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Unformatted text preview: IE 300/GE 331 Homework Solution #7 1 IE 300 / GE 331 Spring 2009 Homework Solution #7 Problem 1. (a) Let X denote the random wait time. We have the observation X = 30. Using Bayes’ rule, the posterior PDF is ?  ¡ 30 ¢ = ? ( ) ?  30 ¡¢ £ ? ( ′ ) ?  30 ¡′¢?′ . Using the given prior, ? ¢ = 10 for ∈ [0,1/5] , we obtain ?  ¡ 30 ¢ = ¤ 10 ?  30 ¡¢ £ 10 ′?  30 ¡′¢?′ 1/5 , ¥? ∈ [0, 1 5 ] 0, ¦§ℎ?¨©¥ª? . « We also have ?  30 ¡¢ = ? − 30 , so the posterior is ?  ¡ 30 ¢ = ¬ 2 ? − 30 £ ′ 2 ? − 30 ′ ?′ 1/5 , ¥? ∈ [0, 1 5 ] 0, ¦§ℎ?¨©¥ª? . « The MAP rule selects that maximize the posterior (or equivalently its numerator, since the denominator is a positive constant). By setting the derivative of the numerator to 0, we obtain ? ? ® 2 ? − 30 ¯ = 2 ? − 30 − 30 2 ? − 30 = 2 − 30 ¢? − 30 = 0. IE 300/GE 331 Homework Solution #7 2 Therefore, = 2/30 ....
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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