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UIUC STAT 400 - 400Ex3_2ans

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STAT 400 Lecture AL1 Answers for 3.2 (Part 1) Spring 2015 Dalpiaz Uniform Distribution over an interval [ a , b ] : For Uniform distribution, P ( c  X  d ) = d cb a, a  c  d  b. E ( X ) = a b2, Var ( X ) = 12ab2)( . 0.25. Let X be a random variable distributed uniformly over the interval [ a , b ]. Find the moment-generating function of X. M X ( t ) = E ( e t X ) =  - dxxfxte = baxtdxabe 1 = abtabxte 1  =   abtatbtee, t  0. M X ( 0 ) = 1.Exponential Distribution:  otherwise00forθ1θxxf x e  otherwise00forλλ xxf x e E ( X ) = , Var ( X ) =  2. E ( X ) = 1/, Var ( X ) = 1/2. 0.50. Find the moment generating function of an exponential random variable. M X ( t ) = E ( e t X ) =  - dxxfxte = 0 θ1 θdxxxtee =  0 1 θ 1 θdxtxe =  0 θθ1θ1 1ttxe = θ 11t, t < 1/. OR M X ( t ) = E ( e t X ) =  - dxxfxte = 0 λ λ dxxxtee =  0 λ λ dxtxe =  0 λλ λ ttxe = t λλ, t < .0.75. Suppose X has Exponential distribution with mean .  f X ( x ) = θ 1θxe, x > 0. a) Find P ( X > t ) for t > 0. P ( X > t ) = txdxe 1 θθ = txe θ = θte, t > 0. b) Show that for positive t and s, P ( X > t + s | X > t ) = P ( X > s ) ( memoryless property ). P ( X > t + s | X > t ) =    X P X X P ttst =    X P X P tst =  θθtstee = θse = P ( X > s ). Compare to Geometric distribution: a) For Geometric ( p ), P ( X > a ) = ( 1 – p ) a, a = 0, 1, 2, 3, … . b) For positive integers a and b, P ( X > a + b | X > a ) = P ( X > b ) ( memoryless property ).1. Suppose the lifetime of a particular brand of light bulbs is exponentially distributed with mean of 400 hours. Mean = 400   = 400,  = 1/400. a) Find the probability that a randomly selected light bulb would last over 500 hours. Exponential,  = 400. P ( T > 500 ) = e – 500/400 = e – 1.25  0.2865. OR P ( T > 500 ) = 500400 4001 dxxe = … b) Find the probability that 3 out of 7 randomly selected light bulbs would last over 500 hours. Binomial, n = 7, p = 0.2865. P ( X = 3 ) = 43 7135.02865.037  0.2133. c) Find the probability that a randomly selected light bulb would last between 300 hours and 800 hours. Exponential,  = 400. P ( 300 < T < 800 ) = P ( T > 300 ) – P ( T > 800 ) = e – 300/400 – e – 800/400 = e – 0.75 – e – 2.0  0.337. OR P ( 300 < T < 800 ) 800300400 4001 dxxe = …2. An insurance policy reimburses a loss up to a benefit limit of C and has a deductible of d. The policyholder’s loss, X, follows a distribution with density function f ( x ). Find the expected value of the benefit paid under the insurance policy? Benefit Paid = dxdxddxdxCCC0 E ( Benefit Paid ) =      dCdddCCdxxfdxxfdxdxxf 0 XX0X. For example, if X has an Exponential distribution with mean , E ( Benefit Paid ) =     dCddCCdxxfdxxfdx XX =  dxCddxCCdxdxdx ee 1 1 θθθθ =  dxCddxxCCeeedx   θθθθ =  θθ θθdd Cee. For example, if d = 2, C = 10, X has an Exponential distribution with mean  = 5, E ( Benefit Paid ) =      12X122X 10 2 dxxfdxxfx =  1251225 5110 512 dxdxxxxee =  12 10212 52555  xxxeeex = 51252 5 5 ee = 5 e – 0.4 – 5 e – 2.4  2.898.Fact: Let X be a nonnegative continuous random variable with p.d.f. f ( x ) and c.d.f. F ( x ). Then E ( X ) =   0 1 xx dF. Proof: E ( X ) =  0 xxfx d =   0 0 xxfy dxd =   0 0 xyxf dxd    0 0 xyxf dxd =   0 yxxf dyd  E ( X ) =   0 yxxf dyd =  0 XP yy d =   0 1 yy dF. Example: ( 2. from Examples for 3.3 ) Let X be a continuous random variable with the cumulative distribution function F ( x ) = 0, x < 0, F ( x ) = 83  x, 0  x  2, F ( x ) = 1 – 2 1x, x > 2. c) E ( X ) =   0 F1 xx d =  2220 1831 xxxx d d = 21 02283 22 xx   = 21 43 2  = 47 = 1.75.Example: Find the expected value of an Exponential (  ) distribution. For Exponential (  ), 1 – F ( x ) = P ( X > x ) = θxe, t > 0. E ( X ) =   0 1 xx dF = 0 θxdxe = . Fact: Let X be a random variable of the discrete type with pmf p ( x ) that is positive on the nonnegative integers and is equal to zero elsewhere. Then E ( X ) =   0 F1 xx, where F ( x ) is the cdf of X. “Proof”: 1 – F ( x ) = P ( X > x ) = p ( x + 1 ) + p ( x + 2 ) + p ( x + 3 ) + p ( x + 4 ) + … 1 – F ( 0 ) p ( 1 ) + p ( 2 ) + p ( 3 ) + p


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UIUC STAT 400 - 400Ex3_2ans

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