elemprob-fall2010-page38

elemprob-fall2010-page38 - X 1 and X 2 are independent. Let...

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Just as in the one-dimensional case, there is a change of variables formula. Let us recall how the formula goes in one dimension. If X has a density f X and y = g ( X ), then F Y ( y ) = P ( Y y ) = P ( g ( X ) y ) = P ( X g - 1 ( y )) = F X ( g - 1 ( y )) . Taking the derivative, using the chain rule, and recalling that the derivative of g - 1 ( y ) is 1 /g 0 ( y ), we have f Y ( y ) = f X ( g - 1 ( y )) 1 g ( y ) . The higher dimensional case is very analogous. Suppose Y 1 = g 1 ( X 1 ,X 2 ) and Y 2 = g 2 ( X 1 ,X 2 ). Let h 1 and h 2 be such that X 1 = h 1 ( Y 1 ,Y 2 ) and X 2 = h 2 ( Y 1 ,Y 2 ). (This plays the role of g - 1 .) Let J be the Jacobian of the mapping ( x 1 ,x 2 ) ( g 1 ( x 1 ,x 2 ) ,g 2 ( x 1 ,x 2 ), so that J = ∂g 1 ∂x 1 ∂g 2 ∂x 2 - ∂g 1 ∂x 2 ∂g 2 ∂x 1 . (This is the analogue of g 0 ( y ).) Using the change of variables theorem from multivariable calculus, we have f Y 1 ,Y 2 ( y 1 ,y 2 ) = f X 1 ,X 2 ( x 1 ,x 2 ) | J | - 1 . An example. Suppose X 1 is N (0 , 1), X 2 is N (0 , 4), and
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Unformatted text preview: X 1 and X 2 are independent. Let Y 1 = 2 X 1 + X 2 ,Y 2 = X 1-3 X 2 . Then y 1 = g 1 ( x 1 ,x 2 ) = 2 x 1 + x 2 ,y 2 = g 2 ( x 1 ,x 2 ) = x 1-x 3 , so J = 2 1 1-3 =-7 . (In general, J might depend on x , and hence on y .) Some algebra leads to x 1 = 3 7 y 1 + 1 7 y 2 , x 2 = 1 7 y 1-2 7 y 2 . Since X 1 and X 2 are independent, f X 1 ,X 2 ( x 1 ,x 2 ) = f X 1 ( x 1 ) f X 2 ( x 2 ) = 1 2 e-x 2 1 / 2 1 8 e-x 2 2 / 8 . Therefore f Y 1 ,Y 2 ( y 1 ,y 2 ) = 1 2 e-( 3 7 y 1 + 1 7 y 2 ) 2 / 2 1 8 e-( 1 7 y 1-2 7 y 2 ) 2 / 8 1 7 . 38...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.

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