410Hw01ans - STAT 410 Summer 2009 Homework #1 (due Friday,...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
STAT 410 Summer 2009 Homework #1 (due Friday, June 19, by 4:00 p.m.) 1. 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 = 0 ! 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 = 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 ) ( = ( 29 = 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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 0 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 0 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 = 0 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 .
Background image of page 2
2. Suppose a random variable X has the following probability density function: = otherwise 0 1 1 ) ( C x x x f 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) ( 29 1 = - dx x f . ( 29 C C C dx x dx x f ln ln ln 1 1 1 1 = = = = - - . Therefore, C = e . b) Find P ( X < 2 ). P ( X < 2 ) = ( 29 1 2 1 ln ln 2 1 2 - = = - dx x dx x f = ln 2 . c) Find P ( X < 3 ). P ( X < 3 ) = ( 29 1 1 ln ln 1 3 - = = - e e dx x dx x f = 1 . d) Find μ X = E ( X ). μ X = E ( X ) = ( 29 = = - e e dx dx x x dx x f x 1 1 1 1 = e – 1 . e) Find σ X 2 = Var ( X ). E ( X 2 ) = ( 29 = = - e e dx x dx x x dx x f x 1 1 2 2 1 = 2 1 2 - e . σ X 2 = Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 2 3 4 2 - + - e e 0.242 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
3. An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder’s loss, Y, follows a distribution with density function: f ( y ) = otherwise 0 1 if 2 3 y y a) What is the expected value and the variance of the policyholder’s loss?
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 13

410Hw01ans - STAT 410 Summer 2009 Homework #1 (due Friday,...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online