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# Hw01ans - STAT 410 Homework#1(due Friday January 25 by 3:00...

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STAT 410 Spring 2008 Homework #1 (due Friday, January 25, by 3:00 p.m.) 1. Consider a continuous random variable X with probability density function f X ( x ) = ° ± ° ² ³ < < o.w. 0 1 0 3 2 x x Find the moment-generating function of X, M X ( t ) . M X ( t ) = E ( e t X ) = ( ) ´ - dx x f x t e = ´ 1 0 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 0 2 3 dx x x t e = ´ µ · ¸ ¹ º - µ · ¸ ¹ º 1 0 2 6 1 0 1 1 3 dx x e t e t x x t x t = ´ µ · ¸ ¹ º - 1 0 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 0 6 1 3 dx x e t e t x t t = ´ µ µ · ¸ ¸ ¹ º - µ µ · ¸ ¸ ¹ º - 1 0 2 2 6 1 0 1 1 6 3 dx e t e t x e t x t x t t = 0 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 ( 0 ) = 1.

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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 = ( ) » = 0 ! 3 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 = ( ) ( ) » = 0 ! 2 2 ln ln k k k = 2 ln 2 . c) Find the moment-generating 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 = » µ · ¸ ¹ º = 1 ! 2 ln k k t k e = 1 2 ln - t e e = 1 2 - t e . d) Use M X ( t ) to find μ X = E ( X ) . ( ) t e t e t 2 2 M ln ' X = , E ( X ) = ( ) 0 M ' X = 2 ln 2 .
3. Suppose a random variable X has the following probability density function: ° ± ° ² ³ = - otherwise 0 1 0 ) ( 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 0 1 0 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 0 1 1 . Therefore, µ · ¸ ¹ º - = 1 e e C 1.5819767 .

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