Comparing Frequency and Aggregate-Loss Standards

Why Compare the Two Standards?

A risk may be judged credible by examining how stable its claim count is, or by examining how stable its total claim cost is. These are different questions. Claim counts vary because the number of claims is random; aggregate losses vary because both the number of claims and their severities are random. When severity is variable, a risk can meet a frequency-based standard while still falling short of the aggregate-loss standard.

This tutorial compares the two full-credibility exposure requirements for the same risk. Both use a normal approximation and the same confidence level and relative tolerance, so the effect of severity variation can be isolated. The standards are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: A frequency-based full-credibility standard requires the observed claim count to be within a specified relative tolerance of its expected value with a specified probability. An aggregate-loss full-credibility standard applies the same type of requirement to total claim losses. A full-credibility threshold is the exposure volume at which the selected standard is met under its stated assumptions and approximation.

Model and Threshold Formulas

Suppose there are $n$ equal exposure units. Each unit has an independent Poisson claim count with mean $\lambda$, so the total claim count $N_n$ is Poisson with mean and variance $n\lambda$. Let the claim severities $X_1,X_2,\ldots$ be independent and identically distributed, independent of the count, with mean $\mu>0$ and variance $\sigma^2$. Assume these moments are finite.

The aggregate loss is $S_n=\sum_{j=1}^{N_n}X_j$, where the sum is zero if $N_n=0$. Let $k$ be the relative tolerance, and let $p$ be the desired central probability. Write $z$ for the standard normal critical value satisfying $P(-z\le Z\le z)=p$ for $Z\sim N(0,1)$. For example, $z$ is approximately 1.645 for 90% central confidence and 1.96 for 95% central confidence.

Standard: Using a normal approximation, the frequency standard requires $P(|N_n-n\lambda|\le k n\lambda)$ to be at least $p$. The aggregate-loss standard requires $P(|S_n-E[S_n]|\le kE[S_n])$ to be at least $p$. The formulas below are approximate normal-based thresholds, not exact probability guarantees for every count or severity distribution.

For frequency, the count has mean and variance $n\lambda$. Its approximate standard deviation is therefore $\sqrt{n\lambda}$. Requiring the tolerance to be at least $z$ standard deviations gives the frequency threshold.

k n\lambda \ge z\sqrt{n\lambda} \quad\Longrightarrow\quad n_{\mathrm{freq}} \ge \frac{z^2}{k^2\lambda}.

For aggregate loss, the corresponding result depends on the second moment of severity. The aggregate mean is $n\lambda\mu$, and its variance is $n\lambda(\sigma^2+\mu^2)$. The aggregate threshold follows by requiring $k$ times the mean to be at least $z$ times the standard deviation.

n_{\mathrm{agg}} \ge \frac{z^2}{k^2\lambda} \frac{\sigma^2+\mu^2}{\mu^2} = \frac{z^2}{k^2\lambda} \left(1+\frac{\sigma^2}{\mu^2}\right).

Why Severity Variation Changes the Threshold

The frequency threshold depends on the expected number of claims per exposure unit, $\lambda$, but not on claim size. The aggregate threshold includes the factor $1+\sigma^2/\mu^2$, which is one plus the squared coefficient of variation of severity. Since $\sigma^2\ge 0$, the aggregate threshold is at least as large as the frequency threshold under this model. The thresholds are equal when severity is constant, so $\sigma^2=0$.

This comparison is about required volume, not about which standard is universally preferable. If the rate indication is primarily driven by claim counts, a frequency standard may be relevant. If the quantity being stabilized is total loss cost, severity variation matters and the aggregate standard is the direct measure.

Derivation of the Aggregate-Loss Variance

The aggregate variance explains the severity adjustment. Conditional on $N_n=m$, the sum contains $m$ independent severities, so its conditional mean is $m\mu$ and its conditional variance is $m\sigma^2$. Apply the law of total variance:

\begin{aligned} \operatorname{Var}(S_n) &=E[\operatorname{Var}(S_n\mid N_n)] +\operatorname{Var}(E[S_n\mid N_n])\\ &=E[N_n\sigma^2]+\operatorname{Var}(N_n\mu)\\ &=n\lambda\sigma^2+n\lambda\mu^2\\ &=n\lambda(\sigma^2+\mu^2). \end{aligned}

The first variance contribution, $n\lambda\sigma^2$, comes from random severities given the number of claims. The second, $n\lambda\mu^2$, comes from the randomness of the claim count. Dividing the variance by the square of the aggregate mean gives the relative variance used in the standard:

\frac{\operatorname{Var}(S_n)}{(E[S_n])^2} = \frac{n\lambda(\sigma^2+\mu^2)}{(n\lambda\mu)^2} = \frac{1+\sigma^2/\mu^2}{n\lambda}.

The aggregate standard deviation, relative to the mean, is therefore larger than the frequency standard deviation relative to the mean count whenever severity varies. This is why the aggregate threshold must increase by the factor $1+\sigma^2/\mu^2$.

Worked Comparisons

Worked Example: Variable Severity Requires More Exposure

A hypothetical portfolio has Poisson frequency $\lambda=1.00$ claim per exposure unit. Claim severity has mean $\mu=\$1{,}000$ and standard deviation $\sigma=\$500$. Find the frequency and aggregate-loss thresholds using $z=2.00$ and a 10% relative tolerance. Report the thresholds in exposure units.

Method and setup. Use $k=0.10$. The stated critical value $z=2.00$ corresponds to central confidence of about 95.45% under the normal approximation. First calculate the frequency threshold:

n_{\mathrm{freq}} = \frac{2.00^2}{0.10^2(1.00)} = \frac{4}{0.01} = 400.

For aggregate loss, the severity variance is $\sigma^2=\$250{,}000$, and the squared mean is $\mu^2=\$1{,}000{,}000$. Thus the severity multiplier is $1+250{,}000/1{,}000{,}000=1.25$. Therefore,

n_{\mathrm{agg}} = 400(1.25) = 500.

Check and interpretation. At 400 exposure units, expected claim count is 400 and its standard deviation is 20; twice the standard deviation is 40, exactly 10% of the mean count. At 500 units, expected aggregate loss is $\$500{,}000$. Its variance is $500(1.00)(\$1{,}000{,}000+\$250{,}000)=\$625{,}000{,}000$, so its standard deviation is $\$25{,}000$; twice that is $\$50{,}000$, exactly 10% of the mean. The aggregate standard needs 100 more units because severity is variable. Common mistake avoided: using the frequency threshold for total losses would omit the 1.25 severity multiplier.

Worked Example: Constant Severity Gives Equal Thresholds

A hypothetical equipment program has Poisson frequency $\lambda=0.50$ claim per annual exposure unit. Every claim has the same severity, $\$600$, so $\mu=\$600$ and $\sigma=0$. Find both thresholds for $z=2.00$ and a 20% tolerance, then verify the standards at the threshold.

Frequency calculation. Here $k=0.20$, so

n_{\mathrm{freq}} = \frac{2.00^2}{0.20^2(0.50)} = \frac{4}{0.02} = 200.

Aggregate calculation. Since severity has zero variance, the severity multiplier is $1+0/600^2=1$. Thus $n_{\mathrm{agg}}=200(1)=200$ exposure units.

Check. At 200 units, expected claim count is $200(0.50)=100$, with standard deviation $\sqrt{100}=10$. The permitted count deviation is $0.20(100)=20$, equal to $2(10)$. Expected aggregate loss is $100(\$600)=\$60{,}000$. Because each claim costs exactly $\$600$, aggregate-loss standard deviation is $\$600(10)=\$6{,}000$; the permitted dollar deviation is $\$12{,}000=2(\$6{,}000)$. The thresholds agree. Common mistake avoided: treating fixed claim size as a source of additional severity variance would incorrectly make the aggregate threshold larger.

Worked Example: Comparing Two Different Risk Profiles

Two hypothetical portfolios have the same Poisson frequency, $\lambda=0.80$ claims per exposure unit. Portfolio A has constant severity of $\$100$. Portfolio B has severity mean $\$100$ and standard deviation $\$50$. For each portfolio, calculate the frequency and aggregate thresholds at $z=2.00$ and a 10% tolerance. Explain which standard determines the larger volume.

Step 1: Frequency threshold. Frequency is the same in both portfolios:

n_{\mathrm{freq}} = \frac{2.00^2}{0.10^2(0.80)} = \frac{4}{0.008} = 500.

Step 2: Portfolio A aggregate threshold. Its severity variance is zero. The multiplier is $1+0/100^2=1$, so $n_{\mathrm{agg,A}}=500$ units.

Step 3: Portfolio B aggregate threshold. Its severity variance is $50^2=2{,}500$, and its squared mean is $100^2=10{,}000$. The multiplier is $1+2{,}500/10{,}000=1.25$, so $n_{\mathrm{agg,B}}=500(1.25)=625$ units.

Check and interpretation. Portfolio B needs 125 more units than Portfolio A under the aggregate standard, while their frequency thresholds remain identical. The difference is entirely due to Portfolio B's severity variability. Common mistake avoided: comparing the two aggregate thresholds without first noting that the frequency threshold is the same can obscure which part of the model causes the difference.

Conditions, Uses, and Common Traps

These formulas assume independent exposure units with a stable Poisson frequency, independent claim severities with a common distribution, and independence between count and severity. They also use a normal approximation to the count or aggregate loss. For very small expected counts, strongly skewed severities, dependence, changing risk conditions, or overdispersed counts, the approximation or model may be unsuitable. The thresholds should not be treated as exact probability statements in those settings.

In an application, the exposure unit must be consistent with $\lambda$. If $\lambda$ is an annual claim frequency per policy-year, $n$ is measured in policy-years. A different time basis requires a correspondingly converted frequency. Also keep the tolerance relative to the correct expected value: $kn\lambda$ for counts and $kn\lambda\mu$ for aggregate losses.

Do not omit severity variation. For aggregate loss, multiply the frequency threshold by $1+\sigma^2/\mu^2$.
Do not confuse standard deviation and variance. The formula uses $\sigma^2/\mu^2$, not $\sigma/\mu^2$.
Do not change the confidence convention. Use the critical value for the specified central probability and a two-sided tolerance.
Handle integer exposure plans transparently. The formulas give a continuous threshold. If a plan must use whole exposure units and must meet or exceed the threshold, round upward; do not round a threshold down and call it sufficient.
Check whether the standard matches the quantity of interest. Claim-count stability alone does not establish aggregate-loss stability when claim sizes vary.

Key takeaway: Under the stated compound Poisson assumptions, the aggregate-loss threshold equals the frequency threshold multiplied by $1+\sigma^2/\mu^2$. The standards coincide for constant severity; otherwise, aggregate-loss credibility requires more exposure.

Check Your Understanding

For each problem, use the stated normal critical value and round a required whole-unit volume upward when the continuous threshold is not an integer.

A hypothetical risk has $\lambda=2.00$ claims per exposure unit and constant severity. Find both thresholds for $z=1.645$ and $k=0.10$. (Answer: 136 units for each standard.)
A hypothetical service plan has $\lambda=0.25$, severity mean $\$400$, and severity standard deviation $\$300$. Find both thresholds for $z=2.00$ and $k=0.20$. (Answer: Frequency 400 units; aggregate loss 625 units.)
A hypothetical portfolio has $\lambda=1.50$, severity mean $\$200$, and standard deviation $\$100$. Find both whole-unit thresholds for $z=1.96$ and $k=0.05$. (Answer: Frequency 1,025 units; aggregate loss 1,281 units.)
Two hypothetical risks each have $\lambda=0.60$. Risk A has constant severity; Risk B has severity mean $\$500$ and standard deviation $\$500$. Find their frequency thresholds and aggregate thresholds for $z=2.00$ and $k=0.15$. (Answer: Frequency 297 units for each; aggregate 297 units for A and 593 units for B.)
A hypothetical risk has $\lambda=0.40$, severity mean $\$800$, and standard deviation $\$800$. Find both thresholds for $z=2.00$ and $k=0.10$, and state why they differ. (Answer: Frequency 1,000 units; aggregate loss 2,000 units, because the squared severity coefficient of variation is 1.)

What You'll Learn