14. Negative Binomial Distribution: Properties, Applications, and Examples

This document provides a comprehensive overview of the Negative Binomial Distribution, covering its fundamental properties, probability density function, mean and variance, practical applications in business statistics, and illustrative examples.

14.1 Properties of the Negative Binomial Distribution

The Negative Binomial Distribution is a discrete probability distribution that models the number of Bernoulli trials needed to achieve a specified number of successes. Key properties include:

• Discrete: It deals with countable outcomes (number of trials).
• Memoryless Property (related): While not strictly memoryless like the geometric distribution, it focuses on the number of trials to reach a fixed number of successes, which is a crucial distinction.
• Relationship to Binomial Distribution: It can be viewed as the inverse of the Binomial distribution in some contexts, focusing on trials rather than successes in a fixed number of trials.
• Parametrization: Typically defined by two parameters:
  • r: The number of successes.
  • p: The probability of success on a single trial.
  • (Alternatively, k for the number of failures before the r-th success, and p).

14.2 Probability Density Function (PDF) of the Negative Binomial Distribution

The Negative Binomial Distribution can be defined in two common ways, leading to slightly different PDF formulations. We will focus on the version that models the number of trials (k) until the r-th success.

Definition: The number of Bernoulli trials, $k$, required to get $r$ successes, where the probability of success on each trial is $p$. The last trial must be a success.

The probability mass function (PMF) for this definition is:

$$ P(X=k; r, p) = \binom{k-1}{r-1} p^r (1-p)^{k-r} $$

Where:

• $k$ is the total number of trials ($k = r, r+1, r+2, \ldots$).
• $r$ is the number of successes.
• $p$ is the probability of success on each trial.
• $\binom{k-1}{r-1}$ is the binomial coefficient, representing the number of ways to arrange the $r-1$ successes in the first $k-1$ trials.

Alternative Definition (Number of Failures): Another common formulation models the number of failures, $k$, before the $r$-th success. In this case, the PMF is:

$$ P(Y=k; r, p) = \binom{k+r-1}{r-1} p^r (1-p)^k $$

Where:

• $k$ is the number of failures ($k = 0, 1, 2, \ldots$).
• $r$ is the number of successes.
• $p$ is the probability of success on each trial.

For consistency and broader application in many statistical contexts, we will primarily refer to the first definition (number of trials) unless otherwise specified.

14.3 Mean and Variance of the Negative Binomial Distribution

For the Negative Binomial distribution modelling the number of trials ($k$) to achieve $r$ successes with probability of success $p$ on each trial:

• Mean (Expected Value): The expected number of trials to achieve $r$ successes is given by: $$ E[X] = \frac{r}{p} $$

• Variance: The variance of the number of trials is: $$ Var(X) = \frac{r(1-p)}{p^2} $$

Intuition:

• The mean increases with r (more successes require more trials) and decreases with p (higher success probability means fewer trials).
• The variance also increases with r and the probability of failure $(1-p)$, and decreases with p.

14.4 Applications of the Negative Binomial Distribution in Business Statistics

The Negative Binomial Distribution is highly valuable in business and statistics for modelling events that occur over time or trials where the number of successes is fixed, and we are interested in the number of trials or failures.

• Customer Retention and Churn: Modelling the number of purchases a customer makes before becoming inactive (churning). The r-th purchase could be the one before they stop buying.
• Quality Control: Determining the number of items that need to be inspected until a certain number of defective items are found. For example, finding the number of products to test until 3 defective units are detected.
• Sales and Marketing: Analysing the number of sales calls required to close a specific number of deals.
• Reliability Engineering: Estimating the number of operational cycles of a device before a certain number of failures occur.
• Count Data Analysis: When count data exhibits overdispersion (variance is greater than the mean), the Negative Binomial distribution is often a better fit than the Poisson distribution. This is common in many business contexts where events are not independent.

14.5 Examples of the Negative Binomial Distribution

Here are some practical examples illustrating the use of the Negative Binomial Distribution:

Example 1: Quality Control of Manufactured Parts

A factory produces electronic components. The probability that a component is defective is $p = 0.05$. We want to know the probability that it takes exactly 15 components to be inspected to find the 3rd defective component.

Here, $k = 15$ (number of trials), $r = 3$ (number of successes/defects), and $p = 0.05$ (probability of defect).

Using the PMF: $$ P(X=15; r=3, p=0.05) = \binom{15-1}{3-1} (0.05)^3 (1-0.05)^{15-3} $$ $$ P(X=15; r=3, p=0.05) = \binom{14}{2} (0.05)^3 (0.95)^{12} $$ $$ P(X=15; r=3, p=0.05) = \frac{14 \times 13}{2 \times 1} \times 0.000125 \times (0.95)^{12} $$ $$ P(X=15; r=3, p=0.05) = 91 \times 0.000125 \times 0.54036 $$ $$ P(X=15; r=3, p=0.05) \approx 0.00615 $$

So, there is approximately a 0.615% chance that exactly 15 components need to be inspected to find the 3rd defective one.

Example 2: Customer Sales Calls

A sales representative makes sales calls. The probability of making a sale on any given call is $p = 0.2$. The representative wants to know the expected number of calls required to make 5 sales.

Here, $r = 5$ (number of successes/sales), and $p = 0.2$ (probability of sale).

Using the mean formula: $$ E[X] = \frac{r}{p} = \frac{5}{0.2} = 25 $$

The sales representative can expect to make 25 calls to achieve 5 sales.

Example 3: Website Visitor Conversion

A website owner wants to understand how many visitors it takes to get 10 newsletter sign-ups. The probability that a visitor signs up for the newsletter is $p = 0.02$. What is the variance in the number of visitors needed to get 10 sign-ups?

Here, $r = 10$ (number of successes/sign-ups), and $p = 0.02$ (probability of sign-up).

Using the variance formula: $$ Var(X) = \frac{r(1-p)}{p^2} = \frac{10(1-0.02)}{(0.02)^2} $$ $$ Var(X) = \frac{10(0.98)}{0.0004} = \frac{9.8}{0.0004} = 24500 $$

The variance in the number of visitors needed to obtain 10 sign-ups is 24,500. This indicates a wide spread in the number of visitors required, which is typical for low probabilities.