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# solution2 - Homework 2 solution 1 Problem 1 We need to show...

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Homework 2 solution 1 Problem 1 We need to show that P ( X x ) = Φ( x ) , where the function Φ( x ) = R x -∞ 1 2 π exp ( - x 2 2 ) is the CDF of N (0 , 1). P ( X x ) = P ( X x, U 0 . 5) + P ( X x, U > 0 . 5) = P ( X x | U 0 . 5) P ( U 0 . 5) + P ( X x | U > 0 . 5) P ( U > 0 . 5) = P ( X x | U 0 . 5) 1 2 + P ( X x | U > 0 . 5) 1 2 = 1 2 P ( Y 1 x | Y 2 > (1 - Y 1 ) 2 2 ) + 1 2 P ( - Y 1 x | Y 2 > (1 - Y 1 ) 2 2 ) = 1 2 P ( Y 1 x | Y 2 > (1 - Y 1 ) 2 2 ) + 1 2 P ( Y 1 ≥ - x | Y 2 > (1 - Y 1 ) 2 2 ) 1

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If x 0, then P ( Y 1 x | Y 2 > (1 - Y 1 ) 2 2 ) = 0 because Y 1 follows exp (1). Since the joint pdf of Y 1 and Y 2 is e - y 1 - y 2 , we have P ( X x ) = 1 2 P ( Y 1 ≥ - x | Y 2 > (1 - Y 1 ) 2 2 ) = 1 2 P ( Y 1 ≥ - x, Y 2 > (1 - Y 1 ) 2 2 ) P ( Y 2 > (1 - Y 1 ) 2 2 ) = 1 2 R - x R (1 - y 1 ) 2 2 e - y 1 - y 2 dy 2 dy 1 R 0 R (1 - y 1 ) 2 2 e - y 1 - y 2 dy 2 dy 1 = 1 2 R - x e - y 1 e - (1 - y 1 ) 2 2 dy 1 R 0 e - y 1 e - (1 - y 1 ) 2 2 dy 1 = 1 2 R - x e - 1+ y 2 1 2 dy 1 R 0 e - 1+ y 2 1 2 dy 1 = 1 2 R - x e - y 2 1 2 dy 1 R 0 e - y 2 1 2 dy 1 = 1 2 R - x e - y 2 1 2 dy 1 2 π 2 = 1 2 π Z - x e - y 2 1 2 dy 1 = 1 2 π Z x -∞ e - t 2 2 dt = Φ( x ) .
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solution2 - Homework 2 solution 1 Problem 1 We need to show...

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