Proof that negative binomial distribution is a distribution function?

Solution 1:

It's evident that $\Bbb{P}(N=n)\ge 0$ for $n\ge r$. So you have to prove that $\sum_{n\ge r}\Bbb{P}(N=n)=1$: $$ \begin{align} \sum_{n\ge r}\Bbb{P}(N=n)&=\sum_{n\ge r} \binom {n-1} {r-1} p^r \left({1-p}\right)^{n-r}\\ &=\sum_{n\ge r} \binom {n-1} {n-r} p^r \left({1-p}\right)^{n-r}\;\;\quad\quad\text{(symmetry})\\ &=p^r\sum_{j\ge 0} \binom {r+j-1} {j} \left({1-p}\right)^{j}\qquad\text{(substituting }j=n-r)\\ &=p^r\sum_{j\ge 0} (-1)^j \binom{-r}{j}\left({1-p}\right)^{j}\qquad\text{(identity}\tbinom{j+r-1}{j}=(-1)^j \tbinom{-r}{j})\\ &=p^r\sum_{j\ge 0} \binom{-r}{j}\left({p-1}\right)^{j}\\ &=p^r\left(1+(p-1)\right)^{-r} \qquad\qquad\qquad\text{(binomial theorem) }\\ &=1 \end{align} $$ using the identity $$ \begin{align} \binom{j+r-1}{j}&=\frac{(j+r-1)(j+r-2) \cdots r}{j!}\\ &=(-1)^j \frac{(-r-(j-1))(-r-(j-2)) \cdots (-r)}{j!} \\&=(-1)^j \frac{(-r)(-r-1) \cdots (-r-(j-1))}{j!} \\&=(-1)^j \binom{-r}{j} \end{align} $$

Solution 2:

You can show this directly with generating functions: $$\left({z\over 1-z}\right)^r=\left( \sum_{\alpha\geq 1} z^\alpha \right)^r =\sum_{\alpha_1,\dots, \alpha_r \geq 1} z^{\alpha_1+\cdots +\alpha_r}=\sum_{n\geq r} {n-1\choose r-1} z^n.$$ The last equation follows from "stars and bars" which gives you the number of compositions of $n$ in $r$ pieces. Now substitute in $z=1-p$ to get the result.