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 1ttxe = 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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