Why Use a Negative Binomial Model?
A claim-count model describes the number of claims in a specified exposure period. The Poisson model is a useful starting point, but it has a restrictive feature: its variance equals its mean. In a hypothetical portfolio, counts may vary more than that equality permits. The negative binomial model provides a count distribution whose variance can exceed its mean, with a parameter that controls the amount of extra dispersion.
The model does not assert that actual claims occur as a sequence of Bernoulli trials. A repeated-trial interpretation defines one standard parameterization, while the resulting probability distribution can be used directly to model claim counts. The exposure period and the event counted must be defined consistently before applying the model.
For integer \(r\), the repeated-trial description gives an intuitive construction. To have exactly \(k\) failures before the \(r\)th success, the first \(k+r-1\) trials must contain \(k\) failures and \(r-1\) successes, in any order, and the last trial must be a success. There are \(\binom{k+r-1}{k}\) such arrangements. Each has probability \(q^k p^r\), giving the stated formula.
The negative binomial distribution is also defined for any real \(r>0\) by replacing the binomial coefficient with its gamma-function extension. This broader parameter range is useful when estimating a model from a desired mean and variance. The repeated-trial story then serves as an interpretation for integer \(r\), not as a literal trial construction.
Mean, Variance, and Dispersion
Under this parameterization, the mean and variance are
The variance can be rewritten to show the source of overdispersion:
Since \(E[N]^2/r>0\) for positive \(r\) and positive mean, the variance exceeds the mean. At a fixed mean, decreasing \(r\) increases the extra variance. As \(r\) grows, the extra term shrinks; in the large-\(r\) limit at fixed mean, the model approaches the Poisson distribution.
The mean-dispersion form is convenient when an application supplies a target mean and variance. If the variance exceeds the mean, solve
These equations require \(\mu>0\) and \(\operatorname{Var}(N)>\mu\) for a finite positive \(r\). If the variance equals the mean, the Poisson model is the limiting case rather than a negative binomial model with finite \(r\). If the variance is below the mean, this overdispersed negative binomial parameterization cannot match both moments.
Why the Moment Formulas Hold
For integer \(r\), write \(N=G_1+\cdots+G_r\), where each \(G_j\) is the number of failures before one success. The \(G_j\) are independent because they count failures in separate successive blocks of the independent Bernoulli trials. For one block,
Its probability generating function is \(H(z)=E[z^{G_j}]=p/(1-qz)\). Differentiating and evaluating at \(z=1\) gives
For a probability generating function, \(\operatorname{Var}(G_j)=H''(1)+H'(1)-[H'(1)]^2\). Substitution yields
Independence makes the expectation and variance of the sum equal to the sums of the component expectations and variances. Therefore \(E[N]=r q/p\) and \(\operatorname{Var}(N)=r q/p^2\). Rewriting the latter in terms of the mean gives \(\operatorname{Var}(N)=E[N]+E[N]^2/r\). The real-\(r\) extension uses the same moment formulas for the generalized negative binomial distribution.
Worked Probability and Parameter Calculations
Worked Example: Calculating a Claim-Count Tail Probability
A hypothetical repair program models the number \(N\) of covered service claims per contract-year as negative binomial with \(r=3\) and \(p=0.6\). Calculate the mean, variance, and probability of at least three claims. Report probabilities to four decimal places.
Method. Use \(q=1-p=0.4\), the negative binomial moments, and the probability mass function. This approach avoids confusing the parameter \(p\) with the probability of a claim.
The mean is $$ E[N]=\frac{3(0.4)}{0.6}=2. $$ The variance is $$ \operatorname{Var}(N)=\frac{3(0.4)}{0.6^2} =\frac{1.2}{0.36}=3.3333. $$ The variance is larger than the mean, as expected.
It is simpler to find \(P(N\geq 3)\) by subtracting the probabilities for \(N=0,1,2\) from one. Those probabilities are $$ P(N=0)=\binom{2}{0}(0.6)^3(0.4)^0=0.216, $$ $$ P(N=1)=\binom{3}{1}(0.6)^3(0.4)^1 =3(0.216)(0.4)=0.2592, $$ $$ P(N=2)=\binom{4}{2}(0.6)^3(0.4)^2 =6(0.216)(0.16)=0.20736. $$ Thus $$ P(N\geq 3)=1-(0.216+0.2592+0.20736)=0.31744. $$ Rounded to four decimal places, the answer is \(P(N\geq3)=0.3174\). The three probabilities below 3 sum to \(0.68256\), and \(0.68256+0.31744=1\), checking the calculation.
Worked Example: Recovering the Parameters From Moments
For a hypothetical class of small-business policies, the annual claim count is modeled as negative binomial with mean 4 and variance 12. Find \(r\) and \(p\), then calculate the probability of at least two claims. Report probabilities to four decimal places.
Method. First recover the size parameter from the mean and variance, then convert it to \(p\). This avoids using a Poisson model that would force the variance to equal 4.
Using \(\operatorname{Var}(N)=\mu+\mu^2/r\), $$ 12=4+\frac{4^2}{r}, \qquad 8=\frac{16}{r}, \qquad r=2. $$ Then $$ p=\frac{r}{r+\mu}=\frac{2}{2+4}=\frac{1}{3}, \qquad q=1-p=\frac{2}{3}. $$ The model's mean and variance check as $$ E[N]=\frac{2(2/3)}{1/3}=4, \qquad \operatorname{Var}(N)=\frac{2(2/3)}{(1/3)^2}=12. $$
Calculate the probabilities below 2: $$ P(N=0)=\left(\frac{1}{3}\right)^2=\frac{1}{9}=\frac{3}{27}, $$ $$ P(N=1)=\binom{2}{1}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right) =\frac{4}{27}. $$ Therefore $$ P(N\geq2)=1-\frac{3}{27}-\frac{4}{27} =\frac{20}{27}=0.7407. $$ The probability of zero or one claim is \(7/27\); adding \(7/27\) and \(20/27\) gives 1.
Worked Example: Using the Mean-and-Dispersion Parameterization
A hypothetical annual coverage segment is assigned mean claim count \(\mu=3\) and size parameter \(r=3\). Calculate its dispersion parameter \(\alpha\), variance, and probability of at least two claims. Report the probability to four decimal places.
Method. Use \(\alpha=1/r\) and \(p=r/(r+\mu)\), then apply the probability mass function. The main trap is to treat the mean count as the probability of at least one claim.
First, $$ \alpha=\frac{1}{3}=0.3333, \qquad p=\frac{3}{3+3}=\frac{1}{2}, \qquad q=\frac{1}{2}. $$ The variance is $$ \operatorname{Var}(N)=\mu+\alpha\mu^2 =3+\left(\frac{1}{3}\right)(9)=6. $$ The probabilities for zero and one claim are $$ P(N=0)=\left(\frac{1}{2}\right)^3=\frac{1}{8}, $$ $$ P(N=1)=\binom{3}{1}\left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)=\frac{3}{16}. $$ Thus $$ P(N\geq2)=1-\frac{1}{8}-\frac{3}{16} =\frac{16}{16}-\frac{2}{16}-\frac{3}{16} =\frac{11}{16}=0.6875. $$ The expected number of claims is 3, not a probability; the probability of at least two claims is \(0.6875\). The complement calculation checks because \(P(N\leq1)=5/16\) and \(5/16+11/16=1\).
Assumptions, Uses, and Common Traps
A negative binomial model is applied to counts over a specified, consistently defined exposure period. Its count support is \(0,1,2,\ldots\), and its parameters must satisfy \(r>0\) and \(0<p<1\). When the repeated-trial interpretation is used, the trials are independent and have a constant success probability. In actuarial applications, the distribution is often used as a probability model for counts without claiming that claims themselves arise from literal trials.
The model permits variance greater than the mean, but that does not establish that it is appropriate for every portfolio. A single mean and variance do not describe every feature of a count distribution. The model is not suitable for matching an underdispersed count distribution using this parameterization, and a fitted model still needs to be assessed in context.
A frequent parameter trap is to switch conventions without converting. Some sources use \(r,p\) for failures before successes; others use a mean and a dispersion parameter. Under the convention here, \(E[N]=rq/p\), not \(rp/q\), and \(\alpha=1/r\), not \(r\). State the parameterization before substituting values.
| Quantity | Negative binomial formula | Interpretation |
|---|---|---|
| Mean | \(\mu=r(1-p)/p\) | Expected count in the defined exposure period |
| Variance | \(\mu+\mu^2/r\) | Exceeds the mean for finite positive \(r\) |
| Dispersion form | \(\mu+\alpha\mu^2\), where \(\alpha=1/r\) | Smaller \(r\) or larger \(\alpha\) means more extra variance |
| Probability of zero | \(p^r\) | One mass in the full count distribution, not a substitute for the mean |
- Using the wrong tail complement. To calculate \(P(N\geq k)\), subtract \(P(N=0)+\cdots+P(N=k-1)\) from 1.
- Confusing frequency with probability. The expected count can exceed 1. It is not the probability of at least one claim.
- Assuming variance equals the mean. That is the Poisson restriction. For a finite-size negative binomial model, \(\operatorname{Var}(N)=\mu+\mu^2/r>\mu\).
- Ignoring the exposure definition. Parameters describe counts for the specified exposure period; a model for annual counts cannot be applied to a different period without an appropriate model adjustment.
Check Your Understanding
For each problem, use the parameterization \(P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k\), unless the mean-dispersion form is specified.
- A hypothetical policy group has \(r=4\) and \(p=0.5\). Find the mean, variance, and probability of zero claims. (Answer: mean \(4\), variance \(8\), \(P(N=0)=0.0625\))
- A count model has mean \(\mu=2\) and size \(r=2\). Find \(p\), the variance, and the probability of zero claims. (Answer: \(p=0.5\), variance \(4\), \(P(N=0)=0.2500\))
- A hypothetical count has \(r=3\) and \(p=0.75\). Calculate \(P(N\leq1)\), to four decimal places. (Answer: \(0.7383\))
- A negative binomial count has mean 5 and variance 15. Find \(r\), \(p\), and \(\alpha\), using the real-\(r\) extension if needed. (Answer: \(r=2.5\), \(p=1/3\), \(\alpha=0.4\))
- For \(r=2\) and \(p=0.5\), calculate the probability of at least three claims by summing the probabilities for counts zero, one, and two and taking the complement. (Answer: \(P(N\geq3)=0.3125\))