ECEHW6Soln_SP10 1

# ECEHW6Soln_SP10 1 - ECE 302 Homework 6 Prof Saul Gelfand...

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Unformatted text preview: ECE 302: Homework 6 Prof. Saul Gelfand Spring 2010 Due April 9, 2010 Problem 1 (text 5.93) We have that X and Y are geometric so that p X ( x ) = (1- p 1 ) x p 1 , for x = 0 , 1 , 2 ,... and p Y ( x ) = (1- p 2 ) y p 2 , for y = 0 , 1 , 2 ,... Since E [ X ] = 1- p 1 p 1 = 2, and E [ Y ] = 1- p 2 p 2 = 4, we have p 1 = 1 3 and p 2 = 1 5 . Since X and Y are independent, we have that the joint PMF p X,Y ( x,y ) = p X ( x ) p Y ( y ) = 1 15 parenleftbigg 2 3 parenrightbigg x parenleftbigg 4 5 parenrightbigg y , for x,y = 0 , 1 , 2 ,... a. We have Z = X- Y has PMF p Z ( z ) = Pr [ Z = z ] = Pr [ X- Y = z ] = summationdisplay { ( x,y ): x- y = z } p X,Y ( x,y ) = ∞ summationdisplay x =-∞ p X,Y ( x,x- z ) = ∞ summationdisplay x =-∞ p X ( x ) p Y ( x- z ) . Note that this sum is the convolution sum, i.e. p Z ( z ) = p X ( x ) * p Y (- y ) . 1 We will have in general that if two independent random variables A and B are added, their sum C = A + B has pmf that is the convolution of the pdfs, i.e. p C ( c ) = p A ( a ) * p B ( b ). We now carry out the summation to find p Z ( z ). p Z ( z ) = ∞ summationdisplay x =-∞ p X ( x ) p Y ( x- z ) = braceleftBigg ∑ ∞ x =0 1 15 ( 2 3 ) x ( 4 5 ) x- z = 1 15 ( 5 4 ) z ∑ ∞ x =0 ( 8 15 ) x = 1 7 ( 5 4 ) z , z < ∑ ∞ x = z 1 15 ( 2 3 ) x ( 4 5 ) x- z = 1 15 ( 5 4 ) z ∑ ∞ x = z ( 8 15 ) x = 1 7 ( 2 3 ) z , z ≥ . b. We have that the Bulldogs beat the Flames when X > Y , or equivalently when Z > 0. Therefore Pr[Bulldogs win] = Pr [ Z > 0] = ∞ summationdisplay z =1 p Z ( z ) = ∞ summationdisplay z =1 1 7 parenleftbigg 2 3 parenrightbigg z = 2 7 . Similarly, the teams tie when X = Y , or equivalently when Z = 0: Pr[teams tie] = Pr [ Z = 0] = p Z (0) = 1 7 . c. E [ Z ] = E [ X- Y ] = E [ X ]- E [ Y ] = 2- 4 =- 2 Problem 2 (text 5.96) We have X 1 , X 2 and X 3 are uniform in [- 1 , 1] so that f X 1 ( x ) = f X 2 ( x ) = f X 3 ( x ) = braceleftbigg 1 2 ,- 1 ≤ x ≤ 1 , , otherwise a. Since X 1 and X 2 are independent, we have that f X 1 ,X 2 ( x 1 ,x 2 ) = f X 1 ( x 1 ) f X 2 ( x 2 ) . 2 We have then that the CDF of Y = X 1 + X 2 is F Y ( y ) = Pr [ Y ≤ y ] = Pr [ X 1 + X 2 ≤ y ] = integraldisplay { ( x 1 ,x 2 ): x 1 + x 2 ≤ y } f X 1 ,X 2 ( x 1 ,x 2 ) dx 1 dx 2 = integraldisplay ∞-∞ integraldisplay y- x 1-∞ f X 1 ,X 2 ( x 1 ,x 2 ) dx 2 dx 1 = y ≤ - 2 integraltext...
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## This note was uploaded on 02/07/2011 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.

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ECEHW6Soln_SP10 1 - ECE 302 Homework 6 Prof Saul Gelfand...

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