Dr. Hackney STA Solutions pg 55

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

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Unformatted text preview: Second Edition 4-11 = = Therefore, r+x-1 x r+x-1 x ( + ) 1 (r+)-1 p (1 - p)(x+)-1 dp ()() 0 ( + ) (r + )(x + ) x = 0, 1, . . . ()() (r + x + + ) r(1 - P ) r = , P -1 EX = E[E(X|P )] = E since E 1-P P 1 = 0 1-P P 1 0 ( + ) -1 p (1 - p)-1 dp ()() p(-1)-1 (1 - p)(+1)-1 dp = ( + ) ( - 1)( + 1) ()() ( + ) = = ( + ) ()() . -1 Var(X) = E(Var(X|P )) + Var(E(X|P )) = E ( + - 1) ( + 1)( + ) + r2 , ( - 1) ( - 1)2 ( - 2) r(1 - P ) + Var P2 r(1 - P ) P = r since E 1-P P2 = = and Var 1-P P ( + ) (-2)-1 p (1 - p)(+1)-1 dp = 0 ()() ( + 1)( + ) ( - 1) 1 ( + ) ( - 2)( + 1) ()() ( + - 1) = = E 1-P P 2 - E 1-P P 2 = ( + 1) 2 -( ) ( - 2)( - 1) -1 ( + - 1) , ( - 1)2 ( - 2) where E 1-P P 2 1 = 0 ( + ) (-2)-1 p (1 - p)(+2)-1 dp ()() = ( + 1) . ( - 2)( - 1) = ( + ) ( - 2)( + 2) ()() ( - 2 + + 2) 4.35 a. Var(X) = E(Var(X|P )) + Var(E(X|P )). Therefore, Var(X) E[nP (1 - P )] + Var(nP ) = n + n2 VarP ( + )( + + 1) ( + + 1 - 1) = n + n2 VarP ( + 2 )( + + 1) = ...
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