Panjer Recursion Workshop With Different Claim Counts

Why Compare Frequency Models With Panjer Recursion?

An aggregate loss is the sum of a random number of claim amounts. Even if two portfolios have the same severity distribution and the same expected claim count, different frequency distributions can produce different aggregate probabilities. Panjer recursion provides a direct way to calculate those probabilities when claim amounts lie on an integer lattice and the frequency distribution belongs to the Panjer class.

This workshop uses one severity distribution for both a Poisson count and a negative binomial count. Comparing them on the same lattice isolates the effect of the frequency model. The calculations are for hypothetical portfolios valued at a common date; there is no interest discounting or mortality assumption because neither is part of these aggregate claim models.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $Z_1,Z_2,\ldots$ be independent, identically distributed claim amounts, independent of $N$. The aggregate loss is $S=\sum_{i=1}^{N}Z_i$, with $S=0$ when $N=0$. For a lattice unit $h>0$, write $Z_i=hJ_i$, where $J_i$ is a nonnegative integer, and set $p_j=P(J_i=j)$ and $g_k=P(S=kh)$.

For the main comparison, the lattice unit is one claim unit, so the severity probabilities are $p_1=0.5$, $p_2=0.3$, and $p_3=0.2$, with $p_j=0$ for all other $j$. In particular, $p_0=0$: every claim has positive severity. The probabilities sum to one. The aggregate can nevertheless be zero when no claims occur.

The Recursion and Its Frequency Parameters

A frequency distribution is in the $(a,b,0)$ Panjer class if its probabilities $f_n=P(N=n)$ satisfy $f_n=(a+b/n)f_{n-1}$ for $n\geq 1$. For such a count and a lattice severity distribution, the aggregate probabilities satisfy the following recursion. The version shown allows a severity atom at zero; in the examples here $p_0=0$.

Theorem: If $N$ is in the $(a,b,0)$ class, $p_j=P(Z_i=jh)$, and $g_k=P(S=kh)$, then

g_0=G_N(p_0), \qquad g_k=\frac{1}{1-ap_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)p_jg_{k-j},\quad k\geq1,

where $G_N(t)=E[t^N]$ is the probability generating function of $N$. The formula requires $1-ap_0\ne0$. For positive claim severities, $p_0=0$, so the denominator is one and $g_0=P(N=0)$. Terms with $p_j=0$ contribute nothing.

For a Poisson count with parameter $\lambda$, $f_n=e^{-\lambda}\lambda^n/n!$, so $f_n=(\lambda/n)f_{n-1}$. Its Panjer parameters are $a=0$, $b=\lambda$, and its recursion is $g_k=(\lambda/k)\sum_{j=1}^k j p_jg_{k-j}$.

For the negative binomial parameterization used here, $N$ has probability generating function $G_N(t)=((1-q)/(1-qt))^r$, where $r>0$ and $0<q<1$. Its mean is $rq/(1-q)$, and its probabilities satisfy $f_n=q(n+r-1)f_{n-1}/n$. Thus $a=q$, $b=q(r-1)$, and $g_0=(1-q)^r$ when $p_0=0$. The parameterization matters: $q$ is not the mean.

Why the Recursion Holds

A short generating-function derivation explains both the $j/k$ factor and why the frequency parameters enter the aggregate calculation. Let $P(z)=\sum_{j\geq0}p_jz^j$ be the severity probability generating function and $H(z)=E[z^{S/h}]=G_N(P(z))=\sum_{k\geq0}g_kz^k$ the aggregate generating function.

The frequency recurrence implies a differential equation. Multiplying $f_n=(a+b/n)f_{n-1}$ by $n$, then summing over $n\geq1$, gives $G_N'(t)=atG_N'(t)+(a+b)G_N(t)$, or $(1-at)G_N'(t)=(a+b)G_N(t)$. Applying the chain rule to $H(z)=G_N(P(z))$ therefore gives

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

For $k\geq1$, equate the coefficients of $z^{k-1}$. The coefficient from $H'$ is $kg_k$. In the product $P(z)H'(z)$, the term involving $p_j$ is $p_j(k-j)g_{k-j}$; in $H(z)P'(z)$, it is $jp_jg_{k-j}$. Isolating the term with $j=0$, which is $akp_0g_k$, yields

k(1-ap_0)g_k =\sum_{j=1}^{k}\bigl(a(k-j)+(a+b)j\bigr)p_jg_{k-j} =\sum_{j=1}^{k}(ak+bj)p_jg_{k-j}.

Dividing by $k(1-ap_0)$ gives the stated recursion. This derivation also shows why the claim severity index $j$ is weighted: it arises from differentiating the severity generating function.

Worked Calculations for the Two Counts

Worked Example: Poisson Claim Count Through Four Lattice Units

A hypothetical protection pool has $N\sim\operatorname{Poisson}(2)$. Each claim is one, two, or three lattice units, with probabilities $0.5$, $0.3$, and $0.2$, respectively. Calculate $g_0,\ldots,g_4$, then find $P(S>4h)$.

Method and setup. Use the Poisson Panjer parameters $a=0$, $b=\lambda=2$. Since $p_0=0$, $g_0=P(N=0)=e^{-2}$. Substitute the nonzero $p_j$'s into the recursion; keep exact coefficients until the final probability rounding.

For $k=1$, only severity index $j=1$ contributes:

g_1=\frac{2}{1}(1)(0.5)g_0=g_0=e^{-2}.

For $k=2$, the $j=1$ and $j=2$ terms are:

g_2=\frac{2}{2}\bigl((1)(0.5)g_1+(2)(0.3)g_0\bigr) =(0.5+0.6)g_0=\frac{11}{10}g_0.

For $k=3$, the three terms give:

g_3=\frac{2}{3}\bigl((1)(0.5)g_2+(2)(0.3)g_1+(3)(0.2)g_0\bigr) =\frac{2}{3}(0.55+0.6+0.6)g_0 =\frac{7}{6}g_0.

For $k=4$, the possible severity indices are 1, 2, and 3:

g_4=\frac{2}{4}\bigl((1)(0.5)g_3+(2)(0.3)g_2+(3)(0.2)g_1\bigr) =\frac{1}{2}\left(\frac{7}{12}+\frac{33}{50}+\frac{3}{5}\right)g_0 =\frac{553}{600}g_0.

Using $e^{-2}=0.1353352832\ldots$, the probabilities, rounded only at the end to four decimal places, are $g_0=0.1353$, $g_1=0.1353$, $g_2=0.1489$, $g_3=0.1579$, and $g_4=0.1247$. Their sum before rounding is $0.7021645612\ldots$. Therefore

P(S>4h)=1-\sum_{k=0}^{4}g_k =1-0.7021645612\ldots =0.2978.

Check and trap avoided. The five probabilities sum to less than one, as they must because positive probability remains above $4h$; adding the unrounded tail gives one. A common wrong calculation drops the severity index $j$ in the recursion at $k=2$, producing $g_2^{\text{wrong}}=(2/2)(0.5g_1+0.3g_0)=0.8g_0$. The correct recursion has $2p_2=0.6$, not $p_2=0.3$, and gives $g_2=1.1g_0$.

Worked Example: Negative Binomial Count on the Same Lattice

A second hypothetical pool has the same claim severity probabilities, but its count follows the negative binomial model with $r=2$ and $q=0.5$. Calculate the aggregate probabilities through $4h$ and find $P(S>4h)$.

Method and setup. Under the stated parameterization, the mean count is $rq/(1-q)=2$, matching the Poisson example. The Panjer parameters are $a=q=0.5$, $b=q(r-1)=0.5$, and $g_0=(1-q)^r=0.25$. This count has greater variance than the Poisson count with mean 2, but that fact alone does not determine the tail at every particular threshold.

For $k=1$, the coefficient is $a+b=1$:

$$ g_1=(0.5+0.5)(0.5)g_0=0.5g_0=0.125. $$

For $k=2$, the coefficients $a+bj/k$ are $0.75$ for $j=1$ and $1$ for $j=2$:

$$ g_2=(0.75)(0.5)g_1+(1)(0.3)g_0 =0.375(0.125)+0.3(0.25) =0.046875+0.075 =0.121875. $$

For $k=3$, the three coefficients are $2/3$, $5/6$, and $1$:

g_3=\left(\frac{2}{3}\right)(0.5)g_2 +\left(\frac{5}{6}\right)(0.3)g_1 +(1)(0.2)g_0 =0.040625+0.03125+0.05 =0.121875.

For $k=4$, the coefficients are $5/8$, $3/4$, and $7/8$ for the nonzero severity indices:

$$ g_4=(0.625)(0.5)g_3+(0.75)(0.3)g_2+(0.875)(0.2)g_1 =0.0380859375+0.027421875+0.021875 =0.0873828125. $$

Thus $g_0=0.2500$, $g_1=0.1250$, $g_2=0.1219$, $g_3=0.1219$, and $g_4=0.0874$, to four decimal places. The unrounded cumulative probability through $4h$ is

$$ 0.25+0.125+0.121875+0.121875+0.0873828125 =0.7061328125. $$

Consequently, $P(S>4h)=1-0.7061328125=0.2938671875$, or $0.2939$ to four decimal places. The calculation uses the frequency parameter $b=0.5$, not the mean count 2. Substituting the mean for a Panjer parameter would give incorrect recursion coefficients.

Worked Example: Compare Inclusive and Strict Thresholds

Suppose the lattice unit in the two pools above represents $\$500$. A hypothetical analyst wants both the probability that aggregate claims are at least $\$2{,}000$ and the probability that they strictly exceed $\$2{,}000$. Calculate both probabilities under each count model.

Method. Convert the money thresholds into lattice units before selecting the probabilities. Since $\$2{,}000=4h$, at least $\$2{,}000$ means $S/h\geq4$, while strictly more than $\$2{,}000$ means $S/h\geq5$. These are different events because $g_4=P(S=4h)$ is positive.

For the Poisson model, the inclusive event is the complement of aggregate values 0 through $3h$. Using the unrounded probabilities from the first calculation:

P(S\geq4h)=1-(g_0+g_1+g_2+g_3) =1-0.5774305418\ldots =0.4226.

The strict event removes the probability at $4h$, or equivalently takes the complement through $4h$:

P(S>4h)=P(S\geq5h) =0.4225694582\ldots-0.1247340194\ldots =0.2978.

For the negative binomial model, the probability through $3h$ is $0.25+0.125+0.121875+0.121875=0.61875$. Therefore

P(S\geq4h)=1-0.61875=0.3813, \qquad P(S>4h)=0.38125-0.0873828125=0.2939.

The inclusive answer $0.38125$ rounds to $0.3813$. At this threshold, both tails are smaller for the negative binomial model than for the Poisson model, despite the negative binomial count having greater variance. Tail ordering depends on the threshold and the severity distribution; it should be calculated rather than inferred from frequency variance alone.

Check and trap avoided. For each model, $P(S\geq4h)-P(S>4h)=P(S=4h)=g_4$. The difference is $0.1247$ for the Poisson model and $0.0874$ for the negative binomial model, matching the calculated probabilities. Treating “at least” as “strictly greater than” would omit this entire mass.

Practical Checks Before Reporting a Tail

Panjer recursion is sequential: to find $g_k$, use probabilities $g_0,\ldots,g_{k-1}$ already calculated. The relevant severity index is bounded by $1\leq j\leq k$; a claim larger than $kh$ cannot contribute to an aggregate of exactly $kh$. These details make it possible to check each term before moving on.

Verify the count parameterization. For the negative binomial model here, write down $r$, $q$, the mean $rq/(1-q)$, and the resulting $a,b$ before using the recursion.
Keep lattice units consistent. If one unit is $\$500$, a $\$2{,}000$ threshold is four units. Do not mix dollar thresholds with unit indices.
Use the correct event boundary. $P(S>x)=1-P(S\leq x)$, whereas $P(S\geq x)=1-P(S<x)$. On a lattice, the probability at the boundary is generally nonzero.
Check probability mass. A partial sum $\sum_{k=0}^{m}g_k$ must be between zero and one. Its complement is the strict tail $P(S>mh)$.

For the common severity distribution in this tutorial, the Poisson model gives $P(S>4h)=0.2978$, while the negative binomial model gives $0.2939$. This is a comparison at one specified threshold, not a general theorem that one model has a larger aggregate tail. Panjer recursion makes the comparison precise by preserving the same severity probabilities and changing only the frequency mechanism.

Check Your Understanding

For each problem, state the relevant recursion or event boundary before calculating. Answers are rounded to four decimal places when they are probabilities.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.5)$, with $p_1=0.4$, $p_2=0.6$, and no other positive severity probabilities. Find $g_0$, $g_1$, and $g_2$. (Answer: $g_0=0.2231,\ g_1=0.1339,\ g_2=0.2410$)
A hypothetical count follows the negative binomial model with $r=3$, $q=0.25$, and severity probabilities $p_1=0.7,\ p_2=0.3$. Find $g_0$ and $g_1$. (Answer: $g_0=0.4219,\ g_1=0.2215$)
For $N\sim\operatorname{Poisson}(2)$ and severity probabilities $p_1=0.25,\ p_2=0.75$, calculate $g_0$, $g_1$, and $g_2$. (Answer: $g_0=0.1353,\ g_1=0.0677,\ g_2=0.2199$)
Two portfolios share $p_1=0.5,p_2=0.3,p_3=0.2$. One has Poisson mean 2; the other has negative binomial parameters $r=2,q=0.5$. Using the calculations in this tutorial, compare the probabilities of aggregate loss at least $4h$. (Answer: Poisson $0.4226$; negative binomial $0.3813$)
A hypothetical pool uses the negative binomial model $r=2,q=0.4$, with $p_1=0.6,p_2=0.4$. Find $g_0$ and $g_1$, and identify the mean claim count. (Answer: $g_0=0.3600,\ g_1=0.1728$; mean count $4/3$)

What You'll Learn