Error Tolerance Sensitivity in Credibility Standards

Why Error Tolerance Changes the Credibility Standard

A full-credibility standard specifies how much experience is needed for an observed statistic to fall within a chosen relative error of its expected value, at a chosen confidence level. The relative error tolerance is a measure of permitted imprecision: a smaller tolerance demands a closer estimate and therefore more experience. Understanding this sensitivity is useful when a ratemaking standard must be recalculated for a different precision requirement.

This tutorial focuses on changing the relative error tolerance while holding the model and confidence level fixed. It covers full-credibility requirements for claim counts, average severity, and aggregate loss. These are limited-fluctuation standards based on a normal approximation; they do not, by themselves, determine a partial credibility weight.

Definition: A relative error tolerance $r$ is the permitted absolute deviation from an expected value, divided by that expected value. A central confidence level $p$ is represented by $z_p$, where a standard normal random variable $Z$ satisfies $P(-z_p \leq Z \leq z_p)=p$. A full-credibility requirement is the volume at which the normal-approximation probability of being within relative error $r$ is at least $p$.

The Inverse-Square Relationship

Suppose a statistic $T_n$, based on volume $n$, has expectation $\theta>0$ and variance of the form $\operatorname{Var}(T_n)=A\theta^2/n$, where $A$ is a constant under the assumed model. The normal approximation gives the standard deviation of $T_n$ as $\theta\sqrt{A/n}$. A central confidence requirement for relative error $r$ is therefore

P\left(\left|T_n-\theta\right|\leq r\theta\right)\approx P\left(|Z|\leq \frac{r\theta}{\theta\sqrt{A/n}}\right) = P\left(|Z|\leq r\sqrt{\frac{n}{A}}\right).

To make this probability at least $p$, the normal-score condition is $r\sqrt{n/A}\geq z_p$. Squaring both nonnegative sides and rearranging gives the full-credibility requirement

n\geq \frac{z_p^2 A}{r^2}.

This derivation explains the key sensitivity result: with the confidence level and model fixed, required volume is proportional to $1/r^2$. If tolerance is multiplied by a factor $c$, the unrounded requirement is multiplied by $1/c^2$. In particular, halving tolerance multiplies the required volume by four. This result is about the unrounded threshold; integer requirements are rounded up to the next whole observation or claim.

Assumptions and scope: The formulas below use the stated sampling model, finite moments where needed, and the usual normal approximation for fluctuations about the expected value. The observations or claims are assumed to follow the relevant homogeneous model. A normal approximation can be poor at small volumes or in highly skewed settings; exact model calculations may then give a different standard. The inverse-square relationship describes the standard under this approximation, not a guarantee that every realized sample will meet the tolerance.

Which Variance Factor Applies?

The general formula has different values of $A$ for different measures. For a Poisson claim count with expected count $\lambda$, the variance is $\lambda$, so its coefficient of variation squared is $1/\lambda$; here $A=1$ and the required expected count is $z_p^2/r^2$. For an average of $n$ independent claim severities with mean $\mu$, standard deviation $\sigma$, and coefficient of variation $\mathrm{CV}=\sigma/\mu$, the variance of the average is $\sigma^2/n$, so $A=\mathrm{CV}^2$.

For aggregate loss, suppose the claim count is Poisson with mean $\lambda$, and claim severities are independent and identically distributed, independent of the count, with mean $\mu>0$ and variance $\sigma^2$. If $S$ is the sum of the severities, then $E[S]=\lambda\mu$ and $\operatorname{Var}(S)=\lambda(\sigma^2+\mu^2)$. Thus the relative variance of aggregate loss is $(1+\mathrm{CV}^2)/\lambda$, giving $A=1+\mathrm{CV}^2$ when the volume is expected claim count.

Measure	Model-based variance factor $A$	Unrounded full-credibility requirement
Poisson claim count	1	$z_p^2/r^2$ expected claims
Average severity	$\mathrm{CV}^2$	$z_p^2\mathrm{CV}^2/r^2$ claims
Poisson aggregate loss	$1+\mathrm{CV}^2$	$z_p^2(1+\mathrm{CV}^2)/r^2$ expected claims

Worked Requirements at Alternative Tolerances

Worked Example: Tightening a Severity Tolerance

A hypothetical equipment insurer models individual covered repair costs as independent observations with coefficient of variation $1.2$. It uses a 95% central confidence standard, with $z_p=1.96$. Find the full-credibility claim requirements at relative error tolerances of 10%, 5%, and 2.5%.

Method and assumptions. This is an average-severity standard. The observations are assumed to have a common finite mean and variance, and the normal approximation is used. Here $A=\mathrm{CV}^2=1.2^2=1.44$, and $z_p^2=1.96^2=3.8416$.

At $r=0.10$, the unrounded requirement is $3.8416(1.44)/(0.10)^2=5.532? $. Compute the numerator first: $3.8416(1.44)=5.532? $. More exactly, $3.8416(1+0.44)=3.8416+1.690304=5.531904$. Thus $n\geq 5.531904/0.01=553.1904$, requiring 554 claims when rounded up.

At $r=0.05$, $n\geq5.531904/0.0025=2212.7616$, requiring 2,213 claims. At $r=0.025$, $n\geq5.531904/0.000625=8851.0464$, requiring 8,852 claims.

Check. Each time the tolerance is halved, the unrounded requirement is multiplied by four: $553.1904(4)=2212.7616$ and $2212.7616(4)=8851.0464$. The final whole-claim requirements are rounded upward so they meet, rather than fall short of, each threshold. The common mistake avoided here is treating a smaller tolerance as only a proportional increase in volume.

Worked Example: Aggregate Loss at Two Tolerances

A hypothetical specialty insurer models annual claim counts as Poisson. Claim severity has mean $\$1{,}000$ and standard deviation $\$1{,}500$. At a 90% central confidence level, use $z_p=1.645$ to calculate the full-credibility expected claim counts for relative aggregate-loss tolerances of 8% and 4%.

Method and assumptions. Use the compound Poisson aggregate-loss standard. The frequency and severity assumptions are as stated above, and claim severities are independent of the claim count. First calculate $\mathrm{CV}=1500/1000=1.5$, so $1+\mathrm{CV}^2=1+2.25=3.25$. Also, $z_p^2=1.645^2=2.706025$. The numerator is $2.706025(3.25)=8.79458125$.

At $r=0.08$, $r^2=0.0064$, so the unrounded expected claim requirement is $8.79458125/0.0064=1374.1533203125$. Rounding up gives 1,375 expected claims.

At $r=0.04$, $r^2=0.0016$, so the unrounded requirement is $8.79458125/0.0016=5496.61328125$. Rounding up gives 5,497 expected claims.

Check. Halving the tolerance multiplies the unrounded requirement by four: $1374.1533203125(4)=5496.61328125$. These requirements are for aggregate loss, not merely for the average severity; the factor $1+\mathrm{CV}^2=3.25$ includes frequency as well as severity variation. Confusing the aggregate-loss factor with the severity-only factor would understate the required volume.

Worked Example: Separating Tolerance and Severity Effects

Two hypothetical repair programs use the same 90% central confidence level, with $z_p=1.645$. Program A has severity $\mathrm{CV}=0.8$, and Program B has severity $\mathrm{CV}=1.2$. Calculate each program’s average-severity full-credibility requirement at tolerances of 5% and 4%.

Method and assumptions. Apply $n\geq z_p^2\mathrm{CV}^2/r^2$ to independent, identically distributed severities with finite variance, using a normal approximation. The common factor $z_p^2$ is $2.706025$.

For Program A, $2.706025(0.8^2)=2.706025(0.64)=1.731856$. At 5%, $n\geq1.731856/0.0025=692.7424$, or 693 claims rounded up. At 4%, $n\geq1.731856/0.0016=1082.41$, or 1,083 claims.

For Program B, $2.706025(1.2^2)=2.706025(1.44)=3.896676$. At 5%, $n\geq3.896676/0.0025=1558.6704$, or 1,559 claims. At 4%, $n\geq3.896676/0.0016=2435.4225$, or 2,436 claims.

Check. For each program, changing tolerance from 5% to 4% multiplies the unrounded requirement by $(5/4)^2=1.5625$. At either tolerance, Program B’s unrounded requirement is $1.44/0.64=2.25$ times Program A’s because severity requirements are proportional to $\mathrm{CV}^2$. This separates the effect of precision from the effect of claim-size variability.

Common Exam Trap: Scaling Volume Linearly

A frequent error is to assume that halving a tolerance doubles the required volume. The full-credibility formula contains $r^2$ in the denominator, so this calculation is wrong.

\text{Wrong: } n_{\text{new}}=2n_{\text{old}} \qquad \text{Correct: } n_{\text{new}}=n_{\text{old}}\left(\frac{r_{\text{old}}}{r_{\text{new}}}\right)^2.

For example, at 95% confidence, the Poisson frequency requirement at a 10% tolerance is $1.96^2/0.10^2=384.16$. Reducing tolerance to 5% gives $1.96^2/0.05^2=1536.64$, not twice 384.16. The thresholds are multiplied by four; rounding up gives 385 and 1,537 expected claims. Always convert percentages to decimals before squaring, and round an integer volume requirement upward.

Using Sensitivity Results Correctly

A tolerance comparison is meaningful only when the other inputs are held fixed. If the confidence level also changes, $z_p^2$ changes; if the severity distribution changes, its coefficient of variation may change as well. Recalculate the full expression rather than applying a tolerance multiplier to a requirement that used different assumptions.

Sensitivity analysis also helps explain why a seemingly small demand for greater precision can be expensive in experience volume. Moving from a 5% tolerance to a 4% tolerance increases the unrounded requirement by 56.25%, since $(5/4)^2=1.5625$. Moving from 5% to 2.5% multiplies it by four. These are consequences of the variance decreasing in proportion to the inverse of volume.

The standards remain approximations under their stated models. For instance, claim counts are discrete, and severity distributions can be skewed; a normal approximation may not describe tail behavior accurately at low volumes. A full-credibility requirement therefore describes a prescribed fluctuation standard, not a claim that the experience is identical to the expected value or that the realized error cannot exceed the tolerance.

Key takeaway: Under the usual limited-fluctuation normal approximation, a full-credibility requirement has the form $n\geq z_p^2A/r^2$. Holding confidence and the model fixed, the required volume varies with the inverse square of relative error tolerance. Halving tolerance requires four times the unrounded volume.

Check Your Understanding

For each problem, use the stated normal approximation and round whole-claim requirements upward.

A hypothetical portfolio has Poisson claim counts and uses a 90% central confidence level with $z_p=1.645$. What expected claim count is required at a 10% relative error tolerance? (Answer: 271 expected claims)
For independent claim severities with $\mathrm{CV}=0.75$, a hypothetical insurer uses a 95% central confidence level with $z_p=1.96$. Find the full-credibility requirement at a 5% tolerance. (Answer: 865 claims)
A Poisson frequency standard uses 95% confidence and $z_p=1.96$. Calculate the expected claim requirements at 8% and 4% tolerances, and state the factor by which the unrounded requirement changes. (Answer: 601 claims at 8%, 2,401 at 4%; the unrounded requirement is multiplied by 4)
A hypothetical aggregate-loss model has Poisson frequency and severity $\mathrm{CV}=1.2$. At 90% confidence with $z_p=1.645$, find the expected claim requirement at a 6% relative error tolerance. (Answer: 1,835 expected claims)
A severity standard uses $\mathrm{CV}=1.1$ and 95% confidence with $z_p=1.96$. For a volume of 1,200 claims, calculate the normal-approximation relative error tolerance corresponding to the full-credibility boundary. (Answer: 6.2238%)

Measure	Model-based variance factor \(A\)	Unrounded full-credibility requirement
Poisson claim count	1	\(z_p^2/r^2\) expected claims
Average severity	\(\mathrm{CV}^2\)	\(z_p^2\mathrm{CV}^2/r^2\) claims
Poisson aggregate loss	\(1+\mathrm{CV}^2\)	\(z_p^2(1+\mathrm{CV}^2)/r^2\) expected claims

What You'll Learn