Adding Independent Counts Across Risk Groups

Why Combine Counts Across Risk Groups?

An insurer may model claim counts separately for distinct groups of risks—for example, two portfolios with different expected claim frequencies. If the subgroup counts are independent, their sum is the total number of claims across the groups. The important question is whether that total still follows one of the same familiar frequency models, and, if so, what its parameters are.

The answer depends on the model and its parameters. Independent Poisson counts can be combined even when their rates differ. Binomial counts combine into a binomial count when they have the same claim probability per trial. Negative binomial failure counts combine into a negative binomial count when they have the same success probability. Independence alone does not guarantee that the sum belongs to the same family.

Definition: Let \(N_1,\ldots,N_k\) be independent subgroup counts, and let \(N=\sum_{j=1}^k N_j\) be their aggregate count. They combine into a specified frequency model when the distribution of \(N\) is that same model, with parameters determined by the subgroup parameters. Independence is required for the addition rules below.

The Three Addition Rules

For Poisson counts, let \(N_j\sim\operatorname{Poisson}(\lambda_j)\), where \(\lambda_j\) is the expected count in the stated exposure period. The subgroup rates do not have to be equal, but the exposures must be interpreted consistently. For example, counts over the same year can be added using their annual expected counts.

Theorem: If \(N_1,\ldots,N_k\) are independent Poisson counts with respective parameters \(\lambda_1,\ldots,\lambda_k\), then \(N=\sum_{j=1}^k N_j\sim\operatorname{Poisson}\!\left(\sum_{j=1}^k\lambda_j\right)\).

For binomial counts, each subgroup count represents the number of successes among a fixed number of independent trials. To combine the counts into one binomial count, all trials must have the same success probability \(p\). The numbers of trials can differ.

Theorem: If \(N_j\sim\operatorname{Binomial}(n_j,p)\) independently, with the same \(p\) for every subgroup, then \(N=\sum_{j=1}^k N_j\sim\operatorname{Binomial}\!\left(\sum_{j=1}^k n_j,p\right)\).

For the negative binomial rule here, use the convention that \(N_j\) counts the number of failures before the \(r_j\)th success, with each trial having success probability \(p\). Thus \(N_j\) can be zero or any nonnegative integer. Combining these counts requires the same success probability across the subgroups.

Theorem: If the independent counts \(N_j\) each count failures before \(r_j\) successes, and have the same success probability \(p\), then \(N=\sum_{j=1}^k N_j\) counts failures before \(\sum_{j=1}^k r_j\) successes. In particular, \(N\) has a negative binomial distribution with parameters \(\sum_{j=1}^k r_j\) and \(p\).

Why the Addition Rules Hold

A probability generating function (PGF) for a nonnegative integer-valued count \(N\) is \(G_N(s)=E[s^N]\). For independent counts, the PGF of their sum is the product of their PGFs: independence allows the expectation of the product \(s^{N_1}\cdots s^{N_k}\) to factor into the product of the expectations.

G_{\sum_j N_j}(s)=\prod_j G_{N_j}(s).

For a Poisson count with parameter \(\lambda_j\), the PGF is \(\exp(\lambda_j(s-1))\). Multiplying these expressions gives \(\exp((\sum_j\lambda_j)(s-1))\), the PGF of a Poisson count with parameter \(\sum_j\lambda_j\).

For a binomial count with \(n_j\) trials and probability \(p\), the PGF is \((1-p+ps)^{n_j}\). With the same \(p\) in each group, multiplying gives \((1-p+ps)^{\sum_j n_j}\), the PGF of the stated aggregate binomial count.

For a negative binomial count of failures before \(r_j\) successes, the PGF is \(\left(\frac{p}{1-(1-p)s}\right)^{r_j}\). If \(p\) is common, multiplying the subgroup PGFs gives the same expression raised to the power \(\sum_j r_j\). That is the PGF of the aggregate negative binomial count. These arguments establish the rules; if a required shared parameter does not exist, the matching factorization does not apply.

Worked Applications

Worked Example: Combining Poisson Counts From Two Portfolios

A hypothetical insurer models the number of claims in one year from two independent portfolios as \(N_A\sim\operatorname{Poisson}(1.2)\) and \(N_B\sim\operatorname{Poisson}(1.8)\). Find the probability of exactly two claims in total. Give the probability to four decimal places.

Method and model check. Both counts are Poisson counts for the same one-year period, and the groups are independent. Their parameters are expected counts for that period, so add them:

N_A+N_B\sim\operatorname{Poisson}(1.2+1.8) =\operatorname{Poisson}(3).

For a Poisson count with parameter \(3\), the probability of exactly two is

P(N_A+N_B=2) =e^{-3}\frac{3^2}{2!} =e^{-3}\frac{9}{2} =4.5e^{-3}.

Using \(e^{-3}\approx0.0497871\), the probability is \(4.5(0.0497871)=0.22404195\). Rounded only at the final step to four decimal places, the answer is 0.2240. As a check, it is a valid probability between 0 and 1. The common mistake avoided here is averaging the two Poisson parameters instead of adding them.

Worked Example: Combining Binomial Counts With a Shared Probability

A hypothetical warranty program has two independent groups. In group A, each of 6 warranties has probability \(0.1\) of generating a claim; in group B, each of 9 warranties also has probability \(0.1\). Each warranty can generate at most one claim. Find the probability of exactly two claims across both groups, to four decimal places.

Method and model check. The trial counts differ, but the individual claim probabilities are both \(p=0.1\). The total number of trials is \(6+9=15\), so the total count is \(\operatorname{Binomial}(15,0.1)\).

P(N=2)=\binom{15}{2}(0.1)^2(0.9)^{13}.

Calculate the factors: \(\binom{15}{2}=15(14)/2=105\), \((0.1)^2=0.01\), and \((0.9)^{13}\approx0.2541866\). Therefore,

P(N=2)=105(0.01)(0.2541866) =1.05(0.2541866) \approx0.2668959.

Rounded to four decimal places, the probability is 0.2669. The result is between 0 and 1, as required. The common mistake avoided is requiring the two groups to have the same number of trials; the necessary shared parameter is the probability \(p\), not \(n\).

Worked Example: Adding Negative Binomial Failure Counts

Two independent hypothetical processes each continue until a specified number of successes occurs. Process A counts failures before its second success, and process B counts failures before its third success. Each trial in both processes has success probability \(p=0.5\). Find the probability that there are exactly two failures in total.

Method and model check. Both variables count failures before a fixed number of successes, and they share the same success probability. The total failure count has a negative binomial distribution with \(r=2+3=5\) successes and \(p=0.5\).

Under this convention, the probability of \(k\) failures before the \(r\)th success is \(\binom{k+r-1}{r-1}p^r(1-p)^k\). For \(k=2\) and \(r=5\),

P(N=2)=\binom{2+5-1}{5-1}(0.5)^5(1-0.5)^2 =\binom{6}{4}(0.5)^5(0.5)^2.

The coefficient is \(\binom{6}{4}=15\), and \((0.5)^5(0.5)^2=(0.5)^7=1/128=0.0078125\). Thus \(P(N=2)=15/128=0.1171875\), which rounds to 0.1172 to four decimal places. The probability is valid and uses the failure-count convention specified in the problem. A common mistake is to add the \(r\) parameters while silently switching to a model that counts successes instead of failures.

Worked Example: Independent Binomial Counts With Different Probabilities

A hypothetical monitoring system has two independent groups. Group A has 2 devices, each with probability \(0.2\) of triggering an alert; group B has 1 device with probability \(0.5\) of triggering an alert. Find the probability of exactly one alert, and decide whether the total is an ordinary binomial count.

Method and model check. Since the probabilities differ, the binomial addition rule does not apply. Calculate the two disjoint ways to get exactly one alert: one alert in A and none in B, or none in A and one in B.

$$ P(N=1) =P(A=1)P(B=0)+P(A=0)P(B=1). $$

Group A has probability \(P(A=1)=\binom{2}{1}(0.2)(0.8)=0.32\) and \(P(A=0)=(0.8)^2=0.64\). Group B has \(P(B=0)=0.5\) and \(P(B=1)=0.5\). Independence gives

$$ P(N=1)=(0.32)(0.5)+(0.64)(0.5) =0.16+0.32 =0.48. $$

Thus the probability is 0.4800. The total is not an ordinary binomial count because its three trials do not share a common probability. The calculation is a convolution: it adds the probabilities of the separate combinations that produce the desired total. Treating the count as \(\operatorname{Binomial}(3,0.2)\), for example, would give \(\binom{3}{1}(0.2)(0.8)^2=0.384\), which is not the correct probability.

When a Single-Model Addition Rule Does Not Apply

The parameter conditions are part of each result, not optional details. In particular, different binomial probabilities prevent the ordinary binomial addition rule, and different negative binomial success probabilities prevent the stated negative binomial addition rule. The sum still has a well-defined distribution, but it generally must be found by combining the subgroup probabilities rather than by assigning one of the same-family models.

For independent integer-valued counts, this direct combination is called convolution. For a target total \(n\) from two groups, it adds the probabilities over every possible split of that total:

P(N_1+N_2=n)=\sum_{j=0}^{n}P(N_1=j)P(N_2=n-j).

A useful exam check is therefore to identify the model for each group, verify independence, and then check the relevant parameter condition before combining. For Poisson counts, add the expected counts. For binomial counts, add trial counts only when the probabilities match. For negative binomial failure counts, add success targets only when the success probabilities match. If a condition fails, do not force the total into a familiar model.

Check independence.
The addition rules stated here assume the subgroup counts are independent.

Identify the parameter condition.
Poisson parameters may differ; binomial probabilities must agree; negative binomial success probabilities must agree.

Combine parameters or use convolution.
Apply the matching addition rule only after its conditions are met; otherwise calculate the needed aggregate probability from subgroup outcomes.

Key takeaway: Independent subgroup counts do not automatically form a familiar single count model. Poisson parameters add; binomial trial counts add under a common success probability; negative binomial success targets add under a common success probability. Check the model convention and parameter compatibility before calculating.

Check Your Understanding

For each problem, first check whether the stated family addition rule applies.

Two independent groups have Poisson counts with parameters \(0.7\) and \(1.3\) over the same period. Find the probability of no claims in total. (Answer: \(e^{-2}\approx0.1353\))
Independent Poisson counts have parameters \(2.4\) and \(0.6\). Find the probability of exactly one claim in total, to four decimal places. (Answer: \(3e^{-3}\approx0.1494\))
Two independent groups have binomial counts with \(n_1=4\), \(n_2=8\), and common probability \(p=0.25\). Find the probability of exactly two successes, to four decimal places. (Answer: \(\binom{12}{2}(0.25)^2(0.75)^{10}\approx0.2323\))
Independent negative binomial counts each count failures before a fixed number of successes. Their success targets are \(r_1=2\) and \(r_2=1\), and both have \(p=0.4\). Find the probability of two failures in total, to four decimal places. (Answer: \(\binom{4}{2}(0.4)^3(0.6)^2=0.1382\))
Group A has two independent Bernoulli trials with success probability \(0.2\); group B has one independent Bernoulli trial with success probability \(0.5\). Find the probability of exactly one success in total and state whether the total is binomial. (Answer: \(0.4800\); it is not an ordinary binomial count because the trial probabilities differ.)

What You'll Learn