Negative Binomial Dispersion And Parameter Roles

Why Use a Negative Binomial Claim-Count Model?

A Poisson model is useful when a claim count has a specified mean and its variance equals that mean. Some claim-count data are more variable: observations may include more very small and very large counts than a Poisson model with the same mean would predict. The negative binomial model provides a count distribution with this greater variability.

The parameters have distinct roles. The size parameter governs the amount of dispersion, while the success-probability parameter helps set both the mean and variance. To interpret the size parameter's effect clearly, comparisons should hold the mean fixed. Increasing size in that comparison reduces the extra dispersion.

Definition: A count $N$ has a negative binomial distribution with size $r>0$ and success-probability parameter $p$, where $0<p<1$, if $$ P(N=k)=\frac{\Gamma(k+r)}{\Gamma(r)\,k!}p^r(1-p)^k,\qquad k=0,1,2,\ldots. $$ Here $\Gamma(\cdot)$ is the gamma function. If $r$ is a positive integer, the coefficient is $\binom{k+r-1}{k}$. The parameters describe the count distribution; in particular, $p$ is not generally the probability that an individual insured has a claim.

The model assigns probabilities to nonnegative integer counts. Its parameters must be specified for the exposure unit and risk group being modeled. This tutorial concerns claim counts, not claim severities or the aggregate amount of claims. The probability function also makes the parameterization explicit: other sources may use different negative binomial conventions, so formulas should not be transferred without checking their definitions.

What the Parameters Control

Write $q=1-p$ for convenience. For fixed $r$, increasing $p$ decreases $q/p$, and therefore decreases the mean count. For fixed $p$, increasing $r$ increases both the mean and the variance. Thus, it is not accurate to say that a larger size parameter always means a smaller variance in absolute terms. Its dispersion effect is clearest when comparing models with the same mean.

The mean and variance are

E[N]=\frac{r(1-p)}{p},\qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}.

The variance is greater than the mean for every allowed $p$, since

\operatorname{Var}(N)-E[N] =\frac{r(1-p)^2}{p^2}>0.

This property is called overdispersion relative to the Poisson model: for a Poisson variable, the variance equals the mean. The negative binomial variance-to-mean ratio is $1/p$, which exceeds 1.

Why the Mean and Variance Have These Forms

A short derivation makes the parameter roles easier to remember. The probability generating function of $N$ is $G(z)=E[z^N]$. Applying the negative binomial series to the probability function gives

G(z)=\left(\frac{p}{1-qz}\right)^r,\qquad q=1-p.

Differentiating and evaluating at $z=1$ yields the first factorial moment and the second factorial moment:

G'(z)=rq\,p^r(1-qz)^{-r-1}, \qquad G'(1)=\frac{rq}{p}=E[N],

G''(z)=r(r+1)q^2p^r(1-qz)^{-r-2}, \qquad G''(1)=\frac{r(r+1)q^2}{p^2}=E[N(N-1)].

Since $N^2=N(N-1)+N$, variance equals the second factorial moment plus the mean minus the square of the mean. Substituting the expressions above and simplifying gives

\begin{aligned} \operatorname{Var}(N) &=\frac{r(r+1)q^2}{p^2}+\frac{rq}{p} -\left(\frac{rq}{p}\right)^2\\ &=\frac{rq^2}{p^2}+\frac{rq}{p} =\frac{rq(q+p)}{p^2} =\frac{rq}{p^2}. \end{aligned}

The final equality uses $q+p=1$. This derivation also shows why the variance exceeds the mean: the variance has one more factor of $1/p$.

Interpreting Size at a Fixed Mean

To isolate the role of size, suppose two models have the same mean $\mu$. The mean equation implies $p=r/(r+\mu)$. Substitution into the variance gives

\operatorname{Var}(N) =\mu+\frac{\mu^2}{r}.

At fixed $\mu$, a larger $r$ reduces the additional term $\mu^2/r$, bringing the variance closer to the Poisson value $\mu$. As $r$ increases without bound while $\mu$ stays fixed, the variance approaches $\mu$. This is a comparison of dispersion at equal means; it does not mean that increasing $r$ while keeping $p$ unchanged reduces the variance.

Worked Example: Calculating Probabilities and Moments

A hypothetical portfolio is modeled with a negative binomial claim count having $r=2$ and $p=0.5$ for the stated exposure period. Calculate the mean, variance, and probabilities of zero, one, and two claims. The method is direct substitution into the probability function and moment formulas; it avoids treating $p$ as the probability of a claim for each insured.

Mean and variance. Here $q=1-p=0.5$, so

E[N]=\frac{2(0.5)}{0.5}=2,\qquad \operatorname{Var}(N)=\frac{2(0.5)}{(0.5)^2}=4.

Zero claims. At $k=0$, the coefficient is 1:

$$ P(N=0)=(0.5)^2=0.25. $$

One claim. The coefficient is $\binom{2}{1}=2$:

$$ P(N=1)=2(0.5)^2(0.5)=0.25. $$

Two claims. The coefficient is $\binom{3}{2}=3$:

$$ P(N=2)=3(0.5)^2(0.5)^2=0.1875. $$

The probabilities are nonnegative and each is at most 1. Their partial sum is $0.25+0.25+0.1875=0.6875$, which is less than 1 because counts greater than two remain possible. The variance of 4 exceeds the mean of 2, consistent with overdispersion.

Worked Example: Comparing Equal Means and Different Sizes

Two hypothetical portfolios have the same expected claim count of 3 per exposure period. Portfolio A uses $r=3,\ p=0.5$; Portfolio B uses $r=6,\ p=2/3$. Compare their variances and their probabilities of exactly three claims. The method is to verify the means first, then compare dispersion and substitute in the probability function. This avoids attributing a size effect to models whose means differ.

For Portfolio A, $q=0.5$, so

E[N_A]=\frac{3(0.5)}{0.5}=3,\qquad \operatorname{Var}(N_A)=\frac{3(0.5)}{(0.5)^2}=6.

For Portfolio B, $q=1/3$, so

E[N_B]=\frac{6(1/3)}{2/3}=3,\qquad \operatorname{Var}(N_B)=\frac{6(1/3)}{(2/3)^2}=4.5.

The means agree, while Portfolio B has the larger size and smaller variance. For exactly three claims in Portfolio A,

P(N_A=3)=\binom{5}{2}(0.5)^3(0.5)^3 =10(0.5)^6=0.15625.

For Portfolio B,

P(N_B=3)=\binom{8}{5}\left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^3 =\frac{3584}{19683}\approx 0.1821.

Thus the probability of exactly three claims is about 0.1563 for A and 0.1821 for B, to four decimal places. The probabilities need not move in the same direction at every individual count when dispersion changes; the variance comparison is the direct summary of spread.

Worked Example: A Multi-Count Probability

A hypothetical equipment-insurance block is modeled with $r=4$ and $p=0.4$. Find its mean, variance, and probability of no more than two claims during the modeled period. Use the probability function for each count and add the disjoint probabilities. This method avoids calculating only the zero-claim probability when a cumulative event is requested.

Here $q=0.6$, giving

E[N]=\frac{4(0.6)}{0.4}=6,\qquad \operatorname{Var}(N)=\frac{4(0.6)}{(0.4)^2}=15.

For zero claims,

$$ P(N=0)=(0.4)^4=0.0256. $$

For one claim, the coefficient is $\binom{4}{1}=4$:

$$ P(N=1)=4(0.4)^4(0.6)=4(0.0256)(0.6)=0.06144. $$

For two claims, the coefficient is $\binom{5}{2}=10$:

$$ P(N=2)=10(0.4)^4(0.6)^2 =10(0.0256)(0.36)=0.09216. $$

The events $N=0$, $N=1$, and $N=2$ are mutually exclusive, so

P(N\leq 2)=0.0256+0.06144+0.09216=0.1792.

The answer is 0.1792, or 0.1792 to four decimal places. As a check, the variance 15 is greater than the mean 6, as required by this model.

Common Parameter and Calculation Traps

A frequent error is to use the wrong exponent or interchange $p$ and $1-p$ in the probability function. For example, with $r=2,\ p=0.5$, an incorrect calculation of $P(N=1)$ might use $2p^2$, giving 0.5. The correct formula includes $p^r(1-p)^k$, so

P(N=1)=\binom{2}{1}(0.5)^2(0.5)^1=0.25.

The incorrect result is not the negative binomial probability at $k=1$; it omitted the factor $1-p$. When $p=0.5$, swapping $p$ and $1-p$ can be especially hard to notice, so check the exponent $r$ on $p$ and the exponent $k$ on $1-p$ explicitly.

Do not call $p$ the individual claim probability. It is a parameter in the count distribution's probability function.
Do not describe size as lowering variance in every comparison. At fixed $p$, both mean and variance increase with $r$; the variance decreases with size when the mean is held fixed.
Do not omit the coefficient. For $k>0$, the factor $\binom{k+r-1}{k}$, or its gamma-function counterpart, is generally not 1.
Check the parameterization. A negative binomial formula using different parameter definitions may assign different meanings to its symbols.

Key takeaway: Under the probability function used here, $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$. The variance exceeds the mean, and at a fixed mean, larger size $r$ means less extra dispersion.

Check Your Understanding

For each problem, use the probability function and moment formulas stated above. Probabilities in the answers are rounded to four decimal places where needed.

A hypothetical claim count has $r=2$ and $p=0.25$. Find its mean and variance. (Answer: mean $6$, variance $24$)
A count has $r=5$ and $p=0.5$. Calculate the probability of zero claims. (Answer: $0.0313$)
A count has $r=3$ and $p=0.6$. Calculate the probability of exactly one claim. (Answer: $0.2592$)
Two models both have mean 4. Model A has $r=2$; Model B has $r=8$. For each model, find its variance. (Answer: Model A variance $12$; Model B variance $6$)
A hypothetical count has $r=2,\ p=0.5$. Calculate the probability of at most one claim by summing the probabilities for zero and one claim. (Answer: $0.5000$)

What You'll Learn