Relative Error Standards for Compound Poisson Loss

Why Relative Precision Depends on Severity

A count standard alone does not determine how stable an aggregate loss total will be. If claim amounts vary widely, two portfolios with the same expected number of claims can have very different aggregate-loss variability. A relative-error standard accounts for this by comparing the aggregate loss's standard deviation with its expected value. The severity distribution enters through its first two moments.

This tutorial considers a hypothetical portfolio with a Poisson claim count and independent claim severities. The resulting standard is a normal-approximation full-credibility standard: it estimates how large the expected claim count must be for the aggregate loss to fall within a specified percentage of its mean with a specified probability. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda$ is the expected claim count over the period of interest. Let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$, with $E[X]=\mu>0$ and $\operatorname{Var}(X)=\sigma^2<\infty$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. For relative tolerance $k>0$ and central confidence level $p$, the normal-approximation standard requires $$ P\bigl(|S-E[S]|\le kE[S]\bigr)\approx p. $$

Here, $k=0.10$ means a tolerance of 10% of the expected aggregate loss, not a tolerance of a fixed dollar amount. The confidence level $p$ is two-sided: for example, a central 90% standard allows deviations both above and below the mean. This tutorial uses $z_{(1+p)/2}$, the standard normal quantile with cumulative probability $(1+p)/2$.

Deriving the Expected Claim Count Standard

For a compound Poisson aggregate loss, the mean is $\lambda\mu$, while its variance is $\lambda E[X^2]$. These are the moments used by the normal approximation. In particular, the variance involves the second raw moment $E[X^2]$, not just the severity variance $\sigma^2$.

Result: Under the model above, the normal-approximation relative-error standard is $$ \lambda\ \ge\ \frac{z_{(1+p)/2}^{\,2}E[X^2]}{k^2\mu^2} \ =\ \frac{z_{(1+p)/2}^{\,2}(\sigma^2+\mu^2)}{k^2\mu^2}. $$ Since $E[X^2]=\operatorname{Var}(X)+(E[X])^2$, severity variability increases the required expected count beyond the amount implied by the mean alone.

The justification follows by standardizing the aggregate loss. If the normal approximation is adequate, then $S$ is approximated by a normal random variable with mean $\lambda\mu$ and variance $\lambda E[X^2]$. Thus the standardized half-width of the allowed interval is

\frac{kE[S]}{\sqrt{\operatorname{Var}(S)}} = \frac{k\lambda\mu}{\sqrt{\lambda E[X^2]}} = k\sqrt{\frac{\lambda\mu^2}{E[X^2]}}.

For a central probability $p$, the corresponding standard normal interval has half-width $z_{(1+p)/2}$. Requiring the standardized half-width to be at least that quantile gives

k\sqrt{\frac{\lambda\mu^2}{E[X^2]}}\ge z_{(1+p)/2} \quad\Longleftrightarrow\quad \lambda\ge \frac{z_{(1+p)/2}^{\,2}E[X^2]}{k^2\mu^2}.

This is a normal-approximation criterion, not an exact probability guarantee for every compound Poisson distribution. It can be less reliable when the expected count is small or the severity distribution is highly skewed. An exact calculation, when feasible, can be used to assess the approximation.

Worked Applications

Worked Example: Exponential Severity and a 90% Standard

A hypothetical warranty portfolio has a Poisson claim count. Each claim severity is exponentially distributed with mean $800, so the severity variance is $640,000. Find the approximate expected claim count required for aggregate loss to be within 10% of its mean with central 90% confidence. Use $z_{0.95}=1.6449$.

Method. Apply the compound Poisson relative-error standard. The exponential distribution has mean $\mu=800$ and variance $\sigma^2=800^2=640{,}000$, so $$ E[X^2]=\sigma^2+\mu^2=640{,}000+640{,}000=1{,}280{,}000. $$

The confidence level is central 90%, so the required normal quantile is $z_{(1+0.90)/2}=z_{0.95}=1.6449$. The tolerance is $k=0.10$. Substitution gives $$ \lambda\ge \frac{1.6449^2(1{,}280{,}000)}{0.10^2(800^2)} = \frac{2.70569601(1{,}280{,}000)}{0.01(640{,}000)} = \frac{3{,}463{,}290.8928}{6{,}400} = 541.1392. $$

The cutoff is approximately 541.1 expected claims when reported to one decimal place. The check is that at this cutoff the approximate standardized half-width is $0.10\sqrt{541.1392(640{,}000)/1{,}280{,}000}=1.6449$, as required. The common mistake avoided here is substituting only $\sigma^2$ for $E[X^2]$; that would omit the claim-count contribution to aggregate variance.

Worked Example: A 95% Standard with a Different Severity Scale

A hypothetical equipment-protection program has Poisson claim frequency and exponentially distributed severity with mean $1,200. Determine the expected claim count for aggregate loss to be within 8% of its mean at central 95% confidence. Use $z_{0.975}=1.959964$.

Method. For an exponential severity, $\mu=1{,}200$, $\sigma^2=1{,}200^2=1{,}440{,}000$, and $$ E[X^2]=1{,}440{,}000+1{,}440{,}000=2{,}880{,}000. $$ The tolerance is $k=0.08$, and the two-sided 95% quantile is $1.959964$. Therefore, $$ \lambda\ge \frac{1.959964^2(2{,}880{,}000)}{0.08^2(1{,}200^2)} = \frac{3.8414588(2{,}880{,}000)}{0.0064(1{,}440{,}000)} = \frac{11{,}063{,}401.3}{9{,}216} = 1{,}200.456. $$

The calculated cutoff is about 1,200.46 expected claims, or approximately 1,200 claims when stated to the nearest whole claim. The standardized half-width check is $0.08\sqrt{1{,}200.456(1{,}440{,}000)/2{,}880{,}000}\approx1.959964$. The common mistake avoided is using a 90% quantile because it is familiar; the central 95% requirement calls for the 97.5th percentile.

Worked Example: Using the Coefficient of Variation

A hypothetical travel-insurance portfolio has lognormal claim severities with mean $1,000 and coefficient of variation 0.5. Thus the lognormal parameters are $\sigma_{\log}^2=\ln(1.25)$ and $\mu_{\log}=\ln(1000)-\tfrac12\ln(1.25)$; these give the stated mean and coefficient of variation. Find the approximate expected claim count for a 10% relative tolerance at central 99% confidence, using $z_{0.995}=2.5758$.

Method. The coefficient of variation is $CV=\sigma/\mu=0.5$, so $\sigma^2=0.25\mu^2$. Consequently, $$ \frac{E[X^2]}{\mu^2} = \frac{\sigma^2+\mu^2}{\mu^2} = 1+CV^2 = 1+0.25 = 1.25. $$ Using the relative-error formula in this ratio form, $$ \lambda\ge \frac{2.5758^2(1.25)}{0.10^2} = \frac{6.63474864(1.25)}{0.01} = \frac{8.2934358}{0.01} = 829.34358. $$

The cutoff is approximately 829.3 expected claims to one decimal place. Checking the standardized half-width gives $0.10\sqrt{829.34358/1.25}=2.5758$. This also shows why the mean amount, $1,000, cancels when the coefficient of variation is supplied: the standard is relative rather than absolute.

How to Apply the Standard and Avoid Traps

A reliable calculation separates the confidence requirement from the severity calculation. First identify whether the probability is central or one-sided; for a central level $p$, use $z_{(1+p)/2}$. Then calculate $E[X^2]$, either directly or as $\sigma^2+\mu^2$. Finally, use the relative tolerance $k$ as a decimal and evaluate the expected-count cutoff.

Specify the model.
Confirm that the claim count is Poisson and that severities are independent, identically distributed, independent of the count, and have finite mean and variance.

Identify the inputs.
Record $\mu$, $\sigma^2$ or $CV$, the relative tolerance $k$, and the central confidence level $p$.

Calculate the cutoff.
Find $E[X^2]=\sigma^2+\mu^2$, select $z_{(1+p)/2}$, and substitute into the expected-count formula.

Check the result.
Substitute the calculated cutoff into $k\sqrt{\lambda\mu^2/E[X^2]}$; it should equal the selected normal quantile before rounding.

The most consequential exam trap is to use the severity variance alone in the numerator. For the first worked example, that incorrect calculation would give $$ \lambda_{\text{wrong}} = \frac{1.6449^2(640{,}000)}{0.10^2(800^2)} = 270.5696, $$ rather than the correct cutoff of about 541.1. The error occurs because aggregate variance is $\lambda E[X^2]$, not $\lambda\operatorname{Var}(X)$. In addition to variation in individual severity, a Poisson number of claims contributes variation to the aggregate total.

A second trap is to enter a percentage as a whole number. A 10% tolerance means $k=0.10$, so the denominator contains $0.10^2$, not $10^2$. A third is to confuse this standard with an absolute-error standard: the allowed deviation here is $kE[S]=k\lambda\mu$, which grows with the expected aggregate loss.

Key takeaway: For compound Poisson aggregate loss, relative full credibility depends on the severity second moment relative to the square of the severity mean. The normal-approximation cutoff is $z_{(1+p)/2}^{\,2}E[X^2]/(k^2\mu^2)$. State the confidence convention, use the second raw moment, and verify the result with the standardized half-width.

Check Your Understanding

For each problem, use the normal-approximation compound Poisson standard and report the cutoff to the indicated precision.

A hypothetical repair plan has exponential severity with mean $500. Find the expected claim count for 20% relative tolerance and central 90% confidence, using $z_{0.95}=1.6449$. (Answer: 135.3 expected claims)
A hypothetical liability portfolio has severity mean $1,000 and coefficient of variation 0.8. Find the expected claim count for 15% relative tolerance and central 95% confidence, using $z_{0.975}=1.959964$. (Answer: 280.0 expected claims)
A hypothetical device policy has severity mean $2,000 and variance $1{,}000{,}000$ dollars squared. Find the expected claim count for 5% relative tolerance and central 90% confidence, using $z_{0.95}=1.6449$. (Answer: 1,352.8 expected claims)
A hypothetical event-cancellation portfolio has severity coefficient of variation 0.3. Find the expected claim count for 8% relative tolerance and central 99% confidence, using $z_{0.995}=2.5758$. (Answer: 1,130.0 expected claims)
A hypothetical portfolio has severity mean $900 and coefficient of variation 0.6. Its expected claim count is 600. Does it meet the approximate standard for 10% relative tolerance at central 95% confidence, using $z_{0.975}=1.959964$? (Answer: Yes; the cutoff is approximately 522.4 expected claims)

What You'll Learn