I am trying to find the moment generating function for the negative binomianl distribution. I know that it's baiscly finding <e^ty>. I therefore put e^ty into the sum from y=0 to infinity. I tried a few ways of solving it but I am stuck for quite sime time now. Here is what I tried:

- Expanding e^ty and use it's summation form. But this only led to showing how the moment generating function works.
- Use different substitutions to try and get the original form or one of it's varience back with a factor outside the summation. But this is not working too.

Am I going in the right direction?

$\displaystyle P\left (Y=y \right )=\binom{y+r-1}{r-1}p^r\left (1-p \right )^y; Supp\left (Y \right )=\left \{0,1,2,3,... \right \}$

$\displaystyle M_{Y}(t)=E(e^{ty})=\sum_{y=0}^{\infty }e^{ty}\binom{y+r-1}{r-1}p^r\left (1-p \right )^y$

$\displaystyle =\sum_{y=0}^{\infty }\binom{y+r-1}{y}p^r\left (1-p \right )^y\sum_{n=0}^{\infty}

\frac{(ty)^n}{n!}$

$\displaystyle =\sum_{n=0}^{\infty}\frac{t^n}{n!}\sum_{y=0}^{\inf ty }\frac{(y+r-1)!}{y!(r-1)!}p^r\left (1-p \right )^yy^n$

$\displaystyle =\sum_{n=0}^{\infty}\frac{t^n}{n!}\ E(y^n)$