University of California, Berkeley - EECS 126hw12-sol

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Department of Electrical Engineering and Computer Sciences EECS 126: Probability and Random Processes Problem Set 12 Spring 2018 Self-Graded Scores Due: 5 PM, Friday May 4, 2018 Submit your self-... graded scores via the Google form: https://goo.gl/forms/HYw8QwfrHNKf0czx2. Make sure you use your SORTABLE NAME on CalCentral. 1. Geometric MMSE Let N be a geometric random variable with parameter 1 − p, and (Xi)i∈N be i.i.d. exponential random variables with parameter λ. Let T = X1 + · · · + XN . Compute the LLSE and MMSE of N given T. Solution: First, we calculate P(N = n | T = t), for t > 0 and n ∈ Z+. P(N = n | T = t) = P(N = n)fT|N (t | n) P∞ k=1 P(N = k)fT|N (t | k) = (1 − p)p n−1λ n t n−1 e −λt/(n − 1)! P∞ k=1(1 − p)p k−1λkt k−1e−λt/(k − 1)! = λ(λpt) n−1/(n − 1)! λ P∞ k=1(λpt) k−1/(k − 1)! = (λpt) n−1 e λpt(n − 1)!, n ∈ Z+. Next, we calculate E[N | T = t]. E[N | T = t] = X∞ n=1 n (λpt) n−1 e λpt(n − 1)! = X∞ n=1 (λpt) n−1 e λpt(n − 1)! + X∞ n=1 (n − 1) (λpt) n−1 e λpt(n − 1)! = 1 + λpt e λpt X∞ n=2 (λpt) n−2 (n − 2)! = 1 + λpt e λpt e λpt = 1 + λpt. Hence, the MMSE is E[N | T] = 1 + λpT. The MMSE is linear, so it is also the LLSE. In terms of a Poisson process, T represents the first arrival of a marked Poisson process with rate λ, where arrivals are marked independently with probability 1 − p. The marked Poisson process has rate λ(1 − p). The unmarked points form a Poisson process of rate λp. In time T, the expected number of unmarked points is λpT, so the conditional expectation of the number of points at time T, N, is 1 + λpT. 1 2. Property of MMSE Let X, Y1, . . . , Yn be square integrable random variables. Argue that E (X − E[X | Y1, . . . , Yn])2 ≤ E X − Xn i=1 E[X | Yi ] 2 . Solution: [Show More]

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