09_16ans - STAT 410 Examples for 09/16/2011 Fall 2011 Let X...

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STAT 410 Examples for 09/16/2011 Fall 2011 Let X 1 and X 2 have joint p.d.f. f ( x 1 , x 2 ) . S = { ( x 1 , x 2 ) : f ( x 1 , x 2 ) > 0 } – support of ( X 1 , X 2 ) . F ( x 1 , x 2 ) = P ( X 1 x 1 , X 2 x 2 ) . f ( x 1 , x 2 ) = 2 F ( x 1 , x 2 ) / x 1 x 2 . Let Y 1 = u 1 ( X 1 , X 2 ) and Y 2 = u 2 ( X 1 , X 2 ) . y 1 = u 1 ( x 1 , x 2 ) y 2 = u 2 ( x 1 , x 2 ) one-to-one transformation maps S onto T – support of ( Y 1 , Y 2 ) . x 1 = w 1 ( y 1 , y 2 ) x 2 = w 2 ( y 1 , y 2 ) J = 2 2 1 2 2 1 1 1 y x y x y x y x The joint p.d.f. g ( y 1 , y 2 ) of ( Y 1 , Y 2 ) is g ( y 1 , y 2 ) = f ( w 1 ( y 1 , y 2 ) , w 2 ( y 1 , y 2 ) ) | J | ( y 1 , y 2 ) T 0 elsewhere. 1. Let X 1 and X 2 have joint p.d.f. f ( x 1 , x 2 ) = 2 e ( x 1 + x 2 ) , 0 < x 1 < x 2 . a) Find the joint p.d.f. g ( y 1 , y 2 ) of the variables Y 1 = X 2 – X 1 and Y 2 = X 1 . Y 2 = X 1 X 1 = Y 2 Y 1 = X 2 – X 1 X 2 = Y 1 + X 1 = Y 1 + Y 2 J = 1 1 1 0 = – 1 x 1 > 0 y 2 > 0 x 2 > x 1 y 1 > 0
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f Y 1 , Y 2 ( y 1 , y 2 ) = 2 e ( y 2 + y 1 + y 2 ) × | 1 | = 2 e ( y 1 + 2 y 2 ) , y 1 > 0, y 2 > 0. Note that f Y 1 ( y 1 ) = e y 1 , y 1 > 0, f Y 2 ( y 2 ) = 2 e 2 y 2 , y 2 > 0, Y 1 and Y 2 are independent. b)
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This note was uploaded on 10/31/2011 for the course MATH 464 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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09_16ans - STAT 410 Examples for 09/16/2011 Fall 2011 Let X...

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