Large-Frequency Limits for Compound Aggregate Loss

Why Large Claim Counts Produce a Normal Limit

An aggregate loss is a sum of claim amounts, but the number of terms in that sum may itself be random. When the expected claim count becomes large, the standardized aggregate can approach a normal distribution. This limiting result explains the normal approximation from aggregate moments: it is not merely a convenient curve fitted to a mean and variance, but can arise from summing many appropriately behaved losses.

The distinction between a fixed and random claim count matters. With a fixed number of independent severities, aggregate variance depends on the severity variance. With a Poisson claim count, variability in the number of claims adds another component, so the aggregate variance depends on the severity second raw moment. Using the wrong variance gives the wrong standardization even when the aggregate mean is correct.

Definition: In a compound Poisson model, $N_\lambda\sim\operatorname{Poisson}(\lambda)$ is the number of claims, and $X_1,X_2,\ldots$ are independent, identically distributed claim severities, independent of $N_\lambda$. The aggregate loss is $S_\lambda=\sum_{j=1}^{N_\lambda}X_j$, with $S_\lambda=0$ if $N_\lambda=0$. Write $m_1=E[X]$ and $m_2=E[X^2]$. For the standard large-frequency result, assume $0<m_2<\infty$.

The parameter $\lambda$ is the expected number of claims, not necessarily the realized count. In this model, the exact aggregate mean and variance are

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

These formulas use the severity's first and second raw moments. The second raw moment is $m_2=\operatorname{Var}(X)+(E[X])^2$, not just $\operatorname{Var}(X)$.

Why the Compound Poisson Variance Uses the Second Moment

Conditioning on the claim count shows where the two sources of variation enter. Given $N_\lambda=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $nm_1$ and its conditional variance is $n\operatorname{Var}(X)$. The law of total expectation and the law of total variance then give

E[S_\lambda]=E[N_\lambda]m_1=\lambda m_1,

\begin{aligned} \operatorname{Var}(S_\lambda) &=E[\operatorname{Var}(S_\lambda\mid N_\lambda)] +\operatorname{Var}(E[S_\lambda\mid N_\lambda])\\ &=\lambda\operatorname{Var}(X)+\lambda m_1^2\\ &=\lambda\bigl(\operatorname{Var}(X)+m_1^2\bigr) =\lambda m_2. \end{aligned}

The first term represents variation in severities when the count is given. The second represents variation from the random count itself. A fixed-count model does not have this second component.

The Large-Frequency Limit

For a compound Poisson model with a fixed severity distribution and increasing $\lambda$, the standardized aggregate satisfies

\frac{S_\lambda-\lambda m_1}{\sqrt{\lambda m_2}} \ \xrightarrow{d}\ N(0,1) \qquad\text{as }\lambda\to\infty.

Here, convergence in distribution means that probabilities at continuity points of the limiting normal distribution converge to the corresponding normal probabilities. The aggregate itself is not asserted to be exactly normal for any finite $\lambda$.

Start with the compound Poisson characteristic function.
For a real number $u$, the characteristic function of $S_\lambda$ is $\exp\{\lambda(E[e^{iuX}]-1)\}$. This follows by conditioning on $N_\lambda$ and using the Poisson probability generating function.

Center and scale the aggregate.
Let $Z_\lambda=(S_\lambda-\lambda m_1)/\sqrt{\lambda m_2}$. Its characteristic function at $t$ is $\exp\{\lambda(E[e^{itX/\sqrt{\lambda m_2}}]-1-itm_1/\sqrt{\lambda m_2})\}$.

Expand near zero.
Finite $E[X^2]$ gives $E[e^{iuX}]=1+iu m_1-u^2m_2/2+o(u^2)$ as $u\to0$. Substituting $u=t/\sqrt{\lambda m_2}$, the exponent becomes $-t^2/2+o(1)$.

Identify the limiting distribution.
The characteristic function therefore tends to $e^{-t^2/2}$, the characteristic function of $N(0,1)$. The continuity theorem for characteristic functions yields the stated convergence in distribution.

The finite second-moment assumption is essential to this standard argument: it supplies a finite scaling variance and the quadratic expansion used in the derivation. If the severity distribution has infinite second moment, this normal limit with this centering and scaling is not justified by the result above.

Using the Limit for Aggregate Probabilities

For a large expected claim count, approximate the aggregate by a normal random variable with the compound Poisson mean and variance:

S_\lambda\ \approx\ N(\lambda m_1,\lambda m_2).

Thus, for a threshold $s$, standardize using $z=(s-\lambda m_1)/\sqrt{\lambda m_2}$, then approximate $P(S_\lambda\le s)$ by $\Phi(z)$, where $\Phi$ is the standard normal distribution function. The approximation is a limiting statement, not an exact identity. If the aggregate is lattice-valued, a continuity correction may be appropriate when translating an inequality into a normal probability.

Worked Example: A Compound Poisson Portfolio

A hypothetical portfolio has a Poisson claim count with mean $\lambda=400$. Each claim is independently \$500 or \$1,500, each with probability 0.5. Approximate the probability that aggregate claims exceed \$445,000. Use a continuity correction for the aggregate's lattice.

Method and moments. The severities are independent of the Poisson count, and $m_1=0.5(500)+0.5(1{,}500)=1{,}000$ dollars. Also, $m_2=0.5(500^2)+0.5(1{,}500^2)=1{,}250{,}000$ dollars squared.

Aggregate parameters. The mean is $400(1{,}000)=\$400{,}000$. The variance is $400(1{,}250{,}000)=500{,}000{,}000$ dollars squared, so the standard deviation is $\sqrt{500{,}000{,}000}=\$22{,}360.68$.

Continuity correction and standardization. Every aggregate is a multiple of \$500. Exceeding \$445,000 means the aggregate is at least \$445,500, so use the boundary \$445,250 for a continuous normal approximation. Then $z=(445{,}250-400{,}000)/22{,}360.68=2.0236$.

Probability and check. The upper-tail probability is approximately $1-\Phi(2.0236)=0.0215$, to four decimal places. The standardized boundary is positive, as expected for a threshold above the mean; the resulting upper-tail probability is less than 0.5.

Common mistake avoided: Using $400\operatorname{Var}(X)$ would omit the variation in the Poisson claim count. Here $\operatorname{Var}(X)=250{,}000$, which would produce a variance of only $100{,}000{,}000$, not the correct $500{,}000{,}000$.

Fixed Exposure Counts and Their Different Scaling

The large-frequency idea also applies to a fixed number of independent policy losses, but the standardization must match that model. If $W_1,\ldots,W_n$ are independent and identically distributed with finite mean $\mu_W$ and finite, positive variance $\sigma_W^2$, then the ordinary central limit theorem gives

\frac{\sum_{j=1}^n W_j-n\mu_W}{\sqrt{n\sigma_W^2}} \ \xrightarrow{d}\ N(0,1).

For example, if each policy has a claim with probability $p$, and a claim, when it occurs, has severity $X$, then one policy's loss is $W=IX$, where $I$ is Bernoulli with parameter $p$, independent of $X$. Its mean is $pE[X]$, while its variance is $pE[X^2]-p^2(E[X])^2$. The fixed number of policies creates a binomial-type claim count, not a Poisson count.

Worked Example: Independent Policies With Claim Occurrence Risk

A hypothetical group contains 1,000 independent policies. Each policy produces a \$5,000 claim with probability 0.20 and no claim otherwise. Approximate the probability that total claims exceed \$1,100,000.

Method and one-policy moments. Let $W$ be one policy's loss. Then $P(W=5{,}000)=0.20$ and $P(W=0)=0.80$. Its mean is $E[W]=0.20(5{,}000)=\$1{,}000$, and $E[W^2]=0.20(5{,}000^2)=\$5{,}000{,}000$. Therefore $\operatorname{Var}(W)=5{,}000{,}000-1{,}000^2=4{,}000{,}000$ dollars squared.

Aggregate parameters. Independence gives mean $1{,}000(1{,}000)=\$1{,}000{,}000$ and variance $1{,}000(4{,}000{,}000)=4{,}000{,}000{,}000$. The standard deviation is $\sqrt{4{,}000{,}000{,}000}=\$63{,}245.55$.

Standardize and calculate. The threshold gives $z=(1{,}100{,}000-1{,}000{,}000)/63{,}245.55=1.5811$. Hence the approximate probability of exceeding the threshold is $1-\Phi(1.5811)=0.0569$, to four decimal places.

Check and common mistake avoided. The threshold is above the mean, and the estimated upper tail is correspondingly below 0.5. Treating the count as Poisson with mean 200 would instead give variance $200(5{,}000^2)=5{,}000{,}000{,}000$; that is not the variance of the stated fixed group with independent Bernoulli claim occurrences.

Worked Example: A Higher-Mean Claim Count

A hypothetical exposure has a Poisson claim count with mean $\lambda=900$. Claim severities are independent exponential random variables with mean \$1,000, independent of the count. Approximate the probability that aggregate claims exceed \$970,000.

Specify the severity moments. For an exponential severity with mean \$1,000, $E[X]=\$1{,}000$ and $\operatorname{Var}(X)=1{,}000{,}000$ dollars squared. Thus $m_2=\operatorname{Var}(X)+(E[X])^2=2{,}000{,}000$ dollars squared.

Find the aggregate mean and standard deviation. The mean is $900(1{,}000)=\$900{,}000$. The variance is $900(2{,}000{,}000)=1{,}800{,}000{,}000$, and the standard deviation is $\sqrt{1{,}800{,}000{,}000}=\$42{,}426.41$.

Standardize and calculate. The threshold is $70{,}000$ above the mean, so $z=70{,}000/42{,}426.41=1.6499$. The normal upper tail is $1-\Phi(1.6499)=0.0495$, to four decimal places.

Check and common mistake avoided. The standardized threshold is positive and about 1.65 standard deviations above the mean, consistent with an upper-tail estimate near 0.05. Using $900\operatorname{Var}(X)$ would give $900{,}000{,}000$ for the aggregate variance; it omits the random-count contribution $900(E[X])^2=900{,}000{,}000$.

Assumptions, Interpretation, and Exam Use

A limiting argument is most informative when its assumptions match the modeled portfolio. For the compound Poisson result above, the severity distribution is held fixed as $\lambda$ increases; severities are identically distributed and independent of one another and of the count; and the severity second moment is finite. The result does not say that any aggregate with a large mean is normal. A large mean caused by very large individual claims is not, by itself, the same as a large number of ordinary independent contributions.

In applications, check the claim-count model before selecting the variance. For a Poisson count use $\lambda E[X^2]$. For a fixed number of independent policy losses use the number of policies times the variance of one policy's loss. Then standardize the threshold with the matching mean and standard deviation. A calculation that uses a correct normal tail function but the wrong aggregate variance remains incorrect.

These limiting results are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In exam problems, state the model, show the aggregate moments, and make clear whether the approximation is justified by a growing random claim count or by a growing number of independent fixed-exposure losses.

Key takeaway: For compound Poisson aggregate loss, as $\lambda$ grows, $(S_\lambda-\lambda E[X])/\sqrt{\lambda E[X^2]}$ approaches $N(0,1)$, provided $0<E[X^2]<\infty$ and the compound Poisson assumptions hold. The second raw severity moment—not the severity variance alone—sets the aggregate variance.

Check Your Understanding

For each problem, identify the appropriate aggregate moments before applying the normal limit. Probabilities are rounded to four decimal places.

A hypothetical compound Poisson portfolio has $\lambda=250$, and every claim is exactly \$2,000. Approximate the probability that aggregate loss exceeds \$550,000. (Answer: 0.0569)
A hypothetical compound Poisson portfolio has $\lambda=100$, with exponential severities of mean \$500. Approximate the probability that aggregate loss exceeds \$60,000. (Answer: 0.0786)
A group of 500 independent policies each has probability 0.10 of producing a \$1,000 claim, with no other possible loss. Approximate the probability that total claims exceed \$60,000. (Answer: 0.0680)
A hypothetical compound Poisson model has $\lambda=625$ and exponential severities with mean \$4,000. Approximate the probability that aggregate loss exceeds \$2,800,000. (Answer: 0.0169)
A fixed group has 1,600 independent policy losses, each with mean \$250 and variance $90{,}000$ dollars squared. Approximate the probability that total loss exceeds \$424,000. (Answer: 0.0228)

What You'll Learn