hw9-sol.pdf - UC Berkeley Department of Electrical...

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UC Berkeley Department of Electrical Engineering and Computer Sciences EECS 126: Probability and Random Processes Problem Set 9 Spring 2018 Self-Graded Scores Due: Monday, March 26, 2018 Submit your self-graded scores via the Google form: . Make sure you use your SORTABLE NAME on CalCentral. 1 . Basketball II Team A and Team B are playing an untimed basketball game in which the two teams score points according to independent Poisson processes. Team A scores points according to a Poisson process with rate λ A and Team B scores points according to a Poisson process with rate λ B . The game is over when one of the teams has scored k more points than the other team. Find the probability that Team A wins. Solution: We consider the merged process with rate λ A + λ B and notice that each point is a point for A with probability p = λ A / ( λ A + λ B ) and a point for B with probability 1 - p = λ B / ( λ A + λ B ). Now, we consider the following Markov chain. Let the state of the chain be the number of additional points Team B needs to score to win. Thus, we have the transition probabilities: P 0 , 0 = 1 P i,i +1 = p, where 0 < i < 2 k P i,i - 1 = 1 - p, where 0 < i < 2 k P 2 k, 2 k = 1 0 · · · i · · · 2 k 1 1 1 - p p p p 1 - p 1 2 We are looking for the probability that a random walk, starting at state k hits state 2 k (as this indicates that Team A won). Now, let P i be the probability that starting at state i , Team A will win (equivalently, the probability of ending in state 2 k when starting at state i ). We let ρ = (1 - p ) /p . We also claim that P i - P i - 1 = ρ i - 1 P 1 for i > 1. We shall show this by induction. Consider the base case P 2 - P 1 . Note that P 1 = pP 2 + (1 - p ) P 0 = pP 2 since P 0 = 0. 1
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We thus see that P 2 - P 1 = ρP 1 . Now, we assume for all i , 0 < i < n , that P i - P i - 1 = ρ i - 1 P 1 . We would like to show that P n - P n - 1 = ρ n - 1 P 1 . We note that P n - 1 = pP n + (1 - p ) P n - 2 . This implies that: P n = 1 p ( P n - 1 - P n - 2 + pP n - 2 ) = 1 p ( ρ n - 2 P 1 + pP n - 2 ) .
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