Aggregation Period and Pure Premium Distribution

Why the Aggregation Period Matters

A policy’s claims can be recorded one claim at a time, summed over a month, or combined over the full policy term. These are different response variables: they have different probability distributions, including different chances of a zero, and different variances. A model observation therefore has to be interpreted together with the period over which its loss response was assembled.

The distinction matters when a portfolio contains policies with different term lengths or exposure amounts. For example, a six-month aggregate loss is not directly comparable to a twelve-month aggregate loss unless the period represented by each response is accounted for. Dividing aggregate loss by exposure changes the response from aggregate dollars to dollars per exposure unit, but does not make the underlying observation period disappear.

Definition: Let $N_t$ be the number of claims in a period of length $t$, measured in exposure years, and let $X_1,\ldots,X_{N_t}$ be the corresponding claim amounts. The aggregate loss for that period is $L_t=\sum_{j=1}^{N_t}X_j$, with $L_t=0$ when $N_t=0$. The pure premium per exposure year is $U_t=L_t/t$, for $t>0$.

For the results below, assume a homogeneous Poisson frequency model: $N_t\sim\operatorname{Poisson}(\lambda t)$, where $\lambda$ is the expected claim count per exposure year. Assume claim amounts are positive, identically distributed, and independent of the count and one another, with finite mean $E[X]$ and variance $\operatorname{Var}(X)$. Claims in disjoint periods are independent. These assumptions make frequency proportional to period length and allow the aggregate loss to be treated as a compound Poisson sum. If frequency varies within the period, periods are dependent, or claim amounts depend on the count, the formulas need not apply as written.

Zero Loss and Period Length

A positive claim amount means aggregate loss is zero exactly when there are no claims. Since the probability of zero claims from a Poisson count with mean $\lambda t$ is $e^{-\lambda t}$, the zero-loss probability is

P(L_t=0)=P(N_t=0)=e^{-\lambda t}.

As the aggregation period gets longer, the expected number of claims increases and the chance of observing no claims decreases. For two independent periods of lengths $t_1$ and $t_2$, zero aggregate loss over their combined period requires zero claims in both:

P(L_{t_1+t_2}=0) =e^{-\lambda t_1}e^{-\lambda t_2} =e^{-\lambda(t_1+t_2)}.

This formula concerns zero loss as well as zero claims because claim amounts are assumed strictly positive. If a claim can have a zero payment, a policy could have claims but still have zero paid loss; then the probability of zero aggregate payment is not simply $e^{-\lambda t}$.

Worked Example: A Half-Year Aggregate Response

A hypothetical book has Poisson claim frequency $\lambda=0.6$ claims per exposure year. Claim amounts have an exponential distribution with mean $1,200, so their variance is $1,440,000. A policy observation covers $t=0.5$ exposure year. Find its probability of zero aggregate loss, expected aggregate loss, aggregate-loss variance, and expected pure premium per exposure year. No interest discounting or mortality assumption is involved.

Method and assumptions. Use a compound Poisson model with mean count $\lambda t$, positive exponential claim amounts, and independent claims. The exponential distribution’s mean is $1,200 and its variance is $(1,200)^2=\$1,440,000$. Its second moment is $E[X^2]=\operatorname{Var}(X)+(E[X])^2=1,440,000+1,440,000=\$2,880,000$.

Step 1: Find the expected claim count. $\lambda t=0.6(0.5)=0.3$. Thus $N_t\sim\operatorname{Poisson}(0.3)$.

Step 2: Find the zero-loss probability. $P(L_t=0)=e^{-0.3}=0.7408$, to four decimals. The half-year observation has a 74.08% chance of zero aggregate loss under these assumptions.

Step 3: Find the expected aggregate loss. $E[L_t]=\lambda t E[X]=0.3(1,200)=\$360$.

Step 4: Find the aggregate-loss variance. $\operatorname{Var}(L_t)=\lambda t E[X^2]=0.3(2,880,000)=\$864,000$.

Step 5: Convert to pure premium per exposure year. $E[U_t]=E[L_t]/t=360/0.5=\$720$ per exposure year. Check: $\lambda E[X]=0.6(1,200)=\$720$, the same expected rate.

The common error is to use $e^{-0.6}=0.5488$ for the zero probability, treating the half-year period as a full year. The correct Poisson mean is $0.6(0.5)=0.3$, so the correct probability is $e^{-0.3}=0.7408$.

How Variance Changes With the Period

Under the stated compound Poisson assumptions, aggregate-loss variance is proportional to the period length. A short justification also shows why the second moment of claim amount appears in the formula. Write $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$. Given $N_t=n$, the sum has mean $n\mu$ and variance $n\sigma^2$. Applying the law of total variance and using the Poisson mean and variance, both equal to $\lambda t$, gives:

\begin{aligned} \operatorname{Var}(L_t) &=E[\operatorname{Var}(L_t\mid N_t)] +\operatorname{Var}(E[L_t\mid N_t])\\ &=E[N_t]\sigma^2+\operatorname{Var}(N_t\mu)\\ &=\lambda t\sigma^2+\lambda t\mu^2\\ &=\lambda t E[X^2]. \end{aligned}

The first contribution reflects variation in claim amounts when a count is given; the second reflects variation in the number of claims. Their sum is $\lambda t E[X^2]$, not $\lambda t\operatorname{Var}(X)$. In particular, random claim counts create aggregate variation even if each claim amount is constant.

Because $U_t=L_t/t$, dividing by $t$ scales variance by $1/t^2$. Consequently:

E[U_t]=\lambda E[X], \qquad \operatorname{Var}(U_t) =\frac{\operatorname{Var}(L_t)}{t^2} =\frac{\lambda E[X^2]}{t}.

The expected pure premium per exposure year does not depend on the period length under this stationary model. Its variance does: more exposure averages over more claim experience, so the variance of loss per exposure unit decreases as $1/t$. This does not mean that a longer period has less aggregate-loss variance. Aggregate-loss variance increases with $t$; it is the variance after dividing by the period exposure that decreases.

Worked Example: Combining Four Quarterly Periods

A hypothetical risk has Poisson frequency $\lambda=1.2$ claims per exposure year. Each claim amount is exponentially distributed with mean $500 and variance $250,000; therefore $E[X^2]=250,000+500^2=\$500,000$. Compare a quarter-year aggregate response with the aggregate response formed by combining four independent quarters into one year. Find the zero-loss probabilities, aggregate-loss variances, and variances of pure premium per exposure year. No interest discounting or mortality assumption is involved.

Quarter-year response. Here $t=0.25$, so the expected count is $1.2(0.25)=0.3$. Its zero-loss probability is $e^{-0.3}=0.7408$. Its aggregate-loss variance is $1.2(0.25)(500,000)=\$150,000$. Since quarterly pure premium is $L_{0.25}/0.25$, its variance is $150,000/(0.25)^2=\$2,400,000$.

Combined annual response. Four quarters give $t=1$ and expected count $1.2$. Its zero-loss probability is $e^{-1.2}=0.3010$. Equivalently, all four quarters must have zero loss, so the probability is $(e^{-0.3})^4=e^{-1.2}$. Annual aggregate-loss variance is $1.2(500,000)=\$600,000$.

Annual pure premium is annual aggregate loss divided by 1, so its variance is also $\$600,000$. As a check on aggregation, four independent quarterly aggregate losses have total variance $4(150,000)=\$600,000$, matching the annual formula. The annual pure-premium variance is lower than the quarterly pure-premium variance because the annual amount averages over four times the exposure.

The common mistake is to add quarterly pure-premium variances to obtain the annual pure-premium variance. That would give $4(2,400,000)$, which is not the variance of the annual rate. First sum the quarterly aggregate losses, then divide by the full-year exposure: the resulting variance is $\$600,000$.

What Period Does a Model Observation Represent?

The response definition determines the observation unit. In a policy-level aggregate pure-premium model, one row commonly represents the aggregate loss for one policy over a specified term or reporting period. If the response is dollars per exposure unit, it is that period’s aggregate loss divided by its exposure. A row with a six-month response and a row with a twelve-month response do not represent equal durations, even if both appear as one row in a data file.

In a frequency-severity approach, the frequency response may be the number of claims for a policy-period, while severity observations may be individual claim amounts. Thus the frequency row’s period is the policy-period being counted, but a severity row represents a claim, not the whole policy-period. A claim-level data set does not turn each claim into a full policy-year observation. The exposure attached to the frequency response must correspond to the time during which the claims could occur.

Response recorded in a row	What one row represents	Period or unit to verify
Aggregate loss	One risk or policy’s total paid or incurred loss	The policy term or reporting period over which claims are summed
Pure premium	Aggregate loss divided by exposure	Both the loss period and the exposure unit used in the division
Claim count	Number of claims for a risk or policy	The period during which claims are counted and its exposure
Claim severity	Amount for one claim	One claim; not a policy-period aggregate

Worked Example: Unequal Policy Terms in One Data Set

A hypothetical insurer records aggregate loss for two policies. Policy A has $t=0.25$ exposure year and policy B has $t=1.5$ exposure years. Both follow a Poisson frequency model with $\lambda=0.8$ claims per exposure year. Each claim has a constant amount of $1,000, so $E[X^2]=\$1,000,000$. For each policy, calculate the probability of zero aggregate loss, expected aggregate loss, aggregate-loss variance, and expected pure premium per exposure year. No interest discounting or mortality assumption is involved.

Policy A: quarter-year term. Its expected claim count is $0.8(0.25)=0.2$, so $P(L_{0.25}=0)=e^{-0.2}=0.8187$. Expected aggregate loss is $0.2(1,000)=\$200$. Aggregate-loss variance is $0.2(1,000,000)=\$200,000$. Expected pure premium is $200/0.25=\$800$ per exposure year.

Policy B: 1.5-year term. Its expected claim count is $0.8(1.5)=1.2$, so $P(L_{1.5}=0)=e^{-1.2}=0.3010$. Expected aggregate loss is $1.2(1,000)=\$1,200$. Aggregate-loss variance is $1.2(1,000,000)=\$1,200,000$. Expected pure premium is $1,200/1.5=\$800$ per exposure year.

Both pure premiums have the same expected value, but the aggregate response for policy B covers six times as much exposure as policy A. Accordingly, its aggregate-loss variance is six times as large, and its chance of zero loss is smaller. The calculation checks against $\lambda E[X]=0.8(1,000)=\$800$ for each expected rate. Treating both aggregate losses as if they represented one year would misstate their period and make their dollar responses incomparable.

Common Period-Matching Traps

Using an unscaled frequency rate. For a $t$-year observation, the Poisson mean is $\lambda t$, not $\lambda$, unless $t=1$.
Confusing aggregate dollars with dollars per exposure. Aggregate-loss variance scales with $t$; the variance of aggregate loss divided by $t$ scales with $1/t$.
Assuming one row always means one policy-year. Check the response construction and policy term; a row may cover a fraction of a year or multiple years.
Confusing severity and aggregate loss. A claim-level severity is one claim amount. A policy-level aggregate loss sums the claim amounts across the specified period.
Applying the Poisson zero formula without its conditions. It requires a Poisson claim count and, for zero loss to mean zero claims, strictly positive claim payments.

Key takeaway: Always name the period before calculating or comparing a response. Under a homogeneous compound Poisson model, $P(L_t=0)=e^{-\lambda t}$, aggregate-loss variance is $\lambda tE[X^2]$, and the variance of loss per exposure unit is $\lambda E[X^2]/t$. These results are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, identify the response period before using the model.

A hypothetical portfolio has Poisson frequency $\lambda=0.5$ claims per exposure year and strictly positive claim amounts. What is the probability of zero aggregate loss for a six-month observation? (Answer: $e^{-0.25}=0.7788$.)
A hypothetical risk has $\lambda=1$ claim per exposure year and constant claim amount $2,000. For a two-year observation, find expected aggregate loss and aggregate-loss variance. (Answer: $4,000 and $\$8,000,000$.)
For a compound Poisson model, $\lambda=0.4$ per exposure year and $E[X^2]=\$900,000$. What is the variance of pure premium per exposure year for a three-year aggregate response? (Answer: $\$120,000$.)
A hypothetical claim count is Poisson with rate $0.8$ per year. Independent quarterly periods are combined. What is the annual probability of zero loss, and how does it relate to the quarterly zero probability? (Answer: $e^{-0.8}=0.4493$, equal to $(e^{-0.2})^4$.)
A data file has one row per policy and records total loss over each policy’s full term. One term is 0.5 year and another is 1.25 years. Under stable frequency and severity assumptions, do these rows represent equal periods, and what exposure should be used to express each as a pure premium per exposure year? (Answer: No; divide each policy’s full-term aggregate loss by its own exposure, 0.5 and 1.25 exposure years, respectively.)

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