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CAS Exam 8 · Frequency Model Specification · Tutorial 119 of 989

Repeated Observations And Claim-Count Dependence

Learn to recognize shared-risk dependence across repeated policy periods, quantify its effect on claim-count variance, and choose inference that respects the policy as a cluster.

Intermediate 10 min read

What You'll Learn

  • Distinguish repeated observations from independent observational units
  • Describe conditional independence and marginal dependence under a shared risk effect
  • Derive the variance of a policy’s aggregate claim count
  • Quantify how ignoring within-policy dependence affects standard errors
  • Recognize why zero covariance does not establish independence
  • Identify when cluster-robust standard errors are appropriate

Why Repeated Policy-Period Rows Need Attention

A frequency data set may contain one row for each policy-period rather than one row per policy. A policy observed over several years therefore contributes several counts. Treating every row as independent can be inappropriate when the periods share risk characteristics that are not represented in the model. A policy with persistently higher-than-average risk, for example, may tend to have more claims in each of its observed years.

This issue is separate from whether an exposure offset is correctly specified. An offset can scale the expected count for the length of a policy-period; it does not make repeated counts from one policy independent. The relevant questions are what is the observational unit for the dependence structure, what assumptions are being made about counts within that unit, and how the dependence affects uncertainty.

Definition: A cluster is a group of observations that may be dependent because they share a unit or source, such as repeated policy-periods from one policy. Within-cluster observations are dependent when their joint distribution does not factor into the product of their marginal distributions. Nonzero covariance is one possible sign of dependence, but dependence does not require nonzero covariance.

A common model makes counts independent conditional on a policy-specific risk effect, while allowing them to be dependent after that unobserved effect is averaged out. Conditional independence is not the same as marginal independence. This distinction is useful for understanding why repeated rows can contain less independent information than the row count suggests.

A Shared-Risk Count Model

Consider a policy \(i\) observed in periods \(t=1,\ldots,T_i\). Let \(N_{it}\) be its claim count, \(e_{it}\) its exposure, and \(\lambda\) the baseline claim rate. Suppose the policy has a positive shared risk multiplier \(\Theta_i\). The following model assumes \(E[\Theta_i]=1\), \(\operatorname{Var}(\Theta_i)=\alpha\), and conditional independence of its period counts given \(\Theta_i\):

$$ N_{it}\mid\Theta_i\ \sim\ \operatorname{Poisson}(e_{it}\lambda\Theta_i), \qquad N_{i1},\ldots,N_{iT_i}\text{ are conditionally independent given }\Theta_i. $$

For the gamma-frailty version of this model, take \(\Theta_i\sim\operatorname{Gamma}(1/\alpha,\text{ rate }1/\alpha)\), with \(\alpha>0\). This parameterization gives the stated mean and variance. Different policies are assumed independent of one another, including their risk effects. This is a model for illustrating shared heterogeneity; it is not a necessary description of every form of claim-count dependence.

Conditional independence means that after fixing the policy’s risk effect, the period counts are independent Poisson variables. But a high value of \(\Theta_i\) raises the expected count in every period, and a low value lowers each one. Averaging over the distribution of \(\Theta_i\) therefore generally creates dependence between the period counts.

Deriving the Marginal Variance and Covariance

Write \(\mu_{it}=e_{it}\lambda\). By the law of total expectation, the marginal expected count is \(E[N_{it}]=\mu_{it}\), since \(E[\Theta_i]=1\). The law of total variance separates the Poisson variation conditional on risk from the variation in the conditional mean:

$$ \begin{aligned} \operatorname{Var}(N_{it}) &= E[\operatorname{Var}(N_{it}\mid\Theta_i)] +\operatorname{Var}(E[N_{it}\mid\Theta_i])\\ &= E[\mu_{it}\Theta_i] +\operatorname{Var}(\mu_{it}\Theta_i)\\ &= \mu_{it}+\alpha\mu_{it}^2. \end{aligned} $$

For two distinct periods \(s\ne t\), conditional independence gives a zero conditional covariance. The law of total covariance then leaves the covariance of the conditional means:

$$ \begin{aligned} \operatorname{Cov}(N_{is},N_{it}) &= E[\operatorname{Cov}(N_{is},N_{it}\mid\Theta_i)] +\operatorname{Cov}(E[N_{is}\mid\Theta_i],E[N_{it}\mid\Theta_i])\\ &= 0+\operatorname{Cov}(\mu_{is}\Theta_i,\mu_{it}\Theta_i)\\ &= \alpha\mu_{is}\mu_{it}. \end{aligned} $$

Thus, in this shared-multiplier model, distinct periods have positive covariance when \(\alpha>0\) and both have positive expected counts. This result is specific to the model. More generally, a dependence structure can be present without positive covariance, so checking covariance alone cannot establish independence.

Let \(N_i=\sum_t N_{it}\) be the policy’s total count and \(M_i=\sum_t\mu_{it}\) its expected total count. Adding the period variances and all pairwise covariance terms yields:

$$ \operatorname{Var}(N_i) =\sum_t(\mu_{it}+\alpha\mu_{it}^2) +2\sum_{s<t}\alpha\mu_{is}\mu_{it} =M_i+\alpha M_i^2. $$

The equality follows because \((\sum_t\mu_{it})^2=\sum_t\mu_{it}^2+2\sum_{s<t}\mu_{is}\mu_{it}\). The same result follows by conditioning directly on \(\Theta_i\): given the shared multiplier, the total is Poisson with mean \(M_i\Theta_i\).

Worked Example: Two Periods With a Shared Risk Effect

Method and assumptions. A hypothetical policy is observed for one year and then two years. Its baseline annual claim rate is \(\lambda=0.20\). Assume \(\Theta_i\sim\operatorname{Gamma}(4,\text{ rate }4)\), so \(E[\Theta_i]=1\) and \(\operatorname{Var}(\Theta_i)=\alpha=1/4=0.25\). Given \(\Theta_i\), the two period counts are independent Poisson variables with means \(1(0.20)\Theta_i\) and \(2(0.20)\Theta_i\), respectively. Find the marginal means, variances, covariance, and total-count variance.

First period. Its expected count before applying the risk effect is \(\mu_{i1}=1(0.20)=0.20\). Its marginal mean is \(0.20\), and its variance is \(0.20+0.25(0.20)^2=0.20+0.01=0.21\).

Second period. Its expected count before applying the risk effect is \(\mu_{i2}=2(0.20)=0.40\). Its marginal mean is \(0.40\), and its variance is \(0.40+0.25(0.40)^2=0.40+0.04=0.44\).

Covariance. The shared-risk formula gives \(\operatorname{Cov}(N_{i1},N_{i2})=0.25(0.20)(0.40)=0.02\).

Total. The mean total count is \(0.20+0.40=0.60\). Its variance, including twice the covariance, is \(0.21+0.44+2(0.02)=0.69\). The aggregate formula checks this result: \(0.60+0.25(0.60)^2=0.60+0.09=0.69\).

Check and common mistake. The computed total variance exceeds the mean by \(0.09\), as expected under this gamma shared-risk model. Treating the two rows as independent would omit the \(2(0.02)=0.04\) covariance contribution and give \(0.65\), not \(0.69\). That independence calculation is the mistake this method avoids.

Implications for Standard Errors

Suppose the target is a common annual rate and there are \(m\) independent policies, each observed for two one-year periods under the shared-risk model. The rate estimator is total claims divided by total exposure, \(\widehat{\lambda}=\sum_i(N_{i1}+N_{i2})/(2m)\). If the model’s mean is correct, its expectation is \(\lambda\). Dependence within a policy affects its variance even though different policies are independent.

For each policy, the total expected count is \(2\lambda\), so its total-count variance is \(2\lambda+\alpha(2\lambda)^2\). Summing over independent policies and dividing by the squared total exposure gives:

$$ \operatorname{Var}(\widehat{\lambda}) =\frac{m\{2\lambda+4\alpha\lambda^2\}}{(2m)^2} =\frac{\lambda}{2m}+\frac{\alpha\lambda^2}{m}. $$

The first term is the variance that would arise under independent Poisson counts in all policy-period rows. The second term is the extra contribution from shared risk. This difference matters for uncertainty statements and tests even when the rate estimate itself remains centered on the true rate under the stated mean model.

Worked Example: Comparing Rate Standard Errors

Method and assumptions. Consider \(m=100\) independent hypothetical policies, each observed for two one-year periods. The true annual rate is \(\lambda=0.20\), and the common shared multiplier has a gamma distribution with mean \(1\) and variance \(\alpha=0.25\). Conditional on each policy’s multiplier, the two counts are independent Poisson with mean \(0.20\Theta_i\) each. Compare the theoretical standard error of the exposure-based rate estimate with the standard error obtained by incorrectly treating all 200 rows as independent Poisson observations. These are model-based theoretical standard errors, not estimates from a realized portfolio.

Correct cluster calculation. For one policy, the two-period total has mean \(2(0.20)=0.40\) and variance \(0.40+0.25(0.40)^2=0.44\). Across 100 independent policies, total claims have variance \(100(0.44)=44\). Total exposure is \(200\), so the estimator’s variance is \(44/200^2=44/40{,}000=0.0011\). Its standard error is \(\sqrt{0.0011}=0.033166\), or \(0.0332\) to four decimal places.

Incorrect row-independent calculation. Under independent Poisson counts, the total mean and variance would both be \(200(0.20)=40\). The resulting variance estimate for the rate is \(40/40{,}000=0.0010\), giving standard error \(\sqrt{0.0010}=0.031623\), or \(0.0316\) to four decimal places.

Comparison and check. The standard-error ratio is \(\sqrt{0.0011/0.0010}=\sqrt{1.1}=1.0488\), so the cluster-aware standard error is about \(4.88\%\) larger. The variance difference, \(0.0011-0.0010=0.0001\), equals \(\alpha\lambda^2/m=0.25(0.20)^2/100=0.0001\). Treating the two years for each policy as independent omits this contribution.

Dependence Is Not the Same as Covariance

A common exam trap is to define dependent observations as observations with nonzero covariance. That definition is too narrow: zero covariance does not, in general, imply independence. The following discrete example makes the distinction explicit.

Worked Example: Dependent Counts With Zero Covariance

Method and assumptions. Let \(X\) be equally likely to take the count values \(0,1,2\). Define the second count \(Y\) to be \(1\) when \(X\) is \(0\) or \(2\), and \(0\) when \(X=1\). Calculate the covariance and determine whether \(X\) and \(Y\) are independent.

The means are \(E[X]=(0+1+2)/3=1\) and \(E[Y]=(1+0+1)/3=2/3\). The product \(XY\) takes values \(0,0,2\) across the three equally likely outcomes, so \(E[XY]=2/3\).

Therefore, \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=2/3-(1)(2/3)=0\). Nevertheless, the variables are dependent: \(P(Y=1\mid X=0)=1\), whereas \(P(Y=1)=2/3\). The conditional probability differs from the marginal probability, so independence fails.

Check and common mistake. The zero covariance is correct, but concluding independence from it would be wrong. Independence always implies zero covariance when the relevant moments exist; the converse is not generally true. Use the joint distribution or conditional probabilities to assess independence rather than relying on covariance alone.

Choosing an Inference Strategy

When policy-period rows are clustered, the standard error must reflect the unit across which observations may be dependent. Two common approaches are to specify a joint model for the counts, such as a shared-risk model, or to retain a suitable mean model and use a cluster-robust variance estimator. The choice depends on the intended assumptions and analysis; neither approach makes arbitrary dependence disappear.

In a generalized linear model, a cluster-robust sandwich variance uses policy-level score contributions rather than treating each period’s score as an independent contribution. In schematic form, if \(U_i\) is the total score contribution from policy \(i\), and \(A\) is the negative derivative of the total score with respect to the model parameters, the unadjusted sandwich estimator is:

$$ \widehat{\operatorname{Var}}(\widehat{\beta}) =A^{-1}\left(\sum_{i=1}^{m}U_iU_i^{\mathsf T}\right)A^{-1}. $$

This estimator allows dependence within a policy cluster, but relies on independence across clusters and generally requires a sufficiently large number of independent clusters for its usual large-sample justification. With few clusters, its finite-sample performance may be poor, and a small-sample correction or a more explicitly specified model may be needed. A sandwich estimator adjusts estimated uncertainty; it does not by itself correct a misspecified mean structure or supply a model for the dependence.

Before fitting or interpreting repeated policy-period data, make the observational and dependence units explicit. If rows are years and clusters are policies, ask whether the working likelihood treats years from one policy as independent, whether that assumption is credible, and whether the standard errors account for the clustering. This is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus because ratemaking conclusions depend on both the fitted mean and an appropriate assessment of its uncertainty.

Common Mistakes and Practical Checks

  • Counting rows as independent policies. Repeated years from one policy may share risk effects. Identify the cluster before interpreting a row-based standard error.
  • Assuming an offset resolves dependence. An exposure offset adjusts the expected count for exposure length; it does not change the dependence structure among repeated observations.
  • Adding period variances but omitting covariance terms. For a total count, include twice each distinct-period covariance. In the shared-gamma model, this produces \(M_i+\alpha M_i^2\).
  • Equating dependence with nonzero covariance. Nonzero covariance implies dependence when defined, but dependent variables can have zero covariance.
  • Assuming a robust standard error repairs every model problem. Cluster-robust inference addresses within-cluster dependence under its conditions; it does not automatically repair an incorrect mean model or a small number of clusters.
Key takeaway: Repeated observations from one policy may be conditionally independent under a model yet dependent marginally because they share unobserved risk. Account for the resulting covariance when calculating uncertainty, and do not use zero covariance as proof of independence.

Check Your Understanding

For each problem, identify the dependence implication before relying on an independent-row calculation.

  1. A policy has two periods with exposures \(1\) and \(2\), baseline rate \(\lambda=0.30\), and a shared multiplier with mean \(1\) and variance \(\alpha=0.50\). Given the multiplier, counts are independent Poisson. What is the marginal covariance of the two counts? (Answer: \(0.50(1\cdot0.30)(2\cdot0.30)=0.0900\).)
  2. There are \(50\) independent policies, each observed for two one-year periods. Assume a shared gamma multiplier with mean \(1\) and variance \(0.40\), and conditional Poisson counts with annual rate \(\lambda=0.10\). What is the variance of the total-claims-per-total-exposure rate estimator? (Answer: Each policy’s total variance is \(0.20+0.40(0.20)^2=0.216\); the estimator variance is \(50(0.216)/100^2=0.00108\).)
  3. Two count variables have covariance zero. Does this establish that they are independent? (Answer: No. Zero covariance does not generally imply independence; check the joint distribution or conditional probabilities.)
  4. A frequency model has many policy-period rows but only 18 policies, with possible dependence within policy. Is the usual large-number-of-independent-clusters justification for a cluster-robust standard error automatically convincing? (Answer: No. The independent-cluster count is 18, so finite-sample performance may be poor and should be considered.)
  5. Under a shared multiplier model, a policy has total expected count \(M_i=0.80\) and multiplier variance \(\alpha=0.25\). What is its total-count variance, and what term would be omitted by an independent-row calculation? (Answer: \(\operatorname{Var}(N_i)=0.80+0.25(0.80)^2=0.96\); the shared-risk excess over the Poisson variance is \(0.16\).)
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