Dr. Hackney STA Solutions pg 53

Dr. Hackney STA Solutions pg 53 - P ( Y = y | N = n, ) P (...

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Second Edition 4-9 4.31 a. E Y = E { E( Y | X ) } = E nX = n 2 . Var Y = Var (E( Y | X )) + E (Var( Y | X )) = Var( nX ) + E nX (1 - X ) = n 2 12 + n 6 . b. P ( Y = y,X x ) = ± n y ² x y (1 - x ) n - y , y = 0 , 1 ,...,n, 0 < x < 1 . c. P ( y = y ) = ± n y ² Γ( y + 1)Γ( n - y + 1) Γ( n + 2) . 4.32 a. The pmf of Y , for y = 0 , 1 ,... , is f Y ( y ) = Z 0 f Y ( y | λ ) f Λ ( λ ) = Z 0 λ y e - λ y ! 1 Γ( α ) β α λ α - 1 e - λ/β = 1 y !Γ( α ) β α Z 0 λ ( y + α ) - 1 exp - λ ³ β 1+ β ´ = 1 y !Γ( α ) β α Γ( y + α ) ± β 1+ β ² y + α . If α is a positive integer, f Y ( y ) = ± y + α - 1 y ²± β 1+ β ² y ± 1 1+ β ² α , the negative binomial( α, 1 / (1 + β )) pmf. Then E Y = E(E( Y | Λ)) = EΛ = αβ Var Y = Var(E( Y | Λ)) + E(Var( Y | Λ)) = VarΛ + EΛ = αβ 2 + αβ = αβ ( β + 1) . b. For y = 0 , 1 ,... , we have P ( Y = y | λ ) = X n = y
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Unformatted text preview: P ( Y = y | N = n, ) P ( N = n | ) = X n = y n y p y (1-p ) n-y e- n n ! = X n = y 1 y !( n-y )! p 1-p y [(1-p ) ] n e- = e- X m =0 1 y ! m ! p 1-p y [(1-p ) ] m + y (let m = n-y ) = e- y ! p 1-p y [(1-p ) ] y &quot; X m =0 [(1-p ) ] m m ! # = e- ( p ) y e (1-p ) = ( p ) y e-p y ! ,...
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