Compound Negative Binomial Aggregate Probabilities by Recursion

Why Negative Binomial Frequency Changes Aggregate Probabilities

A compound aggregate loss is the sum of a random number of claim amounts. Even when two models have the same expected claim count and the same severity distribution, their aggregate probabilities can differ because the claim-count variances differ. Negative binomial frequency is useful when the count model allows more variability than a Poisson model.

For discrete severities on a common integer lattice, Panjer recursion converts the frequency model and severity probabilities into aggregate probabilities one lattice point at a time. The key is to identify the negative binomial parameterization correctly: different conventions attach different meanings to its parameters, and those differences change the recursion coefficients.

Definition: In this tutorial, $N$ is the number of claims and has a negative binomial distribution in the failures-before-$r$-successes convention: $$ P(N=n)=\binom{n+r-1}{n}p^r q^n,\qquad n=0,1,2,\ldots,\quad q=1-p, $$ where $r>0$ and $0<p<1$. Its mean is $E[N]=rq/p$, and its variance is $\operatorname{Var}(N)=rq/p^2$. The claim amounts $X_1,X_2,\ldots$ are assumed independent and identically distributed, independent of $N$, and supported on nonnegative integer multiples of a chosen unit. The aggregate is $S=\sum_{i=1}^N X_i$, with $S=0$ when $N=0$.

Write $f_j=P(X_i=j)$ for the severity probability at lattice point $j$, and $g_k=P(S=k)$. These probabilities describe amounts in the chosen lattice unit; if the unit is thousands of dollars, for example, $k=3$ represents an aggregate of $3,000. Panjer recursion applies when the claim count is in the $(a,b,0)$ class and the severity is discrete on that lattice. Here, the negative binomial coefficients are $a=q$ and $b=q(r-1)$.

Theorem: For the negative binomial model above, the initial aggregate probability is $$ g_0=\left(\frac{p}{1-qf_0}\right)^r. $$ For $k\geq 1$, the compound aggregate probabilities satisfy $$ g_k=\frac{1}{1-qf_0}\sum_{j=1}^{k}\left(q+\frac{q(r-1)j}{k}\right)f_jg_{k-j}. $$ If claim severities are strictly positive, then $f_0=0$, so $g_0=p^r$ and the denominator in the recursion is 1.

Why These Recursion Coefficients Work

The negative binomial probability ratio provides the first step. For $n\geq1$, $$ \frac{P(N=n)}{P(N=n-1)} =q\frac{n+r-1}{n} =q+\frac{q(r-1)}{n}. $$ Thus $P(N=n)=(a+b/n)P(N=n-1)$ with $a=q$ and $b=q(r-1)$. The corresponding Panjer recursion uses these same coefficients.

The recursion can also be justified directly from probability generating functions. Let $F(z)=\sum_{j\geq0}f_jz^j$ be the severity probability generating function and $H(z)=\sum_{k\geq0}g_kz^k$ the aggregate probability generating function. Independence and the negative binomial probability generating function give $$ H(z)=\left(\frac{p}{1-qF(z)}\right)^r. $$ Differentiating and rearranging yields $$ (1-qF(z))H'(z)=rqF'(z)H(z). $$ Equating the coefficients of $z^{k-1}$ gives $$ k(1-qf_0)g_k=\sum_{j=1}^{k}\bigl(q(k-j)+rqj\bigr)f_jg_{k-j}. $$ Dividing by $k(1-qf_0)$ produces the stated recursion. The zero-aggregate probability follows separately from $g_0=H(0)=E[f_0^N]$. This derivation also shows why the factor $1-qf_0$ is required when a claim can have zero severity.

Worked Aggregate Calculations

Worked Example: Unit Claims With Negative Binomial Frequency

A hypothetical coverage has claim count $N$ following the failures-before-$r$-successes negative binomial model with $r=2$ and $p=0.5$. Every claim is exactly one lattice unit, so $f_1=1$ and all other $f_j=0$. Calculate $g_0,g_1,g_2,g_3$, then check the result directly from the count distribution.

Method and parameters. Here $q=1-p=0.5$, $a=q=0.5$, $b=q(r-1)=0.5$, and $f_0=0$. The initial probability is $g_0=p^r=(0.5)^2=0.25$. Because every claim is one unit, an aggregate of $k$ units requires exactly $k$ claims.

For $k=1$, only $j=1$ contributes: $$ g_1=\left(0.5+\frac{0.5(1)}{1}\right)(1)(0.25)=1(0.25)=0.25. $$

For $k=2$, only $j=1$ contributes: $$ g_2=\left(0.5+\frac{0.5(1)}{2}\right)(1)(0.25) =0.75(0.25)=0.1875. $$

For $k=3$, again only $j=1$ contributes: $$ g_3=\left(0.5+\frac{0.5(1)}{3}\right)(1)(0.1875) =\frac{2}{3}(0.1875)=0.125. $$

Check. The negative binomial probabilities are $P(N=0)=0.25$, $P(N=1)=2(0.5)^2(0.5)=0.25$, $P(N=2)=3(0.5)^2(0.5)^2=0.1875$, and $P(N=3)=4(0.5)^2(0.5)^3=0.125$. They match the recursion because $S=N$. A common mistake is to divide the recursion sum by $k$ a second time; the $j/k$ term is already part of its coefficient.

Worked Example: Two-Point Severity and Multiple Recursion Terms

A hypothetical portfolio has $N$ following the negative binomial model with $r=2$, $p=0.5$, and $q=0.5$. Each claim is one lattice unit with probability $0.75$ or two units with probability $0.25$. Find $g_0$ through $g_4$.

Method and parameters. The severity probabilities are $f_1=0.75$, $f_2=0.25$, and $f_j=0$ for $j\geq3$. Since $f_0=0$, $g_0=(0.5)^2=0.25$. The recursion coefficients are $q=0.5$ and $q(r-1)=0.5$.

At $k=1$, only $j=1$ contributes: $$ g_1=(0.5+0.5)(0.75)(0.25)=0.1875. $$ At $k=2$, the $j=1$ and $j=2$ terms both contribute: $$ \begin{aligned} g_2 &=(0.5+0.5/2)(0.75)(0.1875) +(0.5+0.5\cdot2/2)(0.25)(0.25)\\ &=0.75(0.140625)+1(0.0625)\\ &=0.10546875+0.0625=0.16796875. \end{aligned} $$

At $k=3$, $j=1,2$ contribute; larger severities have probability zero: $$ \begin{aligned} g_3 &=(0.5+0.5/3)(0.75)(0.16796875) +(0.5+0.5\cdot2/3)(0.25)(0.1875)\\ &=\frac{2}{3}(0.1259765625)+\frac{5}{6}(0.046875)\\ &=0.083984375+0.0390625=0.123046875. \end{aligned} $$ At $k=4$: $$ \begin{aligned} g_4 &=(0.5+0.5/4)(0.75)(0.123046875) +(0.5+0.5\cdot2/4)(0.25)(0.16796875)\\ &=0.625(0.09228515625)+0.75(0.0419921875)\\ &=0.05767822265625+0.031494140625\\ &=0.08917236328125. \end{aligned} $$

Check and trap. For aggregate 2, direct conditioning gives a useful check: either there is one claim of size 2, with probability $0.25(0.25)=0.0625$, or there are two claims both of size 1, with probability $0.1875(0.75)^2=0.10546875$. The sum is $0.16796875=g_2$. Do not use just the one-claim contribution: different claim counts can produce the same aggregate amount.

Overdispersion at a Fixed Mean

For the failures-before-$r$-successes convention, the negative binomial count variance exceeds its mean whenever $0<p<1$: $$ \frac{\operatorname{Var}(N)}{E[N]}=\frac{rq/p^2}{rq/p}=\frac{1}{p}>1. $$ This is the frequency overdispersion relative to a Poisson count, whose variance equals its mean. It affects the shape of the aggregate distribution, not just its mean. In particular, more count variability can increase both the probability of no claims and the probability of some large counts; the exact aggregate effect also depends on the severity distribution.

Worked Example: Comparing Negative Binomial and Poisson Tails

A hypothetical portfolio has unit-sized claims, so its aggregate $S$ equals its claim count $N$. Compare $P(S\geq4)$ for a negative binomial model with $r=4$, $p=2/3$, and for a Poisson model with mean 2. Confirm that the two frequency models have the same mean and compare their variances.

Method and negative binomial parameters. Here $q=1/3$, so $E[N]=rq/p=4(1/3)/(2/3)=2$, and $\operatorname{Var}(N)=rq/p^2=4(1/3)/(4/9)=3$. The Panjer coefficients are $a=q=1/3$ and $b=q(r-1)=1$. Since severity is exactly one unit, $g_0=p^r=16/81$, and $$ g_k=\left(\frac{1}{3}+\frac{1}{k}\right)g_{k-1}. $$

The recursion gives $$ \begin{aligned} g_1&=\left(\frac{1}{3}+1\right)\frac{16}{81}=\frac{64}{243},\\ g_2&=\left(\frac{1}{3}+\frac{1}{2}\right)\frac{64}{243}=\frac{160}{729},\\ g_3&=\left(\frac{1}{3}+\frac{1}{3}\right)\frac{160}{729}=\frac{320}{2187}. \end{aligned} $$ Writing all four probabilities with denominator 2187 gives $g_0=432/2187$, $g_1=576/2187$, $g_2=480/2187$, and $g_3=320/2187$. Therefore $$ P(S\geq4)=1-\sum_{k=0}^3g_k =1-\frac{1808}{2187} =\frac{379}{2187} \approx 0.1733. $$

Poisson comparison. A Poisson count with mean 2 has variance 2, and $$ P(S\geq4)=1-e^{-2}\left(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}\right) =1-\frac{19}{3}e^{-2} \approx 0.1429. $$ The negative binomial model has the larger variance, 3 rather than 2, and the larger probability of at least four unit claims in this example. The method is to calculate the tail as the complement of the probabilities below 4. A common mistake is to compare models at different means; here both means are 2, so the contrast reflects their different frequency distributions.

Applying the Recursion Reliably

A calculation should begin by fixing the lattice unit and recording the severity probabilities at every relevant point. Then identify $r,p$ using the stated negative binomial convention, calculate $q=1-p$, and obtain $a=q$ and $b=q(r-1)$. Only after those steps should the recursion be applied. The following sequence keeps the setup and computation distinct.

Confirm the model and units.
Check that claims are independent and identically distributed, independent of $N$, and represented on a nonnegative integer lattice. Translate dollar amounts into the chosen unit before assigning the $f_j$.

Set the frequency coefficients.
For the convention used here, compute $q=1-p$, $a=q$, and $b=q(r-1)$. Do not substitute parameters from a negative binomial convention without first reconciling its definition.

Calculate the initial value and proceed upward.
Use $g_0=(p/(1-qf_0))^r$, then compute $g_1,g_2,\ldots$ in order, using only previously calculated aggregate probabilities.

Check the result.
For small targets, compare with direct conditioning on claim count. Also check that the calculated probabilities are nonnegative and that their partial sum does not exceed 1; a partial sum below 1 is expected when the remaining aggregate tail has not been calculated.

The method requires a discrete severity distribution on the specified lattice; it does not directly apply to an unmodified continuous severity distribution. A continuous model must first be represented on a lattice for this recursion. Likewise, the coefficients above depend on the stated negative binomial convention. If a problem specifies a different convention, derive or verify its probability ratio before using Panjer recursion.

Key takeaway: Under the failures-before-$r$-successes parameterization, negative binomial frequency uses $a=q$ and $b=q(r-1)$ in Panjer recursion. Correct parameterization, severity lattice, and initial value are essential; the resulting probabilities reflect the frequency model’s overdispersion as well as the severity distribution.

Check Your Understanding

For each problem, use the failures-before-$r$-successes convention and the stated lattice units.

A claim count has $r=1$, $p=0.4$, and every claim is one unit. Find $P(S=2)$. (Answer: 0.1440)
A claim count has $r=2$, $p=0.6$. Claim severity is 1 or 2 units, each with probability 0.5. Find $P(S=2)$, accounting for every possible claim count. (Answer: 0.1872)
A claim count has $r=3$, $p=0.5$, and every claim is one unit. Find $P(S\leq2)$. (Answer: 0.5000)
A claim count has $r=2$, $p=0.5$, and severity probabilities $f_1=f_2=0.5$. Use Panjer recursion to find $P(S=3)$. (Answer: 0.1094)
For unit-sized claims, compare negative binomial frequency with $r=4,p=2/3$ to Poisson frequency with mean 2. State which has the larger variance and give the negative binomial probability of zero aggregate. (Answer: Negative binomial variance 3 exceeds Poisson variance 2; $P(S=0)=16/81\approx0.1975$.)

What You'll Learn