hw6_sol - EE351K Probability, Statistics and Random...

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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 - λ λ k k ! + (1 - p ) e - μ μ k k ! , for k 0. Hence, M X ( θ ) = E ( e θX ) = k =0 ( p e - λ λ k k ! × e + (1 - p ) e - μ μ k k ! × e ) = pe - λ k =0 ( λe θ ) k k ! + (1 - p ) e - μ k =0 ( μe θ ) k k ! = pe - λ × e λe θ + (1 - p ) e - μ × e μe θ , since e z = k =0 z k k ! , for all complex numbers z = pe λ ( e θ - 1) + (1 - p ) e μ ( e θ - 1) This expression for M X ( θ ) is valid for all complex numbers θ . 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 ( θ ) ± ± ± θ =0 = E [ X ] M X ( θ ) = ae θ + be 4( e θ - 1) . M 0 X ( θ ) = ae θ + be 4( e θ - 1) × 4 e θ . M 0 X (0) = a + 4 b. Hence,
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This note was uploaded on 10/26/2009 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.

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hw6_sol - EE351K Probability, Statistics and Random...

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