Poisson Aggregation Across Independent Exposures

Why Aggregate Poisson Counts?

Risk models often record counts separately by policy type, region, or time period. An aggregate question then asks about the total number of events across those groups. If the counts are independent and each has a Poisson distribution, their sum also has a Poisson distribution. The key calculation is to add the segment means, not the segment rates by themselves.

For example, a hypothetical insurer might observe different numbers of claims in two groups of policies. If the groups have different exposure amounts and different claim rates, each group's expected claim count depends on both quantities. Aggregation combines those expected counts. It does not generally combine rates by simple addition or by an unweighted average.

Definition: Let \(N_1,\ldots,N_k\) be independent counts with \(N_j\sim\operatorname{Poisson}(\lambda_j)\), where \(\lambda_j\geq 0\) is the expected count for segment \(j\). Their aggregate count is \(N=\sum_{j=1}^{k}N_j\). Then \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda=\sum_{j=1}^{k}\lambda_j\). In particular, \(E[N]=\operatorname{Var}(N)=\lambda\).

When segment \(j\) has a constant event rate \(r_j\) per unit of exposure and exposure amount \(e_j\), its count is modeled as \(N_j\sim\operatorname{Poisson}(e_jr_j)\). Thus its mean is \(\lambda_j=e_jr_j\), and the aggregate mean is the sum of the segment means. Exposure units must match the units used for the rates.

Why the Sum Is Poisson

A probability generating function provides a short derivation. For a nonnegative integer-valued random variable \(N\), its probability generating function is \(G_N(s)=E[s^N]\). For a Poisson random variable with mean \(\lambda\), the probability mass function gives

G_N(s)=\sum_{n=0}^{\infty}s^n e^{-\lambda}\frac{\lambda^n}{n!} =e^{-\lambda}\sum_{n=0}^{\infty}\frac{(\lambda s)^n}{n!} =e^{\lambda(s-1)}.

For independent counts, the generating function of their sum is the product of their generating functions. Independence matters here: it lets the expectation of the product factor into a product of expectations.

G_{\sum_{j=1}^{k}N_j}(s) =\prod_{j=1}^{k}G_{N_j}(s) =\prod_{j=1}^{k}e^{\lambda_j(s-1)} =e^{(\sum_{j=1}^{k}\lambda_j)(s-1)}.

The final expression is the generating function of a Poisson random variable with mean \(\sum_j\lambda_j\). Since a probability distribution is determined by its generating function, the sum has that Poisson distribution. Its mean and variance both equal its Poisson parameter.

Result: If independent segment counts satisfy \(N_j\sim\operatorname{Poisson}(e_jr_j)\), then

N=\sum_{j=1}^{k}N_j \sim\operatorname{Poisson}\left(\sum_{j=1}^{k}e_jr_j\right), \qquad E[N]=\operatorname{Var}(N)=\sum_{j=1}^{k}e_jr_j.

If the rates are different, an overall rate can still be reported when total exposure is positive and the exposure units are the same. It is the total expected count divided by the total exposure:

r_{\mathrm{agg}} =\frac{\sum_{j=1}^{k}e_jr_j}{\sum_{j=1}^{k}e_j}.

This is an exposure-weighted average of the segment rates. Multiplying \(r_{\mathrm{agg}}\) by total exposure recovers the aggregate expected count. If all segment rates are equal to a common rate \(r\), then \(r_{\mathrm{agg}}=r\).

Applying Exposure and Rate

Use a consistent sequence: express each exposure in the rate's units, calculate the expected count for each segment, and add those means. Only after obtaining the total mean should you use the Poisson probability formula for the aggregate count.

Check the model conditions.
Each segment count must be Poisson, and the segment counts must be independent for the stated aggregation result to apply.

Calculate segment means.
For a constant rate \(r_j\) per exposure unit, compute \(\lambda_j=e_jr_j\), after converting exposure to the correct unit.

Aggregate the means.
Set \(\lambda=\sum_j\lambda_j\). The total count has a Poisson distribution with parameter \(\lambda\).

Answer the requested question.
Use the aggregate Poisson distribution for a probability, or report \(\lambda\) as both the total expected count and the total-count variance.

Worked Example: Two Independent Policy Groups

Method. Convert each group's exposure and rate into a Poisson mean, then aggregate the means. In a hypothetical portfolio, Group A has 40 policy-years and a claim rate of 0.02 per policy-year. Group B has 30 policy-years and a claim rate of 0.04 per policy-year. Assume the groups' claim counts are independent Poisson variables. Find the probability of exactly two claims in total.

Group A mean. The exposure and rate use the same unit, so \(\lambda_A=40(0.02)=0.80\).

Group B mean. Similarly, \(\lambda_B=30(0.04)=1.20\).

Aggregate mean. The sum is \(0.80+1.20=2.00\), so \(N_A+N_B\sim\operatorname{Poisson}(2.00)\).

Probability. For a Poisson random variable with mean 2, the probability of exactly two is

P(N_A+N_B=2) =e^{-2}\frac{2^2}{2!} =e^{-2}\frac{4}{2} =2e^{-2} \approx 0.2706706.

Rounded to four decimal places, the probability is 0.2707. The aggregate mean and variance are both 2.00 claims. The independence assumption is what supports the Poisson distribution for the sum.

Common mistake avoided: Adding the rates, \(0.02+0.04=0.06\), does not give the aggregate mean or an aggregate rate. The actual total mean is 2.00 claims because each rate must first be multiplied by its own exposure.

Worked Example: Finding an Exposure-Weighted Aggregate Rate

Method. Find the segment means, add them, and divide by total exposure to obtain the overall rate. Suppose a hypothetical book has 120 exposure units at rate 0.015 per unit and 80 exposure units at rate 0.030 per unit. Assume independent Poisson counts. Find the aggregate mean, aggregate rate, and probability of exactly three events.

Segment means. The first segment has mean \(120(0.015)=1.80\). The second has mean \(80(0.030)=2.40\).

Aggregate distribution. The mean is \(1.80+2.40=4.20\), so the sum is Poisson with parameter 4.20. Total exposure is \(120+80=200\) units.

Aggregate rate. Dividing the total expected count by total exposure gives \(r_{\mathrm{agg}}=4.20/200=0.021\) per unit. As a check, the exposure-weighted calculation is \((120(0.015)+80(0.030))/200=4.20/200=0.021\).

Probability of exactly three. Substituting the aggregate mean into the Poisson probability formula gives

P(N=3) =e^{-4.2}\frac{4.2^3}{3!} =e^{-4.2}\frac{74.088}{6} =12.348e^{-4.2} \approx 0.1851654.

Rounded to four decimal places, \(P(N=3)=\mathbf{0.1852}\). The calculation uses the aggregate mean 4.20, not either segment's rate or the sum of the two rates.

Exposure Units and Independence

Exposure is a measure of how long or how much a group is at risk, expressed in units such as policy-years, vehicle-years, or device-months. A rate per year must be multiplied by exposure measured in years. If the exposure is supplied in months, convert it before multiplying by an annual rate. The product of rate and exposure is an expected count, so it is the quantity that can be added across segments.

Worked Example: Converting Months to Years Before Aggregation

Method. Convert all exposures into policy-years, calculate each mean, and then use the aggregate distribution. In a hypothetical portfolio, 24 contracts are observed for six months each at a rate of 0.10 claims per policy-year. Another 10 contracts are observed for three months each at a rate of 0.20 claims per policy-year. Assume the segment counts are independent Poisson variables. Find the probability of no claims in total and the aggregate rate per policy-year.

First segment exposure. Six months is \(6/12=0.5\) years per contract. Thus the exposure is \(24(0.5)=12\) policy-years, giving mean \(12(0.10)=1.20\).

Second segment exposure. Three months is \(3/12=0.25\) years per contract. Thus the exposure is \(10(0.25)=2.5\) policy-years, giving mean \(2.5(0.20)=0.50\).

Aggregate mean and rate. The mean is \(1.20+0.50=1.70\) claims, and total exposure is \(12+2.5=14.5\) policy-years. Therefore the aggregate rate is \(1.70/14.5\approx0.1172\) claims per policy-year. The aggregate count is Poisson with mean 1.70.

Probability of zero claims.

P(N=0)=e^{-1.7}\frac{1.7^0}{0!} =e^{-1.7} \approx0.1826835.

Rounded to four decimal places, the probability is 0.1827. A common unit error would multiply the annual rates by 24 and 10 without accounting for the observation periods. That would incorrectly treat each contract as exposed for a full year.

What Can Break the Aggregation Result?

The result requires each segment count to have a Poisson distribution and the counts to be independent. Separate labels for segments do not by themselves guarantee independence. For example, a shared event affecting several groups could make their counts dependent. In that case, the expected total still equals the sum of the expected counts, because expectation adds whether or not counts are independent. However, the total is not necessarily Poisson.

Dependence also affects variance. In general,

\operatorname{Var}\left(\sum_{j=1}^{k}N_j\right) =\sum_{j=1}^{k}\operatorname{Var}(N_j) +2\sum_{j<\ell}\operatorname{Cov}(N_j,N_\ell).

For independent counts, every covariance in this expression is zero. If each count is Poisson, the sum of the segment variances is then \(\sum_j\lambda_j\), which equals the aggregate mean. Without independence, it is not valid to assume the aggregate count has a Poisson distribution merely because each segment count individually does.

Key takeaway: For independent Poisson counts, add the means to obtain the aggregate Poisson parameter. When a mean is specified through a rate, calculate it as rate multiplied by exposure in matching units. An aggregate rate, when useful, is the total mean divided by total exposure.

Check Your Understanding

For each problem, assume independent Poisson counts unless the question says otherwise. Give probabilities to four decimal places.

A hypothetical portfolio has two segments with Poisson means 0.7 and 1.3. Find the probability that the total count is zero. (Answer: \(e^{-2}=0.1353\))
A group has 40 exposure units at rate 0.03 per unit, and another has 60 exposure units at rate 0.01 per unit. Find the total expected count and variance. (Answer: mean \(=1.8\), variance \(=1.8\))
A hypothetical event rate is 0.24 per policy-year. Ten policies are observed for 1.5 months each. Find the expected event count and the probability of exactly one event. (Answer: mean \(=0.30\); probability \(=0.30e^{-0.30}=0.2222\))
Three independent segments have Poisson means 1.5, 2.5, and 0. Find the probability that the aggregate count is at least one. (Answer: \(1-e^{-4}=0.9817\))
A hypothetical portfolio has 100 exposure units at rate 0.012 per unit and 50 units at rate 0.024 per unit. Find the exposure-weighted aggregate rate and the probability of exactly two total events. (Answer: aggregate rate \(=0.016\) per unit; probability \(=e^{-2.4}(2.4^2/2)=0.2613\))

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