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In my textbook, a clear proof that the Geometric Distribution is a distribution function is given, namely

$$\sum_{n=1}^{\infty} \Pr(X=n)=p\sum_{n=1}^{\infty} (1-p)^{n-1} = \frac{p}{1-(1-p))}=1.$$

Then the textbook introduces the Negative Binomial Distribution; it gives a fairly clear explanation for why the PMF of a Negative Binomial random variable $N$ with parameter $r$ is

$$p\binom{n-1}{r-1}p^{r-1}(1-p)^{n-r} = \binom{n-1}{r-1}p^{r}(1-p)^{n-r} $$

But to show

$$\sum_{n=r}^{\infty} \Pr(N=n)=\sum_{n=r}^{\infty}\binom{n-1}{r-1}p^{r}(1-p)^{n-r}=1$$

the textbook gives (in my opinion) a wordy and informal argument that is nowhere near as clear.

What is a straightforward algebraic way to prove the above statement; that the Negative Binomial is a distribution function? I also looked at a different probability textbook, plus wolfram.com's definition before asking.

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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 $$ \begin{align} \binom{j+r-1}{j}&=\frac{(j+r-1)(j+r-2) \cdots r}{j!}\\ &=(-1)^j \frac{(-r-(j-1))(-r-(j-2)) \cdots (-r)}{j!} \\&=(-1)^j \frac{(-r)(-r-1) \cdots (-r-(j-1))}{j!} \\&=(-1)^j \binom{-r}{j} \end{align} $$

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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.

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  • $\begingroup$ In my Probability theory undergrad course, and to the extent of my math curriculum so far, we have yet to cover generating functions. Exam is tomorrow, so I'll look at this approach later. $\endgroup$ – NaN Nov 12 '13 at 5:37

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