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UIUC STAT 400 - 400Ex2_3_2

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STAT 400 Lecture AL1 Example 2.3 (Bonus) Spring 2015 Dalpiaz 1. Let X be a discrete random variable with p.m.f. p ( x ) that is positive on the odd non-negative integers { 1, 3, 5, 7, 9, … } and is zero elsewhere. Suppose p ( 1 ) = c ( unknown ), p ( k ) = k 21, k = 3, 5, 7, 9, … . a) Find the value of c that makes this is a valid probability distribution. Must have  xxp all = 1. 1 = p ( 1 ) + p ( 3 ) + p ( 5 ) + p ( 7 ) + p ( 9 ) + … = c + 9753 21212121 + … = c + 41181 = c + 61.  c = 65. b) Find E ( X ). E ( X ) = x xpxall)( = 9753 2927252365 + … 41 E ( X ) = 975 272523245 + …  43 E ( X ) = 975 2222222458365 + … = 1 + 411322 = 1213.  E ( X ) = 913. ORM X ( t ) = E ( e t X ) = ( )( )∑∞=++⋅⋅+112 12 1 21 65kktktee = ∑∞=+⋅⋅1 2 4 265kkttteee = 414265 2 2 tttteeee−+⋅⋅ = tttteeee 2 2 4265−+⋅⋅ = ttteee 2 3 2865−+⋅, t < ln 2. M 'X ( t ) = ()()( )2 2 2 3 2 3 28 428 365tttttteeeeee−−−−+⋅ = ( )2 2 5 3 28 22465tttteeee−−+⋅, t < ln 2. E ( X ) = M 'X ( 0 ) = 362265+ = 3652 = 913. OR M X ( t ) = E ( e t X ) = ( )( )∑∞=++⋅⋅ +112 12 1 21 65kktktee = ∑∞=+⋅⋅1 2 4 265kkttteee = −−+⋅⋅1411265 2 ttteee = ttteee 2 28431−+⋅, t < ln 2. M 'X ( t ) = ()()( )2 2 2 2 28 4428 431tttttteeeeee−−−−+⋅ = ( )2 2 3 28 83231tttteeee−++⋅, t < ln 2. E ( X ) = M 'X ( 0 ) = 364031+ = 3652 = 913.2. Let X be a discrete random variable with p.m.f. p ( k ) = kc 43⋅, k = 5, 6, 7, 8, … . a) Find the value of c that makes this is a valid probability distribution. Must have ()∑xxp all = 1. ⇒ ∑∞=5 43 kkc = ∑∞=5 43 kkc = 1. ∑∞=5 43 kk = basetermfirst−1 = 431435 − = 41 1024243 = 256243. OR ∑∞=5 43 kk = 25681642716943143 0 −−−−−∞∑=kk = 256812561082561442561922562564311−−−−−− = 2567814 − = 2567811024− = 256243. ⇒ c = 243256 = 54 34. b) Find P ( X is even ). P ( X is even ) = p ( 6 ) + p ( 8 ) + p ( 10 ) + p ( 12 ) + … = 121086 43434343+++⋅⋅⋅⋅ cccc + … = basetermfirst−1 = 26 43143−⋅c = 167 163 = 73 ≈ 0.42857.OR P ( X is even ) = 121086 43434343+++⋅⋅⋅⋅ cccc + … P ( X is odd ) = 11975 43434343+++⋅⋅⋅⋅ cccc + … ⇒ P ( X is even ) = 43 ⋅ P ( X is odd ). P ( X is odd ) = 34 ⋅ P ( X is even ). ⇒ 1 = P ( X is odd ) + P ( X is even ) = 37 ⋅ P ( X is even ). ⇒ P ( X is even ) = 73 ≈ 0.42857. c) Find the moment-generating function of X, M X ( t ). For which values of t does it exist? M X ( t ) = E ( e t X ) = ∑∞=⋅5 43243256 kktke = ∑∞=⋅5 4 3 243256kkte = 4 314 3243256 5ttee−⋅ = ttee 7681024 243243256 5−⋅ = ttee 34 5−, t < ln 34. d) Find E ( X ). M 'X ( t ) = ()()()2 5 5 34 334 5ttttteeeee−−−− = ()2 6 5 34 1220ttteee−−, t < ln 34.E ( X ) = M 'X ( 0 ) = 8. OR E ( X ) = ∑⋅x xpxall)( = 8765 432432568432432567432432566432432565+++⋅⋅⋅⋅⋅⋅⋅⋅ + … 43 E ( X ) = 876 432432567432432566432432565++⋅⋅⋅⋅⋅⋅ + … ⇒ 41 E ( X ) = 87655 43243256432432564324325643243256432432564++++⋅⋅⋅⋅⋅⋅ + … = 1 + ∑∞=⋅5 43 243256kk = 1 + 1 = 2. ⇒ E ( X ) = 8. OR E ( X ) = ∑⋅x xpxall)( = ∑∞=⋅⋅5 43243256 kkk = ∑∞=−⋅⋅⋅⋅51 4341 3243256kkk = −−−−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅∑∞=− 3211 4341443413434124114341 81256kkk = ( )−−−−⋅6427642716641YE81256 = = ( )−⋅6494YE81256 , where Y has a Geometric distribution with probability of “success” p = 41. ⇒ E ( X ) = ( )−⋅6494YE81256 = −⋅6494481256 = 6416281256⋅ = 8.OR p ( k ) = kc 43⋅, k = 5, 6, 7, 8, … , c = 54 34. ⇒ p ( k ) = 5 4341−⋅k, k = 5, 6, 7, 8, … . ⇒ X = Geometric=41p + 4. Let Y have a Geometric=41p distribution. X = Y + 4. c) M Y ( t ) = ttee 431 41− = ttee 34 −, t < – ln 43 = ln 34. Theorem: Let V = a W + b. Then M V ( t ) = e b t M W ( a t ). Proof: M V ( t ) = E ( e t V ) = E ( e t ( a W + b ) ) = E ( e a t W e b t ) = e b t E ( e a t W ) = e b t M W ( a t ). ⇒ M X ( t ) = M Y + 4 ( t ) = e 4 t M Y ( t ) = ttteee 34 4−⋅ = = ttee 34 5−, t < ln 34. d) E ( X ) = E ( Y + 4 ) = 4 + 4 = 8.3. Suppose a discrete random variable X has the following probability distribution: P( X = k ) = ( )! 2lnkk, k = 1, 2, 3, … . a) Verify that this is a valid probability distribution. • p ( x ) ≥ 0 ∀ x  • ()∑xxp all = 1 ( )∑∞=1! 2lnkkk = ( )∑∞=0! 2lnkkk – 1 = e ln 2 – 1 = 2 – 1 = 1.  b) Find µ X = E( X ) by finding the sum of the infinite series. E ( X ) = ∑⋅x xpx all)( = ( )∑∞=⋅1 ! 2lnkkkk = ( )( )∑∞−=1! 12lnkkk = ( )( )(


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UIUC STAT 400 - 400Ex2_3_2

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