Deriving Moment Generating Function of the Negative Binomial? My textbook did the derivation for the binomial distribution, but omitted the derivations for the Negative Binomial Distribution.
I know it is supposed to be similar to the Geometric, but it is not only limited to one success/failure. (i.e the way I understand it is that the negative binomial is the sum of independent geometric random variables). For example: $Y_1 +Y_2 +Y_3+\cdots $ where $Y_i$ is a geometric parameter. I can't seem to find online one for the negative binomial and am having trouble with even doing the geometric. 
Can anyone show me a derivation of the negative binomial? 
Edit: My book calls the negative binomial as the distribution of the number of trials needed to get a specified number r of successes.
 A: To derive the mgf of the negative binomial distribution we are going to use the following identity:
$$\binom{-r}{y}=\left( -1 \right)^y \binom{r+y-1}{y}  $$
We can prove that in the following way:
$$\begin{align}
\binom{-r}{y} & = \frac{ \left( -r \right) \left(-r-1 \right) \ldots \left(-r-y+1 \right)}{y!}\\
& = \left(-1 \right)^y \frac{ \left(r+y-1 \right) \ldots \left( r+1 \right)r}{y!} \\
& = \left(-1 \right)^y \binom{r+y-1}{y} 
\end{align}$$
Now
$$M \left( t \right)=\sum_{y=0}^{\infty} e^{ty} \binom{y+r-1}{r-1} \left( 1-p \right)^y \times p^r $$
Grouping terms and using the above idenity we get:
$$\begin{align} M \left( t \right)& =p^r \sum_{y=0}^{\infty} \binom{y+r-1}{r-1} \left[ e^t \left( 1-p \right) \right]^y \\&=p^r\sum_{y=0}^{\infty} \binom{-r}{y}\left( -1 \right)^y\left[ e^t \left( 1-p \right) \right]^y \\& =p^r\sum_{y=0}^{\infty} \binom{-r}{y}\left[ -e^t \left( 1-p \right) \right]^y \end{align} $$
Then using Newton's Binomial Theorem: $\left( x+1 \right)^r= \sum_{i=0}^\infty {r\choose i}x^i$ provided that $|x|<1$, the last term becomes:
$$M \left(t \right)= \frac{p^r}{\left[ 1- \left(1-p \right)e^t \right]^r}$$
provided that $t<-\log(1-p)$
Note that the negative binomial distribution can come with a slightly different parameterization as well, as it has been pointed out in the comments. I leave it to you to derive the mgf for the other case.
Hope this helps.
A: The m.g.f. of a sum of independent random variables is just the product of their m.g.f.s, so
$$
M_{Y_1+\cdots+Y_r}(t) = \left( M_{Y_1}(t) \right)^r.
$$
\begin{align}
M_{Y_1}(t) & = \operatorname E(e^{tY_1}) = \sum_{y=1}^\infty e^{ty} \Pr(Y_1=y) \\[10pt]
& = \sum_{y=1}^\infty e^{ty} p(1-p)^{y-1} = \frac{p}{1-p} \sum_{y=1}^\infty \left(e^t(1-p)\right)^y \\[10pt]
& = \frac{p}{1-p} \cdot \frac{\text{first term}}{1-\text{common ratio}} \\[10pt]
& = \frac{p}{1-p}\cdot\frac{e^t(1-p)}{1-e^t(1-p)} \\[10pt]
& = \frac{e^tp}{1-e^t(1-p)}.
\end{align}
(Then remember to raise the whole thing to the power $r$.)
