Negative Binomial Distribution Gives Rise to Probability Mass Function/First Form

Theorem

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.

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

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

Proof

By definition:

$\Img X = \set {0, 1, 2, \ldots}$

$\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$

Then:

\(\ds \map \Pr \Omega\)

\(=\)

\(\ds \sum_{k \mathop \ge n} \binom {n + k - 1} {n - 1} p^k \paren {1 - p}^n\)

\(\ds \)

\(=\)

\(\ds \paren {1 - p}^n \sum_{k \mathop \ge n} \binom {n + k - 1} k p^k\)

Symmetry Rule for Binomial Coefficients

\(\ds \)

\(=\)

\(\ds \paren {1 - p}^n \sum_{j \mathop \ge 0} \binom {-n} k p^k\)

Negated Upper Index of Binomial Coefficient

\(\ds \)

\(=\)

\(\ds \paren {1 - p}^n p^{-n}\)

Binomial Theorem

\(\ds \)

\(=\)

\(\ds 1\)

So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.

$\blacksquare$