410Hw06ans - STAT 410 Summer 2009 Homework #6 (due...

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STAT 410 Summer 2009 Homework #6 (due Wednesday, July 15, 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 0 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.
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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.
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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 ) = - 0 2 2 2 2 dx x x e = u = 2 2 x du = x dx = - 0 2 2 du u e = 2 2 = 2 . E ( X 2 ) = - 0 2 2 2 2 2 dx x x e = - - dx x x e 2 2 2 2 1 = E ( Z 2 ) = 1. ( Z ~ N ( 0, 1 ) ) Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = 2 1 - = 2 - . 4. 3.4.10 If e 3 t + 8 t 2 is the mgf of the random variable X, find P ( 1 < X < 9 ). M X ( t ) = + 2 2 2 exp σ μ t t = exp { 3 t + 8 t 2 }. Normal distribution, μ = 3, σ 2 = 16, σ = 4. P ( 1 < X < 9 ) = - < < - - 4 3 9 Z 4 3 1 P = P ( 1.00 < Z < 1.50 ) = Φ ( 1.50 ) – Φ ( 1.00 ) = 0.9332 – 0.1587 = 0.7745 .
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5. 3.4.28 Let X 1 and X 2 be independent with normal distributions N ( 6, 1 ) and N ( 7, 1 ), respectively. Find P ( X 1 > X 2 ). Hint:
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This note was uploaded on 10/29/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.

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410Hw06ans - STAT 410 Summer 2009 Homework #6 (due...

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