Panjer Recursion with Estimated Inputs

Why Panjer Recursion Is Useful with Fitted Inputs

When a frequency model and a discrete severity model have been fitted, the aggregate loss is a compound distribution: the number of losses is random, and each loss amount is random. Exact convolution can calculate aggregate probabilities, but its repeated summations become cumbersome as the number of possible aggregate amounts grows. Panjer recursion uses a relationship between neighboring frequency probabilities to calculate aggregate probabilities successively.

The word fitted matters. In practice, the frequency and severity probabilities in the calculation may be estimates from hypothetical or observed data. The recursion then gives plug-in aggregate probabilities: it treats the estimated parameters and probabilities as fixed. It does not, by itself, account for uncertainty in those estimates.

Definition: Let $N$ be the number of claims, with probabilities $p_n=P(N=n)$. Suppose the individual claim payment $X$ is positive and discrete on a lattice with unit $h$: $P(X=jh)=f_j$, for $j=1,2,\ldots$, where $\sum_{j\geq 1} f_j=1$. For independent severities with this distribution, define $S=\sum_{r=1}^{N}X_r$, with $S=0$ when $N=0$, and let $g_k=P(S=kh)$. The assumptions include independence between frequency and severities, and identical independent severities conditional on $N$.

The lattice unit $h$ is a measurement scale, not an extra probability. If payments are recorded in thousands of dollars, for example, $h=\$1{,}000$; then $g_2$ is the probability of an aggregate payment of $\$2{,}000$. The severity distribution must use the same lattice throughout the calculation. Payments that do not lie on the chosen lattice cannot simply be rounded without changing the model.

Frequency Classes and the Recursion

The Panjer classes used here are specified by a recurrence for the frequency probabilities. In the $(a,b,0)$ class, the relation holds for $n\geq1$. In the $(a,b,1)$ class, it holds for $n\geq2$, so $p_0$ and $p_1$ must be supplied separately.

Definition: A frequency distribution is in the $(a,b,0)$ class if $p_n=(a+b/n)p_{n-1}$ for $n\geq1$. It is in the $(a,b,1)$ class if the same relation holds for $n\geq2$, with $p_0$ and $p_1$ specified separately.

For the common $(a,b,0)$ models, Poisson frequency with mean $\lambda$ has $a=0$, $b=\lambda$. A zero-based geometric frequency with $p_n=pq^n$, where $q=1-p$, has $a=q$, $b=0$. A binomial frequency with parameters $m$ and $p$ has $a=-p/(1-p)$ and $b=(m+1)p/(1-p)$. These values describe the frequency distribution; they are not fitted severity probabilities.

Theorem: If the frequency is in the $(a,b,0)$ class and the severity is strictly positive on the lattice, then $g_0=p_0$, and for $k\geq1$, $$ g_k=\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}. $$ For a frequency in the $(a,b,1)$ class, let $c=p_1-(a+b)p_0$. Then, under the same positive-severity assumption, $g_0=p_0$ and $$ g_k=\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}+c f_k,\qquad k\geq1. $$

The extra term for the $(a,b,1)$ case accounts for the fact that the frequency recurrence does not determine $p_1$ from $p_0$. It should not be dropped merely because the later probabilities follow a Panjer-class relation. The formulas here assume $f_0=0$. If a claim can produce a zero payment, the zero-severity probability affects the recursion and the aggregate mass at zero; use the version that explicitly includes $f_0$, rather than applying these simplified formulas unchanged.

Why the Recursion Holds

A short generating-function argument explains the weighted severity terms. For an $(a,b,0)$ frequency distribution, define its probability generating function $P(z)=\sum_{n\geq0}p_nz^n$. Multiplying $np_n=(an+b)p_{n-1}$ by $z^{n-1}$, summing over $n\geq1$, and collecting terms gives

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

Let $F(z)=\sum_{j\geq1}f_jz^j$ be the severity probability generating function. Independence gives the aggregate generating function $H(z)=P(F(z))=\sum_{k\geq0}g_kz^k$. Substituting $F(z)$ for $z$ in the frequency identity and using the chain rule yields

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

To obtain the coefficient of $z^{k-1}$, the left side contributes $kg_k-a\sum_{j=1}^{k}f_j(k-j)g_{k-j}$. The right side contributes $(a+b)\sum_{j=1}^{k}j f_jg_{k-j}$. Equating these coefficients and moving the sum on the left to the right gives

kg_k=\sum_{j=1}^{k}\bigl(a(k-j)+(a+b)j\bigr)f_jg_{k-j} =\sum_{j=1}^{k}(ak+bj)f_jg_{k-j}.

Dividing by $k$ gives the stated $(a,b,0)$ recursion. For the $(a,b,1)$ class, the generating-function identity instead has the additional constant $c=p_1-(a+b)p_0$; after composition with $F(z)$, that contributes $c f_k$ to the coefficient equation. This justifies the correction term in that case.

Applying the Recursion to Fitted Inputs

A reliable calculation separates estimation from recursion. First determine the fitted frequency parameters and the severity probabilities on the chosen lattice. Then identify the Panjer class and its $a,b$ values, initialize $g_0$, and calculate each $g_k$ using only already-computed probabilities. Keep unrounded values through the recursion and round the requested probabilities at the end.

Set the lattice and severity probabilities.
Express each possible payment as $jh$, and record $f_j$. Verify that the positive severity probabilities sum to one when $f_0=0$.

Fit and classify frequency.
Use the stated frequency model and estimated parameters to obtain $p_0$, any needed $p_1$, and the corresponding $a,b$.

Initialize and recurse.
Set $g_0=p_0$, then calculate $g_1,g_2,\ldots$ in order. For the $(a,b,1)$ class, include $c f_k$.

Interpret and check.
Translate $g_k$ back to the amount $kh$, and independently check selected values by conditioning on small claim counts.

Worked Example: Poisson Frequency with Empirical Severity

A hypothetical portfolio has 24 claims over 24 exposure units. Fit a Poisson frequency using claims per exposure unit. In a separate set of eight complete claim records, two payments are $\$1{,}000$ and six are $\$2{,}000$. Using $h=\$1{,}000$, calculate the aggregate probabilities through $g_3$.

Method and estimates. The fitted Poisson mean is $\hat{\lambda}=24/24=1$, so $p_0=e^{-1}=0.3678794412$, $a=0$, and $b=1$. The empirical severity probabilities are $f_1=2/8=0.25$ and $f_2=6/8=0.75$; all other $f_j$ are zero. The $(a,b,0)$ recursion is $g_k=(1/k)\sum_{j=1}^k j f_jg_{k-j}$.

Initialize. Since severity is positive, $g_0=p_0=0.3678794412$.

Calculate $g_1$. $$ g_1=1(0.25)g_0=0.25(0.3678794412)=0.0919698603. $$

Calculate $g_2$. $$ g_2=\frac{1}{2}\left(1(0.25)g_1+2(0.75)g_0\right) =\frac{1}{2}(0.0229924651+0.5518191618) =0.2874058135. $$

Calculate $g_3$. $$ g_3=\frac{1}{3}\left(1(0.25)g_2+2(0.75)g_1\right) =\frac{1}{3}(0.0718514534+0.1379547904) =0.0699354146. $$

Check and interpretation. An aggregate of $\$2{,}000$ can arise from one $\$2{,}000$ claim or two $\$1{,}000$ claims. Their probabilities are $e^{-1}(0.75)=0.2759095809$ and $(e^{-1}/2)(0.25^2)=0.0114962325$; their sum is $0.2874058134$, matching $g_2$ up to displayed precision. To four decimal places, $g_0=0.3679$, $g_1=0.0920$, $g_2=0.2874$, and $g_3=0.0699$. This method avoids treating the severity probabilities as frequency probabilities or omitting the factor $1/k$.

Worked Example: Binomial Frequency and the Frequency-Class Parameters

A hypothetical contract permits at most two covered events per exposure. A fitted binomial model uses two independent event opportunities per exposure; 50 events were recorded across 100 exposures, giving $\hat p=50/(2\cdot100)=0.25$. Suppose fitted severity probabilities are $f_1=f_2=0.5$, in units of $\$500$. Find the probability that aggregate payment is $\$1{,}000$.

Method and frequency parameters. Here $N\sim\operatorname{Binomial}(2,0.25)$, so $p_0=0.75^2=0.5625$, $p_1=2(0.25)(0.75)=0.375$, and $p_2=0.25^2=0.0625$. The binomial parameters for the recursion are $$ a=-\frac{0.25}{0.75}=-\frac{1}{3},\qquad b=\frac{(2+1)(0.25)}{0.75}=1. $$

Initialize and calculate $g_1$. We have $g_0=0.5625$. For $k=1$, the coefficient is $a+b=-1/3+1=2/3$, so $$ g_1=\frac{2}{3}(0.5)(0.5625)=0.1875. $$

Calculate $g_2$. $$ g_2=\left(-\frac{1}{3}+\frac{1}{2}\right)(0.5)g_1 +\left(-\frac{1}{3}+1\right)(0.5)g_0 $$ $$ =\frac{1}{6}(0.5)(0.1875)+\frac{2}{3}(0.5)(0.5625) =0.015625+0.1875 =0.203125. $$

Check and interpretation. An aggregate of two units results either from one claim of two units, with probability $p_1f_2=(0.375)(0.5)=0.1875$, or from two one-unit claims, with probability $p_2f_1^2=(0.0625)(0.25)=0.015625$. The sum is $0.203125$, so the requested probability is $0.2031$ to four decimals. The check also catches a common error: using the frequency probabilities as though they were severity probabilities.

Worked Example: Geometric Frequency Fitted from a Mean

A hypothetical book is fitted with a zero-based geometric frequency model whose estimated mean is 1.5 claims per exposure. The fitted severity is $\$1{,}000$ with probability 0.6 and $\$2{,}000$ with probability 0.4. Find the probabilities of aggregate payments from zero through $\$3{,}000$.

Method and fitted parameters. For $p_n=pq^n$, the mean is $q/p$ and $p+q=1$. Thus $q/p=1.5$, so $p=1/(1+1.5)=0.4$ and $q=0.6$. The frequency recurrence has $a=0.6,b=0$, and $p_0=0.4$. The severity probabilities are $f_1=0.6$ and $f_2=0.4$, with $h=\$1{,}000$.

Initialize. $g_0=p_0=0.4$.

Calculate $g_1$ and $g_2$. $$ g_1=0.6(0.6)g_0=0.6(0.6)(0.4)=0.144. $$ $$ g_2=0.6\left(f_1g_1+f_2g_0\right) =0.6\left((0.6)(0.144)+(0.4)(0.4)\right) =0.6(0.0864+0.16)=0.14784. $$

Calculate $g_3$. $$ g_3=0.6\left(f_1g_2+f_2g_1\right) =0.6\left((0.6)(0.14784)+(0.4)(0.144)\right) =0.6(0.088704+0.0576)=0.0877824. $$

Check and interpretation. An aggregate of $\$2{,}000$ arises from one claim of $\$2{,}000$, probability $P(N=1)f_2=(0.24)(0.4)=0.096$, or two claims of $\$1{,}000$, probability $P(N=2)f_1^2=(0.144)(0.36)=0.05184$. Their sum is $0.14784$, confirming $g_2$. To four decimals, the probabilities for $\$0,\$1{,}000,\$2{,}000,\$3{,}000$ are $0.4000,0.1440,0.1478,0.0878$. The method avoids confusing the geometric parameter $p=0.4$ with its mean of 1.5.

Common Traps and Scope of the Result

Panjer recursion is efficient only after the inputs and model class are correctly identified. A fitted mean alone does not specify every frequency distribution: the assumed family matters. Likewise, severity probabilities must describe payment amounts after the relevant coverage modifications have been applied. If the model is intended to describe covered payments, applying a deductible or limit again inside the recursion would change the severity model and count the modification twice.

Omitting the lattice scale. If $h=\$500$, then $k=2$ means $\$1{,}000$, not $\$2$. State the unit when interpreting each $g_k$.
Using the wrong class parameters. For binomial frequency, $a$ is negative. Replacing it with a Poisson value changes the recursion.
Rounding at every step. Early rounding is propagated to later $g_k$. Retain full precision until the requested final probabilities.
Ignoring the $(a,b,1)$ adjustment. When the recurrence starts at $n=2$, use $c=p_1-(a+b)p_0$ and include $c f_k$.
Interpreting plug-in output as parameter certainty. Recursion evaluates the fitted model; it does not quantify the sampling uncertainty in estimated parameters or severity probabilities.

A useful final check is to compare selected aggregate amounts with a direct conditioning argument, as in the examples. This is not a substitute for checking all outputs, but it can quickly reveal a wrong frequency parameter, a misindexed severity value, or an arithmetic error. Panjer recursion with estimated inputs is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated positive lattice severity and fitted frequency model. Round probabilities to four decimals.

A fitted Poisson model has mean 0.5 claims per exposure. Severity is one lattice unit with probability 0.8 and two units with probability 0.2. Calculate $g_0$, $g_1$, and $g_2$. (Answer: $g_0=0.6065,\ g_1=0.2426,\ g_2=0.1092$)
A fitted frequency is $\operatorname{Binomial}(3,0.2)$, and every claim payment is exactly one lattice unit. Find the aggregate probability at two units, using the binomial frequency probabilities or the Panjer class. (Answer: $0.0960$)
A zero-based geometric frequency model has fitted mean 0.75. Severity is one unit or three units, each with probability 0.5. Use the geometric frequency recurrence to find the probability of an aggregate of three units. (Answer: $0.1281$)
A fitted Poisson frequency has mean 1.2 claims per exposure. Severity is one unit with probability 0.5 and two units with probability 0.5. Calculate $g_1$ and $g_2$, retaining $g_0=e^{-1.2}$ until the final rounding. (Answer: $g_1=0.1807,\ g_2=0.2349$)

What You'll Learn