# So p x 0 y y e 7 y also note that y is exponentially

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So P ( X = 0 | Y = y ) = e - 7 y . Also note that Y is exponentially distributed with rate 5, so f Y ( y ) = 5 e - 5 y . Therefore, P ( X = 0) = Z 0 P ( X = 0 | Y = y ) f Y ( y ) dy = Z 0 5 e - 12 y dy = - 5 12 e - 12 y 0 = 5 12 . (We saw another way to solve this problem in class. See Example 2 in Section 5.2 and note that X = 0 is just the event that the first woman arrives before the first man.) 5. [8 points] Let U 1 , U 2 , U 3 , U 4 , U 5 be independently and uniformly chosen from (0 , 1). Let Y be the largest and X be the third largest. (In the book’s notation, X = U (3) and Y = U (5) .) Find the joint density of ( X, Y ). Solution: For 0 < x < y < 1, P ( X dx and Y dy ) = 5 · 4 · P ( U 1 is the third largest and U 1 dx, U 2 is the largest and U 2 dy ) = 20 · 3 · P ( U 3 , U 4 (0 , x ) , U 1 dx, U 5 ( x, y ) , U 2 dy ) = 60[ x 2 ][1 · dx ][ y - x ][1 · dy ] = 60 x 2 ( y - x ) dx dy. Therefore, the joint density of X and Y is f ( x, y ) = 60 x 2 ( y - x ), for 0 < x < y < 1. Page 4

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Math 431: Final Exam 6. Let X be a random variable with exponential λ distribution. Given X = x , let Y be uniformly distributed on (0 , x ). (a) [3 points] What is E ( Y | X )? Solution: The mean of a uniform distribution on an interval is the midpoint, so E ( Y | X = x ) = x/ 2. More compactly, E ( Y | X ) = X/ 2. (b) [3 points] What is E ( Y )? Solution: The expectation of X is 1 , so E ( Y ) = E [ E ( Y | X )] = E X 2 = 1 2 λ . (c) [6 points] What the joint density of X and Y ? Solution: f X ( x ) = λe - λx , for x > 0, f Y ( y | X = x ) = 1 x , for y (0 , x ), so f X,Y ( x, y ) = f X ( x ) f Y ( y | X = x ) = λe - λx x , when 0 < y < x .
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