This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View Full
Document
 Fall '08
 GELFAND

Click to edit the document details