# 10_20ans - STAT 400 p.m.f or p.d.f 1 Fall 2011 Examples for...

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STAT 400 Examples for 10/20/2011 Fall 2011 p.m.f. or p.d.f. f ( x ; θ ) , θ Ω . Ω – parameter space. 1. Suppose Ω = { 1, 2, 3 } and the p.m.f. f ( x ; θ ) is θ = 1: f ( 1 ; 1 ) = 0.6, f ( 2 ; 1 ) = 0.1, f ( 3 ; 1 ) = 0.1, f ( 4 ; 1 ) = 0.2. θ = 2: f ( 1 ; 2 ) = 0.2, f ( 2 ; 2 ) = 0.3, f ( 3 ; 2 ) = 0.3, f ( 4 ; 2 ) = 0.2. θ = 3: f ( 1 ; 3 ) = 0.3, f ( 2 ; 3 ) = 0.4, f ( 3 ; 3 ) = 0.2, f ( 4 ; 3 ) = 0.1. What is the maximum likelihood estimate of θ ( based on only one observation of X ) if … a) X = 1; f ( 1 ; 1 ) = 0.6 f ( 1 ; 2 ) = 0.2 θ ˆ = 1 . f ( 1 ; 3 ) = 0.3 b) X = 2; f ( 2 ; 1 ) = 0.1 f ( 2 ; 2 ) = 0.3 θ ˆ = 3 . f ( 2 ; 3 ) = 0.4 c) X = 3; f ( 3 ; 1 ) = 0.1 f ( 3 ; 2 ) = 0.3 θ ˆ = 2 . f ( 3 ; 3 ) = 0.2 d) X = 4. f ( 4 ; 1 ) = 0.2 f ( 4 ; 2 ) = 0.2 θ ˆ = 1 or 2 . f ( 4 ; 3 ) = 0.1 (maximum likelihood estimate may not be unique)

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Likelihood function: L ( θ ) = L ( θ ; x 1 , x 2 , … , x n ) = = n i 1 f ( x i ; θ ) = f ( x 1 ; θ ) f ( x n ; θ ) It is often easier to consider ln L ( θ ) = = n i 1 ln f ( x i ; θ ) . 1½. Let X 1 , X 2 , … , X n be a random sample of size n from a Poisson distribution with mean λ , λ > 0. a) Obtain the maximum likelihood estimator of λ , λ ˆ . ( ) ( ) = - = = = n i i n i i e i f 1 X 1 X λ ; X L ! λ λ λ . ( ) ( ) = = - - = n i i n i i n 1 1 ! λ λ λ X ln ln X L ln . ( ) n n i i d d - = = 1 X 1 L ln λ λ λ = 0. X X 1 λ ˆ = = = n n i i .
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