solutions07

# solutions07 - Z ∞-∞ f X,Y x,y d y = Z ∞ x 2 e-x-y d y...

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Math 302.102 Fall 2010 Solutions to Assignment #7 1. (a) We must choose the value of c so that Z -∞ Z -∞ f X,Y ( x,y ) d x d y = 1. Since Z b a Z y a ( y - x ) d x d y = - Z b a ( y - x ) 2 2 ± ± ± ± x = y x = a d y = Z b a ( y - a ) 2 2 d y = ( y - a ) 3 6 ± ± ± ± y = b y = a = ( b - a ) 3 6 we conclude that c = 6 ( b - a ) 3 . 1. (b) By deﬁnition, f Y ( y ) = Z -∞ f X,Y ( x,y ) d x = Z y a c ( y - x ) d x = - c ( y - x ) 2 2 ± ± ± ± x = y x = a = c ( y - a ) 2 2 = 3( y - a ) 2 ( b - a ) 3 for a y b . Thus, E ( Y ) = Z -∞ yf Y ( y ) d y = 3 ( b - a ) 3 Z b a y ( y - a ) 2 d y = 3 ( b - a ) 3 Z b a ( y 3 - 2 ay 2 + a 2 y ) d y = 3 ( b - a ) 3 ² y 4 4 - 2 ay 3 3 + a 2 y 2 2 ³ b a = 3 ( b - a ) 3 ² b 4 4 - 2 ab 3 3 + a 2 b 2 2 - a 4 4 + 2 a 4 3 - a 4 2 ³ = 3 b 4 - 8 ab 3 + 6 a 2 b 2 - a 4 4( b - a ) 3 = (3 b + a )( b - a ) 3 4( b - a ) 3 = 3 b + a 4 . 1. (c) By deﬁnition, f X ( x ) = Z -∞ f X,Y ( x,y ) d y = Z b x c ( y - x ) d y = c ( y - x ) 2 2 ± ± ± ± y = b y = x = c ( b - x ) 2 2 = 3( b - x ) 2 ( b - a ) 3 for a y b . Thus, E ( X ) = Z -∞ xf X ( x ) d x = 3 ( b - a ) 3 Z b a x ( b - x ) 2 d x = 3 ( b - a ) 3 Z b a ( x 3 - 2 bx 2 + b 2 x ) d x. Notice that this is the same computation as E ( Y ) except that b appears in the integrand instead of a . Hence, if we write 3 ( b - a ) 3 Z b a ( x 3 - 2 bx 2 + b 2 x ) d x = 3 ( a - b ) 3 Z a b ( x 3 - 2 bx 2 + b 2 x ) d x, (one minus sign from switching the limits of integration and one minus sign from changing ( b - a ) 3 to ( a - b ) 3 ), then it is the same computation as above with a and b switched. Thus, E ( X ) = 3 a + b 4 .

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2. (a) By deﬁnition, f X ( x ) =
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Unformatted text preview: Z ∞-∞ f X,Y ( x,y ) d y = Z ∞ x 2 e-x-y d y = 2 e-x Z ∞ x e-y d y = 2 e-x e-x = 2 e-2 x for x > 0. By deﬁnition, f Y ( y ) = Z ∞-∞ f X,Y ( x,y ) d x = Z y 2 e-x-y d x = 2 e-y Z y e-x d x = 2 e-y (1-e-y ) for y > 0. 2. (b) Since f X,Y ( x,y ) 6 = f X ( x ) f Y ( y ), we conclude that X and Y are not independent. 2. (c) By deﬁnition, E ( X ) = Z ∞-∞ xf X ( x ) d x = Z ∞ x · 2 e-2 x d x = Z ∞ 2 xe-2 x d x = 1 2 . By deﬁnition, E ( Y ) = Z ∞-∞ yf Y ( y ) d y = Z ∞ y · 2 e-y (1-e-y ) d y = Z ∞ 2 ye-y d y-Z ∞ 2 ye-2 y d y = 2-1 2 = 3 2 ....
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solutions07 - Z ∞-∞ f X,Y x,y d y = Z ∞ x 2 e-x-y d y...

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