How Claim Size Variability Changes Required Volume

Why Severity Variability Changes the Required Volume

A credibility standard asks how much experience is needed for an observed result to be sufficiently stable. For average claim severity, the relevant result is the average of a specified number of claim amounts. If claim sizes are more variable relative to their mean, that average fluctuates more from one sample to another. More claims are therefore needed to meet the same accuracy and confidence requirements.

The key measure is not variance by itself, but the coefficient of variation: standard deviation divided by mean. It expresses variability relative to the scale of claim amounts. Two portfolios can have very different average claim sizes yet have the same relative variability and, under the same credibility standard, the same required claim count.

Definition: Let $X$ be the amount of one claim, with mean $\mu=E[X]>0$, variance $\sigma^2=\operatorname{Var}(X)$, and standard deviation $\sigma$. Its coefficient of variation is $CV=\sigma/\mu$. For $N$ independent claim amounts $X_1,\ldots,X_N$ from the same severity distribution, the observed average severity is $\bar X_N=(X_1+\cdots+X_N)/N$.

A full credibility standard specifies a relative error tolerance $r>0$ and a two-sided confidence level $p$, where $0<p<1$. Meeting the standard means that the observed average is within $r$ of the true mean with probability at least $p$, using the normal approximation described below. The requirement concerns the number of claims, not directly the number of policies or years of exposure.

Deriving the Severity Full Credibility Standard

Assume the claim amounts are independent and identically distributed, with finite mean $\mu>0$ and finite variance $\sigma^2$. For sufficiently large $N$, the central limit theorem gives the approximation $\bar X_N\sim N(\mu,\sigma^2/N)$. Here the second parameter is the variance. Let $z_{(1+p)/2}$ be the standard normal quantile that leaves probability $(1-p)/2$ in each tail.

P\left(\left|\bar X_N-\mu\right|\le r\mu\right) \approx P\left( \left| \frac{\bar X_N-\mu}{\sigma/\sqrt{N}} \right| \le \frac{r\mu\sqrt{N}}{\sigma} \right).

The standardized quantity is approximately standard normal. Its probability of falling between $-z_{(1+p)/2}$ and $z_{(1+p)/2}$ is $p$. Thus, to attain the stated confidence approximately, the half-width in standardized units must be at least that quantile:

\frac{r\mu\sqrt{N}}{\sigma}\ge z_{(1+p)/2} \quad\Longleftrightarrow\quad N\ge \left(\frac{z_{(1+p)/2}\sigma}{r\mu}\right)^2 = \left(\frac{z_{(1+p)/2}CV}{r}\right)^2.

Theorem: Under the independent, identical-severity and normal-approximation assumptions, the minimum integer claim count for full credibility is $N_0=\left\lceil\left(z_{(1+p)/2}CV/r\right)^2\right\rceil$. The ceiling is necessary because a claim count must be an integer. The standard is an approximation, not an exact guarantee based only on the mean and variance; for highly skewed severity distributions, the normal approximation may be inaccurate at the calculated volume.

The formula explains the effect of variability directly. Holding the confidence level and error tolerance fixed, required claim count is proportional to $CV^2$. If the coefficient of variation is multiplied by $c$, the unrounded standard is multiplied by $c^2$. Since $CV^2=\sigma^2/\mu^2$, the standard is proportional to severity variance relative to the squared mean. It is not proportional to standard deviation alone.

Worked Example: Calculating a Severity Standard from Variance

A hypothetical equipment insurer models claim severity with mean $\$2{,}000$ and variance $\$4{,}000{,}000$. Determine the full credibility claim count for a 90% confidence level and a 20% relative error tolerance. Use $z_{0.95}=1.645$.

Method and assumptions. Use the normal-approximation severity standard. Assume claim amounts are independent and identically distributed, and that their mean and variance are finite. The specified relative tolerance is $r=0.20$, and the confidence level is $p=0.90$.

Step 1: Find the standard deviation and coefficient of variation. The standard deviation is the square root of the variance:

\sigma=\sqrt{4{,}000{,}000}=\$2{,}000, \qquad CV=\frac{2{,}000}{2{,}000}=1.

Step 2: Substitute into the standard. For a two-sided 90% confidence level, the relevant quantile is $z_{(1+0.90)/2}=z_{0.95}=1.645$.

N_0= \left\lceil \left(\frac{1.645(1)}{0.20}\right)^2 \right\rceil = \left\lceil 8.225^2\right\rceil = \left\lceil 67.650625\right\rceil =68.

Check. At $N=68$, the standardized half-width is $0.20\sqrt{68}/1$, approximately $1.649$, which exceeds $1.645$. At $N=67$, it is $0.20\sqrt{67}$, approximately $1.637$, which falls short. Thus 68 claims is the minimum integer under the stated approximation. A common mistake is to round 67.650625 down; doing so would fail the standard.

Comparing Portfolios with Different Variability

When two risks have the same credibility target, compare their coefficients of variation. The portfolio with the larger $CV$ requires more claims. If their means are equal, this is equivalent to comparing their standard deviations or variances. If their means differ, comparing standard deviations alone can give the wrong conclusion because variability must be measured relative to the mean.

Worked Example: How a Higher Coefficient of Variation Changes Volume

Two hypothetical specialty coverages each have a 95% confidence target and a 10% relative error tolerance. Coverage A has $CV=0.80$; Coverage B has $CV=1.20$. Find each full credibility claim count using $z_{0.975}=1.96$, and compare the requirements.

Method. Apply $N_0=\lceil(zCV/r)^2\rceil$ separately to each coverage. The assumptions are independent and identically distributed claim amounts within each coverage and use of the normal approximation.

Coverage A.

N_{0,A} = \left\lceil \left(\frac{1.96(0.80)}{0.10}\right)^2 \right\rceil = \left\lceil 15.68^2\right\rceil = \left\lceil 245.8624\right\rceil =246.

Coverage B.

N_{0,B} = \left\lceil \left(\frac{1.96(1.20)}{0.10}\right)^2 \right\rceil = \left\lceil 23.52^2\right\rceil = \left\lceil 553.1904\right\rceil =554.

Comparison and check. Coverage B has $1.20/0.80=1.5$ times the coefficient of variation. Before rounding, its required volume is $1.5^2=2.25$ times Coverage A’s. Indeed, $553.1904/245.8624=2.25$. The integer standards are 554 and 246 claims. A common mistake is to multiply 246 by 1.5, treating claim volume as proportional to $CV$. The correct factor is the square of the change in $CV$, because the sample-average variance decreases in proportion to $1/N$.

Working Backward from Available Claim Volume

The same relationship can be rearranged to assess whether an available claim count is sufficient for a proposed severity distribution. Solving the standard for $CV$ gives the largest coefficient of variation that meets the target at a specified volume. This is useful when the volume is known but the severity variability is still being evaluated.

CV\le \frac{r\sqrt{N}}{z_{(1+p)/2}}.

Worked Example: Finding the Maximum Supported Variability

A hypothetical portfolio has 400 independent claim amounts available for estimating average severity. The target is 95% confidence with a 10% relative error tolerance. Find the maximum coefficient of variation that meets the full credibility standard, using $z_{0.975}=1.96$. Then determine whether a severity model with $CV=1.05$ meets it.

Method and assumptions. Rearrange the normal-approximation standard for $CV$. The available number of observations is fixed at $N=400$; the confidence, tolerance, and independence assumptions are as stated.

Step 1: Calculate the limiting coefficient of variation.

CV_{\max} = \frac{0.10\sqrt{400}}{1.96} = \frac{0.10(20)}{1.96} = \frac{2}{1.96} \approx 1.0204.

Step 2: Assess the proposed model. Since $1.05>1.0204$, a model with $CV=1.05$ does not meet the target with 400 claims. The corresponding unrounded required volume is

\left(\frac{1.96(1.05)}{0.10}\right)^2 = 20.58^2 = 423.5364,

so at least $\lceil423.5364\rceil=424$ claims are required under the approximation. This confirms that 400 claims are insufficient.

Scope, Interpretation, and Common Traps

This standard is specifically for estimating average claim severity from a fixed number of claim amounts. It treats the number of claims as given and focuses on variation among claim sizes. A standard for aggregate loss or for a loss ratio can also involve variation in claim count or exposure; those are different quantities and should not be substituted into this severity formula.

The required claim count does not depend separately on the dollar scale of severity. If every claim amount and its standard deviation are multiplied by the same positive constant, the coefficient of variation is unchanged, so the full credibility claim count is unchanged. By contrast, increasing variance while keeping the mean fixed increases $CV^2$ and therefore increases the required count.

Common errors include using a one-sided normal quantile for a two-sided tolerance, omitting the square in the formula, and rounding the required count down. For a 95% two-sided standard, use $z_{0.975}$, not $z_{0.95}$. Also distinguish a variance from a standard deviation: take the square root of variance to obtain $\sigma$, then divide by the mean to obtain $CV$. The normal-approximation result is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Key takeaway: For average severity, the normal-approximation full credibility count is the ceiling of $(zCV/r)^2$. At fixed confidence and error tolerance, required claim volume grows with the square of the coefficient of variation.

Check Your Understanding

Use the normal-approximation severity standard in each problem, and round required claim counts up to the next whole claim.

A hypothetical repair portfolio has mean severity $\$800$ and standard deviation $\$400$. Find the claim count for 95% confidence and an 8% relative error tolerance, using $z_{0.975}=1.96$. (Answer: 151 claims)
A hypothetical coverage has mean severity $\$1{,}200$ and variance $\$1{,}440{,}000$. Find the claim count for 90% confidence and a 25% relative error tolerance, using $z_{0.95}=1.645$. (Answer: 44 claims)
Two hypothetical classes have coefficients of variation 0.70 and 1.40. For each, calculate the standard for 95% confidence and a 12% relative error tolerance, using $z_{0.975}=1.96$. By what factor does the unrounded requirement for the second class exceed that for the first? (Answer: 131 claims and 523 claims, respectively; factor 4)
A portfolio has 625 independent claims available. For a 95% confidence level and an 8% relative error tolerance, find the maximum $CV$ supported by this volume, using $z_{0.975}=1.96$. Does a model with $CV=1.05$ meet the standard? (Answer: $CV_{\max}\approx1.0204$; no)
A severity model has mean $\$1{,}000$ and standard deviation $\$600$. For 95% confidence and a 10% relative error tolerance, find the required volume. Then suppose the variance increases by 50% while the mean stays fixed; find the new required volume. Use $z_{0.975}=1.96$ and do not round intermediate values. (Answer: 139 claims initially and 208 claims after the variance increase)

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