Moment Generating Function - Negative Binomial I am trying to find the MGF of
$$ P(X=x)=\binom{r+x-1}{x} p^r (1-p)^x $$
where $x=0,1,..., 0<p<1$, and $r>0$ is an integer.
The answer should be
$$E[e^{tx}]=\left( \frac{p}{1-(1-p) e^t} \right)^r$$
where $t < -ln(1-p)$
However, I just don't get it right. I am in need for an explanation for dummies ;)
(Question from "Statistical Inference"' Casella & Berger)
 A: I think one way to work through it would be by considering the following:
\begin{equation}
E[e^{tx}]= \sum_{x=0}^{\infty}[e^{tx}\binom{r+x-1}{x} p^r (1-p)^x]
\end{equation}
\begin{equation}
=\frac{p^r}{(r-1)!}\sum_{x=0}^{\infty}[e^t(1-p)]^x\frac{(x+r-1)!}{x!}\frac{(r-1)!}{(r-1)!}
\end{equation}
\begin{equation}
= p^r\sum_{x=0}^{\infty}[e^t(1-p)]^x\binom{x+r-1}{r-1}
\end{equation}
\begin{equation}
=p^r[1−(1−p)e^t]^{-r}=\left( \frac{p}{1-(1-p) e^t} \right)^r
\end{equation}
The first line holds by the definition of mgfs, the second equality you open up $\binom{r+x-1}{x}$ and then multiply and divide by $(r-1)!$ and the last equality holds because $(1-w)^{-r}=\sum\limits_{x=0}^\infty \dbinom{x+r-1}{r-1} w^x$, where $w= (1 − p)e^t$ in our case.
To see why the latter holds, consider the pmf of the negative binomial, which is $P(X=x)=\dbinom{x-1}{r-1}(1-p)^{x-r}p^r>0$ when $0<p<1$ and when summed over the support is equal to 1.
\begin{equation}
\sum\limits_{x=r}^\infty \dbinom{x-1}{r-1}(1-p)^{x-r}p^r=1
\end{equation}
Now let $k=x-r$, hence $x=k+r$, then:
$\sum\limits_{k=0 }^\infty \dbinom{k+r-1}{r-1}(1-p)^{k}p^r=p^r\sum\limits_{k=0 }^\infty\dbinom{k+r-1}{r-1}(1-p)^{k}=p^r [1-(1-p)]^{-r}=1$.
The summation bit after $p^r$ is the negative binomial series which is equal to $[1-(1-p)]^{-r}$.
Let me know if it makes sense!
