Why Independent Poisson Counts Add

A portfolio may contain several distinct groups of risks, each with its own claim count. If the group counts are independent Poisson random variables, one useful result lets us find the distribution of the total without listing every possible combination of group outcomes. The result is simple, but its independence condition matters: adding Poisson counts does not automatically produce a Poisson count when the counts are dependent.

This tutorial develops the addition result from the Poisson probability formula and applies it to hypothetical groups with specified mean counts. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N_1,\ldots,N_m$ be independent random variables with $N_j\sim\operatorname{Poisson}(\lambda_j)$, where each $\lambda_j\geq 0$. Their aggregate count is $N=N_1+\cdots+N_m$. Under these assumptions, $N\sim\operatorname{Poisson}(\lambda_1+\cdots+\lambda_m)$. Each parameter is the mean count for its group over the period specified in the problem.

Here, a “count” is a nonnegative integer: $0,1,2,\ldots$. The group counts must refer to the same period and must be independent. The parameters add, not the observed counts from a particular scenario and not the probabilities of the groups having claims. If a parameter is zero, that group contributes zero claims with probability one and does not change the aggregate parameter.

Deriving the Addition Result

First consider two independent counts, $N_1\sim\operatorname{Poisson}(\lambda_1)$ and $N_2\sim\operatorname{Poisson}(\lambda_2)$. For the total $N=N_1+N_2$ to equal a fixed nonnegative integer $n$, the first count could be $k$ while the second is $n-k$, for any $k$ from 0 through $n$. These possibilities are mutually exclusive. Independence lets us multiply the probabilities of each pair and then add them.

\begin{aligned} P(N=n) &=\sum_{k=0}^{n}P(N_1=k,N_2=n-k)\\ &=\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 last equality is the binomial theorem. The result is exactly the Poisson probability formula with parameter $\lambda_1+\lambda_2$. This derivation explains both why the parameters add and why independence is essential: without independence, the joint probability in the second line need not factor into a product.

For more than two independent counts, apply the two-count result repeatedly. For example, $N_1+N_2$ is Poisson with parameter $\lambda_1+\lambda_2$, and it is independent of $N_3$ when $N_1,N_2,N_3$ are mutually independent. Adding $N_3$ gives a Poisson total with parameter $\lambda_1+\lambda_2+\lambda_3$. Repeating this argument gives the stated result for $m$ counts.

Theorem: If $N_1,\ldots,N_m$ are mutually independent and $N_j\sim\operatorname{Poisson}(\lambda_j)$, then for every nonnegative integer $n$, $$ P(N=n)=e^{-\Lambda}\frac{\Lambda^n}{n!}, \qquad \Lambda=\sum_{j=1}^{m}\lambda_j. $$ Thus $N=\sum_{j=1}^{m}N_j\sim\operatorname{Poisson}(\Lambda)$.

Applying the Result to Separate Groups

For separate risk groups, begin by identifying the count for each group, its Poisson parameter, and the period to which that parameter applies. Then check that the group counts are independent. Once those assumptions hold, add the parameters and use the Poisson formula for the total. A group’s parameter is not necessarily an observed count; it describes the distribution of the count.

The addition theorem does not itself establish that a group count is Poisson or that groups are independent. Those are model assumptions that must be supplied or justified separately. In particular, the theorem cannot be used to turn dependent counts into an aggregate Poisson count merely because each group count has a Poisson distribution.

Worked Example: Exact Aggregate Count Across Two Groups

A hypothetical insurer models the number of claims from Group A during a specified quarter as $N_A\sim\operatorname{Poisson}(1.1)$, and from Group B as $N_B\sim\operatorname{Poisson}(1.6)$. The counts are independent. Find the probability of exactly three claims in total.

Method and assumptions. The counts are independent Poisson variables over the same quarter, with parameters $\lambda_A=1.1$ and $\lambda_B=1.6$. No interest or mortality assumption is involved. By the addition theorem, $N=N_A+N_B\sim\operatorname{Poisson}(1.1+1.6)=\operatorname{Poisson}(2.7)$.

For exactly three claims, substitute $n=3$ and $\Lambda=2.7$ into the Poisson formula:

P(N=3)=e^{-2.7}\frac{2.7^3}{3!} =e^{-2.7}\frac{19.683}{6} =e^{-2.7}(3.2805) \approx 0.2204676845.

To verify the parameter combination directly, the possible splits of three claims are $(0,3),(1,2),(2,1),(3,0)$. In the convolution, their coefficients after factoring out $e^{-2.7}$ are:

\begin{aligned} \frac{1.6^3}{3!} +\frac{1.1(1.6^2)}{1!2!} +\frac{1.1^2(1.6)}{2!1!} +\frac{1.1^3}{3!} &=0.6826666667+1.408+0.968+0.2218333333\\ &=3.2805. \end{aligned}

Therefore, the probability of exactly three aggregate claims is $0.2205$, rounded to four decimal places. The convolution check agrees with the aggregate Poisson calculation. The common mistake this method avoids is calculating only one split, such as one claim from A and two from B, instead of including every split that totals three.

Worked Example: At Least Two Claims in Three Groups

For a hypothetical six-month period, three independent groups have claim counts $N_1\sim\operatorname{Poisson}(0.4)$, $N_2\sim\operatorname{Poisson}(0.9)$, and $N_3\sim\operatorname{Poisson}(1.7)$. Find the probability of at least two claims in total.

Method and assumptions. Each group count is Poisson for the same six-month period, and the three counts are mutually independent. The aggregate parameter is $0.4+0.9+1.7=3.0$, so $N=N_1+N_2+N_3\sim\operatorname{Poisson}(3.0)$.

It is shorter to use the complement of “at least two,” namely zero or one claim. The Poisson probabilities for those outcomes are:

\begin{aligned} P(N\geq 2) &=1-P(N=0)-P(N=1)\\ &=1-e^{-3}\frac{3^0}{0!}-e^{-3}\frac{3^1}{1!}\\ &=1-e^{-3}(1+3)\\ &=1-4(0.0497870683679)\\ &=0.8008517265284. \end{aligned}

Thus the probability is $0.8009$, rounded to four decimal places. The result is a probability between zero and one, as required. A tempting but incorrect approach is to calculate each group’s probability of at least two claims and add those probabilities; that does not describe whether the combined total is at least two, because claims in different groups can also produce a total of two or more.

Worked Example: Distinguishing the Total From Its Group Composition

In another hypothetical portfolio, independent group counts are $N_A\sim\operatorname{Poisson}(0.5)$ and $N_B\sim\operatorname{Poisson}(1.5)$ for the same period. Find (a) the probability of exactly two claims in total, and (b) the probability that exactly two total claims are both from Group A.

Method and assumptions. The parameters are $0.5$ and $1.5$, and independence allows the addition theorem. Thus $N=N_A+N_B\sim\operatorname{Poisson}(2)$. Part (a) uses the aggregate distribution. Part (b) requires the particular split $N_A=2,N_B=0$, so use the group probabilities and independence.

For part (a):

P(N=2)=e^{-2}\frac{2^2}{2!} =2e^{-2} \approx 0.2706705665.

For part (b), independence gives:

\begin{aligned} P(N_A=2,N_B=0) &=P(N_A=2)P(N_B=0)\\ &=\left(e^{-0.5}\frac{0.5^2}{2!}\right)(e^{-1.5})\\ &=e^{-2}\left(\frac{0.25}{2}\right)\\ &=0.125e^{-2}\\ &\approx 0.0169169104. \end{aligned}

Rounded to four decimal places, the answers are $0.2707$ for exactly two claims in total and $0.0169$ for exactly two claims, both from A. As a check, the other ways to total two have coefficients $0.75$ for one claim from each group and $1.125$ for zero from A and two from B. Their coefficients sum to $0.125+0.75+1.125=2$, matching the coefficient in $P(N=2)=2e^{-2}$. The common mistake is to treat “two in total” as if it meant “two from a specified group.”

Independence and Common Exam Traps

The addition result is useful because it replaces many combinations of group outcomes with one aggregate Poisson model. But the model conditions determine whether that shortcut is valid. Counts from separate groups may be dependent if a shared event affects both groups, or if the categories overlap so the same claim is counted in more than one group. In such cases, adding the marginal parameters and declaring the total Poisson is not justified by this theorem.

Even if group counts are independent, check that they refer to the same time period and that their parameters are expressed on compatible terms. A count for one quarter cannot simply be combined with a count for a full year without first resolving the period mismatch. Determining parameters from rates and exposure is a separate modeling step; here, the addition theorem takes the group parameters as given.

Identify the group counts.
Write the count variable and Poisson parameter for each group, using the period in the problem.

Check the conditions.
Confirm that the counts are Poisson and independent. If either condition is absent, do not apply the theorem without further justification.

Combine parameters.
Set the aggregate parameter equal to the sum of the group parameters.

Answer the event asked for.
Use the aggregate Poisson distribution for a total-count event; use a joint group probability if the question also specifies how the total is divided among groups.

Keep the distinction between the total count and its composition explicit. “Exactly $n$ claims in all groups” is an aggregate event. “Exactly $n$ claims, all from Group A” is a more specific joint event. The aggregate distribution answers the first question, but generally does not by itself supply the second.

Key takeaway: For mutually independent Poisson counts measured over the same period, the sum is Poisson with parameter equal to the sum of the parameters. Independence is the condition that makes the convolution factor correctly; group composition must still be handled separately when it is part of the event.

Check Your Understanding

For each hypothetical situation, first check the assumptions, then find the requested probability. Give probabilities to four decimal places.

Independent counts from two groups are Poisson with parameters $0.7$ and $1.3$ for the same period. What is the probability of zero claims in total? (Answer: $0.1353$)
Independent counts are Poisson with parameters $0.8$ and $1.2$. What is the probability of exactly one claim in total? (Answer: $0.2707$)
Three independent group counts have Poisson parameters $0.5$, $1.0$, and $1.5$ over the same period. What is the probability of at least one claim in total? (Answer: $0.9502$)
Independent counts for two groups have Poisson parameters $0.4$ and $0.6$. What is the probability of exactly two claims in total? (Answer: $0.1839$)
Independent group counts are Poisson with parameters $1.2$ for Group A and $0.8$ for Group B. What is the probability that the total is exactly one and that this one claim is from A? (Answer: $0.1624$)

Why Independent Poisson Counts Add

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