# Thread: [SOLVED] Linear transformation of the Gamma distribution

1. ## [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 $\displaystyle Y= a X+b$. It looks cumbersome, is there any simpler methods? Where $\displaystyle X$ follows this Gamma distribution:

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

Thank you.

2. you can use MGFs but you won't recognize the density.
Note that a cannot be zero but it can be negative.

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

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

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

4. 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........

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

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

$\displaystyle f_Y(y) ={f_X((y-b)/a)\over a}$ when a>0

5. thanks again the Beagle!
yes, i figured this out