Hw3_02_booksol

Hw3_02_booksol - Stochastic Signals and Systems Homework 3...

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Stochastic Signals and Systems Homework 3 Book Solutions Problem Solutions : Yates and Goodman,3.1.1 3.2.1 3.4.1 3.6.7 5.2.1 5.5.4 5.9.3 5.10.1 and 5.10.4 Problem 3.1.1 In this problem, it is helpful to label points with nonzero probability on the X , Y plane: y x P X , Y ( x , y ) c 3 c 2 c 6 c 4 c 12 c 0 1 2 3 4 0 1 2 3 4 (a) We must choose c so the PMF sums to one: x = 1 , 2 , 4 y = 1 , 3 P X , Y ( x , y ) = c x = 1 , 2 , 4 x y = 1 , 3 y = c [ 1 ( 1 + 3 )+ 2 ( 1 + 3 )+ 4 ( 1 + 3 )] = 28 c Thus c = 1 / 28. (b) The event { Y < X } has probability P [ Y < X ] = x = 1 , 2 , 4 y < x P X , Y ( x , y ) = 1 ( 0 )+ 2 ( 1 )+ 4 ( 1 + 3 ) 28 = 18 28 (c) The event { Y > X } has probability P [ Y > X ] = x = 1 , 2 , 4 y > x P X , Y ( x , y ) = 1 ( 3 )+ 2 ( 3 )+ 4 ( 0 ) 28 = 9 28 (d) There are two ways to solve this part. The direct way is to calculate P [ Y = X ] = x = 1 , 2 , 4 y = x P X , Y ( x , y ) = 1 ( 1 )+ 2 ( 0 ) 28 = 1 28 The indirect way is to use the previous results and the observation that P [ Y = X ] = 1 - P [ Y < X ] - P [ Y > X ] = ( 1 - 18 / 28 - 9 / 28 ) = 1 / 28 (e) P [ Y = 3 ] = x = 1 , 2 , 4 P X , Y ( x , 3 ) = ( 1 )( 3 )+( 2 )( 3 )+( 4 )( 3 ) 28 = 21 28 = 3 4 1
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Problem 3.2.1 On the X , Y plane, the joint PMF P X , Y ( x , y ) is y x P X , Y ( x , y ) c 3 c 2 c 6 c 4 c 12 c 0 1 2 3 4 0 1 2 3 4 By choosing c = 1 / 28, the PMF sums to one. (a) The marginal PMFs of X and Y are P X ( x ) = y = 1 , 3 P X , Y ( x , y ) = 4 / 28 x = 1 8 / 28 x = 2 16 / 28 x = 4 0 otherwise P Y ( y ) = x = 1 , 2 , 4 P X , Y ( x , y ) = 7 / 28 y = 1 21 / 28 y = 3 0 otherwise (b) The expected values of X and Y are E [ X ] = x = 1 , 2 , 4 xP X ( x ) = ( 4 / 28 )+ 2 ( 8 / 28 )+ 4 ( 16 / 28 ) = 3 E [ Y ] = y = 1 , 3 yP Y ( y ) = 7 / 28 + 3 ( 21 / 28 ) = 5 / 2 (c) The second moments are E £ X 2 ¤ = x
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Hw3_02_booksol - Stochastic Signals and Systems Homework 3...

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