[SOLVED] Linear transformation of the Gamma distribution

**stokastik** · May 4th 2010, 11:32 AM

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.

**matheagle** · May 4th 2010, 03:36 PM

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 $f_Y(y) = \left({y-b\over a}\right)^{k-1} \frac{e^{-(y-b)/a}}{\Gamma(k)|a|}\text{ for } y > b$

**stokastik** · May 4th 2010, 04:04 PM

thank you matheagle!
how is it called, the method you used here, I'm gonna search some ressources?
good day!

**matheagle** · May 4th 2010, 04:16 PM

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

**stokastik** · May 5th 2010, 01:53 PM

thanks again the Beagle!
yes, i figured this out

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