A Worked Aggregate-Loss Credibility Problem

Why Aggregate-Loss Credibility Needs Two Sources of Variation

When credibility is based on total losses, variability comes from two sources: the number of claims can fluctuate, and the amount paid on each claim can fluctuate. A full-credibility calculation must allow for both. Using only the variation in claim sizes understates aggregate variability when the claim count is random, and can make an experience period appear more credible than it is.

This tutorial uses a compound Poisson model to connect aggregate-loss moments to a limited-fluctuation standard. The standard specifies how close aggregate losses should be to their expected value, and with what probability. Once the full-credibility exposure is found, the square-root rule determines how much weight to place on a shorter experience period.

Model assumptions: Each exposure unit has a claim count $N\sim\operatorname{Poisson}(\lambda)$. Claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, with $E[X]=\mu$ and $\operatorname{Var}(X)=\sigma^2$. The count is independent of the claim amounts. Aggregate loss for one exposure unit is $S=\sum_{j=1}^{N}X_j$, with $S=0$ if $N=0$. Across exposure units, claim counts and claim amounts are independent and follow the same model.

Here $\lambda$ is the expected number of claims per exposure unit, while $\mu$ and $\sigma^2$ describe the severity distribution. An exposure unit might be one hypothetical policy-year, provided the claim frequency and coverage basis are the same for every unit. The formulas below do not apply unchanged if exposure units are heterogeneous, claim counts are dependent, or the count model is not Poisson.

Aggregate-Loss Moments

Conditional on $N=n$, aggregate loss is the sum of $n$ independent claim amounts. Thus $E[S\mid N=n]=n\mu$ and $\operatorname{Var}(S\mid N=n)=n\sigma^2$. Averaging over the random claim count gives the unconditional moments.

Theorem: Under the compound Poisson assumptions, the aggregate loss per exposure unit has $$ E[S]=\lambda\mu,\qquad \operatorname{Var}(S)=\lambda E[X^2] =\lambda(\sigma^2+\mu^2). $$ For $n$ independent exposure units, the total aggregate loss $T_n$ has $$ E[T_n]=n\lambda\mu,\qquad \operatorname{Var}(T_n)=n\lambda E[X^2]. $$

The variance result follows from the law of total variance. In particular, randomness in $N$ contributes a term that is not captured by the severity variance alone.

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

The first variance term, $\lambda\sigma^2$, reflects claim-size variation given the number of claims. The second, $\lambda\mu^2$, reflects variation in the number of claims. This is why replacing $E[X^2]$ by $\sigma^2$ is generally incorrect for a compound Poisson total.

Worked Example: Finding Moments for a Compound Loss

A hypothetical policy-year has a Poisson claim count with mean $\lambda=0.20$. Each claim amount is equally likely to be $1,000 or $3,000, independently of the count and of other claim amounts. Find the mean and variance of aggregate loss for one policy-year.

Method. Find the severity mean and variance, then use the compound Poisson aggregate-moment formulas. The calculation avoids treating severity variance as aggregate-loss variance.

The severity mean is $$ \mu=\frac{1{,}000+3{,}000}{2}=\$2{,}000. $$

The second severity moment is $$ E[X^2]=\frac{1{,}000^2+3{,}000^2}{2} =\frac{1{,}000{,}000+9{,}000{,}000}{2} =5{,}000{,}000. $$

The severity variance is $5{,}000{,}000-2{,}000^2=1{,}000{,}000$. Therefore, $$ E[S]=0.20(2{,}000)=\$400 $$ and $$ \operatorname{Var}(S)=0.20(5{,}000{,}000)=1{,}000{,}000\text{ dollars}^2. $$

As a check, the formula using both sources of variation gives $0.20(1{,}000{,}000+4{,}000{,}000)=1{,}000{,}000$, the same variance. The standard deviation is $\$1{,}000$. Using severity variance alone would instead give $0.20(1{,}000{,}000)=200{,}000$, which omits claim-count variation.

Full Credibility Under a Limited-Fluctuation Standard

A limited-fluctuation standard defines full credibility by requiring total losses to fall within a specified relative distance of their expected value with a specified probability. Let $k$ be the allowed relative deviation, such as $0.20$ for 20%, and let $p$ be the required probability. For $n$ independent exposure units, the criterion is

P\left(\left|T_n-E[T_n]\right|\leq kE[T_n]\right)\geq p.

A common actuarial calculation uses a normal approximation to $T_n$. With $z=\Phi^{-1}((1+p)/2)$, where $\Phi$ is the standard normal cumulative distribution function, the two-sided condition is approximated by

z\sqrt{\operatorname{Var}(T_n)} \leq kE[T_n].

Substituting the compound Poisson moments and squaring both nonnegative sides gives the full-credibility exposure $n_0$:

\begin{aligned} z\sqrt{n\lambda E[X^2]} &\leq kn\lambda\mu,\\ n &\geq \left(\frac{z}{k}\right)^2 \frac{E[X^2]}{\lambda\mu^2}. \end{aligned}

Full-credibility exposure: Under the compound Poisson model and the normal approximation, the minimum exposure for the stated two-sided precision and probability standard is $$ n_0=\left(\frac{z}{k}\right)^2 \frac{E[X^2]}{\lambda\mu^2}. $$ If exposure must be a whole number of policy-years, round $n_0$ up to the next whole number. This is an approximation, not an exact probability guarantee; it can be less accurate for small expected claim counts or highly skewed losses.

For partial credibility, use the limited-fluctuation square-root rule. If actual exposure is $n$, then $Z=\min(1,\sqrt{n/n_0})$. If $P_{\mathrm{exp}}$ is the observed aggregate loss per exposure unit and $P_{\mathrm{manual}}$ is the manual pure premium on the same basis, the credibility-weighted pure premium is $ZP_{\mathrm{exp}}+(1-Z)P_{\mathrm{manual}}$. The two premiums must use the same currency, coverage, and exposure basis.

Worked Example: Calculating Full-Credibility Exposure

Use the hypothetical model from the first example: claim count is Poisson with mean $\lambda=0.20$, and severity is equally likely to be $1,000 or $3,000. Find the full-credibility exposure for a 20% precision standard with probability approximately 95.45%, using $z=2$.

Method. Apply the normal-approximation full-credibility formula using the second severity moment. The common mistake avoided here is substituting $\sigma^2$ for $E[X^2]$.

From the moment calculation, $E[X^2]=5{,}000{,}000$ and $\mu=2{,}000$. The precision is $k=0.20$, and $z=2$. Hence $$ \begin{aligned} n_0 &=\left(\frac{2}{0.20}\right)^2 \frac{5{,}000{,}000}{0.20(2{,}000)^2}\\ &=10^2\frac{5{,}000{,}000}{0.20(4{,}000{,}000)}\\ &=100\frac{5{,}000{,}000}{800{,}000}\\ &=100(6.25)=625. \end{aligned} $$

Thus 625 exposure units meet this normal-approximation standard. The arithmetic also checks dimensionally: $E[X^2]/(\lambda\mu^2)$ is dimensionless, so $n_0$ is measured in exposure units. The chosen $z=2$ corresponds to central probability $2\Phi(2)-1$, approximately 95.45%.

Blending Aggregate Experience With a Manual Pure Premium

The experience pure premium is observed aggregate loss divided by the number of exposure units. It is not the aggregate loss itself. After calculating $Z$, blend that per-unit experience with the manual pure premium. When exposure reaches or exceeds $n_0$, $Z=1$ and the rule gives full weight to experience. For exposure below $n_0$, the square-root factor gives partial weight.

Worked Example: Blending a Partially Credible Aggregate Experience

A hypothetical portfolio has the compound Poisson model used above, so its full-credibility standard is $n_0=625$ policy-years. The portfolio has 100 policy-years of experience, with total observed losses of $43,000. A manual pure premium is $400 per policy-year. Find the credibility-weighted pure premium.

Method. First calculate exposure credibility, then convert total losses to a per-policy-year experience pure premium, and finally blend. This order prevents a total from being blended directly with a per-unit rate.

The credibility factor is $$ Z=\sqrt{\frac{100}{625}}=\sqrt{0.16}=0.40. $$

The observed pure premium is $$ P_{\mathrm{exp}}=\frac{\$43{,}000}{100}=\$430 \quad\text{per policy-year}. $$

The credibility-weighted pure premium is $$ \begin{aligned} P_{\mathrm{cred}} &=0.40(\$430)+0.60(\$400)\\ &=\$172+\$240\\ &=\$412\text{ per policy-year}. \end{aligned} $$

The result lies between the two input premiums, as a weighted average should. The portfolio experience is $30 above the manual premium, and the 40% credibility weight moves the result $12 above the manual premium: $\$412-\$400=\$12=0.40(\$30)$. This is a useful independent check.

Common Exam Traps and Checks

The aggregate-loss calculation combines model assumptions, units, and a probability standard. Before finalizing a result, check each part rather than relying on a memorized formula alone.

Using severity variance instead of the second moment. For a Poisson claim count, aggregate variance per exposure is $\lambda E[X^2]$, not $\lambda\operatorname{Var}(X)$. The missing $\lambda\mu^2$ term represents random claim counts.
Using a one-sided normal quantile. A symmetric, two-sided precision requirement with probability $p$ uses $z=\Phi^{-1}((1+p)/2)$, not $\Phi^{-1}(p)$.
Forgetting the square root. Partial credibility is $\sqrt{n/n_0}$, capped at one, not $n/n_0$.
Mixing total losses and pure premiums. Divide observed aggregate losses by exposure before blending with a manual pure premium.
Mixing exposure bases. The observed losses, manual rate, model parameters, and $n_0$ must all refer to the same type of exposure unit and coverage period.

A further practical check is that more exposure cannot reduce credibility: increasing $n$ increases $\sqrt{n/n_0}$ until it reaches one. A calculated factor above one signals that the cap has been omitted. The normal approximation and square-root rule are tools for the stated limited-fluctuation framework; they are not substitutes for checking whether the assumed frequency, severity, and independence model is appropriate.

Check Your Understanding

For each hypothetical scenario, state the requested result. Use the normal approximation when a probability standard is specified.

A policy-year has $N\sim\operatorname{Poisson}(0.10)$, and each claim is exactly $5,000. Find the aggregate-loss mean and variance for 20 independent policy-years. (Answer: Mean = $10,000; variance = 50,000,000 dollars squared.)
Claim count per exposure is Poisson with mean $0.25$. Severity is equally likely to be $1,000 or $5,000. For $k=0.25$ and $z=1.96$, calculate $n_0$ and round up to a whole exposure unit. (Answer: $n_0\approx355.13$, so 356 exposure units.)
A model gives $n_0=256$ exposure units. A portfolio has 64 units, observed losses of $46,080, and a manual pure premium of $680 per unit. Find the credibility-weighted pure premium. (Answer: $Z=0.50$, experience pure premium = $720, weighted pure premium = $700 per unit.)
For a compound Poisson model with $\lambda=0.20$, severity mean $\mu=\$2{,}000$, and severity variance $\sigma^2=1{,}000{,}000$ dollars squared, find $E[X^2]$ and the aggregate variance per exposure unit. (Answer: $E[X^2]=5{,}000{,}000$ dollars squared; $\operatorname{Var}(S)=1{,}000{,}000$ dollars squared.)
Claim count per exposure is Poisson with mean $0.50$, and every claim is $2,000. Find full-credibility exposure for a 10% precision standard with $z=1.645$. Round up to a whole exposure unit. (Answer: $n_0\approx541.21$, so 542 exposure units.)

What You'll Learn