Additivity Of Independent Poisson Counts

Why Poisson Counts Add Across a Portfolio

An insurer may track claims separately for several independently modeled groups, such as different product segments or regions. A portfolio question often concerns the total number of claims across those groups rather than each group separately. If the group counts satisfy the required assumptions, their total has a simple distribution: it is Poisson, with a rate equal to the sum of the component rates.

The result is useful because it turns several probability calculations into one. It also depends on conditions that must be checked, especially independence and a common observation period. Merely knowing that two counts are each Poisson is not enough to conclude that their sum is Poisson.

Definition: Let $N_1,\ldots,N_m$ be independent Poisson counts observed over the same specified period, with $N_j\sim\operatorname{Poisson}(\lambda_j)$, where $\lambda_j\geq 0$. Their total is $N=\sum_{j=1}^{m}N_j$. Poisson additivity states that $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda=\sum_{j=1}^{m}\lambda_j$.

Each $\lambda_j$ is the expected number of claims in its group during the observation period. For the addition to be meaningful, the component rates must refer to the same period and compatible definitions of a claim. For example, an annual expected count cannot be added directly to a monthly expected count without first putting them on a consistent basis.

Why the Sum Is Poisson

First consider two independent counts $N_1\sim\operatorname{Poisson}(\lambda_1)$ and $N_2\sim\operatorname{Poisson}(\lambda_2)$. For their total to equal $n$, the first count could be $0$ and the second $n$, the first could be $1$ and the second $n-1$, and so on. These mutually exclusive possibilities give a convolution.

\begin{aligned} P(N_1+N_2=n) &=\sum_{k=0}^{n}P(N_1=k)P(N_2=n-k)\\ &=\sum_{k=0}^{n} \left(e^{-\lambda_1}\frac{\lambda_1^k}{k!}\right) \left(e^{-\lambda_2}\frac{\lambda_2^{n-k}}{(n-k)!}\right)\\ &=e^{-(\lambda_1+\lambda_2)} \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k} \lambda_1^k\lambda_2^{n-k}\\ &=e^{-(\lambda_1+\lambda_2)} \frac{(\lambda_1+\lambda_2)^n}{n!}. \end{aligned}

The first equality uses independence: for each possible split $k,n-k$, the joint probability is the product of the two probabilities. The last equality is the binomial theorem. The resulting expression is exactly the probability mass function of a Poisson variable with parameter $\lambda_1+\lambda_2$. Repeating the argument, or applying it successively to the running total and the next independent count, proves the result for any finite number of counts.

Theorem: If $N_1,\ldots,N_m$ are mutually independent Poisson counts with parameters $\lambda_1,\ldots,\lambda_m$, then for every integer $n\geq 0$, $$ P\left(\sum_{j=1}^{m}N_j=n\right) =e^{-\sum_{j=1}^{m}\lambda_j} \frac{\left(\sum_{j=1}^{m}\lambda_j\right)^n}{n!}. $$ Thus $N=\sum_{j=1}^{m}N_j$ is Poisson with parameter $\sum_{j=1}^{m}\lambda_j$.

For this total, the Poisson mean and variance are both the combined parameter:

E[N]=\operatorname{Var}(N)=\sum_{j=1}^{m}\lambda_j.

This also agrees with the general rules for sums: expectations add, and variances add for independent variables. The distributional result is stronger than those two moment statements; it identifies the entire probability distribution of the total.

Applying Additivity to Portfolio Questions

Once the combined parameter is found, use the ordinary Poisson probability formula. In particular, for $N\sim\operatorname{Poisson}(\lambda)$,

P(N=n)=e^{-\lambda}\frac{\lambda^n}{n!}, \qquad P(N\geq 1)=1-P(N=0)=1-e^{-\lambda}.

For an upper-tail probability, it is often efficient to use the complement. For example, $P(N\geq 3)=1-\{P(N=0)+P(N=1)+P(N=2)\}$. Additivity does not change how the Poisson probabilities are calculated; it supplies the correct parameter for the combined count.

Worked Example: Exact and Positive Totals for Two Groups

A hypothetical insurer models the number of water-damage claims in one quarter for two independent portfolios. The first count is Poisson with parameter $0.8$, and the second is Poisson with parameter $1.2$. Find (a) the probability of exactly two claims across both portfolios and (b) the probability of at least one claim. Give probabilities to four decimal places.

Method. The counts are independent and refer to the same quarter, so add their parameters:

\lambda=0.8+1.2=2.0.

Therefore the total count $N$ is Poisson with parameter $2.0$.

(a) Exactly two claims. Substitute $n=2$ and $\lambda=2.0$:

P(N=2)=e^{-2}\frac{2^2}{2!} =e^{-2}\frac{4}{2} =2e^{-2} =0.2707.

(b) At least one claim. The complement is no claims in either portfolio:

P(N\geq 1)=1-P(N=0)=1-e^{-2} =1-0.1353 =0.8647.

Check. The exact-two probability is between zero and one, and the probability of at least one claim exceeds the probability of exactly two claims, as expected. The common mistake avoided here is averaging the component parameters; the total parameter is their sum, not their average.

Worked Example: A Three-Group Upper Tail

For a hypothetical commercial insurance program, three independent divisions have Poisson claim counts over the same six-month period, with parameters $0.4$, $0.9$, and $1.7$. Find the probability that the combined count is at least three. Give the answer to four decimal places.

Step 1: Combine the parameters.

\lambda=0.4+0.9+1.7=3.0.

The total $N$ is therefore Poisson with parameter $3.0$.

Step 2: Use the complement. The complement of at least three claims is zero, one, or two claims:

\begin{aligned} P(N\geq 3) &=1-\{P(N=0)+P(N=1)+P(N=2)\}\\ &=1-e^{-3}\left(1+3+\frac{3^2}{2!}\right)\\ &=1-8.5e^{-3}\\ &=1-(8.5)(0.0497871)\\ &=0.5768. \end{aligned}

Check. The omitted probabilities sum to approximately $0.4232$, so the upper tail is $1-0.4232=0.5768$. The common mistake avoided is using only one component parameter in the tail calculation instead of first combining all three groups.

Assumptions and Limits of the Result

Before adding Poisson parameters, verify the structure of the problem rather than relying only on the names of the distributions.

Independence: The component counts must be independent. A shared event, common environment, or other dependence can make the total distribution different from Poisson.
Same observation period: Each parameter must describe an expected count over the period for which the total is being calculated.
Consistent count definition: The units being counted must be comparable. If a single event is recorded in two groups, summing group counts may count that event twice.
Poisson component models: Each component count must itself have a Poisson distribution. Additivity is not a general rule for arbitrary count distributions.

Independence is essential to the convolution derivation: it is what permits $P(N_1=k,N_2=n-k)$ to be written as $P(N_1=k)P(N_2=n-k)$. If counts are dependent, adding their expected values still gives the expected value of their sum, but that fact alone does not establish a Poisson distribution for the sum.

Worked Example: Combined Mean, Variance, and Capacity Probability

A hypothetical claims team tracks three independent portfolios over one month. Their Poisson parameters are $1.1$, $0.9$, and $0.5$. Find the mean and variance of the combined count, and the probability that the team receives at most one claim during the month. Give the probability to four decimal places.

Step 1: Find the combined parameter.

\lambda=1.1+0.9+0.5=2.5.

The total $N$ is Poisson with parameter $2.5$. Therefore,

E[N]=2.5,\qquad \operatorname{Var}(N)=2.5.

Step 2: Calculate the capacity probability. At most one claim means $N=0$ or $N=1$:

\begin{aligned} P(N\leq 1) &=P(N=0)+P(N=1)\\ &=e^{-2.5}+e^{-2.5}\frac{2.5^1}{1!}\\ &=e^{-2.5}(1+2.5)\\ &=(0.082085)(3.5)\\ &=0.2873. \end{aligned}

Check. The mean and variance match, as required for a Poisson count, and the probability of zero or one claim is less than one. The common mistake avoided is adding component variances but then treating the resulting moments as if they alone proved a Poisson model; here the Poisson distribution follows from independence and Poisson components.

Common Exam Trap

A tempting but incorrect calculation is to average component rates or to add probabilities for the same event across groups without checking how the event is defined. For example, if independent counts have parameters $0.8$ and $1.2$, the probability of zero total claims is not $e^{-0.8}$ or $e^{-1.2}$. Zero total claims requires zero claims from both groups, so independence gives

P(N_1+N_2=0)=P(N_1=0)P(N_2=0) =e^{-0.8}e^{-1.2}=e^{-2}.

More generally, check the count definitions and the time period first, confirm independence, and only then add the Poisson parameters. This disciplined order prevents a correct Poisson formula from being applied to the wrong total.

Key takeaway: For mutually independent Poisson counts measured over the same period, the total is Poisson with parameter equal to the sum of the component parameters. Independence is not optional, and a sum of rates is valid only when the counted events and time basis are consistent.

Check Your Understanding

For each problem, check the assumptions before combining parameters. Probabilities in the answers are rounded to four decimal places.

Two independent Poisson claim counts for the same quarter have parameters $0.6$ and $1.4$. Find the probability of no claims and the probability of at least two claims in total. (Answer: $P(N=0)=0.1353$; $P(N\geq2)=0.5940$.)
Three independent product groups have Poisson parameters $2.2$, $0.8$, and $1.0$ over the same year. Find the combined mean and variance, then find the probability of exactly one claim. (Answer: mean $=4.0$, variance $=4.0$, $P(N=1)=0.0733$.)
Independent Poisson counts from three portfolios have parameters $0.25$, $0.75$, and $1.5$ over the same period. Calculate the probability of at most two claims in total. (Answer: $e^{-2.5}(1+2.5+2.5^2/2)=0.5438$.)
Two independent Poisson counts are recorded for the same period. The first has parameter $1.3$, and the probability of zero claims in their combined count is $0.1000$. Find the parameter of the second count. (Answer: $-\ln(0.1000)-1.3=1.0026$.)
Two counts are each modeled as Poisson for the same period, but a shared catastrophe can generate claims in both groups. Is their sum necessarily Poisson with parameter equal to the sum of their parameters? State the issue. (Answer: No; the shared catastrophe can make the counts dependent, so additivity does not follow from the Poisson marginal models alone.)

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