Support and Admissibility Conditions

Why Support Conditions Matter

The Panjer probability recursion is useful only when it describes a genuine probability distribution. A proposed pair of parameters may produce negative terms, fail to terminate at a claimed finite endpoint, or generate nonnegative terms whose total probability is not finite. Checking the support and admissibility conditions prevents these errors before the recursion is used in a frequency calculation.

Let \(N\) be a nonnegative-integer-valued claim count, and write \(p_n=P(N=n)\). In the \((a,b,0)\) class, the probabilities satisfy a recursion beginning at zero:

The coefficient \(a+b/n\) is the ratio \(p_n/p_{n-1}\) when \(p_{n-1}>0\). The starting probability \(p_0\) must be positive for a distribution in this class. The recursion determines probabilities relative to \(p_0\); the probabilities must also be nonnegative and sum to one.

Infinite Support: Nonnegative and Summable Terms

For support extending through every nonnegative integer, all probabilities \(p_n\) are positive. Therefore every recurrence coefficient must be positive. Since \(c_n=a+b/n\) approaches \(a\) as \(n\) grows, an infinite sequence of positive probabilities can be summable only if \(a<1\). The exact conditions for strictly positive probabilities at every count are:

These conditions ensure that \(c_n>0\) for every \(n\). If \(b\ge0\), the coefficients are at least \(a\), and \(a+b>0\) handles the case \(a=0\). If \(b<0\), then \(a+b/n\) is smallest at \(n=1\), so \(a+b>0\) guarantees positivity at every index.

They also ensure summability. When \(0<a<1\), choose a number \(q\) with \(a<q<1\). Since \(c_n\) tends to \(a\), all sufficiently late coefficients are less than \(q\). The tail probabilities are then bounded by a constant times a geometric sequence with ratio \(q\). When \(a=0\), the conditions require \(b>0\), and the recursion gives terms proportional to \(b^n/n!\), whose sum is finite. In either case a positive \(p_0\) can be chosen to make the total probability one.

The condition \(a<1\) is substantive: if \(a>1\), the recurrence coefficients eventually exceed one, so probabilities cannot form a summable positive sequence. If \(a=1\) and \(a+b>0\), the positive terms do not have a summable tail. Thus nonnegative coefficients alone do not establish admissibility; normalization must be possible as well.

Finite Support: The Endpoint Must Be Exact

Suppose the intended support is exactly \(0,1,\ldots,m\), where \(m\ge1\). Then \(p_m>0\), but \(p_{m+1}=0\). Applying the recursion at \(n=m+1\) forces the endpoint coefficient to be zero:

For every \(n\) from 1 through \(m\), \(p_{n-1}>0\), so the coefficients before the endpoint must be positive. Substituting \(b=-(m+1)a\) gives

The expression in parentheses is negative for every such \(n\). Thus \(a<0\) is necessary for all pre-endpoint coefficients to be positive, and it is sufficient when combined with the endpoint equation. At \(n=m+1\), the coefficient is zero; at every later index, the recursion multiplies a previous zero probability by a finite coefficient, so all later probabilities remain zero.

Once the coefficients have been checked, normalization is straightforward. Starting from an unnormalized weight \(w_0=1\), apply the recursion to find \(w_1,\ldots,w_m\). Then set \(p_0=1/\sum_{j=0}^{m}w_j\), and multiply every weight by this same factor. This produces a probability distribution without changing the recurrence ratios.

When a Proposed Pair Is Not Admissible

For a proposed infinite-support model, a single negative coefficient at an index reached from a positive probability makes the sequence invalid. A coefficient that is zero can instead terminate the sequence, but it must occur at the claimed endpoint. For finite support \(0,\ldots,m\), the endpoint coefficient must be zero exactly at \(m+1\), while all earlier coefficients must be positive. These checks are stronger than inspecting a few initial probabilities or checking only whether \(a\) is below one.

There is one degenerate exception worth separating from the nondegenerate finite-support condition. If \(b=-a\), then \(c_1=0\); with \(p_0=1\), the recursion gives \(p_n=0\) for every \(n\ge1\). This is the distribution concentrated at zero, with support \(\{0\}\). It is not a distribution with endpoint \(m\ge1\), so it should not be confused with the finite-endpoint parameter condition above.

Common Checks Before Using the Recursion

A reliable admissibility check follows the support that is actually claimed. For infinite support, verify \(0\le a<1\) and \(a+b>0\), so coefficients stay positive and the sequence can be normalized. For exact finite support through \(m\ge1\), verify \(a<0\), \(b=-(m+1)a\), and positivity at every earlier step. The finite case is normalized by summing its relative weights; the infinite case requires a summable sequence of weights.

Do not infer support from the sign of \(a\) alone. A negative \(a\) can be valid for finite support only when the endpoint relation is satisfied. Conversely, a pair that gives positive initial terms may still fail later, as the noninteger-endpoint example demonstrates. The relevant check is the coefficient at every count reached while the preceding probability is positive.