How Confidence and Tolerance Affect Thresholds

Why Confidence and Tolerance Change the Standard

A limited-fluctuation standard asks how much experience is needed for observed results to stay sufficiently close to their expected value. Two choices determine how demanding the standard is: the confidence level, or probability of being within the specified bounds, and the relative tolerance, or size of those bounds as a proportion of the expected value. Requiring more confidence or narrower bounds raises the required volume.

This tutorial quantifies those effects for claim counts and aggregate losses. The calculations use the standard normal approximation underlying the usual limited-fluctuation thresholds. They are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let $p$ be a central confidence level, with $0<p<1$, and let $r>0$ be the relative tolerance. If a quantity has positive mean $m$, the tolerance bounds are $m(1-r)$ and $m(1+r)$. The standard normal critical value for a central probability $p$ is $z_p=\Phi^{-1}((1+p)/2)$, where $\Phi$ is the standard normal cumulative distribution function.

The central probability matters: a 95% requirement here means 95% in the interval around the mean, leaving 2.5% in each tail. It does not mean a one-sided 95% requirement. For the calculations below, $z_{0.90}$ is approximately 1.645 and $z_{0.95}$ is approximately 1.960. A more demanding confidence level gives a larger critical value.

Claim-Count Thresholds

For the standard claim-count calculation, model the claim count $N$ as Poisson with mean $\lambda$, so $E[N]=\lambda$ and $\operatorname{Var}(N)=\lambda$. The limited-fluctuation target is that $N$ be within relative tolerance $r$ of $\lambda$. Using a normal approximation to the count, the threshold is obtained by requiring the central probability of meeting those bounds to be at least $p$.

\lambda_{\min}=\frac{z_p^2}{r^2}

This is a required expected claim count under the approximation. If the result is not an integer, round up to the next whole claim when stating a whole-number threshold. The formula is not an exact discrete Poisson probability calculation; it is the usual normal-based limited-fluctuation standard.

Choose the confidence level.
Use a central probability and find $z_p=\Phi^{-1}((1+p)/2)$.

Express the tolerance as a proportion.
For example, a 10% tolerance means $r=0.10$, not 10.

Calculate and round up.
Evaluate $z_p^2/r^2$, then round up if a whole-number claim threshold is required.

Worked Example: Raising the Confidence Requirement

A hypothetical portfolio needs a claim-count standard with a 10% relative tolerance. Calculate the expected claim-count threshold at 90% and 95% central confidence, then compare the results. Use $z_{0.90}=1.645$ and $z_{0.95}=1.960$.

Method. Apply the Poisson claim-count threshold $\lambda_{\min}=z_p^2/r^2$. The tolerance is $r=0.10$, so $r^2=0.0100$.

At 90% confidence: $z_p^2=1.645^2=2.706025$. Therefore, $\lambda_{\min}=2.706025/0.0100=270.6025$, which rounds up to 271 expected claims.

At 95% confidence: $z_p^2=1.960^2=3.841600$. Therefore, $\lambda_{\min}=3.841600/0.0100=384.16$, which rounds up to 385 expected claims.

The unrounded threshold ratio is $384.16/270.6025$, approximately 1.420. Thus, raising confidence from 90% to 95% increases the required count by about 42.0% before integer rounding. Common mistake avoided: using a one-sided critical value for a central interval. The required critical values here correspond to 90% and 95% central confidence, respectively.

Check: The 95% threshold is higher than the 90% threshold, as expected for a more demanding confidence requirement.

How Tolerance Changes the Count

Holding confidence fixed, the claim-count threshold is inversely proportional to the square of the tolerance. Tightening the tolerance therefore has a substantial effect: halving $r$ multiplies the required count by four. More generally, if confidence stays fixed and tolerance changes from $r_1$ to $r_2$, the ratio of unrounded thresholds is $(r_1/r_2)^2$.

Worked Example: Tightening a Relative Tolerance

A hypothetical classification standard requires 95% central confidence. Compare the claim-count thresholds for a 15% relative tolerance and a 10% relative tolerance. Use $z_{0.95}=1.960$.

At 15% tolerance: $r=0.15$, so $r^2=0.0225$. Also, $z_p^2=1.960^2=3.841600$. The threshold is $3.841600/0.0225=170.7378$, which rounds up to 171 expected claims.

At 10% tolerance: $r=0.10$, so $r^2=0.0100$. The threshold is $3.841600/0.0100=384.16$, which rounds up to 385 expected claims.

The unrounded ratio is $384.16/170.7378=2.25$. This agrees with the direct tolerance comparison: $(0.15/0.10)^2=1.5^2=2.25$. Common mistake avoided: multiplying the threshold by 1.5 because the tolerance changed by a factor of 1.5. The threshold changes by the square of that factor, not by the factor itself.

Check: The 10% tolerance is narrower than the 15% tolerance, so its required count is larger.

Aggregate-Loss Thresholds

An aggregate-loss standard concerns the sum of claim payments, not just the number of claims. Suppose $N$ is Poisson with mean $\lambda$, and claim payments $Y_1,Y_2,\ldots$ are independent and identically distributed, independent of $N$, with finite mean $\mu=E[Y]$ and variance $\sigma^2=\operatorname{Var}(Y)$. Let $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$. These assumptions give $E[S]=\lambda\mu$ and $\operatorname{Var}(S)=\lambda E[Y^2]=\lambda(\sigma^2+\mu^2)$.

Key result: Under the normal approximation to aggregate loss, the expected claim-count threshold for a central confidence $p$ and relative tolerance $r$ is $\lambda_{\min}=z_p^2 E[Y^2]/(r^2\mu^2)$. This formula applies to the compound Poisson model just stated; it is not the claim-count threshold, and it should not be used unchanged when the count model or dependence assumptions differ.

The factor $E[Y^2]/\mu^2=1+\sigma^2/\mu^2$ accounts for payment variation as well as random claim counts. Thus, even when confidence and tolerance are unchanged, greater severity variation increases the aggregate-loss threshold. The severity moments are treated as given here; the standard does not estimate them.

Why the Threshold Formulas Hold

For the count standard, the normal approximation gives $N$ an approximate mean $\lambda$ and standard deviation $\sqrt{\lambda}$. The bounds are $\lambda(1-r)$ and $\lambda(1+r)$, or equivalently deviations of at most $r\lambda$ from the mean. Standardizing those deviations gives bounds of $\pm r\sqrt{\lambda}$. To achieve central probability $p$, those bounds must reach at least $\pm z_p$, so $r\sqrt{\lambda}\ge z_p$. Squaring and rearranging yields the claim-count threshold.

For aggregate loss, the same standardization uses the aggregate mean $\lambda\mu$ and standard deviation $\sqrt{\lambda E[Y^2]}$. The tolerance half-width is $r\lambda\mu$. Requiring that half-width to be at least $z_p$ standard deviations gives the following result.

\frac{r\lambda\mu}{\sqrt{\lambda E[Y^2]}}\ge z_p \quad\Longrightarrow\quad \lambda\ge\frac{z_p^2E[Y^2]}{r^2\mu^2}

The compound Poisson variance used here follows by conditioning on the count. Given $N$, the sum has mean $N\mu$ and variance $N\sigma^2$. The law of total variance then gives $\operatorname{Var}(S)=E[N\sigma^2]+\operatorname{Var}(N\mu)=\lambda\sigma^2+\lambda\mu^2=\lambda E[Y^2]$. This is why the aggregate formula includes the second moment, rather than just the severity variance.

Worked Example: Aggregate Loss at Two Confidence Levels

A hypothetical portfolio has Poisson claim frequency with expected count $\lambda$. Each payment has mean $\mu=\$2{,}000$ and standard deviation $\sigma=\$4{,}000$. Find the expected claim-count thresholds for aggregate loss within 20% of its mean at 90% and 95% central confidence. Use $z_{0.90}=1.645$ and $z_{0.95}=1.960$.

Method. Use the compound Poisson aggregate-loss threshold. First calculate the severity second moment: $\sigma^2=\$16{,}000{,}000$, $\mu^2=\$4{,}000{,}000$, and $E[Y^2]=\sigma^2+\mu^2=\$20{,}000{,}000$. Thus, $E[Y^2]/\mu^2=20{,}000{,}000/4{,}000{,}000=5$. The tolerance is $r=0.20$, so $r^2=0.0400$.

At 90% confidence: $z_p^2=1.645^2=2.706025$. The threshold is $2.706025(5)/0.0400=338.2531$, which rounds up to 339 expected claims.

At 95% confidence: $z_p^2=1.960^2=3.841600$. The threshold is $3.841600(5)/0.0400=480.2$, which rounds up to 481 expected claims.

At the respective unrounded thresholds, the corresponding expected aggregate losses are $338.2531(\$2{,}000)=\$676{,}506.20$ and $480.2(\$2{,}000)=\$960{,}400.00$. These are the means associated with the continuous threshold calculations, not additional requirements. Common mistake avoided: using only $\sigma^2/\mu^2$, which would omit the contribution from the random Poisson claim count. Here the correct ratio is $E[Y^2]/\mu^2=5$.

Check: The aggregate-loss threshold increases from about 338.25 to 480.20 before rounding, in line with the higher confidence level.

Comparing the Effects and Avoiding Traps

Both formulas contain $z_p^2/r^2$, so the direction of the confidence and tolerance effects is the same for claim counts and aggregate losses. The aggregate standard also has the multiplier $E[Y^2]/\mu^2$, which reflects severity variation under the compound Poisson assumptions.

Change to the standard	Effect on the unrounded threshold
Increase confidence from $p_1$ to $p_2$, holding tolerance fixed	Multiply by $z_{p_2}^2/z_{p_1}^2$
Change tolerance from $r_1$ to $r_2$, holding confidence fixed	Multiply by $(r_1/r_2)^2$
Use aggregate loss rather than claim count under the compound Poisson model	Include the additional factor $E[Y^2]/\mu^2$

A frequent exam trap is to put a percentage into the formula as a whole number. A 10% tolerance is $0.10$, and its squared value is $0.0100$; entering 10 and squaring it would produce a threshold wrong by a factor of 10,000. Another trap is to round down: a calculated threshold of 384.16 requires 385 whole expected claims, not 384. Keep the intermediate calculation unrounded and apply the upward rounding only at the end.

These thresholds describe approximate probability standards under stated models. They do not guarantee an exact probability for every discrete claim count or every severity distribution. In particular, use the aggregate-loss formula only when the compound Poisson assumptions and the required severity moments are appropriate.

Check Your Understanding

For each problem, use the central standard normal critical value shown, keep intermediate values unrounded, and round a whole-number claim threshold upward.

A hypothetical count standard requires 90% central confidence and 8% relative tolerance. Use $z=1.645$. What is the minimum expected claim count? (Answer: 423)
A hypothetical count standard requires 99% central confidence and 10% relative tolerance. Use $z=2.5758$. What is the minimum expected claim count? (Answer: 664)
At 12% relative tolerance, compare the claim-count thresholds for 90% and 95% central confidence. Use $z_{0.90}=1.645$ and $z_{0.95}=1.960$. Give both rounded thresholds and the approximate unrounded ratio of the 95% threshold to the 90% threshold. (Answer: 188 and 267 claims; ratio approximately 1.420)
A hypothetical compound Poisson portfolio has severity mean $\$500$, severity standard deviation $\$1{,}500$, 90% central confidence, and 25% relative aggregate-loss tolerance. Use $z=1.645$. What is the minimum expected claim count? (Answer: 433)
A hypothetical compound Poisson portfolio has severity mean $\$1{,}200$, severity standard deviation $\$1{,}600$, 95% central confidence, and 10% relative aggregate-loss tolerance. Use $z=1.960$. What is the minimum expected claim count? (Answer: 1,068)

Change to the standard	Effect on the unrounded threshold
Increase confidence from \(p_1\) to \(p_2\), holding tolerance fixed	Multiply by \(z_{p_2}^2/z_{p_1}^2\)
Change tolerance from \(r_1\) to \(r_2\), holding confidence fixed	Multiply by \((r_1/r_2)^2\)
Use aggregate loss rather than claim count under the compound Poisson model	Include the additional factor \(E[Y^2]/\mu^2\)

What You'll Learn