Interpreting Negative Binomial Dispersion

Dispersion at a Fixed Mean

Two claim-frequency models can have the same expected number of claims but differ in how much probability they assign to counts far from that mean. The negative binomial model represents this extra variation with a dispersion parameter. Understanding the parameter’s direction is essential: depending on the convention, larger dispersion may be represented by a larger number or a smaller one.

Here we use a mean $m>0$ and a size parameter $r>0$. The reciprocal $\alpha=1/r$ is the dispersion parameter used to express the variance directly. Thus, at a fixed mean, increasing $\alpha$ means greater dispersion, while increasing $r$ means less dispersion. The model is not Poisson unless considered as a limit.

Definition: Let $N$ have a negative binomial distribution with mean $m$ and size $r$, where $m>0$ and $r>0$. Write $p=r/(r+m)$ and $q=m/(r+m)$. Then $$ P(N=k)=\frac{\Gamma(k+r)}{\Gamma(r)k!}p^r q^k,\qquad k=0,1,2,\ldots. $$ The mean and variance are $$ E[N]=m,\qquad \operatorname{Var}(N)=m+\frac{m^2}{r}=m+\alpha m^2,\qquad \alpha=\frac{1}{r}. $$

The parameter $p$ in this expression is a probability parameter, not the mean. Since $p+q=1$, the probabilities sum to one. When $r$ is an integer, this negative binomial parameterization can also describe the number of failures before $r$ successes; the mean-and-size form above also permits non-integer $r$.

The variance exceeds the mean by $m^2/r$, or $\alpha m^2$. At a fixed $m$, a larger $\alpha$ therefore increases variance, while a larger $r$ reduces it. The variance-to-mean ratio is $1+\alpha m$, so it also increases with $\alpha$ when the mean is held fixed.

Why the Variance Has This Form

A useful justification uses the gamma-Poisson representation. Let the random claim intensity $\Lambda$ have a gamma distribution with shape $r$ and rate $r/m$, so $E[\Lambda]=m$ and $\operatorname{Var}(\Lambda)=m^2/r$. Conditional on $\Lambda$, suppose $N\mid\Lambda$ is Poisson with mean $\Lambda$. This mixture gives the negative binomial model.

Find the mean.
The conditional Poisson mean is $\Lambda$, so $E[N]=E[E[N\mid\Lambda]]=E[\Lambda]=m$.

Apply the law of total variance.
$\operatorname{Var}(N)=E[\operatorname{Var}(N\mid\Lambda)]+\operatorname{Var}(E[N\mid\Lambda])$.

Substitute the conditional Poisson moments.
Both the conditional mean and variance are $\Lambda$, so $\operatorname{Var}(N)=E[\Lambda]+\operatorname{Var}(\Lambda)=m+m^2/r$.

The first variance contribution, $m$, is the Poisson variation that remains even at a fixed intensity. The additional contribution, $m^2/r$, comes from variation in the intensity across observations or risks. This explains why two portfolios with the same average frequency need not have the same count variability.

Worked Example: Comparing Variance at One Mean

A hypothetical portfolio has an expected annual claim count of $m=3$ per policy. Compare negative binomial models with size parameters $r=3$ and $r=12$, and compare both with a Poisson model of the same mean. Calculate each variance and standard deviation.

Method. Keep $m=3$ fixed and use $\operatorname{Var}(N)=m+m^2/r$. The corresponding dispersion is $\alpha=1/r$.

For $r=3$, $\alpha=1/3$. The variance is $3+3^2/3=3+3=6$, and the standard deviation is $\sqrt{6}=2.4495$ claims, rounded to four decimals.

For $r=12$, $\alpha=1/12$. The variance is $3+3^2/12=3+9/12=3.75$, and the standard deviation is $\sqrt{3.75}=1.9365$ claims, rounded to four decimals.

For a Poisson model with mean $3$, the variance is $3$ and the standard deviation is $\sqrt{3}=1.7321$ claims, rounded to four decimals. The mean is the same in all three cases, but the $r=3$ negative binomial model has the greatest variance, the $r=12$ model is closer to Poisson, and Poisson has the least variance.

Check. The variance-to-mean ratios are $6/3=2$, $3.75/3=1.25$, and $3/3=1$. They are consistent with $1+\alpha m$: the first is $1+3/3=2$, and the second is $1+3/12=1.25$. The Poisson ratio is $1$.

Common mistake avoided: Treating a larger $r$ as more dispersion. Here $r$ is the size parameter, and dispersion is its reciprocal. The smaller size $r=3$, not $r=12$, gives the larger variance.

How Dispersion Changes Probabilities

Variance summarizes overall spread, but insurance questions often ask about a particular event, such as no claims or at least a specified number of claims. To compare such probabilities, use the negative binomial probability formula with the same mean and each model’s own $r$, $p$, and $q$. Do not change the mean when the question asks for a fixed-mean comparison.

Greater dispersion commonly puts more probability both at zero and in the far right tail, relative to a less dispersed model with the same mean. However, a variance comparison alone does not tell you the probability of every specific event. In particular, do not assume that every upper-tail probability at every cutoff must increase. Calculate the event in question.

Worked Example: Comparing an Upper-Tail Probability

For a hypothetical book of policies, compare the probability of at least three claims when the annual mean is $m=2$. Calculate it for a negative binomial model with $r=2$, another with $r=10$, and a Poisson model with mean $2$. Give probabilities to four decimals.

Method. For each negative binomial model, calculate $P(N\ge3)=1-P(N=0)-P(N=1)-P(N=2)$. With $m=2$ and $r=2$, $p=2/(2+2)=1/2$ and $q=1/2$.

For $r=2$, the first three probabilities are $P(N=0)=(1/2)^2=0.25$, $P(N=1)=2(1/2)^2(1/2)=0.25$, and $P(N=2)=3(1/2)^2(1/2)^2=0.1875$. Thus $P(N\ge3)=1-0.25-0.25-0.1875=0.3125$.

For $r=10$, $p=10/(10+2)=5/6$ and $q=2/12=1/6$. The probabilities are $P(N=0)=(5/6)^{10}=0.1615$, $P(N=1)=10(1/6)(5/6)^{10}=0.2692$, and $P(N=2)=55(1/6)^2(5/6)^{10}=0.2467$, with intermediate values rounded here to four decimals. Using unrounded values in the subtraction gives $P(N\ge3)=1-0.1615056-0.2691760-0.2467446=0.3226$.

For the Poisson model, $P(N=k)=e^{-2}2^k/k!$. Therefore $P(N=0)=e^{-2}$, $P(N=1)=2e^{-2}$, and $P(N=2)=2e^{-2}$. The requested probability is $1-5e^{-2}=0.3233$, rounded to four decimals.

Check. The results are $0.3125$ for $r=2$, $0.3226$ for $r=10$, and $0.3233$ for Poisson. The means are all $2$, and the negative binomial model with larger $r$ is closer to the Poisson result.

Common mistake avoided: Calculating $P(N\le2)$ and reporting it as the requested tail. The complement is required: $P(N\ge3)=1-P(N\le2)$.

The Poisson Limit

As $r$ grows while $m$ stays fixed, $\alpha=1/r$ tends to zero and the negative binomial distribution approaches a Poisson distribution with mean $m$. This is a distributional limit, not a claim that a negative binomial with any finite $r$ is exactly Poisson.

For fixed $k$, rewrite the negative binomial probability using $p=r/(r+m)$ and $q=m/(r+m)$:

P(N=k)=\frac{r(r+1)\cdots(r+k-1)}{k!} \left(\frac{m}{r+m}\right)^k \left(\frac{r}{r+m}\right)^r.

The first $k$ factors, divided by $r^k$, tend to $1$ as $r$ grows. The other factors can be rearranged as $(m^k/k!)\left[r(r+1)\cdots(r+k-1)/r^k\right](1+m/r)^{-(r+k)}$. The bracket tends to $1$, and $(1+m/r)^{-(r+k)}$ tends to $e^{-m}$. Hence

\lim_{r\to\infty}P(N=k)=e^{-m}\frac{m^k}{k!},

the probability mass function of a Poisson random variable with mean $m$. The variance also confirms the limit: $m+m^2/r$ tends to $m$, the Poisson variance.

Worked Example: More Dispersion and the Zero-Claim Probability

Two hypothetical groups each have mean annual claim count $m=4$. Group A is modeled with negative binomial size $r=4$; Group B has size $r=16$. For each group, find the variance and the probability of at least one claim. State which group is more dispersed. Round probabilities to four decimals.

Method. Calculate the variance from $m+m^2/r$. For the zero count, use $P(N=0)=p^r$, where $p=r/(r+m)$, then take the complement to find $P(N\ge1)$.

For Group A, $\alpha=1/4$, $p=4/(4+4)=1/2$, and $P(N=0)=(1/2)^4=0.0625$. Thus $P(N\ge1)=1-0.0625=0.9375$. Its variance is $4+16/4=8$.

For Group B, $\alpha=1/16$, $p=16/(16+4)=0.8$, and $P(N=0)=0.8^{16}=0.0281475$. Thus $P(N\ge1)=1-0.0281475=0.9718525$, or $0.9719$ to four decimals. Its variance is $4+16/16=5$.

Check. Both groups have mean $4$; Group A has larger dispersion and variance. Its probability of zero claims is also larger, $0.0625$ versus $0.0281$. Group B’s probability of at least one claim is $0.9719$, greater than Group A’s $0.9375$.

Common mistake avoided: Concluding that the more dispersed group must have a higher probability of at least one claim. Here the more dispersed group has more zero counts, so its probability of at least one claim is lower.

Interpreting Results Without Overclaiming

The negative binomial dispersion parameter is useful because it separates the mean frequency from extra count variability. When comparing models, state the mean and parameter convention before interpreting results. If a problem gives $\alpha$, use $\operatorname{Var}(N)=m+\alpha m^2$; if it gives $r$, use $\operatorname{Var}(N)=m+m^2/r$. A quick check is that increasing $\alpha$, or decreasing $r$, must increase the variance at a fixed mean.

Keep the mean fixed. A change in the mean and a change in dispersion have different effects; separate them in comparisons.
Check the parameter convention. “Size” often refers to $r$, whereas “dispersion” often refers to $1/r$.
Compute the requested event. Variance indicates overall variability; it does not replace a probability calculation for a specific count or tail.
Use the limit carefully. As $r$ tends to infinity, the model approaches Poisson with the same mean; finite $r$ retains extra variance.

These distinctions matter when claim counts are used to describe hypothetical insurance portfolios: two groups with the same expected frequency can imply different chances of unusually low or high counts. The interpretation is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each question, use the stated parameterization and keep track of whether the parameter is size or dispersion.

A negative binomial claim count has mean $6$ and size $r=3$. Find its variance. (Answer: $18$)
A negative binomial count has mean $4$ and dispersion $\alpha=0.25$. Find its size and variance. (Answer: $r=4$, variance $8$)
Compare the probability of zero claims for a Poisson count with mean $2$ and a negative binomial count with mean $2$, size $r=2$. Find the negative binomial probability and its difference above the Poisson probability. (Answer: Negative binomial $P(N=0)=0.2500$; difference $0.1147$)
A negative binomial count has mean $3$ and size $r=6$. Find the probability of at least one claim. (Answer: $0.9122$)
A negative binomial count has mean $2$ and size $r=4$. Find $P(N\ge2)$. (Answer: $0.5391$)

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