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hw6_sol(1) - EE351K Probability Statistics and Random...

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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 − 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 kθ + (1- p ) e − μ μ k k ! × e kθ ) = 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 ) dθ vextendsingle vextendsingle vextendsingle θ =0 = E [ X ] M X ( θ ) = ae θ + be 4( e θ- 1) . ⇒ M ′ X ( θ ) = ae θ + be 4( e θ- 1) × 4 e θ ....
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hw6_sol(1) - EE351K Probability Statistics and Random...

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