Pooling Independent Portfolios With Different Frequencies

Why Pooling Requires Both Counts and Exposure

Portfolios may have different volumes of exposure and different claim frequencies. Combining their frequency rates by taking an unweighted average can give a misleading result: a small portfolio and a large portfolio should not necessarily have equal influence on the pooled rate. The first step is to calculate each portfolio’s expected count from its own exposure and rate, then add those expected counts.

The distribution of the total is a separate question. If each portfolio count is Poisson and the counts are independent, their total is Poisson with parameter equal to the sum of the individual parameters. Different Poisson rates do not prevent this result. Independence matters: the total expected count remains the sum of expected counts even without independence, but dependence can affect the variance and may mean the total is not Poisson. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose portfolio \(j\) has exposure \(e_j\) measured in a specified exposure unit and frequency rate \(\lambda_j\) expected claims per exposure unit. Its count \(N_j\) has expected value \(E[N_j]=e_j\lambda_j\). For \(m\) portfolios, the pooled count is \(N=\sum_{j=1}^{m}N_j\), and its expected value is \(E[N]=\sum_{j=1}^{m}e_j\lambda_j\). If total exposure is \(e=\sum_{j=1}^{m}e_j>0\), the pooled expected frequency per exposure unit is \(\lambda_{\text{pool}}=E[N]/e\).

The exposure unit must be consistent across the calculation. For example, if one rate is stated per policy-year, every exposure used with that rate must be measured in policy-years. If rates have different numerical scales, convert them to a common unit before calculating expected counts.

The pooled rate is an exposure-weighted average:

\lambda_{\text{pool}} =\frac{\sum_{j=1}^{m}e_j\lambda_j}{\sum_{j=1}^{m}e_j}.

This formula summarizes expected claims per combined exposure unit. It does not, by itself, establish that the pooled count follows a Poisson distribution. That conclusion requires distributional assumptions as well.

Pooling Independent Poisson Counts

For each portfolio, assume that \(N_j\sim\operatorname{Poisson}(\mu_j)\), where \(\mu_j=e_j\lambda_j\), and assume the portfolio counts are mutually independent. The individual Poisson parameters can differ. Under these assumptions, the total has a Poisson distribution with parameter \(\sum_j\mu_j\).

Theorem: If \(N_1,\ldots,N_m\) are mutually independent and \(N_j\sim\operatorname{Poisson}(\mu_j)\), then \(N=\sum_{j=1}^{m}N_j\sim\operatorname{Poisson}\left(\sum_{j=1}^{m}\mu_j\right)\). The result applies to independent Poisson counts over the specified exposures; it does not follow from independence alone when the component counts are not Poisson.

One justification uses probability generating functions. For a Poisson random variable \(N_j\), the probability generating function is \(G_{N_j}(z)=E[z^{N_j}]=\exp(\mu_j(z-1))\). Independence means the generating function of the sum is the product of the component generating functions:

G_N(z) =\prod_{j=1}^{m}G_{N_j}(z) =\prod_{j=1}^{m}\exp\!\left(\mu_j(z-1)\right) =\exp\!\left(\left(\sum_{j=1}^{m}\mu_j\right)(z-1)\right).

The final expression is the probability generating function of a Poisson distribution with parameter \(\sum_j\mu_j\), which proves the result. Since \(\mu_j=e_j\lambda_j\), the parameter for the total is also the expected total count.

Standardize the exposure units.
Convert each exposure and rate to matching units before multiplying them.

Calculate component expected counts.
For each portfolio, calculate \(\mu_j=e_j\lambda_j\).

Add expected counts and exposures.
Calculate \(\mu=\sum_j\mu_j\) and \(e=\sum_je_j\).

Find the pooled rate and check the model conditions.
Calculate \(\lambda_{\text{pool}}=\mu/e\). Use \(N\sim\operatorname{Poisson}(\mu)\) only if the component counts are Poisson and independent.

Worked Example: Pooling Two Portfolios With Different Rates

A hypothetical insurer has two independent portfolios for a one-year period. Portfolio A has 100 exposure units and an expected frequency of 0.02 claims per unit. Portfolio B has 50 exposure units and an expected frequency of 0.04 claims per unit. Each portfolio count is Poisson. Calculate the pooled frequency and the probability of exactly three claims in total.

Method and assumptions. Calculate each Poisson parameter as exposure times frequency. Because the two Poisson counts are independent, their sum is Poisson with parameter equal to the sum of those parameters.

Portfolio A. Its expected count is \(100(0.02)=2\).

Portfolio B. Its expected count is \(50(0.04)=2\).

Total count and exposure. The expected total count is \(2+2=4\), and total exposure is \(100+50=150\) units. The pooled frequency is \(4/150=0.026666\ldots\) claims per exposure unit.

Distribution and probability. Independence and the Poisson assumptions give \(N\sim\operatorname{Poisson}(4)\). Therefore,

P(N=3)=e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} \approx 0.1954.

The check is that the Poisson parameter, 4, equals the sum of the two component expected counts. The probability is rounded to four decimal places. Common mistake avoided: averaging the two rates without weighting them by exposure gives \((0.02+0.04)/2=0.03\), which is not the pooled frequency here. The portfolios have different exposures, so the correct pooled rate is \(4/150\), not the unweighted average.

Rates on Different Scales and Units

A rate may be quoted per 1,000 policy-years while another calculation uses individual policy-years. A rate of 3 claims per 1,000 policy-years is \(3/1{,}000=0.003\) claims per policy-year. Calculate the expected count using either the converted rate and policy-year exposure or the original rate and exposure measured in thousands of policy-years. The two approaches must agree.

Worked Example: Converting Rates Before Pooling

Two independent hypothetical portfolios have Poisson claim counts. Portfolio A has 1,200 policy-years and a rate of 2 claims per 1,000 policy-years. Portfolio B has 800 policy-years and a rate of 5 claims per 1,000 policy-years. Find the expected total count, the pooled rate per 1,000 policy-years, and the probability of no claims in total.

Method and assumptions. Convert each quoted rate into claims per policy-year before multiplying by exposure. Since the component counts are independent Poisson variables, the total is Poisson.

Portfolio A. Its rate per policy-year is \(2/1{,}000=0.002\). Its expected count is \(1{,}200(0.002)=2.4\).

Portfolio B. Its rate per policy-year is \(5/1{,}000=0.005\). Its expected count is \(800(0.005)=4\).

Pool the expected counts. The total expected count is \(2.4+4=6.4\). The total exposure is \(1{,}200+800=2{,}000\) policy-years, so the pooled rate per policy-year is \(6.4/2{,}000=0.0032\), or \(3.2\) claims per 1,000 policy-years.

Calculate the probability of zero claims. The total count is Poisson with parameter \(6.4\), so

P(N=0)=e^{-6.4}\frac{6.4^0}{0!}=e^{-6.4}\approx 0.0017.

The check is that \(2{,}000(0.0032)=6.4\), matching the sum of the component expected counts. The probability is rounded to four decimal places. Common mistake avoided: multiplying 1,200 policy-years directly by 2 claims per 1,000 policy-years would overstate Portfolio A’s expected count by a factor of 1,000; convert the units first.

When the Pooled Total Is Not Poisson

The Poisson-sum result depends on both the Poisson form of each component and independence. Different rates are permitted, but dependence between counts can prevent the total from being Poisson. In general, for two counts,

E[N_1+N_2]=E[N_1]+E[N_2], \qquad \operatorname{Var}(N_1+N_2) =\operatorname{Var}(N_1)+\operatorname{Var}(N_2)+2\operatorname{Cov}(N_1,N_2).

The expected total is the sum of the expected counts regardless of dependence. Dependence can affect the variance through the covariance, which may be positive, negative, or zero. Therefore, it is not correct to say that dependence always changes the variance. Even when the covariance is zero, dependence can still mean the total is not Poisson.

Worked Example: A Shared Shock Breaks the Pooled Poisson Model

Consider two hypothetical portfolios with counts \(N_1=A+C\) and \(N_2=B+C\). Here, \(A\sim\operatorname{Poisson}(1)\), \(B\sim\operatorname{Poisson}(2)\), and \(C\sim\operatorname{Poisson}(0.5)\), and \(A\), \(B\), and \(C\) are mutually independent. Interpret \(C\) as a shared event that generates one claim in each portfolio. Find the mean and variance of the total count, then determine whether it is Poisson.

Method and assumptions. Calculate the two marginal means and variances, account for their shared component when finding the covariance, and compare the total variance with the mean. Each component \(A\), \(B\), and \(C\) is Poisson, but the portfolio counts \(N_1\) and \(N_2\) are dependent because both contain \(C\).

Marginal means and variances. For a Poisson variable, mean and variance equal its parameter. Thus \(E[N_1]=1+0.5=1.5\) and \(\operatorname{Var}(N_1)=1+0.5=1.5\). Likewise, \(E[N_2]=2+0.5=2.5\) and \(\operatorname{Var}(N_2)=2+0.5=2.5\).

Total mean. The total is \(N=N_1+N_2=A+B+2C\). Its expected value is \(E[N]=1+2+2(0.5)=4\).

Total variance. The shared term gives \(\operatorname{Cov}(N_1,N_2)=\operatorname{Var}(C)=0.5\). Therefore,

\operatorname{Var}(N) =1.5+2.5+2(0.5)=5.

A Poisson distribution with mean 4 must have variance 4. Since the total has mean 4 and variance 5, it is not Poisson. As a further check, \(P(N=0)=P(A=0,B=0,C=0)=e^{-1-2-0.5}=e^{-3.5}\approx0.0302\), whereas a Poisson distribution with mean 4 would give \(e^{-4}\approx0.0183\). The probabilities differ. Both are rounded to four decimal places. Common mistake avoided: treating each portfolio’s marginal Poisson distribution as sufficient to conclude that their sum is Poisson. The counts here are dependent, so the independent-sum theorem does not apply.

Common Pooling Checks

Before using a pooled Poisson model, check what the component counts represent and how they are related. A sum of independent Poisson counts remains Poisson even when portfolio rates differ; a sum of independent counts that are not Poisson is not generally Poisson. Nor does knowing the marginal distributions alone establish independence.

Do not average rates without exposure weights. The pooled frequency is total expected claims divided by total exposure.
Do not mix exposure units. Convert every rate and exposure to a consistent basis before multiplying.
Do not equate a sum of expectations with a distributional result. Expected counts add, but Poisson form requires the appropriate distribution and independence assumptions.
Do not automatically attribute variance changes to dependence. The covariance enters the variance formula and can be zero.

Key takeaway: Calculate each expected count as exposure times rate, add those counts, and divide by total exposure to find the pooled frequency. The total is Poisson with that summed parameter when the portfolio counts are independent Poisson variables; check those assumptions separately from the rate calculation.

Check Your Understanding

For each question, check both the exposure calculation and any assumptions needed to identify the distribution.

Two independent portfolios have Poisson counts, with exposures of 40 and 60 units and rates of 0.03 and 0.01 claims per unit, respectively. Find the expected total count and pooled rate per unit. (Answer: Expected total count \(=1.8\); pooled rate \(=0.018\) per unit.)
Independent portfolio counts are Poisson with parameters 1.2 and 2.8. Find the distribution of their total and the probability of zero claims. (Answer: \(N\sim\operatorname{Poisson}(4)\); \(P(N=0)=e^{-4}\approx0.0183\).)
Two independent Poisson portfolios have exposures of 2,000 and 500 policy-years, with rates of 1.5 and 4 claims per 1,000 policy-years, respectively. Find the expected total count and pooled rate per 1,000 policy-years. (Answer: Expected total count \(=3+2=5\); pooled rate \(=2\) claims per 1,000 policy-years.)
Three independent portfolio counts are Poisson with parameters 0.8, 1.7, and 0.5. Find the probability of exactly two claims in total. (Answer: \(N\sim\operatorname{Poisson}(3)\); \(P(N=2)=e^{-3}3^2/2!\approx0.2240\).)
Two portfolio counts have Poisson marginal distributions with means 2 and 3, and their covariance is 0.4. Find the total mean and variance. Can the total be Poisson? (Answer: Mean \(=5\); variance \(=2+3+2(0.4)=5.8\); it cannot be Poisson because a Poisson variable with mean 5 would have variance 5.)

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