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  • Transformation problem

    f(x,y)=6(2xy-y^2) 0<x<1, 0<y<1
    Z = 3X-6Y

    (I invented this, not from any source)

    What's the density function for Z?

    I'm attempting to do it using x=v and y=(3v-z)/6, and the Jacobian is |-1/6|=1/6; after substituting for x and y, and integrating with respect to v from 0 to 1 I get f(z)=(-z^2)/36+z/3+1/4, and F(z)=-(z^3)/108+(z^2)/6+z/4, and I assume the interval is -6<z<3 (as that's the interval of 3x-6y, assuming (0,1) and (1,0) as endpoints), but the total density doesn't equal 1. What am I doing wrong?

  • #2
    Don't worry about it - got an answer from another source.

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