20105ee131A_1_ps5_2010fall_CW

# 20105ee131A_1_ps5_2010fall_CW - EE 131A Probability...

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Problem Set 5 Probability Wednesday, November 3, 2010 Instructor: Lara Dolecek Due: Wednesday, November 10, 2010 Reading: Chapter 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. Conditional probability refresher Problem 4.173, page 231 of ALG Let p 1 be the probability of a light bulb working at time t = 9 , given that it is working at time t = 1 . Then we have: p 1 = P ( X > 9 | X > 1) = P ( X > 9 X > 1) P ( X > 1) = P ( X > 9) P ( X > 1) = 3 11 Then the probability of at least one light bulb still working at time t = 9 is given by: 1 - P ( all light bulbs fail at t = 9) = 1 - (1 - p 1 ) 3 = 0 . 615 2. Function of rv refresher The cdf of X is F X ( x ) = 1 - e - λx , x 0 . The pmf of Y is: f Y ( y ) = Z y +1 y f X ( x ) dx = F X ( y + 1) - F X ( y ) = e - λy (1 - e - λ ) , y 0 This is a geometric random variable. 3. Markov inequality. Problem 4.97, parts (a), (b) and (d) only, page 224 of ALG The Markov inequality is given by: P ( X c ) E ( X ) c , c 0 (a) For exponential RV, E ( X ) = b 2 . For

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20105ee131A_1_ps5_2010fall_CW - EE 131A Probability...

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