Compound Negative Binomial Aggregates

Why Negative Binomial Frequency Changes the Aggregate

An aggregate loss is a random sum: the number of claims is uncertain, and the payment on each claim is uncertain. When claim counts are more variable than a Poisson count with the same mean, a negative binomial model can represent that extra frequency variation. Its impact does not stop at the count distribution: variability in the number of claims also contributes to variability in total loss.

This tutorial combines a negative binomial count with independent, identically distributed discrete claim severities. It develops the aggregate probability generating function and moments, with attention to parameterization. The calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N$ be the number of failures before the $r$th success in independent Bernoulli trials, where each trial has success probability $p$, and let $q=1-p$. Then $N$ has a negative binomial distribution with parameters $r$ and $p$: $$ P(N=n)=\binom{n+r-1}{n}p^r q^n,\qquad n=0,1,2,\ldots $$ Under this convention, $E[N]=rq/p$, $\operatorname{Var}(N)=rq/p^2$, and its probability generating function is $G_N(z)=[p/(1-qz)]^r$. Here $r$ is positive and $0<p\leq 1$.

The parameterization matters. In this tutorial, $p$ is the success probability and $q=1-p$ is the failure probability. Some references instead parameterize the negative binomial using a mean, or describe $p$ as a failure probability. Before using a formula, match its parameters to the stated count model.

Composing Frequency and Severity

Let $Y_1,Y_2,\ldots$ be independent and identically distributed nonnegative integer-valued severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}Y_j$, with $S=0$ if $N=0$. A discrete severity distribution is needed for the probability generating function in this form. The moment results below also hold for nonnegative severities with finite second moments, even when severity is not integer-valued.

Definition: For a nonnegative integer-valued random variable $X$, its probability generating function is $G_X(z)=E[z^X]$. If $G_Y(z)$ is the severity probability generating function and $N$ has the negative binomial parameterization above, then the compound aggregate probability generating function is $$ G_S(z)=G_N(G_Y(z))=\left[\frac{p}{1-qG_Y(z)}\right]^r. $$

This composition follows by conditioning on the claim count. Given $N=n$, the aggregate is the sum of $n$ independent severities, so its generating function is $G_Y(z)^n$. Averaging that conditional generating function over all possible values of $N$ gives $E[G_Y(z)^N]=G_N(G_Y(z))$. Substituting the negative binomial generating function gives the displayed result.

The generating function describes the entire aggregate distribution, not just its moments. For example, its coefficient of $z^k$ is $P(S=k)$. Computing those coefficients is a separate task; the focus here is on constructing the function and extracting its mean and variance.

Aggregate Mean and Variance

Assume the severity has finite mean and variance. Write $m_1=E[Y]$ and $m_2=E[Y^2]$, where $Y$ denotes one claim severity. Conditional on $N$, the expected aggregate is $N m_1$, and its conditional variance is $N\operatorname{Var}(Y)$, because the severities are independent.

Theorem: For a random sum $S=\sum_{j=1}^{N}Y_j$ with count independent of independent, identically distributed severities, $$ E[S]=E[N]E[Y], $$ $$ \operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2. $$ For the negative binomial count defined above, these become $$ E[S]=\frac{rq}{p}m_1,\qquad \operatorname{Var}(S)=\frac{rq}{p}(m_2-m_1^2)+\frac{rq}{p^2}m_1^2. $$

The variance formula is an application of the law of total variance. It separates variation among severities, given the number of claims, from variation caused by the random number of claims.

Condition on the count.
Given $N=n$, independence gives $E[S\mid N=n]=nm_1$ and $\operatorname{Var}(S\mid N=n)=n\operatorname{Var}(Y)$.

Average the conditional mean.
$E[S]=E[E[S\mid N]]=E[Nm_1]=E[N]m_1$.

Apply total variance.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)m_1^2$.

Substitute the negative binomial moments.
Use $E[N]=rq/p$, $\operatorname{Var}(N)=rq/p^2$, and $\operatorname{Var}(Y)=m_2-m_1^2$.

An equivalent expression is useful as a check. Since $m_2=\operatorname{Var}(Y)+m_1^2$, the general formula can be rearranged to $$ \operatorname{Var}(S)=E[N]m_2+\bigl(\operatorname{Var}(N)-E[N]\bigr)m_1^2. $$ For the negative binomial model, $\operatorname{Var}(N)-E[N]=rq^2/p^2$, so another equivalent form is $$ \operatorname{Var}(S)=\frac{rq}{p}m_2+\frac{rq^2}{p^2}m_1^2. $$ These expressions are algebraically identical; using either correctly should give the same result.

Worked Applications

Worked Example: Construct the Aggregate Function and Moments

A hypothetical coverage has a negative binomial claim count with $r=2$ and success probability $p=0.4$. Each claim severity is 1 loss unit with probability 0.25 and 3 loss units with probability 0.75. Find the aggregate probability generating function, mean, and variance.

Method and count moments. Under the failures-before-successes convention, $q=1-0.4=0.6$. Therefore, $E[N]=2(0.6)/0.4=3$, and $\operatorname{Var}(N)=2(0.6)/(0.4)^2=1.2/0.16=7.5$.

Severity function and moments. The severity generating function is $G_Y(z)=0.25z+0.75z^3$. The first moment is $E[Y]=0.25(1)+0.75(3)=2.5$. The second moment is $E[Y^2]=0.25(1^2)+0.75(3^2)=0.25+6.75=7$, so $\operatorname{Var}(Y)=7-(2.5)^2=7-6.25=0.75$.

Aggregate generating function. Substitute the severity function into the count generating function: $$ G_S(z)=\left[\frac{0.4}{1-0.6(0.25z+0.75z^3)}\right]^2 =\left[\frac{0.4}{1-0.15z-0.45z^3}\right]^2. $$

Aggregate moments. The mean is $E[S]=3(2.5)=7.5$ loss units. The variance is $$ \operatorname{Var}(S)=3(0.75)+7.5(2.5)^2 =2.25+7.5(6.25) =2.25+46.875 =49.125 $$ squared loss units.

Check and common mistake avoided. The equivalent second-moment calculation gives $3(7)+(7.5-3)(2.5)^2=21+4.5(6.25)=49.125$, confirming the variance. Multiplying the frequency variance by the severity variance alone would give $7.5(0.75)=5.625$, which is not the aggregate variance: that calculation omits both the conditional severity contribution and the effect of variation in the count on the aggregate mean.

Worked Example: A Different Severity Distribution

A hypothetical portfolio has $N$ negative binomial with $r=3$ and $p=0.6$. Each claim pays 2 loss units with probability 0.7 and 5 loss units with probability 0.3. Find the aggregate generating function, mean, and variance.

Count moments. Here $q=0.4$, so $E[N]=3(0.4)/0.6=2$, and $\operatorname{Var}(N)=3(0.4)/(0.6)^2=1.2/0.36=10/3$.

Severity moments. The severity generating function is $G_Y(z)=0.7z^2+0.3z^5$. Its mean and second moment are $E[Y]=0.7(2)+0.3(5)=1.4+1.5=2.9$, and $E[Y^2]=0.7(4)+0.3(25)=2.8+7.5=10.3$. Thus $\operatorname{Var}(Y)=10.3-(2.9)^2=10.3-8.41=1.89$.

Aggregate generating function. $$ G_S(z)=\left[\frac{0.6}{1-0.4(0.7z^2+0.3z^5)}\right]^3 =\left[\frac{0.6}{1-0.28z^2-0.12z^5}\right]^3. $$

Aggregate moments. The mean is $2(2.9)=5.8$ loss units. The variance is $$ 2(1.89)+\frac{10}{3}(2.9)^2 =3.78+\frac{10}{3}(8.41) =3.78+28.033333\ldots =31.8133 $$ squared loss units, rounded to four decimal places.

Check and common mistake avoided. Using the equivalent form gives $2(10.3)+(10/3-2)(8.41)=20.6+(4/3)(8.41)=31.8133$ squared loss units. The frequency variance is $10/3$, not the frequency mean $2$; substituting the mean for the variance in the second term would understate the aggregate variance.

Worked Example: Recover Frequency Parameters From Moments

A hypothetical negative binomial claim count has mean 6 and variance 18. Individual claim severity is 1 loss unit or 3 loss units, each with probability 0.5. Recover $r$ and $p$, then find the aggregate mean, variance, and generating function.

Recover the count parameters. For this negative binomial convention, $\operatorname{Var}(N)-E[N]=rq^2/p^2$ and $E[N]=rq/p$. Dividing the square of the mean by the variance-minus-mean gives $r$: $$ r=\frac{(E[N])^2}{\operatorname{Var}(N)-E[N]} =\frac{6^2}{18-6} =\frac{36}{12}=3. $$ Since $E[N]=rq/p$ and $q=1-p$, equivalently $E[N]=r(1-p)/p$. Hence $6=3(1-p)/p$, so $2p=1-p$, $3p=1$, and $p=1/3$. Thus $q=2/3$. Substitution checks the moments: $3(2/3)/(1/3)=6$ and $3(2/3)/(1/3)^2=18$.

Severity and aggregate moments. The severity generating function is $G_Y(z)=0.5z+0.5z^3$. Its mean is $0.5(1)+0.5(3)=2$, its second moment is $0.5(1)+0.5(9)=5$, and its variance is $5-2^2=1$. The aggregate mean is $6(2)=12$ loss units, and its variance is $$ 6(1)+18(2)^2=6+72=78 $$ squared loss units.

Aggregate generating function and check. $$ G_S(z)=\left[\frac{1/3}{1-(2/3)(0.5z+0.5z^3)}\right]^3 =\left[\frac{1/3}{1-z/3-z^3/3}\right]^3. $$ The moment check using the second-moment form is $6(5)+(18-6)(2)^2=30+12(4)=78$. A common error is to treat the reported count mean and variance as though they uniquely were the negative binomial parameters $r$ and $p$; they are moments, so first solve for the parameters using the model equations.

Exam Traps and Model Checks

The most important check is that the count convention, severity moments, and aggregate formula agree. Negative binomial parameterizations differ, and a correct formula paired with the wrong meaning of $p$ can produce incorrect frequency moments. A quick count check is to verify that $E[N]=rq/p$ and $\operatorname{Var}(N)=rq/p^2$ reproduce the stated model.

Confusing a second moment with a variance. The term $E[Y^2]$ is not $\operatorname{Var}(Y)$. They are related by $E[Y^2]=\operatorname{Var}(Y)+(E[Y])^2$.
Leaving out count variability. The full aggregate variance includes $\operatorname{Var}(N)(E[Y])^2$. The formula $E[N]\operatorname{Var}(Y)$ alone is only the expected conditional variance, not the aggregate variance.
Using severity probabilities as count probabilities. First calculate the count moments from $r,p$, then calculate the severity moments from the severity distribution.
Forgetting the independence assumptions. The composition and moment formulas here require the count to be independent of the severities, with severities independent and identically distributed. If severity depends on claim order or on $N$, these formulas do not automatically apply.

Key takeaway: For a negative binomial count with $E[N]=rq/p$ and $\operatorname{Var}(N)=rq/p^2$, the compound generating function is $[p/(1-qG_Y(z))]^r$. The aggregate mean is $E[N]E[Y]$, while its variance includes both $E[N]\operatorname{Var}(Y)$ and $\operatorname{Var}(N)(E[Y])^2$.

Check Your Understanding

For each problem, use the failures-before-successes convention and assume the count and severities are independent.

A negative binomial count has $r=2$ and $p=0.25$. Every claim pays exactly 2 loss units. Find the aggregate mean and variance. (Answer: mean 12 loss units; variance 96 squared loss units.)
A count has $r=4$ and $p=0.5$. Severity is 1 loss unit with probability 0.4 and 2 loss units with probability 0.6. Find the aggregate mean and variance. (Answer: mean 6.4 loss units; variance 21.44 squared loss units.)
A negative binomial count has mean 8 and variance 24. Severity is 1 loss unit with probability 0.75 and 4 loss units with probability 0.25. Find $r,p$, then the aggregate mean and variance. (Answer: $r=4,\ p=1/3$; mean 14 loss units; variance 87 squared loss units.)
A count has $r=1$, $p=0.5$, and severity is 1 or 2 loss units, each with probability 0.5. Write the aggregate probability generating function and find its mean and variance. (Answer: $G_S(z)=0.5/(1-0.25z-0.25z^2)$; mean 1.5 loss units; variance 4.75 squared loss units.)

What You'll Learn