hw6soln - ORIE 3500/5500, Fall ’10 HW 6 Solutions...

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Unformatted text preview: ORIE 3500/5500, Fall ’10 HW 6 Solutions Assignment 6 Solutions Problem 1 Var( X ) = EX 2- ( EX ) 2 = 1 3- 1 4 = 1 12 . E ( X θ- X ) 2 = E ( X 2 θ- 2 X θ +1 + X 2 ) = 1 2 θ + 1- 2 θ + 2 + 1 3 . E ( X θ- X ) = 1 θ + 1- 1 2 . So Var( Y ) = 1 2 θ + 1- 2 θ + 2 + 1 3- 1 θ + 1- 1 2 2 = 1 2 θ + 1- 2 θ + 2- 1 ( θ + 1) 2 + 1 θ + 1 + 1 12 . Cov( X,Y ) = EXY- EXEY = E ( X θ +1- X 2 )- EXE ( X θ- X ) = 1 θ + 2- 1 3- 1 2( θ + 1) + 1 4 = 1 θ + 2- 1 2( θ + 1)- 1 12 . Since ρ X,Y = Cov( X,Y ) p Var( X ) p Var( Y ) , we get ρ X,Y = - . 7746 θ = 0 . 5 . 33443 θ = 1 . 2 . 18898 θ = 1 . 5- . 24672 θ = 3 Hence θ = 1 . 2 makes the correlation the largest. Problem 2 (a) The sign of the correlation is negative since to satisfy the condition X + Y < 2, if X increases, Y tends to take on smaller values. (b) The joint distribution P ( X = i,Y = j ) is expressed in the table below: i/j 1 2 P ( X = i ) 1 6 1 6 1 6 1 2 1 1 6 1 6 1 3 2 1 6 1 6 P ( Y = j )...
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This note was uploaded on 11/11/2010 for the course ORIE 3500 at Cornell.

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hw6soln - ORIE 3500/5500, Fall ’10 HW 6 Solutions...

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