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Math Help - [SOLVED] Linear transformation of the Gamma distribution

  1. #1
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    [SOLVED] Linear transformation of the Gamma distribution

    Thanks to Mr Fantastic, I got the linear transformation of a simple exponential distribution. Now I'm trying to obtain the pdf of  Y= a X+b  . It looks cumbersome, is there any simpler methods? Where  X follows this Gamma distribution:

    f(x;k) = x^{k-1} \frac{e^{-x}}{\Gamma(k)}\text{ for } x > 0

    Thank you.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you can use MGFs but you won't recognize the density.
    Note that a cannot be zero but it can be negative.

    Use f_Y(y)=f_X(x)\biggl|{dx\over dy}\biggr|

    So f_Y(y) = \left({y-b\over a}\right)^{k-1} \frac{e^{-(y-b)/a}}{\Gamma(k)|a|}\text{ for } y > b
    Last edited by matheagle; May 4th 2010 at 04:47 PM.
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  3. #3
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    thank you matheagle!
    how is it called, the method you used here, I'm gonna search some ressources?
    good day!
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  4. #4
    MHF Contributor matheagle's Avatar
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    It's just calculus one
    I'm sure it's in your probability book, look for change of variables
    You use the CDFs and then differentiate to obtain the relationship between the densities
    see Dependent variables and change of variables
    in http://en.wikipedia.org/wiki/Probabi...nsity_function
    It's trivial to prove........

    F_Y(y) =P(Y\le y)  =P(aX+b\le y) =P(X\le (y-b)/a)) =F_X((y-b)/a) when a>0

    Now differentiate wrt y, the variable. Y is the RV.

    f_Y(y) ={f_X((y-b)/a)\over a} when a>0
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  5. #5
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    thanks again the Beagle!
    yes, i figured this out
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