20105ee131A_1_ps8_2010fall_Sol

20105ee131A_1_ps8_2010fall_Sol - EE 131A Probability...

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EE 131A Problem Set 8 Probability Wednesday, November 24, 2010 Instructor: Lara Dolecek Due: Wednesday, December 1, 2010 Last Assignment! Reading: Chapters 5, 6, 7 and 8 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. Jointly Gaussian. Problem 5.110, page 299 of ALG Solution : X and Y are jointly Gaussian random variable with pdf: f X,Y ( x,y ) = e ( - 2 x 2 - y 2 / 2) 2 πc Comparing the coefficients in jointly Gaussian distribution, we have: ρ X,Y = 0 m 1 = 0 m 2 = 0 - 2 σ 2 1 = - 1 / 2 2 σ 2 2 = 2 Therefore, we have σ 2 1 = 1 / 4 and σ 2 2 = 1. V ar ( X ) = 1 4 V ar ( Y ) = 1 Cov ( X,Y ) = 0 2. Mixture Gaussian. Problem 5.117, page 300 of ALG Solution : h ( x,y ) = e - ( x 2 - 2 ρ 1 xy + y 2 ) / 2(1 - ρ 2 1 ) 2 p 1 - ρ 2 1 g ( x,y ) = e - ( x 2 - 2 ρ 2 xy + y 2 ) / 2(1 - ρ 2 2 ) 2 p 1 - ρ 2 2 (a) The marginal pdfs for X and Y are given by: f X ( x ) = 1 2 Z -∞ ( h ( x,y ) + g ( x,y ))d y = 1 2 ( e - x 2 / 2 2 π + e - x 2 / 2 2 π ) = e - x 2 / 2 2 π f Y ( y ) = 1 2 Z -∞ ( h ( x,y ) + g ( x,y ))d x = 1 2 ( e - y 2 / 2 2 π + e - y 2 / 2 2 π ) = e - y 2 / 2 2 π (b) However, f XY ( x,y ) = p 1 - ρ 2 2 e - ( x 2 - 2 ρ 1 xy + y 2 ) / 2(1 - ρ 2 1 ) + p 1 - ρ 2 1 e - ( x 2 - 2 ρ 2 xy + y 2 ) / 2(1 - ρ 2 2 ) 2 π p 1 - ρ 2 1 p 1 - ρ 2 2 does not have the form required for jointly Gaussian random variables. 1
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3. Joint cdf, pdf refresher. Problem 5.129, page 301 of ALG Solution :(a)To find the value of the constant c, we have: Z π/ 2 0 Z π/ 2 0 c sin( x + y )d x d y = 1 Solving for c, we can find: c = 1 R π/ 2 0 R π/ 2 0 sin( x + y )d x d y = 1 2 (b) The joint cdf of X and Y is given by: F XY ( x,y ) = Z Z f X,Y ( u,v )d u d v = 1 2 Z x 0 Z y 0 sin( u + v )d u d v = 1 2 [sin x +sin y - sin( x + y )] (c) The marginal pdfs of X and Y are given by: f X ( x ) = Z
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20105ee131A_1_ps8_2010fall_Sol - EE 131A Probability...

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