Adding Independent Poisson Frequency Components

Why Combine Poisson Frequency Components?

A portfolio may contain distinct groups of policies, locations, or types of covered events. If each group has a Poisson claim count, an actuary may need the probability distribution of the total number of claims across all groups. The key calculation is simple, but it depends on a key condition: the component counts must be independent. When that condition holds, their rates add.

The rates must also refer to the same observation period. A rate stated per year must be converted before it is combined with a count modeled over a shorter period. The total count then has a Poisson distribution whose parameter is the sum of the component parameters for that period.

Definition: A count $N$ has a Poisson distribution with parameter $\lambda$, written $N\sim\operatorname{Poisson}(\lambda)$, if $\lambda\geq 0$ and $$ P(N=n)=e^{-\lambda}\frac{\lambda^n}{n!},\qquad n=0,1,2,\ldots. $$ The parameter $\lambda$ is both the mean and the variance of the count. It represents the expected count over the stated observation period.

The Sum of Independent Poisson Counts

Suppose $N_1,\ldots,N_m$ are mutually independent counts, with $N_j\sim\operatorname{Poisson}(\lambda_j)$. “Mutually independent” means that the joint probabilities factor into the product of the component probabilities. The sum $N=N_1+\cdots+N_m$ is Poisson with parameter $\lambda_1+\cdots+\lambda_m$.

Theorem: If $N_1,\ldots,N_m$ are mutually independent Poisson counts with parameters $\lambda_1,\ldots,\lambda_m$, all measured over the periods being combined, then $$ N_1+\cdots+N_m\sim\operatorname{Poisson}(\lambda_1+\cdots+\lambda_m). $$ Independence is essential to this result. Having Poisson marginal distributions alone does not guarantee that the sum is Poisson.

For a Poisson process with a constant rate $\lambda_j$ per unit of time, the count during an interval of length $t$ has parameter $\lambda_jt$. Thus, if several independent components have constant rates per year $r_1,\ldots,r_m$, their total count over $t$ years has parameter $t(r_1+\cdots+r_m)$. This scaling assumes the stated constant-rate Poisson model applies over that interval.

Why the Rates Add

For two independent components, the total equals $n$ when one component has $k$ claims and the other has $n-k$, for some $k$ from 0 to $n$. These disjoint possibilities can be added. Independence lets us multiply the probabilities for each pair of component counts.

Partition the event.
The event $N_1+N_2=n$ consists of the disjoint cases $(N_1,N_2)=(k,n-k)$, for $k=0,\ldots,n$.

Use independence and the Poisson probabilities.
For each case, multiply the probability that $N_1=k$ by the probability that $N_2=n-k$, then sum over $k$.

Combine the terms.
The binomial theorem turns the sum into the Poisson probability with parameter $\lambda_1+\lambda_2$.

\begin{aligned} P(N_1+N_2=n) &=\sum_{k=0}^{n}P(N_1=k)P(N_2=n-k)\\ &=e^{-(\lambda_1+\lambda_2)} \sum_{k=0}^{n} \frac{\lambda_1^k\lambda_2^{n-k}}{k!(n-k)!}\\ &=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 expression is the probability that a Poisson count with parameter $\lambda_1+\lambda_2$ equals $n$. For more than two components, apply the two-component result repeatedly. Mutual independence ensures that each step can be applied to the sum already formed and the next component.

Once the combined parameter is known, standard Poisson probabilities apply. For example, $P(N\leq r)$ is the sum of the exact probabilities from 0 through $r$, and $P(N\geq r)=1-P(N\leq r-1)$. A probability recursion can also be useful for calculating successive exact-count probabilities; whichever method is used, it is the combined parameter that belongs in the Poisson formula.

Worked Example: Combining Two Claim Sources

A hypothetical insurer models claims from two independent policy groups over the same one-year period. The first group has count $N_1\sim\operatorname{Poisson}(1.2)$; the second has count $N_2\sim\operatorname{Poisson}(0.8)$. Find the probability of exactly two claims in total.

Method. Both parameters refer to one year, and the counts are independent, so the total $N=N_1+N_2$ is Poisson with parameter $1.2+0.8=2.0$.

Using the total-count formula, $$ P(N=2)=e^{-2}\frac{2^2}{2!} =e^{-2}\frac{4}{2} =2e^{-2} \approx 0.2707. $$ The probability is 0.2707 to four decimal places.

As a check, the component-count possibilities are $(0,2)$, $(1,1)$, and $(2,0)$. Their probabilities sum to $$ e^{-2}\left(\frac{0.8^2}{2!}+1.2(0.8)+\frac{1.2^2}{2!}\right) =e^{-2}(0.32+0.96+0.72)=2e^{-2}. $$ This agrees with the combined-parameter calculation. The common mistake avoided here is adding the two expected counts but then using just one component’s rate in the probability formula.

More Than Two Components and Cumulative Probabilities

The same rule applies to any finite number of mutually independent components. Add every component parameter for the observation period, then calculate the desired probability using the distribution of the total. For a cumulative probability, include every integer count in the stated range; “at most two,” for example, includes 0, 1, and 2.

Worked Example: Probability of At Most Two Claims Across Three Groups

Three independent hypothetical claim counts over one quarter are Poisson with parameters $0.5$, $1.0$, and $1.5$, respectively. Find the probability that the combined count is at most two.

Method. The component counts are mutually independent and refer to the same quarter. The total $N$ is therefore Poisson with parameter $$ \lambda=0.5+1.0+1.5=3.0. $$

“At most two” means $N=0$, $N=1$, or $N=2$. Thus, $$ \begin{aligned} P(N\leq 2) &=e^{-3}\left(\frac{3^0}{0!}+\frac{3^1}{1!}+\frac{3^2}{2!}\right)\\ &=e^{-3}(1+3+4.5)\\ &=8.5e^{-3}\\ &\approx 0.4232. \end{aligned} $$ The probability is 0.4232 to four decimal places.

The result is plausible because the mean total count is 3, and the event includes counts below that mean. The common mistake avoided is interpreting “at most two” as exactly two and using only $P(N=2)$.

Match Each Rate to the Observation Period

Parameters can be added only after they have been expressed for the period of interest. If a component follows a constant-rate Poisson process with annual rate $r$, its count over half a year has parameter $r(0.5)$, not $r$. This rate conversion is a frequent source of errors in multi-component questions.

Worked Example: Converting Annual Rates for a Half-Year Count

Two independent hypothetical claim sources have constant annual rates of 1.2 and 2.4 claims per year. Let $N_A$ and $N_B$ be their counts during the next six months. Find (1) the probability of exactly two claims in total and (2) the probability of exactly two in total with at least one from source A.

Method and period conversion. Under the constant-rate Poisson assumptions, six months is $0.5$ years. The half-year parameters are $$ \lambda_A=1.2(0.5)=0.6,\qquad \lambda_B=2.4(0.5)=1.2. $$ The independent total has parameter $0.6+1.2=1.8$.

For (1), $$ P(N_A+N_B=2)=e^{-1.8}\frac{1.8^2}{2!} =e^{-1.8}(1.62) \approx 0.2678. $$

For (2), the eligible component-count pairs are $(1,1)$ and $(2,0)$. The pair $(0,2)$ is excluded because it has no claims from source A. By independence, $$ \begin{aligned} P(N_A+N_B=2,\ N_A\geq 1) &=P(N_A=1)P(N_B=1)+P(N_A=2)P(N_B=0)\\ &=e^{-0.6}(0.6)e^{-1.2}(1.2) +e^{-0.6}\frac{0.6^2}{2}e^{-1.2}\\ &=e^{-1.8}(0.72+0.18)\\ &=0.9e^{-1.8}\\ &\approx 0.1488. \end{aligned} $$ Both probabilities are rounded to four decimal places. The second is smaller than the first because it excludes the $(0,2)$ case. The common mistake avoided is using the annual parameter sum $3.6$ for a six-month probability.

Independence Is Not Optional

The addition rule is not a claim that any two Poisson-distributed counts have a Poisson sum. It is a result about independent Poisson counts. If two counts tend to rise and fall together because they share the same random events, multiplying their individual probabilities does not give the joint probabilities needed in the derivation.

Worked Example: A Dependent Pair That Does Not Have a Poisson Sum

Let $Z\sim\operatorname{Poisson}(0.5)$, and define two hypothetical component counts by $N_1=Z$ and $N_2=Z$. Each component count is marginally Poisson with parameter 0.5, but the two counts are identical and therefore dependent. Determine whether their total is Poisson.

The total is $N_1+N_2=2Z$, so it can only be an even number. In particular, $$ P(N_1+N_2=1)=0. $$ Its mean is $E[N_1+N_2]=2E[Z]=1$. A Poisson distribution with mean 1 would assign probability $e^{-1}$ to exactly one, which is positive. Thus the total is not Poisson.

This example does not contradict the sum theorem: its independence condition is not met. The common mistake avoided is applying the rate-addition rule merely because each component separately has a Poisson distribution.

Common Exam Traps and a Practical Check

Adding probabilities instead of parameters. Independent Poisson parameters add; the component probabilities do not. After finding the total parameter, calculate the requested probability from the total distribution.
Mixing observation periods. Convert every rate or parameter to the same period before combining. In particular, an annual rate is not automatically the parameter for a quarter or half-year.
Ignoring dependence. Check that the component counts are independent, not merely that each has a Poisson distribution.
Misreading the event. Translate phrases such as “at most,” “at least,” and “exactly” into the corresponding set of integer counts before calculating.
Rounding too early. Keep the combined parameter and intermediate probabilities unrounded until the final reported probability.

A reliable solution states the component distributions, confirms independence and the common observation period, adds the parameters, and only then evaluates the requested event. This sequence makes the assumptions visible and helps expose errors in units or event interpretation.

Key takeaway: For mutually independent Poisson counts over the periods being combined, the total is Poisson with parameter equal to the sum of the component parameters. If the counts are described by constant annual rates, first scale each rate to the observation period. If independence is absent, do not assume that the total is Poisson.

Check Your Understanding

For each problem, identify the total parameter and use the stated independence and time-period assumptions. Probabilities are rounded to four decimal places.

Two independent hypothetical claim counts over the same quarter have Poisson parameters 0.7 and 1.3. What is the probability that the total is exactly one? (Answer: 0.2707)
Two independent sources have constant annual claim rates of 0.9 and 1.5. What is the probability of no claims from either source during the next three months? (Answer: 0.5488)
Three independent hypothetical counts over a two-year period have Poisson parameters 0.4, 0.6, and 1.0 per year. What is the probability that the total count over those two years is exactly three? (Answer: 0.1954)
Independent counts $N_A\sim\operatorname{Poisson}(0.5)$ and $N_B\sim\operatorname{Poisson}(1.5)$ apply to the same period. What is the probability of exactly two claims in total with at least one claim from A? (Answer: 0.1184)
Let $Z\sim\operatorname{Poisson}(0.5)$, with $N_1=Z$ and $N_2=Z$. Are $N_1$ and $N_2$ independent, and is their sum Poisson? (Answer: They are dependent, and their sum is not Poisson.)

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