Negative Binomial Distribution

Modeling the number of trials needed to achieve a specified number of successes.

A Generalization of the Geometric Distribution

The Negative Binomial distribution answers the question: "How many trials will it take to get my $r$ -th success?" It is a generalization of the Geometric distribution, which is just the special case where $r=1$ .

In finance, a trader might use this to model how many trades it will take to achieve 10 winning trades. A venture capitalist could model how many startups they need to fund to get 3 successful exits.

Interactive Negative Binomial Distribution

Adjust the required number of successes (

r

) and the probability (

p

) to see how the distribution changes.

Number of Successes (r): 5

Probability of Success (p): 0.50

Mean (

\mu

): 10.00

Variance (

\sigma^2

): 10.00

Core Concepts

Probability Mass Function (PMF)

The PMF gives the probability that the r-th success occurs on exactly the k-th trial.

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

For the $r$ -th success to happen on trial $k$ , two things must be true:

In the first $k-1$ trials, there must have been exactly $r-1$ successes. The number of ways this can happen is $\binom{k-1}{r-1}$ .
The $k$ -th trial itself must be a success (with probability $p$ ).
The overall probability combines the ways the previous successes could happen with the probabilities of those successes and failures: $\binom{k-1}{r-1} \times p^{r-1} \times (1-p)^{(k-1)-(r-1)}$ , and then all of that is multiplied by the probability of the final success on trial $k$ , which gives $p^r (1-p)^{k-r}$ .

Mean (

\mu

): 10.00

Variance (

\sigma^2

): 10.00

Key Derivations

Deriving the Mean and Variance

The moments are most intuitively derived by viewing the Negative Binomial as a sum of Geometric random variables.

Deriving the Expected Value (Mean)

Step 1: Decompose into Geometric Variables

Let $X$ be the total number of trials to get $r$ successes. We can think of $X$ as the sum of $r$ independent random variables, where each $Y_i$ is the number of trials to get the next success after the previous one.

X = Y_1 + Y_2 + \dots + Y_r

Each $Y_i$ follows a Geometric distribution with probability $p$ . We know from the Geometric distribution page that $E[Y_i] = 1/p$ .

Step 2: Use Linearity of Expectation

The expectation of a sum is the sum of the expectations.

E[X] = E[Y_1 + Y_2 + \dots + Y_r] = E[Y_1] + E[Y_2] + \dots + E[Y_r]

Step 3: Sum the Geometric Means

Since each $Y_i$ has the same mean, we are just adding $1/p$ to itself $r$ times.

E[X] = \sum_{i=1}^{r} \frac{1}{p} = r \cdot \frac{1}{p}

Final Mean Formula

E[X] = \frac{r}{p}

Deriving the Variance

We use the same decomposition as above. The variance of a sum of *independent* random variables is the sum of their variances.

Step 1: Sum the Variances of Geometric Variables

The variance of a Geometric distribution is $Var(Y_i) = (1-p)/p^2$ .

Var(X) = Var(Y_1 + \dots + Y_r) = Var(Y_1) + \dots + Var(Y_r)

Step 2: Final Result

We are adding the same variance to itself $r$ times.

Var(X) = \sum_{i=1}^{r} \frac{1-p}{p^2} = r \cdot \frac{1-p}{p^2}

Final Variance Formula

Var(X) = \frac{r(1-p)}{p^2}

Applications

Quantitative Finance: Structuring Products

An investment bank is structuring a "first-to-default" credit-linked note on a basket of 10 bonds. They need to achieve their first (`r=1`) default to trigger a payout. If the annual probability of default for any bond is `p`, they can use the Geometric distribution (a special case of Negative Binomial with r=1) to model the waiting time for this event and price the note accordingly.