solution08 - Solutions to Homework 08 1. Y = e X . We...

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Unformatted text preview: Solutions to Homework 08 1. Y = e X . We denote the pdf of X by f X ( x ) and its cdf by F X ( x ). Also, Y = e X X = ln Y . a. Note first that for all values of X , i.e.- < x < , Y > 0. Thus, P [ Y 0] = 0. For Y > 0, F Y ( y ) = P [ Y y ] = P [ e X y ] = P [ X ln y ] = F X (ln y ) Therefore F Y ( y ) = braceleftBigg y F X (ln y ) y > Now, f Y ( y ) = d dy F Y ( y ) = d dy F X (ln y ) = F prime X (ln y ) d dy ln y = 1 y f X (ln y ) b. If X is a Gaussian random variable, its pdf is given by f X ( x ) = 1 2 2 e- 1 2 ( x- m ) 2- < x < Then, f Y ( y ) is given by f Y ( y ) = y 1 y 2 2 e- 1 2 ( ln y- m ) 2 y > 2. Y = A cos( t ) + c E [ Y ] = E [ A cos( t ) + c ] = integraldisplay - ( A cos( t ) + c ) f A ( a ) da = cos( t ) integraldisplay - Af A ( a ) da + c integraldisplay - f A ( a ) da = cos( t ) E [ A ] + c = m cos( t ) + c 1 V AR [ Y ] = E [( Y- E [ Y ]) 2 ] = integraldisplay...
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This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.

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solution08 - Solutions to Homework 08 1. Y = e X . We...

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