Specifying A Negative Binomial Frequency Model

Why Specify an NB2 Frequency Model?

A claim-frequency model predicts a count for each policy or other rating unit. In a Poisson model, the conditional variance equals the conditional mean. When a frequency model instead uses an NB2 distribution, it allows the variance to increase quadratically with the mean. The NB2 specification therefore needs both a model for the expected count and a dispersion parameter describing the extra conditional variation.

For ratemaking, exposure belongs in the expected count: twice the exposure should imply twice the expected claims when the other rating characteristics are unchanged. A log link makes the expected count positive, while an exposure offset incorporates exposure with its coefficient fixed at one. The resulting mean is the quantity used for fitted counts; the NB2 dispersion parameter determines how variable observed counts can be around that mean.

Definition: For policy $i$, let $N_i$ be its claim count, $e_i>0$ its exposure measured in a consistent unit, and $x_{ij}$ its rating variables. An NB2 frequency model specifies $$ N_i \mid x_i,e_i \sim \operatorname{NB2}(\mu_i,\alpha), \qquad \log(\mu_i)=\log(e_i)+\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}, $$ where $\mu_i=E[N_i\mid x_i,e_i]$, $\alpha>0$ is the dispersion parameter, and $\beta_0,\ldots,\beta_p$ are coefficients. Under this parameterization, $$ \operatorname{Var}(N_i\mid x_i,e_i)=\mu_i+\alpha\mu_i^2. $$ The term $\log(e_i)$ is an offset: its coefficient is fixed at one, not estimated as an ordinary predictor coefficient.

Parameterization, Mean, and Variance

A negative binomial distribution can be written using a size parameter $r>0$ and a success-probability parameter $p$. In the NB2 form above, $r=1/\alpha$ and $p=r/(r+\mu_i)$. Equivalently, its probability mass function is

P(N_i=k\mid x_i,e_i) = \frac{\Gamma(k+r)}{\Gamma(r)k!} \left(\frac{r}{r+\mu_i}\right)^r \left(\frac{\mu_i}{r+\mu_i}\right)^k, \qquad k=0,1,2,\ldots,\quad r=\frac{1}{\alpha}.

The mean $\mu_i$ is the fitted expected number of claims for the policy's actual exposure. It is not necessarily an integer and is not a guarantee of the count that will be observed. The parameter $\alpha$ controls the additional variability: larger $\alpha$ means greater variance at a fixed mean. The NB2 label refers specifically to the quadratic variance relationship $\mu_i+\alpha\mu_i^2$; it should not be confused with other negative binomial parameterizations that use a different mean-variance relationship.

For interpretation, holding other rating variables fixed, a one-unit increase in $x_{ij}$ multiplies the expected count by $\exp(\beta_j)$. If a variable changes by $\Delta x$, the multiplier is $\exp(\beta_j\Delta x)$. Because the link is logarithmic, coefficients act multiplicatively on the mean, not additively in claim counts.

Why the NB2 Variance Has This Form

One justification for NB2 is a Poisson-Gamma mixture. Suppose that, conditional on an unobserved policy-specific rate $\Lambda_i$, the count is Poisson with mean $\Lambda_i$. Let $\Lambda_i$ have a Gamma distribution with shape $r=1/\alpha$ and scale $\mu_i/r$. This choice gives $E[\Lambda_i]=\mu_i$ and $\operatorname{Var}(\Lambda_i)=\mu_i^2/r=\alpha\mu_i^2$. Integrating out $\Lambda_i$ produces the negative binomial probability mass function above.

The variance follows directly from the law of total variance. Conditional on $\Lambda_i$, a Poisson count has mean and variance $\Lambda_i$. Therefore,

\begin{aligned} E[N_i] &= E\{E[N_i\mid\Lambda_i]\}=E[\Lambda_i]=\mu_i,\\ \operatorname{Var}(N_i) &=E\{\operatorname{Var}(N_i\mid\Lambda_i)\} +\operatorname{Var}\{E[N_i\mid\Lambda_i]\}\\ &=E[\Lambda_i]+\operatorname{Var}(\Lambda_i) =\mu_i+\alpha\mu_i^2. \end{aligned}

The first variance contribution is the ordinary Poisson variation conditional on the rate. The second is variation in the latent rate across otherwise comparable units. This explanation motivates the NB2 form; it does not mean that a particular fitted model has identified the actual cause of excess variation.

Building the Fitted Mean

The most reliable calculation is to keep the exposure and rating factors visible. Write the fitted mean as

\widehat{\mu}_i =e_i\exp\left(\widehat{\beta}_0+\sum_{j=1}^{p}\widehat{\beta}_jx_{ij}\right).

The exponential term is the fitted frequency per unit of exposure. Multiplying it by $e_i$ converts that rate into the expected count over the policy's exposure. In a factor representation, coefficients that are logarithms of rating factors can be converted back into multiplicative relativities. Apply the exposure once, in the unit for which the base rate is defined.

Worked Example: Fitted Counts for Unequal Exposures

A hypothetical insurer models claim counts using an NB2 model. The base annual frequency is $0.20$, the vehicle-class relativity is $1.50$, and the territory relativity is $0.80$. The model's dispersion parameter is $\alpha=0.60$. Calculate the fitted mean and variance for a policy with $0.50$ years of exposure, vehicle-class relativity $1.50$, and territory relativity $0.80$, and for a second policy with $1.25$ years of exposure, vehicle-class relativity $0.75$, and territory relativity $1.20$.

Method and assumptions. Use the NB2 mean $e\times$ annual base frequency $\times$ rating relativities. Conditional on the stated model variables, the count has variance $\mu+0.60\mu^2$. No interest or mortality assumption is involved; the count distribution is NB2 with $\alpha=0.60$.

For the first policy, the fitted annual frequency is $0.20\times1.50\times0.80=0.24$. Its fitted expected count is $0.50\times0.24=0.12$. The squared mean is $0.12^2=0.0144$, so the variance is $0.12+0.60\times0.0144=0.12+0.00864=0.12864$.

For the second policy, the fitted annual frequency is $0.20\times0.75\times1.20=0.18$. Its fitted expected count is $1.25\times0.18=0.225$. The squared mean is $0.225^2=0.050625$, so the variance is $0.225+0.60\times0.050625=0.225+0.030375=0.255375$.

Check. The second policy has $1.25/0.50=2.5$ times the exposure of the first, but its annual frequency is $0.18/0.24=0.75$ times as large. Its fitted count is therefore $2.5\times0.75=1.875$ times the first policy's count, and $0.12\times1.875=0.225$. The calculation uses exposure once and gives a variance greater than the mean for each policy.

Writing the Linear Predictor and Interpreting Coefficients

An equivalent way to write the model is to display each term on the link scale. The exposure appears as $\log(e_i)$, alongside the intercept and rating-variable terms. Exponentiating the full linear predictor recovers the fitted mean. This form is useful when coefficients are supplied directly rather than as rating factors.

Worked Example: From Coefficients to a Policy Mean

For a hypothetical portfolio, suppose the specified NB2 model has $\beta_0=\log(0.10)$, $\beta_1=\log(2)$, and $\beta_2=\log(0.50)$, with $\alpha=0.40$. A policy has exposure $e=1.5$, $x_1=1$, and $x_2=1$. Calculate its fitted count mean and variance. Also state the effect of changing $x_1$ from zero to one while holding the other inputs fixed.

Method and assumptions. Substitute all inputs into the log-link model, including $\log(e)$ as a fixed offset. The count is NB2 with dispersion $\alpha=0.40$; no interest or mortality assumption is relevant.

The linear predictor is $\log(1.5)+\log(0.10)+\log(2)+\log(0.50)$. Exponentiating gives $\widehat{\mu}=1.5\times0.10\times2\times0.50=0.15$. The squared mean is $0.15^2=0.0225$. Thus $\operatorname{Var}(N)=0.15+0.40\times0.0225=0.15+0.009=0.159$.

Changing $x_1$ from zero to one multiplies the expected count by $\exp(\beta_1)=\exp(\log 2)=2$, with the other inputs unchanged. The policy's current fitted mean of $0.15$ includes $x_1=1$; at $x_1=0$, its mean would be $0.15/2=0.075$. Doubling $0.075$ gives $0.15$, which checks the coefficient interpretation.

Using the Specified Distribution for Count Probabilities

A fitted mean alone does not determine a probability of zero claims or of a particular count; the dispersion parameter is also needed. Once $\mu_i$ and $\alpha$ are specified, set $r=1/\alpha$ and substitute into the NB2 probability mass function. For example, the zero-count probability is $(r/(r+\mu_i))^r$. These probabilities describe the specified conditional count distribution, not additional fitted rating factors.

Worked Example: Probability of Zero or One Claim

A hypothetical policy has an NB2 fitted annual frequency of $0.20$, exposure of $2$ years, and dispersion parameter $\alpha=0.50$. Calculate the fitted mean, the probabilities of zero and exactly one claim, and the probability of at most one claim. Give probabilities to four decimals.

Method and assumptions. The count for the full two-year exposure follows NB2 with the stated mean and dispersion. No interest or mortality assumption applies. First, the mean is $\mu=2\times0.20=0.40$. The size parameter is $r=1/\alpha=2$, so $r/(r+\mu)=2/(2+0.40)=2/2.40=5/6$, while $\mu/(r+\mu)=0.40/2.40=1/6$.

For zero claims, the probability mass function gives $P(N=0)=(5/6)^2=25/36=0.6944$, rounded to four decimals.

For one claim, the coefficient in the probability mass function is $\Gamma(3)/(\Gamma(2)1!)=2$. Therefore, $P(N=1)=2(5/6)^2(1/6)=2(25/36)(1/6)=25/108=0.2315$, rounded to four decimals.

Adding the two disjoint outcomes gives $P(N\leq1)=25/36+25/108=75/108+25/108=100/108=25/27=0.9259$, rounded to four decimals.

Check. The probabilities are nonnegative, and $0.6944+0.2315=0.9259$ to four decimals. The probability of at most one claim is greater than the probability of zero claims because it includes the one-claim outcome.

Common Specification Traps

The NB2 mean and dispersion play different roles. The link and rating coefficients determine $\mu_i$; $\alpha$ determines how the conditional variance increases beyond $\mu_i$. Avoid changing one quantity to compensate for a mistake in the other.

Using an estimated exposure coefficient. The offset is $\log(e_i)$ with coefficient fixed at one. Treating it as an ordinary predictor changes the exposure relationship specified by this model.
Multiplying by exposure twice. If the model's linear predictor already includes $\log(e_i)$, exponentiating already includes exposure. Do not multiply the result by $e_i$ again.
Confusing a rate and a count. The exponentiated rating terms, excluding exposure, give frequency per exposure unit. The fitted count mean also includes exposure.
Using the wrong variance expression. For NB2, the variance is $\mu+\alpha\mu^2$, not $\mu+\alpha\mu$. For instance, with $\mu=2$ and $\alpha=0.50$, the incorrect expression gives $2+0.50(2)=3$. The correct variance is $2+0.50(2^2)=4$.
Confusing dispersion with size. Under this parameterization, $r=1/\alpha$. A larger $\alpha$ means a smaller $r$, not a larger one.

The model presumes positive exposure, a correctly specified mean structure, and the stated NB2 conditional distribution. The same dispersion parameter is used across policies in the specification given here. An exposure of zero cannot be entered into $\log(e_i)$; such records require separate treatment rather than substituting zero into this formula. This tutorial specifies the model and calculates its implications; it does not determine whether NB2 is preferable to another fitted model.

Key takeaway: Specify $\log(\mu_i)=\log(e_i)+x_i^\mathsf{T}\beta$, calculate the fitted count mean by exponentiating the full predictor, and then use $\mu_i+\alpha\mu_i^2$ for its NB2 variance. Keep exposure, the rating factors, and dispersion in their distinct roles.

Check Your Understanding

For each hypothetical model, calculate the requested quantity using the NB2 convention in this tutorial.

A policy has annual base frequency $0.12$, relativity $1.50$, and exposure $0.50$. What is its fitted count mean? (Answer: $0.12\times1.50\times0.50=0.09$.)
An NB2 count has mean $0.30$ and dispersion $\alpha=0.40$. What is its variance? (Answer: $0.30+0.40(0.30^2)=0.336$.)
A model has $\beta_0=\log(0.08)$, a binary-variable coefficient $\beta_1=\log(1.25)$, exposure $2$, and $x_1=1$. What is the fitted count mean, and what is the multiplier for changing $x_1$ from zero to one? (Answer: $\widehat{\mu}=2(0.08)(1.25)=0.20$; the multiplier is $1.25$.)
An NB2 model has $\mu=0.50$ and $\alpha=0.25$. What are its size parameter $r$ and variance? (Answer: $r=1/0.25=4$, and $\operatorname{Var}(N)=0.50+0.25(0.50^2)=0.5625$.)
A hypothetical policy has fitted annual frequency $0.10$, exposure $1.5$, and $\alpha=1$. Calculate the probability of zero claims to four decimals. (Answer: $\mu=1.5(0.10)=0.15$, $r=1$, so $P(N=0)=(1/(1+0.15))^1=1/1.15=0.8696$.)

What You'll Learn