solutions1_v1 - Statistics 351 Fall 2008 Midterm #1...

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Unformatted text preview: Statistics 351 Fall 2008 Midterm #1 Solutions 1. (a) By definition, f X ( x ) = Z x e- y d y = (- e- y ) x = e- x , x > . Note that X Exp(1) so that E ( X ) = 1. 1. (b) By definition, f Y ( y ) = Z y e- y d x = ye- y , y > . Note that Y (2 , 1). 1. (c) By definition, f Y | X = x ( y ) = f X,Y ( x, y ) f X ( x ) = e- y e- x = e x- y , < x < y < . 1. (d) By definition, E ( Y | X = x ) = Z - y f Y | X = x ( y ) d y = Z x y e x- y d y. Let u = y- x so that d u = d y and the integral above becomes Z x y e x- y d y = Z ( u + x ) e- u d u = Z ue- u d u + x Z e- u d u = (2)+ x (1) = 1+ x and so E ( Y | X ) = 1 + X . 1. (e) Using (d) we find E ( Y ) = E ( E ( Y | X )) = E (1 + X ) = 1 + E ( X ). However, from (a) we know that E ( X ) = 1 and so E ( Y ) = 1 + E ( X ) = 1 + 1 = 2 . 1. (f) If a > 1, then we find P { aX < Y } = ZZ { ax<y } f X,Y ( x, y ) d x d y = Z Z ax e- y d y d x = Z (- e- y ) ax d x = Z e- ax d...
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solutions1_v1 - Statistics 351 Fall 2008 Midterm #1...

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