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Unformatted text preview: STAT 410 Fall 2009 Homework #2 (due Friday, September 11, by 3:00 p.m.) 1. The p.d.f. of X is f X ( x ) = x 1 , 0 < x < 1, 0 < < . Let Y = 2 ln X. How is Y distributed? a) Determine the probability distribution of Y by finding the c.d.f. of Y F Y ( y ) = P ( Y y ) = P ( 2 ln X y ). Hint: Find F X ( x ) first. F X ( x ) = x , 0 < x < 1. 0 < x < 1 y = 2 ln x y > 0 F Y ( y ) = P ( Y y ) = P ( 2 ln X y ) = P ( X e y / 2 ) = 1 e y / 2 , y > 0. f Y ( y ) = F Y ' ( y ) = 2 1 e y / 2 , y > 0. Y has Exponential distribution with mean 2. b) Determine the probability distribution of Y by finding the m.g.f. of Y M Y ( t ) = E ( e Y t ) = E ( e 2 ln X t ). M Y ( t ) = E ( e Y t ) = E ( e 2 ln X t ) = E ( X 2 t ) = ( 29  1 1 2 dx x x t = 1 1 2 dx x t = t 2= t 2 1 1, t < 2 1 . Y has Exponential distribution with mean 2. c) Determine the probability distribution of Y by finding the p.d.f. of Y, f Y ( y ), using Theorem 1.7.1. y = g ( x ) = 2 ln x x = g 1 ( y ) = = e y / 2 d x / d y = 2 1e y / 2 f Y ( y ) = f X ( g 1 ( y ) ) y x d d = 2 1 2 1 2 y y e e = 2 1 e y / 2 , y > 0. Y has Exponential distribution with mean 2. 2. A fair 6sided die is rolled repeatedly. Find the probability that a) the first 6 occurs on an evennumbered attempt; P(even) = P(2) + P(4) + P(6) + = + + + 5 3 1 6 5 6 1 6 5 6 1 6 5 6 1 = 11 5 == = = = + 36 25 1 1 36 5 36 25 36 5 6 5 6 1 1 2 n n k k . OR P(even) = + + + 5 3 1 6 5 6 1 6 5 6 1 6 5 6 1 P(odd) = + + + 4 2 6 5 6 1 6 5 6 1 6 5 6 1 P(even) = 6 5 P(odd). P(odd) = 5 6 P(even). 1 = P(odd) + P(even) = 5 11 P(even). P(even) = 11 5 . b) 6 occurs before an odd number is rolled. P ( 6 before odd ) = = + + + + + = 2 6 1 6 2 ......
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This note was uploaded on 10/14/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.
 Fall '08
 AlexeiStepanov
 Probability

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