This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STAT 410 Fall 2008 Homework #4 (due Friday, September 26, by 3:00 p.m.) 1. Let X and Y have the joint probability density function f X, Y ( x , y ) = & ± & ² ³ < < < + otherwise 1 4 x y y x a) Find f Y ( y ). f Y ( y ) = ( ) ´ + 1 4 y dx y x = y y x x 1 2 4 2 µ µ ¶ · ¸ ¸ ¹ º + = 2 2 9 4 2 1 y y+ , 0 < y < 1. b) Find f Y  X ( y  x ). f X ( x ) = ( ) ´ + x dy y x 4 = ( ) 2 2 x y y x + = 3 x 2 , 0 < x < 1. f Y  X ( y  x ) = 2 3 4 x y x + , 0 < y < x , 0 < x < 1. f Y  X ( y  x ) is undefined for x < 0 or x > 1. c) Find E ( Y  X ). E ( Y  X = x ) = ´ + ⋅ x dy x y x y 2 3 4 = 3 2 2 3 4 2 3 1 x y y x x µ µ ¶ · ¸ ¸ ¹ º + ⋅ = 18 11 x , 0 < x < 1. E ( Y  X = x ) is undefined for x < 0 or x > 1. E ( Y  X ) = 18 X 11 . d) 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 . e) 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 . From the textbook: 2.3.3 Let f ( x 1 , x 2 ) = 3 2 2 1 21 x x , 0 < x 1 < x 2 < 1, zero elsewhere, be the joint pdf of X 1 and X 2 . (a) Find the conditional mean and variance of X 1 , given X 2 = x 2 , 0 < x 2 < 1. (b) Find the distribution of Y = E ( X 1  X 2 ). (c) Determine E ( Y ) and Var ( Y ) and compare these to E ( X 1 ) and Var ( X 1 ), respectively. f ( x 1 , x 2 ) = 3 2 2 1 21 x x , 0 < x 1 < x 2 < 1. (a) f 2 ( x 2 ) = & 2 1 3 2 2 1 21 x dx x x = 6 2 7 x , 0 < x 2 < 1. f 1  2 ( x 1  x 2 ) = 3 2 2 1 3 x x , 0 < x 1 < x 2 , 0 < x 2 < 1. E ( X 1  X 2 = x 2 ) = & ⋅ 2 1 3 2 2 1 1 3 x dx x x x = 2 4 3 x ⋅ , 0 < x 2 < 1. E ( X 1 2  X 2 = x 2 ) = & ⋅ 2 1 3 2 2 1 2 1 3 x dx x x x = 2 2 5 3 x ⋅ , 0 < x 2 < 1. Var ( X 1  X 2 = x 2 ) = 2 2 2 2 16 9 5 3 x x ⋅ ⋅= 2 2 80 3 x ⋅ , 0 < x 2 < 1....
View
Full Document
 Fall '08
 AlexeiStepanov
 Probability, Probability theory, probability density function, X1

Click to edit the document details