{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 08_31ans - STAT 410 Examples for Fall 2011 Let X and Y be...

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

STAT 410 Examples for 08/31/2011 Fall 2011 Let X and Y be two discrete random variables. The joint probability mass function p ( x , y ) is defined for each pair of numbers ( x , y ) by p ( x , y ) = P ( X = x and Y = y ) . Let A be any set consisting of pairs of ( x , y ) values. Then P ( ( X, Y ) A ) = ( ) ( ) ∑ ∑ y x A y x p , , . Let X and Y be two continuous random variables. Then f ( x , y ) is the joint probability density function for X and Y if for any two-dimensional set A P ( ( X, Y ) A ) = ( ) ∫∫ A dy dx y x f , . 1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: x \ y 0 1 2 1 0.15 0.10 0 2 0.25 0.30 0.20 a) Find P ( X > Y ) . P ( X > Y ) = p ( 1, 0 ) + p ( 2, 0 ) + p ( 2, 1 ) = 0.15 + 0.25 + 0.30 = 0.70 . b) Find P ( X + Y = 2 ) . P ( X + Y = 2 ) = p ( 1, 1 ) + p ( 2, 0 ) = 0.10 + 0.25 = 0.35 . The marginal probability mass functions of X and of Y are given by p X ( x ) = ( ) y y x p all , , p Y ( y ) = ( ) x y x p all , . The marginal probability density functions of X and of Y are given by f X ( x ) = ( ) - , dy y x f , f Y ( y ) = ( ) - , dx y x f .

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

View Full Document
c) Find the (marginal) probability distributions p X ( x ) of X and p Y ( y ) of Y. y p Y ( y ) x p X ( x ) 0 0.40 1 0.25 1 0.40 2 0.75 2 0.2 If p ( x , y ) is the joint probability mass function of ( X, Y ) OR f ( x , y ) is the joint probability density function of ( X, Y ) , then discrete continuous E ( g ( X, Y ) ) = ∑ ∑ x y y x p y x g all all ) , ( ) , ( E ( g ( X, Y ) ) = - - dy dx y x f y x g ) , ( ) , ( d) Find E ( X ) , E ( Y ) , E ( X + Y ) , E ( X Y ) . E ( X ) = 1 × 0.25 + 2 × 0.75 = 1.75 . E ( Y ) = 0 × 0.40 + 1 × 0.40 + 2 × 0.20 = 0.8 . E ( X + Y ) = 1 × 0.15 + 2 × 0.25 + 2 × 0.10 + 3 × 0.30 + 3 × 0 + 4 × 0.20 = 2.55 . OR E ( X + Y ) = E ( X ) + E ( Y ) = 1.75 + 0.8 = 2.55 . E ( X Y ) = 0 × 0.15 + 0 × 0.25 + 1 × 0.10 + 2 × 0.30 + 2 × 0 + 4 × 0.20 = 1.5 . Moment-generating function M X Y ( t 1 , t 2 ) = E ( e t 1 X + t 2 Y ) , if it exists for | t 1 | < h 1 , | t 2 | < h 2 . M X Y ( t 1 , 0 ) = M X ( t 1 ) , M X Y ( 0, t 2 ) = M Y ( t 2 ) .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

08_31ans - STAT 410 Examples for Fall 2011 Let X and Y be...

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

View Full Document
Ask a homework question - tutors are online