# soln_HW6 - OR 2700, Spring09 Homework 6 Solutions to HW 6...

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OR 2700, Spring’09 Homework 6 Solutions to HW 6 Problem 1 (a) Since X and Y are independent, P ( X 4 ,Y < 3) = P ( X 4) P ( Y < 3) = e - 4 (1 - e - 3 ) = 0 . 0174 (b) Once again, the indepence of X and Y yields E ( X 2 Y ) = E ( X 2 ) E ( Y ) = Z 0 x 2 e - x dx Z 0 ye - y dy = Γ(3)Γ(2) = 2 Problem 2 (a) Since ( X,Y ) is uniformly distributed on the square, P ( X 2 + Y 2 < 1) = 1 4 area of { ( x,y ) : x 2 + y 2 < 1 , | x | ≤ 1 , | y | ≤ 1 } = π 4 (b) Using the same idea, P (2 X - Y > 0) = 1 4 area of { ( x,y ) : 2 x - y > 0 , | x | ≤ 1 , | y | ≤ 1 } = 1 2 1

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OR 2700, Spring’09 Homework 6 (c) Note that | X | ≤ 1 and | Y | ≤ 1 are sure events and hence | X + Y | ≤ | X | + | Y | ≤ 1 + 1 = 2. Thus, P ( | X + Y | ≤ 2) = 1 Problem 3 Note that P ( U = 1 ,V = 2) = 1 4 whereas P ( U = 1) = 1 2 and P ( V = 2) = 1 4 . Thus clearly P ( U = 1 ,V = 2) 6 = P ( U = 1) P ( V = 2) and hence U and V are dependent. Problem 4 Since the area of the set { ( x,y ) : 0 < x < 1 ,x < y < x + 1 } is 1, the density of ( X,Y ) is f ( x,y ) = 1 { 0 <x< 1 ,x<y<x +1 } (a) Clearly X takes values in [0 , 1] (since f ( x,y ) = 0 for x / [0 , 1]). Let 0 x 1. Z
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## This note was uploaded on 04/05/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell University (Engineering School).

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soln_HW6 - OR 2700, Spring09 Homework 6 Solutions to HW 6...

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