Pricing Aggregate Covers from Aggregate Loss Distributions

Why Aggregate Covers Require an Aggregate Loss Model

An aggregate cover responds to the total covered loss over a period, commonly a year, rather than to each event separately. Its expected recovery therefore depends on the distribution of the total, including the chance of having several losses in the same period. Knowing only the expected number of claims or the expected total loss is not enough to price the cover.

The core calculation is direct once the annual aggregate-loss distribution is available: apply the contract terms to each possible aggregate outcome, then average the resulting recoveries. In practice, the distribution can come from an explicit frequency-severity model, a discrete simulation, or another suitable aggregate-loss model. The calculation must match the contract’s period, covered loss definition, and limit basis.

Definition: Let $S$ be the total covered loss during the contract period. An aggregate cover with attachment $A$ and limit $L$, where $A \geq 0$ and $L>0$, has recovery $$ R=\min\bigl((S-A)_+,L\bigr), \qquad (x)_+=\max(x,0). $$ The attachment is the aggregate loss retained before recovery begins. The limit is the maximum recovery above that attachment. The expected recovery is $E[R]$. The cover is assumed to have no reinstatements or other terms changing the stated limit.

The recovery is zero when $S\leq A$, increases dollar for dollar as $S$ rises from $A$ to $A+L$, and is capped at $L$ when $S\geq A+L$. This makes it a nonlinear function of aggregate loss. In particular, the recovery at the mean aggregate loss is generally not the mean recovery.

Calculating Expected Recovery from Aggregate Outcomes

For a discrete aggregate-loss model, list the possible values of $S$, their probabilities, and the recovery at each value. The expected recovery is the probability-weighted average:

E[R]=\sum_s \min\bigl((s-A)_+,L\bigr)P(S=s).

This method is exact when the listed outcomes and probabilities describe the aggregate distribution. A frequency-severity model is one way to construct that distribution. Let $N$ be the number of claims in the period and let $X_1,X_2,\ldots$ be their severities. Then

S=\sum_{j=1}^{N}X_j,

with $S=0$ when $N=0$. A common compound model assumes that claim severities are identically distributed, mutually independent, and independent of $N$. The chosen frequency and severity distributions must be appropriate to the covered business and period. If the model allows dependence, changing severity by claim count, or additional contract features, the aggregate distribution must reflect those assumptions before the cover is applied.

Worked Example: A Binomial Claim Count with Fixed Severity

A hypothetical annual portfolio has $N\sim\operatorname{Binomial}(4,0.25)$. Each claim is exactly $100,000, and claim amounts are fully covered before the aggregate treaty applies. The treaty attaches at $A=$$200,000 and has an aggregate limit of $L=$$100,000. Find the expected annual recovery.

Method and assumptions. The claim count has four independent opportunities, each with claim probability 0.25. Severity is fixed, so $S=100{,}000N$. For $N=0,1,2$, aggregate losses do not exceed $200,000 and recovery is zero. For $N=3$, loss is $300,000 and recovery is $100,000. For $N=4$, loss is $400,000, but recovery remains capped at $100,000.

The probability of reaching at least three claims is

P(N\geq 3)=P(N=3)+P(N=4) =\binom{4}{3}(0.25)^3(0.75)+(0.25)^4 =0.046875+0.00390625 =0.05078125.

Only these outcomes produce recovery, and each produces $100,000. Thus

E[R]=100{,}000(0.05078125)=\$5{,}078.125.

Rounded to the nearest cent, expected recovery is $5,078.13. The result is between $0 and the treaty limit of $100,000, as it must be.

Common mistake avoided: Treating every loss above the attachment as fully recoverable would ignore the limit. Here the cap happens not to change the recovery for the possible outcomes with three or four claims, but in a model with larger severities it could materially reduce the expected payment.

A Survival-Probability Formula and Its Justification

For a nonnegative aggregate loss, the recovery can also be represented through the probability that losses exceed each point in the covered band from $A$ to $A+L$. This gives a useful formula whether $S$ is discrete or continuous:

Theorem: If $S\geq 0$, the recovery $R=\min((S-A)_+,L)$ satisfies $$ E[R]=\int_A^{A+L}P(S>x)\,dx. $$ The integral is with respect to loss amount. For a discrete distribution, the survival probability is a step function, so the same integral gives the exact expected recovery.

To see why, fix a particular aggregate loss $S=s$. For each $x$ between $A$ and $A+L$, the indicator $I(s>x)$ equals 1 when the loss exceeds $x$, and 0 otherwise. The length of the part of that interval lying below $s$ is exactly the recovery: zero if $s\leq A$, $s-A$ if $A<s<A+L$, and $L$ if $s\geq A+L$. Therefore,

\min\bigl((s-A)_+,L\bigr) =\int_A^{A+L}I(s>x)\,dx.

Taking expectations on both sides and interchanging expectation and integration for this bounded, nonnegative quantity gives

E[R]=\int_A^{A+L}E[I(S>x)]\,dx =\int_A^{A+L}P(S>x)\,dx.

This formulation shows exactly which part of the aggregate distribution matters: the exceedance probabilities within the cover’s attachment-to-exhaustion band. It does not mean that one can substitute a single expected-loss value for the whole distribution.

Worked Example: A Compound Poisson Model with a Limit

For a hypothetical portfolio, $N\sim\operatorname{Poisson}(1)$ claims per year, and every claim has fixed severity $50,000. Thus $S=50{,}000N$. An annual aggregate cover attaches at $100,000 and has a limit of $100,000. Calculate expected recovery.

Method and assumptions. The Poisson count has mean 1, and severity is fixed and independent of the count. The cover pays $0 for $N\leq2$, $50,000 for $N=3$, and its $100,000 maximum for $N\geq4$. Hence

E[R]=50{,}000P(N=3)+100{,}000P(N\geq4).

For a Poisson random variable with parameter 1, $P(N=n)=e^{-1}/n!$. Therefore,

P(N=3)=\frac{e^{-1}}{3!}=0.06131324, $$ $$ P(N\geq4)=1-\sum_{n=0}^{3}\frac{e^{-1}}{n!} =1-e^{-1}\left(1+1+\frac{1}{2}+\frac{1}{6}\right) =0.01898816.

Substitute these probabilities:

E[R] =50{,}000(0.06131324)+100{,}000(0.01898816) =\$3{,}065.66+\$1{,}898.82 =\$4{,}964.48.

The probability that the cover pays at all is $P(N\geq3)=0.08030140$, but multiplying that probability by the limit would overstate the result: at exactly three claims, recovery is only $50,000. The calculated expected recovery is below the maximum and includes that partial-limit outcome.

Sensitivity to Frequency and Severity

Frequency and severity assumptions shape different parts of the aggregate distribution. A higher expected claim count will often increase the probability of reaching the attachment and exhaustion point. Larger claim severities can move more probability into the layer, but may also push outcomes beyond exhaustion, where additional loss no longer increases recovery. Consequently, expected recovery need not change in proportion to expected aggregate loss.

The following example isolates severity sensitivity while holding the frequency model and cover fixed. This is a model comparison, not a claim that every severity increase must have the same effect for every treaty.

Worked Example: Changing Severity Under a Fixed Cover

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$. Compare two fixed-severity assumptions: $50,000 per claim and $60,000 per claim. In both cases, use an annual aggregate attachment of $100,000 and a limit of $100,000. Find the expected recovery under each assumption.

Method and assumptions. The count distribution is unchanged, and each claim has the stated fixed severity. Under $50,000 severity, the recoveries for $N=0,1,2,3,N\geq4$ are respectively $0, $0, $0, $50,000, and $100,000. The probabilities from the preceding Poisson calculation are $P(N=3)=0.06131324$ and $P(N\geq4)=0.01898816$. Thus the expected recovery is $4,964.48.

Under $60,000 severity, $N=2$ gives aggregate loss $120,000 and recovery $20,000. $N=3$ gives loss $180,000 and recovery $80,000. $N\geq4$ gives loss at least $240,000 and recovery capped at $100,000. Therefore,

E[R] =20{,}000P(N=2)+80{,}000P(N=3)+100{,}000P(N\geq4).

The needed probabilities are

P(N=2)=\frac{e^{-1}}{2}=0.18393972,\quad P(N=3)=0.06131324,\quad P(N\geq4)=0.01898816.

Substitution gives

E[R] =20{,}000(0.18393972)+80{,}000(0.06131324)+100{,}000(0.01898816) $$ $$ =\$3{,}678.79+\$4{,}905.06+\$1{,}898.82 =\$10{,}482.67.

The expected recovery rises from $4,964.48 to $10,482.67. The increase comes from outcomes that were below attachment at $50,000 severity moving into the covered band at $60,000 severity. The limit still constrains outcomes with four or more claims. Both results are positive and below the $100,000 limit, which is a basic reasonableness check.

Common mistake avoided: Scaling the first expected recovery by $60{,}000/50{,}000$ would give $5,957.38, not the correct $10,482.67. Such proportional scaling fails because the severity change moves the two-claim outcome from no recovery to a $20,000 payment, while the treaty’s cap also makes high-count outcomes nonlinear.

Practical Checks and Common Traps

A sound aggregate-cover indication depends on matching the model to the contract and applying the payment function to the aggregate outcomes. Before relying on a result, check the following:

Confirm the period and loss basis. The random variable $S$ must represent losses accumulated over the treaty’s stated period and use the covered-loss definition.
Distinguish aggregate terms from occurrence terms. An aggregate attachment is applied to the period total. Applying it separately to each claim prices a different structure.
Apply the attachment before the limit. The recovery is $\min((S-A)_+,L)$, not $\min(S,L)-A$. The latter can produce an invalid negative recovery.
Retain partial-layer outcomes. Losses between $A$ and $A+L$ generate partial recovery and should not be grouped with either zero-payment or full-limit outcomes.
Check the cap. Recovery cannot exceed $L$, regardless of how large the aggregate loss becomes.

A particularly tempting shortcut is to apply the treaty formula to expected aggregate loss. In the compound Poisson example, $E[S]=50{,}000E[N]=\$50{,}000$, so that shortcut would produce $\min((50{,}000-100{,}000)_+,100{,}000)=\$0$. But the exact expected recovery is $4,964.48. The shortcut is wrong because the treaty responds to the distribution of $S$, including the chance of three or more claims, rather than only to its mean.

Key takeaway: Specify the aggregate-loss model, calculate or approximate its distribution, and apply the aggregate attachment and limit to each outcome before averaging. Reassess the expected recovery when frequency or severity assumptions change, since either can move probability into or out of the covered band.

Check Your Understanding

For each problem, calculate the expected recovery under the stated assumptions. Treat all amounts as covered losses and use the aggregate cover formula.

A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.20)$, with fixed severity $40,000 per claim. The aggregate cover attaches at $80,000 and has a $40,000 limit. What is expected recovery? (Answer: $320.00)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.5)$, with fixed severity $25,000. The cover attaches at $25,000 and has a $50,000 limit. Calculate expected recovery, rounding to the nearest cent. (Answer: $2,614.79)
A portfolio has exactly two claims. Independently, each claim is $60,000 or $120,000 with probability 0.50 each. An aggregate cover attaches at $150,000 and has a $100,000 limit. Find expected recovery. (Answer: $37,500.00)
For $N\sim\operatorname{Poisson}(1)$ and fixed severity $50,000 per claim, an aggregate cover attaches at $50,000 and has a $100,000 limit. Calculate expected recovery to the nearest cent. (Answer: $17,227.13)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$, with fixed severity $100,000 per claim. The aggregate cover attaches at $200,000 and has a $100,000 limit. Find expected recovery to the nearest cent. (Answer: $32,332.36)

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