Deriving the Claim-Count Full-Credibility Standard

Why Derive a Claim-Count Standard?

A claim-count full-credibility standard answers a volume question: how large must the expected number of claims be so that the observed count is likely to stay close to its expected value? “Close” is expressed as a relative tolerance, and “likely” is expressed as a specified probability. The limited-fluctuation criterion puts those two requirements together; the derivation below shows how they produce a simple threshold for the expected claim count.

This is a model-based standard, not a universal guarantee. Its familiar form depends on the count having variance equal to its mean and on using a normal approximation. If those conditions are unsuitable, the resulting threshold may not describe the actual count fluctuations well. The standard is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let \(N\) be a claim count with expected value \(m>0\). For relative tolerance \(r\), where \(0<r<1\), the count is within tolerance when \(m(1-r)\leq N\leq m(1+r)\), equivalently \(|N-m|\leq rm\). A two-sided confidence level \(p\) is the target probability of being within those bounds.

Model and Assumptions

For the standard derivation, assume the claim count has mean \(m\) and variance \(m\), and use the normal approximation \(N\approx N(m,m)\). Here the second parameter is the variance, consistent with the convention \(X\sim N(\mu,\sigma^2)\). The approximation is most defensible when the expected count is sufficiently large; it treats an integer-valued count as continuous and can assign a small probability to impossible negative counts.

The variance-equals-mean assumption is important: it makes the standard deviation \(\sqrt m\). The derivation does not follow merely from knowing the mean. If the actual count variance differs from \(m\), this particular threshold does not account for that difference. The probability target here is two-sided: departures above and below the mean both count as outside the tolerance.

Deriving the Full-Credibility Threshold

Standardize the count by subtracting its mean and dividing by its standard deviation. Under the stated normal approximation, the standardized count is approximately standard normal. The allowed distance from the mean, \(rm\), therefore becomes a standardized distance of \(r\sqrt m\).

Write the limited-fluctuation condition.
The target is \(P(|N-m|\leq rm)\geq p\), with \(m\) the expected count and \(r\) the relative tolerance.

Standardize the count.
Since the standard deviation is \(\sqrt m\), divide the deviation and the tolerance bound by \(\sqrt m\). The standardized bound is \(r\sqrt m\).

Use the central normal quantile.
Let \(z_{(1+p)/2}\) be the standard normal quantile with cumulative probability \((1+p)/2\). A central probability \(p\) requires \(r\sqrt m\geq z_{(1+p)/2}\).

Solve for the expected count.
Both sides are nonnegative, so square the inequality and divide by \(r^2\). The minimum continuous threshold is the squared quantile divided by the squared tolerance.

\begin{aligned} P(|N-m|\leq rm) &\approx P\left(\left|\frac{N-m}{\sqrt m}\right|\leq r\sqrt m\right)\\ &\approx 2\Phi(r\sqrt m)-1. \end{aligned}

The central probability is at least \(p\) when \(r\sqrt m\) is at least \(z_{(1+p)/2}\). Rearranging gives the claim-count full-credibility standard:

Theorem: Under the normal approximation \(N\approx N(m,m)\), the minimum expected claim count for a two-sided relative tolerance \(r\) and target probability \(p\) is

m_{\min}=\left(\frac{z_{(1+p)/2}}{r}\right)^2.

This is a minimum for the expected count \(m\), not a requirement that the realized count equal \(m_{\min}\). If the application requires the threshold to be an integer number of expected claims, round the calculated value up to the next whole number. Rounding down can leave the criterion unmet. The formula itself is a normal-approximation result, so extra decimal places do not make it an exact probability guarantee.

Worked Applications

Worked Example: A 90% Standard With a 10% Tolerance

A hypothetical portfolio wants a claim-count standard under which the observed count is within 10% of its expected count with 90% probability. Assume the count has variance equal to its mean and use the normal approximation. Find the minimum expected count, then check the whole-number threshold.

Method. The confidence level is two-sided, so the central 90% standard normal quantile is \(z_{0.95}=1.6449\). The relative tolerance is \(r=0.10\). Apply the derived formula:

\begin{aligned} m_{\min} &=\left(\frac{1.6449}{0.10}\right)^2\\ &=(16.449)^2\\ &=270.569601. \end{aligned}

The continuous expected-count threshold is 270.569601. If a whole-number expected-count threshold is required, round up to 271 expected claims. To check the rounding, at \(m=270\) the standardized tolerance is \(0.10\sqrt{270}\approx1.6432\), which is below 1.6449. At \(m=271\), it is \(0.10\sqrt{271}\approx1.6462\), which is above 1.6449. Thus 271 meets the approximate criterion and 270 does not.

Common mistake avoided. Do not round 270.569601 down to 270 just because the decimal part is less than one; the requirement is a minimum, and rounding down fails the check. Also, \(1.6449/0.10=16.449\), and \(16.449^2=270.569601\), not 270.571.

Worked Example: A 95% Standard With a 10% Tolerance

A hypothetical specialty book is to be assessed using a two-sided 95% count standard and a relative tolerance of 10%. Under the same variance and normal-approximation assumptions, calculate the minimum expected count and verify the integer threshold.

Method. The central 95% standard normal quantile is \(z_{0.975}=1.96\). Substitute \(p=0.95\) and \(r=0.10\):

\begin{aligned} m_{\min} &=\left(\frac{1.96}{0.10}\right)^2\\ &=(19.6)^2\\ &=384.16. \end{aligned}

Rounding up gives 385 expected claims as the whole-number threshold. Check the two adjacent integers: at \(m=384\), \(0.10\sqrt{384}\approx1.9596\), less than 1.96; at \(m=385\), \(0.10\sqrt{385}\approx1.9621\), greater than 1.96. The rounded threshold therefore meets the criterion, while 384 does not.

Common mistake avoided. The 95% two-sided criterion uses \(z_{0.975}=1.96\), not the one-sided 95% quantile \(z_{0.95}=1.6449\). Using the latter would give \((1.6449/0.10)^2=270.569601\), which is a threshold for a different central probability.

Worked Example: Finding the Tolerance for a Given Count

A hypothetical segment has an expected count of \(m=400\). The analyst wants the relative tolerance that corresponds to a two-sided 90% probability under the standard assumptions. Find the tolerance as a percentage and the count interval implied by it.

Method. Start with the threshold relationship \(r\sqrt m=z_{(1+p)/2}\) and solve for \(r\). For 90% central probability, \(z_{0.95}=1.6449\), while \(\sqrt{400}=20\):

\begin{aligned} r&=\frac{1.6449}{\sqrt{400}}\\ &=\frac{1.6449}{20}\\ &=0.082245. \end{aligned}

The relative tolerance is \(0.082245\), or 8.2245%. The permitted absolute deviation is \(rm=0.082245(400)=32.898\) claims. Thus the normal-approximation interval is from \(400-32.898=367.102\) to \(400+32.898=432.898\). Its standardized half-width is \(32.898/20=1.6449\), confirming the intended central 90% probability.

Common mistake avoided. Do not divide the quantile by \(m=400\). The standard deviation is \(\sqrt m=20\); using \(400\) would give the wrong scale and an unjustifiably tiny tolerance.

Interpreting the Standard and Avoiding Traps

The formula displays two useful relationships. A higher confidence level requires a larger normal quantile and therefore a larger expected count. A tighter tolerance means a smaller \(r\); because the tolerance is squared in the denominator, halving it multiplies the required count by four, all else equal. For example, changing a tolerance from 10% to 5% at the same confidence level requires four times the expected count under this standard.

Keep the probability statement precise. For a central probability \(p\), the tails each have probability \((1-p)/2\), so the quantile is \(z_{(1+p)/2}\). A one-sided probability uses a different quantile and answers a different question. Also keep the reference value straight: “within \(r\) of the mean” means a bound of \(rm\), not \(rN\) or a percentage of the observed count.

Using the wrong variance: the standard formula presumes \(\operatorname{Var}(N)=m\). If that assumption is not appropriate, substituting the mean anyway does not repair the model.
Using a one-sided quantile: a two-sided tolerance condition includes both unusually low and unusually high counts.
Rounding to nearest: a minimum threshold should be rounded upward when an integer expected count is required.
Claiming exactness: this derivation gives a normal-approximation standard. It is not an exact discrete probability calculation.

The standard is useful because it translates a statement about acceptable frequency fluctuation into a volume requirement. It does not, by itself, determine how many policies or exposure units are needed; that conversion requires a claim-frequency assumption in addition to this count threshold.

Key takeaway: For a count with mean and variance \(m\), the normal-approximation full-credibility threshold is \(m_{\min}=(z_{(1+p)/2}/r)^2\). Identify the two-sided quantile, use the relative tolerance as a decimal, and round a required whole-number threshold upward.

Check Your Understanding

For each problem, use the normal-approximation count standard with variance equal to the mean. Round a required whole-number threshold upward.

A hypothetical book requires 90% central probability within 8% of its expected count. Use \(z_{0.95}=1.6449\). Find the continuous threshold and the whole-number threshold. (Answer: \(422.7650016\), so 423 expected claims.)
A hypothetical portfolio requires 95% central probability within 5% of its expected count. Use \(z_{0.975}=1.96\). Find the minimum expected count. (Answer: \(1{,}536.64\), so 1,537 expected claims.)
A segment has expected count 625. What relative tolerance corresponds to 90% central probability, using \(z_{0.95}=1.6449\)? (Answer: \(1.6449/25=0.065796\), or 6.5796%.)
A claim count has expected value 900. Find the relative tolerance for 95% central probability using \(z_{0.975}=1.96\). (Answer: \(1.96/30=0.0653333\), or approximately 6.5333%.)
Find the whole-number expected-count threshold for 99% central probability within 10%, using \(z_{0.995}=2.5758\). (Answer: \((2.5758/0.10)^2=663.474564\), so 664 expected claims.)

What You'll Learn