The Panjer (a,b,0) Class Definition

Why the Recursion Matters

A frequency model describes the number of claims, not the amounts paid for those claims. Let $N$ be a nonnegative integer-valued claim count, and write $p_n=P(N=n)$. Many frequency models can be described by a rule connecting each probability $p_n$ to the preceding probability $p_{n-1}$. The Panjer $(a,b,0)$ class is defined by one such rule.

The definition is useful because, once $p_0$ and the two parameters are specified, the recursion generates the count probabilities in order. It also makes the support—the count values that can have positive probability—part of the calculation rather than an afterthought. This tutorial focuses on that definition and its direct consequences, aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: A nonnegative integer-valued random variable $N$ belongs to the Panjer $(a,b,0)$ class if its probabilities $p_n=P(N=n)$ satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\ge 1, $$ for constants $a$ and $b$, with $p_0>0$. The probabilities must also be valid probabilities: they are nonnegative and sum to one over the support of $N$.

Reading the Definition

The recursion relates adjacent probabilities. At count $n$, first calculate the multiplier $a+b/n$, then multiply the preceding probability $p_{n-1}$ by that multiplier. The index $n$ starts at 1: the first calculation is $p_1=(a+b)p_0$, not a calculation at $n=0$.

The parameters $a$ and $b$ are constants for the distribution. The term $a$ contributes the same amount to the multiplier at every step. The contribution $b/n$ depends on the current count and becomes smaller in magnitude as $n$ increases. Neither parameter, by itself, is a probability. A multiplier must be interpreted together with $p_{n-1}$; it is not required to be at most one, since a probability at one count can exceed the probability at the preceding count.

The starting probability $p_0$ anchors the sequence. The recursion determines later probabilities from it, but $p_0$ cannot be chosen independently of the requirement that the probabilities sum to one. In worked calculations, either a valid $p_0$ will be supplied or it will be found by summing the probabilities generated by the recursion.

The support is the set of count values with positive probability. It may be finite or unbounded. For a finite support, the recursion can reach a multiplier of zero at the first count beyond the largest possible count. All subsequent probabilities then remain zero. A proposed parameter pair and starting probability are not automatically a valid model: the resulting probabilities must be nonnegative and normalized.

Why the Recursion Determines the Sequence

The recursion is a step-by-step rule, so its implications can be seen by iteration. Starting at $p_0$, each new probability is obtained by multiplying the preceding one by the multiplier for that count. For every $n\ge 1$, this gives

p_n=p_0\prod_{j=1}^{n}\left(a+\frac{b}{j}\right).

This product follows by repeated substitution: $p_1=p_0(a+b)$; then $p_2=p_0(a+b)(a+b/2)$; each subsequent step appends the factor for its own index. It is therefore a consequence of the recursion, not an additional assumption. If one of the factors is zero, that probability and every later probability are zero. The product also makes clear that a negative factor before termination would produce a negative probability when the preceding probability is positive, which is not permissible for a probability model.

The product expression is useful for checking calculations, but the recursion is usually more efficient: calculate in order, keep exact fractions where possible, and check the final sequence against both the support and the total probability.

Worked Example: Finding a Finite-Support Distribution

A hypothetical claim count is specified to follow the $(a,b,0)$ recursion with $a=-\frac12$ and $b=2$. Its support is contained in $\{0,1,2,3\}$. Determine $p_0,p_1,p_2,p_3$, and calculate $P(N\ge 2)$.

Method. Generate each probability from its predecessor, use the stated support to find the zero tail, and normalize the resulting probabilities.

At $n=1$, the multiplier is $-\frac12+2=\frac32$, so $p_1=\frac32p_0$. At $n=2$, it is $-\frac12+1=\frac12$, so $p_2=\frac12p_1=\frac34p_0$. At $n=3$, it is $-\frac12+\frac23=\frac16$, so $p_3=\frac16p_2=\frac18p_0$.

At $n=4$, the multiplier is $-\frac12+\frac24=0$. Thus $p_4=0$, consistent with the stated support, and later probabilities remain zero. The total probability is

p_0+p_1+p_2+p_3 =p_0\left(1+\frac32+\frac34+\frac18\right) =p_0\frac{27}{8}=1.

Therefore $p_0=\frac{8}{27}$. Substitution gives $p_1=\frac{12}{27}=\frac49$, $p_2=\frac{6}{27}=\frac29$, and $p_3=\frac{1}{27}$. The requested probability is

P(N\ge 2)=p_2+p_3=\frac{6}{27}+\frac{1}{27}=\frac{7}{27}.

Check. The probabilities sum to $(8+12+6+1)/27=1$, and the next recursion multiplier is zero. A common mistake would be to stop after computing $p_3$ without checking the boundary or the total; here both checks agree with a valid distribution.

Interpreting the Support Carefully

A finite support does not mean that the recursion changes. The same formula applies at every positive integer $n$. Instead, the parameters can make the multiplier zero at the first count excluded from the support. Because the preceding probability is then multiplied by zero, the probability at that count is zero; applying the recursion again to a zero probability leaves the rest of the tail at zero.

In contrast, if no such zero factor occurs, the recursion does not itself announce a finite endpoint. The sequence may continue across all nonnegative integer counts. In either case, support and normalization are essential: calculating several terms does not establish that the sequence defines a probability distribution.

Worked Example: Using the Boundary to Check the Support

A hypothetical count model has $a=-\frac13$, $b=1$, and $p_0=\frac{9}{16}$. Find the positive count probabilities up to the first zero, verify the distribution, and calculate $P(N=2\mid N>0)$.

Method. Apply the recursion in increasing order, identify where the multiplier vanishes, then use the conditional-probability definition.

At $n=1$, the multiplier is $-\frac13+1=\frac23$. Thus $p_1=\frac23\cdot\frac{9}{16}=\frac{6}{16}=\frac38$. At $n=2$, the multiplier is $-\frac13+\frac12=\frac16$. Thus $p_2=\frac16\cdot\frac{6}{16}=\frac{1}{16}$. At $n=3$, the multiplier is $-\frac13+\frac13=0$, so $p_3=0$, and all later probabilities are zero.

The probabilities sum to $\frac{9}{16}+\frac{6}{16}+\frac{1}{16}=1$, so the support is $\{0,1,2\}$. Since $P(N>0)=p_1+p_2=\frac{7}{16}$,

P(N=2\mid N>0)=\frac{P(N=2)}{P(N>0)} =\frac{1/16}{7/16}=\frac17.

Check. The conditional probability is between zero and one, and the denominator includes every positive count in the support. A common mistake is to use $p_2$ alone as the conditional probability; conditioning on $N>0$ requires dividing by the probability of that condition.

What the Parameters Do—and Do Not—Tell You

The pair $(a,b)$ specifies the multiplier at each count, while $p_0$ gives the starting level. Together they determine the sequence through the recursion. To see the changing contributions, the first multipliers are $a+b$, $a+b/2$, and $a+b/3$. Thus the same $a$ and $b$ may produce different probabilities at different counts, even though the rule itself is unchanged.

The parameters alone should not be read as the probability of zero claims, the expected count, or a claim probability. Those are different quantities. For this definition, their direct role is to determine the successive probability multipliers. Similarly, a positive multiplier does not mean the probability itself is positive if the preceding probability is zero; the recursion must be read as an equation for probabilities, with the support kept in view.

When checking a proposed $(a,b,0)$ model, use three separate questions:

Does the stated sequence obey $p_n=(a+b/n)p_{n-1}$ at each positive count?
Are all probabilities nonnegative, including at the support boundary?
Do the probabilities over the full support sum to one?

These checks are distinct. A sequence can follow the recurrence for several counts yet fail at a stated boundary, or it can have nonnegative terms that do not sum to one. The recursion defines the relationship between neighboring probabilities; the probability axioms determine whether the resulting sequence is a distribution.

Worked Example: A Longer Finite Support

A hypothetical claim count has $a=-\frac14$, $b=\frac54$, and support contained in $\{0,1,2,3,4\}$. Determine the normalized probabilities and find $P(N\ge 3)$.

Method. Express each probability as a multiple of $p_0$, use the zero multiplier at the support boundary, and then normalize.

The multipliers at $n=1,2,3,4$ are respectively \[ -\frac14+\frac54=1,\quad -\frac14+\frac58=\frac38,\quad -\frac14+\frac{5}{12}=\frac16,\quad -\frac14+\frac{5}{16}=\frac{1}{16}. \] It follows successively that $p_1=p_0$, $p_2=\frac38p_0$, $p_3=\frac{1}{16}p_0$, and $p_4=\frac{1}{256}p_0$. At $n=5$, the multiplier is $-\frac14+\frac{5}{20}=0$, so the tail is zero.

Normalize by summing all five probabilities:

1=p_0\left(1+1+\frac38+\frac{1}{16}+\frac{1}{256}\right) =p_0\frac{625}{256}.

Thus $p_0=\frac{256}{625}$. The probabilities, in order from zero through four, are $p_0=\frac{256}{625}$, $p_1=\frac{256}{625}$, $p_2=\frac{96}{625}$, $p_3=\frac{16}{625}$, and $p_4=\frac{1}{625}$. Therefore

P(N\ge 3)=p_3+p_4=\frac{16}{625}+\frac{1}{625}=\frac{17}{625}.

Check. The numerators sum to $256+256+96+16+1=625$, and the multiplier at $n=5$ is zero. A common mistake is to treat the support endpoint 4 as the count at which the probability must become zero. Here the probability at 4 is positive; termination occurs at count 5.

Common Errors

Dropping the divisor $n$. The multiplier is $a+b/n$, not $a+b$ at every step. The two agree at $n=1$ but generally differ thereafter.
Starting at the wrong index. The rule begins at $n=1$, so it generates $p_1$ from $p_0$. It is not a formula for $p_0$.
Assuming every parameter pair is valid. The recursion must produce nonnegative probabilities that sum to one. A parameter pair by itself does not certify a distribution.
Ignoring the tail. For a finite support, check the first count beyond the endpoint. A zero probability there should be consistent with the recursion.
Confusing a multiplier with a probability. The quantity $a+b/n$ multiplies $p_{n-1}$; it is not $p_n$ and need not lie between zero and one.

Key takeaway: The Panjer $(a,b,0)$ class is defined by $p_n=(a+b/n)p_{n-1}$ for every $n\ge1$. Start with $p_0$, apply the count-specific multiplier, track the support, and verify nonnegativity and total probability before accepting the sequence as a frequency distribution.

Check Your Understanding

For each problem, use the recursion at the stated count and keep the support in view.

A model has $a=-\frac12$, $b=2$, and $p_0=\frac{8}{27}$. Find $p_2$. (Answer: $\frac{2}{9}$)
For the model in Question 1, the support is $\{0,1,2,3\}$. Find $P(N\ge2)$. (Answer: $\frac{7}{27}$)
A model has $a=-\frac13$, $b=1$, and $p_0=\frac{9}{16}$. Find $P(N=2\mid N>0)$, using the recursion to establish the positive-count probabilities. (Answer: $\frac17$)
A model has $a=-\frac14$, $b=\frac54$, and $p_0=\frac{256}{625}$. Find the probability of a count of at least 3, using the recursion through the boundary. (Answer: $\frac{17}{625}$)
A proposed model has $a=-\frac25$, $b=2$, and support $\{0,1,2,3,4\}$. At what count does the recursion first give a zero multiplier, and what does that imply for probabilities from that count onward? (Answer: At $n=5$, since $-\frac25+\frac25=0$; $p_5=0$ and every later probability is zero.)

What You'll Learn