Pricing Per-Claim Excess-of-Loss Under Tail Constraints

Why Per-Claim Excess-of-Loss Changes the Decision

A per-claim excess-of-loss treaty caps the insurer’s payment on each individual loss above an attachment. Because it changes each claim before the claims are aggregated, it affects both expected retained losses and the distribution of the retained aggregate. A retention that reduces tail capital may also increase expected retained loss, so a pricing decision must consider both effects.

The calculations below focus on losses retained by the insurer. They do not include the reinsurer’s quoted premium, expenses, commissions, or other contract costs. A full commercial premium indication would need to account for those items as well. All examples are hypothetical, undiscounted, and based on the stated one-period frequency and severity assumptions.

Definition: Let $X$ be the loss on one claim, $d$ the per-claim attachment, and $u$ the maximum amount recoverable from the reinsurer per claim. The ceded payment is $C=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The insurer’s retained payment is $R=X-C$. If the layer has no upper limit, take $u=\infty$, so $C=(X-d)_+$ and $R=\min(X,d)$.

The attachment $d$ is not the same as a policy limit. The insurer retains the first $d$ dollars of a claim; the reinsurer then pays up to $u$ dollars above that point. For example, with $d=\$100$ and $u=\$200$, a claim of $\$250$ produces a $\$150$ recovery, while a claim of $\$400$ produces the maximum $\$200$ recovery.

Expected Layer Payments

Write $S_X(x)=P(X>x)$ for the severity survival function. For a nonnegative loss, the expected ceded payment under a layer attaching at $d$ and capped at $u$ is

E[C]=\int_0^u S_X(d+t)\,dt =\int_d^{d+u}S_X(x)\,dx.

For an unlimited layer, this becomes $E[C]=\int_d^\infty S_X(x)\,dx$. Since $R=X-C$, the expected retained payment is $E[R]=E[X]-E[C]$. These formulas require a nonnegative severity and finite expectations for the quantities being calculated.

Justification: For any nonnegative random variable $Z$, the tail-integral identity $E[Z]=\int_0^\infty P(Z>t)\,dt$ follows by writing $Z=\int_0^\infty \mathbf{1}_{\{Z>t\}}\,dt$ and interchanging expectation and integration. Here $C=\min((X-d)_+,u)$. For $0\le t<u$, $C>t$ exactly when $X>d+t$, while for $t\ge u$, $C>t$ is impossible. Integrating those probabilities gives the stated formula.

Worked Example: Expected Payments for a Capped Layer

A hypothetical portfolio has a Poisson claim count with mean $\lambda=12$. Claim severities are independent and exponential with mean $\theta=\$100$, so $S_X(x)=e^{-x/100}$ for $x\ge0$. A per-claim layer attaches at $d=\$100$ and has maximum recovery $u=\$200$. Find the expected ceded and retained aggregate losses. Assume no discounting.

Method. For this exponential model, $E[X]=\theta=\$100$. Apply the survival-integral formula to one claim:

E[C]=\int_{100}^{300}e^{-x/100}\,dx =100(e^{-1}-e^{-3}) =100(0.3678794-0.0497871) =\$31.81.

Thus the expected retained payment per claim is

E[R]=100-31.8092=\$68.19.

For a compound Poisson aggregate, expected total payments equal the expected claim count times the expected payment per claim. Therefore, the expected ceded aggregate is $12(31.8092)=\$381.71$, and the expected retained aggregate is $12(68.1908)=\$818.29$. As a check, their sum is $\$1,200.00$, equal to $12E[X]$, the expected gross aggregate loss. The common mistake this method avoids is treating the $\$200$ layer limit as a maximum claim severity: it limits the recovery, not the gross loss.

Aggregate Tail Capital Under a Compound Poisson Model

Let $N\sim\operatorname{Poisson}(\lambda)$ be independent of identically distributed claim severities $X_1,X_2,\ldots$. The aggregate retained loss is $S_R=\sum_{j=1}^{N}R_j$, where each $R_j$ is the retained payment on claim $j$. Provided the relevant moments are finite,

E[S_R]=\lambda E[R], \qquad \operatorname{Var}(S_R)=\lambda E[R^2].

The second formula uses the second moment of the payment per claim, not its variance. To see why, condition on $N$: $E[S_R\mid N]=NE[R]$ and $\operatorname{Var}(S_R\mid N)=N\operatorname{Var}(R)$. The law of total variance gives

\operatorname{Var}(S_R) =\lambda\operatorname{Var}(R)+\lambda(E[R])^2 =\lambda E[R^2],

because a Poisson variable has both mean and variance $\lambda$. This result applies to a compound Poisson model with independent frequency and severities; it is not a general variance formula for every frequency model.

For a tail probability level $q$, define tail capital above the mean as $K_q=\operatorname{VaR}_q(S_R)-E[S_R]$, where $\operatorname{VaR}_q$ is the $q$-quantile. A commonly used approximation is

K_q\approx z_q\sqrt{\lambda E[R^2]},

where $z_q$ is the standard normal $q$-quantile. This is a normal approximation, not an exact compound Poisson quantile. It can be unreliable for small expected claim counts, highly skewed severities, or tail constraints that require precise extreme-quantile estimates. In those settings, use a suitable aggregate distribution method or simulation and verify the approximation before relying on it.

Worked Example: Tail Capital with an Unlimited Attachment

A hypothetical portfolio has $N\sim\operatorname{Poisson}(16)$ claims, independent exponential severities with mean $\theta=\$30$, and an unlimited per-claim excess-of-loss layer attaching at $d=\$30$. Estimate 95% tail capital above the mean using $z_{0.95}=1.64485$. Assume no discounting.

Method. The retained payment is $R=\min(X,30)$. For any nonnegative $R$, $E[R^2]=2\int_0^\infty tP(R>t)\,dt$. Here $P(R>t)=e^{-t/30}$ for $0\le t<30$, and it is zero for $t\ge30$. Hence

E[R^2]=2\int_0^{30}t e^{-t/30}\,dt =2(30)^2[1-2e^{-1}] =1,800(1-2e^{-1}) =475.6340.

The aggregate retained variance is

\operatorname{Var}(S_R)=16(475.6340)=7,610.1442, \qquad \operatorname{SD}(S_R)=\sqrt{7,610.1442}=87.2361.

The normal-approximation capital is $1.64485(87.2361)=\$143.49$, rounded to the nearest cent. As a check, the retained mean is $16(30(1-e^{-1}))=\$303.42$; the approximate 95% VaR is therefore $\$303.42+\$143.49=\$446.91$. The calculation avoids the trap of using $\lambda\operatorname{Var}(R)$ for aggregate variance, which would omit the contribution from random claim count.

Finite Limits and Retained-Loss Moments

A finite recovery limit changes the retained payment above $d+u$: once the recovery reaches $u$, additional severity remains with the insurer. For a nonnegative ceded payment, its second moment can be calculated by the same tail-integral argument:

E[C^2]=2\int_0^u tS_X(d+t)\,dt.

The retained second moment must reflect the full retained-payment function $R=X-\min((X-d)_+,u)$. In general, calculate it from that function and the severity distribution rather than substituting the unlimited-layer formula. The following example makes the pieces explicit for an exponential severity.

Worked Example: Tail Capital with a Finite Recovery Limit

A hypothetical portfolio has $N\sim\operatorname{Poisson}(20)$ claims. Each severity is exponential with mean $\theta=\$50$, and the per-claim layer attaches at $d=\$50$ with maximum recovery $u=\$100$. Estimate 95% retained aggregate tail capital using $z_{0.95}=1.64485$. Assume no discounting.

Method. The retained payment equals $X$ when $X\le50$, equals $50$ when $50<X\le150$, and equals $X-100$ when $X>150$. Thus its second moment is the sum of three contributions. The first is the truncated exponential integral:

\int_0^{50}x^2\frac{1}{50}e^{-x/50}\,dx =2(50)^2[1-(1+1+1/2)e^{-1}] =401.5070.

The middle interval contributes $50^2P(50<X\le150)=2,500(e^{-1}-e^{-3})=795.2309$. For the last interval, the exponential memoryless property means that conditional on $X>150$, $X-150$ has exponential mean $50$; hence $X-100=50+(X-150)$. Its conditional second moment is $50^2+2(50)(50)+2(50)^2=12,500$, and the unconditional contribution is $12,500e^{-3}=622.3384$. Therefore,

$$ E[R^2]=401.5070+795.2309+622.3384=1,819.0763. $$

The aggregate variance is $20(1,819.0763)=36,381.5254$, giving standard deviation $\sqrt{36,381.5254}=\$190.74$. The estimated 95% capital is $1.64485(190.74)=\$313.70$, rounded to the nearest cent. The method accounts for the retained severity above $d+u$; treating this contract as unlimited would incorrectly assume the insurer retains nothing beyond $\$50$.

Using a Capital Constraint to Select a Retention

A retention decision can be framed as a constrained comparison. For example, specify a maximum permitted tail capital and a pricing indication that charges a stated proportion of that capital. One simple hypothetical risk-loaded indication is

P(d)=E[S_R]+\rho K_q,

where $\rho$ is a stated capital-load factor. This is a chosen pricing convention, not a universal formula for an insurance premium. If a proposed retention violates the capital constraint, a lower retention may reduce capital but also change the expected retained loss. The direction and size of each effect should be calculated rather than assumed.

Worked Example: Choosing Among Retentions Under a Capital Limit

A hypothetical portfolio has a Poisson claim count with mean $\lambda=16$ and independent exponential claim severities with mean $\theta=\$30$. Consider unlimited per-claim layers with retentions $d=\$20,\$30,\$40$. Estimate 95% capital using the normal approximation with $z_{0.95}=1.64485$. The capital constraint is $K_{0.95}\le\$150$, and the retained-loss indication is $P(d)=E[S_R]+0.20K_{0.95}$. Assume no discounting.

Method. For $R=\min(X,d)$ under an exponential severity,

E[R]=30(1-e^{-d/30}), \qquad E[R^2]=2(30)^2[1-(1+d/30)e^{-d/30}].

For $d=\$20$, $E[R]=30(1-e^{-2/3})=\$14.5975$, so $E[S_R]=16(14.5975)=\$233.56$. Also, $E[R^2]=1,800[1-(5/3)e^{-2/3}]=259.7486$. The aggregate standard deviation is $\sqrt{16(259.7486)}=\$64.47$, and capital is $1.64485(64.47)=\$106.05$. The indication is $233.56+0.20(106.05)=\$254.77$.

For $d=\$30$, the calculations give $E[S_R]=16(30(1-e^{-1}))=\$303.42$ and $E[R^2]=475.6340$. The aggregate standard deviation is $\sqrt{16(475.6340)}=\$87.24$, and capital is $\$143.49$. The indication is $303.42+0.20(143.49)=\$332.12$.

For $d=\$40$, $E[S_R]=16(30(1-e^{-4/3}))=\$353.47$. Also, $E[R^2]=1,800[1-(7/3)e^{-4/3}]=692.8920$. The aggregate standard deviation is $\sqrt{16(692.8920)}=\$105.29$, so capital is approximately $\$173.18$, which exceeds the $\$150$ limit. Therefore, $d=\$30$ is the largest of the three tested retentions that satisfies the constraint. The indicated premiums for the two feasible choices are $\$254.77$ and $\$332.12$, respectively. The check is that higher retention raises both expected retained loss and capital in these options; the feasible choice must still be determined by the specified constraint, not by the premium alone.

Common Traps and Practical Checks

Confusing attachment and recovery limit. The ceded payment is capped at $u$; the gross claim is not. For a finite layer, losses above $d+u$ can still increase retained loss.
Using the wrong aggregate variance. For compound Poisson payments, use $\lambda E[R^2]$, not $\lambda\operatorname{Var}(R)$.
Applying an aggregate stop-loss formula to a per-claim treaty. A per-claim layer transforms each severity before summation. It is not a layer applied to the total aggregate loss.
Calling a normal approximation an exact VaR. State the approximation and check whether the frequency and severity make it credible for the requested tail.
Ignoring what the price includes. An expected retained-loss indication plus a capital load does not automatically include reinsurance premium or operating expenses.

Key takeaway: First transform each claim into its ceded and retained payments. Then calculate the retained aggregate mean and, for compound Poisson frequency, its variance as $\lambda E[R^2]$. Apply a stated tail approximation and pricing rule only after checking that its assumptions match the decision.

Check Your Understanding

For each hypothetical problem, identify the layer or retained-payment method before calculating. Answers are rounded as stated.

A portfolio has Poisson claim count with mean 5 and exponential severity with mean $\$80$. An unlimited per-claim layer attaches at $\$80$. Find the expected ceded and retained aggregate losses. (Answer: ceded $\$147.15$; retained $\$252.85$.)
For one exponential claim with mean $\$40$, a layer attaches at $\$40$ and has maximum recovery $\$80$. Find the expected ceded payment per claim. (Answer: $\$12.72$.)
A portfolio has Poisson claim count with mean 9 and exponential severities with mean $\$20$. An unlimited per-claim layer attaches at $\$20$. Estimate 95% capital above the retained aggregate mean using $z_{0.95}=1.64485$. (Answer: approximately $\$71.74$.)
A portfolio has Poisson claim count with mean 16 and exponential severities with mean $\$30$. For unlimited retention $d=\$30$, estimate 95% capital using $z_{0.95}=1.64485$ and the compound Poisson normal approximation. (Answer: approximately $\$143.49$.)
For the portfolio in question 4, compare retentions $d=\$20$ and $d=\$30$ using $P=E[S_R]+0.20K_{0.95}$. Which has the lower indication, and what is that indication? (Answer: $d=\$20$, approximately $\$254.77$; the $d=\$30$ indication is approximately $\$332.12$.)

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