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HW5_solution_to_POST

# HW5_solution_to_POST - #10 abc f(x = f(y = 1(6-5 = 1 a f(x...

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#10 abc f ( x ) = f ( y ) = 1/(6-5) = 1 a) f ( x , y ) = 0 1 otherwise y x 6 5 , 6 5 since f ( x ) = f ( y ) = 1/(6-5) = 1, for 5 x 6, 5 y 6 b) P(5.25 X 5.75, 5.25 Y 5.75) = P(5.25 X 5.75) P(5.25 Y 5.75) = (.5)(.5) = .25 (by independence) c) P((X,Y) A) = ∫ ∫ A dxdy 1 = area of A = 1 – (area of I + area of II ) = 306 . 36 11 36 25 1 = = - Q1 :   [First read xeroxed material on joint conditional distributions.] The joint probability density for rainfall at two places on rainy days could be described by ( 29 3 2/ 1 , 0 ( , ) 0 otherwise XY x y x y f x y x + + = < If you haven’t finished Math 293 and are having trouble with this problem, see TA or instructor. For integrations, use ∫ [x+b] –n dx = – [x+b] –(n-1) /(n-1) + constant. > Calculate and provide a rough graph of: a) The joint cumulative distribution of X and Y, F XY (x,y). ( 4 points) ( 29 ( 29 , , , x y XY XY x y F f x y dydx - = ( 29 3 0 0 2 1 y y x x x y dy dx x y = = = = = + + % % 2 = 1 2 - ( 29 ( 29 2 2 0 0 0 0 1 1 1 y y x x dx dx x y x y = - + + + + % % % % ( 29 ( 29 2 2 0 1 1 1 1 x dx x y x - = + + + + % % = - 1 1 - ( 29 ( 29 0 0 1 1 1 1 1 x x x x y + - + + + % % ( 29 ( 29 ( 29 ( 29 , 1 1 1 1 1 1 1 XY x y F x y y x = - - - + + + + I II 6 / 1 + = x y 6 / 1 - = x y 5 5 6 6

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( 29 ( 29 ( 29 ( 29 , 1 1 1 1 1 1 1 XY x y F x y y x = - - + + + + + b) The marginal cumulative dist. of X, F
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HW5_solution_to_POST - #10 abc f(x = f(y = 1(6-5 = 1 a f(x...

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