solution10

# solution10 - EL630 Homework10 Solutions 1 This is...

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EL630: Homework10 Solutions 1. This is equation (12) in Lecture 10. 2. This is exactly example 10.3 in lecture 10 except that a constant c is added to Z . This just changes the mean of the random variable. 3. The domain of the joint density function is the shaded region shown below. To find k we use ∫ ∫ = 1 0 0 1 dy dx x k y . We get 6 / 2 / 1 1 0 2 1 0 0 k dy y k dy dx x k y = = = ∫ ∫ . Therefore, 6 = k . 2 / 1 12 / 6 3 / 6 6 ) 6 ( 1 0 3 1 0 0 2 1 0 0 = = = = = ∫ ∫ ∫ ∫ dy y dy dx x dy dx x x y y x μ . 4 / 3 8 / 6 2 / 6 6 ) 6 ( 1 0 3 1 0 0 1 0 0 = = = = = ∫ ∫ ∫ ∫ dy y dy dx x y dy dx x y y y y μ [ ] ∫ ∫ ∫ ∫ + - = - = - = 1 0 0 2 3 1 0 0 2 2 ) 4 / ( 6 ) 6 ( ) 2 / 1 ( ) 2 / 1 ( ) ( dy dx x x x dy dx x x X E X Var y y 1 Y X 1

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05 . 0 ) 24 / 1 12 / 1 20 / 1 ( 6 ) 8 / 3 / 4 / ( 6 ) ( 1 0 2 3 4 = + - = + - = dy y y y X Var . [ ] ∫ ∫ ∫ ∫ + - = - = - = 1 0 0 2 1 0 0 2 2 ) 16 / 9 2 / 3 ( 6 ) 6 ( ) 4 / 3 ( ) 4 / 3 ( ) ( dy dx x y y dy dx x y Y E Y Var y y 0375 . 0 ) 32 / 3 16 / 3 10 / 1 ( 6 ) 32 / 9 4 / 3 2 / ( 6 ) ( 1 0 2 3 4 = + - = + - = dy y y y Y Var . [ ] ∫ ∫ ∫ ∫ - - = - - = - - = 1 0 0 2 1 0 0 ) 4 / 3 )( 2 / ( 6 ) 6 ( ) 4 / 3 )( 2 / 1 ( ) 4 / 3 )( 2 / 1 ( ) , ( dy dx y x x dy dx x y x Y X E Y X Cov y y + - = - - = 1 0 2 3 4 1 0 2 3 ) 16 / 3 2 / 3 / ( 6 ) 4 / 3 / )( 4 / 3 ( 6 ) , ( dy y y y dy y y y Y X
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