Negative Binomial Distribution Gives Rise to Probability Mass Function

Theorem

Let $X$ be a discrete random variable on a probability space $\left({\Omega, \Sigma, \Pr}\right)$.

Let $X$ have the negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).

Then $X$ gives rise to a probability mass function.

Proof

By definition:

• $\operatorname{Im} \left({X}\right) = \left\{{n, n+1, n+2, \ldots}\right\}$

• $\displaystyle \Pr \left({X = k}\right) = \binom {k-1} {n-1} p^n \left({1-p}\right)^{k-n}$

Then:

	\(\displaystyle \)	\(\displaystyle \Pr \left({\Omega}\right)\)	\(=\)	\(\displaystyle \sum_{k \ge n} \binom {k-1} {n-1} p^n \left({1-p}\right)^{k-n}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle p^n \sum_{j \ge 0} \binom {n+j-1} {j} \left({1-p}\right)^j\)	\(\displaystyle \)	by substituting $j = k - n$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle p^n \sum_{j \ge 0} \binom {-n} {j} \left({p-1}\right)^j\)	\(\displaystyle \)	Negated Upper Index of Binomial Coefficient
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle p^n \left({1 - \left({p-1}\right)}\right)^{-n}\)	\(\displaystyle \)	Directly from the Binomial Theorem
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)

So $X$ satisfies $\Pr \left({\Omega}\right) = 1$, and hence the result.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 2.2$