hw7_solns - ORIE 4580/5580/5581 Solutions for Homework...

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ORIE 4580/5580/5581 Solutions for Homework Assignment 7 Question 1. The likelihood function is L ( λ ) = e - λ λ X 1 X 1 ! e - λ λ X 2 X 2 ! .. . e - λ λ X n X n ! = e - λ ( X 1 + X 2 + ...X n ) X 1 ! X 2 ! .. . X n ! . We will maximize ln L ( λ ): ln L ( λ ) = - + ( X 1 + X 2 + .. . X n ) ln( λ ) - ln( X 1 ! X 2 ! .. .X n !) . Taking the derivative of ln L ( λ ) with respect to λ and setting it equal to 0, we get d ln L ( λ ) = - n + X 1 + X 2 + .. .X n λ = 0 . Thus ˆ λ , the MLE of λ , is ˆ λ = X 1 + X 2 + . .. X n n . Question 2. The likelihood function is L ( p ) = ( 7 X 1 ) p X 1 (1 - p ) 7 - X 1 ( 7 X 2 ) p X 2 (1 - p ) 7 - X 2 .. . ( 7 X n ) p X n (1 - p ) 7 - X n = ( 7 X 1 )( 7 X 2 ) ... ( 7 X n ) p [ X 1 + ... + X n ] (1 - p ) 7 n - [ X 1 + ... + X n ] . We will maximize ln L ( p ): ln L ( p ) = ln {( 7 X 1 )( 7 X 2 ) .. . ( 7 X n )} + [ X 1 + .. . + X n ] ln( p ) + [7 n - [ X 1 + .. . + X n ]] ln(1 - p ) . Taking the derivative of ln L ( p ) with respect to p and setting it equal to 0, we get d ln L ( p ) dp = X 1 + .. . + X n p - 7 n - [ X 1 + .. . + X n ] 1 - p = 0 Thus ˆ p , the MLE of p , is ˆ p = X 1 + .. . + X n 7 n . 1
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Question 3. We start by solving part (b): b) The likelihood of the sample is the product of the individual likelihoods. Thus, L ( k ) = ( k + 1 2 k +1 ) n n i =1 X k i ln L ( k ) = n [ln( k
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This note was uploaded on 03/08/2012 for the course ORIE 4580 at Cornell.

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hw7_solns - ORIE 4580/5580/5581 Solutions for Homework...

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