Panjer Frequency Classes And Recursion Eligibility

Why Frequency-Class Eligibility Matters

For a compound aggregate, the claim count determines how many severity amounts contribute to the total. Some count distributions have a useful structure: each probability can be related to the preceding one by a simple expression in the count value. Recognizing that structure identifies the Panjer $(a,b,0)$ frequency class, an important eligibility check before applying aggregate calculation methods.

This tutorial focuses on the frequency distribution. It does not require a particular severity distribution, and eligibility is a property of the claim count alone. A distribution’s mean and variance, by themselves, do not establish that it belongs to this class.

Definition: Let $N$ be a nonnegative integer-valued claim count, with $p_n=P(N=n)$. The distribution belongs to the Panjer $(a,b,0)$ class if constants $a$ and $b$, independent of $n$, satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 1. $$ When $p_{n-1}>0$, this can be written as $p_n/p_{n-1}=a+b/n$. For a finite-support distribution, the recurrence equation remains the appropriate statement after probabilities become zero; a ratio with zero denominator is not defined.

The notation $(a,b,0)$ refers to this probability recurrence, with the initial probability $p_0$ supplied by the frequency model. The parameters $a$ and $b$ must be fixed across the count values; they cannot be chosen separately for each $n$. In addition, the resulting probabilities must be valid probabilities: nonnegative and summing to one.

The condition is useful because it compresses the relationship among all the count probabilities into two constants and an initial value. For aggregate calculations, that relationship can be exploited when the frequency model is eligible. The aggregate recursion itself is a separate result; here the task is to establish whether the count model has the required form.

The Three Standard Frequency Models

Poisson, binomial, and negative binomial counts all satisfy the $(a,b,0)$ recurrence under their usual nondegenerate parameterizations. The values of $a$ and $b$ differ, so it is important to derive them from the model’s probability formula rather than memorize a ratio without checking the parameter convention.

Poisson model: If $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, then $p_n=e^{-\lambda}\lambda^n/n!$, and $$ \frac{p_n}{p_{n-1}}=\frac{\lambda}{n}. $$ Thus $a=0$ and $b=\lambda$.

Binomial model: If $N\sim\operatorname{Binomial}(m,\theta)$, where $m$ is a positive integer and $0<\theta<1$, then $$ \frac{p_n}{p_{n-1}}=\frac{m-n+1}{n}\frac{\theta}{1-\theta} =-\frac{\theta}{1-\theta} +\frac{(m+1)\theta}{(1-\theta)n},\qquad 1\leq n\leq m. $$ Thus $a=-\theta/(1-\theta)$ and $b=(m+1)\theta/(1-\theta)$. The support is $\{0,1,\ldots,m\}$; in particular, $p_m$ can be positive. The recurrence gives $p_{m+1}=0$, and probabilities are zero beyond $m$.

Negative binomial model: Let $N$ count failures before the $r$th success, with success probability $\pi$, where $r>0$ and $0<\pi<1$. For integer $r$, this is the usual failures-before-$r$-successes model. Its probabilities satisfy $$ \frac{p_n}{p_{n-1}}=(1-\pi)\frac{n+r-1}{n} =(1-\pi)+\frac{(r-1)(1-\pi)}{n}. $$ Thus $a=1-\pi$ and $b=(r-1)(1-\pi)$. Its initial probability is $p_0=\pi^r$.

Why the Probability Ratios Have This Form

Each result follows by dividing consecutive probabilities. That division cancels the common factors in the probability formula and leaves a constant plus a constant divided by $n$. For example, for the binomial model,

p_n={m\choose n}\theta^n(1-\theta)^{m-n}, \qquad p_{n-1}={m\choose n-1}\theta^{n-1}(1-\theta)^{m-n+1}.

Taking the ratio cancels the powers common to the two expressions. The binomial-coefficient ratio is ${m\choose n}/{m\choose n-1}=(m-n+1)/n$, so

\frac{p_n}{p_{n-1}} =\frac{m-n+1}{n}\frac{\theta}{1-\theta} =-\frac{\theta}{1-\theta} +\frac{(m+1)\theta}{(1-\theta)n}.

This is exactly the required form. The same cancellation explains the Poisson and negative binomial cases: their consecutive-probability ratios reduce, respectively, to $\lambda/n$ and $(1-\pi)(n+r-1)/n$. This argument also shows why the parameters must be constant: the remaining expression must have the same $a$ and $b$ at every count value where the ratio is defined.

Applying the Eligibility Check

Worked Example: Checking a Poisson Frequency Model

A hypothetical equipment portfolio models the number of covered breakdown claims in one year as Poisson with mean $\lambda=2.4$. Determine whether the count model is eligible for the $(a,b,0)$ class, identify its parameters, and check the first probabilities. Use probabilities rounded to four decimal places in the final values.

Method. Use the Poisson probability formula and compare consecutive probabilities. The assumed frequency distribution is $N\sim\operatorname{Poisson}(2.4)$; no severity or interest assumption is needed for this frequency-class check.

First, $p_0=e^{-2.4}$, so $p_0\approx0.0907$. For $n\geq1$,

\frac{p_n}{p_{n-1}} =\frac{e^{-2.4}(2.4)^n/n!}{e^{-2.4}(2.4)^{n-1}/(n-1)!} =\frac{2.4}{n}.

This has the required form with $a=0$ and $b=2.4$. At $n=1$, $p_1=2.4p_0\approx0.2177$. At $n=2$, $p_2=(2.4/2)p_1=1.2p_1\approx0.2613$. The ratios are $p_1/p_0=2.4$ and $p_2/p_1=1.2$, matching $2.4/n$ at each count.

Check. Both ratios use the same $a=0$ and $b=2.4$, with only the $1/n$ term changing. The model is eligible. This avoids the mistake of treating the changing numerical ratio itself as evidence that the distribution is ineligible.

Worked Example: Checking a Binomial Model at Its Endpoint

A hypothetical annual policy covers at most five scheduled inspections, and each inspection independently produces a reportable event with probability $\theta=0.3$. The count is $N\sim\operatorname{Binomial}(5,0.3)$. Find the $(a,b,0)$ parameters, check the probability at the endpoint, and determine what happens beyond the endpoint. Round final probabilities to four decimal places.

Method. Apply the binomial consecutive-probability ratio with $m=5$. The model parameters are $m=5$ and $\theta=0.3$.

The constants are

a=-\frac{0.3}{0.7}=-\frac{3}{7}, \qquad b=\frac{(5+1)(0.3)}{0.7}=\frac{18}{7}.

At $n=1$, the recurrence multiplier is $-3/7+18/7=15/7$, which agrees with $p_1/p_0=5(0.3)/0.7=15/7$. At the endpoint $n=5$, the multiplier is

a+\frac{b}{5} =-\frac{3}{7}+\frac{18}{35} =\frac{3}{35}.

The endpoint probability is $p_5=(0.3)^5=0.00243$, or $0.0024$ to four decimal places. Also, $p_4=5(0.3)^4(0.7)=0.02835$, and $p_5/p_4=0.00243/0.02835=3/35$, as required.

At $n=6$, the multiplier is $-3/7+(18/7)/6=0$, so the recurrence gives $p_6=0$. For $n>6$, $p_{n-1}=0$, and the recurrence equation still gives $p_n=(a+b/n)0=0$. The model is eligible.

Check. The support endpoint $5$ has positive probability; zeros begin at $6$, beyond the endpoint. Setting $p_5=0$ would incorrectly remove a valid endpoint outcome and contradict the binomial probability formula.

Worked Example: Checking a Negative Binomial Parameterization

A hypothetical roadside-assistance model counts the number of unsuccessful calls before the third successful resolution. Each call succeeds independently with probability $\pi=0.6$. Determine the frequency class and parameters, then calculate $p_0$ through $p_3$ to check the recurrence. Round final probabilities to four decimal places.

Method. Use the failures-before-$r$-successes convention, with $r=3$, $\pi=0.6$, and $1-\pi=0.4$. This convention matters: $N=0$ means the first three calls all succeed.

The parameters are

a=1-\pi=0.4,\qquad b=(r-1)(1-\pi)=2(0.4)=0.8.

The initial probability is $p_0=\pi^r=(0.6)^3=0.2160$. The recurrence multipliers for $n=1,2,3$ are, respectively,

0.4+\frac{0.8}{1}=1.2,\qquad 0.4+\frac{0.8}{2}=0.8,\qquad 0.4+\frac{0.8}{3}=\frac{2}{3}.

Therefore $p_1=1.2(0.216)=0.2592$, $p_2=0.8(0.2592)=0.20736$, and $p_3=(2/3)(0.20736)=0.13824$. Rounded to four decimal places, these are $0.2592$, $0.2074$, and $0.1382$. The direct probability formula gives the same values: $p_n={n+2\choose n}(0.6)^3(0.4)^n$.

Check. The consecutive ratios are $1.2$, $0.8$, and $2/3$, agreeing with $0.4+0.8/n$. The model is eligible. Using $\pi$ as the probability of a failure instead would describe a different parameterization and produce incorrect $a$, $b$, and $p_0$.

Common Eligibility Traps

Using a different $a$ or $b$ at each count. The class requires constants that work throughout the distribution, not a separate fit to each pair of probabilities.
Confusing the endpoint with values beyond it. For support $\{0,\ldots,m\}$, $p_m$ may be positive. The probabilities that must be zero are $p_n$ for $n>m$.
Dividing by zero after a finite-support model terminates. Once $p_{n-1}=0$, the ratio $p_n/p_{n-1}$ is undefined. Check the recurrence equation $p_n=(a+b/n)p_{n-1}$ instead.
Changing the negative binomial convention without changing the formulas. Here $N$ counts failures before $r$ successes, so $p_0=\pi^r$ and $a=1-\pi$.
Assuming matching moments proves eligibility. The class is defined by the probability recurrence, not by a particular mean or variance.

Key takeaway: To verify $(a,b,0)$ eligibility, write the count probabilities, form the consecutive-probability ratio wherever its denominator is positive, and check whether it equals $a+b/n$ for fixed constants. For finite support, separately verify the recurrence at the endpoint and the zero probabilities beyond it.

Check Your Understanding

For each hypothetical frequency model, identify or test the $(a,b,0)$ form. Answers are provided after each question for checking.

A portfolio has $N\sim\operatorname{Poisson}(4.2)$. State $a$ and $b$, and give $p_1/p_0$. (Answer: $a=0$, $b=4.2$, and $p_1/p_0=4.2$.)
A count is $N\sim\operatorname{Binomial}(7,0.25)$. Find $a$, $b$, and the recurrence multiplier at $n=8$. (Answer: $a=-1/3$, $b=8/3$, and $a+b/8=0$, so $p_8=0$.)
A count $N$ records failures before four successes, with success probability $\pi=0.7$. Find $p_0$, $a$, and $b$, and calculate the recurrence multiplier at $n=3$. (Answer: $p_0=0.7^4=0.2401$, $a=0.3$, $b=0.9$, and $a+b/3=0.6$.)
A proposed count model has $p_n=(1/2)^{n+1}$ for every $n\geq0$. Check whether it is in the class and state its parameters. (Answer: Yes. $p_n/p_{n-1}=1/2$, so $a=1/2$ and $b=0$.)
A proposed model has $p_n/p_{n-1}=0.2+1.2/n$ for every $n\geq1$, with $p_0=0.8^7$. Identify a standard frequency model and verify the parameters. (Answer: It is negative binomial with $r=7$, success probability $\pi=0.8$, $a=0.2$, and $b=1.2$; its initial probability is $p_0=\pi^r=0.8^7$.)

What You'll Learn