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Dr. Hackney STA Solutions pg 55

# Dr. Hackney STA Solutions pg 55 - Second Edition 4-11 = =...

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Second Edition 4-11 = r + x - 1 x Γ( α + β ) Γ( α )Γ( β ) 1 0 p ( r + α ) - 1 (1 - p ) ( x + β ) - 1 dp = r + x - 1 x Γ( α + β ) Γ( α )Γ( β ) Γ( r + α )Γ( x + β ) Γ( r + x + α + β ) x = 0 , 1 , . . . Therefore, E X = E[E( X | P )] = E r (1 - P ) P = α - 1 , since E 1 - P P = 1 0 1 - P P Γ( α + β ) Γ( α )Γ( β ) p α - 1 (1 - p ) β - 1 dp = Γ( α + β ) Γ( α )Γ( β ) 1 0 p ( α - 1) - 1 (1 - p ) ( β +1) - 1 dp = Γ( α + β ) Γ( α )Γ( β ) Γ( α - 1)Γ( β + 1) Γ( α + β ) = β α - 1 . Var( X ) = E(Var( X | P )) + Var(E( X | P )) = E r (1 - P ) P 2 + Var r (1 - P ) P = r ( β + 1)( α + β ) α ( α - 1) + r 2 β ( α + β - 1) ( α - 1) 2 ( α - 2) , since E 1 - P P 2 = 1 0 Γ( α + β ) Γ( α )Γ( β ) p ( α - 2) - 1 (1 - p ) ( β +1) - 1 dp = Γ( α + β ) Γ( α )Γ( β ) Γ( α - 2)Γ( β + 1) Γ( α + β - 1) = ( β + 1)( α + β ) α ( α - 1) and Var 1 - P P = E 1 - P P 2 - E 1 - P P 2 = β ( β + 1) ( α - 2)( α - 1) - ( β α - 1 ) 2 = β ( α + β - 1) ( α - 1) 2 ( α - 2) , where E 1 - P P 2 = 1 0 Γ( α + β ) Γ( α
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