Full Credibility for Average Claim Severity

Why Severity Needs Its Own Credibility Standard

A collection of claims can have a stable claim count but an unstable average severity. A few unusually large claims may move the average substantially, particularly when the number of claims is small. A full-credibility severity standard answers a specific planning question: how many independent claims are needed for the observed average severity to be close to the expected severity with a specified probability?

The calculation depends on the variability of individual claim severities, the target confidence level, and the permitted error. It does not by itself establish that the expected severity is known, that the data are relevant to a rating group, or that an indicated rate is adequate. This treatment is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: Let $X_1,\ldots,X_n$ be independent claim severities drawn from the same population, each with finite mean $E[X]=\mu$ and variance $\operatorname{Var}(X)=\sigma^2$. The average severity is $\bar{X}=(X_1+\cdots+X_n)/n$. A full-credibility standard is the smallest whole number of claims $n$ for which the specified probability that $\bar{X}$ is within the stated tolerance of $\mu$ is at least the target confidence level, using the selected model or approximation.

The Normal-Approximation Standard

For an error tolerance expressed as a fraction $r$ of the expected severity, the required event is $|\bar{X}-\mu|\leq r\mu$. For example, $r=0.05$ means that the sample average is within 5% of the expected severity. The confidence level $p$ is two-sided: the average may be above or below $\mu$, but must be within the tolerance in either direction.

When the claim severities are independent and identically distributed with finite variance, the central limit theorem gives the approximation $\bar{X}\approx N(\mu,\sigma^2/n)$ for a sufficiently large $n$. This is an approximation for the sample average; it does not require each individual claim severity to be normally distributed. Let $z_{(1+p)/2}$ be the standard normal quantile with cumulative probability $(1+p)/2$. Under this approximation, the full-credibility standard is:

n \geq \left(\frac{z_{(1+p)/2}\sigma}{r\mu}\right)^2 \qquad\Longrightarrow\qquad n_{\text{full}}=\left\lceil\left(\frac{z_{(1+p)/2}\sigma}{r\mu}\right)^2\right\rceil .

The ceiling notation means round upward to the next whole claim. A fractional claim count cannot meet the standard; rounding to the nearest integer could leave the result below the required threshold.

Why the Formula Holds

Under the normal approximation, the average has mean $\mu$ and standard deviation $\sigma/\sqrt{n}$. Standardizing the deviation from $\mu$ gives:

P(|\bar{X}-\mu|\leq r\mu) \approx P\left(\left|\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right| \leq \frac{r\mu\sqrt{n}}{\sigma}\right).

The standardized quantity is approximately standard normal. For a central confidence level $p$, the probability between $-z_{(1+p)/2}$ and $z_{(1+p)/2}$ is $p$. Thus the target is met when $r\mu\sqrt{n}/\sigma\geq z_{(1+p)/2}$. Squaring both nonnegative sides and rearranging gives $n\geq(z_{(1+p)/2}\sigma/(r\mu))^2$, which yields the stated integer standard.

If the allowed error is an absolute amount $a$, measured in dollars, use $|\bar{X}-\mu|\leq a$ instead. The corresponding standard is:

n_{\text{full}}=\left\lceil\left(\frac{z_{(1+p)/2}\sigma}{a}\right)^2\right\rceil .

The two forms agree when $a=r\mu$. Use the relative form when the requirement is stated as a percentage of expected severity, and the absolute form when the requirement is stated as a fixed dollar amount.

Applying the Standard

A reliable calculation follows the same sequence: identify whether the tolerance is relative or absolute, use the standard deviation rather than the variance, select the two-sided normal quantile for the confidence level, calculate the unrounded claim standard, and round upward.

Set the target.
Write the permitted deviation as $r\mu$ for a relative tolerance, or as $a$ for an absolute tolerance.

Identify severity variability.
Use $\sigma=\sqrt{\operatorname{Var}(X)}$, and confirm the mean and variance refer to the same severity basis.

Use the correct quantile.
For a central confidence level $p$, use $z_{(1+p)/2}$, not a one-sided quantile.

Calculate and round up.
Evaluate the formula without intermediate rounding, then take the ceiling to obtain a whole-claim standard.

Worked Example: Relative Tolerance for a Specialty Repair Portfolio

A hypothetical insurer models individual repair-claim severities as independent and identically distributed gamma random variables with shape 4 and scale $250. For a gamma distribution with shape $\alpha$ and scale $\theta$, the mean is $\alpha\theta$ and the variance is $\alpha\theta^2$. Find the full-credibility claim count for a two-sided 90% confidence level and a tolerance of 10% of expected severity. No interest rate or mortality assumption applies to this severity calculation.

Method and parameters. The gamma model has $\mu=4(250)=\$1{,}000$ and $\sigma^2=4(250^2)=\$250{,}000$, so $\sigma=\$500$. The relative tolerance is $r=0.10$, and the central 90% normal quantile is $z_{0.95}=1.645$.

Substitute these values into the relative-tolerance formula:

n \geq \left(\frac{1.645(500)}{0.10(1{,}000)}\right)^2 =\left(\frac{822.5}{100}\right)^2 =(8.225)^2 =67.650625.

Rounding upward gives $n_{\text{full}}=68$ claims. As a boundary check, 67 claims give a standardized tolerance of $100\sqrt{67}/500\approx1.637$, below 1.645. For 68 claims it is $100\sqrt{68}/500\approx1.649$, above 1.645. Thus 68 is the smallest whole-number standard under this approximation.

Common mistake avoided: Using the variance, $250,000, in place of the standard deviation, $500, would put the wrong quantity in the formula and produce a meaningless claim count. The formula uses $\sigma$, not $\sigma^2$.

Worked Example: A Tighter Target for Equipment Claims

A hypothetical equipment insurer assumes independent and identically distributed claim severities with a gamma distribution of shape 25 and scale $50. Determine the claim standard for a two-sided 95% confidence level when the average must be within 5% of its expected value. No interest rate or mortality assumption applies.

Method and parameters. The mean is $\mu=25(50)=\$1{,}250$, and the variance is $\sigma^2=25(50^2)=\$62{,}500$, giving $\sigma=\$250$. The tolerance is $r=0.05$, and $z_{0.975}=1.96$.

The allowable deviation is $r\mu=0.05(\$1{,}250)=\$62.50$. Therefore:

n \geq \left(\frac{1.96(250)}{62.50}\right)^2 =\left(\frac{490}{62.50}\right)^2 =(7.84)^2 =61.4656.

The full-credibility standard is $n_{\text{full}}=62$ claims. At 62 claims, the standardized tolerance is $62.50\sqrt{62}/250\approx1.968$, which exceeds 1.96. At 61 claims, it is $62.50\sqrt{61}/250\approx1.952$, which is below 1.96. The upward rounding is therefore necessary.

Common mistake avoided: A 95% confidence level here is two-sided, so the required quantile is 1.96. Using 1.645 would treat the target as a one-sided 95% probability and understate the needed number of claims.

Worked Example: An Absolute Dollar Tolerance

A hypothetical cargo insurer models independent claim severities using a gamma distribution with shape 4 and scale $150. Find the claim standard for a two-sided 99% confidence level if the average must be within $120 of expected severity. No interest rate or mortality assumption applies.

Method and parameters. The gamma mean is $\mu=4(150)=\$600$, and the variance is $\sigma^2=4(150^2)=\$90{,}000$, so $\sigma=\$300$. The error tolerance is the absolute amount $a=\$120$, and $z_{0.995}=2.576$.

Because the tolerance is stated in dollars, use the absolute-tolerance formula:

n \geq \left(\frac{2.576(300)}{120}\right)^2 =\left(\frac{772.8}{120}\right)^2 =(6.44)^2 =41.4736.

The standard is $n_{\text{full}}=42$ claims. For 41 claims, the standardized tolerance is $120\sqrt{41}/300\approx2.561$, below 2.576. For 42 claims, it is $120\sqrt{42}/300\approx2.592$, above 2.576. Thus 42 is the smallest qualifying whole number.

Common mistake avoided: Converting an absolute tolerance to a percentage is not necessary. If converted, $120 divided by the mean of $600 gives 20%, and the relative formula produces the same answer; dividing by another amount would change the target.

Interpretation and Limits

A full-credibility standard is a precision rule for the average severity under a stated model. It is not a guarantee that any particular sample average will land within the tolerance, nor does reaching the standard eliminate random fluctuation. The confidence statement is a probability over repeated samples under the assumed severity process.

The normal approximation is most defensible when observations are independent, come from a stable population with a finite variance, and the sample size is large enough for the sample mean to be approximately normal. A strongly skewed severity distribution or a heavy upper tail can make convergence slow. In such circumstances, a calculated standard should not be treated as a substitute for checking the data and assumptions. If observations are dependent, or the severities do not share a common distribution, the stated standard does not apply directly without accounting for that structure.

The standard also depends on how severity is defined. The expected value and variance must relate to the same claim basis—for example, the same coverage and treatment of limits or deductibles. Changing that basis changes the severity distribution and may change the required claim count. The earlier material on credibility complements and data quality is also relevant when deciding whether the observations represent the population for which the estimate will be used.

Key takeaway: For a two-sided relative tolerance $r$, calculate $n_{\text{full}}=\left\lceil(z_{(1+p)/2}\sigma/(r\mu))^2\right\rceil$. For an absolute tolerance $a$, replace $r\mu$ with $a$. Use the standard deviation, the correct two-sided quantile, and an upward final rounding.

Check Your Understanding

For each problem, assume independent claim severities with finite variance and use the normal approximation. Round the final claim standard upward.

A hypothetical severity population has mean $400 and standard deviation $200. Find the standard for a two-sided 90% confidence level and a relative tolerance of 8%. (Answer: 106 claims)
A hypothetical portfolio has mean severity $900 and standard deviation $300. Find the standard for a two-sided 95% confidence level and a relative tolerance of 10%. (Answer: 43 claims)
A hypothetical severity population has mean $2,000 and variance 1,000,000. Find the standard for a two-sided 99% confidence level and a relative tolerance of 5%. (Answer: 664 claims)
A hypothetical claim type has mean severity $750 and standard deviation $150. Find the standard for a two-sided 95% confidence level with an absolute tolerance of $75. (Answer: 16 claims)
A hypothetical severity population has mean $500 and standard deviation $400. Find the standard for a two-sided 95% confidence level and a relative tolerance of 12.5%. (Answer: 158 claims)

What You'll Learn