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131A_1_ds2_2010fall_Sol

# 131A_1_ds2_2010fall_Sol - EE 131A Probability Instructor...

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EE 131A Discussion Set 2 Probability Wednesday, October 6, 2010 Instructor: Lara Dolecek and Friday, October 8, 2010 Reading: Chapter 2 of Probability, Statistics, and Random Processes by A. Leon-Garcia 1. Conditional Probability. Problem 2.73, page 88 of ALG (a) If P ( A B ) = 0, then P ( A | B ) = P ( A B ) P ( B ) = 0 If P ( A B ) = 0, then P ( A | B ) = P ( A B ) P ( B ) = P ( A ) P ( B ) If P ( A B ) = 0, then P ( A | B ) = P ( A B ) P ( B ) = P ( B ) P ( B ) = 1 (b) If P ( A | B ) > P ( A ), then P ( A | B ) = P ( A B ) P ( B ) > P ( A ) P ( B ) < P ( A B ) P ( A ) = P ( B | A ) 2. Binary Communication and Malicious Interferer . Problem 2.4, page 81 of ALG (a) S = { 2 , 1 , 0 , - 1 , - 2 } (b) T = { 2 , 1 } . Note that T does NOT include 0. (c) The two tosses of the coin being both heads will result in the outcome of Y = 0. Problem 2.78, page 89 of ALG (a) The tree diagram is shown as the following: 1/2 1/2 1/2 1/2 1/4 1/4 1/4 1/4 +2 -2 +2 +1 0 -1 -2 Channel Coin tosses 1

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(b) P ( X = +2 , Y = +2) = P ( Y = +2 | X = +2) × P ( X = +2) = 1 / 8 P ( X = +2 , Y = +1) = P ( Y = +1 | X = +1) × P ( X = +2) = 1 / 4 P ( X = +2 , Y = 0) = P ( Y = 0 | X = +2) × P ( X = +2) = 1 / 8 P ( X = - 2 , Y = 0) = P ( Y = 0 | X = - 2) × P ( X = - 2) = 1 / 8 P ( X = - 2 , Y = - 1) = P ( Y = - 1 | X = - 2) × P ( X = - 2) = 1 / 4 P ( X = - 2 , Y = - 2) = P ( Y = - 2 | X = - 2) × P ( X = - 2) = 1 / 8 (c) P
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