What a Panjer Classification Must Prove
A list of probabilities can resemble a familiar frequency model without belonging to its Panjer class. The defining test is not whether a few adjacent probabilities happen to produce plausible parameters. It is whether the candidate probability mass function yields one fixed pair of constants \(a\) and \(b\) in the required probability-ratio relation at every count where that relation applies. This tutorial develops that proof method for the Panjer \((a,b,0)\) and \((a,b,1)\) classes, including the special checks needed at zero and at a finite endpoint.
Write \(p_k=P(N=k)\) for a nonnegative integer-valued count \(N\). First establish that the candidate \(p_k\) is a valid probability mass function: its probabilities are nonnegative and sum to one. Then derive the ratio from the mass function itself. Numerical checks can confirm the algebra, but a few matching ratios alone do not prove membership for an entire support.
Where \(p_{k-1}>0\), the same condition can be written as
This form is useful for classification because the dependence on \(k\) has a specified shape: a constant plus a constant divided by \(k\). The values of \(a\) and \(b\) must not change from one count to another. If a candidate has zeros inside its positive support, the ratio may be undefined there; such a gap must be examined against the class condition rather than ignored.
A Reliable Derivation and Checking Method
For a candidate given by a formula, take the ratio \(p_k/p_{k-1}\) symbolically. Simplify factorials or binomial coefficients before comparing the result with \(a+b/k\). This symbolic step shows why the same parameters apply at every required count. Substituting two or three counts is then a useful arithmetic check, not a substitute for the derivation.
Record the formula for \(p_k\), the permitted integer values of \(k\), and whether the support is infinite or ends at a finite maximum.
For each count where the denominator is positive, simplify \(p_k/p_{k-1}\) directly from the candidate mass function.
Rewrite the ratio as \(a+b/k\), with \(a\) and \(b\) independent of \(k\). If no such constants work throughout the required range, the proposed classification fails.
For \((a,b,0)\), include the relation at \(k=1\). For \((a,b,1)\), start at \(k=2\). If positive probabilities end at \(n\), check the required relation at \(k=n+1\), where \(p_{n+1}=0\).
The endpoint check follows directly from the relation. If \(p_n>0\) but \(p_{n+1}=0\), then
Since \(p_n\) is positive, the multiplier must be zero. A negative multiplier would predict a negative next probability, and a positive multiplier would predict a positive next probability. Neither agrees with a finite support ending at \(n\).
Worked Classifications
Worked Example: A Binomial Count With a Finite Endpoint
A hypothetical inspection program records the number \(N\) of defective seals among four independently inspected seals. Each seal is defective with probability \(q=1/3\). Determine whether this count belongs to the \((a,b,0)\) class and verify its finite endpoint.
Model and support. The count has a binomial distribution with \(n=4\) and \(q=1/3\). Thus, for \(k=0,1,2,3,4\),
These probabilities are nonnegative and sum to one by the binomial theorem. To derive the ratio for \(1\leq k\leq4\), divide the formula at \(k\) by the formula at \(k-1\):
The same constants \(a=-1/2\) and \(b=5/2\) apply for every \(k=1,2,3,4\), so the distribution satisfies the \((a,b,0)\) relation. For an arithmetic check, the probabilities are \(p_0=16/81\), \(p_1=32/81\), \(p_2=24/81\), \(p_3=8/81\), and \(p_4=1/81\). Their successive ratios are \(2,\ 3/4,\ 1/3,\ 1/8\); the formula gives those same values at \(k=1,2,3,4\).
At the first unsupported count, \(k=5\), the multiplier is \[ -\frac{1}{2}+\frac{5}{2(5)}=0. \] Thus the relation gives \(p_5=0\), as required by the finite support. Method and trap avoided: derive the general binomial ratio before identifying the parameters; dropping the factor \(5-k\) or checking only the first ratio can give an incorrect \(b\) and a false endpoint conclusion.
Worked Example: A Negative Binomial Count
A hypothetical monitoring process counts the number \(N\) of unsuccessful checks before the third successful check. Each check succeeds independently with probability \(q=1/2\). Determine the Panjer parameters and check them at several counts.
Model and parameters. Here \(N\) has a negative binomial distribution with \(r=3\) required successes and success probability \(q=1/2\). The count has infinite support \(k=0,1,2,\ldots\), and
The probabilities are nonnegative and sum to one because this is the count of failures before the third success under independent Bernoulli trials. Taking the adjacent ratio gives
Therefore \(a=1/2\) and \(b=1\), and the relation holds for every \(k\geq1\). For a direct check, \(p_0=1/8\), \(p_1=3/16\), \(p_2=3/16\), and \(p_3=5/32\). The ratios are \(p_1/p_0=3/2\), \(p_2/p_1=1\), and \(p_3/p_2=5/6\). The expression \(1/2+1/k\) gives \(3/2,\ 1,\ 5/6\) at those counts. Method and trap avoided: specify whether the negative binomial counts failures or successes; the chosen parameterization determines the mass function and hence the ratio.
Worked Example: A Zero-Truncated Poisson Count
A hypothetical device is included in a study only if it registers at least one alert during a fixed period. Before selection, the alert count is Poisson with parameter \(\lambda=2\). Determine the relevant Panjer classification after selection.
Model and selection condition. Before selection, \(P(N=k)=e^{-2}2^k/k!\). Conditioning on \(N\geq1\) gives the zero-truncated probabilities
The denominator is \(P(N\geq1)=1-e^{-2}\), so the positive probabilities sum to one. The \((a,b,1)\) relation begins at \(k=2\), since \(p_0=0\) and the ratio from \(p_0\) to \(p_1\) is not defined. For every \(k\geq2\), the common normalizing factor cancels:
Thus \(a=0\), \(b=2\), and one pair works at every required count \(k\geq2\). For example, \(p_1=2e^{-2}/(1-e^{-2})\), \(p_2=2e^{-2}/(1-e^{-2})\), and \(p_3=4e^{-2}/[3(1-e^{-2})]\). The ratios \(p_2/p_1=1\) and \(p_3/p_2=2/3\) agree with \(2/k\) at \(k=2\) and \(k=3\). Method and trap avoided: conditioning changes the boundary probabilities through a common normalization factor, but the ratio among positive counts is unchanged; do not impose the \((a,b,0)\) condition at \(k=1\).
Interpreting a Failed or Successful Check
A successful classification requires both a valid candidate distribution and a ratio proof over the entire required range. For an infinite-support candidate, the algebra should establish the ratio for every required integer \(k\), not merely for a finite sample of counts. For a finite-support candidate with positive probabilities through \(n\), the relation must also be compatible with \(p_{n+1}=0\) if \(n+1\) is in the required range.
A common exam trap is to derive constants from selected ratios, then assume the endpoint must work. For example, suppose a candidate has \(p_6>0\), claims \(p_7=0\), and proposes \(a=-1/3,\ b=2\). At \(k=7\), the proposed multiplier is
The relation would therefore give \(p_7=-(1/21)p_6<0\), not zero. The classification fails. The correct response is no: the proposed relation predicts a negative probability at the first unsupported count. This is different from an endpoint multiplier of zero, which is exactly what is needed when positive support ends there.
Another frequent error is reversing the ratio. The class condition uses \(p_k/p_{k-1}\), not \(p_{k-1}/p_k\). Reversing it generally changes the dependence on \(k\) and invalidates the parameter match. Keep the numerator at the current count \(k\), and substitute into the candidate formula before simplifying.
This method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. It supports reliable identification of frequency models and helps prevent incorrect classifications from a missed index, reversed ratio, or mishandled finite endpoint.
Check Your Understanding
For each candidate, derive or inspect the adjacent ratio, identify the relevant constants, and decide whether the stated class condition holds. Answers are given for checking.
- A hypothetical count has geometric probabilities \(p_k=(1/3)(2/3)^k\) for \(k\geq0\). Find \(p_k/p_{k-1}\) for \(k\geq1\) and identify \(a,b\) for the \((a,b,0)\) class. (Answer: \(p_k/p_{k-1}=2/3\), so \(a=2/3,\ b=0\); it is in \((a,b,0)\).)
- A count is binomial with \(n=3\) and success probability \(q=1/4\). Derive the ratio for \(1\leq k\leq3\), identify \(a,b\), and check the multiplier at \(k=4\). (Answer: \(p_k/p_{k-1}=(4-k)/(3k)=-1/3+4/(3k)\), so \(a=-1/3,\ b=4/3\); at \(k=4\) the multiplier is \(0\), consistent with the endpoint.)
- A Poisson count with parameter \(\lambda=2\) is conditioned to be positive. State the ratio range and the \((a,b,1)\) parameters; explain why \(k=1\) is not tested. (Answer: \(p_k/p_{k-1}=2/k\) for \(k\geq2\), so \(a=0,\ b=2\); \(k=1\) is not tested because \(p_0=0\).)
- A negative binomial count records failures before \(r=2\) successes, with success probability \(q=1/3\). Starting from \(p_k=\binom{k+1}{1}q^2(1-q)^k\), derive the ratio and classify it. (Answer: \(p_k/p_{k-1}=(2/3)(k+1)/k=2/3+2/(3k)\), so \(a=2/3,\ b=2/3\) in \((a,b,0)\).)
- A candidate has \(p_6>0\), \(p_7=0\), and is claimed to follow the \((a,b,0)\) relation with \(a=-1/3,\ b=2\). Does the claim pass the endpoint test? (Answer: No. At \(k=7\), the multiplier is \(-1/3+2/7=-1/21\), so the relation predicts \(p_7=-(1/21)p_6<0\), not zero.)