Deriving Severity Standards from Claim Size Variance

Why Severity Variability Determines the Claim Count

A sample average of claim severities is more stable when individual claim sizes vary less. It is also more stable when more claims are observed. A full-credibility claim-count standard makes that relationship operational: given a confidence level and an acceptable error around the expected severity, it identifies the minimum number of claims required by a specified probability rule.

The standard is conditional on the number of claims. It concerns the average size of $n$ claims, not the uncertainty in how many claims will occur. Keeping those sources of variation separate is essential: a claim-count requirement for severity is not automatically an exposure-unit requirement.

Definition: Let $X_1,\ldots,X_n$ be independent, identically distributed individual claim severities, with finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$. The sample average severity is $\bar X=(X_1+\cdots+X_n)/n$. For confidence level $p$ and relative error tolerance $k$, the full-credibility claim-count standard is the smallest integer $n$ for which the probability that $\bar X$ is within $k\mu$ of $\mu$ is at least $p$, under the chosen probability model or approximation.

The usual limited-fluctuation calculation uses a normal model for the standardized sample average. If individual severities are normally distributed, the standardized average is exactly standard normal. For other severity distributions, the calculation uses a normal approximation to the sample average; its quality depends on the distribution and the claim count. The formula is not a guarantee that the approximation is accurate for a small sample from a highly skewed or heavy-tailed distribution.

Deriving the Severity Claim-Count Standard

Under the assumptions above, the sample average has mean $\mu$ and variance $\sigma^2/n$. Thus its standard error is $\sigma/\sqrt n$. Let $\Phi$ denote the standard normal cumulative distribution function, and define $z_p=\Phi^{-1}((1+p)/2)$. The probability that the sample average is within the relative tolerance is approximated by the probability that its standardized value falls between $-z_p$ and $z_p$.

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

The standardized sample average is approximately standard normal, so the probability on the right is at least $p$ when its cutoff is at least $z_p$. This gives a direct derivation of the required count:

\frac{k\mu\sqrt n}{\sigma}\geq z_p \quad\Longleftrightarrow\quad n\geq \frac{z_p^2\sigma^2}{k^2\mu^2}.

Since the number of claims must be an integer, round the calculated requirement up. Rounding down can leave the probability below the target. If the tolerance is instead an absolute amount $\varepsilon$, so that the target is $|\bar X-\mu|\leq\varepsilon$, the corresponding requirement is:

n\geq\frac{z_p^2\sigma^2}{\varepsilon^2}.

These formulas require $\mu>0$ for a relative tolerance and a meaningful finite variance. They also presume the claim severities being averaged are drawn from the stated population under comparable conditions. If severities are dependent, systematically changing over time, or drawn from materially different populations, the independent and identically distributed model may not describe the uncertainty.

Worked Applications

Worked Example: A 95% Standard With a Relative Tolerance

A hypothetical specialty insurer models individual claim severity as Gamma with shape parameter 4 and scale parameter $150. For this parameterization, the mean is shape times scale and the variance is shape times scale squared. How many claims are required for the sample average to be within 10% of the expected severity with approximately 95% probability?

Method and parameters. The Gamma model has mean $\mu=4(150)=\$600$ and variance $\sigma^2=4(150^2)=\$90{,}000$, so $\sigma=\$300$. The relative tolerance is $k=0.10$. For a two-sided 95% target, use $z_p=1.96$. Because the severity distribution is Gamma rather than normal, this is a normal approximation for the sample average.

Calculate the unrounded requirement. The allowed absolute deviation is $k\mu=0.10(\$600)=\$60$. Substitution gives:

n\geq \left(\frac{1.96(\$300)}{\$60}\right)^2 = (9.8)^2 = 96.04.

The count must be rounded up, so the standard is 97 claims. To check the boundary, the normal cutoff for $n=96$ is $0.10(\$600)\sqrt{96}/\$300=0.2\sqrt{96}\approx1.95959$, which is below 1.96. For $n=97$, it is $0.10(\$600)\sqrt{97}/\$300=\sqrt{97}/5\approx1.96977$, which exceeds 1.96. Thus 97, not 96, meets the stated normal-approximation criterion.

Common mistake avoided. The expression $1.96(300)/60=9.8$ is not itself the required claim count; it must be squared. Using 9.8 claims, or rounding it to 10, would omit the square-root relationship between standard error and claim count.

Worked Example: A 90% Standard With a Different Severity Scale

A hypothetical equipment insurer assumes that individual claim severity follows a Gamma distribution with shape 9 and scale $200. Determine the claim count required for the sample average to be within 5% of expected severity with approximately 90% probability.

Method and parameters. The mean is $\mu=9(200)=\$1{,}800$. The variance is $9(200^2)=\$360{,}000$, giving $\sigma=\$600$. The allowed deviation is $0.05(\$1{,}800)=\$90$. Use $z_p=1.645$ for a two-sided 90% normal probability target.

Calculate and round up.

n\geq \left(\frac{1.645(\$600)}{\$90}\right)^2 = \left(\frac{987}{90}\right)^2 \approx (10.96667)^2 \approx120.26778.

The required standard is 121 claims. The boundary check uses the standardized tolerance $k\mu\sqrt n/\sigma=0.15\sqrt n$: at $n=120$, it is $0.15\sqrt{120}\approx1.64317<1.645$; at $n=121$, it is $0.15\sqrt{121}=1.65>1.645$. The integer result therefore meets the target under the stated approximation.

Common mistake avoided. A 90% two-sided target uses the standard normal cutoff 1.645, not the 95% cutoff 1.96. Selecting the wrong confidence cutoff changes the count even if the mean, variance, and tolerance are used correctly.

Worked Example: An Absolute Dollar Tolerance

For a hypothetical repair-coverage portfolio, model individual claim severity as Gamma with shape 0.25 and scale $4,000. Find the claim count required for the sample average to be within $100 of expected severity with approximately 90% probability.

Method and parameters. The mean is $\mu=0.25(4{,}000)=\$1{,}000$. The variance is $0.25(4{,}000^2)=\$4{,}000{,}000$, and the standard deviation is $\sigma=\$2{,}000$. The target is an absolute tolerance $\varepsilon=\$100$, so use the absolute-tolerance formula and $z_p=1.645$.

Calculate and round up.

n\geq \left(\frac{1.645(\$2{,}000)}{\$100}\right)^2 = (32.9)^2 = 1{,}082.41.

The standard is 1,083 claims. The cutoff check is $100\sqrt n/2{,}000=0.05\sqrt n$. At $n=1{,}082$, this is approximately $1.64469<1.645$; at $n=1{,}083$, it is approximately $1.64545>1.645$.

Common mistake avoided. The $100 tolerance is not 10% of the mean merely because the mean happens to be $1,000. The problem specifies an absolute error, so the calculation uses $\varepsilon=\$100$ directly. Here it happens to equal 10% of the mean; that numerical coincidence should not replace reading the condition.

Interpretation, Limits, and Exam Traps

The standard grows with the square of the severity coefficient of variation, $\sigma/\mu$, and with the square of the normal cutoff. It falls with the square of the allowed relative error. This follows directly by rearranging the relative-tolerance formula:

n\geq z_p^2\left(\frac{\sigma}{k\mu}\right)^2.

Consequently, halving the allowed error requires four times as many claims under this model, not twice as many. A more variable severity distribution also requires more claims even when its mean is unchanged. Those relationships help explain why a severity credibility standard cannot be inferred from the mean alone.

A full-credibility count is a probability-based benchmark, not a finding that the underlying data are representative, stable, or adequate for a particular pricing decision. It does not address changes in coverage, claim handling, inflation, or mix. The earlier material on credibility concepts distinguishes precision from data quality and rate adequacy; this count addresses only the stated sampling-variation criterion.

Do not confuse variance and standard deviation. The formula may use $\sigma^2$, but a calculation written with a standard error uses $\sigma$. Do not substitute the variance where a standard deviation is required.
Do not forget to square. The ratio $z_p\sigma/(k\mu)$ is the square root of the required count. The claim-count formula squares the entire ratio.
Do not round down. The full standard is the smallest integer satisfying the inequality, so a noninteger result is rounded upward.
Do not treat claims as exposures. This tutorial’s $n$ is the number of claim severities in the average. Converting it into exposure units requires a separate frequency relationship.
Check the approximation. With strongly skewed severities, the normal approximation may be weak at moderate counts. The formula is a model-based standard, not an exact distribution-free guarantee.

Key takeaway: For a relative tolerance $k$, use $n\geq z_p^2\sigma^2/(k^2\mu^2)$; for an absolute tolerance $\varepsilon$, use $n\geq z_p^2\sigma^2/\varepsilon^2$. State the assumptions, use the correct two-sided normal cutoff, and round the resulting claim count up.

Check Your Understanding

For each problem, determine the full-credibility claim-count standard under the stated normal model or normal approximation. Round the final count up to the next whole claim.

A hypothetical liability severity model is Gamma with shape 16 and scale $50. Find the count for a 95% probability target and a relative tolerance of 5%. (Answer: 97 claims)
Individual severity is modeled as normal with mean $500 and standard deviation $250. Find the count for a 90% probability target and a relative tolerance of 8%. (Answer: 106 claims)
A hypothetical parts-coverage severity model is Gamma with shape 4 and scale $250. Find the count for a 95% probability target and an absolute tolerance of $125. (Answer: 62 claims)
In a hypothetical portfolio, severity is $400 with probability 0.8 and $2,400 with probability 0.2, independently from claim to claim. Find the count for a 90% probability target and a relative tolerance of 10%, using the normal approximation for the sample average. (Answer: 271 claims)

What You'll Learn