Annual Claim Counts and Full Credibility
A full-credibility count standard asks how much claim-count experience is needed for the observed count to be sufficiently close to its expected value with a specified probability. For annual claim counts, the key inputs are the expected number of claims in one year, the permitted relative error, and the probability standard. The result is first a threshold for the expected total number of claims; it can then be translated into a minimum number of years.
This tutorial focuses on a homogeneous Poisson model for annual counts. The standard is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. No interest or mortality assumption is needed: the model concerns claim counts, not the timing or present value of payments.
The standard Poisson model assumes a constant expected annual count over the period and independent counts from year to year. It is not appropriate without adjustment if the annual exposure changes, the underlying frequency changes over time, or counts exhibit dependence or overdispersion. Those situations require a model or standard that accounts for the difference.
The Full-Credibility Count Threshold
Let \(m=n\lambda\) denote the expected total number of claims. A two-sided relative-error criterion with tolerance \(k\) and target probability \(p\) is \(P(|N-m|\leq km)\geq p\). Under the normal approximation to a Poisson count, the usual full-credibility threshold is \(m_0=(z/k)^2\), where \(z\) is the standard normal quantile appropriate to the probability criterion. This is a threshold on the expected total count, not a requirement that the realized count equal a particular number.
For a two-sided central probability \(p\), use \(z=z_{(1+p)/2}\). For a one-sided probability \(p\), use \(z=z_p\), with the direction of the one-sided error stated in the criterion. The corresponding minimum number of whole years under a constant annual mean is \(n_{\min}=\left\lceil m_0/\lambda\right\rceil\). Keep \(m_0\) unrounded when calculating the years, and round the number of years upward because a fractional year is not a whole annual observation period.
Why the Threshold Has This Form
The reason for the square in the threshold is the relationship between a Poisson mean and its variance. If the total expected count is \(m\), then \(\operatorname{Var}(N)=m\), so the standard deviation is \(\sqrt m\). Under the normal approximation, a two-sided error band of \(km\) has standardized width \(km/\sqrt m=k\sqrt m\). To make that width at least \(z\), require \(k\sqrt m\geq z\), which gives \(m\geq(z/k)^2\).
The annual conversion follows from additivity of independent Poisson counts. For example, if one year contributes a Poisson count with mean \(\lambda\), the sum over \(n\) independent years has a Poisson distribution with mean \(n\lambda\). Thus the full-credibility count threshold can be compared directly with the expected count accumulated over whole years.
Write the annual mean \(\lambda\), the probability standard, and whether the criterion is one-sided or two-sided.
Select the corresponding \(z\) and evaluate \(m_0=(z/k)^2\).
Calculate \(\lceil m_0/\lambda\rceil\), then verify that this many years meet the threshold and one fewer year does not.
Worked Applications
Worked Example: A Two-Sided 90% Standard
A hypothetical roadside-assistance portfolio is modeled with independent annual counts \(N_j\sim\operatorname{Poisson}(12)\). Find the minimum whole number of years for a two-sided 90% full-credibility standard with relative error \(k=20\%\). Use \(z_{0.95}=1.645\). Report the whole-claim threshold as well.
Method. Use the two-sided normal quantile because the criterion allows deviations both above and below the expected count. Here \(\lambda=12\), \(k=0.20\), and \(z=1.645\).
The unrounded expected-claim threshold is \(m_0=(1.645/0.20)^2=8.225^2=67.650625\). The requested whole-claim threshold is \(\lceil67.650625\rceil=68\) claims.
For the whole-year requirement, use the unrounded threshold: \(n_{\min}=\lceil67.650625/12\rceil=\lceil5.637552\ldots\rceil=6\) years. Five years provide an expected \(5(12)=60\) claims, below 67.650625. Six years provide \(6(12)=72\) expected claims, which meets the standard. Therefore, the minimum is 6 years, or a whole-claim threshold of 68 expected claims.
The check confirms that the rounding is upward: five years are insufficient and six years are sufficient. The count threshold is based on expected claims, not on a promise that exactly 68 claims will be observed.
Worked Example: A One-Sided 95% Standard
A hypothetical equipment insurer has independent annual claim counts \(N_j\sim\operatorname{Poisson}(24)\). A one-sided 95% standard allows an upward deviation no greater than 10% of the expected total count, with probability at least 95%. Find the minimum whole number of years. Use \(z_{0.95}=1.645\).
Method. Because the stated criterion is one-sided, use \(z_{0.95}=1.645\), not the two-sided 95% quantile \(1.96\). The relative error is \(k=0.10\).
The threshold is \(m_0=(1.645/0.10)^2=16.45^2=270.6025\) expected claims. The minimum number of years is \(n_{\min}=\lceil270.6025/24\rceil=\lceil11.275104\ldots\rceil=12\).
Eleven years have expected count \(11(24)=264\), which is less than 270.6025. Twelve years have expected count \(12(24)=288\), which exceeds it. The minimum is therefore 12 years.
The common mistake avoided here is treating every 95% criterion as two-sided. That would substitute 1.96 and produce a larger threshold, even though the question specifies only one direction of deviation.
Worked Example: Check the Error Tolerance Before Rounding
A hypothetical portfolio has independent annual counts \(N_j\sim\operatorname{Poisson}(18)\). For a two-sided 95% standard with relative error \(k=15\%\), find the whole-claim threshold and the minimum number of years. Use \(z_{0.975}=1.96\).
Method. The probability standard is two-sided, so use \(z=1.96\). Convert 15% to the decimal \(k=0.15\).
The expected-claim threshold is \(m_0=(1.96/0.15)^2=13.066666\ldots^2=170.737777\ldots\). Thus the whole-claim threshold is \(\lceil170.737777\ldots\rceil=171\) claims.
The minimum number of years is \(n_{\min}=\lceil170.737777\ldots/18\rceil=\lceil9.485432\ldots\rceil=10\). Nine years have expected count \(9(18)=162\), which is below the unrounded threshold. Ten years have expected count \(10(18)=180\), which is above it. The answer is 171 claims as the rounded threshold and 10 years.
Common mistake. If the 15% tolerance is accidentally replaced by 10%, the calculation becomes \((1.96/0.10)^2=384.16\), or 385 whole claims. That is not the stated standard. Using the given 15% tolerance gives 170.737777… and 171 whole claims; using the annual mean then gives 10 years, not 22. The check with nine and ten years confirms the minimum.
Common Checks and Interpretation
A full-credibility threshold is a planning standard under the specified model and normal approximation. It does not say that a particular year's realized count must exceed the threshold, nor does it guarantee that every possible sample will fall within the error band. The criterion concerns the probability of the count being within the specified band.
- Use the right quantile. Match the quantile to whether the requirement is one-sided or two-sided.
- Use the relative error as a decimal. For example, 15% means \(k=0.15\), not 15.
- Do not round down years. The number of annual periods must be an integer that meets or exceeds the expected-count threshold.
- Do not round too early. Use the unrounded \(m_0\) in the years calculation, then check the expected counts for the selected number of years and one fewer year.
- Keep the model assumptions in view. A changing annual mean or dependent annual counts can invalidate the simple \(n\lambda\) calculation.
A reliable solution therefore identifies the criterion and model first, calculates the threshold with the matching quantile, and verifies the integer-year result against both sides of the cutoff. These steps prevent a plausible-looking answer from being wrong because of a quantile choice, a percentage conversion, or a rounding decision.
Check Your Understanding
For each problem, assume the stated annual counts are independent and Poisson with a constant mean. Use the specified normal quantile and round whole years upward.
- A hypothetical ferry operator has an expected 10 claims per year. Find the minimum years for a two-sided 90% standard with 25% relative error, using \(z_{0.95}=1.645\). Give the whole-claim threshold too. (Answer: threshold 44 expected claims; 5 years)
- A hypothetical storage insurer has an expected 16 claims per year. Find the minimum years for a two-sided 95% standard with 20% relative error, using \(z_{0.975}=1.96\). (Answer: threshold 97 expected claims; 7 years)
- A hypothetical delivery fleet has an expected 30 claims per year. For a one-sided 95% standard with 12% relative error, use \(z_{0.95}=1.645\). Find the whole-claim threshold and minimum years. (Answer: threshold 188 expected claims; 7 years)
- A hypothetical event-cancellation portfolio has an expected 8.5 claims per year. Find the minimum years for a two-sided 90% standard with 10% relative error, using \(z_{0.95}=1.645\). (Answer: threshold 271 expected claims; 32 years)
- A hypothetical small appliance portfolio has an expected 25 claims per year. Under a two-sided 95% standard with 12% relative error, using \(z_{0.975}=1.96\), determine whether 10 years are sufficient and give the minimum whole number of years. (Answer: 10 years are not sufficient; minimum 11 years)