Adjusting Panjer Calculations for Zero-Severity Claims

Why Zero-Severity Claims Need Special Treatment

A claim can occur without producing a payment. For example, a hypothetical policy may have a deductible, so a reported loss below the deductible results in zero payment. In an aggregate loss model, that zero payment contributes nothing to total loss, but the claim still belongs to the original claim count. A Panjer calculation must handle this distinction consistently.

There are two equivalent approaches. One can retain zero-severity claims in the severity distribution and use the original claim count, or remove them by thinning the count to the number of positive payments and conditioning severity on a positive amount. The aggregate distribution is the same if each approach is applied correctly.

Definition: Suppose the claim count is \(N\), and individual payments \(X_1,X_2,\ldots\) are independent and identically distributed, independent of \(N\). The aggregate payment is \(S=\sum_{i=1}^{N}X_i\), with \(S=0\) when \(N=0\). For severity on the lattice \(0,h,2h,\ldots\), write \(p_j=P(X=jh)\), so \(p_0\) is the probability of zero payment. The aggregate probabilities are \(f_k=P(S=kh)\).

The probability generating function (PGF) of the severity is \(P_X(z)=\sum_{j\geq 0}p_jz^j\). If \(G_N(z)=E[z^N]\) is the PGF of the claim count, independence gives the aggregate PGF \(F_S(z)=G_N(P_X(z))\). In particular,

$$ f_0=P(S=0)=G_N(p_0). $$

This includes both the possibility of no claims and the possibility that every claim has zero payment. It is generally not just \(P(N=0)\).

Retaining Zero Payments in the Panjer Recursion

For a claim count in the \((a,b,0)\) Panjer class, its probabilities satisfy \(P(N=n)=(a+b/n)P(N=n-1)\) for \(n\geq1\). Let the severity be supported on the nonnegative lattice and let \(1-ap_0\neq0\). The aggregate probabilities are initialized at \(f_0=G_N(p_0)\), then for \(k\geq1\) satisfy:

f_k=\frac{1}{1-ap_0}\sum_{j=1}^{k}\left(a+b\frac{j}{k}\right)p_j f_{k-j}.

The sum begins at \(j=1\): a zero payment does not increase the aggregate amount. Its probability \(p_0\) instead appears in the denominator and in the initial value \(f_0\). For a Poisson count with mean \(\lambda\), \(a=0\), \(b=\lambda\), and \(f_0=\exp\{\lambda(p_0-1)\}\). The recursion reduces to \(f_k=(\lambda/k)\sum_{j=1}^{k}j p_jf_{k-j}\).

Derivation: The PGF of an \((a,b,0)\) count satisfies \((1-at)G_N'(t)=(a+b)G_N(t)\), as follows by multiplying its probability recurrence by powers of \(t\) and summing. Substitute \(t=P_X(z)\) and use \(F_S(z)=G_N(P_X(z))\). The chain rule gives \((1-aP_X(z))F_S'(z)=(a+b)P_X'(z)F_S(z)\). Equating coefficients of \(z^{k-1}\) yields \(k(1-ap_0)f_k=\sum_{j=1}^{k}(ak+bj)p_jf_{k-j}\). Dividing by \(k(1-ap_0)\) gives the stated recursion. The \(p_0\) term is part of the coefficient multiplying \(f_k\), which explains the denominator.

Worked Example: Poisson Count With Two Positive Payment Amounts

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(2)\) claims. Each claim pays 0 with probability \(p_0=1/2\), 1 unit with probability \(p_1=1/3\), and 2 units with probability \(p_2=1/6\). Find \(P(S=0)\), \(P(S=1)\), \(P(S=2)\), and \(P(S\leq2)\), by both methods.

Method 1: retain zero payments. Here \(f_0=\exp\{2(1/2-1)\}=e^{-1}\). For \(k=1\), the Poisson recursion gives

f_1=2p_1f_0=2\left(\frac13\right)e^{-1}=\frac23e^{-1}.

For \(k=2\), both positive severity probabilities contribute:

\begin{aligned} f_2 &=\frac{2}{2}\left(p_1f_1+2p_2f_0\right)\\ &=\left(\frac13\cdot\frac23+\;2\cdot\frac16\right)e^{-1}\\ &=\left(\frac29+\frac13\right)e^{-1} =\frac59e^{-1}. \end{aligned}

Therefore \(P(S\leq2)=f_0+f_1+f_2=(1+2/3+5/9)e^{-1}=(20/9)e^{-1}\), which is \(0.8175\) to four decimal places.

Method 2: remove zero payments. A claim has positive payment with probability \(1-p_0=1/2\), so Poisson thinning gives \(M\sim\operatorname{Poisson}(2\cdot1/2)=\operatorname{Poisson}(1)\) positive payments. Conditional on a positive payment, the probabilities are \(r_1=p_1/(1-p_0)=2/3\) and \(r_2=p_2/(1-p_0)=1/3\). Thus \(g_0=e^{-1}\), \(g_1=1\cdot(2/3)g_0=(2/3)e^{-1}\), and

g_2=\frac{1}{2}\left(r_1g_1+2r_2g_0\right) =\frac12\left(\frac23\cdot\frac23+2\cdot\frac13\right)e^{-1} =\frac59e^{-1}.

The two methods agree at each aggregate amount, and their cumulative probability is \(0.8175\). This check also illustrates the key adjustment: after removing zero payments, use the conditional probabilities \(r_1,r_2\), not the original probabilities \(p_1,p_2\).

Removing Zero Payments by Thinning

Define \(M\) as the number of claims with positive payment. Conditional on each original claim independently having positive payment with probability \(1-p_0\), the PGF of \(M\) is:

G_M(z)=G_N\bigl(p_0+(1-p_0)z\bigr).

For \(j\geq1\), the severity conditional on positive payment has probabilities \(r_j=p_j/(1-p_0)\). Then \(M\) and the positive-payment amounts produce the same aggregate \(S\) as the original count and severity. The initialization after thinning is \(P(M=0)=G_M(0)=G_N(p_0)=f_0\). The conditional severity has zero probability at zero, so its recursion has no \(p_0\) denominator adjustment.

Thinning is especially direct for Poisson and binomial counts. If \(N\sim\operatorname{Poisson}(\lambda)\), then \(M\sim\operatorname{Poisson}(\lambda(1-p_0))\). If \(N\sim\operatorname{Binomial}(n,\theta)\), then \(M\sim\operatorname{Binomial}(n,\theta(1-p_0))\). In other frequency models, the PGF formula remains valid, but the resulting count distribution and any applicable recursion must be identified rather than assumed.

Worked Example: A Binomial Count With Zero-Payment Claims

A hypothetical contract has \(N\sim\operatorname{Binomial}(3,0.4)\) claims. Each claim pays either 0 or 1 unit, with \(p_0=0.25\) and \(p_1=0.75\). Find the probabilities \(f_0,f_1,f_2,f_3\) for the aggregate payment.

Method 1: retain zero payments. For the binomial count, \(a=-\theta/(1-\theta)=-2/3\) and \(b=(n+1)\theta/(1-\theta)=8/3\). The initial probability is

f_0=G_N(0.25)=(0.6+0.4\cdot0.25)^3=0.7^3=0.343.

The denominator in the recursion is \(1-ap_0=1-(-2/3)(1/4)=7/6\). For \(k=1\),

f_1=\frac{(a+b)p_1f_0}{1-ap_0} =\frac{2\cdot0.75\cdot0.343}{7/6} =0.441.

For \(k=2\), only \(j=1\) contributes because \(p_2=0\):

f_2=\frac{(a+b/2)p_1f_1}{7/6} =\frac{(2/3)(0.75)(0.441)}{7/6} =0.189.

Similarly, for \(k=3\), again only \(j=1\) contributes:

f_3=\frac{(a+b/3)p_1f_2}{7/6} =\frac{(2/9)(0.75)(0.189)}{7/6} =0.027.

Method 2: remove zero payments. Thinning gives \(M\sim\operatorname{Binomial}(3,0.4\cdot0.75)=\operatorname{Binomial}(3,0.3)\). Every positive payment is 1 unit, so \(P(S=k)=P(M=k)\). The probabilities are \(0.7^3=0.343\), \(3(0.3)(0.7)^2=0.441\), \(3(0.3)^2(0.7)=0.189\), and \((0.3)^3=0.027\), for \(k=0,1,2,3\), respectively. The check \(0.343+0.441+0.189+0.027=1\) confirms the full distribution. The calculation avoids treating the original expected claim count as the count of positive payments.

Keeping the Lattice and the Probabilities Straight

The recursion index refers to lattice units, not necessarily dollars. If payment amounts are \(0,2,4,\ldots\), then \(k=1\) corresponds to an aggregate of 2 units of currency and \(k=2\) to 4. The severity probabilities must use the same indexing throughout. Zero payments still have index 0; they are not a positive payment at the first lattice point.

Worked Example: Comparing Both Methods on a Coarser Lattice

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(1.5)\). A claim pays 0 with probability \(1/3\), 2 units with probability \(1/2\), or 4 units with probability \(1/6\). Find the probability that the aggregate payment is at most 4 units.

Index the lattice in steps of \(h=2\). Thus \(p_0=1/3\), \(p_1=1/2\), and \(p_2=1/6\). Retaining zero payments gives \(f_0=\exp\{1.5(1/3-1)\}=e^{-1}\). Then \(f_1=1.5(1/2)f_0=(3/4)e^{-1}\), and

\begin{aligned} f_2 &=\frac{1.5}{2}\left(p_1f_1+2p_2f_0\right)\\ &=\frac34\left(\frac12\cdot\frac34+2\cdot\frac16\right)e^{-1}\\ &=\frac34\left(\frac38+\frac13\right)e^{-1} =\frac{17}{32}e^{-1}. \end{aligned}

The event \(S\leq4\) includes the lattice amounts 0, 2, and 4. Hence

P(S\leq4)=f_0+f_1+f_2 =\left(1+\frac34+\frac{17}{32}\right)e^{-1} =\frac{73}{32}e^{-1} =0.8392.

As a check using positive payments, the count is Poisson with mean \(1.5(2/3)=1\), and the conditional severity probabilities at amounts 2 and 4 are \(3/4\) and \(1/4\). This gives \(g_0=e^{-1}\), \(g_1=(3/4)e^{-1}\), and \(g_2=\frac12[(3/4)(3/4)+2(1/4)]e^{-1}=(17/32)e^{-1}\), the same terms and the same cumulative probability.

Common Mistakes and a Practical Check

Using the original severity after thinning. Once zero payments are removed from the count, divide each positive severity probability by \(1-p_0\). In the first example, \(p_1=1/3\) becomes \(r_1=2/3\); using \(1/3\) with the thinned Poisson mean would give the wrong \(g_1\).
Changing the count mean but forgetting to change severity. Thinning and conditional severity are a paired adjustment. Do not thin the count and also keep the unconditional positive probabilities.
Forgetting the zero aggregate probability. With zero severities, \(f_0=G_N(p_0)\), not generally \(P(N=0)\). In the first example, \(P(N=0)=e^{-2}\), whereas \(P(S=0)=e^{-1}\).
Mixing lattice units. If the severity step is 2 units, the recursion index \(k=2\) represents an aggregate of 4 units, not 2.

A reliable calculation records \(p_0\), the positive probabilities, the original frequency model, and the chosen approach before starting the recursion. If both approaches are available, compare \(f_0\) and at least one positive aggregate probability. Agreement is a strong check that the thinning probability and conditional severity have been handled consistently.

Check Your Understanding

For each problem, state whether the zero-severity probability is retained or removed as specified, and keep the aggregate lattice consistent.

\(N\sim\operatorname{Poisson}(3)\). Each claim pays 0 with probability 0.2 and 1 unit with probability 0.8. Using the positive-payment count, find \(P(S=0)\) and \(P(S=1)\). (Answer: \(e^{-2.4}=0.0907\) and \(2.4e^{-2.4}=0.2177\), respectively.)
\(N\sim\operatorname{Binomial}(4,0.25)\). Each claim pays 0 with probability 0.4 and 1 unit with probability 0.6. Find \(P(S=0)\) and \(P(S=1)\). (Answer: \(0.85^4=0.5220\) and \(4(0.15)(0.85)^3=0.3685\), respectively.)
\(N\sim\operatorname{Poisson}(1)\), with severity probabilities \(p_0=0.25,p_1=0.5,p_2=0.25\) on the unit lattice. Find \(P(S\leq2)\) using positive-payment thinning. (Answer: \(1.875e^{-0.75}=0.8857\).)
\(N\sim\operatorname{Poisson}(2)\). A claim pays 0 with probability 0.6 and 1 unit with probability 0.4. Find the aggregate probabilities at 0 and 1, using the positive-payment count. (Answer: \(e^{-0.8}=0.4493\) and \(0.8e^{-0.8}=0.3595\), respectively.)
A Poisson count has mean 5, and the severity probabilities are \(p_0=0.3,p_1=0.35,p_2=0.35\). After zero payments are removed, give the mean count and the conditional probabilities of payments at 1 and 2 units. (Answer: mean count \(3.5\); conditional probabilities \(0.5000\) and \(0.5000\).)

What You'll Learn