solution9 - EL630: Homework9 Solutions 1. X+Y=z Y z 1 X FZ...

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EL630: Homework9 Solutions 1. ) ( ) ( z Y X P z F Z + = . Refer to the diagram, in which the value of z is between 0 and 1, and the line z Y X = + is shown. The event } { z Y X + is all points below and to the left of the line. Therefore we see that = + = ) ( ) ( z Y X P z F Z the integral of ) , ( y x f XY over the region below and to the left of the line. The integral just mentioned is just the area of the blue-shaded below and to the left of the line. Thus for 1 0 < < z , 2 2 1 ) ( 2 z z F Z + = . By taking a value of z between –1 and 0 and using a similar idea, it is not hard to see that for –1< z <0, 2 2 1 ) ( 2 z z F Z - = . So, < < + < < - - = 1 0 , 2 2 1 0 1 , 2 2 1 ) ( 2 2 z z z z z F Z . Taking the derivative yields 1 1 , ) ( < < - = z for z z f Z . 2. (a) You may easily show that the density of the random variable Y W 3 = is ) 3 / ( 3 / 1 ) ( w f w f Y W = . Then the density of W X Z + = is just z 1 X Y X+Y=z
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- - - - - - = - = = dx x z U e x U e dx x z f x f z f z f z f x z x W X W X Z ) ( ) ( 3 1 ) ( ) ( ) ( ) ( ) ( 3 / ) ( β α ( 29 0 , 3 / 3 / 3 ) ( 3 / 0 ) 3 / ( 3 / - - = = - - - - z for e e dx e e z f z z z x z Z αβ (b) Let’s do this one the direct way. Let ) 3 , max( Y X Z = . Then ) ) 3 , (max( ) ( z Y X P z F Z = . Now the event } ) 3 , {max( z Y X may be rewritten as } 3 / , { } 3 , { } ) 3 , {max( z Y z X z Y
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This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.

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solution9 - EL630: Homework9 Solutions 1. X+Y=z Y z 1 X FZ...

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