Dr. Hackney STA Solutions pg 126

Dr. Hackney STA Solutions pg 126 - Second Edition 8-5...

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Second Edition 8-5 Noting that ˆ a ˆ θ = ˆ σ 2 , we obtain λ ( x ) = L ( ˆ θ R | x ) L a, ˆ θ | x ) = Q n i =1 1 2 π ˆ θ R e - ( x i - ˆ θ R ) 2 / (2 ˆ θ R ) Q n i =1 1 2 π ˆ a ˆ θ e - ( x i - ˆ θ ) 2 / (2ˆ a ˆ θ ) = ± 1 / (2 π ˆ θ R ) ² n/ 2 e - Σ i ( x i - ˆ θ R ) 2 / (2 ˆ θ R ) (1 / (2 π ˆ σ 2 )) n/ 2 e - Σ i ( x i - ¯ x ) 2 / (2ˆ σ 2 ) = ± ˆ σ 2 / ˆ θ R ² n/ 2 e ( n/ 2) - Σ i ( x i - ˆ θ R ) 2 / (2 ˆ θ R ) . b. In this case we have log L ( a,θ | x ) = n X i =1 ³ - 1 2 log(2 πaθ 2 ) - 1 2 2 ( x i - θ ) 2 ´ . Thus log L ∂a = n X i =1 µ - 1 2 a + 1 2 a 2 θ 2 ( x i - θ ) 2 = - n 2 a + 1 2 a 2 θ 2 n X i =1 ( x i - θ ) 2 set = 0 . log L ∂θ = n X i =1 ³ - 1 θ + 1 3 ( x i - θ ) 2 + 1 2 ( x i - θ ) ´ = - n θ + 1 3 n X i =1 ( x i - θ ) 2 + 1 2 n X i =1 ( x i - θ ) set = 0 . Solving the first equation for a in terms of θ yields a = 1 2 n X i =1 ( x i - θ ) 2 . Substituting this into the second equation, we get - n θ + n θ + n i ( x i - θ ) i ( x i - θ ) 2 = 0 . So again, ˆ θ = ¯ x and ˆ a = 1 n ¯ x 2 n X i =1 ( x i - ¯ x ) 2 = ˆ σ 2 ¯ x 2 in the unrestricted case. In the restricted case, set
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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