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CAS Exam 5 · Claim Count and Frequency · Tutorial 801 of 989

Adding Independent Poisson Claim Counts

Combine independent Poisson counts correctly, including counts from exposure groups with different claim frequencies.

Intermediate 10 min read

What You'll Learn

  • State the conditions under which independent Poisson claim counts have a Poisson sum
  • Derive the distribution of the combined count from the individual count distributions
  • Convert exposure and frequency into a Poisson mean for each group
  • Calculate total-count probabilities using the combined mean
  • Avoid errors from averaging frequencies without weighting by exposure
  • Recognize when dependence makes the Poisson-sum result inapplicable

Why Combining Claim Counts Requires Care

A portfolio count is often the total of counts from several groups: separate territories, coverage segments, or policy cohorts. If each group has a Poisson claim count, the total can also have a Poisson distribution—but only when the group counts are independent. The total mean is the sum of the group means, even when the groups have different claim frequencies.

The important calculation is to get each group’s mean right before combining. A group’s expected claim count depends on both its exposure and its frequency. Averaging the groups’ frequencies without accounting for their different exposure sizes can produce the wrong total mean and therefore the wrong count probabilities.

Definition: Let \(N_j\) be the claim count in group \(j\). If the counts \(N_1,\ldots,N_m\) are mutually independent and \(N_j\sim\operatorname{Poisson}(\lambda_j)\), then their total \(N=\sum_{j=1}^m N_j\) has distribution \(N\sim\operatorname{Poisson}(\lambda)\), where \(\lambda=\sum_{j=1}^m\lambda_j\). Here, \(\lambda_j\) is the expected count in group \(j\).

Deriving the Distribution of a Sum

For two independent counts, \(N_1\sim\operatorname{Poisson}(\lambda_1)\) and \(N_2\sim\operatorname{Poisson}(\lambda_2)\), the event \(N_1+N_2=n\) can occur through any allocation of the \(n\) claims between the groups. The possible allocations are \(N_1=k\) and \(N_2=n-k\), for \(k=0,\ldots,n\). Independence lets us multiply the probabilities for each allocation and then add them.

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

The last equality is the binomial theorem. The expression is exactly the Poisson probability formula with mean \(\lambda_1+\lambda_2\), which justifies the result for two groups. For more groups, apply the two-group result repeatedly: the sum of the first two counts is Poisson with mean \(\lambda_1+\lambda_2\), and it remains independent of each other group count. The resulting total is Poisson with mean \(\sum_j\lambda_j\).

This derivation relies on independence. If groups share a source of claim variation—for example, a common event that can affect both groups—then the counts need not be independent, and this argument does not establish a Poisson distribution for their total. The result also requires each group count to have a Poisson distribution; having the same average count alone is not sufficient.

Different Frequencies and Unequal Exposures

Suppose group \(j\) has exposure \(E_j\), measured in a specified exposure unit, and a constant expected frequency \(\nu_j\) per exposure unit. Under a Poisson count model, the group mean is

$$ \lambda_j=E_j\nu_j. $$

The frequency may differ among groups. The total count still has a Poisson distribution if the group counts are independent, with mean

$$ \lambda_{\text{total}}=\sum_{j=1}^m E_j\nu_j. $$

If total exposure is positive, the corresponding combined expected frequency per exposure unit is

$$ \nu_{\text{total}}= \frac{\sum_{j=1}^m E_j\nu_j}{\sum_{j=1}^m E_j}. $$

This is an exposure-weighted average of the group frequencies. A group with more exposure contributes more to the portfolio’s expected count. The equation describes the expected frequency implied by the model; it does not require the observed count to equal its expected value.

Worked Example: Combining Counts From Two Independent Service Areas

In a hypothetical insurance program, the annual claim counts for two service areas are independent. The first area has count \(N_1\sim\operatorname{Poisson}(1.2)\), and the second has count \(N_2\sim\operatorname{Poisson}(2.8)\). Find the distribution of the total count and the probability of exactly two claims. Verify the probability by adding the possible allocations between areas.

Method. Apply the independent Poisson sum result. The combined mean is

$$ \lambda=1.2+2.8=4.0, \qquad N_1+N_2\sim\operatorname{Poisson}(4.0). $$

The direct calculation for exactly two total claims is

$$ P(N_1+N_2=2)=e^{-4}\frac{4^2}{2!} =e^{-4}\frac{16}{2} =8e^{-4} \approx 0.1465. $$

To verify, the total of two can be allocated as \((N_1,N_2)=(0,2),(1,1)\), or \((2,0)\). By independence, the three probabilities are

$$ \begin{aligned} P(0,2)&=e^{-4}\frac{2.8^2}{2}=3.92e^{-4},\\ P(1,1)&=e^{-4}(1.2)(2.8)=3.36e^{-4},\\ P(2,0)&=e^{-4}\frac{1.2^2}{2}=0.72e^{-4}. \end{aligned} $$

Adding them gives \((3.92+3.36+0.72)e^{-4}=8e^{-4}\), the same result. As a further check, the expected total is \(1.2+2.8=4.0\), consistent with the combined Poisson mean. The common mistake this method avoids is treating exactly two total claims as only one particular allocation, such as one claim in each area.

Worked Example: Different Frequencies Across Unequal Exposures

Two hypothetical policy groups have independent Poisson claim counts. Group A has 60 policy-years and an expected frequency of 0.04 claims per policy-year. Group B has 80 policy-years and an expected frequency of 0.02 claims per policy-year. Find the total count distribution, the combined expected frequency per policy-year, and the probability of no claims or one claim in total.

Step 1: Calculate each group’s Poisson mean.

$$ \lambda_A=60(0.04)=2.4, \qquad \lambda_B=80(0.02)=1.6. $$

Step 2: Combine the independent counts. The total mean is \(2.4+1.6=4.0\), so the total count \(N\) has distribution \(N\sim\operatorname{Poisson}(4.0)\).

Step 3: Calculate the combined expected frequency. Total exposure is \(60+80=140\) policy-years. Therefore,

$$ \nu_{\text{total}}=\frac{4.0}{140} =0.0285714\ldots \approx 0.02857 \quad\text{claims per policy-year}. $$

Step 4: Calculate the probability of at most one claim. The outcomes are zero claims and one claim:

$$ \begin{aligned} P(N\leq 1) &=P(N=0)+P(N=1)\\ &=e^{-4}\frac{4^0}{0!}+e^{-4}\frac{4^1}{1!}\\ &=e^{-4}(1+4)=5e^{-4}\\ &\approx 0.0916. \end{aligned} $$

The combined frequency is not the unweighted average \((0.04+0.02)/2=0.03\), because the groups have different amounts of exposure. That incorrect average would imply a mean of \(140(0.03)=4.2\), rather than the correctly calculated mean of 4.0. Weighting the rates by their exposures gives the appropriate combined expected frequency.

Worked Example: Three Groups and a Tail Probability

A hypothetical portfolio has three independent segments. Their annual claim counts are Poisson with means 1.5, 0.75, and 2.25. Find the distribution of the portfolio total, the probability of exactly two claims, and the probability of at least two claims.

Method. Add the independent Poisson means:

$$ \lambda=1.5+0.75+2.25=4.5, \qquad N\sim\operatorname{Poisson}(4.5). $$

For exactly two claims,

$$ P(N=2)=e^{-4.5}\frac{4.5^2}{2!} =e^{-4.5}\frac{20.25}{2} \approx 0.1125. $$

For at least two claims, use the complement of zero or one claim:

$$ \begin{aligned} P(N\geq 2) &=1-P(N=0)-P(N=1)\\ &=1-e^{-4.5}\left(1+4.5\right)\\ &=1-5.5e^{-4.5}\\ &\approx 0.9389. \end{aligned} $$

The probability of exactly two is contained in the probability of at least two, as it must be: \(0.1125<0.9389\). A common error in the tail calculation is to subtract only \(P(N=0)\), which would give \(P(N\geq1)\), not \(P(N\geq2)\). The excluded outcomes must be all counts below two: both zero and one.

What the Combined Mean Does—and Does Not—Tell You

For independent Poisson counts, the mean and variance of the total both equal the sum of the group means:

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

The mean follows from adding expectations. For the variance, independence makes every covariance between distinct group counts zero, so the variance of the sum is the sum of the variances. Each Poisson count has variance equal to its mean, giving the displayed result. This equality is a useful consistency check, but it does not prove that a count is Poisson: other count distributions can have a specified mean or variance without satisfying the independent Poisson model.

When applying the result, check the model at the group level before adding means. Confirm that the counts being combined are Poisson and that they are mutually independent. Then convert each exposure and frequency to a mean in consistent units, add the means, and use the Poisson probability formula or its cumulative form. If there is dependence, do not use this result automatically; the total’s distribution must be assessed under the appropriate dependence model.

Key takeaway: Independent Poisson counts add to a Poisson count whose mean is the sum of the individual means. For groups with different frequencies, calculate each mean as exposure times frequency first; the combined frequency is exposure-weighted, not a simple average.

Check Your Understanding

For each problem, first identify the total Poisson mean. Round probabilities to four decimal places.

  1. Two independent hypothetical account groups have annual count distributions \(\operatorname{Poisson}(0.7)\) and \(\operatorname{Poisson}(1.3)\). Find the total count distribution and the probability of zero claims. (Answer: \(N\sim\operatorname{Poisson}(2.0)\); \(P(N=0)=0.1353\).)
  2. Two independent groups have 30 and 50 exposure units, respectively, with frequencies of 0.08 and 0.02 claims per exposure unit. Find the total mean and the probability of exactly two claims. (Answer: \(\lambda=30(0.08)+50(0.02)=3.4\); \(P(N=2)=e^{-3.4}(3.4^2/2!)=0.1929\).)
  3. Three independent segments have Poisson means 0.5, 1.25, and 2.25. Find the total distribution and the probability of exactly four claims. (Answer: \(N\sim\operatorname{Poisson}(4.0)\); \(P(N=4)=e^{-4}4^4/4!=0.1954\).)
  4. Two independent hypothetical groups have exposures of 100 and 200 policy-years and frequencies of 0.03 and 0.015 claims per policy-year. Find the combined expected frequency and the probability of no claims. (Answer: \(\lambda=3+3=6\), combined frequency \(6/300=0.020\) claims per policy-year; \(P(N=0)=e^{-6}=0.0025\).)
  5. Independent fleets have Poisson counts with means \(20(0.10)\) and \(60(0.025)\). Find the total mean and the probability of at least one claim. (Answer: \(\lambda=2+1.5=3.5\); \(P(N\geq1)=1-e^{-3.5}=0.9698\).)

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