Panjer Recursion for (a,b,1) Claim Counts

Why the Initial Claim Probabilities Matter

Panjer recursion calculates aggregate-loss probabilities one value at a time rather than summing over every possible claim count separately. For frequency distributions in the $(a,b,0)$ class, one frequency recurrence applies starting with a count of one. In the $(a,b,1)$ class, the recurrence applies only from a count of two onward. The probabilities of zero and one claim are exceptional, so the aggregate recursion must retain their effect.

This distinction is important even when all the later claim-count probabilities follow a simple pattern. Using the $(a,b,0)$ aggregate formula without adjustment can give plausible-looking values that are nevertheless wrong.

Definition: A claim count $N$ belongs to the $(a,b,1)$ class if its probabilities $p_n=P(N=n)$ satisfy $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq 2. $$ The values $p_0$ and $p_1$ are specified separately, subject to the probabilities being nonnegative and summing to one. The recurrence does not require $p_1=(a+b)p_0$.

Let the aggregate loss be $S=\sum_{i=1}^N X_i$, where the claim amounts $X_i$ are independent and identically distributed, and independent of $N$. For the recursion, severity must be discrete on the nonnegative integers. Write $f_j=P(X_i=j)$ and $g_k=P(S=k)$. A positive amount of loss is measured in the chosen integer units; for example, one unit could represent one hundred dollars.

Model assumptions: The claim count follows the $(a,b,1)$ recurrence for $n\geq2$; claim amounts are independent and identically distributed nonnegative integer-valued random variables; and claim amounts are independent of the claim count. The recursion below is not directly a recursion for continuous severity. Continuous severities must first be represented on an integer lattice for this method to apply.

The Aggregate Recursion

Define the probability generating functions $P_N(z)=\sum_{n=0}^{\infty}p_nz^n$ and $F(z)=\sum_{j=0}^{\infty}f_jz^j$. Independence implies that the aggregate-loss generating function is $G(z)=P_N(F(z))$. The $(a,b,1)$ frequency recurrence gives a differential equation for $P_N$, and substituting $F(z)$ into it leads to the aggregate recursion.

Theorem: For an $(a,b,1)$ frequency distribution and independent, identically distributed nonnegative integer-valued severities, let $$ C=p_1-(a+b)p_0. $$ If $1-af_0\neq0$, then for $k\geq1$, $$ g_k=\frac{\displaystyle\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}+C f_k}{1-af_0}. $$ The aggregate probability at zero is $g_0=P_N(f_0)=\sum_{n=0}^{\infty}p_nf_0^n$.

The correction $C f_k$ is the part that accounts for the exceptional initial probabilities. In the special case $p_1=(a+b)p_0$, $C=0$, and the correction disappears. That condition should not be assumed for a general $(a,b,1)$ distribution.

Why the Correction Appears

The derivation shows exactly where the exceptional probabilities enter. From the frequency recurrence, for $n\geq2$, $np_n=(an+b)p_{n-1}$. Summing these terms in the derivative of $P_N$, while excluding the $n=1$ term, gives $$ P_N'(z)-p_1=a\bigl(zP_N'(z)+P_N(z)-p_0\bigr)+b\bigl(P_N(z)-p_0\bigr). $$ Rearranging, $$ (1-az)P_N'(z)=(a+b)P_N(z)+p_1-(a+b)p_0. $$

Set $G(z)=P_N(F(z))$ and differentiate using the chain rule. Then $$ (1-aF(z))G'(z)=\bigl((a+b)G(z)+C\bigr)F'(z). $$ To obtain the recursion for $g_k$, equate the coefficients of $z^{k-1}$. On the left, the coefficient is $$ kg_k-a\sum_{j=0}^{k-1}f_j(k-j)g_{k-j}. $$ On the right, it is $$ (a+b)\sum_{j=1}^{k}j f_jg_{k-j}+Ck f_k. $$ The $j=0$ term on the left is $a f_0kg_k$. Moving it to the other side and dividing by $k$ gives $$ (1-af_0)g_k=\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}+C f_k. $$ This derivation also explains why $f_0$ appears in the denominator: a zero-sized claim can leave aggregate loss unchanged, so the coefficient of $g_k$ must be collected before solving for it.

For valid frequency and severity distributions, the recursion is used with a nonzero denominator. In standard nondegenerate cases with $0\leq f_0\leq1$, this denominator is positive. If it vanishes, this rearranged recursion cannot be used to solve for $g_k$.

Worked Calculations

Worked Example: A Finite Claim-Count Distribution

A hypothetical portfolio has claim count probabilities $p_0=1/5$, $p_1=8/15$, and $p_2=4/15$, with no probability of more than two claims. Its claim-count recurrence is $(a,b,1)=(-1,3,1)$: for $n=2$, $a+b/n=-1+3/2=1/2$, so $p_2=p_1/2=4/15$; at $n=3$, the recurrence multiplier is zero. Each claim is either 1 or 2 loss units, with $f_1=f_2=1/2$ and $f_0=0$. Find $g_1$ and $g_2$.

Method. Compute the exceptional-probability correction first. Here $a+b=2$, so $$ C=p_1-(a+b)p_0=\frac{8}{15}-2\left(\frac{1}{5}\right)=\frac{8}{15}-\frac{6}{15}=\frac{2}{15}. $$ The denominator is $1-af_0=1-(-1)(0)=1$. Since $f_0=0$, $g_0=p_0=1/5$.

For $k=1$, the sum has only $j=1$: $$ g_1=\left(a+b\right)f_1g_0+C f_1 =2\left(\frac12\right)\left(\frac15\right)+\left(\frac{2}{15}\right)\left(\frac12\right) =\frac15+\frac{1}{15}=\frac{4}{15}. $$

For $k=2$, the $j=1$ coefficient is $a+b/2=-1+3/2=1/2$; the $j=2$ coefficient is $a+b=-1+3=2$. Therefore $$ \begin{aligned} g_2 &=\left(\frac12\right)\left(\frac12\right)g_1 +2\left(\frac12\right)g_0+C f_2\\ &=\frac14\left(\frac{4}{15}\right)+1\left(\frac15\right) +\left(\frac{2}{15}\right)\left(\frac12\right)\\ &=\frac{1}{15}+\frac{3}{15}+\frac{1}{15} =\frac{1}{3}. \end{aligned} $$

Check. Aggregate loss 1 occurs with one claim of size 1, with probability $p_1f_1=(8/15)(1/2)=4/15$. Aggregate loss 2 occurs either with one claim of size 2, probability $ (8/15)(1/2)=4/15$, or with two claims of size 1, probability $(4/15)(1/2)^2=1/15$. Their sum is $5/15=1/3$, matching the recursion. This direct check also shows why the exceptional correction must be included.

Worked Example: Positive Probability of a Zero-Size Claim

For a hypothetical count model, $a=-1$, $b=3$, $p_0=1/5$, $p_1=8/15$, and $p_2=4/15$, with no larger counts. Each claim has $f_0=1/5$, $f_1=1/2$, and $f_2=3/10$. Find $g_1$ and $g_2$.

Method. The frequency correction remains $C=2/15$, and the denominator is $$ 1-af_0=1-(-1)\left(\frac15\right)=\frac65. $$ First calculate $g_0$. The aggregate is zero if there are no claims, or if every claim has size zero: $$ g_0=p_0+p_1f_0+p_2f_0^2 =\frac15+\frac{8}{15}\left(\frac15\right) +\frac{4}{15}\left(\frac1{25}\right) =\frac{75+40+4}{375} =\frac{119}{375}. $$

For $k=1$, $$ \begin{aligned} g_1 &=\frac{(a+b)f_1g_0+C f_1}{1-af_0}\\ &=\frac{2\left(\frac12\right)\left(\frac{119}{375}\right) +\left(\frac{2}{15}\right)\left(\frac12\right)}{\frac65}\\ &=\frac{\frac{119}{375}+\frac{25}{375}}{\frac65} =\frac{\frac{144}{375}}{\frac65} =\frac{48}{125}\cdot\frac56 =\frac{8}{25}=0.32. \end{aligned} $$

For $k=2$, the coefficients for $j=1,2$ are $1/2$ and $2$, respectively: $$ \begin{aligned} g_2 &=\frac{\left(\frac12\right)\left(\frac12\right)g_1 +2\left(\frac{3}{10}\right)g_0+C f_2}{\frac65}\\ &=\frac{\frac14\left(\frac{8}{25}\right) +\frac35\left(\frac{119}{375}\right) +\frac{2}{15}\left(\frac{3}{10}\right)}{\frac65}\\ &=\frac{\frac{2}{25}+\frac{119}{625}+\frac{1}{25}}{\frac65} =\frac{\frac{50+119+25}{625}}{\frac65}\\ &=\frac{194}{625}\cdot\frac56 =\frac{97}{375}\approx0.2587. \end{aligned} $$

Check. A zero aggregate can arise from zero claims, one zero-size claim, or two zero-size claims; these contributions are exactly those used in $g_0$. The factor $1-af_0=6/5$ is also essential: omitting it would leave both positive-loss probabilities too small.

Worked Example: A Geometric-Tail Count Model

A hypothetical count distribution has $p_0=1/5$, $p_1=2/5$, and $p_n=(1/2)p_{n-1}$ for $n\geq2$, so it is in the $(a,b,1)$ class with $a=1/2$ and $b=0$. The probabilities sum to one because $\sum_{n=1}^{\infty}(2/5)(1/2)^{n-1}=4/5$, in addition to $p_0=1/5$. Let $f_0=0$, $f_1=3/5$, and $f_2=2/5$. Find $g_1$ and $g_2$.

Method. The exceptional correction is $$ C=p_1-(a+b)p_0=\frac25-\frac12\left(\frac15\right) =\frac25-\frac1{10}=\frac3{10}. $$ The denominator is $1-af_0=1$, and $g_0=p_0=1/5$.

For $k=1$, $a+b=1/2$, so $$ g_1=\frac12\left(\frac35\right)\left(\frac15\right) +\frac3{10}\left(\frac35\right) =\frac{3}{50}+\frac{9}{50} =\frac{6}{25}=0.24. $$ For $k=2$, the coefficients are $a+b/2=1/2$ and $a+b=1/2$: $$ \begin{aligned} g_2 &=\frac12\left(\frac35\right)g_1 +\frac12\left(\frac25\right)g_0 +C f_2\\ &=\frac{3}{10}\left(\frac{6}{25}\right) +\frac15\left(\frac15\right) +\frac3{10}\left(\frac25\right)\\ &=\frac{9}{125}+\frac{1}{25}+\frac{3}{25} =\frac{9+5+15}{125} =\frac{29}{125}=0.232. \end{aligned} $$

Check. Aggregate loss 1 can come from one claim of size 1, or from two claims when one has size 1 and the other size 0. Their probabilities are $p_1f_1=(2/5)(3/5)=0.24$ and $2p_2f_1f_0=0$, respectively. For loss 2, a single claim of size 2 contributes $(2/5)(2/5)=0.16$, and two claims both of size 1 contribute $p_2(3/5)^2=(1/5)(9/25)=0.072$. The sum is $0.232$, as calculated.

Common Mistake: Dropping the Exceptional Term

A tempting but incorrect shortcut is to use the $(a,b,0)$ recursion unchanged. That formula omits $C f_k$, which is justified only when $p_1=(a+b)p_0$. In the first worked example, if the correction were omitted, the calculation of $g_1$ would be $$ g_1^{\mathrm{wrong}}=(a+b)f_1g_0 =2\left(\frac12\right)\left(\frac15\right)=\frac15, $$ rather than the correct $g_1=4/15$. Continuing with that incorrect value would give $$ g_2^{\mathrm{wrong}}=\left(\frac12\right)\left(\frac12\right)\left(\frac15\right) +2\left(\frac12\right)\left(\frac15\right) =\frac{1}{20}+\frac15=\frac14, $$ rather than $1/3$. The direct count-and-severity check identifies the error: the probability of one claim of size 2 alone is $4/15$, already greater than the incorrect aggregate probability $1/4$.

The reverse mistake is to include the correction twice—for example, by adjusting $p_0$ or $p_1$ and also adding $C f_k$. Keep the original frequency probabilities in the model, calculate $C=p_1-(a+b)p_0$ once, and use the stated recursion.

Key takeaway: In an $(a,b,1)$ model, the count recurrence starts at $n=2$, not $n=1$. Consequently, aggregate probabilities require the additional term $\bigl[p_1-(a+b)p_0\bigr]f_k$. Check the initial probabilities and include this term unless it is exactly zero.

Check Your Understanding

For each problem, use the $(a,b,1)$ recursion and keep the exceptional-probability correction distinct from the summation.

A count distribution has $a=-1$, $b=3$, $p_0=1/5$, $p_1=8/15$, $p_2=4/15$, and no larger counts. Severity has $f_0=0$, $f_1=f_2=1/2$. Find $g_1$ and $g_2$. (Answer: $g_1=0.2667,\ g_2=0.3333$)
A count distribution has $a=1/2$, $b=0$, $p_0=1/5$, and $p_n=(1/2)p_{n-1}$ for $n\geq2$. Severity has $f_0=0$, $f_1=3/5$, and $f_2=2/5$. Find $g_2$. (Answer: $0.2320$)
A hypothetical portfolio has $p_0=1/4$, $p_1=3/4$, and no larger claim counts, with $a=-1$, $b=2$. Severity has $f_0=1/10$, $f_1=3/5$, and $f_2=3/10$. Calculate $g_0$ and $g_1$. (Answer: $g_0=0.3250,\ g_1=0.4500$)
Use $a=-1$, $b=3$, $p_0=1/5$, $p_1=8/15$, $p_2=4/15$, and no larger counts. For severity $f_0=1/5$, $f_1=1/2$, $f_2=3/10$, calculate $g_0$, $g_1$, and $g_2$. (Answer: $g_0=0.3173,\ g_1=0.3200,\ g_2=0.2587$)
If $p_0=0.10$, $p_1=0.30$, $a=0.4$, and $b=1.2$, calculate $C$. State whether the exceptional correction vanishes. (Answer: $C=0.1400$; it does not vanish)

What You'll Learn