Compound Poisson Loss Simulation

Why the Poisson frequency model matters

A compound loss simulation has two sources of randomness: the number of claims and the size of each claim. In a compound Poisson model, the claim count is Poisson, while each claim severity is drawn from a specified severity distribution. A simulated total is therefore not obtained by simulating a fixed number of claims; the frequency draw determines how many severity draws are needed.

This tutorial focuses on simulating that Poisson count and combining it with severities. The method is useful when a model describes claims arriving at a constant average rate over an exposure period. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose $N\sim\operatorname{Poisson}(\lambda t)$ is the number of claims in an exposure period of length $t$, where $\lambda$ is the expected number of claims per unit of exposure. Let $X_1,X_2,\ldots$ be independent, identically distributed claim severities with distribution function $F_X$, independent of $N$. The aggregate loss is $$ S=\sum_{j=1}^{N}X_j, $$ with $S=0$ when $N=0$. This is a compound Poisson model.

The model assumes a constant claim rate over the exposure period, independent claim arrivals, and severities that are independent of the count and of one another. The parameter of the Poisson distribution is the expected count for the period being simulated, $\lambda t$, not automatically the annual rate. Each generated severity and the resulting aggregate must use consistent monetary units.

Generating the claim count

For $N\sim\operatorname{Poisson}(\theta)$, the probability of exactly $n$ claims is $P(N=n)=e^{-\theta}\theta^n/n!$, for $n=0,1,2,\ldots$. Here $\theta=\lambda t$. To generate a count from a uniform variate $U$ on $[0,1)$, choose the smallest integer $n$ for which $P(N\le n)\ge U$. Thus, the cumulative probability intervals—not the probability of each individual count—determine the inverse-transform result.

A convenient way to compute the probabilities is to start with $p_0=e^{-\theta}$ and use the recurrence $p_n=p_{n-1}\theta/n$. Add the probabilities in order until their cumulative sum reaches the supplied uniform variate. In software, a Poisson random-number generator can perform this step directly; the cumulative method makes the role of the model parameter explicit.

Set the exposure parameter.
Convert the frequency rate to the exposure period and calculate $\theta=\lambda t$.

Generate the count.
Draw $U$ and find the smallest $n$ with Poisson cumulative probability at least $U$.

Generate severities.
If $N=n$, draw exactly $n$ independent values from $F_X$, using a suitable generator for that severity model.

Sum the losses.
Set $S=X_1+\cdots+X_n$; if $n=0$, set $S=0$.

Worked simulations

Worked Example: Annual Poisson Count With Exponential Severities

A hypothetical portfolio is modeled for one year. The annual claim count is Poisson with mean 2. Each claim severity is exponential with mean $100, independently of the count and other severities. Use count uniform $U_N=0.73$, followed, if needed, by severity uniforms $0.20,0.50,0.80$. Simulate the aggregate. Report money to the nearest cent.

Method: inverse-transform Poisson count, followed by inverse-transform severity generation. The one-year Poisson parameter is $\theta=2$. The count probabilities, to four decimals, begin $P(N=0)=0.1353$, $P(N=1)=0.2707$, $P(N=2)=0.2707$, and $P(N=3)=0.1804$. The cumulative probabilities through 0, 1, 2, and 3 are respectively $0.1353$, $0.4060$, $0.6767$, and $0.8571$. Since $0.6767<0.73\le0.8571$, the simulated count is $N=3$.

For an exponential severity with mean 100, $F_X(x)=1-e^{-x/100}$ for $x\ge0$, so the inverse transform is $X=-100\ln(1-U)$. The first severity is $-100\ln(0.80)=\$22.31$ (unrounded value \$22.3144). The second is $-100\ln(0.50)=\$69.31$ (unrounded value \$69.3147). The third is $-100\ln(0.20)=\$160.94$ (unrounded value \$160.9438).

Summing the unrounded severities gives $S=22.3144+69.3147+160.9438=\$252.5729$, so the simulated aggregate is $252.57 to the nearest cent. The count selected three severity uniforms, as required.

Common mistake avoided: stopping after generating $N=3$ would give a frequency result, not an aggregate-loss result. The aggregate requires three severity draws and their sum.

Worked Example: Converting an Annual Rate to a Quarter

A hypothetical insurer models claims with a Poisson rate of 12 claims per year. Each claim severity is $500 with probability 0.60, $1,000 with probability 0.30, or $2,000 with probability 0.10. Simulate one quarter using count uniform $U_N=0.61$ and severity uniforms $0.10,0.70,0.95$, in that order. Give the aggregate in dollars.

Method: exposure adjustment, inverse-transform count generation, and cumulative-category severity sampling. One quarter is $t=0.25$ years, so the Poisson parameter is $\theta=12(0.25)=3$. For this parameter, the cumulative probabilities through counts 0, 1, 2, and 3 are $0.0498$, $0.1991$, $0.4232$, and $0.6472$. Since $0.4232<0.61\le0.6472$, $N=3$.

Use cumulative severity intervals: $0\le U<0.60$ selects $500; $0.60\le U<0.90$ selects $1,000; and $0.90\le U<1$ selects $2,000. Thus, uniform $0.10$ selects $500, $0.70$ selects $1,000, and $0.95$ selects $2,000. The aggregate is $S=500+1{,}000+2{,}000=\mathbf{\$3{,}500}$.

Common mistake avoided: using $\theta=12$ would simulate a full year, not a quarter. The correct parameter is the annual rate multiplied by the fraction of a year covered.

Worked Example: A Quarterly Simulation With Pareto Severities

A hypothetical book has a Poisson claim rate of 4 claims per year. Simulate one quarter. Severity follows a Pareto distribution with minimum value $100 and shape parameter $\alpha=2$, so $F_X(x)=1-(100/x)^2$ for $x\ge100$. Use count uniform $U_N=0.80$, then severity uniforms $0.36$ and $0.75$ if required. Report the aggregate to the nearest cent.

Method: exposure-adjusted Poisson inversion and Pareto inverse transformation. The quarterly parameter is $\theta=4(0.25)=1$. The Poisson cumulative probabilities through counts 0, 1, and 2 are $0.3679$, $0.7358$, and $0.9197$. Since $0.7358<0.80\le0.9197$, the simulated count is $N=2$.

To invert the Pareto distribution, set $U=1-(100/x)^2$. Rearranging gives $x=100/\sqrt{1-U}$. The first severity is $100/\sqrt{1-0.36}=100/0.8=\$125.00$. The second is $100/\sqrt{1-0.75}=100/0.5=\$200.00$. Therefore $S=125+200=\mathbf{\$325.00}$.

Common mistake avoided: using the exponential inverse transform for a Pareto severity would generate losses from the wrong distribution. The count generator and severity generator must each match their specified model.

Why frequency and severity both matter

The count determines how many terms enter the aggregate sum, while the severity distribution determines the size of each term. Two simulations with the same frequency rate can produce different totals when their severity distributions differ. Likewise, holding the severity model fixed does not make the total fixed: the Poisson count can be zero or can take several positive values.

The compound Poisson model also gives a useful theoretical check on the simulation design. Suppose $E[X]=\mu_X$ and $\operatorname{Var}(X)=\sigma_X^2$, with finite variance. Conditional on $N=n$, the sum has mean $n\mu_X$ and variance $n\sigma_X^2$, because it contains $n$ independent severities. Taking expectations gives

E[S]=E\bigl[E[S\mid N]\bigr]=E[N]\mu_X=\theta\mu_X.

For the variance, apply the conditional-variance identity $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. A Poisson variable has $E[N]=\operatorname{Var}(N)=\theta$, so

\operatorname{Var}(S) =E[N\sigma_X^2]+\operatorname{Var}(N\mu_X) =\theta\sigma_X^2+\theta\mu_X^2 =\theta E[X^2].

This derivation explains why a change in frequency affects both the expected total and its variability, while severity moments also contribute to both. The formulas describe the model distribution; they do not mean that any one simulated aggregate must be close to its expected value.

Common implementation traps

Using the wrong time unit. A rate of 12 claims per year gives a Poisson parameter of 3 for a quarter, not 12. Match the rate and exposure units before generating the count.
Using a probability instead of a cumulative probability. In inverse transform sampling, compare $U$ with $P(N\le n)$, not with $P(N=n)$ alone.
Generating a fixed number of severities. Generate exactly $N$ severities for each simulated period. If $N=0$, the aggregate is zero and no severity draw is used.
Reusing a count uniform as a severity draw. Under the model, count and severity draws are independent. Use separate independent uniforms (or a random-number generator that produces independent draws).
Ignoring severity units or the severity model. Keep amounts in consistent monetary units and use the inverse transform or sampling method appropriate to $F_X$.

Key takeaway: For a compound Poisson simulation, first set $\theta=\lambda t$, then generate $N\sim\operatorname{Poisson}(\theta)$, draw $N$ independent severities from $F_X$, and sum them. Exposure, count, severity, and monetary units must all match the model being simulated.

Check Your Understanding

For each problem, use the stated uniforms in order and assume all required draws are independent.

A portfolio has a rate of 5 claims per year. For 0.4 years of exposure, what is the Poisson parameter? If the count uniform is $0.50$, what count is generated? (Answer: $\theta=2$, $N=2$)
A hypothetical cover has a rate of 8 claims per year and is simulated for three months. If $U_N=0.30$, how many claims are generated? (Answer: $N=1$)
For $N\sim\operatorname{Poisson}(1.5)$, use $U_N=0.70$ to generate a count. (Answer: $N=2$)
There are two simulated claims, each exponential with mean $50. Use severity uniforms $0.20$ and $0.60$. What is the aggregate, to the nearest cent? (Answer: $56.97)
A hypothetical portfolio has a rate of 6 claims per year, is simulated for half a year, and has severities of $100 or $300, each with probability 0.50. Use $U_N=0.70$, then severity uniforms $0.10,0.40,0.60,0.90$, as needed. What is the aggregate? (Answer: $800)

What You'll Learn