Avoiding Frequency and Severity Variance Mix-Ups

Why the Quantity Being Tested Matters

A limited-fluctuation standard asks whether observed experience is sufficiently stable around its expected value. The calculation depends on what is being observed. A claim count fluctuates because the number of claims varies. An average severity fluctuates because claim amounts vary. Aggregate loss fluctuates because both the number and amounts of claims vary. A variance formula suited to one of these quantities cannot automatically be used for another.

This distinction is especially important when an exposure is described in terms of expected claims. The same expected claim count can support different conclusions about frequency and aggregate loss if the severity distribution changes. The methods below use the normal approximation commonly used in limited-fluctuation standards; the approximation should be recognized as such, especially for small expected counts.

Definition: Let $T$ be the experience quantity being tested, with positive mean $\theta=E[T]$ and finite variance. A relative-error standard with tolerance $r$ and approximate central probability $p$ requires $P(|T-\theta|\leq r\theta)\geq p$. Under a normal approximation, let $z=\Phi^{-1}((1+p)/2)$. The standard is met when $\displaystyle z\frac{\sqrt{\operatorname{Var}(T)}}{\theta}\leq r$. The ratio $\sqrt{\operatorname{Var}(T)}/\theta$ is the coefficient of variation of $T$.

The formula is not a distribution-free guarantee: the probability statement uses a normal approximation. In particular, a small Poisson claim count can be notably skewed. The expected-count thresholds below are the usual approximate limited-fluctuation standards, not exact Poisson tail calculations.

Three Quantities, Three Variance Calculations

Suppose claim counts follow a Poisson model, and severity amounts are independent and identically distributed, independent of the count, with mean $\mu$ and variance $\sigma^2$. These assumptions make the differences among the three quantities explicit.

Quantity	Model or condition	Relative standard deviation
Claim count $N$	$N\sim\operatorname{Poisson}(m)$, where $m=E[N]$	$1/\sqrt{m}$
Average severity $\bar X$	Exactly $n$ independent severities are averaged	$\sigma/(\mu\sqrt{n})$
Aggregate loss $S$	Poisson count with expected count $m$; independent severities	$\sqrt{1+(\sigma/\mu)^2}/\sqrt{m}$

For claim frequency, if $N$ is the number of claims in exposure $e$, then the observed frequency is $N/e$. When $N\sim\operatorname{Poisson}(\lambda e)$, its mean is $\lambda$ and its variance is $\lambda/e$. Its relative standard deviation is therefore $1/\sqrt{\lambda e}$, or $1/\sqrt{m}$ where $m=\lambda e$ is the expected claim count.

For an average severity based on a fixed number $n$ of claims, the variance is $\sigma^2/n$, so its relative standard deviation is the severity coefficient of variation divided by $\sqrt n$. This is a different source of variation from the Poisson variation in the number of claims.

Why Aggregate Loss Has an Additional Variance Term

Let $S=X_1+\cdots+X_N$ be aggregate loss, where $N\sim\operatorname{Poisson}(m)$ and each $X_j$ has mean $\mu$ and variance $\sigma^2$. Conditional on $N$, the sum has mean $N\mu$ and variance $N\sigma^2$. The law of total variance gives the aggregate variance:

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

The first variance term, $m\sigma^2$, reflects variation in severity conditional on the count. The second, $m\mu^2$, reflects variation in the count itself. Since $E[X^2]=\sigma^2+\mu^2$, the same result can be written $\operatorname{Var}(S)=mE[X^2]$. Dividing the variance by the square of the mean gives $\operatorname{Var}(S)/(E[S])^2=(1+(\sigma/\mu)^2)/m$.

Thus, if $c=\sigma/\mu$ is the severity coefficient of variation, an approximate full-credibility standard for aggregate loss requires $m\geq z^2(1+c^2)/r^2$. The corresponding claim-count standard requires only $m\geq z^2/r^2$. They agree only in the special case $c=0$, where severities do not vary.

Key distinction: Use the Poisson count variance for claim counts or frequency. Use $\sigma^2/n$ for an average of a fixed number of independent severities. For compound Poisson aggregate loss, include both count and severity variation: $\operatorname{Var}(S)=m(\sigma^2+\mu^2)$.

Worked Standards and the Errors They Prevent

Before inserting values into a limited-fluctuation formula, name the experience quantity and its model. Then calculate its mean, variance, and relative standard deviation. The examples use hypothetical insurance portfolios and illustrate how the selected quantity changes the answer.

Worked Example: Claim-Frequency Credibility and Exposure

A hypothetical portfolio has a Poisson claim frequency of $0.8$ claims per policy-year. For a claim-count standard, the actuary requires the count to be within 25% of its mean with approximate central probability 95.45%, for which $z=2$. Find the minimum expected claim count and the corresponding exposure.

Method and model. The quantity is the claim count $N$, not average severity or aggregate loss. For a Poisson count with mean $m$, $\operatorname{Var}(N)=m$, so its relative standard deviation is $1/\sqrt m$.

Step 1: Find the required expected count. The approximate standard requires $z/\sqrt m\leq r$. Thus $m\geq(z/r)^2=(2/0.25)^2=8^2=64$. The minimum is 64 expected claims.

Step 2: Check the boundary. At $m=64$, the standard deviation of the count is $\sqrt{64}=8$, and the relative standard deviation is $8/64=0.125$. The approximate 95.45% margin is $2(0.125)=0.25$, exactly the allowed relative error.

Step 3: Convert expected claims to exposure. At $0.8$ expected claims per policy-year, exposure $e$ satisfies $0.8e=64$. Therefore $e=64/0.8=80$ policy-years.

Check and trap avoided. The frequency and exposure are consistent: $0.8(80)=64$. Applying a severity coefficient of variation here would introduce a severity assumption that this claim-count standard does not use. Conversely, this count-only result cannot be carried over as an aggregate-loss standard without considering severity variability.

Worked Example: Credibility of an Average Severity

A hypothetical equipment insurer observes individual claim amounts with mean $\$400$ and standard deviation $\$800$. How many independent claims are needed for the sample average severity to be within 20% of its mean with approximate central probability 90%, using $z=1.645$?

Method and model. The quantity is the average of a fixed number $n$ of independent severities. Its mean is $\$400$, and its variance is $800^2/n$. The severity coefficient of variation is $c=800/400=2$.

Step 1: Apply the average-severity standard. Its relative standard deviation is $c/\sqrt n$. Require $1.645(2/\sqrt n)\leq0.20$, which gives $n\geq(1.645\cdot2/0.20)^2=16.45^2=270.6025$. Because the number of claims must be an integer, round the required count up to 271.

Step 2: Check the rounded answer. At $n=271$, the approximate relative margin is $1.645(2/\sqrt{271})\approx3.29/16.4621\approx0.1999$, which is within 20%. At $n=270$, it is $3.29/\sqrt{270}\approx0.2002$, which is just above 20%. Thus 271 is the smallest integer meeting the approximate standard.

Trap and correction. Using the Poisson count formula alone would give $(1.645/0.20)^2=67.6506$, or 68 claims. That calculation incorrectly assumes the relative standard deviation is $1/\sqrt n$. Here the severity coefficient of variation is 2, so the standard deviation of the average is four times the count-only variance factor on a relative-variance scale. The correct requirement is 271 claims, not 68.

Worked Example: Aggregate-Loss Credibility With Variable Severities

A hypothetical portfolio is expected to produce 87 claims during the period being assessed. Individual claim severity has mean $\$600$ and standard deviation $\$600$. Assess whether aggregate loss meets a 25% relative-error standard with approximate central probability 90%, using $z=1.645$.

Method and model. The aggregate is compound Poisson: the count is Poisson with mean $m=87$, and severities are independent of the count and of one another. Here $c=\sigma/\mu=1$.

Step 1: Calculate the aggregate mean. $E[S]=m\mu=87(\$600)=\$52{,}200$.

Step 2: Calculate the aggregate variance. $E[X^2]=\sigma^2+\mu^2=\$600^2+\$600^2=\$720{,}000$. Hence $\operatorname{Var}(S)=87(\$720{,}000)=\$62{,}640{,}000$.

Step 3: Calculate relative standard deviation and margin. The relative standard deviation is $\sqrt{62{,}640{,}000}/52{,}200=\sqrt{2/87}\approx0.1516$. Multiplying the unrounded relative standard deviation by $1.645$ gives approximately $0.2494$. This is within the 25% tolerance, so the standard is met under the normal approximation.

Check and trap avoided. The aggregate mean and variance use the same dollar units appropriately: the mean is in dollars and the variance is in squared dollars. A claim-count-only calculation would give $1.645/\sqrt{87}\approx0.1764$, understating the aggregate margin because it omits severity variation. The correct aggregate calculation includes the factor $1+c^2=2$.

A Reliable Exam Check

A compact way to avoid a variance mix-up is to write the random quantity explicitly before choosing a formula. If it is a count, identify its count model. If it is an average severity, identify whether the number of severities is fixed. If it is an aggregate, account for both the random count and the random severities when the model includes both. Only after those choices should the tolerance and probability requirement be applied.

Name the tested quantity.
State whether it is a count, frequency, average severity, or aggregate loss. Do not treat these quantities as interchangeable.

State the model and assumptions.
For example, specify a Poisson count, a fixed number of independent severities, or an independent compound Poisson model.

Calculate mean and variance for that quantity.
Use the resulting relative standard deviation, not a formula recalled from a different quantity.

Apply the probability and tolerance.
Check whether $z\sqrt{\operatorname{Var}(T)}/E[T]\leq r$, and round a required count or exposure upward when necessary.

A common exam trap is to see an expected claim count and immediately use the frequency standard $z/\sqrt m$. That is appropriate for a Poisson count, but it is not automatically appropriate for a severity average or aggregate loss. Another trap is to use only the severity variance for aggregate loss; doing so omits the count-variation term $m\mu^2$. The model and the quantity, rather than the word “claims” in the problem statement, determine the variance.

Key takeaway: Frequency variability, severity variability, and aggregate-loss variability are different. For a compound Poisson aggregate, the relative variance is $(1+c^2)/m$, not $1/m$. Always match the variance formula to the quantity whose credibility is being assessed.

Check Your Understanding

For each problem, identify the quantity before applying its variance formula. Answers use the stated normal approximation where applicable.

A hypothetical portfolio has a Poisson expected count of 100 claims. For a claim-count standard with $z=1.96$, what is the approximate relative margin? Does it meet a 20% tolerance? (Answer: relative margin $0.1960$; yes.)
Individual severities have mean $\$250$ and standard deviation $\$500$. How many independent claims are required for an average-severity standard with $z=1.645$ and tolerance 25%? (Answer: $174$ claims.)
A compound Poisson aggregate has expected count $m=60$, severity mean $\$800$, and severity standard deviation $\$600$. Find its relative standard deviation and its approximate margin for $z=1.645$. Does it meet a 27% tolerance? (Answer: relative standard deviation $0.1614$, margin $0.2655$; yes.)
A compound Poisson aggregate has expected count 40 and severity coefficient of variation 1.5. For $z=1.645$, calculate the approximate relative margin and decide whether a 40% tolerance is met. (Answer: relative standard deviation $\sqrt{3.25/40}\approx0.2850$, margin $\approx0.4689$; no.)
For compound Poisson aggregate loss, severity coefficient of variation is 0.5. How many expected claims are required for a 20% tolerance with $z=1.645$? (Answer: at least $85$ expected claims.)

Quantity	Model or condition	Relative standard deviation
Claim count \(N\)	\(N\sim\operatorname{Poisson}(m)\), where \(m=E[N]\)	\(1/\sqrt{m}\)
Average severity \(\bar X\)	Exactly \(n\) independent severities are averaged	\(\sigma/(\mu\sqrt{n})\)
Aggregate loss \(S\)	Poisson count with expected count \(m\); independent severities	\(\sqrt{1+(\sigma/\mu)^2}/\sqrt{m}\)

What You'll Learn