Lecture_24

# Lecture_24 - 1 MLE FOR 2-DIMENSIONAL PARAMETERS Lecture 24...

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Unformatted text preview: 1 MLE FOR 2-DIMENSIONAL PARAMETERS Lecture 24 ORIE3500/5500 Summer2009 Chen Class Today • MLE for 2-dimensional parameters • Method of Moments Estimators • Interval Estimation 1 MLE for 2-dimensional parameters While computing MLE for 2-dimensional parameters, the following (trivial) result often proves helpful: Fact: If f is a function from R 2 to R with the following properties: 1. There is x so that for all y , max x f ( x,y ) = f ( x ,y ) 2. There is y so that max y f ( x ,y ) = f ( x ,y o ) Then max x,y f ( x,y ) = f ( x ,y ) Proof. For any x,y , by property 1, f ( x,y ) ≤ f ( x ,y ) By property 2, f ( x ,y ) ≤ f ( x ,y ) Combining, for all x,y f ( x,y ) ≤ f ( x ,y ) which is the statement of the theorem. To apply the above theorem, one has to be careful about deciding which variable to fix so that there will be an unique point where the maximum in the other variable is achieved. 1 1 MLE FOR 2-DIMENSIONAL PARAMETERS Example If for some λ > 0 and α ∈ R , X 1 ,...,X n are i.i.d. with density f ( x ) = λe- λ ( x- α ) , x ≥ α. Find MLE of α and λ . The joint pdf of X 1 ,...,X n is f X 1 ...X n ( x 1 ,...,x n ) = λ n exp ˆ- λ n X i =1 ( x i- α ) ! , x i ≥ α Hence the likelihood function is L ( λ,α ; ~ X ) = λ n exp ˆ- λ n X i =1 ( X i- α ) ! 1 { X i ≥ α ∀ i } = λ n exp ˆ- λ n X i =1 ( X i- α ) ! 1 { X (1) ≥ α } where X (1) is the smallest X i . Clearly the likelihood function is not even continuous and hence naively differentiating won’t help. Let’s fix λ and view the likelihood as a function of α . Then we could show that (the same as Q1 in HW10) for every λ , l ( α,λ ) is maximized at α = X (1) and hence ˆ α MLE = X (1) ....
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## This note was uploaded on 09/22/2009 for the course ORIE 3500 taught by Professor Weber during the Summer '08 term at Cornell.

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Lecture_24 - 1 MLE FOR 2-DIMENSIONAL PARAMETERS Lecture 24...

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