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Unformatted text preview: STAT 410 Fall 2008 Homework #1 (due Friday, September 5, by 3:00 p.m.) 1. Consider a continuous random variable X with probability density function f X ( x ) = & ± & ² ³ < < o.w. 1 3 2 x x Find the momentgenerating function of X, M X ( t ). M X ( t ) = E ( e t X ) = ( ) ´ ∞ ∞ ⋅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 = ´ µ ¶ · ¸ ¹ ºµ ¶ · ¸ ¹ º ⋅ ⋅ 1 2 6 1 1 1 3 dx x e t e t x x t x t = ´ µ ¶ · ¸ ¹ º⋅ 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 ) = ´ µ ¶ · ¸ ¹ º⋅ 1 6 1 3 dx x e t e t x t t = ´ µ µ ¶ · ¸ ¸ ¹ ºµ µ ¶ · ¸ ¸ ¹ º⋅ ⋅ 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 µ µ ¶ · ¸ ¸ ¹ º +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 ) = ( ) ! 2 ln k k , k = 1, 2, 3, … . a) Verify that this is a valid probability distribution. • p ( x ) ≥ 0 ∀ x & • ( ) & x x p all = 1 ( ) & ∞ = 1 ! 2 ln k k k = ( ) & ∞ = ! 2 ln k k k – 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 x p x all ) ( = ( ) & ∞ = ⋅ 1 ! 2 ln k k k k = ( ) ( ) & ∞= 1 ! 1 2 ln k k k = ( ) ( ) ( ) & ∞=⋅ 1 1 ! 1 2 2 ln ln k k k = ( ) ( ) & ∞ = ⋅ ! 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 ) ( = ( ) & ∞ = ⋅ 1 ! 2 ln k k k t k e = & ∞ ± ² ³ ´ µ ¶ = 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 ). ( ) t e t e t 2 2 M ln ' X ⋅ ⋅ = , E ( X ) = ( ) M ' X = 2 ln 2. 3. Suppose a random variable X has the following probability density function: & ± & ² ³ ≤ ≤ ⋅ =otherwise 1 ) ( x C x f x e a) What must the value of C be so that f ( x ) is a probability density function? For f ( x ) to be a probability density function, we must have: 1) f ( x ) ≥ 0, 2) ( ) 1 d = ´ ∞ ∞x x f . ( ) ´ ´ ´∞ ∞⋅ = ⋅ = = 1 1 d d d 1 x C x C x x f x x e e ( ) ( ) µ ¶ · ¸ ¹ º⋅ =⋅ =⋅ =e e e e C C C x 1 1 1 1 . Therefore, µ ¶ · ¸ ¹ º= 1 e e C ≈ 1.5819767 . & & ± & & ² ³ ≤ ≤ ⋅ µ ¶ · ¸ ¹ º=otherwise 1 1 ) ( x x f x e e e b) Find the cumulative distribution function F ( x ) = P( X ≤ x ). ( ) ( ) ( ) ´ ∞= ≤ = x y y f x x F X d P . F ( x ) = 0 for x < 0....
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This note was uploaded on 10/22/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.
 Fall '08
 AlexeiStepanov
 Probability

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