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Exam FAM · Frequency Probabilities and Approximations · Tutorial 553 of 988

Fixed-Trial And Unbounded Frequency Models

Learn how trial structure determines whether a frequency count has a fixed maximum or unbounded support, and calculate probabilities without confusing the models.

Intermediate 11 min read

What You'll Learn

  • Identify the assumptions that give a binomial count a fixed upper bound.
  • State the support of binomial, Poisson, and negative binomial counts.
  • Calculate exact and tail probabilities while respecting each model’s support.
  • Explain why Poisson and negative binomial probabilities can be positive above a finite trial count.
  • Avoid applying a fixed-trial ceiling to a model that allows repeated claims.

Why a Frequency Model Has a Particular Support

A frequency model describes the number of events in a specified group or period. Before calculating a probability, determine which counts are possible under the model. A count of claims from a fixed number of policies, each of which can produce at most one claim, has a natural maximum. A count of events that can recur within an exposure period need not have such a ceiling.

The support of a discrete random variable is the set of values that can have positive probability. Support is not merely a technical detail: it reflects the structure assumed for the count. A model can assign a small probability to a large count, but it cannot assign positive probability to a count outside its support.

Definition: If \(N\) is a frequency count, its support is the set of integer values \(k\) for which \(P(N=k)>0\). In the standard nondegenerate binomial model, the support is \(0,1,\ldots,n\). In the standard Poisson and negative binomial models, it is \(0,1,2,\ldots\), with no finite upper bound.

Three Models and Their Supports

A binomial model requires a fixed number \(n\) of trials. Each trial has two outcomes, such as “claim” and “no claim”; the probability of a claim is the same \(p\) on every trial; and the trial outcomes are independent. The count \(N\) is the number of claim outcomes. Under these assumptions,

$$ N\sim\operatorname{Bin}(n,p),\qquad P(N=k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,\ldots,n. $$

For \(0<p<1\), each listed count has positive probability. No more than \(n\) claims can occur because there are only \(n\) trials and each trial contributes at most one claim. The endpoint cases \(p=0\) and \(p=1\) are degenerate: the count is always 0 or always \(n\), respectively.

A Poisson model is commonly used for a count over a specified exposure when events occur singly, at a constant rate, and independently over disjoint portions of exposure. In particular, the model does not impose a fixed number of opportunities with at most one event per opportunity. For \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\),

$$ P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}, \qquad k=0,1,2,\ldots. $$

Every nonnegative integer has positive probability because each factor in this expression is positive. The Poisson model has unbounded support, even if a very large count is unlikely. If \(\lambda=0\), the distribution is instead degenerate at zero.

One standard negative binomial parameterization counts the number \(N\) of failures before the \(r\)th success in independent Bernoulli trials, where each trial has success probability \(p\). Here \(r\) is a positive integer and \(0<p<1\). Its probability function is

$$ P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k, \qquad k=0,1,2,\ldots. $$

For any finite \(k\), the binomial coefficient and both probability factors are positive, so every nonnegative integer is possible. This parameterization is also used to model frequency counts. When interpreting a negative binomial model in an insurance setting, be clear about the parameterization being used; here \(p\) is the probability of success, and \(N\) counts failures before \(r\) successes.

Why the Binomial Ceiling Holds

The fixed upper bound follows directly from what the binomial count measures, rather than from an approximation or a calculation. Let \(I_j\) equal 1 if trial \(j\) produces a claim and 0 otherwise. Then \(N=I_1+\cdots+I_n\). Since each \(I_j\leq 1\),

$$ N=\sum_{j=1}^{n} I_j\leq \sum_{j=1}^{n}1=n. $$

Thus \(P(N>n)=0\). In contrast, the Poisson and negative binomial probability formulas assign positive probability to each finite nonnegative integer. This is the key structural difference: a binomial count records successes in a fixed set of one-event trials, whereas the other two models do not place that same finite-trial ceiling on the count.

1
Identify what is being counted.
Decide whether the count is a number of successful trials, events over exposure, or failures before a specified number of successes.
2
Check for a genuine fixed maximum.
A fixed number \(n\) of independent trials, each contributing at most one event, implies a ceiling of \(n\). A finite exposure period alone does not create that binomial ceiling.
3
Use the probability formula on its support.
For a binomial count, values above \(n\) have probability zero. For Poisson or negative binomial counts, any finite nonnegative integer is in the support.

Worked Probability Calculations

Worked Example: A Four-Policy Portfolio and an Impossible Count

A hypothetical portfolio has four policies. During a specified period, assume each policy independently has probability \(0.25\) of producing exactly one claim and cannot produce more than one claim during that period. Let \(N\) be the number of claims. Find \(P(N=2)\) and \(P(N\geq 5)\). Then compare the latter event with a Poisson model having \(\lambda=np\).

Method. The four policies are four independent Bernoulli trials with the same claim probability, so the exact model is \(N\sim\operatorname{Bin}(4,0.25)\). The binomial support ends at 4.

For exactly two claims,

$$ P(N=2)=\binom{4}{2}(0.25)^2(0.75)^2 =6(0.0625)(0.5625) =0.2109375. $$

To five decimal places, \(P(N=2)=0.21094\). Since \(N\leq 4\), \(P(N\geq 5)=0\) under the exact binomial model.

For the requested comparison, \(\lambda=np=4(0.25)=1\). A Poisson model with this parameter gives

$$ P(N\geq 5) =1-\sum_{k=0}^{4}e^{-1}\frac{1^k}{k!} =1-e^{-1}\left(1+1+\frac12+\frac16+\frac1{24}\right) =1-e^{-1}\left(\frac{65}{24}\right) \approx 0.0037. $$

This positive Poisson tail is possible because the Poisson model has no four-claim ceiling. It is not the exact probability for the stated four-policy setup. Common mistake avoided: using a Poisson approximation and then treating its positive probability above four as though the fixed-trial portfolio could actually produce five claims.

Worked Example: A Poisson Count Above a Chosen Threshold

A hypothetical repair service models the number \(N\) of qualifying breakdown calls in a month by a Poisson distribution with \(\lambda=2.4\). Find \(P(N\geq4)\) and \(P(N=6)\), giving probabilities to four decimal places.

Method. Use the Poisson probability function and complement the probabilities for counts 0 through 3 to find the upper tail. No fixed number of trials is specified, so there is no binomial ceiling.

$$ \begin{aligned} P(N\geq4) &=1-\sum_{k=0}^{3}e^{-2.4}\frac{2.4^k}{k!}\\ &=1-e^{-2.4}\left(1+2.4+\frac{2.4^2}{2}+\frac{2.4^3}{6}\right)\\ &=1-e^{-2.4}(1+2.4+2.88+2.304)\\ &=1-e^{-2.4}(8.584)\\ &\approx 1-0.7787229\\ &\approx 0.2213. \end{aligned} $$

For exactly six calls,

$$ \begin{aligned} P(N=6) &=e^{-2.4}\frac{2.4^6}{6!}\\ &=e^{-2.4}\frac{191.102976}{720}\\ &=e^{-2.4}(0.2654208)\\ &\approx 0.0241. \end{aligned} $$

Both events have positive probability. In particular, six calls are possible under the Poisson model even though six exceeds the mean parameter \(2.4\). The parameter is not a maximum count. Common mistake avoided: treating the rate parameter as an upper bound or rounding the model’s support to a “typical” range.

Worked Example: A Negative Binomial Tail

In a hypothetical trial-based process, independent trials continue until the third success. Each trial succeeds with probability \(p=0.60\). Let \(N\) count failures before the third success. Find \(P(N\geq4)\), and then find \(P(N\geq5)\).

Method. This is a negative binomial count with \(r=3\) and \(p=0.60\), so \(1-p=0.40\). First calculate the probabilities for \(N=0,1,2,3\) and subtract their sum from 1.

$$ \begin{aligned} P(N=0)&=\binom{2}{0}(0.60)^3(0.40)^0=0.216,\\ P(N=1)&=\binom{3}{1}(0.60)^3(0.40)^1=3(0.216)(0.40)=0.2592,\\ P(N=2)&=\binom{4}{2}(0.60)^3(0.40)^2=6(0.216)(0.16)=0.20736,\\ P(N=3)&=\binom{5}{3}(0.60)^3(0.40)^3=10(0.216)(0.064)=0.13824. \end{aligned} $$

The sum through 3 is \(0.216+0.2592+0.20736+0.13824=0.8208\). Therefore,

$$ P(N\geq4)=1-0.8208=0.1792. $$

To obtain \(P(N\geq5)\), remove \(P(N=4)\) from this tail:

$$ \begin{aligned} P(N=4) &=\binom{6}{4}(0.60)^3(0.40)^4\\ &=15(0.216)(0.0256)=0.082944,\\ P(N\geq5)&=0.1792-0.082944=0.096256. \end{aligned} $$

Thus, to four decimal places, \(P(N\geq4)=0.1792\) and \(P(N\geq5)=0.0963\). Counts of four or more failures are possible because the process can require more trials to reach three successes. Common mistake avoided: assuming that the target of three successes limits the number of failures to three.

Choosing the Correct Ceiling

A maximum count must come from the model’s structure, not from the size of an average, the length of an observation period, or the fact that the count is usually small. For example, a fixed group of \(n\) policies supports a binomial count if each policy can contribute at most one claim and the binomial assumptions are appropriate. If a policy can produce multiple claims, the count of claims is not bounded by the number of policies in that way. A Poisson or negative binomial model may then be appropriate depending on the stated assumptions; neither is limited to a fixed number of claims.

A Poisson approximation to a binomial distribution also deserves careful interpretation. The approximation can assign positive probability above \(n\), even though the exact binomial probability there is zero. This does not change the exact support of the binomial model. It is a consequence of using a different distribution to approximate probabilities. Whether the approximation is sufficiently accurate is a separate question from which counts the exact model permits.

Key takeaway: A fixed number of one-event trials gives a binomial count with maximum \(n\). Poisson and negative binomial counts have unbounded support under their standard nondegenerate parameterizations. Check the event-generating structure first, then calculate only with the model and support that structure warrants. This distinction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the support issue before calculating. Probabilities should be rounded to four decimal places where needed.

  1. A hypothetical group has six independent members, each of whom either files one claim or files none, with claim probability \(0.10\). Under the binomial model, what is the probability of more than six claims? (Answer: \(0\))
  2. A hypothetical machine station has a Poisson count of service interruptions with \(\lambda=1.5\). Calculate the probability of exactly four interruptions. (Answer: \(e^{-1.5}(1.5^4/4!)\approx0.0471\))
  3. Let \(N\) be the number of failures before the second success in independent trials with success probability \(0.50\). Calculate \(P(N\geq2)\). (Answer: \(0.5000\))
  4. A hypothetical three-unit portfolio has independent one-claim-or-no-claim outcomes, each with claim probability \(0.40\). Calculate the probability of at least two claims, and state why a count of four is impossible. (Answer: \(0.3520\); four is impossible because the binomial ceiling is three)
  5. A hypothetical call count has a Poisson distribution with \(\lambda=2\). Calculate the probability of more than three calls. (Answer: \(1-e^{-2}(1+2+2+8/6)\approx0.1429\))

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