Negative Binomial Count Distributions

Why Use a Negative Binomial for Claim Counts?

For a Poisson count model, the conditional mean and variance are equal. Claim counts across policyholders or rating cells may vary more than that model permits, even after relevant risk characteristics have been accounted for. The negative binomial is a count distribution that allows this additional variation while retaining nonnegative integer outcomes.

Here the response $N$ is the number of claims in a specified observation period for a specified risk or group. The distribution is a model for counts, not claim severities or dollar losses. Its parameters are not self-explanatory: different sources use different conventions, so identify the parameterization before substituting values.

Definition: In this tutorial, $N$ has a negative binomial distribution with mean $\mu>0$ and shape parameter $\theta>0$ if, for $y=0,1,2,\ldots$, $$ P(N=y)=\frac{\Gamma(y+\theta)}{\Gamma(\theta)y!} \left(\frac{\theta}{\theta+\mu}\right)^\theta \left(\frac{\mu}{\theta+\mu}\right)^y. $$ Here $\Gamma(\cdot)$ is the gamma function, which satisfies $\Gamma(k)=(k-1)!$ for positive integers $k$. This mean-and-shape parameterization has $E[N]=\mu$ and $\operatorname{Var}(N)=\mu+\mu^2/\theta$.

A second common convention writes the model in terms of a mean $\mu$ and a dispersion parameter $\alpha>0$. In this tutorial, “mean-and-dispersion” means the specific convention $ \operatorname{Var}(N)=\mu+\alpha\mu^2 $, sometimes called the NB2 parameterization. It follows that $\alpha=1/\theta$. The name “dispersion” is not used identically in every reference; do not assume a parameter has the same meaning just because it has a similar name.

Parameter conversion: For the mean-and-shape model, $\alpha=1/\theta$, so $$ \operatorname{Var}(N)=\mu+\frac{\mu^2}{\theta} =\mu+\alpha\mu^2. $$ A smaller $\theta$, equivalently a larger $\alpha$, means more variation at the same mean. As $\theta$ grows without bound, the distribution approaches a Poisson distribution with mean $\mu$.

Failures Before a Fixed Number of Successes

The classical negative binomial convention describes the number of failures before a fixed number $r$ of successes. Let $r>0$ and $0<p<1$, where $p$ is the probability of success on each independent trial. Under this convention, $N$ counts failures, so its support includes zero and has no finite upper bound. Its probability mass function is

P(N=y)=\frac{\Gamma(y+r)}{\Gamma(r)y!}p^r(1-p)^y, \qquad y=0,1,2,\ldots.

For positive integer $r$, the coefficient is the number of ways to arrange $y$ failures before the final success, with the other $r-1$ successes placed among those trials. In actuarial count modeling, $r$ need not be an integer; the gamma function extends the formula to any positive real shape.

The failures-before-successes convention and the mean-and-shape convention describe the same family of distributions when their parameters are matched. The translation is $r=\theta$ and $p=\theta/(\theta+\mu)$. Thus $1-p=\mu/(\theta+\mu)$. In particular, $p$ is not the expected claim count and is not the probability of a claim.

Why the Two Parameterizations Agree

The parameter conversion can be checked directly by substituting it into the classical probability mass function. Starting with the failures-before-successes form, set $r=\theta$ and $p=\theta/(\theta+\mu)$. Then $1-p=\mu/(\theta+\mu)$, and the resulting probability at count $y$ is

\frac{\Gamma(y+\theta)}{\Gamma(\theta)y!} \left(\frac{\theta}{\theta+\mu}\right)^\theta \left(\frac{\mu}{\theta+\mu}\right)^y.

This is exactly the mean-and-shape probability mass function. The same conversion also gives the stated mean: the failures-before-successes mean is $r(1-p)/p$, which becomes $\theta[\mu/(\theta+\mu)]/[\theta/(\theta+\mu)]=\mu$. The variance under that convention is $r(1-p)/p^2$; substituting the same parameters gives $\mu+\mu^2/\theta$. The algebra verifies that the shape parameter controls the amount of variation without changing the specified mean.

A useful interpretation comes from a mixed Poisson model. Suppose that, conditional on a latent claim rate $\Lambda$, $N$ is Poisson with mean $\Lambda$, and $\Lambda$ has a gamma distribution with shape $\theta$ and rate $\theta/\mu$. Integrating the conditional Poisson probabilities over that gamma distribution gives

\begin{aligned} P(N=y) &=\int_0^\infty \frac{e^{-\lambda}\lambda^y}{y!} \frac{(\theta/\mu)^\theta}{\Gamma(\theta)} \lambda^{\theta-1}e^{-(\theta/\mu)\lambda}\,d\lambda\\ &=\frac{\Gamma(y+\theta)}{\Gamma(\theta)y!} \left(\frac{\theta}{\theta+\mu}\right)^\theta \left(\frac{\mu}{\theta+\mu}\right)^y. \end{aligned}

The integral uses the gamma-function identity $\int_0^\infty \lambda^{a-1}e^{-b\lambda}\,d\lambda=\Gamma(a)/b^a$ for $a,b>0$. This representation explains a practical source of extra variation: risks may have different underlying rates, even within a group modeled with the same average rate. The negative binomial is then the unconditional count distribution after that rate variation is averaged out. The representation does not require that every real claim-count process literally follow a gamma-distributed latent rate.

Worked Probability Calculations

Worked Example: Translate a Failures Model and Find a Range Probability

A hypothetical warranty portfolio is modeled by the failures-before-successes convention, with $r=3$ and $p=0.60$ for the number of claims per contract-year. Find the mean, variance, and probability of at most two claims.

Model and assumptions. The count $N$ follows a negative binomial distribution with $r=3$, $p=0.60$, and independent trials in the classical interpretation. Interest and mortality assumptions are not applicable to this count model.

Mean and variance. First, $1-p=0.40$. Therefore, $E[N]=r(1-p)/p=3(0.40)/0.60=2$. The variance is $r(1-p)/p^2=3(0.40)/(0.60)^2=1.20/0.36=3.3333$.

Probability. Use the failures-before-successes probability mass function for $y=0,1,2$:

\begin{aligned} P(N=0)&=(0.60)^3=0.216,\\ P(N=1)&=3(0.60)^3(0.40)=0.2592,\\ P(N=2)&=\frac{3(4)}{2}(0.60)^3(0.40)^2=6(0.216)(0.16)=0.20736. \end{aligned}

Adding gives $P(N\leq 2)=0.216+0.2592+0.20736=0.68256$, or 0.6826 to four decimals. As a check, the mean of 2 is consistent with a model whose mass is concentrated around small integer counts, while its variance exceeds its mean. A common mistake is to use $p=0.60$ as the claim probability; here it is the success probability in the failures-counting convention.

Worked Example: Calculate Probabilities From Mean and Shape

For a hypothetical equipment-insurance segment, the annual claim count is modeled with mean $\mu=4$ and shape $\theta=4$. Find the dispersion parameter, variance, and probability of zero or one claim.

Model and assumptions. Use the negative binomial mean-and-shape distribution with $\mu=4$, $\theta=4$, and independent annual counts under the model. Interest and mortality assumptions are not applicable.

Dispersion and variance. The mean-and-dispersion parameter is $\alpha=1/\theta=1/4=0.25$. The variance is $4+4^2/4=4+4=8$.

Probability. The equivalent success parameter is $p=\theta/(\theta+\mu)=4/(4+4)=0.5$. The zero-claim probability is $P(N=0)=p^\theta=(0.5)^4=0.0625$. For one claim, $P(N=1)=\theta p^\theta(1-p)=4(0.0625)(0.5)=0.1250$. Thus $P(N\leq1)=0.0625+0.1250=\mathbf{0.1875}$.

The check $E[N]=4$ and $\operatorname{Var}(N)=8$ confirms that this model has twice the mean as its variance. A common error is to use $\theta$ where the variance formula requires $\alpha$, or vice versa: here $1/\theta=0.25$, not 4.

Worked Example: Infer Shape and Zero-Claim Probability From Moments

A hypothetical class of small commercial accounts has a fitted annual claim-count mean of $2.4$ and variance of $5.28$. Under the mean-and-dispersion negative binomial model, determine $\alpha$, $\theta$, and the modeled probability of no claims.

Model and assumptions. Treat the stated mean and variance as the fitted moments for a negative binomial count $N$, with a common mean and shape for this calculation. Interest and mortality assumptions are not applicable.

Recover dispersion. From $5.28=2.4+\alpha(2.4)^2$, the excess variance is $5.28-2.4=2.88$. Since $(2.4)^2=5.76$, $\alpha=2.88/5.76=0.50$. Therefore $\theta=1/\alpha=2$.

Calculate the zero probability. The equivalent success parameter is $p=\theta/(\theta+\mu)=2/(2+2.4)=2/4.4=5/11$. Thus $P(N=0)=p^\theta=(5/11)^2=25/121=0.206611\ldots$, or 0.2066 to four decimals.

Check. Substituting the recovered shape gives $2.4+(2.4)^2/2=2.4+2.88=5.28$, matching the supplied variance. The calculation would be wrong if the variance were below the mean: this negative binomial parameterization requires $\alpha>0$, so its variance cannot be less than $\mu$.

Interpreting the Model and Avoiding Parameter Traps

The negative binomial is useful when a Poisson model's equal-mean-and-variance restriction is too narrow for the count variation being modeled. It does not by itself explain why counts vary: omitted rating characteristics, genuine differences in latent claim rates, or other features of the data may contribute. A fitted negative binomial distribution describes the count variation under its assumptions; it does not establish a particular causal explanation.

In an applied count model, define the observation period and exposure basis before interpreting $\mu$. A mean of 2 might represent expected claims per policy-year, per vehicle-year, or per another defined unit. Those are different quantities. This tutorial treats $\mu$ as the mean for the stated observation unit; it does not introduce exposure offsets or rating-model link functions.

Use these checks when reading or calculating a negative binomial model:

State whether the parameters are $(r,p)$, $(\mu,\theta)$, or $(\mu,\alpha)$.
For failures before successes, $p$ is the success probability and the count is the number of failures.
For mean-and-shape, $\theta>0$; for the specified mean-and-dispersion convention, $\alpha>0$ and $\alpha=1/\theta$.
Check that the claimed variance exceeds the mean for positive $\alpha$. A variance equal to the mean is approached in the Poisson limit as $\alpha$ approaches zero.
Use the probability mass function for integer count values $y\geq0$, not for dollar losses or continuous outcomes.

Common exam trap. Suppose a problem gives $\mu=3$ and $\theta=2$, then asks for the variance. The incorrect calculation $3+3^2(2)=21$ treats shape as dispersion. The correction is to use $\alpha=1/\theta=0.5$, giving $\operatorname{Var}(N)=3+0.5(3^2)=7.5$, equivalently $3+9/2=7.5$. Always convert the parameter before applying the variance expression.

Key takeaway: The negative binomial models nonnegative integer counts with extra variation relative to a Poisson count at the same mean. In the convention used here, mean-and-shape uses $\theta$, mean-and-dispersion uses $\alpha=1/\theta$, and the failures-before-successes form uses $r=\theta$ and $p=\theta/(\theta+\mu)$. Naming the convention is essential to a correct calculation.

Check Your Understanding

For each problem, identify the parameterization before calculating.

A hypothetical policy group follows a failures-before-successes negative binomial count model with $r=2$ and $p=0.50$. Find the mean and variance. (Answer: mean $=2$, variance $=4$.)
A count has mean $\mu=5$ and mean-and-shape parameter $\theta=5$. Find $\alpha$ and the variance. (Answer: $\alpha=0.20$, variance $=10$.)
For a negative binomial count with mean $\mu=2$ and shape $\theta=2$, find the success parameter $p$ and $P(N=0)$. (Answer: $p=0.5000$, $P(N=0)=0.2500$.)
A hypothetical account class has modeled mean 1.5 and variance 3.75 under the mean-and-dispersion convention. Find $\alpha$ and $\theta$. (Answer: $\alpha=1.00$, $\theta=1.00$.)
A count model has $\mu=4$ and $\alpha=0.25$. Find the shape parameter, variance, and zero-count probability. (Answer: $\theta=4$, variance $=8$, $P(N=0)=0.0625$.)

What You'll Learn