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Math Help - Transforming pdf, distribution question

  1. #1
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    Transforming pdf, distribution question

    A book I am reading has a section on probability density function.

    Random var Xi is drawn from a pdf p(x)

    Random var Yi is drawn from y(Xi), a one to one function with strictly positive or negative derivative.

    The book I read claims
    Pr( Y <= y(x) ) = Pr( X <= x )

    Which I am not understanding how it is derived. Notice It does not make much distinction of X, Xi, and Y, Yi.

    Pr is the CDF. The goal of this section is to find the pdf of a transformed random var. The final formula is

    p(y) = |dy/dx|^-1 * q(x)

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    What you do is differentiate the two CDFs. That produces the densities.

    For example if X\sim U(0,1) and Y=aX+b

    then F_Y(y)= P(Y\le y)=P(aX+b\le y)=P(X\le (y-b)/a)=F_X((y-b)/a)

    so f_y(y)= {d\over dy} F_Y(y)= {d\over dy}F_X((y-b)/a)= {1\over a} f_X((y-b)/a)={1\over a}

    where 0<{y-b\over a}<1 in this case.
    Last edited by matheagle; May 1st 2009 at 04:38 PM.
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  3. #3
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    Thank you for providing the example.

    Perhaps it is simple, but what I didn't get is how the claim
    Pr( Y<=y(x) ) = Pr( X <=x )
    is reasoned from the facts that y(x) is one to one and y(x)'s derivatives are all negative or all positive.

    Also, though the absolute value around dy/dx enforces the transformed pdf being non-negative (because dy may be negative), why may the math not work out without it?


    Quote Originally Posted by matheagle View Post
    What you do is differentiate the two CDFs. That produces the densities.

    For example if X\sim U(0,1) and Y=aX+b

    then F_Y(y)= P(Y\le y)=P(aX+b\le y)=P(X\le (y-b)/a)=F_X((y-b)/a)

    so f_y(y)= {d\over dy} F_Y(y)= {d\over dy}F_X((y-b)/a)= {1\over a} f_X((y-b)/a)={1\over a}

    where 0<{y-b\over a}<1 in this case.
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