Common Negative Binomial Parameterizations

Why Negative Binomial Parameterizations Need Care

A negative binomial claim-count model allows variance to exceed the mean, unlike a Poisson model, where the variance equals the mean. The phrase “negative binomial” does not identify one unique parameter convention: the same distribution can be described by a size parameter and a probability, or by its mean and a dispersion parameter. In addition, the labels NB1 and NB2 refer to different ways variance can grow with the mean. Translating the parameters correctly matters because the two variance functions can imply materially different uncertainty as expected claim counts change.

The focus here is parameterization and interpretation, not fitting a frequency model. In the formulas below, \(N\) is a claim count for a specified observation unit, and \(\mu=E[N]\) is its expected count. The mean and variance refer to the same unit and exposure basis.

Definition: In the NB1 mean-dispersion convention, \(\operatorname{Var}(N)=\mu+\phi\mu=\mu(1+\phi)\), where \(\phi\geq 0\). In the NB2 convention, \(\operatorname{Var}(N)=\mu+\alpha\mu^2\), where \(\alpha\geq 0\). The Poisson case is recovered when the applicable dispersion parameter is zero.

The Size-Probability Form

One common negative binomial convention specifies a positive size parameter \(r\) and a probability \(p\), with \(0<p<1\). In this convention, \(N\) counts failures before the \(r\)th success. The probability mass function, mean, and variance are:

P(N=n)=\frac{\Gamma(n+r)}{\Gamma(r)n!}p^r(1-p)^n,\qquad E[N]=\frac{r(1-p)}{p},\qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2},\quad n=0,1,2,\ldots

This definition fixes what \(p\) means. Some references instead use the probability of a failure, or describe a rate rather than a probability. Such formulas can look different even when they describe the same distribution. Always check the stated probability convention before translating parameters.

For a specified mean \(\mu\), the probability in this size-probability convention is obtained by solving the mean equation:

\mu=\frac{r(1-p)}{p} \quad\Longrightarrow\quad p=\frac{r}{r+\mu}.

A fixed \(r\) and this value of \(p\) give variance \(\mu+\mu^2/r\), which is the NB2 form with \(\alpha=1/r\). For NB1, \(r\) is not held fixed as the mean varies: setting \(r=\mu/\phi\) and \(p=1/(1+\phi)\) gives the NB1 mean and variance. Thus NB1 and NB2 are not interchangeable labels for a single variance function.

Why the NB2 Variance Is Quadratic

A useful justification for the NB2 form comes from a Poisson-Gamma mixture. Suppose, conditionally on a latent rate \(\Lambda\), the count satisfies \(N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)\). Let \(\Lambda\) have a Gamma distribution with shape \(r\) and rate \(r/\mu\). Its mean is \(\mu\), and its variance is \(\mu^2/r\). The unconditional mean and variance follow from the laws of total expectation and total variance:

E[N]=E[E(N\mid\Lambda)]=E[\Lambda]=\mu,

\begin{aligned} \operatorname{Var}(N) &=E[\operatorname{Var}(N\mid\Lambda)] +\operatorname{Var}(E[N\mid\Lambda])\\ &=E[\Lambda]+\operatorname{Var}(\Lambda)\\ &=\mu+\frac{\mu^2}{r}. \end{aligned}

Consequently, \(\alpha=1/r\) gives \(\operatorname{Var}(N)=\mu+\alpha\mu^2\). This mixture derivation justifies the NB2 variance and its interpretation: variation in the latent rate contributes a term proportional to the square of the mean. It does not establish NB1; NB1 is a different variance specification.

Comparing the Variance Functions

NB1 adds \(\phi\mu\) to the Poisson variance. Its variance-to-mean ratio is constant at \(1+\phi\), so the excess variance grows linearly with the mean. NB2 adds \(\alpha\mu^2\), so its variance-to-mean ratio is \(1+\alpha\mu\); the relative overdispersion grows as the mean increases.

Feature	NB1	NB2
Variance function	\(\mu+\phi\mu\)	\(\mu+\alpha\mu^2\)
Excess over Poisson variance	\(\phi\mu\)	\(\alpha\mu^2\)
Variance-to-mean ratio	\(1+\phi\)	\(1+\alpha\mu\)
Size-probability translation	\(r=\mu/\phi,\ p=1/(1+\phi)\)	\(r=1/\alpha,\ p=r/(r+\mu)\)

The two variance functions can agree at a particular positive mean even when they are not the same function. Equating them gives \(\phi\mu=\alpha\mu^2\), so for \(\mu>0\) they match at \(\mu=\phi/\alpha\). This one-point match does not imply agreement at other means. If either dispersion is zero, its model reduces to the Poisson variance; if both are zero, the variance functions agree for every mean.

Worked Example: Translating an NB2 Model

A hypothetical portfolio segment has an expected count of \(\mu=12\) claims per policy-year. An NB2 description gives \(\alpha=0.25\). Find the corresponding size and probability, then calculate the variance and the probability of zero claims. Use the failures-before-successes convention for \(p\).

Method. For NB2, \(r=1/\alpha\). Then find \(p\) from the negative binomial mean equation and use the NB2 variance function.

First, \(r=1/0.25=4\). The probability is \(p=r/(r+\mu)=4/(4+12)=4/16=0.25\). Checking the translated mean gives \(r(1-p)/p=4(0.75)/0.25=12\), as required.

The variance is \(\mu+\alpha\mu^2=12+0.25(12^2)=12+0.25(144)=12+36=48\). Using the mass function at \(n=0\), the probability of zero claims is \(p^r=0.25^4=0.00390625\), which is \(0.0039\) to four decimal places.

Check. In size-probability form, the variance is \(r(1-p)/p^2=4(0.75)/(0.25^2)=3/0.0625=48\), matching the NB2 calculation. The common mistake avoided is treating the \(p\) in this convention as the probability of a failure; here it is the success probability.

Worked Example: Comparing NB1 and NB2 at Two Means

For a hypothetical class of risks, consider NB1 with \(\phi=0.25\) and NB2 with \(\alpha=0.0625\). Calculate both variances at means \(\mu=8\) and \(\mu=16\), and determine whether the models agree at either mean.

Method. Substitute each mean separately into its own variance function. Do not assume that equal or similar dispersion numbers make the models equivalent.

At \(\mu=8\), NB1 has variance \(8+0.25(8)=8+2=10\). NB2 has variance \(8+0.0625(8^2)=8+0.0625(64)=8+4=12\). They do not agree at \(\mu=8\).

At \(\mu=16\), NB1 has variance \(16+0.25(16)=16+4=20\). NB2 has variance \(16+0.0625(16^2)=16+0.0625(256)=16+16=32\). They do not agree at \(\mu=16\), either.

The positive mean at which the variance functions would match is \(\phi/\alpha=0.25/0.0625=4\). That is neither of the two means tested. The common mistake avoided is calculating \(0.0625(8^2)\) as 2: it equals 4, so the NB2 variance at 8 is 12, not 10. The same check at 16 gives an excess variance of 16, not 4.

Worked Example: Recovering Dispersion from a Mean and Variance

Suppose a hypothetical rating group has mean claim count \(\mu=20\) and a specified variance of \(50\). Treat these two moments as exact for this calculation. Find the NB1 and NB2 dispersion parameters that reproduce them. For each model, also give its size-probability representation under the stated convention.

Method. Subtract the Poisson variance \(\mu\) to find the excess variance, then solve each model’s variance equation. This is moment matching at this mean; it does not show that the two models have identical variance functions at other means.

The excess variance is \(50-20=30\). For NB1, \(30=\phi(20)\), so \(\phi=30/20=1.5\). Its size is \(r=\mu/\phi=20/1.5=40/3\), and \(p=1/(1+\phi)=1/2.5=0.4\).

For NB2, \(30=\alpha(20^2)=400\alpha\), so \(\alpha=30/400=0.075\). Its size is \(r=1/\alpha=1/0.075=40/3\), and \(p=r/(r+\mu)=(40/3)/(40/3+20)=40/100=0.4\).

Both translated forms have \(r=40/3\) and \(p=0.4\), which gives mean \((40/3)(0.6)/0.4=20\) and variance \((40/3)(0.6)/(0.4^2)=50\). They represent the same negative binomial distribution at this specified mean. However, as the mean changes, NB1 requires \(r=\mu/1.5\), while NB2 keeps \(r=40/3\). The common mistake avoided is assuming that matching moments at one mean makes NB1 and NB2 the same model across all means.

Exam Traps and Interpretation

A disciplined calculation begins by identifying the variance function, rather than trying to infer it from a parameter’s name. When given size and probability, first record the convention and derive the mean. When given NB1 or NB2 dispersion, write its variance function explicitly before substituting values.

Confusing NB1 and NB2. The NB1 excess term is \(\phi\mu\); the NB2 excess term is \(\alpha\mu^2\). The second has an extra factor of the mean.
Using the wrong size translation. NB2 has \(r=1/\alpha\). NB1 has \(r=\mu/\phi\), which changes with the specified mean.
Assuming a match at one mean is universal. If \(\mu=\phi/\alpha\), the variances coincide at that positive mean only; their growth patterns can still differ elsewhere.
Switching probability conventions unnoticed. In this tutorial’s size-probability form, \(p\) is the probability of success and \(N\) counts failures before \(r\) successes.

This distinction is useful when interpreting excess claim-count variation across rating cells with different expected counts. A constant variance-to-mean ratio corresponds to NB1, while a ratio that increases linearly with the mean corresponds to NB2. Neither variance function should be selected merely because its parameterization is familiar; the model’s assumptions and the evidence about variability matter.

Key takeaway: Recognize NB1 by \(\operatorname{Var}(N)=\mu(1+\phi)\) and NB2 by \(\operatorname{Var}(N)=\mu+\alpha\mu^2\). In the failures-before-successes convention, their size-probability translations are respectively \(r=\mu/\phi,\ p=1/(1+\phi)\) and \(r=1/\alpha,\ p=r/(r+\mu)\).

Check Your Understanding

For each item, use the failures-before-successes convention when translating to size \(r\) and probability \(p\).

An NB2 count has mean \(\mu=10\) and \(\alpha=0.1\). Find its variance, size, and probability. (Answer: variance \(20\), \(r=10\), \(p=0.5000\).)
An NB1 count has mean \(\mu=15\) and \(\phi=0.2\). Find its variance and size-probability parameters. (Answer: variance \(18\), \(r=75\), \(p=5/6\).)
At \(\mu=6\), compare NB1 with \(\phi=0.5\) and NB2 with \(\alpha=0.1\). Which has the larger variance, and by how much? (Answer: NB1 variance \(9\); NB2 variance \(9.6\); NB2 is larger by \(0.6\).)
A count has mean \(24\) and variance \(96\). Find the NB2 dispersion, size, and probability that match these moments. (Answer: \(\alpha=0.125\), \(r=8\), \(p=0.2500\).)
NB1 has \(\phi=0.3\), while NB2 has \(\alpha=0.05\). Find their positive mean of equal variances and the variance there. (Answer: \(\mu=6\), and both variances are \(7.8\).)

What You'll Learn