Panjer Recursion With A Zero-Modified Frequency

Why Zero Modification Changes the Recursion

Panjer recursion turns a count distribution’s probability recurrence into a calculation for the aggregate loss distribution. For an \((a,b,0)\) frequency distribution, the recursion at positive aggregate amounts has no extra correction term. A zero-modified frequency distribution follows the same count recurrence only from count one to count two onward. Its probability at count one therefore need not be the value implied by extending the recurrence back to zero. That difference changes the aggregate recursion.

The key is to keep three quantities distinct: the probability \(p_0\) of zero claims, the aggregate probability \(g_0\) of zero total payment, and the correction determined by \(p_1\) and \(p_0\). When every claim pays a positive amount, \(g_0=p_0\). If a claim can pay zero, then positive claim counts can also produce zero aggregate payment, so \(g_0\) is generally larger than \(p_0\).

Definition: Let \(N\) be the claim count, with \(p_n=P(N=n)\), and suppose its probabilities satisfy \[ n p_n=(a n+b)p_{n-1},\qquad n\ge 2. \] This is an \((a,b,1)\) frequency recurrence: it is required from \(n=2\), not from \(n=1\). Define the zero-modification correction \[ C=p_1-(a+b)p_0. \] The correction is zero precisely when the recurrence also holds at \(n=1\), as in the \((a,b,0)\) case.

For the aggregate model, let \(X_1,X_2,\ldots\) be independent and identically distributed claim payments, independent of \(N\). Assume the payments are nonnegative integers, with \(f_j=P(X_1=j)\), and write \(S=\sum_{r=1}^{N}X_r\), where an empty sum is zero. Let \(g_k=P(S=k)\). The probability generating function of a claim payment is \(F(z)=\sum_{j\ge0}f_jz^j\); the count and aggregate probability generating functions are \(P(z)=\sum_{n\ge0}p_nz^n\) and \(G(z)=\sum_{k\ge0}g_kz^k\). Independence and conditioning on \(N\) give \(G(z)=P(F(z))\).

Deriving the Zero-Modified Aggregate Recursion

The derivation starts with the count recurrence only where it is assumed to hold: for \(n\ge2\). Multiply by \(z^n\) and sum over those counts. The left side is \(zP'(z)-p_1z\). Re-indexing the right side gives

zP'(z)-p_1z =z\left[a zP'(z)+(a+b)\{P(z)-p_0\}\right].

Dividing by \(z\) and rearranging yields the generating-function identity

$$ (1-az)P'(z)=p_1+(a+b)P(z)-(a+b)p_0. $$

Substitute \(F(z)\) for \(z\), use \(G(z)=P(F(z))\), and apply the chain rule \(G'(z)=P'(F(z))F'(z)\). The result is

(1-aF(z))G'(z)=\{C+(a+b)G(z)\}F'(z), \qquad C=p_1-(a+b)p_0.

Equating coefficients of \(z^{k-1}\) gives the recursion. In the product \(F G'\), the \(f_0\) term contributes \(f_0 k g_k\) on the left. For the terms with \(j\ge1\), combining the coefficients from \(aFG'\) and \((a+b)GF'\) gives \(ak+bj\). The term \(C F'\) contributes \(Ck f_k\). Thus, for \(k\ge1\),

(1-af_0)k g_k =\sum_{j=1}^{k}(ak+bj)f_jg_{k-j}+Ck f_k,

or, equivalently,

g_k=\frac{1}{1-af_0} \left[ \sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j} +C f_k \right],\qquad k\ge1.

Theorem: For a discrete \((a,b,1)\) count distribution and independent nonnegative integer claim payments, the aggregate probabilities satisfy the recursion above whenever \(1-af_0\ne0\). The required initial value is \(g_0=P(F(0))=P(f_0)\), where the final \(P\) denotes the count probability generating function. In particular, if \(f_0=0\), then \(g_0=p_0\). If \(f_0>0\), use \(g_0=P(f_0)\), not \(p_0\).

The result also shows exactly how zero modification enters. When \(C=0\), the extra term disappears and the \((a,b,0)\) recursion is recovered. When \(C\ne0\), omitting \(C f_k\) generally gives incorrect aggregate probabilities, even if the remaining sum is evaluated correctly.

Worked Calculations

Worked Example: A Zero-Modified Geometric Count

A hypothetical portfolio has count probabilities \(p_0=0.2\) and \(p_n=0.4(0.5)^{n-1}\) for \(n\ge1\). Each claim pays 1 or 2 units, each with probability \(0.5\). Calculate \(g_0,g_1,g_2,\) and \(g_3\).

Method and count check. For \(n\ge2\), \(p_n=0.5p_{n-1}\), so \(n p_n=(0.5n)p_{n-1}\). The parameters are \(a=0.5,\ b=0\). The count probabilities are normalized because \[ p_0+\sum_{n=1}^{\infty}p_n=0.2+\frac{0.4}{1-0.5}=1. \] The correction is \(C=p_1-(a+b)p_0=0.4-0.5(0.2)=0.3\). Since payments are positive, \(f_0=0\) and \(g_0=p_0=0.2\).

Calculate \(g_1\). Here \(f_1=f_2=0.5\). Only \(j=1\) contributes: \[ g_1=(0.5)(0.5)(0.2)+(0.3)(0.5)=0.05+0.15=0.20. \]

Calculate \(g_2\). \[ g_2=(0.5)(0.5)(0.20)+(0.5)(0.5)(0.20)+(0.3)(0.5) =0.05+0.05+0.15=0.25. \] The first two terms correspond to \(j=1\) and \(j=2\); the last is the zero-modification correction.

Calculate \(g_3\). There is no \(f_3\) term, so \[ g_3=(0.5)(0.5)(0.25)+(0.5)(0.5)(0.20)=0.0625+0.05=0.1125. \] As a check by direct conditioning, \(S=3\) can occur with two claims of sizes 1 and 2, contributing \(p_2(2)(0.5)(0.5)=0.2(0.5)=0.1\), or with three claims all of size 1, contributing \(p_3(0.5)^3=0.1(0.125)=0.0125\). Their sum is \(0.1125\), as calculated. The method avoids incorrectly extending the count recurrence to \(n=1\).

Worked Example: A Zero-Modified Poisson Count

Suppose a hypothetical claim count has \(p_n=0.25/n!\) for \(n\ge1\), with \(p_0\) chosen to make the probabilities sum to one. This is a zero-modified Poisson-type count with \(\lambda=1\). Claim payments are 1 unit with probability \(0.6\) and 2 units with probability \(0.4\). Find \(g_1\) and \(g_2\).

Method and count check. For \(n\ge2\), \(n p_n=p_{n-1}\), so \(a=0,\ b=1\). Normalization gives \[ p_0=1-\sum_{n=1}^{\infty}\frac{0.25}{n!} =1-\frac{e-1}{4} =0.57042955\ldots. \] Therefore \(C=p_1-p_0=0.25-p_0=-0.32042955\ldots\). Payments are positive, hence \(g_0=p_0\).

Calculate \(g_1\). \[ g_1=p_1f_1=0.25(0.6)=0.15. \] The recursion gives the same result: \(g_1=p_0f_1+C f_1=(p_0+C)0.6=0.25(0.6)\).

Calculate \(g_2\). In the recursion, the \(j=1\) contribution is \((1/2)(0.6)(0.15)=0.045\), the \(j=2\) contribution is \((0.4)p_0\), and the correction is \((0.4)C\). Thus \[ g_2=0.045+0.4p_0+0.4(0.25-p_0)=0.145. \] Direct conditioning checks this: aggregate 2 arises from one claim of size 2, with probability \(0.25(0.4)=0.1\), or two claims both of size 1, with probability \(p_2(0.6)^2=(0.25/2)(0.36)=0.045\). The total is \(0.145\). The cancellation also illustrates why a negative \(C\) is not itself a problem; it corrects the recursion while the resulting aggregate probability remains nonnegative.

Worked Example: Initialization When a Claim Can Pay Zero

Use the geometric count model \(p_0=0.2,\ p_n=0.4(0.5)^{n-1}\) for \(n\ge1\), with \(a=0.5,\ b=0\). Now suppose each claim payment is 0, 1, or 2 units with probabilities \(0.2,\ 0.5,\ 0.3\), respectively. Calculate \(g_0,g_1,\) and \(g_2\).

Method and initialization. The correction remains \(C=0.4-0.5(0.2)=0.3\). The aggregate is zero if the count is zero or if every claim pays zero. Hence \[ g_0=P(0.2)=0.2+\sum_{n=1}^{\infty}0.4(0.5)^{n-1}(0.2)^n =0.2+\frac{0.4(0.2)}{1-0.5(0.2)} =0.2+\frac{0.08}{0.9} =\frac{13}{45}. \]

Calculate \(g_1\). The denominator is \(1-af_0=1-(0.5)(0.2)=0.9\). Therefore \[ g_1=\frac{(0.5)(0.5)(13/45)+(0.3)(0.5)}{0.9} =\frac{2/9}{9/10} =\frac{20}{81} \approx0.246914. \]

Calculate \(g_2\). \[ g_2=\frac{ (0.5)(0.5)(20/81) +(0.5)(0.3)(13/45) +(0.3)(0.3) }{0.9} =\frac{79/405}{9/10} =\frac{158}{729} \approx0.216735. \] Probabilities are rounded to six decimals in the final approximations. The essential check is that \(g_0=13/45\), not \(p_0=0.2\): claims can occur while all their payments are zero. The method avoids treating the zero-claim probability as the only way to obtain zero aggregate payment.

Common Exam Trap and Practical Checks

A frequent error is to use the \((a,b,0)\) formula unchanged for a zero-modified count. In the first worked example, that wrong calculation would set \(C=0\). It would produce \(g_1=(0.5)(0.5)(0.2)=0.05\), even though one claim alone contributes \(p_1f_1=0.4(0.5)=0.20\). The correction is to calculate \(C=p_1-(a+b)p_0=0.3\) and include \(C f_1\), giving \(g_1=0.20\).

A second error is to initialize \(g_0\) as \(p_0\) without checking whether individual payments can be zero. If \(f_0>0\), the correct initialization is \(g_0=P(f_0)\). A third is to assert \((a,b,1)\) membership after checking only that probabilities sum to one. Check the defining relation \(n p_n=(an+b)p_{n-1}\) for counts \(n\ge2\), and separately check that the probabilities are valid and normalized.

A useful calculation sequence is: identify \(a,b,p_0,p_1\); verify the count recurrence and normalization; calculate \(C\); determine \(g_0\) from the payment model; then apply the recursion in increasing order of \(k\). When possible, check an early value by conditioning on the count, as in the examples. This topic is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state or use the relevant zero-modification correction and show enough work to check the final probability.

A hypothetical count has \(p_0=0.2\) and \(p_n=0.6(0.25)^{n-1}\) for \(n\ge1\). Claims pay 1 unit with probability \(0.7\) and 2 units with probability \(0.3\). Find \(g_1\) and \(g_2\). (Answer: \(g_1=0.42,\ g_2=0.2535\).)
A zero-modified Poisson-type count has \(p_n=0.25/n!\) for \(n\ge1\), with \(p_0\) determined by normalization. Claims pay 1 unit with probability \(0.8\) and 2 units with probability \(0.2\). Find \(g_1\) and \(g_2\). (Answer: \(g_1=0.2000,\ g_2=0.1300\).)
A count has \(p_0=0.25\) and \(p_n=0.45(0.4)^{n-1}\) for \(n\ge1\). Payments are 0, 1, and 2 units with probabilities \(0.1,\ 0.6,\ 0.3\). Find \(g_0\) and \(g_1\). (Answer: \(g_0=0.296875,\ g_1=0.293750\).)
A count has \(p_0=0.1,\ p_n=0.45(0.5)^{n-1}\) for \(n\ge1\). Payments are 1 or 2 units, each with probability \(0.4\) and \(0.6\), respectively. Calculate \(g_2\) using the recursion. (Answer: \(g_2=0.3060\).)

What You'll Learn