Why a Membership Checklist Matters
A frequency model is not ready for aggregate-loss recursion merely because its probabilities look familiar or several consecutive ratios fit a pattern. The count support, the range where the probability recurrence holds, and the boundary probabilities must all agree. A finite-support model also has an endpoint to check; a zero-modified model may have special boundary probabilities; and a recurrence for positive counts does not automatically determine the probability at zero.
The checklist in this tutorial separates two decisions. First, does the frequency distribution belong to a Panjer class? Second, does it meet the conditions for the ordinary compound-distribution recursion? Keeping those decisions separate avoids applying a valid-looking formula outside the range for which it was derived. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The labels identify where the recurrence starts, not just a convenient name for the distribution. For an \((a,b,0)\) model, \(p_0\) and the recurrence generate the probabilities from \(n=1\) onward. For an \((a,b,1)\) model, the recurrence starts at \(n=2\), so both \(p_0\) and \(p_1\) must be supplied or otherwise established. The latter class can include distributions with a modified probability at zero or at one.
The Membership Checklist
Apply the steps in order. A later step does not repair a failure at an earlier one: for example, probabilities that sum to one can still fail the Panjer ratio test.
Identify the possible values of \(N\), including whether zero is possible and whether there is a finite maximum. The recurrence uses probabilities of consecutive counts, so do not form a ratio across a zero probability.
For each relevant positive \(n\) with \(p_{n-1}>0\), calculate \(r_n=p_n/p_{n-1}\). Test whether these ratios can be written as \(a+b/n\) using the same constants throughout the required range.
Check \(n\geq1\) for \((a,b,0)\) or \(n\geq2\) for \((a,b,1)\). For finite support ending at \(m\), the recurrence must also be consistent with \(p_{m+1}=0\), where \(p_m>0\).
Use the correct boundary values for the proposed class. Confirm that all probabilities are nonnegative and that their sum over the full support is one.
For the ordinary recursion, verify that the frequency recurrence is \((a,b,0)\), and that the severity probabilities are represented on the required lattice. If the model is only \((a,b,1)\), use a recursion derived for that class rather than treating it as \((a,b,0)\).
Why the Ratio Test Works
The defining recurrence says that the ratio of consecutive probabilities is an affine function of \(1/n\): \(r_n=a+b/n\). Thus, if a proposed model is in the class, the values of \(r_n\) must lie on the same straight-line relationship as \(1/n\) varies. Two ratios can determine candidate parameters, but additional ratios and the support endpoints must still be checked. Fitting two points is not proof of membership.
For an \((a,b,0)\) model, the ordinary aggregate recursion follows from this recurrence. Let the individual claim payment \(Y\) take nonnegative integer values with probabilities \(q_j=P(Y=j)\), and let \(S\) be the sum of \(N\) independent payments, with the payments identically distributed and independent of \(N\). Write \(g_s=P(S=s)\). Define the probability generating functions \(G(t)=\sum_{n\geq0}p_nt^n\) and \(Q(z)=\sum_{j\geq0}q_jz^j\). The generating function of \(S\) is \(H(z)=G(Q(z))=\sum_{s\geq0}g_sz^s\).
To justify the recursion, multiply \(np_n=(an+b)p_{n-1}\) by \(t^{n-1}\) and sum over \(n\geq1\). The left side is \(G'(t)\). On the right, setting \(k=n-1\) gives \(a(tG'(t)+G(t))+bG(t)\). Therefore,
Substitute \(t=Q(z)\), use \(H(z)=G(Q(z))\), and compare coefficients of \(z^s\). This yields the ordinary recursion below. Its denominator must be nonzero; when \(q_0=0\), it equals one. The starting value is \(g_0=G(q_0)\), which reduces to \(p_0\) when payments are strictly positive.
This derivation also shows why the class check matters: the ordinary formula follows from a recurrence valid starting at \(n=1\). If the probability relation is known only for \(n\geq2\), that derivation has an extra boundary contribution. The ordinary formula cannot be assumed just by fitting ratios from \(n=2\) onward.
Worked Examples
Worked Example: Checking a Finite Binomial Model
A hypothetical portfolio has a claim count \(N\sim\operatorname{Binomial}(3,0.5)\). Determine its Panjer parameters and decide whether it is ready for the ordinary aggregate recursion when each claim has a strictly positive integer payment.
Method. Calculate all probabilities, test consecutive ratios including the first one, then check the finite endpoint and normalization.
The binomial probabilities are \(p_n=\binom{3}{n}(0.5)^n(0.5)^{3-n}\). Thus \(p_0=1/8\), \(p_1=3/8\), \(p_2=3/8\), \(p_3=1/8\), and \(p_4=0\). The ratios for \(n=1,2,3\) are
From \(r_1=a+b=3\) and \(r_2=a+b/2=1\), subtraction gives \(b/2=2\), so \(b=4\) and \(a=-1\). The third ratio is consistent: \(a+b/3=-1+4/3=1/3\). At the endpoint, the next ratio is \(p_4/p_3=0\), and the proposed formula gives \(a+b/4=-1+1=0\). Thus the recurrence correctly terminates.
Normalization also holds: \(1/8+3/8+3/8+1/8=1\), and all probabilities are nonnegative. The model is \((a,b,0)=(-1,4,0)\). Because payments are strictly positive, \(q_0=0\), and the ordinary aggregate recursion applies with \(g_0=p_0=1/8\) and denominator \(1-aq_0=1\).
Check. The recurrence gives \(p_1=(a+b)p_0=3/8\), \(p_2=(a+b/2)p_1=3/8\), and \(p_3=(a+b/3)p_2=1/8\), agreeing with the binomial probabilities. The common mistake avoided is stopping after the first two ratios without checking the endpoint.
Worked Example: A Positive Geometric Count Is Not Automatically an (a,b,0) Model
A hypothetical service desk records a positive number of daily requests. Its count distribution is \(P(N=n)=2^{-n}\) for \(n\geq1\), with \(P(N=0)=0\). Classify the model and decide whether the ordinary \((a,b,0)\) aggregate recursion is justified.
Method. Check support first, then test ratios only over the recurrence range relevant to each class, and verify the boundary and total probability.
For every \(n\geq2\),
This is \(a+b/n\) with \(a=1/2\) and \(b=0\). The support is \(n\geq1\), and the required \((a,b,1)\) recurrence holds for every \(n\geq2\). The boundary values are \(p_0=0\) and \(p_1=1/2\). Normalization is verified by the geometric series \(\sum_{n=1}^{\infty}2^{-n}=1\), and every probability is nonnegative. The distribution is therefore in the \((1/2,0,1)\) class.
It is not in the \((a,b,0)\) class: the \(n=1\) recurrence would require \(p_1=(a+b)p_0=0\), but the actual value is \(p_1=1/2\). Therefore the ordinary recursion derived above is not justified. Aggregate-loss recursion would require a formula that accounts for the separate \((a,b,1)\) boundary probabilities.
Check. The positive-count probabilities sum to one, but that does not make the distribution \((a,b,0)\). The common mistake avoided is treating a recurrence that begins at \(n=2\) as though it also determined \(p_1\).
Worked Example: Rejecting a Candidate From Its Ratios
A proposed count model has \(p_0=0.20\), \(p_1=0.20\), \(p_2=0.40\), and \(p_3=0.20\), with no probability above three. Decide whether the probabilities belong to either Panjer class.
Method. Test the ratio form on the stated positive support, then check whether any fitted recurrence can accommodate the finite endpoint.
The probabilities sum to \(0.20+0.20+0.40+0.20=1\), and they are nonnegative. The consecutive ratios through the positive support are \(r_1=1\), \(r_2=2\), and \(r_3=0.5\). For an \((a,b,0)\) fit, the first two equations give \(a+b=1\) and \(a+b/2=2\). Subtracting gives \(b/2=-1\), hence \(b=-2\) and \(a=3\). These parameters predict \(r_3=3-2/3=7/3\), not \(0.5\). So the model is not \((a,b,0)\).
For an \((a,b,1)\) fit, use \(r_2=2\) and \(r_3=0.5\). The equations \(a+b/2=2\) and \(a+b/3=0.5\) give \(b/6=1.5\), so \(b=9\) and \(a=-2.5\). But the finite support requires \(p_4/p_3=0\), whereas this recurrence predicts \(a+b/4=-2.5+2.25=-0.25\). This is not a valid ratio for a positive \(p_3\) followed by \(p_4=0\): it would imply \(p_4=-0.05\). Thus the model is not \((a,b,1)\) either.
Check. Normalization alone is insufficient: the probabilities total one, but neither recurrence is consistent with all the required ratios and the endpoint. The common mistake avoided is declaring membership after fitting only two ratios.
Common Traps and Final Decision
Several checks that seem minor determine whether the recursion is valid. In particular, a probability sequence can be normalized and still fail the Panjer condition, while a sequence can satisfy a ratio pattern on positive counts but have a boundary probability that rules out the proposed class.
- Using a ratio where the denominator is zero. The ratio \(p_n/p_{n-1}\) is not defined if \(p_{n-1}=0\). Check the support and use the stated recurrence range instead of dividing across a zero.
- Ignoring the first count. An \((a,b,1)\) recurrence starts at \(n=2\); it does not determine \(p_1\). Do not force it into the \((a,b,0)\) formula.
- Ignoring a finite endpoint. If \(p_m>0\) and \(p_{m+1}=0\), the recurrence must predict a zero probability at \(m+1\). A nonzero or negative prediction contradicts the proposed finite support.
- Checking only normalization. A sum of one is necessary for a probability distribution, but it does not establish a Panjer class or justify aggregate recursion.
- Using the ordinary aggregate formula for the wrong class. The formula derived here requires the frequency recurrence from \(n=1\). A model whose recurrence starts at \(n=2\) needs its own boundary-adjusted treatment.
Check Your Understanding
For each problem, state the membership or recursion conclusion requested and verify the relevant boundary or endpoint.
- A count model has \(p_0=1/8,\ p_1=3/8,\ p_2=3/8,\ p_3=1/8\), and \(p_n=0\) for \(n\geq4\). Identify its \((a,b,0)\) parameters and confirm the endpoint. (Answer: \((a,b)=(-1,4)\); the recurrence predicts \(p_4/p_3=0\).)
- A positive-count model has \(p_n=2^{-n}\) for \(n\geq1\) and \(p_0=0\). Is it \((a,b,0)\), \((a,b,1)\), or both? (Answer: \((a,b,1)\) with \((a,b)=(1/2,0)\), but not \((a,b,0)\).)
- For \(p_0=0.25\), a proposed \((a,b,0)\) model has \(a=0.2\), \(b=0.6\). Use the recurrence to calculate \(p_1,p_2,p_3,p_4\). (Answer: \(p_1=0.20,\ p_2=0.10,\ p_3=0.04,\ p_4=0.014\).)
- A proposed distribution has \(p_0=0.40,\ p_1=0.40,\ p_2=0.15,\ p_3=0.05\), and no mass above three. Is it in either Panjer class? (Answer: No; fitting \(r_2,r_3\) gives \((a,b)=(0.25,0.25)\), which predicts \(r_1=0.50\), not \(1\); the \((a,b,1)\) fit also fails at the finite endpoint.)
- Suppose \(N\) is Poisson with mean \(2\), payments are strictly positive integer amounts with \(q_1=0.6,\ q_2=0.4\), and all other \(q_j=0\). Using \(p_0=e^{-2}\), apply the ordinary recursion to find \(g_2\). (Answer: \(g_1=1.2e^{-2}\), \(g_2=1.52e^{-2}\approx0.2057\).)