Hw03ans - STAT 410 Fall 2008 Homework#3(due Friday...

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Unformatted text preview: STAT 410 Fall 2008 Homework #3 (due Friday, September 19, by 3:00 p.m.) 1. 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 ). b) What must the value of C be so that f ( x , y ) is a valid joint p.d.f.? Must have ( ) & & ∞ ∞ ∞ ∞--dy dx y x f , = 1. & & ± ± ² ³ ´ ´ µ ¶-1 2 2 dx dy y x x x C = & =-= ± ² ³ ´ µ ¶ 1 2 2 2 2 dx y x x y x y C = ( ) [ ] & ± ² ³ ´ µ ¶--1 2 2 2 2 2 dx x x x C = ( ) &-1 3 2 2 2 dx x x C C = 1 4 3 2 3 2 ± ² ³ ´ µ ¶-x x C C = 6 C = 1. · C = 6 . c) Find P ( X + Y < 1 ). & & ± ± ² ³ ´ ´ µ ¶-5 . 1 2 6 dx dy y x x x = ( ) & =-= 5 . 1 2 2 3 dx y x x y x y = ( ) [ ] ( ) &--5 . 2 2 2 1 3 dx x x x = ( ) &-5 . 3 2 6 3 dx x x = 5 . 4 3 2 3 ± ² ³ ´ µ ¶-x x = 4 3 2 1 2 3 2 1 ± ² ³ ´ µ ¶-± ² ³ ´ µ ¶ = 32 3 8 1-= 32 1 = 0.03125 . 2. Suppose that the random variables X and Y have joint p.d.f. f ( x , y ) given by f ( x , y ) = 6 x 2 y , 0 < x < y , x + y < 2. a) Find the marginal probability density function for X. First, X can only take values in ( , 1 ). f X ( x ) = ( ) & ∞ ∞-, dy y x f = &-x x dy y x 2 2 6 = ( ) 3 2 2 2 x y x y y x =-= = ( ) { } 2 2 2 2 3 x x x--= 12 x 2 – 12 x 3 = 12 x 2 ( 1 – x ), 0 < x < 1. b) Find the marginal probability density function for Y. First, Y can only take values in ( , 2 ). f Y ( y ) = ( ) & ∞ ∞-, dx y x f = ± ± ± ² ± ± ± ³ ´ < < < < & &-2 1 6 1 6 2 2 2 y dx y x y dx y x y y = ( ) ( ) ± ± ± ² ± ± ± ³ ´ < < < < =-= = = 2 1 2 1 2 2 3 3 y y x y y x x y x x y x = ( ) ± ± ² ± ± ³ ´ < <-< < 2 1 2 2 1 2 3 4 y y y y y 3. Suppose that ( X, Y ) is uniformly distributed over the region defined by 0 ≤ y ≤ 1 – x 2 and – 1 ≤ x ≤ 1. a) What is the joint probability density function of X and Y ? & &--± ± ± ² ³ ´ ´ ´ µ ¶ 1 1 1 2 dx dy x = ( ) &--1 1 2 1 dx x = 1 1 3 3-± ± ² ³ ´ ´ µ ¶-x x = 3 4 . · f X , Y ( x , y ) = 4 3 , – 1 ≤ x ≤ 1, 0 ≤ y ≤ 1 – x 2 . b) Find the marginal densities of X and Y. f X ( x ) = &-2 1 4 3 x dy = ( ) 2 1 4 3 x-, – 1 ≤ x ≤ 1. y = 1 – x 2 x = y-± 1 f Y ( y ) = &---y y dy 1 1 4 3 = y-1 2 3 , 0 ≤ y ≤ 1. c) Find the two conditional densities....
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This note was uploaded on 10/22/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

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Hw03ans - STAT 410 Fall 2008 Homework#3(due Friday...

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