Confidence Level Sensitivity in Credibility Standards

Why Confidence Level Changes the Full-Credibility Standard

A full-credibility standard is a volume requirement: it estimates how much experience is needed so that a measured result will fall within a specified relative range of its expected value with a specified probability. The confidence target matters. Keeping the allowed relative deviation fixed but requiring a higher probability of meeting it means the experience must be less variable, which generally requires more claims or exposures.

This tutorial recalculates classical full-credibility requirements at alternative confidence levels. It uses the normal approximation customary in limited-fluctuation standards. These are fluctuation-probability calculations under an assumed model, not statements about a confidence interval for an unknown parameter. The formulas and interpretations are aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Definition: Let $p$ be a central confidence target and $r$ the allowed relative deviation from an expected value $\theta>0$. A full-credibility standard seeks the smallest volume for which $P(|T-\theta|\le r\theta)\ge p$, where $T$ is the experience measure. Define $z_p=\Phi^{-1}((1+p)/2)$, with $\Phi$ the standard normal cumulative distribution function. Thus $z_p$ is the standard-normal quantile that leaves probability $(1-p)/2$ in each tail.

For example, a 90% central target uses approximately $z_{0.90}=1.645$, leaving 5% in each tail. A 95% central target uses $z_{0.95}=1.960$, leaving 2.5% in each tail. These are not one-sided quantiles: for a one-sided probability target the corresponding quantile would be different. Unless stated otherwise, confidence levels below mean central, two-sided targets.

Why the Confidence Quantile Enters the Requirement

Suppose the experience measure is approximately normal with mean $\theta$ and standard deviation $\sigma_T$. After standardizing, the probability of being within the allowed range is approximately $P(|Z|\le r\theta/\sigma_T)$, where $Z$ is standard normal. To make that probability at least $p$, the half-width in standard-normal units must be at least $z_p$.

P(|T-\theta|\le r\theta)\ge p \quad\Longrightarrow\quad \frac{r\theta}{\sigma_T}\ge z_p \quad\Longleftrightarrow\quad \frac{\sigma_T}{\theta}\le \frac{r}{z_p}.

The quantity $\sigma_T/\theta$ is the coefficient of variation of the experience measure. In the usual full-credibility models, its square decreases in inverse proportion to volume. Consequently, the required volume is proportional to $z_p^2$, not merely to $z_p$. Holding the model and $r$ fixed, moving from one central target $p_1$ to another $p_2$ changes the requirement by the factor $(z_{p_2}/z_{p_1})^2$.

Assumptions and scope: These standards use the stated frequency or severity model, independent observations as appropriate, a finite variance, and a normal approximation for the experience measure. They are most useful when the volume is large enough for that approximation to be reasonable. If the model is materially different, the variance does not exist, or the volume is very small and the distribution is highly skewed, the normal-approximation standard may not accurately describe the actual probability.

Changing the Target for Frequency and Severity

For frequency, assume the claim count $N$ over $H$ homogeneous exposure units follows a Poisson distribution with mean $\lambda H$, where $\lambda$ is the expected claim frequency per exposure unit. The observed frequency $N/H$ has mean $\lambda$ and variance $\lambda/H$. Its relative standard deviation is therefore $1/\sqrt{\lambda H}$. For a central target $p$ and relative tolerance $r$, the full-credibility exposure requirement is:

H\ge \frac{z_p^2}{r^2\lambda}.

For severity, assume $n$ independent claim amounts have mean $\mu_X$, standard deviation $\sigma_X$, and coefficient of variation $c=\sigma_X/\mu_X$. The sample average has relative standard deviation $c/\sqrt{n}$. Its full-credibility claim-count requirement is:

n\ge \frac{z_p^2c^2}{r^2}.

In both cases, increasing $z_p$ tightens the permitted standardized fluctuation. The required volume rises by the square of the quantile ratio; it does not change the expected frequency, the severity distribution, or the relative tolerance.

Worked Example: Frequency at Three Confidence Levels

A hypothetical roadside-assistance coverage has Poisson claim frequency $\lambda=0.08$ claims per exposure unit. Find the minimum whole number of exposure units for a 10% relative tolerance at 90%, 95%, and 99% central confidence. Use $z_{0.90}=1.645$, $z_{0.95}=1.960$, and $z_{0.99}=2.576$. There is no interest or mortality assumption in this claim-count calculation.

Method. Apply the Poisson frequency standard $H\ge z_p^2/(r^2\lambda)$, with $r=0.10$. Round each threshold up to a whole exposure unit; the requirement is a minimum, so rounding down would not satisfy a threshold that is not an integer.

At 90% confidence, $z_p^2=1.645^2=2.706025$, and $r^2\lambda=0.10^2(0.08)=0.0008$. Thus $H\ge 2.706025/0.0008=3,382.53125$, requiring 3,383 exposure units.

At 95% confidence, $z_p^2=1.960^2=3.8416$. Thus $H\ge 3.8416/0.0008=4,802$, requiring 4,802 exposure units.

At 99% confidence, $z_p^2=2.576^2=6.635776$. Thus $H\ge 6.635776/0.0008=8,294.72$, requiring 8,295 exposure units.

The results increase as the target becomes more stringent. For a check, the 95%-to-90% threshold ratio before rounding is $3.8416/2.706025$, approximately 1.420, which agrees with the ratio $4,802/3,382.53125$, approximately 1.420. The check uses unrounded thresholds, rather than the rounded whole-unit requirements.

Worked Example: Severity Claim Counts at Alternative Targets

A hypothetical equipment-repair portfolio has independent claim severities following a Gamma distribution with shape $\alpha=4/9$ and scale $\theta=1{,}800$ dollars. Find the minimum number of claims needed for a 15% relative tolerance at 90% and 95% central confidence. The Gamma parameterization here has mean $\alpha\theta$ and variance $\alpha\theta^2$. There is no interest or mortality assumption.

Method. First calculate the severity coefficient of variation, then use the full-credibility requirement $n\ge z_p^2c^2/r^2$. The stipulated Gamma parameters give $\mu_X=(4/9)(1{,}800)=800$ dollars and $\operatorname{Var}(X)=(4/9)(1{,}800^2)=1{,}440{,}000$ dollars squared. Hence $\sigma_X=\sqrt{1{,}440{,}000}=1{,}200$ dollars and $c=1{,}200/800=1.5$.

At 90%, $z_p^2=1.645^2=2.706025$, $c^2=2.25$, and $r^2=0.15^2=0.0225$. Therefore $n\ge (2.706025)(2.25)/0.0225=2.706025(100)=270.6025$. The minimum whole number is 271 claims.

At 95%, $z_p^2=1.960^2=3.8416$. Therefore $n\ge (3.8416)(2.25)/0.0225=3.8416(100)=384.16$. The minimum whole number is 385 claims.

The ratio of the unrounded requirements is $384.16/270.6025$, approximately 1.420, again the squared quantile ratio. The Gamma model specifies the severity distribution, but the calculation uses its mean and variance through $c$; the normal approximation to the sample average remains the full-credibility approximation.

Confidence Sensitivity for Aggregate Loss

Frequency and severity standards treat one component of experience at a time. A compound Poisson aggregate-loss standard accounts for variation in both claim count and claim size. Let $S$ be aggregate loss over $H$ exposure units. Assume the claim count is Poisson with mean $\lambda H$, and claim severities are independent and identically distributed, independent of the count, with mean $\mu_X$, standard deviation $\sigma_X$, and coefficient of variation $c=\sigma_X/\mu_X$. The compound Poisson moments are $E[S]=H\lambda\mu_X$ and $\operatorname{Var}(S)=H\lambda E[X^2]$. Since $E[X^2]=\sigma_X^2+\mu_X^2=\mu_X^2(1+c^2)$, the relative variance of aggregate loss is $(1+c^2)/(H\lambda)$. The resulting standard is:

H\ge \frac{z_p^2(1+c^2)}{r^2\lambda}.

The extra $1$ in $1+c^2$ reflects claim-count fluctuation; $c^2$ reflects severity variation. A higher confidence target multiplies the whole requirement by the corresponding increase in $z_p^2$, while leaving both sources of variability unchanged.

Worked Example: Aggregate-Loss Requirement Near an Integer

A hypothetical small-business coverage has Poisson frequency $\lambda=0.15$ claims per exposure unit. Claim severity follows an exponential distribution with mean $\$1{,}000$, equivalently rate $0.001$ per dollar, so $c=1$. Find the minimum whole number of exposure units for a 20% relative tolerance at 90% and 95% central confidence. Use $z_{0.90}=1.645$ and $z_{0.95}=1.960$. There is no interest or mortality assumption.

Method. Use the compound Poisson aggregate-loss standard. The severity has $\sigma_X=\$1{,}000$, so $c=1$, and $1+c^2=2$. The common denominator is $r^2\lambda=0.20^2(0.15)=0.04(0.15)=0.006$.

At 90%, the numerator is $1.645^2(2)=2.706025(2)=5.41205$. Therefore $H\ge 5.41205/0.006=902.008333\ldots$, requiring 903 exposure units.

At 95%, the numerator is $1.960^2(2)=3.8416(2)=7.6832$. Therefore $H\ge 7.6832/0.006=1{,}280.533333\ldots$, requiring 1,281 exposure units.

Check. At 90%, 902 is below the calculated threshold of 902.008333..., so it fails the stated standard; 903 is the first whole number that meets it. At 95%, 1,280 is below 1,280.533333..., so 1,281 is the first whole number that meets that standard. These checks are important because both thresholds lie close to an integer.

Reading the Results and Avoiding Exam Traps

A useful comparison is the multiplicative change in the required volume, before rounding. Using the stated quantiles, the 95%-to-90% factor is $1.960^2/1.645^2=3.8416/2.706025$, approximately 1.420. Thus changing from 90% to 95% confidence requires about 42.0% more volume under the same model and relative tolerance. For 99% versus 95%, the factor is $2.576^2/1.960^2=6.635776/3.8416$, approximately 1.727. These are approximate percentage comparisons because the quantiles are rounded; whole-number requirements may not preserve the ratios exactly.

Do not use a one-sided quantile for a central target. For a central 90% target, 5% lies in each tail, so the relevant quantile is approximately 1.645, not the 90th percentile of the standard normal.
Do not scale volume linearly with the quantile. Since the standard deviation decreases with the square root of volume, the required volume scales with $z_p^2$. The correct ratio is $(z_{p_2}/z_{p_1})^2$.
Do not round a minimum requirement down. If the formula gives 902.008 exposure units, 902 does not meet the calculated threshold. Round up to the next whole unit, even if the difference appears small.
Keep the model attached to the formula. The frequency, severity, and aggregate-loss standards are not interchangeable. In particular, the compound Poisson aggregate standard includes the claim-count contribution in $1+c^2$.
Do not interpret the target as certainty. Full credibility is a probability standard under the assumed model and normal approximation; it does not guarantee that any particular observed result will be within the tolerance.

Key takeaway: With the model and relative tolerance held fixed, a more stringent central confidence target increases the full-credibility requirement in proportion to the square of its normal quantile. Calculate the threshold with the stated quantile, then round up only at the final step and verify borderline integer results.

Check Your Understanding

For each problem, use a central two-sided target and the normal-approximation full-credibility standards above. Round minimum whole-number requirements up. Each scenario is hypothetical.

A parcel-protection plan has Poisson frequency $\lambda=0.25$ claims per exposure unit. Find the minimum exposures for a 20% relative tolerance at 95% confidence, using $z_{0.95}=1.960$. (Answer: 385 exposure units.)
Independent severities have mean $\$500$ and standard deviation $\$400$. Find the minimum claim count for a 10% relative tolerance at 90% confidence, using $z_{0.90}=1.645$. (Answer: 174 claims.)
A hypothetical compound Poisson risk has frequency $\lambda=0.40$ claims per exposure and severity coefficient of variation $c=0.5$. Find the minimum exposure volume for a 25% relative tolerance at 95% confidence, using $z_{0.95}=1.960$. (Answer: 193 exposure units.)
A frequency standard has $\lambda=0.05$ claims per exposure and a 10% relative tolerance. Calculate the minimum whole exposure volume at both 90% and 99% confidence, using $z_{0.90}=1.645$ and $z_{0.99}=2.576$. (Answer: 5,413 exposures at 90% and 13,272 exposures at 99%.)
Severity has coefficient of variation $c=1.2$, with a 12% relative tolerance. Calculate the minimum whole claim count at 90% and 95% confidence, using $z_{0.90}=1.645$ and $z_{0.95}=1.960$, then give the increase in the unrounded requirement as a percentage of the 90% requirement. (Answer: 271 claims at 90%, 385 claims at 95%, and approximately a 42.0% increase.)

What You'll Learn