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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, 1. a + b = 1 2. a + 4 b = 5 a = -1/3, b = 4/3 . Part b : M X ( θ ) = - 1 3 e θ + 4 3 e 4( e θ - 1) . So, this MGF is a convex combination of two MGFs, namely M 1 ( θ ) = e θ and M 2 ( θ ) = e 4( e θ - 1) . As a result, the PMF of X will be the corresponding convex combination of the PMFs associated with the two MGFs. (This follows from the fact that the Laplace transform of addition of 2 functions is the sum of their
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