# 410Hw03ans - STAT 410 Fall 2009 Homework #3 (due Friday,...

This preview shows pages 1–6. Sign up to view the full content.

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

View Full Document

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

View Full Document

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: STAT 410 Fall 2009 Homework #3 (due Friday, September 18, by 3:00 p.m.) 1. Consider two continuous random variables X and Y with joint p.d.f. f X, Y ( x , y ) = < < < < otherwise , 81 2 2 K K y x y x a) Find the value of K so that f X, Y ( x , y ) is a valid joint p.d.f. 1 = ∫ ∫ K K dy dx y x 0 0 2 81 2 = 243 5 K . ⇒ K = 3 . b) Find P ( X > 3 Y ). P ( X > 3 Y ) = ∫ ∫ 3 3 2 81 2 dx dy y x x = ∫ 3 4 729 1 dx x = 15 1 . OR P ( X > 3 Y ) = ∫ ∫ 1 3 3 2 81 2 dy dx y x y = … = 15 1 . c) Are X and Y independent? Justify your answer. f X ( x ) = 2 9 1 x , 0 < x < 3, f Y ( y ) = y 9 2 , 0 < y < 3. f ( x , y ) = f X ( x ) ⋅ f Y ( y ). ⇒ X and Y are independent . d) Find Cov ( X, Y ). Since X and Y are independent, Cov ( X, Y ) = . 2. Let X and Y have the joint probability density function f X, Y ( x , y ) = < < < + otherwise 1 4 x y y x a) Find the marginal p.d.f.s f X ( x ) and f Y ( y ). f X ( x ) = ( 29 ∫ + x dy y x 4 = ( 29 2 2 x y y x + = 3 x 2 , 0 < x < 1. f Y ( y ) = ( 29 ∫ + 1 4 y dx y x = y y x x 1 2 4 2 + = 2 2 9 4 2 1 y y-+ , 0 < y < 1. b) Are X and Y independent? Justify your answer. The support of ( X, Y ) is not a rectangle. ⇒ X and Y are NOT independent . OR f ( x , y ) ≠ f X ( x ) ⋅ f Y ( y ). ⇒ X and Y are NOT independent . c) Find Cov ( X, Y ). E ( X ) = ∫ ⋅ 1 2 3 dx x x = 4 3 . E ( X 2 ) = ∫ ⋅ 1 2 2 3 dx x x = 5 3 . E ( Y ) = E [ E ( Y | X ) ] = E ( 18 X 11 ) = 4 3 18 11 ⋅ = 24 11 . E ( X Y ) = E [ E ( X Y | X ) ] = E [ X E ( Y | X ) ] = E ( 18 X 11 2 ) = 5 3 18 11 ⋅ = 30 11 . Cov ( X, Y ) = E ( X Y ) – E ( X ) ⋅ E ( Y ) = 24 11 4 3 30 11 ⋅-= 480 11 . 3. Suppose that the random variables X and Y have joint p.d.f. f ( x , y ) given by f ( x , y ) = C x 2 y , 0 < x < y , x + y < 2. a) Sketch the support of ( X , Y ). That is, sketch { 0 < x < y , x + y < 2 }. b) What must the value of C be so that f ( x , y ) is a valid joint p.d.f.? Must have ( 29 ∫ ∫ ∞ ∞ ∞ ∞--dy dx y x f , = 1. ∫ ∫ -1 2 2 dx dy y x x x C = ∫ =-= 1 2 2 2 2 dx y x x y x y C = ( 29 [ ] ∫ --1 2 2 2 2 2 dx x x x C = ( 29 ∫-1 3 2 2 2 dx x x C C = 1 4 3 2 3 2 -x x C C = 6 C = 1. ⇒ C = 6 . c) Find P ( X + Y < 1 ). ∫ ∫ -5 . 1 2 6 dx dy y x x x = ( 29 ∫ =-= 5 . 1 2 2 3 dx y x x y x y = ( 29 [ ] ( 29 ∫--5 . 2 2 2 1 3 dx x x x = ( 29 ∫-5 . 3 2 6 3 dx x x = 5 . 4 3 2 3 -x x = 4 3 2 1 2 3 2 1 - = 32 3 8 1-= 32 1 = 0.03125 ....
View Full Document

## This note was uploaded on 10/14/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

### Page1 / 16

410Hw03ans - STAT 410 Fall 2009 Homework #3 (due Friday,...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online