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Dr. Hackney STA Solutions pg 57

# Dr. Hackney STA Solutions pg 57 - Second Edition 4-13 r = =...

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Second Edition 4-13 = 1 Γ( r )Γ(1 - r ) 0 1 x x λu r e - ( u + x/λ ) du = x r - 1 e - x/λ λ r Γ( r )Γ(1 - r ) 0 1 u r e - u du = x r - 1 e - x/λ Γ( r ) λ r , since the integral is equal to Γ(1 - r ) if r < 1. b. Use the transformation t = ν/λ to get λ 0 p λ ( ν ) = 1 Γ( r )Γ(1 - r ) λ 0 ν r - 1 ( λ - ν ) - r = 1 Γ( r )Γ(1 - r ) 1 0 t r - 1 (1 - t ) - r dt = 1 , since this is a beta( r, 1 - r ). c. d dx log f ( x ) = d dx log 1 Γ( r ) λ r +( r - 1) log x - x/λ = r - 1 x - 1 λ > 0 for some x , if r > 1. But, d dx log 0 e - x/ν ν q λ ( ν ) = - 0 1 ν 2 e - x/ν q λ ( ν ) 0 1 ν e - x/ν q λ ( ν ) < 0 x. 4.39 a. Without loss of generality lets assume that i < j . From the discussion in the text we have that f ( x 1 , . . . , x j - 1 , x j +1 , . . . , x n | x j ) = ( m - x j )! x 1 ! · · · · · x j - 1 ! · x j +1 ! · · · · · x n ! × p 1 1 - p j x 1 · · · · · p j - 1 1 - p j x j - 1 p j +1 1 - p j x j +1 · · · · · p n 1 - p j x n . Then, f ( x i | x j ) = ( x 1 ,...,x i - 1 ,x i +1 ,...,x j - 1 ,x j +1 ,...,x n ) f ( x 1 , . . . , x j - 1 , x j +1 , . . . , x n | x j ) = ( x k = x i ,x j ) ( m - x j )! x 1 ! · · · · · x j - 1
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