HW8sol - X(g – 1(y d dy(g – 1(y where g – 1(y = ln y For – A ≤ r ≤ A = f θ = 1 π Thus f R(r = 1 π A 1 – r A 2 –A ≤ r ≤ A f

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EE/NIS 605-CS505 Probability and Stochastic Processes I Homework 8 Solutions Professor Heffes Textbook: S. Ross, A First Course in Probability, Seventh Edition , Prentice Hall, 2006 Chapter 5 : Theoretical Exercises: 30; Problems: 41 TE 30 : F Y (y) = P[Y y] = P[e X y] = P[X ln y] = F X (ln y) f Y (y) = d dy F Y (y) = d dy F X (ln y) =f X (ln y) 1 y = 1 y 2 π σ e – (ln y – μ ) 2 / 2 σ 2 Also from f Y (y) =f
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Unformatted text preview: X (g – 1 (y)) d dy (g – 1 (y)) where g – 1 (y) = ln y For – A ≤ r ≤ A, = f θ = 1 π . Thus f R (r) = 1 π A 1 – ( r A ) 2 , –A ≤ r ≤ A f R (r) = f θ (sin – 1 r A ) d dr (sin – 1 r A ) P 41 : F R (r) = P[R ≤ r] = P[sin θ ≤ r A ] = P[ θ ≤ sin – 1 r A ] = F θ (sin – 1 r A )...
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This note was uploaded on 04/29/2008 for the course ECE ee605 taught by Professor Sss during the Spring '08 term at Stevens.

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