Tail Risk of Retained Loss Under Excess-of-Loss Reinsurance

Why the Per-Claim Retention Matters

An excess-of-loss treaty changes the insurer’s payment on each covered claim before those payments are added into an aggregate. This order matters: applying a cap to the total ground-up loss is generally not the same as capping each claim and then summing. The difference can change both value-at-risk (VaR) and tail value-at-risk (TVaR), particularly when the aggregate distribution has probability concentrated at its VaR threshold.

Throughout this tutorial, $u$ denotes the insurer’s per-claim retention, or cap on the insurer’s payment. For a ground-up claim amount $X_j$, the insurer retains $\min(X_j,u)$, and the reinsurer pays $(X_j-u)^+$, where $x^+=\max(x,0)$. The examples use hypothetical contracts and losses; no interest or mortality assumptions are needed.

Definition: For $N$ claims with ground-up amounts $X_1,\ldots,X_N$, the ground-up aggregate loss is $S=\sum_{j=1}^N X_j$. Under a per-claim excess-of-loss treaty with retention $u$, the insurer’s retained aggregate loss is $S_R=\sum_{j=1}^N \min(X_j,u)$, and the reinsurer’s aggregate payment is $S_C=\sum_{j=1}^N (X_j-u)^+$. For each outcome, $S=S_R+S_C$.

This is a per-claim treaty: the retention is applied separately to each claim. A treaty with an annual aggregate limit or attachment point would require a different payment formula. When claim frequency is random, the sum has $N$ terms; when $N=0$, both aggregate losses are zero.

VaR and TVaR for Retained Aggregate Loss

For a loss $Z$, its VaR at confidence level $\alpha$ is the smallest threshold whose cumulative probability is at least $\alpha$. Write $F_Z(z)=P(Z\le z)$. Then $\operatorname{VaR}_{\alpha}(Z)=\inf\{z:F_Z(z)\ge\alpha\}$. In a discrete distribution, the VaR can have positive probability at the threshold.

TVaR averages the upper tail beyond the confidence level. To handle both continuous and discrete losses consistently, use the quantile-integral definition below. For a discrete loss, this convention includes only the fraction of probability at the VaR threshold needed to make the upper tail have probability $1-\alpha$; it is not always the same as the conditional mean given $Z>\operatorname{VaR}_{\alpha}(Z)$.

\operatorname{TVaR}_{\alpha}(Z) =\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{p}(Z)\,dp =\operatorname{VaR}_{\alpha}(Z) +\frac{E[(Z-\operatorname{VaR}_{\alpha}(Z))^+]}{1-\alpha}.

The second expression is especially useful when the distribution or stop-loss expectation is available. The excess term counts the amount by which the loss exceeds the VaR threshold; losses equal to or below the threshold add nothing. Under the treaty, calculate these measures for $S_R$ using its own distribution. Substituting the ground-up distribution of $S$ instead answers a different question.

Key result: Since $\min(X_j,u)\le X_j$ for every claim, $S_R\le S$ for every outcome. Consequently, the retained aggregate VaR and TVaR cannot exceed the corresponding ground-up measures at the same confidence level. This comparison does not determine the size of the reduction; the treaty must be applied to the claim-level outcomes.

Worked Discrete Calculations

Worked Example: Two Claims with a Discrete Severity Model

A hypothetical insurer observes exactly two independent claims. Each ground-up claim amount is $4 with probability 0.60 and $9 with probability 0.40. A per-claim excess-of-loss treaty has retention $u=\$5$. Calculate the retained and ground-up VaR and TVaR at $\alpha=0.80$. Use the discrete quantile convention above.

Method and assumptions. The two severities are independent and identically distributed, with the stated probabilities. Apply the treaty to each claim first, enumerate the two-claim outcomes, and then calculate the aggregate tail measures.

Step 1: Transform each claim. A $4 claim is fully retained, so its insurer payment is $\min(4,5)=\$4$. A $9 claim has insurer payment $\min(9,5)=\$5$, with the remaining $4 ceded.

Step 2: Form the outcome probabilities. Both claims are $4 with probability $0.60^2=0.36$; one is $4 and one is $9 with probability $2(0.60)(0.40)=0.48$; both are $9 with probability $0.40^2=0.16$. The probabilities sum to $0.36+0.48+0.16=1.00$.

Step 3: Find the retained aggregate distribution. The corresponding retained totals are $8, $9, and $10. Thus $P(S_R=8)=0.36$, $P(S_R=9)=0.48$, and $P(S_R=10)=0.16$. The cumulative probability through $8 is 0.36, and through $9 it is $0.36+0.48=0.84$. Therefore $\operatorname{VaR}_{0.80}(S_R)=\$9$.

Step 4: Calculate retained TVaR. The excess above $9 is $1 only when both claims are $9, an event of probability 0.16. Therefore $E[(S_R-9)^+]=0.16(1)=\$0.16$, and $\operatorname{TVaR}_{0.80}(S_R)=9+0.16/0.20=\$9.80$.

Step 5: Compare with ground-up risk. The ground-up aggregate totals are $8, $13, and $18 with the same respective probabilities. Its cumulative probability through $8 is 0.36 and through $13 is 0.84, so $\operatorname{VaR}_{0.80}(S)=\$13$. Also, $E[(S-13)^+]=0.16(18-13)=\$0.80$, giving $\operatorname{TVaR}_{0.80}(S)=13+0.80/0.20=\$17.00$.

Check and common mistake. Both retained measures are below their ground-up counterparts, as the claim-by-claim payment rule requires. A mistaken shortcut would subtract the maximum possible ceded amount, $2(9-5)=\$8$, from ground-up TVaR and report $17-8=\$9$. That fails because the amount ceded varies by outcome; the correct retained TVaR is $9.80.

Worked Example: Random Claim Frequency with Fixed Severity

In a hypothetical year, claim frequency $N$ is 0 with probability 0.20, 1 with probability 0.50, and 2 with probability 0.30. Each claim is exactly $7. A per-claim treaty has retention $u=\$4$. Find retained and ground-up VaR and TVaR at $\alpha=0.50$.

Method and assumptions. The frequency probabilities are the complete frequency model; conditional on $N$, each claim has the stated fixed amount. The aggregate is obtained by multiplying the per-claim payment by $N$.

Step 1: Calculate the per-claim payments. The insurer retains $\min(7,4)=\$4$ per claim, while the ground-up amount is $7 per claim.

Step 2: List each aggregate distribution. For retained loss, $S_R=0,4,8$ with probabilities $0.20,0.50,0.30$. For ground-up loss, $S=0,7,14$ with the same probabilities. Each probability set sums to 1.

Step 3: Calculate VaR. For retained loss, the cumulative probability through 0 is 0.20 and through 4 is $0.20+0.50=0.70$, so $\operatorname{VaR}_{0.50}(S_R)=\$4$. For ground-up loss, the cumulative probability through 0 is 0.20 and through 7 is 0.70, so $\operatorname{VaR}_{0.50}(S)=\$7$.

Step 4: Calculate TVaR. The retained excess above $4 occurs only when $N=2$, giving $E[(S_R-4)^+]=0.30(8-4)=\$1.20$. Hence $\operatorname{TVaR}_{0.50}(S_R)=4+1.20/0.50=\$6.40$. For ground-up loss, $E[(S-7)^+]=0.30(14-7)=\$2.10$, so $\operatorname{TVaR}_{0.50}(S)=7+2.10/0.50=\$11.20$.

Check and common mistake. The frequency distribution is used unchanged, but each positive-frequency outcome has a different aggregate amount after reinsurance. Treating the insurer’s retained aggregate as $\min(7N,4)$ would cap the entire year at $4. For $N=2$, that incorrectly gives $4 instead of the correct per-claim retained total of $4+4=\$8$.

A Continuous Severity and the Stop-Loss Calculation

For a continuous claim distribution, the treaty can create a discrete point mass in retained payments: every ground-up claim at or above $u$ produces a retained payment exactly equal to $u$. The stop-loss expression for TVaR handles this mass without needing to condition only on losses strictly greater than the VaR threshold.

Justification: Let $Y=\min(X,u)$ for a single nonnegative claim, and suppose $q<u$. For any outcome, $(Y-q)^+$ is the amount of retained payment above $q$. It is zero when $X\le q$, equals $X-q$ when $q<X<u$, and equals $u-q$ when $X\ge u$. For a nonnegative random variable, $E[(Y-q)^+]=\int_q^\infty P(Y>t)\,dt$. Since $P(Y>t)=P(X>t)$ for $q\le t<u$ and is zero for $t\ge u$, it follows that $E[(Y-q)^+]=\int_q^u S_X(t)\,dt$. This is the amount needed in the TVaR formula; the capped point mass is included through the integral up to $u$.

Worked Example: Exponential Claim with a Per-Claim Cap

A hypothetical single claim $X$ has an exponential distribution with mean $10, so its survival function is $S_X(x)=e^{-x/10}$ for $x\ge0$. The insurer has a per-claim retention $u=\$30$. Calculate retained and ground-up VaR and TVaR at $\alpha=0.90$, rounding dollar amounts to the nearest cent.

Method and assumptions. This is a one-claim aggregate, so $S=X$ and $S_R=\min(X,30)$. The exponential parameter is rate $\lambda=0.10$ per dollar (mean $1/\lambda=\$10$). Find the ground-up quantile, check whether it is below the cap, and then use the stop-loss integral for the retained TVaR.

Step 1: Find the quantile. For an exponential loss, $P(X\le q)=1-e^{-q/10}$. Setting this equal to 0.90 gives $e^{-q/10}=0.10$, and therefore $q=-10\ln(0.10)=\$23.0259$. Because $q<30$, the retained loss has the same 90th-percentile VaR as the ground-up loss: both VaRs are $23.03.

Step 2: Calculate retained TVaR. By the stop-loss justification, $E[(S_R-q)^+]=\int_q^{30}e^{-t/10}\,dt=10(e^{-q/10}-e^{-3})$. Substituting $e^{-q/10}=0.10$ and $e^{-3}=0.0497871$ gives $10(0.1000000-0.0497871)=\$0.502129$. Therefore $\operatorname{TVaR}_{0.90}(S_R)=23.0259+0.502129/0.10=\$28.05$.

Step 3: Calculate ground-up TVaR. For an exponential loss, the conditional mean above $q$ is $q+10$. Equivalently, $E[(X-q)^+]=\int_q^\infty e^{-t/10}\,dt=10e^{-q/10}=\$1.00$, so $\operatorname{TVaR}_{0.90}(X)=q+1.00/0.10=\$33.03$.

Check and common mistake. A claim of $30 or more produces a retained payment of exactly $30, so the retained distribution has a point mass at $30. The stop-loss calculation includes the portion of that mass needed in the upper 10% tail. Using the uncapped exponential TVaR of $33.03 for the insurer’s retained risk would ignore the treaty; the correct retained TVaR is $28.05.

Exam Checks and Practical Interpretation

For any per-claim treaty calculation, keep the payment transformation separate from the aggregate risk calculation. A reliable sequence is to transform each claim, construct the resulting aggregate distribution or stop-loss expectation, and then apply the VaR or TVaR definition at the stated confidence level. For discrete distributions, check the cumulative probability immediately below and at the VaR; for TVaR, account for the exact probability mass needed above the threshold.

State what is retained.
For each claim amount $X_j$, write the insurer payment $\min(X_j,u)$. Do not cap the total unless the contract specifies an aggregate limit.

Build the aggregate.
Sum retained claim payments over the fixed or random number of claims, preserving the relevant probabilities or dependence structure.

Calculate the tail measure.
Find the quantile from the retained aggregate distribution, then calculate TVaR using the stop-loss excess over that quantile.

A treaty reduces each realized aggregate loss or leaves it unchanged, so retained VaR and TVaR should not exceed ground-up values at the same confidence level. A result that violates this check is a signal to revisit the treaty transformation, the frequency calculation, or the handling of probability at the VaR threshold. These distinctions are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical contract, calculate the requested retained and ground-up measures. Apply the treaty per claim, and use the fractional-tail convention for discrete TVaR.

Exactly two independent claims each equal $3 or $8 with probabilities 0.50 and 0.50. The per-claim retention is $5. Find retained and ground-up VaR and TVaR at $\alpha=0.50$. (Answer: Retained VaR $8.00 and TVaR $9.00; ground-up VaR $11.00 and TVaR $13.50.)
Claim frequency $N$ is 0, 1, or 2 with probabilities 0.10, 0.60, and 0.30; each claim is $9 and the per-claim retention is $4. Find retained and ground-up VaR and TVaR at $\alpha=0.75$. (Answer: Retained VaR and TVaR are both $8.00; ground-up VaR and TVaR are both $18.00.)
There is one claim, equal to $2 with probability 0.70 or $12 with probability 0.30. The per-claim retention is $7. Find retained and ground-up VaR and TVaR at $\alpha=0.50$. (Answer: Retained VaR $2.00 and TVaR $5.00; ground-up VaR $2.00 and TVaR $8.00.)
Exactly two independent claims each equal $2 or $6 with probabilities 0.50 and 0.50. The per-claim retention is $4. Find retained and ground-up VaR and TVaR at $\alpha=0.75$. (Answer: Retained VaR $6.00 and TVaR $8.00; ground-up VaR $8.00 and TVaR $12.00.)
One claim $X$ is exponential with mean $5, and the per-claim retention is $12. Find retained and ground-up VaR and TVaR at $\alpha=0.80$, rounding to the nearest cent. (Answer: Both VaRs are $8.05; retained TVaR is $10.78 and ground-up TVaR is $13.05.)

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