How Precision Requirements Change Credibility

Why the Relative Tolerance Matters

A full-credibility standard specifies how much experience is needed before an estimate meets a chosen precision requirement. The relative tolerance controls how close the estimate must be to its expected value, measured against that expected value. A smaller tolerance is a stricter requirement and needs more experience.

For a variance-based standard, the key comparison is between the standard deviation of the estimate and the size of its mean. This tutorial focuses on how changing the tolerance changes the required experience. It does not convert a variance requirement into a probability guarantee; that is a different type of requirement.

Definition: Let $T$ be a nonnegative estimator with $E[T]>0$, and let $r>0$ be the permitted relative tolerance. A variance-based full-credibility standard is met when $$ \frac{\sqrt{\operatorname{Var}(T)}}{E[T]} \le r. $$ The ratio on the left is the coefficient of variation of $T$. The tolerance $r$ is entered as a decimal: for example, a 10% tolerance means $r=0.10$. This criterion compares standard deviation with the mean; by itself, it does not specify a probability that an observed result lies within a particular range.

A full-credibility requirement is a threshold, not a guarantee that a particular observed estimate will be close to its expectation. The formula depends on the model for the experience and on the chosen measure of relative variability. State both before calculating a standard.

Why the Requirement Has an Inverse-Square Form

For claim counts with a Poisson model, experience can be measured by the expected number of claims. Suppose $N\sim\operatorname{Poisson}(m)$, where $m=E[N]$. Then $E[N]=m$ and $\operatorname{Var}(N)=m$. Applying the relative standard deviation criterion to $N$ gives

\frac{\sqrt{\operatorname{Var}(N)}}{E[N]} = \frac{\sqrt{m}}{m} = \frac{1}{\sqrt{m}} \le r.

Both sides are nonnegative, so this inequality is equivalent to $m\ge 1/r^2$. The inverse square explains how precision changes: if the tolerance is halved, the required expected claim count is multiplied by four.

Write the precision criterion.
For the quantity being studied, set its coefficient of variation no greater than the relative tolerance $r$.

Substitute the model moments.
Use the stated mean and variance, with exposure and frequency expressed in consistent units.

Solve for experience.
Rearrange the inequality to find the minimum expected claims or exposure that meets the standard.

Worked Example: Comparing Claim-Count Standards

A hypothetical portfolio has a Poisson claim count over its accumulated experience. Find the full-credibility standards for relative tolerances of 20%, 10%, and 5%, measured using the relative standard deviation of the count. Explain the effect of tightening the tolerance. Use the variance-based count method.

Model and method. Let $N\sim\operatorname{Poisson}(m)$, with expected claim count $m$. Since $E[N]=\operatorname{Var}(N)=m$, the standard requires $1/\sqrt{m}\le r$, or $m\ge 1/r^2$.

At 20%. Here $r=0.20$, so $$ m\ge\frac{1}{(0.20)^2}=\frac{1}{0.04}=25. $$ The standard is at least 25 expected claims.

At 10%. Here $r=0.10$, so $$ m\ge\frac{1}{(0.10)^2}=\frac{1}{0.01}=100. $$ The standard is at least 100 expected claims.

At 5%. Here $r=0.05$, so $$ m\ge\frac{1}{(0.05)^2}=\frac{1}{0.0025}=400. $$ The standard is at least 400 expected claims.

Check and common mistake avoided. Each time the tolerance is halved, the standard is multiplied by four: $100/25=4$ and $400/100=4$. Doubling the required claims when the tolerance halves would be incorrect because the tolerance is squared in the formula.

Compound Poisson Aggregate Loss

A count standard measures variability in claim numbers. For aggregate loss, claim amounts also vary. Let $N\sim\operatorname{Poisson}(m)$, and suppose individual claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, independent of $N$, with finite $E[X]$ and $E[X^2]$. Define aggregate loss as $S=\sum_{j=1}^{N}X_j$, taking $S=0$ when $N=0$.

Conditional on $N$, the mean aggregate loss is $N E[X]$ and its variance is $N\operatorname{Var}(X)$. The law of total variance therefore gives

\begin{aligned} E[S] &= mE[X],\\ \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])\\ &=m\operatorname{Var}(X)+m(E[X])^2\\ &=mE[X^2]. \end{aligned}

The last equality uses $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. In particular, aggregate-loss variance includes the variability from the random claim count as well as variation in claim amounts. Applying the relative standard deviation criterion to $S$ yields

\frac{\sqrt{\operatorname{Var}(S)}}{E[S]} = \frac{\sqrt{mE[X^2]}}{mE[X]} \le r, \qquad m\ge \frac{E[X^2]}{r^2(E[X])^2}.

This form gives the required expected number of claims. If the claim frequency is $\lambda$ per unit of exposure and exposure is $e$, then $m=e\lambda$. Thus the required exposure is

e\ge\frac{E[X^2]}{r^2\lambda(E[X])^2}.

The same relative standard deviation applies to the pure premium estimate $S/e$: dividing both $S$ and its expectation by $e$ does not change their coefficient of variation. These formulas require a compound Poisson model, independent claim amounts with finite second moment, and a positive mean severity. Do not apply them unchanged to a different frequency model.

Worked Example: A Tighter Standard for Aggregate Loss

A hypothetical book has Poisson claim frequency $\lambda=0.20$ claims per policy-year. Claim severity $X$ is $500 or $1,500 with probability $0.5$ each. Find the policy-year exposure and expected claims required at 10% and 5% relative tolerances for aggregate loss. Use the compound Poisson variance standard.

Model and severity moments. Claim counts over $e$ policy-years have mean $m=0.20e$. The severity mean is $$ E[X]=0.5(500)+0.5(1{,}500)=250+750=\$1{,}000. $$ The second moment is $$ E[X^2]=0.5(500^2)+0.5(1{,}500^2) =0.5(250{,}000)+0.5(2{,}250{,}000) =1{,}250{,}000. $$

At 10%. The expected-claim requirement is $$ m\ge\frac{1{,}250{,}000}{(0.10)^2(1{,}000)^2} =\frac{1{,}250{,}000}{10{,}000} =125. $$ Since $m=0.20e$, the exposure requirement is $e\ge125/0.20=625$ policy-years.

At 5%. The expected-claim requirement is $$ m\ge\frac{1{,}250{,}000}{(0.05)^2(1{,}000)^2} =\frac{1{,}250{,}000}{2{,}500} =500. $$ The exposure requirement is $e\ge500/0.20=2{,}500$ policy-years.

Check and common mistake avoided. Tightening the tolerance from 10% to 5% quadruples both required expected claims and required exposure: $500/125=4$ and $2{,}500/625=4$. Using $\operatorname{Var}(X)$ in place of $E[X^2]$ would omit the claim-count contribution to aggregate variance and understate the standard.

How Severity Variation Changes the Standard

At a fixed expected claim count, the compound Poisson relative variance is $$ \frac{\operatorname{Var}(S)}{(E[S])^2} = \frac{E[X^2]}{m(E[X])^2}. $$

The ratio $E[X^2]/(E[X])^2$ is at least 1, because $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. It equals 1 for a constant severity and is larger when severity varies. Consequently, a variable severity distribution requires more expected claims than a constant severity with the same mean, at the same relative tolerance.

Worked Example: Is the Available Experience Sufficient?

Compare two hypothetical books, each with frequency $\lambda=0.25$ claims per policy-year and 400 policy-years of exposure. In Book A every claim is $1,000. In Book B, claim severity is $500 or $1,500 with probability $0.5$ each. Determine whether each book meets a 10% aggregate-loss relative standard. Use the compound Poisson coefficient of variation and identify the mistake of treating equal mean severities as equal precision.

Expected claims in each book. Both have $m=400(0.25)=100$ expected claims.

Book A. Here $E[X]=1{,}000$ and $E[X^2]=1{,}000^2=1{,}000{,}000$. Its coefficient of variation is $$ \frac{\sqrt{mE[X^2]}}{mE[X]} =\sqrt{\frac{1{,}000{,}000}{100(1{,}000)^2}} =\sqrt{0.01}=0.1000. $$ Book A exactly meets the 10% standard.

Book B. Its moments are $E[X]=1{,}000$ and $E[X^2]=1{,}250{,}000$, as calculated from the two equally likely severities. Its coefficient of variation is $$ \sqrt{\frac{1{,}250{,}000}{100(1{,}000)^2}} =\sqrt{0.0125} \approx 0.1118. $$ Book B does not meet the standard because $0.1118>0.10$.

Check and common mistake avoided. Both books have the same frequency, exposure, and mean severity, but only Book A meets the requirement. Matching expected losses alone does not imply matching precision; Book B’s variable severity increases aggregate-loss variance.

Common Exam Trap: Scaling the Tolerance Incorrectly

For a fixed model and a variance-based standard, required experience is proportional to $1/r^2$. If a 10% standard requires 100 expected claims, a 5% standard requires $100(0.10/0.05)^2=400$, not 200. More generally, changing from tolerance $r_1$ to $r_2$ changes the standard by the factor $(r_1/r_2)^2$, provided the model and the quantity being measured stay the same.

Also distinguish a relative standard deviation criterion from a probability requirement. The former sets a limit on a ratio of model moments; it does not assert that a specified percentage of outcomes will fall within the tolerance. Keep the stated criterion intact when comparing standards.

Key takeaway: Under a variance-based full-credibility standard, halving the relative tolerance requires four times the experience. For compound Poisson aggregate loss, the severity second moment matters as well as the severity mean; greater severity variation can therefore increase the required experience.

Check Your Understanding

For each problem, use the stated variance-based relative standard deviation criterion.

A Poisson claim count has expected count $m$. Find the full-credibility count standards at 25% and 12.5% relative tolerances. (Answer: 16 and 64 expected claims, respectively.)
A compound Poisson model has $E[X]=\$2{,}000$, $E[X^2]=5{,}000{,}000$, and a 10% relative tolerance for aggregate loss. Find the required expected claim count. (Answer: 125 expected claims.)
A hypothetical portfolio has Poisson frequency $\lambda=0.40$ claims per policy-year and 200 policy-years of exposure. What is its expected claim count and count coefficient of variation? (Answer: 80 expected claims and $1/\sqrt{80}\approx0.1118$, or 11.18%.)
Claim severity is constant at $800, frequency is $\lambda=0.50$ claims per policy-year, and the aggregate-loss tolerance is 10%. Find the required expected claims and policy-year exposure. (Answer: 100 expected claims and 200 policy-years.)
For an unchanged model, a relative tolerance is tightened from 8% to 4%. By what factor does the required experience change? (Answer: It is multiplied by 4.)

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