# HW29Answers(11.1) - Math331 Spring 2008 Instructor David...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 11.1 Homework Answers Homework: pgs. 465 - 466, #’s 1, 3, 6, 7, 9, 17. 1. We have M X ( t ) = E bracketleftbig e tX bracketrightbig = 5 summationdisplay x =1 e tx P { X = x } = 5 summationdisplay x =1 e tx 1 5 = 1 5 ( e t + e 2 t + e 3 t + e 4 t + e 5 t ) . 3. We are given that for i = 1 , 2 , 3 , . . . p ( X = i ) = p ( i ) = 2 parenleftbigg 1 3 parenrightbigg i , and zero elsewhere. Thus, M X ( t ) = bracketleftbig e tX bracketrightbig = ∞ summationdisplay i =1 e ti P ( X = i ) = 2 ∞ summationdisplay i =1 e ti parenleftbigg 1 3 parenrightbigg i = 2 ∞ summationdisplay i =1 parenleftbigg e t 3 parenrightbigg i = 2 ∞ summationdisplay i =0 parenleftbigg e t 3 parenrightbigg i- 2 . Now, the sum above is only convergent if e t < 3, or t < ln(3). Thus, for such t , M X ( t ) = 2 1 1- e t / 3- 2 = 2 3 3- e t- 2(3- e t ) 3- e t = 2 e t 3- e t ....
View Full Document

## This note was uploaded on 06/24/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.

### Page1 / 3

HW29Answers(11.1) - Math331 Spring 2008 Instructor David...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online