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Unformatted text preview: STAT 410 Fall 2009 Homework #1 (due Friday, September 4, by 3:00 p.m.) 1. Consider a continuous random variable X with probability density function f X ( x ) = < < o.w. 1 3 2 x x Find the momentgenerating function of X, M X ( t ). M X ( t ) = E ( e t X ) = ( 29 ∫ ∞ ∞ ⋅dx x f x t e = ∫ ⋅ 1 2 3 dx x x t e . u = 3 x 2 , dv = e t x dx , du = 6 x dx , v = t 1 e t x . M X ( t ) = ∫ ⋅ 1 2 3 dx x x t e = ∫  ⋅ ⋅ 1 2 6 1 1 1 3 dx x e t e t x x t x t = ∫ ⋅ 1 6 1 3 dx x e t e t x t t u = 6 x , dv = = t 1 e t x dx , du = 6 dx , v = 2 1 t e t x . M X ( t ) = ∫ ⋅ 1 6 1 3 dx x e t e t x t t = ∫  ⋅ ⋅ 1 2 2 6 1 1 1 6 3 dx e t e t x e t x t x t t = 1 6 6 3 3 2 +x t t t e t e t e t = 3 3 2 6 6 6 3 t e t e t e t t t t+, t ≠ 0. M X ( ) = 1. 2. Suppose a discrete random variable X has the following probability distribution: P( X = k ) = ( 29 ! 2 ln k k , k = 1, 2, 3, … . a) Verify that this is a valid probability distribution. • p ( x ) ≥ 0 2200 x c • ( 29 ∑ x x p all = 1 ( 29 ∑ ∞ = 1 ! 2 ln k k k = ( 29 ∑ ∞ = ! 2 ln k k k – 1 = e ln 2 – 1 = 2 – 1 = 1. c b) Find μ X = E ( X ) by finding the sum of the infinite series. E ( X ) = ∑ ⋅ x x p x all ) ( = ( 29 ∑ ∞ = ⋅ 1 ! 2 ln k k k k = ( 29 ( 29 ∑ ∞= 1 ! 1 2 ln k k k = ( 29 ( 29 ( 29 ∑ ∞=⋅ 1 1 ! 1 2 2 ln ln k k k = ( 29 ( 29 ∑ ∞ = ⋅ ! 2 2 ln ln k k k = 2 ln 2. c) Find the momentgenerating function of X, M X ( t ). M X ( t ) = ∑ ⋅ x x t x p e all ) ( = ( 29 ∑ ∞ = ⋅ 1 ! 2 ln k k k t k e = ∑ ∞ = 1 ! 2 ln k k t k e = 1 2 lnt e e = 1 2t e . d) Use M X ( t ) to find μ X = E ( X ). ( 29 t e t e t 2 2 M ln ' X ⋅ ⋅ = , E ( X ) = ( 29 M ' X = 2 ln 2. e) Find σ X 2 = Var ( X ). ( 29 ( 29 t t e e t t e e t 2 2 2 2 M ln ln 2 X ' ' ⋅ ⋅ ⋅ ⋅ + = . E ( X 2 ) = ( 29 M ' ' X = 2 ( ln 2 ) 2 + 2 ln 2. Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = 2 ln 2 – 2 ( ln 2 ) 2 . OR E ( X ( X – 1 ) ) = ( 29 ( 29 ∑ ∞= ⋅ ⋅ 1 ! 2 1 ln k k k k k = ( 29 ( 29 ∑ ∞= ⋅ ⋅ 2 ! 2 1 ln k k k k k = ( 29 ( 29 ∑ ∞= 2 ! 2 2 ln k k k = ( 29 ( 29 ( 29 ∑ ∞=⋅ 2 2 2 ! 2 2 2 ln ln k k k = ( 29 ( 29 ∑ ∞ = ⋅ 2 ! 2 2 ln ln n n n = 2 ( ln 2 ) 2 . E ( X 2 ) = E ( X ( X – 1 ) ) + E ( X ) = 2 ( ln 2 ) 2 + 2 ln 2. Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = 2 ln 2 – 2 ( ln 2 ) 2 . 3. Let a > 2. Suppose a discrete random variable X has the following probability distribution: p ( ) = P ( X = 0 ) = c , p ( k ) = P ( X = k ) = k a 1 , k = 1, 2, 3, … ....
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 Fall '08
 AlexeiStepanov
 Probability

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