Negative binomial distribution coefficient origin The book said that PDF of NBD(negative binomial distribution) is below:
$P_{Y}=C^{r-1}_{y+r-1}p^{r}\left(1-p\right)^{y}$ y=0,1,2,3...
I understand the PDF but the author said the reason why this is called NBD is that the coefficient of the PDF is the same as the coefficient of the following formula's expansion series: 
$p^{r}\left[1-\left(1-p\right)\right]^{-r}$
 , I tried to write the coefficient of the latter formula but can hardly find they are the same. Can someone explain it to me? Many thx.
 A: What was undoubtedly intended is that $\Pr(Y=n)$ is the coefficient of $z^n$ in the expansion of 
$$p^r[1-z(1-p)]^{-r}$$
(note the inserted $z$). 
One can calculate the coefficient of $z^n$ directly by expanding.  
It is somewhat less work to note that $Y$ is the sum of $r$ independent random variables with one of the versions of the geometric distribution. The probability generating function of the sum of independent random variables is the product of the probability generating functions. That is what the $\left(\frac{p}{1-z(1-p)}\right)^r$ comes from. 
A: The generalized binomial theorem writes the expansion of $(x+y)^n$ for any real $n$ as
$$(x+y)^n=\sum_{k=0}^\infty{n\choose k}x^{n-k}y^k\;,$$ with the understanding that $${n\choose k}:={(n)(n-1)\cdots(n-k+1)\over k!}\;,$$ a ratio whose numerator and denominator both contain $k$ factors. If $n$ is a positive integer, then ${n\choose k}$ is zero for $k>n$. If the exponent $n$ is negative, say $n=-r$ for $r>0$, the sum will continue forever:
$$(x+y)^{-r}=\sum_{k=0}^\infty{-r\choose k}x^{n-k}y^k$$
Putting $x=1$, $y=-(1-p)$, this becomes
$$p^{-r} = [1-(1-p)]^{-r}=\sum_{k=0}^\infty{-r\choose k}1^{n-k}[-(1-p)]^k
=\sum_{k=0}^\infty{-r\choose k}(-1)^k(1-p)^k\;.\tag{1}$$
You can check that
$$(-1)^k{-r\choose k} = {k+r-1\choose k}={k+r-1\choose r-1}\,.$$
Substituting this into (1) and rearranging, we've proved that the negative binomial probability mass function sums to 1.
