# HW23Answers(8.2) - Math331, Spring 2008 Instructor: David...

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 8.2 Homework Answers Homework: pgs. 339 - 340, #s 1, 2, 3, 5, 7, 8. 1. We have that p ( x,y ) = 1 25 ( x 2 + y 2 ) , x = 1 , 2 , y = 0 , 1 , 2 , and zero otherwise. Therefore, the marginal probability distributions are given by p X ( x ) = 2 summationdisplay y =0 1 25 ( x 2 + y 2 ) = 1 25 (3 x 2 + 5) , x = 1 , 2 p Y ( y ) = 2 summationdisplay x =1 1 25 ( x 2 + y 2 ) = 1 25 (5 + 2 y 2 ) , y = 0 , 1 , 2 . For X and Y to be independent we need p ( x,y ) = p X ( x ) p Y ( y ) for all x,y R . However, simply picking x = 1 ,y = 0 shows p X (1) p Y (0) = 1 25 (3 1 2 + 5) 1 25 (5 + 2 2 ) = 8 25 5 25 = 40 25 . However, p (1 , 0) = (1 / 25)(1 + 0) = 1 / 25 negationslash = p X (1) p Y (0) and so they are not independent. 2. We have that p ( x,y ) = 1 7 x 2 y, ( x,y ) = (1 , 1) , (1 , 2) , (2 , 1) , and zero otherwise. The marginals are given by p X (1) = p (1 , 1) + p (1 , 2) = 1 7 + 2 7 = 3 7 p X (2) = p (2 , 1) = 1 7 4 = 4 7 p Y...
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## This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at Wisconsin.

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HW23Answers(8.2) - Math331, Spring 2008 Instructor: David...

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