# PS12_sol - University of Illinois Spring 2010 ECE 313...

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Unformatted text preview: University of Illinois Spring 2010 ECE 313: Problem Set 12: Problems and Solutions Functions of random variables, conditional pdfs, covariances Due: Wednesday April 28 at 4 p.m. Reading: Ross, Chapter 6, Sections 1-4; Powerpoint Lecture Slides, Sets 30-34 Noncredit Exercises: Chapter 6: Problems 1-3, 9, 10, 13, 15, 19-23, 40-42; Theoretical Exercises 4, 6; Self-Test Problems 3, 5, 6, 7 1. [Joint Distributions] Suppose two jointly continuous random variables X and Y have joint distribution given by f X , Y ( u,v ) = ( 1 C , if- 1 2 ≤ u ≤ 1 2 ,- 1 2 ≤ v ≤ 1 2 , and u 2 + v 2 > 1 16 , otherwise (a) find C . Solution: since they are uniformly distributed over a region, we know that the area of the region is equal to 1 C . Note that the support is the origin-centered unit rectangle, except for the origin- centered circle of radius 1 4 . Thus the area of the region can be calculated as the area of the unit square minus the area of circle of radius 1 4 . So C = 1 1- π/ 16 . (b) find f X | Y ( u | . 45), f X | Y ( u | 0), f Y | X ( v | . 45), and f Y | X ( v | 0). Are X and Y independent? Solution: note that the shape of the conditional pdf is the same as that of the joint pdf, it is only normalized to integrate to one. Thus we know that given any value of Y = v , X is uniformly distributed over some interval pertaining to the support. So we know that given Y = 0, X is uniform over [- 1 2 ,- 1 4 ] ∪ [ 1 4 , 1 2 ]. However, given Y =- . 45, X is uniform over [- 1 2 , 1 2 ]. Also, from symmetry, the conditional distribution on Y given X = u is the same as the conditional distribution of X given Y = v . Thus, we have: f X | Y ( u | . 45) = ( 1 , u ∈ [- 1 2 , 1 2 ] , otherwise f X | Y ( u | 0) = ( 2 , u ∈ [- 1 2 ,- 1 4 ] ∪ [ 1 4 , 1 2 ] , otherwise f Y | X ( v | . 45) = ( 1 , v ∈ [- 1 2 , 1 2 ] , otherwise f Y | X ( v | 0) = ( 2 , v ∈ [- 1 2 ,- 1 4 ] ∪ [ 1 4 , 1 2 ] , otherwise Lastly, clearly from above, they are not independent: f X | Y (0 | . 45) = 1 whereas f X | Y (0 | 0) = 0. (c) Let Z = X 2 + Y 2 . Find F Z ( π 64 ). Solution: Note that since π < 4, F Z ‡ π 64 · ≤ F Z 1 16 ¶ = P X 2 + Y 2 ≤ 1 16 ¶ = 0 . where the last equality holds from observation of the support of the joint density f X , Y ( u,v ) 2. [Joint Distributions] Let X and Y be two independent random variables where X is exponentially distributed of rate λ 1 and Y is exponentially distributed of rate λ 2 . (a) Let Z = min( X , Y ). Find the density of Z , given by f Z ( a ). ( hint: first find P ( Z > a ) using equivalence of events ). Solution: P ( Z > a ) = P (min( X , Y ) > a ) = P ( X > a, Y > a ) = P ( X > a ) P ( Y > a ) = e- ( λ 1 + λ 2 ) a ....
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PS12_sol - University of Illinois Spring 2010 ECE 313...

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