Panjer Recursion for (a,b,0) Claim Counts

Why the Frequency Recurrence Matters

For a compound aggregate loss, the number of claims determines how many individual losses are added. When the claim count has a particular probability recurrence, the aggregate probabilities can be calculated one value at a time rather than by separately summing over every possible claim count. The $(a,b,0)$ class provides this structure for several standard frequency models.

The method requires a discrete severity distribution on a positive-integer lattice. If one lattice unit represents $100, for example, a severity of two units is a payment of $200. The recursion calculates probabilities for the total number of units; the chosen unit converts those totals back to currency.

Definition: A claim count $N$ belongs to the $(a,b,0)$ class if its probabilities $p_n=P(N=n)$ satisfy $p_n=(a+b/n)p_{n-1}$ for $n\geq 1$, with $p_0>0$. The constants $a$ and $b$ do not depend on $n$. For a finite-support count, the recurrence also gives zero probabilities after the upper endpoint.

Let $Y_1,Y_2,\ldots$ be independent claim severities, identically distributed and independent of $N$. Assume each severity is a strictly positive integer number of lattice units, with $f_j=P(Y=j)$ for $j\geq1$. Define the aggregate $S=\sum_{r=1}^{N}Y_r$, taking $S=0$ when $N=0$, and write $g_s=P(S=s)$. The initial probability is $g_0=p_0$: a positive total cannot arise from one or more strictly positive claims.

Recognizing the Frequency Parameters

The parameter $a$ is not necessarily positive. In particular, the binomial distribution has a negative $a$. Identify parameters by matching the ratio $p_n/p_{n-1}$ with $a+b/n$, and check that the resulting $p_0$ is the probability of no claims.

Claim count model	Parameter convention	$a$	$b$	$p_0$
Poisson	Mean $\lambda$	0	$\lambda$	$e^{-\lambda}$
Binomial	Number of trials $m$, claim probability $q$	$-q/(1-q)$	$(m+1)q/(1-q)$	$(1-q)^m$
Negative binomial	$P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$	$1-p$	$(r-1)(1-p)$	$p^r$

For the negative binomial row, $r$ is a positive integer and $p$ is the success probability in the stated probability function. This convention counts failures before $r$ successes. Other negative binomial parameterizations are possible, so matching the probability function matters.

Theorem: If $N$ is in the $(a,b,0)$ class and severities are independent, identically distributed positive-integer amounts independent of $N$, then for $s\geq1$, $g_s=\sum_{j=1}^{s}(a+bj/s)f_jg_{s-j}$, with initial value $g_0=p_0$.

Why the Aggregate Recursion Holds

A short probability-generating-function argument explains the weights $a+bj/s$. Let $P_N(t)=E[t^N]$ be the count probability-generating function and $F(z)=\sum_{j\geq1}f_jz^j$ the severity probability-generating function. The count recurrence implies \[ (1-at)P_N'(t)=(a+b)P_N(t). \] Indeed, the coefficient of $t^{n-1}$ on the left is $np_n-a(n-1)p_{n-1}$, which equals $(a+b)p_{n-1}$ by $np_n=(an+b)p_{n-1}$.

The aggregate generating function is $G(z)=P_N(F(z))=\sum_{s\geq0}g_sz^s$. Applying the chain rule gives $(1-aF(z))G'(z)=(a+b)F'(z)G(z)$. For $s\geq1$, compare coefficients of $z^{s-1}$. The coefficient on the left is $sg_s-a\sum_{j=1}^{s-1}(s-j)f_jg_{s-j}$, while the right-hand coefficient is $(a+b)\sum_{j=1}^{s}j f_jg_{s-j}$. Rearranging and dividing by $s$ yields \[ g_s=\sum_{j=1}^{s}\left(a+\frac{bj}{s}\right)f_jg_{s-j}. \] For $j<s$, the combined coefficient is $\{a(s-j)+(a+b)j\}/s=a+bj/s$; for $j=s$, it is $a+b$, also as in the formula. This establishes the recursion.

Applying the Recursion

At each target total $s$, include every severity size $j$ from 1 through $s$ that has positive probability. Each term uses the severity probability $f_j$, the previously calculated aggregate probability $g_{s-j}$, and the weight $a+bj/s$. The values $g_{s-j}$ are already available if totals are calculated in increasing order.

Worked Example: Poisson Counts With Two Possible Claim Sizes

Problem. A hypothetical portfolio has a Poisson claim count with mean $\lambda=1$. Each claim independently costs either one or two units, each with probability $0.5$, and the claims are independent of the count. Find $P(S=3)$. One unit represents $100.

Method and parameters. The count is Poisson, so $a=0$, $b=1$, and $g_0=p_0=e^{-1}=0.367879\ldots$. The severity probabilities are $f_1=f_2=0.5$, with all other $f_j=0$. Calculate the preceding aggregate probabilities needed for $g_3$.

For $s=1$, only $j=1$ contributes: $g_1=(0+1)f_1g_0=0.5e^{-1}=0.183940$.

For $s=2$, the two terms are $g_2=(1/2)(0.5)g_1+(2/2)(0.5)g_0=0.25g_1+0.5g_0$. Substituting gives $g_2=0.25(0.5e^{-1})+0.5e^{-1}=0.625e^{-1}=0.229925$.

For $s=3$, only $j=1,2$ contribute: $g_3=(1/3)(0.5)g_2+(2/3)(0.5)g_1=(1/6)g_2+(1/3)g_1$. Thus $g_3=(1/6)(0.625e^{-1})+(1/3)(0.5e^{-1})=(13/48)e^{-1}=0.099634$. The result is $P(S=3)=0.0996$, or an aggregate payment of $300 with probability 0.0996, rounded to four decimals.

Check and trap avoided. The severity-two term uses the weight $2/3$, not $1/3$, because its lattice amount is $j=2$. The contributing count-and-severity combinations for a total of three are two claims of sizes one and two, or three claims all of size one; both are represented in the recursion.

Worked Example: A Binomial Count and a Finite Aggregate Support

Problem. A hypothetical inspection program has $N\sim\operatorname{Binomial}(2,0.5)$. Each inspection generates an independent payment of one or two units, each with probability $0.5$, if it generates a claim. Find $P(S=3)$, and calculate the aggregate probabilities through the maximum possible total.

Method and parameters. Here $m=2$ and $q=0.5$. Therefore $a=-0.5/0.5=-1$, $b=3(0.5)/0.5=3$, and $g_0=p_0=(0.5)^2=0.25$. Since at most two claims can occur and each is at most two units, $S$ cannot exceed four.

At $s=1$, $g_1=(a+b)f_1g_0=2(0.5)(0.25)=0.25$.

At $s=2$, $g_2=(a+b/2)f_1g_1+(a+b)f_2g_0$ $=(0.5)(0.5)(0.25)+(2)(0.5)(0.25)=0.0625+0.25=0.3125$.

At $s=3$, $g_3=(a+b/3)f_1g_2+(a+2b/3)f_2g_1$ $=(0)(0.5)(0.3125)+(1)(0.5)(0.25)=0.125$. Thus $P(S=3)=0.1250$.

At $s=4$, $g_4=(a+b/4)f_1g_3+(a+b/2)f_2g_2$ $=(-0.25)(0.5)(0.125)+(0.5)(0.5)(0.3125)$ $=-0.015625+0.078125=0.0625$.

Check and trap avoided. The aggregate probabilities sum to $g_0+g_1+g_2+g_3+g_4=0.25+0.25+0.3125+0.125+0.0625=1$, as required. The negative term in the calculation of $g_4$ is valid: binomial counts require a negative $a$, and the complete recursion still gives nonnegative aggregate probabilities.

Worked Example: Negative Binomial Counts With a Three-Unit Target

Problem. A hypothetical repair portfolio has a negative binomial claim count with $r=2$ and success probability $p=0.5$, using $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$. Each claim costs one unit with probability $0.75$ or two units with probability $0.25$, independently. Find $P(S=3)$.

Method and parameters. Here $a=1-p=0.5$, $b=(r-1)(1-p)=0.5$, and $g_0=p^r=0.25$. First, $g_1=(a+b)f_1g_0=(1)(0.75)(0.25)=0.1875$.

For $s=2$, $g_2=(a+b/2)f_1g_1+(a+b)f_2g_0$ $=(0.75)(0.75)(0.1875)+(1)(0.25)(0.25)$ $=0.10546875+0.0625=0.16796875$.

For $s=3$, $g_3=(a+b/3)f_1g_2+(a+2b/3)f_2g_1$ $=(2/3)(0.75)(0.16796875)+(5/6)(0.25)(0.1875)$ $=0.083984375+0.0390625=0.123046875$. Therefore $P(S=3)=0.1230$, rounded to four decimals.

Check and trap avoided. A total of three can arise from two claims with sizes one and two, or three claims all of size one. The recursion includes both possibilities. Using the negative binomial formula stated in the problem gives $p_0=0.25$; using a different parameter convention without converting it would change the frequency parameters.

Common Parameter Error

A frequent exam trap is assigning $a$ a positive sign for the binomial count. For the binomial example above, the ratio is $p_n/p_{n-1}=\{(m-n+1)/n\}\{q/(1-q)\}$. Rearranging gives $p_n/p_{n-1}=-q/(1-q)+(m+1)q/\{n(1-q)\}$, so $a=-1$ and $b=3$, not $a=1$.

For instance, using the incorrect $a=1$ while retaining $b=3$ would give $g_1=(1+3)(0.5)(0.25)=0.5$, rather than the correct $g_1=0.25$. Check the count probability ratio at $n=1$: $p_1/p_0=2$, while the correct parameters give $a+b=2$. The incorrect parameters give a ratio of four and therefore cannot describe this binomial count.

Key takeaway: Confirm the frequency model's parameterization, match its probability ratio to $a+b/n$, and set $g_0=p_0$. Then apply the recursion in increasing order of aggregate lattice amount, using the weight $a+bj/s$ for severity size $j$ at target total $s$. The stated formula assumes strictly positive integer severities; a severity that can be zero requires separate treatment.

Check Your Understanding

For each problem, identify the frequency parameters where requested and apply the stated discrete model.

A hypothetical count is Poisson with mean 2, and every claim costs exactly one unit. Find $a$, $b$, and $P(S=2)$. (Answer: $a=0$, $b=2$, $P(S=2)=2e^{-2}=0.2707$.)
A hypothetical count is binomial with $m=3$ and $q=0.25$, and every claim costs one unit. Find $a$, $b$, and $P(S=2)$. (Answer: $a=-1/3$, $b=4/3$, $P(S=2)=0.1406$.)
A hypothetical Poisson count has mean 1. Each claim costs one or two units with probabilities $0.5$ each. Find $P(S=2)$. (Answer: $P(S=2)=(5/8)e^{-1}=0.2299$.)
A hypothetical negative binomial count has $r=3$, $p=0.5$, and probability function $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$. Every claim costs one unit. Find $a$, $b$, and $P(S=1)$. (Answer: $a=0.5$, $b=1$, $P(S=1)=0.1875$.)
A hypothetical count is binomial with $m=2$ and $q=0.4$. Each claim costs one unit with probability $0.25$ and two units otherwise. Find $P(S=1)$. (Answer: $P(S=1)=0.1200$.)

Claim count model	Parameter convention	\(a\)	\(b\)	\(p_0\)
Poisson	Mean \(\lambda\)	0	\(\lambda\)	\(e^{-\lambda}\)
Binomial	Number of trials \(m\), claim probability \(q\)	\(-q/(1-q)\)	\((m+1)q/(1-q)\)	\((1-q)^m\)
Negative binomial	\(P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n\)	\(1-p\)	\((r-1)(1-p)\)	\(p^r\)

What You'll Learn