131A_1_ds4_2010fall_Sol

# 131A_1_ds4_2010fall_Sol - EE 131A Probability Instructor:...

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EE 131A Discussion Set 4 Probability Wednesday, October 20, 2010 Instructor: Lara Dolecek and Friday, October 22, 2010 Reading: Chapters 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia 1. Cdf and pdf and functions. Problem 4.13, page 216 of ALG (a) The cdf is shown below: -1 01 1 x F(x) 0 3/4 This is a mixed typed of random variable. (b) The probabilities for the given regions are shown below: P ( X 2) = F X (2) = 0 . 9954 P ( X = 0) = 0 . 75 P ( X < 0) = 0 P (2 < X < 6) = F X (6) - F X (2) = 0 . 0046 P ( X > 10) = 1 - F X (10) = 5 . 15 × 10 - 10 1

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Problem 4.81, page 223 of ALG Since Y = 2 - 4 X , we have X = 2 - Y 4 Then we have the cdf: F Y ( y ) = P ( Y y ) = P ( X 2 - y 4 ) = 1 - P ( X < 2 - y 4 ) = ± 1 , y 2 1 4 exp - (2 - y ) / 2 , y < 2 Then we have the pdf: f Y ( y ) = d dy F Y ( y ) = 1 8 e - (1 - y/ 2) + 3 4 δ ( y - 2), for y 2, and f Y ( y ) = 0 elsewhere. 2.
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## This note was uploaded on 02/03/2011 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

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131A_1_ds4_2010fall_Sol - EE 131A Probability Instructor:...

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