Why Frequency Recursions Matter
A frequency model assigns probabilities to the number of claims in a specified period. For a Poisson count, for example, each probability can be calculated from the preceding one using a simple ratio. The same is true for binomial and negative binomial counts. This structure is useful when a calculation needs several count probabilities: it avoids evaluating a separate factorial expression for every count and provides a direct way to check the sequence.
The central task is to identify the model, match its parameters to the appropriate recursion, and use the correct starting probability. A zero count may be possible, impossible, or treated separately, so the starting values and the range over which the recurrence applies matter. These are the frequency-distribution results used in calculations aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The Two Panjer Frequency Classes
Let \(N\) be a nonnegative integer-valued frequency random variable, and write \(p_n=P(N=n)\). A frequency recursion expresses \(p_n\) in terms of the probability at the preceding count. The two forms used here differ in where that relationship begins.
The values of \(a\) and \(b\), together with the starting probability or probabilities, must produce nonnegative probabilities that sum to one. In a finite-support model, the recurrence may reach a zero probability at the end of the support. In an infinite-support model, the recurrence generates a tail whose probabilities must sum to the remaining probability mass.
The distinction between the two classes is not just notation. In an \((a,b,0)\) distribution, the recurrence links the zero-count probability to the one-count probability. In an \((a,b,1)\) distribution, that link is absent. For example, conditioning a Poisson count on being positive makes \(p_0=0\); its positive-count probabilities still obey a recurrence from \(n=2\) onward, but \(p_1\) must be supplied separately.
Poisson, Binomial, and Negative Binomial Recursions
Each of the three standard frequency models belongs to the \((a,b,0)\) class under its usual parameter assumptions. Their parameters are not interchangeable: for a binomial model the number of trials is finite, while for the negative binomial model below the count is the number of failures before a specified number of successes.
| Frequency model | Parameters and starting probability | Class parameters |
|---|---|---|
| Poisson | \(N\sim\operatorname{Poisson}(\lambda)\), \(\lambda>0\); \(p_0=e^{-\lambda}\) | \(a=0,\ b=\lambda\) |
| Binomial | \(N\sim\operatorname{Binomial}(m,q)\), positive integer \(m\), \(0<q<1\); \(p_0=(1-q)^m\) | \(a=-q/(1-q),\ b=(m+1)q/(1-q)\) |
| Negative binomial | \(N\) is the number of failures before the \(r\)th success, where \(r\) is a positive integer and success probability is \(\theta\), \(0<\theta<1\); \(p_0=\theta^r\) | \(a=1-\theta,\ b=(r-1)(1-\theta)\) |
For the negative binomial row, requiring integer \(r\) gives the literal repeated-trial interpretation: there can be a \(1\)st success, a \(2\)nd success, and so on. A formula with real \(r>0\) can define a generalized negative binomial distribution, but it does not describe failures before the \(r\)th success in that trial interpretation.
Why the Recursions Hold
A short derivation makes the parameter mappings easier to remember. In each case, divide the probability of count \(n\) by the probability of count \(n-1\). The resulting ratio can be written as \(a+b/n\); multiplying by \(p_{n-1}\) gives the Panjer recurrence.
For \(N\sim\operatorname{Poisson}(\lambda)\), \(p_n=e^{-\lambda}\lambda^n/n!\). Thus, for \(n\geq1\), \(p_n/p_{n-1}=\lambda/n\), so \(a=0\), \(b=\lambda\), and \(p_0=e^{-\lambda}\).
For \(N\sim\operatorname{Binomial}(m,q)\), \(p_n=\binom{m}{n}q^n(1-q)^{m-n}\). For \(1\leq n\leq m\), the ratio is \(\frac{m-n+1}{n}\frac{q}{1-q}=-\frac{q}{1-q}+\frac{(m+1)q}{n(1-q)}\). Hence \(a=-q/(1-q)\), \(b=(m+1)q/(1-q)\), and \(p_0=(1-q)^m\).
For integer \(r\geq1\), \(p_n=\binom{n+r-1}{n}\theta^r(1-\theta)^n\), for \(n\geq0\). The ratio is \(\frac{n+r-1}{n}(1-\theta)=(1-\theta)+\frac{(r-1)(1-\theta)}{n}\). Therefore \(a=1-\theta\), \(b=(r-1)(1-\theta)\), and \(p_0=\theta^r\).
The binomial ratio applies for \(1\leq n\leq m\). At \(n=m+1\), the recurrence coefficient is \(a+b/(m+1)=0\), so it gives \(p_{m+1}=0\), consistent with the model’s finite support. The Poisson and negative binomial models instead have probabilities at every nonnegative integer under the parameter assumptions above.
Worked Example: A Poisson Count From Its Starting Probability
A hypothetical repair portfolio has a Poisson number of covered repairs in a year with parameter \(\lambda=2\). Use the Panjer recursion to calculate \(P(N\leq2)\). State the class and avoid confusing the frequency parameter with a probability of exactly one repair.
Method. A Poisson distribution has \(a=0\), \(b=\lambda=2\), and \(p_0=e^{-2}\). The recurrence is \(p_n=(2/n)p_{n-1}\) for \(n\geq1\).
First, \(p_0=e^{-2}=0.135335\ldots\). For one repair, \(p_1=(2/1)p_0=2e^{-2}=0.270671\ldots\). For two repairs, \(p_2=(2/2)p_1=p_1=0.270671\ldots\).
Therefore, \(P(N\leq2)=p_0+p_1+p_2=5e^{-2}=0.676676\ldots\). Rounded to four decimal places, \(P(N\leq2)=0.6767\).
Check. The recurrence gives \(p_1/p_0=2\) and \(p_2/p_1=1\), as required by \(2/n\). The answer is a cumulative probability, not the single probability \(p_2\).
Worked Example: A Finite-Support Binomial Count
For a hypothetical equipment program, suppose the number of devices that require service is \(N\sim\operatorname{Binomial}(5,0.2)\). Find \(P(N\geq4)\) using the recurrence, and verify that the generated probabilities sum to one. State why the sequence terminates.
Method. Here \(m=5\), \(q=0.2\), and \(1-q=0.8\). The class parameters are \(a=-0.2/0.8=-0.25\) and \(b=6(0.2)/0.8=1.5\). The starting probability is \(p_0=0.8^5=0.32768\).
For \(n=1\), the coefficient is \(-0.25+1.5=1.25\), so \(p_1=1.25(0.32768)=0.40960\). For \(n=2\), the coefficient is \(-0.25+1.5/2=0.5\), so \(p_2=0.5(0.40960)=0.20480\).
For \(n=3\), the coefficient is \(-0.25+1.5/3=0.25\), so \(p_3=0.25(0.20480)=0.05120\). For \(n=4\), the coefficient is \(-0.25+1.5/4=0.125\), so \(p_4=0.125(0.05120)=0.00640\). For \(n=5\), the coefficient is \(-0.25+1.5/5=0.05\), so \(p_5=0.05(0.00640)=0.00032\).
The requested tail probability is \(P(N\geq4)=p_4+p_5=0.00640+0.00032=0.00672\). Rounded to four decimal places, it is \(0.0067\). For the normalization check, \(0.32768+0.40960+0.20480+0.05120+0.00640+0.00032=1.00000\). At \(n=6\), the coefficient is \(-0.25+1.5/6=0\), so \(p_6=0\); counts above five are impossible.
Check. The probabilities are nonnegative, sum to one, and stop after \(m=5\). The zero coefficient at \(n=6\) confirms finite support.
Worked Example: Negative Binomial Counts Before the Third Success
In a hypothetical inspection process, each trial results in a success with probability \(\theta=0.6\), independently of other trials. Let \(N\) be the number of failures before the third success. Calculate \(P(N\leq2)\) using the recurrence. The target is the number of failures, not the total number of trials.
Method. Here \(r=3\), a positive integer, and \(\theta=0.6\). The model is negative binomial with \(a=1-\theta=0.4\), \(b=(r-1)(1-\theta)=0.8\), and \(p_0=\theta^r=0.6^3=0.216\).
For \(n=1\), \(a+b/n=0.4+0.8=1.2\), giving \(p_1=1.2(0.216)=0.2592\). For \(n=2\), the coefficient is \(0.4+0.8/2=0.8\), giving \(p_2=0.8(0.2592)=0.20736\).
Thus \(P(N\leq2)=p_0+p_1+p_2=0.216+0.2592+0.20736=0.68256\), or \(0.6826\) to four decimal places.
Check. The direct formula for \(p_2\) is \(\binom{2+3-1}{2}(0.6)^3(0.4)^2=6(0.216)(0.16)=0.20736\), matching the recurrence. The count \(N=2\) means two failures before the third success; it does not mean two trials in total.
Starting Values for the \((a,b,1)\) Form
A positive-count model is often obtained by conditioning on \(N>0\). If the original count has probabilities \(p_n\) and \(p_0<1\), the conditioned probabilities are \(P(N=n\mid N>0)=p_n/(1-p_0)\) for \(n\geq1\). For \(n\geq2\), dividing both adjacent probabilities by the same factor leaves their ratio unchanged. Thus the recurrence for positive counts remains the same, but the new \(p_1\) must be calculated after conditioning.
For a zero-truncated Poisson model with original parameter \(\lambda>0\), the starting probabilities are \(p_0=0\) and \(p_1=P(N=1\mid N>0)=\lambda e^{-\lambda}/(1-e^{-\lambda})=\lambda/(e^\lambda-1)\). For \(n\geq2\), the recurrence is still \(p_n=(\lambda/n)p_{n-1}\).
Worked Example: A Zero-Truncated Poisson Starting Probability
A hypothetical maintenance program records a positive number of callouts each year. Its count is modeled as a Poisson random variable with parameter \(\lambda=1\), conditional on at least one callout. Calculate the probability of exactly one or two callouts using the \((a,b,1)\) form.
Method. Conditioning makes the zero-count probability \(p_0=0\). The first starting probability is \(p_1=1/(e-1)=0.581976\ldots\). The unchanged Poisson recurrence for \(n\geq2\) has \(a=0,b=1\), so \(p_2=(1/2)p_1=0.290988\ldots\).
Therefore \(P(N=1\text{ or }2)=p_1+p_2=1.5/(e-1)=0.872964\ldots\), which is \(0.8730\) to four decimal places.
Check. The positive-count probabilities sum to one: the full zero-truncated Poisson series is the original Poisson probability of a positive count, \(1-e^{-1}\), divided by \(1-e^{-1}\). The result \(0.8730\) is below one, as the event excludes counts of three or more.
Common Traps and Practical Checks
Frequency recursions are short, but a wrong model or starting value can quietly affect every probability that follows. Before generating a sequence, write down what \(N\) counts, the model parameters, the class, the applicable range of \(n\), and the starting probability or probabilities.
- Using the wrong negative binomial count. In this tutorial’s parameterization, \(N\) counts failures before the \(r\)th success. The total number of trials is \(N+r\), not \(N\).
- Using \(p_0\) to start an \((a,b,1)\) recursion. That recurrence begins at \(n=2\); \(p_1\) is an independent starting value. For a zero-truncated model, compute it after conditioning.
- Ignoring binomial support. A binomial count with \(m\) trials cannot exceed \(m\). The recurrence reaches a zero coefficient at \(m+1\); it does not describe an infinite positive tail.
- Confusing a recurrence coefficient with a probability. The quantity \(a+b/n\) is the ratio \(p_n/p_{n-1}\), not necessarily a probability. It can exceed one, as in the first Poisson step.
- Confusing frequency recursion with aggregate recursion. These formulas generate probabilities for the count \(N\). They are not, by themselves, recursions for the distribution of total claim payments.
The frequency class matters in a compound calculation because the count probabilities are one input to the distribution of total payments. A careful model identification and a correctly initialized recurrence help prevent errors before severity calculations are combined with frequency. They also provide simple diagnostics: verify the parameter mapping, check the first ratio, confirm any finite endpoint, and ensure the probabilities are valid.
Check Your Understanding
For each problem, identify or use the frequency recursion as requested. Answers are shown for checking.
- A hypothetical count is Poisson with parameter \(\lambda=2.5\). Use its recursion to calculate \(P(N=4)\), rounded to four decimal places. (Answer: \(0.1336\))
- A hypothetical number of submitted forms has distribution \(\operatorname{Binomial}(4,0.3)\). Calculate \(P(N=2)\) using the binomial probability formula or its Panjer recursion. (Answer: \(0.2646\))
- Let \(N\) count failures before the second success, with independent trials and success probability \(\theta=0.7\). State \(a,b,p_0\) and calculate \(p_3\). (Answer: \(a=0.3,\ b=0.3,\ p_0=0.49,\ p_3=0.05292\))
- A Poisson count with parameter \(\lambda=1\) is conditioned to be positive. Use the \((a,b,1)\) starting value and recurrence to calculate \(P(N\leq2)\), to four decimal places. (Answer: \(0.8730\))
- For \(N\sim\operatorname{Binomial}(3,0.25)\), use \(a,b,p_0\) and the recursion to calculate \(P(N\geq2)\). (Answer: \(a=-1/3,\ b=4/3,\ p_0=0.421875,\ P(N\geq2)=0.15625\))