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Dr. Hackney STA Solutions pg 82

# Dr. Hackney STA Solutions pg 82 - 5-16 Solutions Manual for...

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5-16 Solutions Manual for Statistical Inference 5.55 Let X denote the number of comparisons. Then E X = k =0 P ( X > k ) = 1 + k =1 P ( U > F y ( y k - 1 )) = 1 + k =1 (1 - F y ( y k - 1 )) = 1 + k =0 (1 - F y ( y i )) = 1 + E Y 5.57 a. Cov( Y 1 , Y 2 ) = Cov( X 1 + X 3 , X 2 + X 3 ) = Cov( X 3 , X 3 ) = λ 3 since X 1 , X 2 and X 3 are independent. b. Z i = 1 if X i = X 3 = 0 0 otherwise p i = P ( Z i = 0) = P ( Y i = 0) = P ( X i = 0 , X 3 = 0) = e - ( λ i + λ 3 ) . Therefore Z i are Bernoulli( p i ) with E[ Z i ] = p i , Var( Z i ) = p i (1 - p i ) and E[ Z 1 Z 2 ] = P ( Z 1 = 1 , Z 2 = 1) = P ( Y 1 = 0 , Y 2 = 0) = P ( X 1 + X 3 = 0 , X 2 + X 3 = 0) = P ( X 1 = 0) P ( X 2 = 0) P ( X 3 = 0) = e - λ 1 e - λ 2 e - λ 3 . Therefore, Cov( Z 1 , Z 2 ) = E[ Z 1 Z 2 ] - E[ Z 1 ]E[ Z 2 ] = e - λ 1 e - λ 2 e - λ 3 - e - ( λ i + λ 3 ) e - ( λ 2 + λ 3 ) = e - ( λ i + λ 3 ) e - ( λ 2 + λ 3 ) ( e λ 3 - 1) = p 1 p 2 ( e λ 3 - 1) . Thus Corr( Z 1 , Z 2 ) = p 1 p 2 ( e λ 3 - 1) p 1 (1 - p 1 ) p 2 (1 - p 2 ) . c. E[ Z 1 Z 2 ] p i , therefore Cov( Z 1 , Z 2 ) = E[ Z 1 Z 2 ] - E[ Z 1 ]E[ Z 2 ] p 1 - p 1 p 2 = p 1 (1 - p 2 ) , and Cov( Z 1 , Z 2 ) p 2 (1 - p 1 ) . Therefore, Corr( Z 1 , Z 2 ) p 1 (1 - p 2 ) p 1 (1 - p 1 ) p 2 (1 - p 2 ) = p 1 (1 - p 2 ) p 2 (1 - p 1 ) and Corr( Z 1 , Z
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