Negative binomial distribution

The negative binomial distribution $𝑋 \sim N B i n (𝑟, 𝑝)$ describes the number of failures of independent Bernoulli trials with success probability $𝑝$ before the $𝑟$ -th success. #m/def/prob It is thus the sum of $𝑟$ independent geometrically distributed variables, whence follows the probability mass function

ℙ (𝑋 = 𝑥) = 𝑝 𝑟 (𝑥 + 𝑟 - 1 𝑟 - 1) (1 - 𝑝) 𝑥

Proof by induction

In the case $𝑋 \sim N B i n (1, 𝑝)$ it reduces to the Geometric distribution $𝑋 \sim G e o m (𝑝)$ . Now assume the probability mass function above is valid for $𝑋 𝑟 \sim N B i n (𝑟, 𝑝)$ . Let $𝑌 \sim G e o m (𝑝)$ be independent so that $𝑋 𝑟 + 1 = 𝑋 𝑟 + 𝑌 \sim N B i n (𝑟 + 1, 𝑝)$ . Then

ℙ (𝑋 𝑟 + 1 = 𝑥) = ℙ (𝑋 𝑟 + 𝑌 = 𝑥) = 𝑥 \sum 𝑖 = 0 ℙ (𝑋 𝑟 + 𝑌 = 𝑥 ∣ 𝑋 𝑟 = 𝑖) ℙ (𝑋 𝑟 = 𝑖) = 𝑥 \sum 𝑖 = 0 ℙ (𝑌 = 𝑥 - 𝑖) ℙ (𝑋 𝑟 = 𝑖) = 𝑥 \sum 𝑖 = 0 (1 - 𝑝) 𝑥 - 𝑖 𝑝 𝑝 𝑟 (𝑖 + 𝑟 - 1 𝑟 - 1) (1 - 𝑝) 𝑖 = 𝑝 𝑟 + 1 (1 - 𝑝) 𝑥 𝑥 \sum 𝑖 = 0 (𝑖 + 𝑟 - 1 𝑟 - 1) = 𝑝 𝑟 + 1 (𝑥 + 𝑟 𝑟) (1 - 𝑝) 𝑥

where on the final line we invoked ^P5.

Properties

Let $𝑋 \sim N B i n (𝑟, 𝑝)$ and $𝑞 = 1 - 𝑝$

Expectation: $𝜇 = 𝔼 [𝑋] = 𝑟 𝑞 𝑝$
Variance: $𝜎 2 = V a r [𝑋] = 𝑟 𝑞 𝑝 2$
Moment-generating function: $𝑀 𝑋 (𝑡) = (𝑝 1 - 𝑞 e 𝑡) 𝑟$ for $𝑞 e 𝑡 < 1$
Probability generating function: $𝑔 𝑋 (𝑡) = (𝑝 1 - 𝑞 𝑡) 𝑟$

Proof of 1–3

^P1 follows from ^P1 by linearity of expectation, while ^P2 follows from ^P2 by ^P3 ^P3 follows from ^P3 by ^P1.

Relationship to other distributions

By the Central limits theorem, $N B i n (𝑛, 𝑝) ⇝ N (𝑛 𝑞 𝑝, 𝑛 𝑞 𝑝 2)$ as $𝑛 \to \infty$

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