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# HW 3 - Homework Set#3 Solutions IE 336 Spring 2011 1 From...

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Homework Set #3 Solutions IE 336 Spring 2011 1. From HW1, we know that f X ( x ) = 2 e x - 2 e - 2 x , 0 x < and f Y ( y ) = 2 e - 2 y , 0 y < . Determine E ( Y ): Y exp (2). Therefore, E ( Y ) = 1 λ = 1 2 . Determine E ( X ): E ( X ) = Z x xf X ( x ) dx = Z 0 x ( 2 e - x - 2 e - 2 x ) dx = Z 0 2 xe - x dx - Z 0 2 xe - 2 x dx = 2 - xe - x 0 + Z 0 e - x dx - 2 - 1 2 xe - 2 x 0 + Z 0 1 2 e - 2 x dx = 2 0 + e 0 - 2 0 + 1 4 e 0 E ( X ) = 3 2 Since the interval [ y, ) was used to find f X ( x ), it may be tempting to use the same interval to find E ( X ). However, this interval was used to integrate away the y values from the joint pdf, f ( x, y ). As a result, f X ( x ) is defined for x [0 , ). Determine E ( Y | x ): From HW1, f ( y | x ) = e - y 1 - e - x , 0 y x < . E ( Y | x ) = Z y yf ( y | x ) dy = Z x 0 y e - y 1 - e - x dy = 1 1 - e - x Z x 0 ye - y dy = 1 1 - e - x - ye - y x 0 + Z x 0 e - y dy = - xe - x - e - x + 1 1 - e - x E ( Y | x ) = 1 - x e x - 1 Again, be careful with the bounds for integration. Since the conditional expectation is asked for, both x and y are part of the formulation. Therefore, the interval 0 y x < must be considered.

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