elemprob-page22

# elemprob-page22 - analogue of g y Using the change of...

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Note that it is not always the case that the sum of two independent random variables will be a random variable of the same type. If X and Y are independent normals, then - Y is also a normal (with E ( - Y ) = - E Y and Var ( - Y ) = ( - 1) 2 Var Y = Var Y ), and so X - Y is also normal. To deﬁne a conditional density in the discrete case, we write p X | Y = y ( x | y ) = P ( X = x | Y = y ) . This is equal to P ( X = x,Y = y ) P ( Y = y ) = p ( x,y ) p Y ( y ) . Analogously, we deﬁne in the continuous case f X | Y = y ( x | y ) = f ( x,y ) f Y ( y ) . 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
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Unformatted text preview: analogue of g ( 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 . Example. Suppose X 1 is N (0 , 1), X 2 is N (0 , 4), and 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 ² =-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 π e-x 2 1 / 2 1 √ 8 π e-x 2 2 / 8 . Therefore f Y 1 ,Y 2 ( y 1 ,y 2 ) = 1 √ 2 π e-( 3 7 y 1 + 1 7 y 2 ) 2 / 2 1 √ 8 π e-( 1 7 y 1-2 7 y 2 ) 2 / 8 1 7 . 22...
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## This note was uploaded on 12/29/2011 for the course MATH 317 taught by Professor Wen during the Spring '09 term at SUNY Stony Brook.

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