solutions1

# solutions1 - Statistics 351 Fall 2007 Midterm#1 –...

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Unformatted text preview: Statistics 351 Fall 2007 Midterm #1 – Solutions 1. (a) By definition, f X ( x ) = Z ∞ x 2 e- x e- y dy = 2 e- x (- e- y ) ∞ x = 2 e- 2 x , x > , and f Y ( y ) = Z y 2 e- x e- y dx = 2 e- y (- e- x ) y = 2 e- y ( 1- e- y ) , y > . 1. (b) Since f X,Y ( x,y ) 6 = f X ( x ) · f Y ( y ), we immediately conclude that X and Y are not independent random variables. 1. (c) By definition, f Y | X = x ( y ) = f X,Y ( x,y ) f X ( x ) = 2 e- x- y 2 e- 2 x = e x- y , < x < y < ∞ . 1. (d) By definition, E ( Y | X = x ) = Z ∞-∞ y · f Y | X = x ( y ) dy = Z ∞ x y · e x- y dy. Let u = y- x so that du = dy and the integral above becomes Z ∞ x y · e x- y dy = Z ∞ ( u + x ) e- u du = Z ∞ ue- u du + x Z ∞ e- u du = Γ(2)+ x Γ(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, E ( X ) = Z ∞-∞ x · f X ( x ) dx = Z ∞ 2 xe- 2 x dx....
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solutions1 - Statistics 351 Fall 2007 Midterm#1 –...

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