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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 β 1 X α β β α x e x x f = , x > 0, where α > 0, β > 0. ( X has a Weibull distribution. ) Consider Y = β X . What is the probability distribution of Y? Y = β X X = g – 1 ( y ) = β 1 Y x > 0 ⇒ y > 0 dy dx = 1 1 β 1 β y . f Y ( y ) = f X ( g – 1 ( y ) ) y x d d = ( 29 1 1 / 1 β β β 1 α β β α ⋅ y y y e = y e α α , y > 0. OR F X ( x ) = ∫ x du u u e 1 β α β β α = β α 1 x e , x > 0. F Y ( y ) = P ( Y ≤ y ) = P ( X ≤ β 1 y ) = F X ( β 1 y ) = y e α 1 , y > 0. f Y ( y ) = y e α α , y > 0. 2. Let X have an exponential distribution with θ = 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 = α + β ln W, where – ∞ < α < ∞ and β > 0. Show that W has a Weibull distribution. t = α + β ln w dw dt = w β f W ( t ) = f T ( α + β ln w ) dw dt = e α + β ln w e – e α + β ln w w β ⋅ = e α β w β – 1 e – e α w β , w > 0. 3. 3.4.11 Let the random variable X have the p.d.f. f ( x ) = 2 2 2 2 x e π , 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 π = … u = 2 2 x du = x dx … = ∫ ∞ 2 2 du u e π = π 2 2 = π 2 ....
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 Fall '08
 Monrad
 Normal Distribution, Probability, Probability theory, Exponential distribution

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