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Unformatted text preview: UC Berkeley Department of Electrical Engineering and Computer Science EE 126: Probablity and Random Processes Problem Set 4: Solutions Fall 2007 Issued: Thursday, September 20, 2007 Due: Friday, September 28, 2007 Reading: Bertsekas & Tsitsiklis, § 2.1 — 2.5 Problem 4.1 Each of n files is saved independently onto either hard drive A (with probability p ) or hard drive B (with probability 1 − p ). Let A be the total number of files selected for drive A and let B be the total number of files selected for drive B. 1. Determine the PMF, expected value, and variance for random variable A . 2. Evaluate the probability that the first file to be saved ends up being the only one on its drive. 3. Evaluate the probability that at least one drive ends up with a total of exactly one file. 4. Evaluate the expectation and the variance for the difference, D = A − B . 5. Assume n ≥ 2. Given that both of the first two files to be saved go onto drive A, find the conditional expectation, variance and probability law for random variable A . Solution: (a) We recognize that A is a binomial random variable (sum of n independent Bernoulli random variables). p A ( a ) = braceleftbigg ( n a ) p a (1 − p ) n a a = 0 , 1 ,... ,n otherwise. So, we have E [ A ] = np , var( A ) = np (1 − p ). (b) Let L denote the desired event, i.e. L : the first file saved ends up being the only one on its drive. Let FA : the first file is saved on drive A. Let FB : the first file is saved on drive B. We apply the law of total probability and obtain P ( L ) = P ( L  FA ) · P ( FA ) + P ( L  FB ) · P ( FB ) = p · (1 − p ) n 1 + (1 − p ) · p n 1 (c) Let EA : drive A ends up with exactly one file. Let EB : drive B ends up with exactly one file. Finally, let E denote the desired event  at least one drive ends up with a total of exactly one file. Notice that E = EA ∪ EB . In general, we know that, P ( E ) = P ( EA ) + P ( EB ) − P ( EA ∩ EB ). Now when n negationslash = 2, P ( EA ∩ EB ) = 0. We distinguish two cases: 1 case (1): n negationslash = 2 . We have P ( E ) = P ( EA )+ P ( EB ) = ( n 1 ) p (1 − p ) n 1 + ( n 1 ) (1 − p ) p n 1 , since both A and B are binomial random variables (with different “success” prob abilities). case (2): n = 2 . We have P ( E ) = P ( EA )+ P ( EB ) − P ( EA ∩ EB ) = 2 p (1 − p )+2(1 − p ) p − 2 p (1 − p ) = 2 p (1 − p ). (d) By linearity of expectation, we have E [ D ] = E [ A ] − E [ B ]. Both A and B are binomial, and we have E [ D ] = np − n (1 − p ) = n (2 p − 1). Now since A and B are not independent (i.e. A and B are dependent) we cannot simply add their variances. Notice however that A + B = n , hence B = n − A . Therefore, var( D ) = var( A − B ) = var( A − n + A ) = var(2 A − n ) = 4var( A ) = 4 np (1 − p )....
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This note was uploaded on 10/19/2008 for the course ECSE 4500 taught by Professor Woods during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 WOODS

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