hwk3_soln - X i !) + 2 n 2 log( ) n X i =1 X i + 2 n X i =1...

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IEOR E4702 Statistical Inference for Financial Engineering HWK Solution 3 1. Under H 0 the likelihood is given by L H 0 ( λ 0 )= n Y i =1 f ( X i | λ 0 )= n Y i =1 e λ 0 ( λ 0 ) X i X i ! , and log L H 0 ( λ 0 )= n λ 0 +log( λ 0 ) n X i =1 X i n X i =1 log( X i !) . Without constraints under H a ,wehave log L H a ( λ )= n λ +log( λ ) n X i =1 X i n X i =1 log( X i !) . Taking the derivative with respect to λ and setting it to zero we have n + P n i =1 X i λ =0 . Therefore, the MLE under H a is ˆ λ = P n i =1 X i n = ¯ X. Thus, log L H a ( ˆ λ )= n ¯ X +log( ¯ X ) n X i =1 X i n X i =1 log( X i !) . Therefore, the test statistic is given by Λ =2 l o g ( L H a ( ˆ λ ) L H 0 ( λ 0 ) )=2log( L H a ( ˆ λ )) 2log( L H 0 ( λ 0 )) = 2 n ¯ X +2log( ¯ X ) n X i =1 X i 2 n X
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Unformatted text preview: X i !) + 2 n 2 log( ) n X i =1 X i + 2 n X i =1 log( X i !) = 2 n ( X ) + 2 log( X ) log( ) n X i =1 X i , which asymptotically has 2 distribution with d.f. one. Thus, we reject the null hypothesis if 2 n ( X ) + 2 log( X ) log( ) n X i =1 X i . 2 , 1 . 2. See the attached excel f le. 1...
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