Why the Endpoint Matters
Panjer recursions generate count probabilities from earlier probabilities. For a distribution with finite support, the sequence must eventually reach zero and remain zero. The endpoint is more than a bookkeeping detail: if the recursion instead produces a negative probability before reaching zero, the proposed sequence is not a probability distribution. This distinction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The binomial distribution provides the standard finite-support example. Its probability at the largest possible count is positive, the next probability is exactly zero, and all later probabilities remain zero. The recursion coefficient can become negative after that point without invalidating the sequence, because it is multiplied by a probability that is already zero.
For the \((a,b,0)\) form, the recursion begins with \(p_0\). For the \((a,b,1)\) form, the first recursion step instead connects \(p_2\) to \(p_1\); the value of \(p_0\) does not determine that step. In either form, probabilities must be nonnegative and sum to one. A recursion formula alone does not guarantee those requirements.
The Endpoint Condition
Suppose a sequence has finite support ending at \(M\), with \(p_M>0\), and the recursion applies to the step from \(M\) to \(M+1\). For the next probability to be zero, the recursion must have a zero multiplier at that step:
The implication follows because \(p_M\) is strictly positive. This condition is necessary for a finite endpoint under that recursion. It is also sufficient to make the subsequent probabilities zero: once \(p_{M+1}=0\), each later recursion step multiplies the preceding zero probability by a finite coefficient. For the \((a,b,1)\) form, this argument applies when \(M+1\geq2\), since that is where its recursion begins.
Let \(p_M>0\) be the final positive term in the proposed sequence.
Evaluate \(a+b/(M+1)\). It must equal zero if the recursion applies at \(M+1\) and the sequence is to terminate there.
With \(p_{M+1}=0\), later terms remain zero. A negative multiplier after this point gives zero times a negative number, not a negative probability.
All terms through \(M\) must be nonnegative, and their sum, including the zero tail, must be one.
This also explains when a negative coefficient is a genuine problem. If \(p_{n-1}>0\) and \(a+b/n<0\), then \(p_n<0\), which is impossible for a probability. If \(p_{n-1}=0\), however, the recursion gives \(p_n=0\) regardless of whether the multiplier is positive, zero, or negative.
Why the Binomial Distribution Terminates
For a binomial count \(N\sim\operatorname{Bin}(m,p)\), assume \(m\) is a positive integer and \(0<p<1\). Its probabilities are \(p_n=\binom{m}{n}p^n(1-p)^{m-n}\) for \(0\leq n\leq m\), and zero for \(n>m\). For \(1\leq n\leq m\), dividing adjacent positive probabilities gives the recursion multiplier:
Thus the binomial parameters in the \((a,b,0)\) recursion are \(a=-p/(1-p)\) and \(b=(m+1)p/(1-p)\). At the next count \(n=m+1\), the multiplier is zero. This produces \(p_{m+1}=0\). At counts beyond \(m+1\), the multiplier is negative, but each such step starts from a zero probability, so the probabilities remain zero.
Worked Example: Following a Binomial Sequence Through Its Endpoint
A hypothetical portfolio model has claim count \(N\sim\operatorname{Bin}(4,0.30)\). Use the Panjer recursion to calculate \(p_0\) through \(p_6\), identify the endpoint, and check the total probability. State why a negative multiplier after the endpoint does not make the model invalid.
Method. For \(m=4\) and \(p=0.30\), the recursion parameters are \(a=-0.30/0.70=-3/7\) and \(b=5(0.30)/0.70=15/7\). Therefore, for \(n\geq1\), the multiplier is \(a+b/n=(5-n)3/(7n)\). The starting probability is \(p_0=(1-0.30)^4=0.7^4=0.2401\).
At \(n=1\), the multiplier is \(12/7\), so \(p_1=(12/7)(0.2401)=0.4116\).
At \(n=2\), the multiplier is \(9/14\), so \(p_2=(9/14)(0.4116)=0.2646\).
At \(n=3\), the multiplier is \(2/7\), so \(p_3=(2/7)(0.2646)=0.0756\).
At \(n=4\), the multiplier is \(3/28\), so \(p_4=(3/28)(0.0756)=0.0081\).
At \(n=5\), the multiplier is zero, so \(p_5=0(0.0081)=0\). At \(n=6\), the multiplier is \(-1/14\), but \(p_6=(-1/14)(0)=0\). The endpoint is \(M=4\), since \(p_4>0\) and all probabilities after it are zero.
Check. The probabilities sum to \(0.2401+0.4116+0.2646+0.0756+0.0081=1.0000\). The displayed probabilities are exact to four decimal places, so no intermediate rounding affects the check.
Common mistake avoided. Treating the negative multiplier at \(n=6\) as evidence of an invalid model ignores its zero predecessor. A negative multiplier is disqualifying only when it acts on a positive preceding probability.
Finite Support in Both Panjer Forms
The endpoint check works for both recursion forms, but their starting conditions differ. In the \((a,b,0)\) form, the recursion links \(p_1\) to \(p_0\), so a finite endpoint at \(M=0\) requires the multiplier at \(n=1\) to be zero if \(p_0>0\). In the \((a,b,1)\) form, \(p_0\) and \(p_1\) are boundary values; if \(p_1>0\), the recursion can generate terms from \(p_2\) onward. A finite endpoint at \(M\geq1\) requires a zero multiplier at \(n=M+1\).
The \((a,b,1)\) form can therefore have a nonzero \(p_0\) alongside a separate positive-count sequence that terminates. Normalization must include both boundary terms and every generated term. Do not infer \(p_1\) from the \((a,b,1)\) recursion: that recursion starts at \(n=2\).
Worked Example: Checking a Finite Endpoint in the (a,b,1) Form
A hypothetical count model has \(a=-1/3\), \(b=1\), and boundary probabilities \(p_0=1/2\) and \(p_1=3/7\). Apply the \((a,b,1)\) recursion, check whether the sequence terminates, and verify its normalization.
Method. The recursion applies for \(n\geq2\), with multiplier \(-1/3+1/n\). At \(n=2\), this is \(-1/3+1/2=1/6\). Thus \(p_2=(1/6)(3/7)=1/14\).
At \(n=3\), the multiplier is \(-1/3+1/3=0\), so \(p_3=0(1/14)=0\). At \(n=4\), the multiplier is \(-1/3+1/4=-1/12\), but \(p_4=(-1/12)(0)=0\); all later probabilities remain zero.
Check. The total probability is \(p_0+p_1+p_2=1/2+3/7+1/14=7/14+6/14+1/14=1\). The support ends at \(M=2\). The endpoint condition holds because the multiplier at \(M+1=3\) is zero.
Common mistake avoided. The boundary probability \(p_0\) is not part of the recursion step that generates \(p_2\). The step uses \(p_1\), as required by the \((a,b,1)\) definition.
When a Negative Probability Shows the Sequence Is Invalid
A zero factor at the correct endpoint is essential. If no such factor occurs while the preceding probability is positive, the sequence may continue into negative values. In particular, a root of \(a+b/n=0\) at a noninteger value of \(n\) cannot terminate a recursion at an integer count. The sign of the multiplier at the integer steps must be checked directly.
Worked Example: Rejecting a Recursion With a Negative Term
Suppose a proposed \((a,b,0)\) count sequence has \(p_0=1/2\), \(a=-1/5\), and \(b=1/2\). Calculate the first terms through \(p_3\) and determine whether this can be a probability distribution.
Method. At \(n=1\), the multiplier is \(-1/5+1/2=3/10\), so \(p_1=(3/10)(1/2)=3/20\).
At \(n=2\), the multiplier is \(-1/5+1/4=1/20\), so \(p_2=(1/20)(3/20)=3/400\).
At \(n=3\), the multiplier is \(-1/5+1/6=-1/30\), so \(p_3=(-1/30)(3/400)=-1/4000\).
Check and conclusion. Since \(p_2=3/400>0\) and the next multiplier is negative, \(p_3=-1/4000<0\). The proposed sequence is invalid as a probability distribution, regardless of whether some choice of normalization could make its terms sum to one. Its zero-multiplier location would be \(n=b/(-a)=(1/2)/(1/5)=5/2\), not an integer count.
Common mistake avoided. A formal root between integer values does not supply a valid endpoint. The recursion is evaluated at integer counts, and here it generates a negative term before any zero probability is reached.
Endpoint Checks to Use Consistently
A reliable check separates the multiplier from the probability it acts on. First identify whether the preceding probability is positive. If it is positive, a negative multiplier creates a negative probability, and a zero multiplier ends the positive sequence. If the preceding probability is already zero, subsequent terms remain zero under the recursion. Finally, verify normalization; termination alone does not establish that the probabilities sum to one.
For a binomial model, the zero multiplier occurs at \(n=m+1\), giving the finite endpoint \(M=m\). For other proposed Panjer sequences, do not assume that the endpoint is valid merely because the parameters resemble a familiar model. Check the actual recursion index, the sign of each multiplier before termination, the boundary probabilities where applicable, and the sum of the sequence.
Check Your Understanding
For each problem, use the recursion index stated in the definition and distinguish a zero tail from a negative probability.
- A hypothetical count is \(N\sim\operatorname{Bin}(3,0.20)\). At what count does its \((a,b,0)\) recursion first produce a zero probability? What happens at subsequent counts? (Answer: \(p_4=0\), and all later probabilities remain zero; the support ends at 3.)
- For an \((a,b,0)\) sequence, let \(a=-1/4\), \(b=1\), and \(p_0\) be chosen to normalize the sequence. Find \(p_0,p_1,p_2,p_3\) and the endpoint. (Answer: \(p_0=64/125,\ p_1=48/125,\ p_2=12/125,\ p_3=1/125\); \(p_4=0\), so the endpoint is 3.)
- An \((a,b,1)\) sequence has \(a=-1/2\), \(b=3/2\), \(p_0=1/5\), and \(p_1=16/25\). Calculate \(p_2\) and identify the endpoint. (Answer: \(p_2=(1/4)(16/25)=4/25\); the multiplier at \(n=3\) is zero, so the endpoint is 2.)
- An \((a,b,0)\) sequence has \(p_0=1/2\), \(a=-1/5\), and \(b=3/10\). Calculate \(p_1\) and \(p_2\), and decide whether it can be a probability distribution. (Answer: \(p_1=(1/10)(1/2)=1/20\), \(p_2=(-1/20)(1/20)=-1/400\); it is invalid because \(p_2<0\).)
- For an \((a,b,0)\) recursion, suppose \(p_4>0\), the multiplier at \(n=5\) is zero, and the multiplier at \(n=6\) is negative. What are \(p_5\) and \(p_6\), and does the negative multiplier invalidate the sequence by itself? (Answer: \(p_5=0\), \(p_6=0\); no, the negative multiplier acts on a zero preceding probability.)