Approximation Risks With Low Frequency and Heavy Skewness

Why Aggregate Moments Can Be Misleading

An aggregate loss model summarizes total claims over a fixed period, but its mean and variance do not describe every feature of its distribution. This matters particularly when the expected claim count is small or when a severity distribution assigns modest probability to very large losses. In those settings, a moment-based approximation can give a reasonable-looking center and still estimate a tail probability poorly.

The issue is not that moments are incorrect. They correctly describe the aggregate mean and variance under the model. The issue is that an approximation, such as a normal distribution, uses those moments to stand in for the full distribution. A compound aggregate may have a large atom at zero, pronounced asymmetry, or rare jumps that a smooth moment-matched curve does not capture.

Definition: Let $N$ be a claim count and let $X_1,X_2,\ldots$ be independent, identically distributed nonnegative claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. In a compound Poisson model, $N\sim\operatorname{Poisson}(\lambda)$. Write $m_r=E[X^r]$ for the $r$th raw severity moment when it exists.

For this tutorial, the normal comparison means approximating $S$ by a normal random variable with the same mean and variance. This approximation has support on the whole real line, whereas $S$ is nonnegative. For lattice-valued losses, a continuity correction can improve a boundary calculation; it does not fix every mismatch in shape.

What the Compound-Poisson Moments Tell Us

Under the compound Poisson assumptions above, provided the required severity moments are finite,

E[S]=\lambda m_1,\qquad \operatorname{Var}(S)=\lambda m_2.

The variance uses the second raw moment $m_2=E[X^2]$, not the severity variance alone. A useful related result is that the third cumulant of $S$ is $\lambda m_3$, so its standardized skewness is

\gamma_S=\frac{\lambda m_3}{(\lambda m_2)^{3/2}} =\frac{m_3}{\sqrt{\lambda}\,m_2^{3/2}}.

This expression shows two potential warnings. A small $\lambda$ can make aggregate skewness large, and a large third severity moment can do the same. A large claim count often helps a normal approximation, but does not by itself guarantee an accurate tail estimate, particularly for heavy-tailed severities.

State the model.
Check whether the count is Poisson, severities are independent and identically distributed, and count and severities are independent.

Compute the relevant moments.
Use raw severity moments for compound-Poisson variance, and check that the moments required by the approximation exist.

Inspect the feature being estimated.
For a tail event, consider the zero atom, lattice spacing, rare severity values, and skewness—not just the aggregate mean and variance.

Compare with structure the model reveals exactly.
For example, in a compound Poisson model with strictly positive severities, $P(S=0)=e^{-\lambda}$.

Why the Compound-Poisson Variance Uses a Raw Moment

The variance formula follows by conditioning on the claim count. Let the severity mean be $m_1$ and its variance be $\sigma_X^2$. Given $N=n$, the aggregate is the sum of $n$ independent severities, so $E[S\mid N=n]=nm_1$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$. The law of total variance gives

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N]\sigma_X^2+\operatorname{Var}(N)m_1^2\\ &=\lambda\sigma_X^2+\lambda m_1^2\\ &=\lambda E[X^2]=\lambda m_2, \end{aligned}

because a Poisson count has both mean and variance $\lambda$, and $E[X^2]=\sigma_X^2+m_1^2$. The second term accounts for variation in the number of claims. Omitting it is a common way to understate aggregate variance.

Worked Example: A Low-Frequency Model With Fixed Claim Size

A hypothetical portfolio has a Poisson claim count with $\lambda=0.4$ during the period. Each claim is exactly $1,000, independently of the count. Estimate the probability of aggregate loss at least $1,000 by a continuity-corrected normal approximation, then compare it with the exact probability.

Method and assumptions. This is a compound Poisson model with deterministic severity $X=1000$. Thus $m_1=1000$ and $m_2=1{,}000{,}000$.

Aggregate moments.

E[S]=0.4(1000)=400,\qquad \operatorname{Var}(S)=0.4(1{,}000{,}000)=400{,}000, \qquad \operatorname{SD}(S)=\sqrt{400{,}000}=632.4555.

The aggregate takes values in multiples of $1,000. Therefore $S\ge 1000$ has continuity-corrected normal boundary $500. Standardizing gives

z=\frac{500-400}{632.4555}=0.1581, \qquad P_{\mathrm{normal}}(S\ge1000)\approx 1-\Phi(0.1581)=0.4372.

For the exact probability, the aggregate is at least $1,000 precisely when at least one claim occurs:

P(S\ge1000)=P(N\ge1)=1-e^{-0.4}=0.3297.

The normal estimate exceeds the exact result by about $0.4372-0.3297=0.1075$, or 10.75 percentage points. The approximation obscures the model’s discrete structure despite using its exact mean and variance.

Check. The exact probability is between 0 and 1 and equals the probability of at least one claim, as required by the fixed positive claim size. The common mistake to avoid is treating a mean-and-variance match as a guarantee of a good tail estimate.

Rare Large Claims Can Control the Tail

A severity distribution with a small probability of a very large payment can have a substantial second or third raw moment. Its aggregate mean may remain modest, while a tail event is driven mainly by whether one of those large claims occurs. A normal approximation spreads probability smoothly around the mean; it may assign too little probability to a rare jump, even when a continuity correction is used.

Worked Example: A Sparse Large-Severity Component

A hypothetical insurer’s count is Poisson with $\lambda=0.25$. Each claim is 1 loss unit with probability 0.8 and 21 loss units with probability 0.2, independently across claims and of the count. Compare the exact probability of aggregate loss at least 21 with a continuity-corrected normal estimate.

Method and assumptions. First calculate severity raw moments, then apply the compound-Poisson formulas. To find the exact tail, split the count into large and small claims. Poisson thinning gives independent counts $B\sim\operatorname{Poisson}(0.05)$ for 21-unit claims and $C\sim\operatorname{Poisson}(0.20)$ for 1-unit claims.

Moments.

\begin{aligned} m_1&=0.8(1)+0.2(21)=5,\\ m_2&=0.8(1^2)+0.2(21^2)=0.8+88.2=89,\\ E[S]&=0.25(5)=1.25,\\ \operatorname{Var}(S)&=0.25(89)=22.25,\\ \operatorname{SD}(S)&=\sqrt{22.25}=4.7170. \end{aligned}

For the exact event, any large claim guarantees $S\ge21$. If there are no large claims, the event occurs only if there are at least 21 small claims. Therefore

\begin{aligned} P(S\ge21) &=1-P(B=0,\ C\le20)\\ &=1-e^{-0.05}P(C\le20)\\ &=1-e^{-0.25}\sum_{c=0}^{20}\frac{0.2^c}{c!}\\ &\approx 0.0488. \end{aligned}

The small-count tail $P(C\ge21)$ is negligible at the displayed precision; the probability is essentially the chance of at least one large claim, $1-e^{-0.05}=0.0488$. For the normal approximation, the lattice boundary for $S\ge21$ is 20.5:

z=\frac{20.5-1.25}{4.7170}=4.081, \qquad P_{\mathrm{normal}}(S\ge21)\approx 1-\Phi(4.081)<0.0001.

The normal estimate is far below the exact tail. The decisive feature is the 0.05 expected count of large claims: at least one such claim occurs with probability about 0.0488. The common mistake to avoid is assuming that a tail event many standard deviations above the mean must be negligible without checking how individual severities can produce it.

Check. The second raw moment is 89, not the severity variance. Using $0.25(89)=22.25$ gives the correct aggregate variance; substituting only $\operatorname{Var}(X)$ would omit claim-count variation.

Atoms and Skewness Are Diagnostic Warnings

Even when no particular rare severity point dominates, low frequency can create a substantial probability of no claims. If severities are strictly positive, $S=0$ exactly when $N=0$, so $P(S=0)=e^{-\lambda}$. A continuous normal distribution has no atom at zero and assigns positive probability to negative losses. These differences can be important when estimating probabilities near zero or other thresholds.

The following example uses a hypothetical lognormal severity to illustrate the contrast. For a lognormal variable with parameters $\mu$ and $\sigma^2$, $\ln X$ is normal with mean $\mu$ and variance $\sigma^2$, and $E[X^r]=\exp(r\mu+r^2\sigma^2/2)$.

Worked Example: A Heavy-Skewness Warning From a Lognormal Severity

Suppose $N\sim\operatorname{Poisson}(0.5)$, and severities are independent lognormal variables with $\mu=-\ln 2$ and $\sigma^2=\ln 4$, independent of $N$. Compare the exact probability of a positive aggregate with the normal approximation, and check whether the fitted normal model can represent the aggregate’s support.

Method and assumptions. Use lognormal raw moments to obtain the compound-Poisson mean and variance. The severity is strictly positive, so $S>0$ if and only if at least one claim occurs.

Severity and aggregate moments.

E[X]=\exp\left(-\ln2+\frac{\ln4}{2}\right)=1, \qquad E[X^2]=\exp\left(-2\ln2+2\ln4\right)=4.

Consequently,

E[S]=0.5(1)=0.5,\qquad \operatorname{Var}(S)=0.5(4)=2,\qquad \operatorname{SD}(S)=\sqrt2=1.4142.

The exact probability of a positive aggregate is

P(S>0)=P(N\ge1)=1-e^{-0.5}=0.3935.

The normal approximation assigns

P_{\mathrm{normal}}(S>0) =1-\Phi\left(\frac{0-0.5}{\sqrt2}\right) =\Phi(0.3536) \approx0.6382.

It also assigns probability $\Phi(-0.3536)\approx0.3618$ to negative aggregate losses, which are impossible under this model. The exact aggregate has an atom $P(S=0)=e^{-0.5}=0.6065$; the continuous normal distribution cannot reproduce that atom. This does not make the normal approximation useless for every threshold, but it is a strong warning when the probability of no claims or a nearby tail is the quantity of interest.

Check. The lognormal parameters give $E[X]=1$ and $E[X^2]=4$, so the aggregate variance is 2. The common mistake to avoid is expecting a continuous, symmetric distribution to reflect both the zero-claim mass and the severe right skew of a low-frequency aggregate.

Exam Traps and Practical Checks

The purpose of a moment approximation is to provide a tractable estimate, not to replace the model’s structural information. When exact recursion or a direct event calculation is available, it can serve as a useful benchmark. Even without a full calculation, the model may reveal an atom, a rare-severity mechanism, or a support restriction that makes an approximation questionable.

Using the severity variance in place of the second raw moment. For a compound Poisson aggregate, $\operatorname{Var}(S)=\lambda E[X^2]$, not $\lambda\operatorname{Var}(X)$.
Ignoring the zero-claim probability. At low frequency, $e^{-\lambda}$ can be large and can strongly affect probabilities near zero.
Trusting a normal tail because the moments were matched. Matching two moments does not match skewness, atoms, or rare jumps.
Applying a continuity correction as a cure-all. It adjusts a lattice boundary; it does not repair a distributional shape mismatch.
Failing to inspect severity-driven event structure. A threshold may be reached mostly by one rare large claim, so its probability can be controlled by that claim’s arrival rate.

Key takeaway: Before relying on an aggregate moment approximation, check the frequency, severity skewness, zero atom, support, and mechanism that produces the tail event. These checks are especially important when the expected count is low or the severity distribution is highly skewed, and are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, calculate the requested quantities and consider what the approximation might miss.

A compound Poisson model has $\lambda=0.3$ and fixed severity $400. Find the aggregate mean, variance, standard deviation, and exact probability of zero loss. (Answer: mean $120, variance 48,000 dollars squared, standard deviation $219.09, and $P(S=0)=0.7408$.)
A compound Poisson model has $\lambda=0.2$. Severity is 2 units with probability 0.9 and 12 units with probability 0.1. Find $m_1$, $m_2$, the aggregate mean, and aggregate variance. (Answer: $m_1=3$, $m_2=18$, $E[S]=0.6$, and $\operatorname{Var}(S)=3.6$ units squared.)
A hypothetical model has $N\sim\operatorname{Poisson}(0.6)$ and fixed severity 5. Find the exact probability $P(S>0)$ and the normal approximation to that probability using the aggregate mean and variance. (Answer: exact $P(S>0)=0.4512$; normal estimate $\Phi(3/\sqrt{15})\approx0.7807$.)
A compound Poisson model has $\lambda=0.4$, with severity 1 unit with probability 0.9 and 31 units with probability 0.1. Find the aggregate mean and variance, and compare the exact probability $P(S\ge31)$ with a continuity-corrected normal estimate. (Answer: mean $1.6$, variance $38.8$; exact probability $1-e^{-0.04}P(C\le30)\approx0.0392$, where $C\sim\operatorname{Poisson}(0.36)$; normal estimate $1-\Phi((30.5-1.6)/\sqrt{38.8})<0.0001$.)

What You'll Learn