Panjer Recursion Initialization And Numerical Checks

Why Initialization and Checks Matter

Panjer recursion replaces repeated convolution with a sequence of calculations for aggregate probabilities. The recurrence is only useful if it starts from the correct probability at zero. That initial value depends on both the probability of zero claims and the probability that an individual claim payment is zero. A mistaken initialization can make every later value wrong while still producing a plausible-looking table.

This tutorial considers a claim count $N$ and independent, identically distributed, nonnegative integer claim payments $Y^L_1,Y^L_2,\ldots$, independent of $N$. The aggregate payment is $S=\sum_{r=1}^{N}Y^L_r$, with $S=0$ when $N=0$. Write $p_j=P(Y^L=j)$ and $g_k=P(S=k)$. The severity distribution is assumed to lie on the stated integer payment scale; the recursion does not itself convert a continuous severity into integer amounts.

Definition: The probability generating functions of the claim count, claim payment, and aggregate are $Q(s)=E[s^N]$, $\phi(z)=E[z^{Y^L}]=\sum_{j\geq 0}p_jz^j$, and $H(z)=E[z^S]=\sum_{k\geq 0}g_kz^k$. Under the independence assumptions, $H(z)=Q(\phi(z))$, and therefore $g_0=Q(p_0)$.

The identity $g_0=Q(p_0)$ has a direct probability interpretation. Conditional on $N=n$, all $n$ payments must equal zero for the aggregate to be zero, which has probability $p_0^n$. Averaging over the claim count gives $g_0=\sum_{n\geq0}P(N=n)p_0^n=Q(p_0)$. This includes the $N=0$ case, since $p_0^0=1$.

Panjer Initialization and Recursion

For the $(a,b,0)$ Panjer frequency class, the count probabilities satisfy $f_n=(a+b/n)f_{n-1}$ for $n\geq1$, where $f_n=P(N=n)$. Common examples are Poisson frequency, with $a=0,b=\lambda$, and binomial frequency with parameters $m,q$, with $a=-q/(1-q)$ and $b=(m+1)q/(1-q)$. For an eligible frequency distribution and integer-valued severity, the aggregate probabilities satisfy the following initialization and recursion.

Theorem: For an $(a,b,0)$ frequency distribution and independent nonnegative integer claim payments with probabilities $p_j$, initialize at $g_0=Q(p_0)$. For $k\geq1$, if $1-ap_0\ne0$, $$ g_k=\frac{1}{1-ap_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)p_jg_{k-j}. $$ Terms with $p_j=0$ contribute zero.

The denominator matters: the severity can have a zero payment probability, so terms involving the unknown $g_k$ arise when the generating functions are combined. For valid standard Panjer models the recursion is used with a nonzero denominator. If a degenerate case makes the formula inapplicable, handle that case directly from the model rather than dividing by zero. In particular, if $p_0=1$, every claim payment is zero and $g_0=1$, $g_k=0$ for $k\geq1$.

Why the Recursion Holds

A short generating-function argument explains both the coefficient and the denominator. For an $(a,b,0)$ frequency distribution, multiply the count recurrence by $n f_{n-1}s^{n-1}$ and sum over $n\geq1$. Equivalently, the recurrence gives $$ Q'(s)=\sum_{n\geq1}(an+b)f_{n-1}s^{n-1} =a\{sQ'(s)+Q(s)\}+bQ(s), $$ so $(1-as)Q'(s)=(a+b)Q(s)$. Now substitute $s=\phi(z)$ and use the chain rule, $H'(z)=Q'(\phi(z))\phi'(z)$. This yields $$ \{1-a\phi(z)\}H'(z)=(a+b)H(z)\phi'(z). $$

To extract the coefficient of $z^{k-1}$, the term $H'$ contributes $kg_k$. In the product $\phi H'$, the contribution involving $p_jg_{k-j}$ is $(k-j)p_jg_{k-j}$, while the corresponding term in $H\phi'$ is $jp_jg_{k-j}$. Collecting them gives $$ kg_k=\sum_{j=1}^{k}(ak+bj)p_jg_{k-j}. $$ Dividing by $k$ and moving the $ap_0g_k$ contribution to the left gives the stated recursion. The initialization is separate: $H(0)=Q(\phi(0))=Q(p_0)$. This derivation also shows why omitting the denominator is generally incorrect.

Worked Example: Poisson Frequency With a Zero-Payment Atom

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$. Each claim payment is independently 0 with probability 0.5, 1 with probability 0.3, and 2 with probability 0.2, in the chosen integer payment units. Calculate $g_0,g_1,g_2,g_3$ and the remaining probability above 3. Use the Panjer recursion and identify the initialization trap.

Method and initialization. Poisson frequency has $a=0,b=1$ and $Q(s)=e^{s-1}$. Thus $g_0=Q(p_0)=e^{0.5-1}=e^{-0.5}=0.606531$. Using $e^{-1}$ instead would calculate $P(N=0)$, not $P(S=0)$.

For $a=0,b=1$, the recurrence is $g_k=(1/k)\sum_{j=1}^{k}j p_jg_{k-j}$. At $k=1$, $g_1=1(0.3)(0.60653066)=0.181959$. At $k=2$, $g_2=\frac12\{(0.3)(0.18195920)+2(0.2)(0.60653066)\} =\frac12(0.05458776+0.24261226)=0.148600$. At $k=3$, $p_3=0$, so $g_3=\frac13\{(0.3)(0.14860001)+2(0.2)(0.18195920)\} =\frac13(0.04458000+0.07278368)=0.039121$.

The computed mass through 3 is $0.60653066+0.18195920+0.14860001+0.03912123=0.97621110$. Therefore the probability above 3 is $1-0.97621110=0.02378890$, or 0.023789 to six decimals. The positive residual is expected: a Poisson count is unbounded and positive claim payments are possible. This is a check on the partial sum, not a claim that the recursion has produced an exact finite-support distribution.

Frequency Initialization Is Not Always the Same

A zero-modified count distribution has a zero probability that may differ from the corresponding unmodified frequency model. It does not generally belong to the $(a,b,0)$ class with the same recurrence parameters. In particular, substituting its $f_0$ into an ordinary Panjer recursion without checking the frequency recurrence can give incorrect aggregate probabilities. One reliable approach for a zero-modified Poisson model is to use a mixture: put probability $q_0$ on $N=0$, and with probability $1-q_0$ use a Poisson($\lambda$) count conditional on being positive.

Let $g_k^{P}$ denote the aggregate probabilities under the unmodified Poisson model with the same severity. Its probability of zero claims is $e^{-\lambda}$, while its probability of zero aggregate is $g_0^{P}=e^{\lambda(p_0-1)}$. For the zero-modified model, $$ g_0=q_0+(1-q_0)\frac{g_0^{P}-e^{-\lambda}}{1-e^{-\lambda}}, \qquad g_k=(1-q_0)\frac{g_k^{P}}{1-e^{-\lambda}}\quad(k\geq1). $$ For $k\geq1$, the unmodified aggregate event already excludes $N=0$; for $k=0$, its zero-claim contribution must be removed before conditioning on $N>0$.

Worked Example: Zero-Modified Poisson Initialization and Partial Sum

A hypothetical coverage has a zero-modified Poisson claim count: $P(N=0)=0.2$, and conditional on $N>0$, $N$ has a Poisson(1) distribution conditioned to be positive. Each payment is 0 or 1, each with probability 0.5. Calculate the aggregate probabilities at 0 through 3 and assess the truncated sum.

Method. First calculate the unmodified Poisson aggregate probabilities. Here $g_0^P=e^{-0.5}=0.60653066$, and the recursion gives $g_1^P=0.5g_0^P=0.30326533$, $g_2^P=(1/2)(0.5g_1^P)=0.07581633$, and $g_3^P=(1/3)(0.5g_2^P)=0.01263606$.

For the modified model, the conditional probability of zero aggregate given $N>0$ is $(0.60653066-e^{-1})/(1-e^{-1}) =(0.60653066-0.36787944)/(1-0.36787944)=0.37754067$. Thus $g_0=0.2+0.8(0.37754067)=0.502033$. For positive aggregate values, the scaling factor is $0.8/(1-e^{-1})=1.26558137$. Therefore $g_1=1.26558137(0.30326533)=0.383807$, $g_2=1.26558137(0.07581633)=0.095952$, and $g_3=1.26558137(0.01263606)=0.015992$.

Using unrounded values, the sum through 3 is 0.99778319, so the remaining probability is 0.00221681 (0.002217 to six decimals). This is consistent with unbounded positive claim counts. The check also confirms that the zero-modified initialization is not obtained by simply replacing the Poisson $f_0$ in the ordinary recursion.

Numerical Checks: Support, Mass, and Truncation

A recursion table should be checked structurally as well as arithmetically. A numerical result can satisfy a local recurrence and still be based on the wrong model, wrong payment unit, or wrong initialization.

Check the payment lattice.
State what one unit means and confirm that every $p_j$ refers to that same unit. The recursion applies to integer-valued payments; it does not justify rounding continuous payments to integers.

Check the support.
If every positive payment is a multiple of an integer $h$, then aggregate payments must also be multiples of $h$; probabilities at other values must be zero. If $N\leq n_{\max}$ and each payment is at most $M$, then $S\leq n_{\max}M$. With unbounded frequency and a positive chance of positive payment, aggregate support is generally unbounded.

Check the zero probability.
Verify $g_0=Q(p_0)$ from the model. In particular, if $p_0=0$, then $g_0=P(N=0)$; if $p_0>0$, the aggregate-zero probability also includes positive claim counts whose payments are all zero.

Check the total mass.
The full distribution must sum to 1. A partial sum through $K$, $\sum_{k=0}^{K}g_k$, should not exceed 1, apart from negligible displayed-rounding effects. Its complement is the probability above $K$, not an error term to discard.

Cross-check a moment when practical.
Under independent frequency and severity with finite means, $E[S]=E[N]E[Y^L]$. A complete distribution should reproduce this mean. A truncated sum of probabilities alone does not provide a complete mean check because omitted outcomes also contribute to the expectation.

Worked Example: Binomial Support and a Moment Check

A hypothetical group policy has $N\sim\operatorname{Binomial}(2,0.4)$. Each claim payment is 0 with probability 0.25 and 1 with probability 0.75. Use Panjer recursion to calculate the full aggregate distribution and check both its total probability and mean.

Method. The frequency parameters are $a=-0.4/0.6=-2/3$ and $b=(2+1)(0.4/0.6)=2$. Its pgf at $p_0=0.25$ gives $g_0=Q(0.25)=(0.6+0.4(0.25))^2=0.7^2=0.49$. The denominator is $1-ap_0=1-(-2/3)(0.25)=7/6$.

At $k=1$, only $p_1=0.75$ contributes: $g_1=\{(-2/3+2)(0.75)(0.49)\}/(7/6) =(0.49)/(7/6)=0.42$. At $k=2$, $p_2=0$, and $g_2=\{(-2/3+2/2)(0.75)(0.42)\}/(7/6) =\{(1/3)(0.315)\}/(7/6)=0.105/(7/6)=0.09$. There can be no aggregate above 2 because there are at most two claims and each payment is at most 1.

The probabilities sum to $0.49+0.42+0.09=1$. The distributional mean is $0(0.49)+1(0.42)+2(0.09)=0.60$. Independently, $E[N]=2(0.4)=0.8$ and $E[Y^L]=0.75$, giving $E[S]=0.8(0.75)=0.60$. Both checks agree.

Common Errors and Exam Checks

The most frequent initialization error is setting $g_0=f_0$ without checking whether a positive number of claims can produce zero aggregate payment. The correct expression is $Q(p_0)$; $g_0=f_0$ follows only when $p_0=0$. Another common error is to omit $1-ap_0$ from the recursion. This changes every calculated probability whenever $ap_0\ne0$.

Truncation and rounding require separate attention. A sum below 1 after calculating only through $K$ is not automatically a failure: it can represent $P(S>K)$. Conversely, forcing a truncated sum to equal 1 by rescaling changes the probabilities and is not a valid substitute for calculating the tail. Keep sufficient precision during the recurrence, sum the unrounded values for checks, and round only the reported results. This initialization and numerical-check discipline is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

Work through each problem, retaining precision until the final reported values.

A hypothetical count is Poisson with mean 0.5. Claim payment is 0 with probability 0.4 and 1 with probability 0.6. Find $g_0$ and explain why it differs from $P(N=0)$. (Answer: $g_0=e^{-0.3}=0.7408$, whereas $P(N=0)=e^{-0.5}=0.6065$.)
A hypothetical count is Poisson with mean 2, and payment probabilities are $p_1=0.25,p_2=0.75$, with $p_0=0$. Use the recursion to find $g_0,g_1,g_2$, and the mass through 2. (Answer: $g_0=0.1353,\ g_1=0.0677,\ g_2=0.2199$, and $\sum_{k=0}^2g_k=0.4229$.)
A hypothetical count is $\operatorname{Binomial}(3,0.2)$. Payments are 0 and 1 with probabilities 0.5 each. Find $g_0$ and $g_1$ using the binomial Panjer parameters. (Answer: $g_0=(0.8+0.2(0.5))^3=0.7290$, $g_1=0.2430$.)
A hypothetical count is $\operatorname{Binomial}(2,0.4)$, and each payment is 1 with probability 0.4 or 3 with probability 0.6. State the aggregate support, $P(S=6)$, and $P(S=5)$. (Answer: support $\{0,1,2,3,4,6\}$, $P(S=6)=0.16(0.6^2)=0.0576$, and $P(S=5)=0$.)
A zero-modified Poisson count has $P(N=0)=0.3$, and conditional on a positive count it is Poisson(1.5) conditioned to be positive. Payments are 0 with probability 0.2 and 1 otherwise. Find the aggregate probability at zero using the mixture initialization. (Answer: $0.3+0.7\{(e^{-1.2}-e^{-1.5})/(1-e^{-1.5})\}=0.3703$.)

What You'll Learn