solutions1_v2 - 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 Y ( y ) = Z ∞ y e- x d x = (- e- x ) ∞ y = e- y , y > . Note that Y ∈ Exp(1) so that E ( Y ) = 1. 1. (b) By definition, f X ( x ) = Z x e- x d y = xe- x , x > . Note that X ∈ Γ(2 , 1). 1. (c) By definition, f Y | X = x ( y ) = f X,Y ( x, y ) f X ( x ) = e- x xe- x = 1 x , < y < x. Note that Y | X = x ∈ U (0 , x ). 1. (d) By definition, E ( Y | X = x ) = Z ∞-∞ y · f Y | X = x ( y ) d y = Z x y · 1 x d y = x 2 . Thus, we conclude E ( Y | X ) = X 2 . 1. (e) Using (d) we find E ( Y ) = E ( E ( Y | X )) = E X 2 = 1 2 E ( X ) . However, from (a) we know that E ( Y ) = 1 and so E ( X ) = 2 . 1. (f) If a < 1, then we find P { Y < aX } = ZZ { y<ax } f X,Y ( x, y ) d x d y = Z ∞ Z ax e- x d y d x = Z ∞ axe- x d x = a Γ(2) = a. 1. (g) If U = X + Y and V = 2 Y , then solving for X and Y gives X = U- V 2 and Y = V 2 ....
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.

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solutions1_v2 - Statistics 351 Fall 2008 Midterm#1 –...

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