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Unformatted text preview: 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 ) = & & < < 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 =  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 ) = ( ) ! 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 = ( ) & = ! 3 ln k k k 1 = e ln 2 1 = 2 1 = 1. & b) Find 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 = & = 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 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: & & = 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 ( ) ( )  = = = e e e e C C C x 1 1 1 1 ....
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 Spring '08
 AlexeiStepanov
 Probability

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