This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STAT 410 Spring 2008 Homework #4 1. Let X and Y be independent random variables, each geometrically distributed with the probability of “success” p , 0 < p < 1. That is, p X ( k ) = p Y ( k ) = ( ) 1 1 ⋅ k p p , k = 1, 2, 3, … , a) Find P ( X > Y ). [ Hint: First, find P ( X = Y ). ] P ( X = Y ) = ( ) ( ) & ∞ = ⋅ 1 Y X k k p k p = ( ) ( ) & ∞ = ⋅ ⋅ ⋅ 1 1 1 1 1 k k k p p p p = ( ) [ ] & ∞ = ⋅ 1 1 2 2 1 k k p p = ( ) [ ] & ∞ = ⋅ 2 2 1 n n p p = ( ) 2 2 1 1 p p = p p 2 . P ( X > Y ) + P ( X = Y ) + P ( X < Y ) = 1. Since P ( X > Y ) = P ( X < Y ), P ( X > Y ) = ( ) ( ) Y X P 1 2 1 = ⋅ = ¡ ¡ ¢ £ ¤ ¤ ¥ ¦ ⋅ 2 1 2 1 p p = p p 2 1 . OR P ( X > Y ) = ( ) ( ) & & ∞ ∞ = + = ⋅ ⋅ ⋅ 1 1 1 1 1 1 y y x y x p p p p = ( ) ( ) & & ∞ ∞ = + = ⋅ ⋅ 1 1 1 1 2 1 1 y y x x y p p p = ( ) ( ) ( ) & ∞ = ⋅ ⋅ 1 1 2 1 1 1 1 y y y p p p p = ( ) & ∞ = ⋅ 1 1 2 1 y y p p = ( ) ( ) [ ] & ∞ = ⋅ ⋅ 2 1 1 n n p p p = ( ) ( ) 2 1 1 1 p p p ⋅ = p p 2 1 . b) Find P ( X + Y = n ), n = 2, 3, 4, … , and P ( X = k  X + Y = n ), k = 1, 2, 3, … , n – 1. P ( X + Y = n ) = ( ) ( ) & = = = ⋅ 1 1 Y P X P n k k n k = ( ) ( ) & = ⋅ ⋅ ⋅ 1 1 1 1 1 1 n k k n k p p p p = ( ) & = ⋅ 1 1 2 2 1 n k n p p = ( ) ( ) 2 2 1 1 ⋅ ⋅ n p p n , n = 2, 3, 4, … . Since X and Y both have Geometric ( p ) distribution and are independent, X + Y has Negative Binomial distribution with r = 2....
View
Full
Document
This note was uploaded on 04/13/2008 for the course STAT 410 taught by Professor Alexeistepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 AlexeiStepanov
 Probability

Click to edit the document details