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Unformatted text preview: STAT 410 Fall 2009 Homework #7 (due Friday, October 16, by 4:00 p.m.) 1. Let X be a continuous random variable with probability density function ( 29 1 X x e x x f = , x > 0, where > 0, > 0. ( X has a Weibull distribution. ) Consider Y = X . What is the probability distribution of Y? Y = X X = g 1 ( y ) = 1 Y x > 0 y > 0 dy dx = 1 1 1  y . f Y ( y ) = f X ( g 1 ( y ) ) y x d d = ( 29 1 1 / 1 1  y y y e = y e  , y > 0. OR F X ( x ) =  x du u u e 1 = 1 x e , x > 0. F Y ( y ) = P ( Y y ) = P ( X 1 y ) = F X ( 1 y ) = y e 1 , y > 0. f Y ( y ) = y e  , y > 0. 2. Let X have an exponential distribution with = 1; that is, the p.d.f. of X is f ( x ) = e x , 0 < x < . Let T be defined by T = ln X. a) Show that the p.d.f. of T is g ( t ) = e t e e t , < t < , which is the p.d.f. of an extreme value distribution. t = ln ( x ) x = e t dt dx = e t x > 0 < t < g ( t ) = f T ( t ) = f X ( e t ) dt dx = e t e e t , < t < . OR F X ( x ) = x e 1 , x > 0. F T ( t ) = P ( T t ) = P ( X e t ) = F X ( e t ) = t e e 1 , < t < . g ( t ) = f T ( t ) = e t e e t , < t < . b) Let W be defined by T = + ln W, where < < and > 0. Show that W has a Weibull distribution. t = + ln w dw dt = w f W ( t ) = f T ( + ln w ) dw dt = e + ln w e e + ln w w = e w 1 e e w , w > 0. 3. 3.4.11 Let the random variable X have the p.d.f. f ( x ) = 2 2 2 2 x e , 0 < x < , zero elsewhere. Find the mean and the variance of X. Hint: Compute E ( X ) directly and E ( X 2 ) by comparing the integral with the integral representing the variance of a random variable that is N ( 0, 1 ). E ( X ) =  2 2 2 2 dx x x e = u = 2 2 x du = x dx =  2 2 du u e = 2 2 = 2 ....
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This note was uploaded on 04/15/2010 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Monrad
 Probability

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