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Fall 2010 HW09 - University of Illinois Fall 2010 ECE 313...

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University of Illinois Fall 2010 ECE 313: Problem Set 9: Solutions Functions of a random variable, failure rate functions 1. [Log random variables] (a) Z takes values in the set [ e a , e b ]; for e a c e b : F Z ( c ) = P { e U c } = P { U ln c } = ln c - a b - a . Differentiating yields f Z ( c ) = 1 c ( b - a ) e a c e b 0 else . (b) By LOTUS, E [ Z ] = R b a e u b - a du = e b - e a b - a . (c) Y is positive valued; for c 0 , F Y ( c ) = P { e X c } = P { X ln c } = F X (ln( c )) . Differentiating and using the chain rule yields f Y ( c ) = ( f X (ln c ) c = 1 c 2 π exp - (ln( c )) 2 2 c 0 0 else . (d) By LOTUS, E [ Y ] = Z -∞ e u 1 2 π exp - u 2 2 du = Z -∞ 1 2 π exp - u 2 - 2 u 2 du = exp 1 2 Z -∞ 1 2 π exp - ( u - 1) 2 2 du = exp 1 2 = e = 1 . 64872 . 2. [Generation of random variables with specified probability density function] Since U and X are both distributed over the interval [0 , 1] , the function g should map the interval [0 , 1] onto interval [0 , 1] . The desired CDF of g ( U ) is given by F X ( c ) = R c 0 2 vdv = c 2 for 0 c 1 , and we let g ( u ) = F - 1 ( u ) . If g ( u ) = c then F X ( c ) = u or c 2 = u or c = u.
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