³ 1 + (1-p ) log p ´ . Alternatively, the mgf can be calculated, M x ( t ) =-1 log p ∞ X x =1 h (1-p ) e t i x = log(1+ pe t-e t ) log p and can be differentiated to obtain the moments....
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