Likelihoods For Grouped Claim Counts

Why Group Claim Counts?

A frequency data set may contain one record per policy, even when many policies share the same rating characteristics. If those policies have the same modeled claim rate per unit of exposure, their counts can be summarized by total claims and total exposure for that rating cell. The key is to preserve the exposure that belongs with the modeled rate: summing counts is not enough.

This tutorial considers Poisson claim counts with known, nonnegative exposures. Its focus is the likelihood for a grouped count and the conditions under which aggregation preserves estimation of the modeled rate. It does not assume that policies with different rates can always be pooled into an ordinary cell.

Definition: Suppose independent counts $Y_i$ satisfy $Y_i\sim\operatorname{Poisson}(e_i\lambda)$, where $e_i$ is known exposure and $\lambda$ is a common claim rate per exposure unit. For a group of observations, define total claims $Y=\sum_iY_i$ and total exposure $E=\sum_i e_i$. The grouped cell likelihood is the Poisson likelihood for $Y$ with mean $E\lambda$: $$ L(\lambda;Y,E)=\frac{\exp(-E\lambda)(E\lambda)^Y}{Y!}. $$

The common-rate condition matters. Exposure may differ from one policy to another, but the modeled rate $\lambda$ must be the same for the observations being grouped. For example, each policy can have a different term length while still sharing a rate per policy-year. The exposures then add. The result assumes independent Poisson counts conditional on the modeled rates and known exposures, with expected count proportional to exposure. If the modeled rates differ, raw exposure totals do not generally produce a valid common-rate cell.

Deriving the Grouped Likelihood and Rate Estimate

For independent policy counts, the joint likelihood is the product of their Poisson likelihoods. Write $Y=\sum_i y_i$ and $E=\sum_i e_i$. Then

\begin{aligned} L_{\mathrm{joint}}(\lambda) &=\prod_i\frac{\exp(-e_i\lambda)(e_i\lambda)^{y_i}}{y_i!}\\ &=\exp(-E\lambda)\lambda^Y \prod_i\frac{e_i^{y_i}}{y_i!}. \end{aligned}

The part of this expression that depends on $\lambda$ is $\exp(-E\lambda)\lambda^Y$. The grouped likelihood can be written as

L_{\mathrm{group}}(\lambda) =\exp(-E\lambda)\lambda^Y\frac{E^Y}{Y!}.

The joint and grouped likelihoods therefore differ only by a factor that does not depend on $\lambda$. They give the same likelihood-based estimate of the common rate. They are not necessarily numerically equal: the grouped likelihood treats the total as one Poisson observation, while the joint likelihood also describes how that total was allocated among policies.

Derivation of the maximum likelihood estimate: Taking the log of the grouped likelihood gives $\ell(\lambda)=-E\lambda+Y\log(E\lambda)-\log(Y!)$. For $Y>0$, its derivative is $-E+Y/\lambda$. Setting this equal to zero gives $\hat{\lambda}=Y/E$. Its second derivative is $-Y/\lambda^2<0$, so this stationary point is a maximum. If $Y=0$ and $E>0$, the likelihood is $\exp(-E\lambda)$, which is maximized at the boundary $\hat{\lambda}=0$.

Thus, for a cell with a common rate, the total count and total exposure contain the information needed to estimate that rate. If the exposure is reported in a different unit, the rate estimate changes units accordingly; the count does not change. The exposure basis must remain consistent with the interpretation of the rate.

Worked Example: Grouping Policies With Unequal Terms

A hypothetical rating cell has three policies with exposures of 0.5, 1.0, and 1.5 policy-years. Their observed claim counts are 0, 1, and 2. Assume independent Poisson counts with one common rate $\lambda$ per policy-year. Find the grouped likelihood and maximum likelihood estimate, and compare the grouped and individual likelihoods at that estimate.

Step 1: Sum claims and exposure. Total claims are $Y=0+1+2=3$. Total exposure is $E=0.5+1.0+1.5=3.0$ policy-years.

Step 2: Write the grouped likelihood. The cell total has a Poisson distribution with mean $3\lambda$, so $L_{\mathrm{group}}(\lambda)=\exp(-3\lambda)(3\lambda)^3/3!$.

Step 3: Estimate the rate. The common-rate estimate is $\hat{\lambda}=Y/E=3/3=1.00$ claim per policy-year.

Step 4: Evaluate the grouped likelihood at the estimate. Substituting $\hat{\lambda}=1$ gives $\exp(-3)3^3/3!=\exp(-3)(27/6)=4.5\exp(-3)\approx0.2240$.

Step 5: Check against the individual likelihood. At $\lambda=1$, the product of the three individual probabilities is $\exp(-3)(0.5^0/0!)(1.0^1/1!)(1.5^2/2!)=1.125\exp(-3)\approx0.0560$. The grouped likelihood is four times this value, but the ratio $4$ does not depend on $\lambda$. Both likelihoods therefore have their maximum at the same rate, $1.00$.

Method and trap avoided. Add both the counts and the exposures, then use $Y/E$. The grouped likelihood is not required to equal the product of the individual likelihoods; it must preserve the parameter-dependent part for rate estimation.

When Fixed Relativities Differ

There is a useful extension when observations have different rates that are known multiples of a shared base rate. Suppose $Y_i\sim\operatorname{Poisson}(e_i r_i\lambda_0)$, independently, where each relativity $r_i$ is fixed and known and $\lambda_0$ is the unknown base rate. The sum is Poisson with mean $\lambda_0\sum_i e_i r_i$. Consequently, the total count can be used to estimate $\lambda_0$, provided the accompanying exposure is the relativity-weighted exposure $E^*=\sum_i e_i r_i$, not the raw exposure $\sum_i e_i$.

This aggregation preserves inference about the one unknown base rate. It does not mean that the original observations had a common rate per unit of raw exposure. If the relativities are unknown parameters, or if the aim is to estimate separate rates for groups, combining counts can discard information about how claims are allocated among groups.

Worked Example: Combining Counts With Fixed Rate Multipliers

In a hypothetical portfolio, group A has exposure 40 policy-years and fixed relativity $r_A=1$. Group B has exposure 20 policy-years and fixed relativity $r_B=1.5$. Both groups share an unknown base rate $\lambda_0$. Their observed claim counts are 25 and 17. Find the effective exposure, the maximum likelihood estimate of $\lambda_0$, and the grouped likelihood contribution at that estimate.

Step 1: Identify the group means. The Poisson means are $40(1)\lambda_0=40\lambda_0$ for A and $20(1.5)\lambda_0=30\lambda_0$ for B.

Step 2: Calculate total claims and effective exposure. Total claims are $Y=25+17=42$. The effective exposure is $E^*=40(1)+20(1.5)=40+30=70$ relativity-weighted policy-years.

Step 3: Form the grouped likelihood and estimate. The total has a Poisson distribution with mean $70\lambda_0$. Thus $L(\lambda_0)=\exp(-70\lambda_0)(70\lambda_0)^{42}/42!$, and $\hat{\lambda}_0=42/70=0.6000$ claims per relativity-weighted policy-year.

Step 4: Substitute and check. At the estimate, the fitted total mean is $70(0.6000)=42$, matching the observed total. The likelihood contribution is $\exp(-42)42^{42}/42!$. The group-specific fitted means are $40(0.6000)=24$ and $30(0.6000)=18$, which sum to 42.

Method and trap avoided. The counts can be represented by one Poisson total for estimating the shared base rate because the multipliers are fixed. The correct exposure is 70, not the raw sum $40+20=60$. Using raw exposure 60 with one common rate per policy-year would specify a different model.

When Separate Rates Must Stay Separate

If two rating groups have separately estimated rates, a total count and raw total exposure do not preserve both rate estimates. The total can be modeled as Poisson under an imposed common-rate assumption, but that assumption is a different model from one allowing separate rates. The allocation of claims between groups is informative about the separate rates and should be retained.

Worked Example: Why Pooling Separate Rates Changes the Model

Two hypothetical groups have exposures 2 and 1 policy-years and observed counts 3 and 1. Assume the groups have distinct, separately estimated Poisson rates $\lambda_A$ and $\lambda_B$. Compare their separate rate estimates with the estimate obtained by incorrectly imposing one common rate.

Step 1: Estimate each rate under the stated model. Group A has $\hat{\lambda}_A=3/2=1.5$ claims per policy-year. Group B has $\hat{\lambda}_B=1/1=1.0$ claim per policy-year.

Step 2: Write the separate likelihood. It is $L(\lambda_A,\lambda_B)=\exp(-2\lambda_A)(2\lambda_A)^3/3!\times\exp(-\lambda_B)\lambda_B$. The separate estimates maximize this likelihood with respect to their own rates.

Step 3: Calculate the pooled estimate under a different assumption. If one instead forces $\lambda_A=\lambda_B=\lambda$, the total count is 4 and the raw total exposure is 3. The imposed common-rate estimate is $\hat{\lambda}=4/3=1.3333$ claims per policy-year.

Step 4: Check the distinction. The pooled estimate $1.3333$ is neither separate estimate, $1.5$ or $1.0$. It answers a common-rate question, not the separate-rate question. The total count alone cannot recover both group-specific estimates.

Method and trap avoided. Add counts and exposures only when doing so preserves the parameters being estimated. Pooling separate rates silently imposes equality of those rates.

Practical Checks and Common Traps

Before calculating a grouped likelihood contribution, verify what the modeled rate represents and which exposure belongs with it. A rating cell can combine policies with different term lengths when their modeled rate is the same. Fixed known multipliers can also be accommodated for estimation of a shared base rate, but only through their weighted exposure. By contrast, a cell total is not a substitute for group counts when the model estimates distinct rates or unknown group effects.

Wrong exposure basis: using raw summed exposure when fixed known relativities differ. Use $\sum_i e_i r_i$ for a shared base rate.
Unintended pooling: combining groups with separate rates and treating them as one common-rate cell. This changes the model.
Likelihood equality confusion: assuming the grouped likelihood must equal the product of individual likelihoods. They can differ by a parameter-independent factor and still give the same rate estimate.
Missing exposure: estimating a rate from total claims alone. For a common Poisson rate, the estimate is total claims divided by the corresponding exposure.

These checks are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus because they connect the Poisson count likelihood with the exposure basis used to model frequency. A correct cell summary preserves the relationship between the count and the rate; a careless one can instead alter the model being fit.

Check Your Understanding

For each problem, identify the appropriate grouping calculation before using the final answer to check your work.

A rating cell has two independent Poisson counts with a common rate per policy-year. Exposures are 1.2 and 0.8 policy-years, and counts are 0 and 2. Find the rate estimate and the grouped likelihood evaluated at that estimate. (Answer: $\hat{\lambda}=2/2=1.000$; $L=\exp(-2)2^2/2!=2\exp(-2)\approx0.2707$.)
Two groups share an unknown base rate, with fixed known relativities 0.8 and 1.4. Exposures are 10 and 5 policy-years, and counts are 7 and 9. Find the effective exposure and base-rate estimate. (Answer: $E^*=10(0.8)+5(1.4)=15$; $\hat{\lambda}_0=16/15\approx1.0667$.)
Three policies in one common-rate cell have exposures 0.25, 0.75, and 0.50 policy-years, with counts 1, 0, and 0. Find the grouped rate estimate and grouped likelihood at the estimate. (Answer: $Y=1$, $E=1.50$, $\hat{\lambda}=2/3$; $L=\exp(-1)(1^1)/1!=\exp(-1)\approx0.3679$.)
Two groups have separate rates, exposures 3 and 2 policy-years, and counts 6 and 2. State the separate estimates and the estimate if one instead imposes a common rate. (Answer: separate estimates are $6/3=2$ and $2/2=1$; the imposed common-rate estimate is $8/5=1.6$.)
Two groups have exposures 10 and 20 policy-years, and counts 4 and 6. Their rates are $\lambda_0$ and $\lambda_0\exp(\beta)$, with both $\lambda_0$ and $\beta$ unknown. Can the total count alone preserve information needed to estimate both parameters? (Answer: No. The group counts must be retained because their allocation contains information about $\beta$.)

What You'll Learn