Why the Starting Index Matters
Panjer frequency classes are useful because a simple relation between successive count probabilities can describe an entire tail. For the \((a,b,1)\) class, that relation begins at count two. This starting index is essential: the recursion specifies how \(p_2,p_3,\ldots\) follow from \(p_1\), but does not prescribe a relation between \(p_0\) and \(p_1\).
That differs from the \((a,b,0)\) class, whose recursion begins at count one and therefore also relates \(p_1\) to \(p_0\). A common error is to carry that count-one condition over to the \((a,b,1)\) class. It can wrongly reject a valid model or assign an incorrect value to a boundary probability.
The phrase “specified separately” does not mean that any values whatsoever give a probability distribution. Every probability must be nonnegative, and all probabilities must sum to one. It means only that the tail recursion does not determine \(p_0\) from \(p_1\), or \(p_1\) from \(p_0\). The conditions for choosing boundary probabilities and normalizing such a sequence are a separate question.
How to Read the Recursion
At each count \(n\ge2\), the multiplier is \(a+b/n\). The first application gives \(p_2=(a+b/2)p_1\); the next gives \(p_3=(a+b/3)p_2\). Continue in order, using the current probability as the input to the next step. If a multiplier is zero, the corresponding probability is zero. If that zero probability is followed by further recursion, those later probabilities are also zero.
This product follows by repeated substitution: \(p_2\) is \(p_1\) times the \(j=2\) multiplier; \(p_3\) is \(p_2\) times the \(j=3\) multiplier, and hence is \(p_1\) times both factors. Repeating the same step gives the displayed expression. The formula describes the tail relative to \(p_1\), but says nothing about \(p_0\).
When \(p_{n-1}>0\), the recursion can also be expressed as a probability ratio: $$ \frac{p_n}{p_{n-1}}=a+\frac{b}{n}. $$ This ratio form is useful for recognizing the class. If a probability is zero, division by it is not valid; use the original recursion instead.
Worked Applications
Worked Example: A Finite Tail with a Zero Multiplier
A hypothetical count model has \(p_0=0\), \(a=-1/5\), \(b=1\), and \(p_1\) is to be chosen so that the recursion-generated probabilities sum to one. Find the nonzero probabilities and identify where the tail terminates.
Method. Apply the recursion at each integer \(n\ge2\), including the count at which the multiplier becomes zero. Do not stop after the first few positive terms.
At \(n=2\), the multiplier is $$ -\frac15+\frac12=\frac{3}{10}, $$ so \(p_2=(3/10)p_1\). At \(n=3\), the multiplier is $$ -\frac15+\frac13=\frac{2}{15}, $$ so \(p_3=(2/15)(3/10)p_1=(1/25)p_1\). At \(n=4\), the multiplier is $$ -\frac15+\frac14=\frac{1}{20}, $$ so \(p_4=(1/20)(1/25)p_1=(1/500)p_1\). The \(n=5\) multiplier is \(-1/5+1/5=0\), giving \(p_5=0\); all later probabilities are zero as well.
The total probability is therefore $$ p_0+p_1+p_2+p_3+p_4 =p_1\left(1+\frac{3}{10}+\frac{1}{25}+\frac{1}{500}\right) =p_1\frac{671}{500}. $$ Setting this equal to one gives \(p_1=500/671\). Thus \(p_2=150/671\), \(p_3=20/671\), and \(p_4=1/671\), while \(p_0=0\) and \(p_n=0\) for \(n\ge5\). The probabilities sum to \((500+150+20+1)/671=1\), as required.
Common mistake avoided: stopping at \(p_3\) because a few terms look small. The \(n=4\) multiplier is positive, not zero, so \(p_4\) must be included.
Worked Example: A Valid Tail with Separately Chosen Boundary Probabilities
A hypothetical frequency model has \(a=0\), \(b=1\), \(p_0=1/2\), and \(p_1=1/[2(e-1)]\), where \(e\) is the base of the natural logarithm. Compute the first three probabilities after \(p_1\), and verify that the full sequence is a probability distribution.
Method. Since \(a=0\) and \(b=1\), the recursion is \(p_n=p_{n-1}/n\) for \(n\ge2\). Repeated application gives \(p_n=p_1/n!\) for all \(n\ge1\). In particular, $$ p_2=\frac{1}{4(e-1)},\qquad p_3=\frac{1}{12(e-1)},\qquad p_4=\frac{1}{48(e-1)}. $$ Numerically, \(p_1\approx0.2910\), \(p_2\approx0.1455\), \(p_3\approx0.0485\), and \(p_4\approx0.0121\).
To check the entire sequence, use the exponential series \(\sum_{n=0}^{\infty}1/n!=e\). The tail beginning at one has total $$ \sum_{n=1}^{\infty}p_n =\frac{1}{2(e-1)}\sum_{n=1}^{\infty}\frac{1}{n!} =\frac{1}{2(e-1)}(e-1) =\frac12. $$ Adding \(p_0=1/2\) gives total probability one. All terms are nonnegative, so the sequence is a valid distribution.
Common mistake avoided: assuming that \(p_0\) must equal \(p_1\) because this tail has the same recursion as a Poisson count with parameter one. Here the independently specified boundary values are \(p_0=1/2\) and \(p_1=1/[2(e-1)]\); the tail recursion does not equate them.
Worked Example: Recovering \(a\) and \(b\) from a Finite Distribution
A hypothetical count distribution has \(p_0=2/5\), \(p_1=18/35\), \(p_2=3/35\), and \(p_3=0\), with all probabilities from count four onward also zero. Determine whether its positive tail can satisfy an \((a,b,1)\) recursion, and, if so, find \(a\) and \(b\).
Method. Use the ratios at counts two and three to obtain two equations. Both denominators are positive, so these ratios are defined: $$ \frac{p_2}{p_1}=\frac{3}{18}=\frac16=a+\frac b2, \qquad \frac{p_3}{p_2}=0=a+\frac b3. $$ Subtracting the second equation from the first gives \(b/6=1/6\), hence \(b=1\). Substitution into \(a+b/3=0\) gives \(a=-1/3\).
Now verify the tail. At \(n=2\), the multiplier is \(-1/3+1/2=1/6\), which takes \(p_1=18/35\) to \(p_2=3/35\). At \(n=3\), the multiplier is \(-1/3+1/3=0\), giving \(p_3=0\). At \(n=4\), the multiplier is \(-1/3+1/4=-1/12\), but \(p_4=(-1/12)p_3=0\); the same reasoning gives zero thereafter. Finally, $$ p_0+p_1+p_2=\frac{14}{35}+\frac{18}{35}+\frac{3}{35}=1. $$ Thus the distribution is valid and has the \((a,b,1)\) tail with \(a=-1/3\), \(b=1\). The recursion is verified even though its multiplier at \(n=4\) is negative, because it multiplies a zero probability.
Common mistake avoided: treating a negative multiplier by itself as proof that a sequence is inadmissible. The probabilities must be checked; here the input probability is zero, so all subsequent terms remain zero.
Distinguishing the Two Panjer Starting Indices
The difference between \((a,b,0)\) and \((a,b,1)\) is the starting index of the recursion, not a change to the expression \(a+b/n\). For \((a,b,0)\), the relation applies at \(n\ge1\), so \(p_1=(a+b)p_0\). For \((a,b,1)\), it applies only at \(n\ge2\), and that count-one equation is not part of the definition.
| Feature | \((a,b,0)\) class | \((a,b,1)\) class |
|---|---|---|
| Recursion | \(p_n=(a+b/n)p_{n-1}\) | \(p_n=(a+b/n)p_{n-1}\) |
| Starting count | \(n\ge1\) | \(n\ge2\) |
| Boundary probabilities linked by the recursion? | Yes: \(p_1=(a+b)p_0\) | No count-one relation is imposed |
| First tail probability generated from | \(p_0\), at count one | \(p_1\), at count two |
Common Exam Traps
- Starting at \(n=1\). This imposes a boundary equation that is not part of the \((a,b,1)\) definition. Check the index written in the model before calculating.
- Skipping a small or zero-looking term. Evaluate every multiplier in sequence. A term is zero only when the multiplier or the preceding probability is exactly zero.
- Using ratios when a denominator is zero. The ratio form requires \(p_{n-1}>0\). The original recursion remains meaningful when \(p_{n-1}=0\).
- Confusing the tail rule with a complete distribution check. Satisfying the recursion does not by itself establish that all terms are nonnegative and sum to one.
These checks matter in frequency modeling because the recursive structure is useful only when applied to the correct probabilities. On problems aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus, explicitly writing the starting index is a simple way to prevent a boundary error from carrying through an otherwise correct calculation.
Check Your Understanding
For each problem, apply the recursion only at the stated counts and distinguish the tail rule from any separate probability check.
- Let \(a=0\), \(b=2\), \(p_0=1/4\), and \(p_1=3/20\). Find \(p_2,p_3,p_4\). (Answer: \(p_2=3/20,\ p_3=1/10,\ p_4=1/20\).)
- Let \(a=-1/5\), \(b=1\), and \(p_0=0\). Choose \(p_1\) so the recursion-generated sequence sums to one. Give the nonzero probabilities. (Answer: \(p_1=500/671,\ p_2=150/671,\ p_3=20/671,\ p_4=1/671\); all later probabilities are zero.)
- A sequence has positive \(p_1,p_2,p_3\), with \(p_2/p_1=3/5\) and \(p_3/p_2=1/2\). If it follows an \((a,b,1)\) recursion, find its parameters. (Answer: \(a=3/10,\ b=3/5\).)
- A candidate sequence has \(a=0\), \(b=1\), \(p_0=1/5\), and \(p_1=1/4\), and follows the stated recursion for \(n\ge2\). Does the recursion require \(p_1=(a+b)p_0\)? (Answer: No. The \((a,b,1)\) recursion starts at \(n=2\); it imposes no count-one relation.)
- Suppose \(p_0=0\), \(p_1=48/61\), and the recursion has \(a=-1/4\), \(b=1\). Find the nonzero probabilities and check the total. (Answer: \(p_2=12/61,\ p_3=1/61\), and \(p_n=0\) for \(n\ge4\); \(p_0+p_1+p_2+p_3=1\).)