20105ee131A_1_ps4_2010fall_ZZsol

# 20105ee131A_1_ps4_2010fall_ZZsol - EE 131A Probability...

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EE 131A Problem Set 4 Probability Wednesday, October 20, 2010 Instructor: Lara Dolecek Due: Wednesday, November 3, 2010 Reading: Chapter 4 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. Cdf and pdf refresher. Problem 4.12, page 216 of ALG Solution : (a) we plot the cdf in the following Figure. Note cdf has jumps on -1 and increases continuously over 0 x 1, so it is mixed type random variable. (b) P { X ≤ - 1 } = F X ( - 1) = 0 . 5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -0.5 0 0.5 1 1.5 x F X (x) P { X = - 1 } = 0 . 5 P { X < 0 . 5 } = F X (0 . 5) = (1 + 0 . 5) / 2 = 0 . 75 P {- 0 . 5 < X < 0 . 5 } = F X (0 . 5) - F X ( - 0 . 5) = 0 . 75 - 0 . 5 = 0 . 25 P { X > - 1 } = 1 - F X ( - 1) = 1 - 0 . 5 = 0 . 5 P { X 2 } = 1 P { X > 3 } = 1 - F X (3) = 0 2. Cdf and pdf of a function. Problem 4.80, page 223 of ALG Solution : The signal X is ampliﬁed and shifted as Y = 2 X + 3, then cdf function is equivalent to F Y ( y ) = P { Y y } = P { 2 X + 3 y } = P { X ( y - 3) / 2 } = F X { ( y - 3) / 2 } F Y ( y ) = 0 y < 1 0 . 5 1 y 3 ( y - 1) / 4 3 y 5 1 y 5 1

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The pdf function is given by: f Y ( y ) = d F Y ( y ) d y = 0 . 5 δ ( y - 1) + 0 . 25 U ( y - 3) U (5 - y ) Note U ( t ) is unit step function and it is equal to 1 if t 0, 0 otherwise. 3. Expectations of RVs and of functions of RVs. Problem 4.56, page 220 of ALG Solution : (a) The mean and variance of Y is given by: E ( Y ) = E (3 X + 2) = 3 E ( X ) + 2 V ar ( Y ) = V ar (3 X + 2) = V ar (3 X ) = 9 V ar ( X ) (b) According to Table 4.1, we have E ( X ) = 0 and V ar ( X ) = 2 2 for Laplacian
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20105ee131A_1_ps4_2010fall_ZZsol - EE 131A Probability...

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