e2_07 - ECE 514 , Fall 2007 Exam 2: Due 12:30pm at ECE...

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Unformatted text preview: ECE 514 , Fall 2007 Exam 2: Due 12:30pm at ECE front office, October 25, 2007 Solutions (version: October 25, 2007, 21:11) 75 mins.; Total 50 pts. 1. (14 pts.) Consider three real random variables X , Y , and Z . a. Suppose that Z = X 2 + Y 2 and that X and Y have a joint density that is uniform on the region { [ x, y ] : x, y [- 1 , 1] } (i.e., the density function is constant in this region and is elsewhere). Find the CDF of Z , F Z ( z ) , for z 1 . b. Suppose X = Z 2 , Y = Z , and Z is uniformly distributed on [0 , 1] . Find the joint CDF of the pair ( X, Y ) , F X,Y ( x, y ) , x, y R . Are X and Y independent? Ans.: a. Because Z a.s., we deduce that if z < , then F Z ( z ) = 0 . So fix z . The value of the uniform density on the given region is 1 / 4 (because thats the area of the region). Now, for z 1 , F Z ( z ) = P { X 2 + Y 2 z } , which is simply the 1 / 4 times the area of the circle { [ x, y ] : x 2 + y 2 z } , which is z . So we have F Z ( z ) = z/ 4 , z 1 . b. We have F X,Y ( x, y ) = P { X x, Y y } = P { Z 2 x, Z y } = P { Z x, Z y 2 } = P { Z min( x, y 2 ) } . If x or y , then F X,Y ( x, y ) = 0 . If x 1 and y 1 , then F X,Y ( x, y ) = 1 . Otherwise, F X,Y ( x, y ) = min( x, y 2 ) . This function cannot be written as the product of two functions, one....
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e2_07 - ECE 514 , Fall 2007 Exam 2: Due 12:30pm at ECE...

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