Why the Starting Value Matters
Panjer recursion calculates aggregate probabilities in order, beginning at total loss zero and then moving to positive amounts on a severity grid. A correct recurrence does not, by itself, guarantee a correct table: the initial probability must match the model, and a zero-valued claim must not be confused with the absence of claims. These distinctions are especially important when the frequency distribution has a separate probability for exactly one claim.
Let \(N\) be the number of claims, with \(p_n=P(N=n)\). Given \(N\), let the claim amounts be independent and identically distributed, independent of \(N\), and let \(S\) be their aggregate. On a lattice with unit \(h\), write \(q_j=P(X=jh)\) for the probability that one claim has severity \(jh\), and \(g_k=P(S=kh)\). The grid must contain all possible severity amounts being modelled.
If all claim severities are strictly positive, total loss zero occurs exactly when there are no claims. If zero severities are possible, a positive number of claims can also produce total loss zero. The starting value must account for both possibilities.
The Zero-Loss Probability
Conditional on \(N=n\), aggregate loss is zero precisely when all \(n\) severities equal zero. By independence, that conditional probability is \(q_0^n\). For \(n=0\), the aggregate is zero by the empty-sum convention, so the conditional probability is one. Averaging over the claim count gives
This is the frequency probability generating function evaluated at \(q_0\). If \(q_0=0\), only the \(n=0\) term contributes, and \(g_0=p_0\). If \(q_0>0\), using \(p_0\) alone omits outcomes in which claims occur but all have zero severity.
Worked Example: Positive Severities and a Binomial Frequency
A hypothetical portfolio has \(N\sim\operatorname{Binomial}(2,0.4)\). Each claim is independent of the others and of \(N\), with severity \(h\) with probability \(0.7\) and \(2h\) with probability \(0.3\). Find \(g_0\), \(g_1\), and \(g_2\) using the recursion for an \((a,b,0)\) frequency.
Method and frequency parameters. Here \(p_0=0.6^2=0.36\), \(p_1=2(0.4)(0.6)=0.48\), and \(p_2=0.4^2=0.16\). The binomial frequency is in the \((a,b,0)\) class, with \(a=-0.4/0.6=-2/3\) and \(b=3(0.4)/0.6=2\). Thus \(a+b=4/3\) and \(a+b/2=1/3\). Severity probabilities are \(q_0=0,\ q_1=0.7,\ q_2=0.3\).
Initial value. Since all claim severities are positive, \(g_0=p_0=0.36\).
At \(k=1\). The \((a,b,0)\) recursion is \(g_k=\sum_{j=1}^{k}(a+bj/k)q_jg_{k-j}\) when \(q_0=0\). Therefore \(g_1=(a+b)q_1g_0=(4/3)(0.7)(0.36)=0.336\).
At \(k=2\). The \(j=1\) coefficient is \(a+b/2=1/3\), and the \(j=2\) coefficient is \(a+b=4/3\). Hence \(g_2=(1/3)(0.7)(0.336)+(4/3)(0.3)(0.36)=0.0784+0.144=0.2224\).
Check. A total of \(h\) is possible only with one claim, so its probability is \(p_1q_1=0.48(0.7)=0.336\). A total of \(2h\) comes from one claim of size \(2h\), or two claims of size \(h\): \(p_1q_2+p_2q_1^2=0.48(0.3)+0.16(0.7^2)=0.144+0.0784=0.2224\). Both recursion values agree. The key trap avoided is treating \(g_0\) as anything other than the zero-claim probability when severities are positive.
Starting Values for the Two Frequency Classes
The \((a,b,0)\) class has probabilities satisfying \(p_n=(a+b/n)p_{n-1}\) for \(n\geq1\). In particular, \(p_1=(a+b)p_0\). The \((a,b,1)\) class instead satisfies that relation for \(n\geq2\); its \(p_0\) and \(p_1\) are not required to obey that first-step relation. That exception creates an extra term in the aggregate recursion.
For either class, when severity may be zero, the \((a,b,0)\) recursion is
This formula requires \(1-aq_0\ne0\). For positive severities, \(q_0=0\), so the denominator is one and \(g_0=p_0\). For a frequency in the \((a,b,1)\) class, the corresponding recursion is
Here the initial value is still \(g_0=\sum_{n=0}^{\infty}p_nq_0^n\). The correction \(p_1-(a+b)p_0\) is zero for an \((a,b,0)\) frequency, but need not be zero for an \((a,b,1)\) frequency. When \(q_0=0\), the \((a,b,1)\) recursion reduces to
Why the Exceptional Correction Appears
A short probability-generating-function argument explains the correction. Let \(P_N(t)=\sum_{n\geq0}p_nt^n\) be the frequency generating function and \(Q(z)=\sum_{j\geq0}q_jz^j\) the severity generating function. Independence gives the aggregate generating function \(G(z)=P_N(Q(z))=\sum_{k\geq0}g_kz^k\).
For an \((a,b,1)\) frequency, apply \(np_n=(an+b)p_{n-1}\) only for \(n\geq2\), leaving the \(n=1\) term separate. Summing the resulting terms gives
Substitute \(t=Q(z)\), use \(G(z)=P_N(Q(z))\), and apply the chain rule. This yields
Equating the coefficients of \(z^k\) gives the \((a,b,1)\) recursion above. The term involving \(p_1-(a+b)p_0\) remains because the frequency relation was not assumed at \(n=1\). For an \((a,b,0)\) frequency, that relation does hold at \(n=1\), so \(p_1=(a+b)p_0\) and the exceptional term vanishes. This is why applying the simpler \((a,b,0)\) formula to an \((a,b,1)\) model can give incorrect aggregate probabilities even when its \(g_0\) is correct.
Worked Example: An Exceptional One-Claim Probability
A hypothetical claim count has \(p_0=0.30,\ p_1=0.60,\ p_2=0.10\), and zero probability for all larger counts. For \(n\geq2\), its probabilities follow the \((a,b,1)\) relation with \(a=-1/3,\ b=1\). Each claim has severity \(h\) or \(2h\), each with probability \(0.5\). Calculate \(g_0,\ g_1,\) and \(g_2\).
Check the frequency specification. At \(n=2\), the relation gives \(p_2=(a+b/2)p_1=(-1/3+1/2)(0.60)=(1/6)(0.60)=0.10\). At \(n=3\), it gives \(p_3=(a+b/3)p_2=0\), consistent with the stated support. Also \(p_0+p_1+p_2=0.30+0.60+0.10=1\).
Initial value and correction. Since \(q_0=0\), \(g_0=p_0=0.30\). The common coefficient is \(a+b=2/3\), while the exceptional correction is \(p_1-(a+b)p_0=0.60-(2/3)(0.30)=0.40\).
At \(k=1\). With \(q_1=0.5\), the recursion gives \(g_1=(2/3)(0.5)(0.30)+(0.40)(0.5)=0.10+0.20=0.30\).
At \(k=2\). The \(j=1\) coefficient is \(a+b/2=1/6\), and the \(j=2\) coefficient is \(a+b=2/3\). Thus \(g_2=(1/6)(0.5)(0.30)+(2/3)(0.5)(0.30)+(0.40)(0.5)=0.025+0.10+0.20=0.325\).
Check. A total of \(h\) arises from exactly one claim of size \(h\), giving \(0.60(0.5)=0.30\). A total of \(2h\) arises from one claim of size \(2h\), or two claims both of size \(h\): \(0.60(0.5)+0.10(0.5^2)=0.30+0.025=0.325\). The values agree. Omitting the exceptional correction would miss part of the contribution from the one-claim probability.
Worked Example: Zero-Severity Claims Change the Initial Value
A hypothetical frequency is \(N\sim\operatorname{Binomial}(2,0.5)\), so \(p_0=0.25,\ p_1=0.50,\ p_2=0.25\). A claim has severity \(0,\ h,\) or \(2h\), with respective probabilities \(q_0=0.20,\ q_1=0.50,\ q_2=0.30\). Use the \((a,b,0)\) recursion to find \(g_0,\ g_1,\) and \(g_2\).
Parameters. For this binomial frequency, \(a=-0.5/0.5=-1\), \(b=3(0.5)/0.5=3\), and \(a+b=2\). The zero-severity denominator is \(1-aq_0=1-(-1)(0.20)=1.20\).
Initial value. A zero total occurs with no claims, or with one or two claims all having zero severity. Therefore \(g_0=p_0+p_1q_0+p_2q_0^2=0.25+0.50(0.20)+0.25(0.20^2)=0.25+0.10+0.01=0.36\).
At \(k=1\). The \((a,b,0)\) recursion gives \(g_1=(a+b)q_1g_0/(1-aq_0)=2(0.50)(0.36)/1.20=0.30\).
At \(k=2\). The coefficients for \(j=1\) and \(j=2\) are \(a+b/2=0.5\) and \(a+b=2\), respectively. Thus \(g_2=\bigl[0.5(0.50)(0.30)+2(0.30)(0.36)\bigr]/1.20=(0.075+0.216)/1.20=0.2425\).
Check. Directly, total \(h\) occurs from one claim of size \(h\), or from two claims with one zero and one \(h\): \(0.50(0.50)+0.25[2(0.20)(0.50)]=0.25+0.05=0.30\). Total \(2h\) occurs from one claim of size \(2h\), or from two claims with severities summing to \(2h\): \(0.50(0.30)+0.25[2(0.20)(0.30)+(0.50)^2]=0.15+0.0925=0.2425\). Both values agree. The trap avoided is setting \(g_0=p_0=0.25\), which ignores positive claim counts whose severities are all zero.
Common Initialization Errors
- Using \(p_0\) when \(q_0>0\). The correct starting value is \(\sum p_nq_0^n\). For example, in the preceding example, \(p_0=0.25\) is not the zero-loss probability; the correct value is \(0.36\).
- Using \(p_1=(a+b)p_0\) in every frequency model. This equality is part of the \((a,b,0)\) relation, but is not required for an \((a,b,1)\) model. Use the exceptional correction for the latter.
- Applying a positive-severity formula without checking \(q_0\). When \(q_0\ne0\), the \((a,b,0)\) recursion has denominator \(1-aq_0\); omitting it changes every recursively calculated probability.
- Starting at the wrong lattice point. \(g_k\) is the probability of aggregate amount \(kh\), not of \(k\) claims. The severity grid determines the index.
A reliable calculation therefore begins by stating the lattice, the severity probabilities including \(q_0\), the frequency class, and its relevant parameters. Calculate \(g_0\) from the model, then select the matching recursion and proceed in increasing order of \(k\). These checks are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each situation, identify the relevant initialization and calculate the requested aggregate probability.
- A claim count has \(p_0=0.2,\ p_1=0.5,\ p_2=0.3\), and all claim severities are strictly positive. What is \(P(S=0)\)? (Answer: \(0.2000\))
- A hypothetical frequency is \(N\sim\operatorname{Binomial}(3,0.25)\), and each claim has positive severity. Calculate \(g_0\). (Answer: \(0.4219\))
- A claim count has \(P(N=n)=0.4(0.6)^n\) for \(n\geq0\). Each claim is zero with probability \(0.2\), independently. Calculate \(P(S=0)\). (Answer: \(0.4545\))
- An \((a,b,1)\) frequency has \(p_0=0.30,\ p_1=0.60,\ a=-1/3,\ b=1\). Severity is \(h\) with probability \(0.4\) and \(2h\) with probability \(0.6\), with no zero severity. Calculate \(g_1\), using the correct exceptional term. (Answer: \(0.2400\))
- A frequency has \(p_0=0.30,\ p_1=0.60,\ p_2=0.10\), with no larger counts. A claim is zero with probability \(0.1\), independently. Calculate \(P(S=0)\). (Answer: \(0.3610\))