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Unformatted text preview: X  Y ( x  y ). f Y ( y ) = &y dx x 1 6 = 3 ( 1 – y ) 2 , 0 < y < 1. f X  Y ( x  y ) = ( ) 2 1 2 y x, 0 < x < 1 – y , 0 < y < 1. b) (3) Find ± ± ² ³ ´ ´ µ ¶ = > 4 1 Y 2 1 X P . f X  Y ( x  4 1 ) = x 9 32 , 0 < x < 4 3 . ± ± ² ³ ´ ´ µ ¶ = > 4 1 Y 2 1 X P = & 4 3 2 1 9 32 dx x = 9 4 1= 9 5 . 1. (continued) c) (3) Find & & ± ² ³ ³ ´ µ = 4 1 Y X E . f X  Y ( x  4 1 ) = x 9 32 , 0 < x < 4 3 . & & ± ² ³ ³ ´ µ = 4 1 Y X E = ¶ ⋅ 4 3 9 32 dx x x = 64 27 27 32 ⋅ = 2 1 = 0.50 ....
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 Fall '08
 AlexeiStepanov
 Probability, Probability distribution, Probability theory, probability density function, joint probability density

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