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Unformatted text preview: EE351K Probability, Statistics and Random Processes Spring 2008 Instructor: Shakkottai/Vishwanath { shakkott/sriram } @ece.utexas.edu Homework 6: Solutions Problem 1 : p X ( k ) = p e k k ! + (1 p ) e k k ! , for k 0. Hence, M X ( ) = E ( e X ) = k =0 ( p e k k ! e k + (1 p ) e k k ! e k ) = pe k =0 ( e ) k k ! + (1 p ) e k =0 ( e ) k k ! = pe e e + (1 p ) e e e , since e z = k =0 z k k ! , for all complex numbers z = pe ( e  1) + (1 p ) e ( e  1) This expression for M X ( ) is valid for all complex numbers . Problem 2 : Part a : To find a and b , we use the following two properties of the MGF: 1. M X (0) = 1 2. dM X ( ) d vextendsingle vextendsingle vextendsingle =0 = E [ X ] M X ( ) = ae + be 4( e  1) . M X ( ) = ae + be 4( e  1) 4 e ....
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This note was uploaded on 06/10/2008 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
 Spring '07
 BARD

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