08_29 - STAT 410 Example 9: Examples for 08/29/2008 Fall...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
STAT 410 Examples for 08/29/2008 Fall 2008 Example 9 : Suppose a discrete random variable X has the following probability distribution: P( X = 0 ) = p , P( X = k ) = ! 2 1 k k , k = 1, 2, 3, … a) Find the value of p that would make this a valid probability distribution. Must have = + 1 ! 2 1 k k k p = 1. Since a k k e k a 0 ! = = , 1 ! 2 1 2 1 1 - = = e k k k . Therefore, p + ( 1 2 1 - e ) = 1 and p = 2 1 2 e - . b) Find E ( X ). E ( X ) = x x p x all ) ( = 0 ( ) 2 1 2 e - + = 1 ! 2 1 k k k k = ( ) = - 1 ! 1 2 1 k k k = ( ) = - - 1 1 ! 1 2 1 2 1 k k k = = 0 ! 2 1 2 1 n n n = 2 2 1 e . c) Find the variance of X, Var ( X ). E ( X ( X – 1 ) ) = ( ) = - 2 ! 2 1 1 k k k k k = ( ) = - 2 ! 2 2 1 k k k = ( ) = - - 2 2 ! 2 2 1 4 1 k k k = = 0 ! 2 1 4 1 n n n = 4 2 1 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
E ( X 2 ) = E ( X ( X – 1 ) ) + E ( X ) = 2 1 4 3 e . Var
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

08_29 - STAT 410 Example 9: Examples for 08/29/2008 Fall...

This preview shows document pages 1 - 3. Sign up to view the full document.

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