Worked Partial Credibility Calculations Across Measures

Why Partial Credibility Depends on the Measure

A portfolio may have too little experience to satisfy a chosen full-credibility standard, but its experience can still provide information. A partial-credibility factor expresses how much of the full standard’s required volume is available. Its calculation depends on what is being estimated: a claim frequency, a claim severity, or aggregate loss. The corresponding variability is different for each measure, so a single full-credibility volume cannot be used for all three.

The calculations below use limited-fluctuation credibility. They start from a specified confidence level and relative error tolerance, find the volume required for full credibility, and then compare the available volume with that standard. The result is capped at 1: experience beyond the full-credibility standard does not produce a factor greater than full credibility.

Definition: Let $p$ be the target probability that an estimate lies within a specified relative error $k$ of its mean, and let $z$ be the corresponding standard-normal quantile. If $V_0$ is the volume required for full credibility and $V$ is the available volume on the same basis, the capped partial-credibility factor is $$ Z=\min\left(1,\sqrt{\frac{V}{V_0}}\right). $$ The square-root relationship applies when the estimate’s variance is inversely proportional to volume. The volume and its full-credibility standard must refer to the same measure.

For example, a count-based standard must be compared with claim-count volume, not with a dollar amount of premium. Likewise, the severity standard uses the number of claims, with claim-size variability reflected in its full-credibility count. All examples are hypothetical. Interest and mortality assumptions are not applicable to these credibility calculations.

Frequency Credibility

Suppose claim counts follow a Poisson model with mean $m$. The frequency estimate is the number of claims divided by exposure. For a fixed exposure volume, its relative variability is governed by the expected claim count. A standard frequency calculation therefore first determines the count needed to meet the selected confidence and error requirements.

N_{0,F}=\left(\frac{z}{k}\right)^2, \qquad Z_F=\min\left(1,\sqrt{\frac{m}{N_{0,F}}}\right).

Here $N_{0,F}$ is the full-credibility claim-count standard and $m$ is the expected claim-count volume associated with the experience. In applications that stipulate a claim count as the volume measure, that stipulated count is used in the comparison. The formula relies on Poisson count variation and a normal approximation to the count for the selected standard; it is not a general formula for an arbitrary count distribution.

Worked Example: Frequency Credibility With a Volume Check

A hypothetical warranty portfolio has 800 exposure units and an expected frequency of 0.08 claims per unit. Use a target confidence of approximately 95.45%, represented by $z=2$, and a relative error tolerance of $k=0.20$. Calculate the capped frequency credibility factor. Assume independent Poisson claim counts with the stated expected frequency.

Step 1: Calculate expected claim volume. The exposure and expected frequency imply $$ m=800(0.08)=64\text{ expected claims}. $$

Step 2: Find the full-credibility count. $$ N_{0,F}=\left(\frac{2}{0.20}\right)^2=10^2=100\text{ claims}. $$

Step 3: Compare the available and required volumes. $$ Z_F=\min\left(1,\sqrt{\frac{64}{100}}\right) =\min(1,0.8)=0.8. $$ The capped partial-credibility factor is 0.800, or 80.0%. The portfolio has 64% of the required count volume, but its factor is 80% because the relationship is the square root of the volume ratio.

Check. Squaring the factor gives $0.8^2=0.64=64/100$, confirming the volume relationship. This calculation avoids the mistake of using $64/100=0.64$ as the credibility factor itself.

Severity Credibility

For severity, the estimate is the average claim size. Assume individual claim sizes are independent and identically distributed, with finite mean $\mu_X$ and standard deviation $\sigma_X$. Define the coefficient of variation as $\operatorname{CV}(X)=\sigma_X/\mu_X$. Under a normal approximation to the sample mean, the full-credibility claim count is

N_{0,S}=\left(\frac{z\operatorname{CV}(X)}{k}\right)^2, \qquad Z_S=\min\left(1,\sqrt{\frac{n}{N_{0,S}}}\right),

where $n$ is the number of observed claims. The severity factor uses claim count, not total dollars of losses. Greater severity variation increases the required number of claims; a larger tolerance $k$ reduces it. The normal approximation is less reliable when claim sizes are extremely skewed or the claim count is small, even though the formula can still be calculated as a prescribed limited-fluctuation standard.

Worked Example: Severity Credibility With High Claim-Size Variation

A hypothetical equipment-repair program has 64 observed claims. Claim sizes are independent draws from a distribution with mean $500 and standard deviation $750; no particular parametric distribution is assumed. Use $z=2$ and a relative error tolerance of $k=0.30$. Find the capped severity factor. The interest rate and mortality assumptions are not applicable.

Step 1: Calculate the coefficient of variation. $$ \operatorname{CV}(X)=\frac{750}{500}=1.5. $$

Step 2: Calculate the full-credibility claim count. $$ N_{0,S}=\left(\frac{2(1.5)}{0.30}\right)^2 =\left(\frac{3}{0.30}\right)^2 =10^2=100\text{ claims}. $$

Step 3: Calculate and cap the factor. $$ Z_S=\min\left(1,\sqrt{\frac{64}{100}}\right) =\min(1,0.8)=0.8. $$ The severity credibility factor is 0.800, or 80.0%.

Check. The observed count is below the 100-claim full standard, so the factor is below 1. Its square, $0.64$, matches $64/100$. This calculation avoids treating the mean claim size as a substitute for the coefficient of variation: the relative dispersion, not the dollar scale alone, determines the standard.

Aggregate-Loss Credibility

Aggregate loss combines claim count and claim size. Under a compound Poisson model, let $N\sim\operatorname{Poisson}(m)$, let claim sizes $X_1,X_2,\ldots$ be independent and identically distributed with finite first and second moments, and assume the claim sizes are independent of $N$. Total loss is $S=\sum_{j=1}^{N}X_j$. The target is relative error around the expected aggregate loss, not around frequency or average severity separately.

Derivation: Conditional on the claim count, aggregate loss has mean $N E[X]$ and variance $N\operatorname{Var}(X)$. Taking expectations over the Poisson count gives $E[S]=mE[X]$ and, by the compound Poisson variance formula, $\operatorname{Var}(S)=mE[X^2]$. Thus the squared coefficient of variation of aggregate loss is $$ \frac{\operatorname{Var}(S)}{E[S]^2} =\frac{mE[X^2]}{m^2(E[X])^2} =\frac{1+\operatorname{CV}(X)^2}{m}, $$ because $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. Requiring the approximate relative standard error to be no more than $k/z$ yields $$ N_{0,A}=\left(\frac{z}{k}\right)^2\left(1+\operatorname{CV}(X)^2\right), \qquad Z_A=\min\left(1,\sqrt{\frac{m}{N_{0,A}}}\right). $$

The extra $1$ in the aggregate standard accounts for claim-count variation. Omitting it would use the severity variability alone and understate the required volume. These formulas assume a compound Poisson model and finite severity variance; they do not automatically apply to dependent claim sizes, overdispersed counts, or a model with a different count distribution.

Worked Example: Aggregate-Loss Credibility Under a Compound Poisson Model

A hypothetical property program has 160 exposure units and an expected frequency of 0.8 claims per unit, giving an expected claim count of 128. Claim sizes have mean $1,000 and standard deviation $1,000, and are independent of the Poisson claim count. Use $z=2$ and $k=0.20$. Calculate the capped aggregate-loss factor. No interest or mortality assumption is involved.

Step 1: Confirm the claim-count volume and severity variation. $$ m=160(0.8)=128, \qquad \operatorname{CV}(X)=\frac{1{,}000}{1{,}000}=1. $$

Step 2: Calculate the full-credibility expected count. $$ N_{0,A}=\left(\frac{2}{0.20}\right)^2(1+1^2) =100(2)=200\text{ expected claims}. $$

Step 3: Calculate the capped factor. $$ Z_A=\min\left(1,\sqrt{\frac{128}{200}}\right) =\min(1,\sqrt{0.64}) =0.8. $$ The aggregate-loss credibility factor is 0.800, or 80.0%.

Check. The aggregate full standard is 200 expected claims, which exceeds the 128 available. The factor squared is $0.64=128/200$. This calculation avoids using the severity-only standard, which would omit the additional variability from the random claim count.

Using the Cap and Choosing the Right Standard

The cap is part of each partial-credibility calculation, not an optional adjustment. If the available volume exceeds the full standard, the raw square root exceeds 1; the capped factor remains 1. For example, with a frequency standard of 100 claims and volume of 144 claims, the raw factor is $\sqrt{144/100}=1.2$, but the capped factor is $\min(1,1.2)=1$. Reporting 1.2 would imply more than full credibility under a standard that defines full credibility as the maximum factor.

A reliable solution keeps the measure and volume aligned. Frequency uses claim-count variation. Severity uses the claim-size coefficient of variation and the number of claims. Aggregate loss uses both count and size variation under the compound Poisson assumptions. The confidence quantile $z$ and relative tolerance $k$ must also be applied consistently: increasing $z$ raises the required volume, while increasing $k$ makes the criterion less stringent and lowers it.

Name the target measure.
Decide whether the estimate concerns frequency, severity, or aggregate loss before selecting a standard.

State the assumptions and inputs.
Identify the count model, severity variation where relevant, confidence quantile, error tolerance, and volume measure.

Calculate the full standard.
Use the formula for the selected measure and keep units clear, such as claims or expected claims.

Calculate and cap the factor.
Take the square root of available volume divided by required volume, then apply the minimum with 1.

Key takeaway: Partial credibility is the square root of the available-to-required volume ratio, capped at 1, only after the appropriate full-credibility standard has been selected. Frequency, severity, and aggregate loss generally have different standards because their sources of variability differ.

Check Your Understanding

For each problem, state the relevant standard, show the volume comparison, and apply the cap.

A hypothetical liability book has 300 exposure units and expected frequency 0.12 claims per unit. Under a Poisson frequency model, use $z=2$ and $k=0.25$. Find the capped frequency factor. (Answer: $N_{0,F}=64$, available volume $=36$, $Z_F=0.750$.)
A hypothetical medical-service portfolio has 100 claims. Claim sizes have mean $400 and standard deviation $600. With $z=1.96$ and $k=0.20$, calculate the severity factor, rounding the factor to four decimals. Assume independent claim sizes with finite variance. (Answer: $\operatorname{CV}=1.5$, $N_{0,S}=216.09$, $Z_S\approx0.6803$.)
For a hypothetical compound Poisson book, claim-size mean is $800 and standard deviation is $600. The expected claim count is 80. Using $z=1.645$ and $k=0.25$, calculate the aggregate-loss factor, rounding to four decimals. (Answer: $\operatorname{CV}=0.75$, $N_{0,A}=67.650625$, so $Z_A=1.0000$ after capping.)
A hypothetical frequency standard requires 100 expected claims. A portfolio has 700 exposure units and expected frequency 0.07. Calculate its capped factor, then explain why the expected count rather than the exposure count is used in the ratio. (Answer: expected count $=49$, $Z_F=\sqrt{49/100}=0.700$; exposure units are not claim-count volume.)
A hypothetical aggregate-loss calculation uses a compound Poisson model, $z=2$, $k=0.20$, severity coefficient of variation 1, and expected claim count 128. Calculate the factor and identify the result of incorrectly omitting count variation. (Answer: correct $N_{0,A}=200$ and $Z_A=0.800$; omitting count variation gives $N_0=100$ and an incorrect factor $\sqrt{128/100}>1$, which caps at 1.)

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