131A_1_ds8_2010fall_Sol

# 131A_1_ds8_2010fall_Sol - EE 131A Probability Instructor...

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EE 131A Discussion Set 8 Probability Wednesday, November 17, 2010 Instructor: Lara Dolecek and Friday, November 19, 2010 Reading: Chapters 5 and 6 of Probability, Statistics, and Random Processes by A. Leon-Garcia 1. Expectation of function of RVs. Problem 5.58, page 295 of ALG. Solution : Since X and Y are independent random variables, then we have: E ( X 2 e Y ) = E [ X 2 ] E [ e Y ] = [ V ar ( X ) + E 2 ( X )] E [ e Y ] = [1 + 0] · E [ e Y ] = Z 3 0 1 3 e y d y = e 3 - 1 3 2. Conditional probability and conditional expectation. Problem 5.75, page 296 of ALG (Problem 5.1 is in Demo 7). Solution : (a) According to the joint pmf of (X,Y) found in Problem 5.1, the condi- tional probabilities of p Y ( y | x ) and p X ( x | y ) are given in Table 1. Table 1: Joint pmf p ( x,y ), conditional probability p Y ( y | x ) and p X ( x | y ) respectively 2 0 0 1/16 1 0 4/16 4/16 0 1/16 4/16 2/16 Y/X 0 1 2 2 0 0 1/7 1 0 1/2 4/7 0 1 1/2 2/7 Y/X 0 1 2 2 0 0 1 1 0 1/2 1/2 0 1/7 4/7 2/7 Y/X 0 1 2 (b) According to the joint pmf of (X,Y) found in Problem 5.1, then the conditional

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131A_1_ds8_2010fall_Sol - EE 131A Probability Instructor...

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