ProbSetFive - Rensselaer Electrical, Computer, and Systems...

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Rensselaer Electrical, Computer, and Systems Engineering Department ECSE 4500 Probability for Engineering Applications PS#5 Solutions March 15, 2010 1. (3.4) Method (1): For y> 0 P [ Y y ]= P [ e X y P [ X ln y F X (ln y ) . Hence f Y ( y )= dF X (ln y ) dy = dF X (ln y ) d ( lny ) d (ln y ) dy = 1 y f X (ln y ) . For y 0 P [ Y y P [ e X y P [ φ ]=0 . Hence f Y ( y 1 2 πσy exp[ 1 2 ( ln y μ σ ) 2 ] u ( y ) . Method (2): A plot of y = g ( x ) is The only solution to y g ( x )=0 is x =ln y ,for 0 f Y ( y r X i =1 f X ( x i ) | g 0 ( x i ) | = f X (ln y ) /e x | x =ln y . y 0 , there is no real solution to y g ( x ; hence for y 0 ,thepd fo f Y equals zero there. Combining solutions for both regions of y ,weget f Y ( y f X (ln y ) /y · u ( y ) . 2. (3.5) Method (1): P [ Y y P [ln X y P [ X e y F X ( e y ) f Y ( y dF d ( e y ) d ( e y ) dy = f X ( e y ) e y = 1 3 e 1 3 e y u ( e y ) e y 1
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But since e y 0 all real y , u ( e y )=1 everywhere. Hence f Y ( y )= 1 3 e 1 3 [exp( y ) 3 y ] Method (2): We plot y = g ( x ) f rst. The only solution to y g ( x )=0 is x =ln y for −∞ <y< . Hence f Y ( y f X ( e y ) dg/dx | x = e y = f X ( e y ) · e y 3. (3.6) Here is the plot of
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ProbSetFive - Rensselaer Electrical, Computer, and Systems...

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