Deriving Panjer Recursion For (a,b,0) Models

Why the Panjer Recursion Is Useful

A compound aggregate loss is the total of a random number of individual losses. Directly conditioning on every possible claim count leads to a sum of convolutions, which can become costly when many aggregate values are needed. For particular claim-count distributions, the Panjer recursion replaces those repeated convolutions with a step-by-step calculation of aggregate probabilities.

This tutorial derives the recursion for an $(a,b,0)$ frequency model. Its inputs are the claim-count probabilities and the probability assigned by the severity model to each nonnegative integer amount. The derivation also shows why a severity probability at zero affects both the starting probability and the recursion denominator.

Definition: Let $N$ be a nonnegative integer-valued claim count with probabilities $p_n=P(N=n)$. It is an $(a,b,0)$ frequency model if $p_0>0$, the probabilities are normalized, and $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1}, \qquad n\geq 1. $$ The constants $a$ and $b$ describe the frequency model; they do not by themselves guarantee that a proposed sequence of $p_n$ values is a probability distribution.

Let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$, and define the aggregate $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. For the recursion below, each severity must take values on the nonnegative integer lattice. Write $f_j=P(X=j)$ for $j=0,1,2,\ldots$, and write $q_s=P(S=s)$. These $f_j$ are the severity-probability inputs; they are not aggregate probabilities.

Model conditions: The Panjer recursion here assumes an independent claim count and severities, identically distributed severities, nonnegative integer claim amounts, and a proper $(a,b,0)$ frequency distribution. For a severity measured continuously, the recursion cannot be applied directly to its density; a suitable discrete representation is required first. If the severity is strictly positive, then $f_0=0$. If zero claim amounts are possible, retain $f_0$ rather than silently removing it.

Deriving the Aggregate Recursion

The probability generating function (PGF) of $N$ is $P_N(z)=\sum_{n=0}^{\infty}p_nz^n$, and the severity PGF is $F(z)=\sum_{j=0}^{\infty}f_jz^j$. The aggregate PGF is $H(z)=\sum_{s=0}^{\infty}q_sz^s$. Conditioning on the count, or equivalently composing PGFs, gives $H(z)=P_N(F(z))$.

First translate the frequency relation into a differential equation. Multiply $p_n=(a+b/n)p_{n-1}$ by $nz^n$ and sum over $n\geq1$. The left-hand side is $zP_N'(z)$. On the right, use $np_n=anp_{n-1}+bp_{n-1}$, giving

zP_N'(z)=az\bigl(zP_N'(z)+P_N(z)\bigr)+bzP_N(z).

For $z\neq0$, rearranging yields

$$ (1-az)P_N'(z)=(a+b)P_N(z). $$

Since $H(z)=P_N(F(z))$, the chain rule gives $H'(z)=P_N'(F(z))F'(z)$. Substitute $F(z)$ for the PGF argument in the differential equation and multiply by $F'(z)$:

$$ (1-aF(z))H'(z)=(a+b)F'(z)H(z). $$

To obtain the probability recursion, compare coefficients of $z^{s-1}$ for $s\geq1$. In $F(z)H'(z)$, the coefficient is $\sum_{j=0}^{s-1}f_j(s-j)q_{s-j}$. In $F'(z)H(z)$, it is $\sum_{j=1}^{s}j f_jq_{s-j}$. The $j=0$ term from $F(z)H'(z)$ is $f_0s q_s$; moving that term to the left gives

s(1-af_0)q_s =\sum_{j=1}^{s}\bigl(a(s-j)+(a+b)j\bigr)f_jq_{s-j} =\sum_{j=1}^{s}(as+bj)f_jq_{s-j}.

Thus, whenever $1-af_0\neq0$, the aggregate probabilities for $s\geq1$ satisfy

Theorem: For a proper $(a,b,0)$ frequency model and independent, identically distributed nonnegative integer-valued severities, $$ q_s=\frac{1}{1-af_0}\sum_{j=1}^{s}\left(a+\frac{bj}{s}\right)f_jq_{s-j}, \qquad s\geq1. $$ The starting value is $q_0=H(0)=P_N(f_0)=\sum_{n=0}^{\infty}p_nf_0^n$. The recursion is not a substitute for this initialization.

The starting value has a direct interpretation: the aggregate is zero if there are no claims, or if every claim that occurs has severity zero. In particular, if claims are strictly positive, then $f_0=0$ and $q_0=p_0$; the recursion denominator is then 1. If $f_0>0$, the aggregate can be zero even when $N>0$, which is why $q_0$ generally exceeds $p_0$.

Worked Applications

Worked Example: Binomial Claim Count and Two-Point Severity

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.3)$. Each claim pays 1 unit with probability 0.7 or 2 units with probability 0.3. Find $a,b,q_0,q_1,$ and $q_2$ using the recursion, and check the aggregate probabilities by direct conditioning.

Method and frequency parameters. For a binomial count with parameters $m,p$, the adjacent-probability ratio is $\frac{p_n}{p_{n-1}}=\frac{m-n+1}{n}\frac{p}{1-p}$. With $m=2,p=0.3$, this is $-\frac{0.3}{0.7}+\frac{3(0.3)}{0.7n}$. Therefore $a=-3/7$ and $b=9/7$. The count probabilities include $p_0=(0.7)^2=0.49$, $p_1=2(0.3)(0.7)=0.42$, and $p_2=(0.3)^2=0.09$, which sum to 1.

Severity inputs and initialization. Here $f_0=0,f_1=0.7,f_2=0.3$. Because every claim is positive, $q_0=p_0=0.49$.

Calculate $q_1$. Only $j=1$ contributes: \[ q_1=(a+b)f_1q_0=\frac{6}{7}(0.7)(0.49)=0.294. \]

Calculate $q_2$. There are two terms: \[ \begin{aligned} q_2 &=\left(a+\frac{b}{2}\right)f_1q_1+(a+b)f_2q_0\\ &=\frac{3}{14}(0.7)(0.294)+\frac{6}{7}(0.3)(0.49)\\ &=0.0441+0.1260=0.1701. \end{aligned} \] Rounded to four decimal places, $q_0=0.4900,\ q_1=0.2940,\ q_2=0.1701$.

Check. With one claim, aggregate 1 has probability $0.42(0.7)=0.294$. Aggregate 2 occurs from one claim of size 2, probability $0.42(0.3)=0.126$, or two claims both of size 1, probability $0.09(0.7)^2=0.0441$. Their sum is 0.1701, as found by recursion. The method avoids trying to treat the $j=1$ contribution alone as the entire probability of aggregate 2.

Worked Example: Compound Poisson Model with a Zero-Loss Atom

For a hypothetical coverage, let $N\sim\operatorname{Poisson}(2)$, and let each claim severity be 0 with probability 0.2, 1 with probability 0.5, or 2 with probability 0.3. Calculate $q_0,q_1,$ and $q_2$, rounding probabilities to four decimal places.

Method and inputs. For a Poisson count with mean $\lambda$, $p_n/p_{n-1}=\lambda/n$, so $a=0,b=2$. The severity probabilities are $f_0=0.2,f_1=0.5,f_2=0.3$. The frequency PGF is $P_N(z)=\exp(2(z-1))$, so the required initialization is \[ q_0=P_N(f_0)=\exp(2(0.2-1))=\exp(-1.6)=0.2018965\ldots. \]

Calculate $q_1$. The recursion reduces to $q_s=\frac{2}{s}\sum_{j=1}^{s}j f_jq_{s-j}$. Hence \[ q_1=2(0.5)q_0=q_0=0.2018965\ldots. \]

Calculate $q_2$. The $j=1$ and $j=2$ terms are \[ q_2=\frac{2}{2}(1)(0.5)q_1+\frac{2}{2}(2)(0.3)q_0 =0.5q_1+0.6q_0=1.1q_0=0.2220861\ldots. \] Thus, to four decimal places, $q_0=0.2019,\ q_1=0.2019,\ q_2=0.2221$.

Check. The aggregate PGF is $\exp(2(F(z)-1))=\exp(-1.6)\exp(z+0.6z^2+\cdots)$. Its constant coefficient is $\exp(-1.6)$, its $z$ coefficient is $\exp(-1.6)$, and its $z^2$ coefficient is $\exp(-1.6)(0.6+1/2)=1.1\exp(-1.6)$. These agree with the recursion. The zero-loss probability changes $q_0$; using $p_0=\exp(-2)$ instead would be incorrect.

Worked Example: Negative Binomial Count with Unit-or-Zero Severities

A hypothetical model has $N$ equal to the number of failures before two successes in independent Bernoulli trials, with success probability 0.5. Each claim pays 1 unit with probability 0.75 and 0 with probability 0.25. Find $q_0,q_1,$ and $q_2$, rounding probabilities to four decimal places.

Method and frequency parameters. The count probabilities are $p_n=(n+1)(0.5)^2(0.5)^n$. Their ratio is \[ \frac{p_n}{p_{n-1}}=\frac{n+1}{n}(0.5)=0.5+\frac{0.5}{n}, \] so $a=0.5,b=0.5$. The severity inputs are $f_0=0.25,f_1=0.75$, with all $f_j=0$ for $j\geq2$.

Initialize. The frequency PGF is $P_N(z)=[0.5/(1-0.5z)]^2$. Therefore \[ q_0=P_N(0.25)=\left(\frac{0.5}{1-0.125}\right)^2 =\left(\frac{4}{7}\right)^2=\frac{16}{49}=0.3265306\ldots. \] Also $1-af_0=1-(0.5)(0.25)=7/8$.

Calculate $q_1$. Only severity value 1 contributes: \[ q_1=\frac{1}{7/8}(a+b)(0.75)q_0 =\frac{6}{7}q_0=\frac{96}{343}=0.2798834\ldots. \]

Calculate $q_2$. Since $f_2=0$, only the $j=1$ term contributes: \[ q_2=\frac{1}{7/8}\left(a+\frac{b}{2}\right)(0.75)q_1 =\frac{9}{14}q_1=\frac{432}{2401}=0.1799250\ldots. \] Thus, to four decimal places, $q_0=0.3265,\ q_1=0.2799,\ q_2=0.1799$.

Check. The $j=1$ severity probability appears in each step, and the denominator is $7/8$, not 1. Omitting either would give a different result. Here $q_0+q_1+q_2=0.7863$ to four decimals; probabilities for aggregate values above 2 remain, so this partial sum is appropriately less than 1.

Interpreting and Checking the Inputs

The recursion is a computational identity, not a guarantee that arbitrary inputs describe a valid model. The frequency probabilities must sum to 1, the severity probabilities must be nonnegative and sum to 1, and the independence and lattice assumptions must match the intended model. A frequency sequence satisfying the adjacent-probability relation but not normalizing cannot be used as a claim-count distribution.

For $f_0=0$, the recursion simplifies to \[ q_s=\sum_{j=1}^{s}\left(a+\frac{bj}{s}\right)f_jq_{s-j},\qquad q_0=p_0. \] For $f_0>0$, both $q_0=P_N(f_0)$ and the factor $1-af_0$ matter. The formula requires $1-af_0\neq0$; if this condition fails, the displayed rearrangement cannot be used, and the model must be examined directly rather than dividing by zero.

Common exam trap: Applying the recursion with $f_0>0$ as though $q_0=p_0$ and the denominator were 1. For the Poisson example above, that mistake would initialize at $\exp(-2)=0.1353$ instead of $q_0=\exp(-1.6)=0.2019$, and would then understate every recursively calculated probability. The correction is to evaluate $q_0=P_N(f_0)$ and retain $1-af_0$ in each subsequent step.

A reliable calculation therefore follows the same sequence: identify the valid frequency parameters, tabulate the severity probabilities, compute the zero-aggregate probability, and then calculate $q_1,q_2,\ldots$ in order. Each step uses only previously calculated aggregate probabilities. As a numerical check, probabilities must be nonnegative, and their cumulative sum through any finite value $s$ cannot exceed 1.

Check Your Understanding

For each problem, identify the relevant inputs and apply the recursion where requested. Answers are rounded to four decimal places where needed.

A hypothetical compound Poisson model has claim-count mean 1.5 and severity probabilities $f_0=0.4,f_1=0.6$. Find $q_0$ and $q_1$. (Answer: $q_0=0.4066,\ q_1=0.3659$)
A claim count is $\operatorname{Binomial}(3,0.2)$, and all claims pay a positive integer amount, with $f_1=0.4,f_2=0.6$. Find $a,b,q_0,$ and $q_1$. (Answer: $a=-0.25,\ b=1,\ q_0=0.5120,\ q_1=0.1536$)
A hypothetical frequency model is geometric, with $p_n=0.4(0.6)^n$, and severity probabilities $f_0=0.5,f_1=0.5$. Find the $(a,b)$ parameters, $q_0,$ and $q_1$. (Answer: $a=0.6,\ b=0,\ q_0=0.5714,\ q_1=0.2449$)
A hypothetical count is $\operatorname{Binomial}(2,0.4)$; severities are 0 with probability 0.2 and 1 with probability 0.8. Find $q_0$ and $q_1$, retaining the zero-severity term in the recursion. (Answer: $q_0=0.4624,\ q_1=0.4352$)
A proposed frequency sequence has $p_0=0.4$ and $p_n=(0.2+0.3/n)p_{n-1}$. Can it be used as a normalized claim-count distribution with this $p_0$? (Answer: No. Its PGF from the recurrence is $0.4(1-0.2z)^{-2.5}$, whose value at 1 is approximately 0.6988, not 1.)

What You'll Learn