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Unformatted text preview: ECE 302: Homework 6 Prof. Saul Gelfand Fall 2009 Due November 20, 2009 Problem 1: Text 5.95 Since X and Y are independent, we have that the joint pdf f XY ( x, y ) = f X ( x ) f Y ( y ) = 1 for ( x, y ) inside the box indicated below: 0.5 1 1.5 0.5 1 1.5 x y We now find the CDF of Z as follows: F Z ( z ) = P [ Z z ] = P [ XY < z ] = P [ Y < z/X ] . The set of points ( X, Y ) satisfying Y < z/X can be visualized as the shaded region in the figure below: Thus, P [ Z z ] is equal to the integration of f XY ( x, y ) over this region, but since f XY ( x, y ) = 1 in the region, this integration is expressed simply as: F Z ( z ) = z + integraldisplay 1 z z x dx = z + [ z ln x ] 1 x = z = z (1 ln z ) . 1 z 0.5 1 1.5 0.5 1 1.5 x y y = z/x Therefore, f Z ( z ) = d dz F Z ( z ) = (1 ln z ) + z parenleftbigg 1 z parenrightbigg = ln parenleftbigg 1 z parenrightbigg . Problem 2: Text 5.99 We have that f Z ( z ) = integraldisplay  f Z ( z  y ) f Y ( y ) dy where, if we fix Y = y , then Z = X + y so that f Z ( z  y ) = f X ( z y ). We are left now with finding the marginal pdfs f Y ( y ) and f X ( x ). We have f Y ( y ) = integraldisplay  f X,Y ( x, y ) dx = integraldisplay 1 y e x e y dx = e y ( e y e 1 ) = e 2 y e ( y +1) , and f X ( x ) = integraldisplay  f X,Y ( x, y ) dy = integraldisplay x e x e y dy = e x (1 e x ) = e x e 2 x . Therefore f Z ( z ) = integraldisplay  f Z ( z  y ) f Y ( y ) dy = integraldisplay 1 ( e y z e 2 y 2 z )( e 2 y e y 1 ) dy = integraldisplay 1 e y z e 2 z e z 1 + e y 2 z 1 dy = e z ( 1 e 1 ) e 2 z e z 1 + e 2 z 1 ( 1 e 1 ) 2 Problem 3: Text 5.106 The symmetry of the joint distribution f XY ( x, y ) gives us by inspection that R is uniformly distributed between r 1 and r 2 , and is uniformly distributed in [0...
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This note was uploaded on 09/14/2010 for the course ECE 302 taught by Professor Gelfand during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 GELFAND

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