Compare Treaty Terms by Expected Cost and Tail Reduction

Why Compare Treaty Terms on a Common Basis?

A treaty with a lower attachment point generally transfers more loss to the reinsurer, but it also has a higher expected recovery and may require a higher premium. Comparing terms therefore requires more than ranking expected recoveries. A useful comparison keeps the portfolio model fixed and reports, for each treaty, the insurer’s expected retained claims, the reinsurer’s expected payment, and an aggregate measure of tail risk.

The examples use hypothetical annual portfolios. All loss amounts are in thousands of dollars unless stated otherwise. We assume a Poisson claim count and independent, identically distributed claim severities, independent of the count. We compare per-claim excess-of-loss treaties; a per-claim payment must be calculated claim by claim before the insurer’s payments are aggregated. This approach is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a claim severity, $a$ a per-claim attachment, and $L$ a layer limit. The reinsurer’s payment on a claim under an excess-of-loss layer is $Y=\min((X-a)_+,L)$, where $(z)_+=\max(z,0)$. The insurer’s retained payment is $Z=X-Y$. For an unlimited layer, $L=\infty$, so $Y=(X-a)_+$ and $Z=\min(X,a)$.

Expected reinsurance recovery is $E[N]E[Y]$ when claim count $N$ is independent of the severities and $E[N]$ is finite. It is not automatically the reinsurance premium: a premium may include a loading for expenses, risk, or profit. In a treaty comparison, state separately the expected recovery and any assumed premium.

Expected Payments for Exponential Severity

Suppose $X$ has an exponential distribution with mean $\theta$, so $S_X(x)=e^{-x/\theta}$ for $x\geq 0$. For an unlimited excess-of-loss layer attaching at $a$, the expected ceded payment is the limited expected value above the attachment:

E[Y]=E[(X-a)_+]=\int_a^\infty S_X(x)\,dx =\theta e^{-a/\theta}.

The insurer retains $Z=\min(X,a)$, so $E[Z]=E[X]-E[Y]=\theta(1-e^{-a/\theta})$. These per-claim expectations can be multiplied by the expected claim count to obtain annual expected payments, under the independence assumptions above.

Worked Example: Compare Expected Payments at Two Attachments

A hypothetical insurer has $N\sim\operatorname{Poisson}(10)$ claims per year. Claim severities are independent exponential random variables with mean $\theta=1$, where amounts are in thousands of dollars. Compare unlimited per-claim excess-of-loss treaties attaching at $a=1$ and $a=2$. Find the annual expected retained claims and expected reinsurance recoveries for each treaty.

Method. For attachment $a$, the expected ceded amount per claim is $e^{-a}$ thousand dollars, and the expected retained amount per claim is $1-e^{-a}$ thousand dollars. Since $E[N]=10$, multiply each per-claim expectation by 10.

For $a=1$, $e^{-1}=0.3678794$. Thus the annual expected recovery is $10(0.3678794)=3.678794$ thousand dollars, and the annual expected retained claims are $10(1-0.3678794)=6.321206$ thousand dollars.

For $a=2$, $e^{-2}=0.1353353$. The annual expected recovery is $10(0.1353353)=1.353353$ thousand dollars, and annual expected retained claims are $10(1-0.1353353)=8.646647$ thousand dollars.

Check. For each treaty, expected retained claims plus expected recovery equals the gross expected claims, $10(1)=10$ thousand dollars. The lower attachment transfers more expected loss to the reinsurer. The common mistake is to treat expected recovery as the premium; no premium information has been supplied. Round only at the final step, to three decimal places in thousands: the recoveries are $3.679$ and $1.353$, and the retained amounts are $6.321$ and $8.647$.

Aggregate Moments and Tail Comparisons

Let $Z_j$ be the insurer’s retained payment on claim $j$, and let $S_R=\sum_{j=1}^{N}Z_j$ be the annual aggregate retained loss. For a Poisson claim count with mean $\lambda$, independent of identically distributed claim payments $Z$, the compound Poisson moments are:

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

The variance uses the second raw moment $E[Z^2]$, not $\operatorname{Var}(Z)$. This distinction matters because the random number of claims contributes variability as well as the randomness in individual payments.

For $Z=\min(X,a)$ and exponential severity with mean $\theta$, the second raw moment is:

E[Z^2] =2\int_0^a xS_X(x)\,dx =2\theta^2\left[1-\left(1+\frac{a}{\theta}\right)e^{-a/\theta}\right].

The first equality follows from the tail-integral identity $E[Z^2]=\int_0^\infty 2xP(Z>x)\,dx$. Here $P(Z>x)=e^{-x/\theta}$ for $0\leq x<a$, and $P(Z>x)=0$ for $x\geq a$. Integrating $2xe^{-x/\theta}$ from 0 to $a$ gives the stated expression. The formula applies to this capped retained payment, not to an arbitrary treaty payment or severity distribution.

Worked Example: Compare Aggregate Variance and Approximate 99th Percentiles

Use the portfolio in the preceding example: $N\sim\operatorname{Poisson}(10)$, exponential severity with mean $\theta=1$ thousand dollars, and unlimited per-claim excess-of-loss. Compare attachments $a=1$ and $a=2$. For a rough tail comparison, approximate each aggregate retained loss by a normal distribution with the same mean and variance, and use $z_{0.99}=2.3263$.

Method. For each attachment, calculate $E[Z]$ and $E[Z^2]$, then use the compound Poisson formulas. The normal approximation gives $\operatorname{VaR}_{0.99}(S_R)\approx E[S_R]+2.3263\sqrt{\operatorname{Var}(S_R)}$. This is an approximation, not an exact compound-distribution percentile.

For $a=1$, $E[Z]=1-e^{-1}=0.6321206$ thousand dollars per claim. Also, $E[Z^2]=2[1-(1+1)e^{-1}]=0.5284822$ thousand-dollars squared per claim. Therefore $E[S_R]=10(0.6321206)=6.321206$ thousand dollars and $\operatorname{Var}(S_R)=10(0.5284822)=5.284822$ thousand-dollars squared. The standard deviation is $\sqrt{5.284822}=2.298874$ thousand dollars. The approximate 99th percentile is $6.321206+2.3263(2.298874)=11.669$ thousand dollars, to three decimal places.

For $a=2$, $E[Z]=1-e^{-2}=0.8646647$ thousand dollars per claim. The second moment is $2[1-(1+2)e^{-2}]=1.1879883$ thousand-dollars squared per claim. Thus $E[S_R]=8.646647$ thousand dollars and $\operatorname{Var}(S_R)=11.879883$ thousand-dollars squared. The standard deviation is $\sqrt{11.879883}=3.446721$ thousand dollars. The approximate 99th percentile is $8.646647+2.3263(3.446721)=16.665$ thousand dollars, to three decimal places.

Check. The higher attachment leaves the insurer with both a higher expected aggregate loss and a higher aggregate variance. For context, the gross aggregate has mean $10$ thousand dollars and variance $10E[X^2]=20$ thousand-dollars squared, giving a normal-approximation 99th percentile of $10+2.3263\sqrt{20}=20.403$ thousand dollars. The treaty reduces this approximate tail measure under either attachment. The calculation does not establish an exact tail probability: aggregate losses can be skewed, so a normal approximation may be inaccurate, particularly at more extreme percentiles. Round only at the final step, to three decimal places in thousands.

Include Premiums Before Making a Decision

Expected recovery measures the reinsurer’s expected claim payment, not the insurer’s total expected outlay under a priced treaty. If the reinsurance premium is specified, compare expected retained claims plus that premium. A risk-sensitive decision can also account for an aggregate tail measure, but it must state how the tail measure enters the decision. There is no universal rule that the treaty with the lowest expected recovery or the lowest approximate percentile is best.

Worked Example: Compare Loaded Expected Outlay and a Risk Criterion

Continue with the Poisson-exponential portfolio above. Suppose each treaty’s annual premium is 115% of its expected annual recovery. Compare the two attachments using the illustrative criterion “expected retained claims plus premium, plus 25% of the approximate 99th-percentile margin above the mean retained aggregate.” Treat the premium as fixed for the year.

Method. For each treaty, calculate premium as $1.15$ times expected recovery. Add it to expected retained claims to get expected total outlay. The risk margin is the approximate 99th percentile of retained aggregate less its mean; adding a fixed premium shifts both by the same amount, so it does not change this margin.

At $a=1$, expected recovery is $3.678794$ thousand dollars, so the premium is $1.15(3.678794)=4.230613$ thousand dollars. Expected total outlay is $6.321206+4.230613=10.551819$ thousand dollars. From the previous calculation, the risk margin is $2.3263(2.298874)=5.3479$ thousand dollars. The criterion is $10.551819+0.25(5.3479)=11.8888$ thousand dollars.

At $a=2$, expected recovery is $1.353353$ thousand dollars, so the premium is $1.15(1.353353)=1.556356$ thousand dollars. Expected total outlay is $8.646647+1.556356=10.203003$ thousand dollars. The risk margin is $2.3263(3.446721)=8.0181$ thousand dollars. The criterion is $10.203003+0.25(8.0181)=12.2075$ thousand dollars.

Conclusion and check. Under this particular criterion, the $a=1$ treaty has the lower value, about $11.889$ thousand dollars versus $12.208$ thousand dollars. The $a=2$ treaty has lower expected outlay, but its larger risk margin outweighs that advantage under the stated 25% weighting. The ranking is a consequence of the supplied premium rule and risk criterion; it is not a general recommendation. Round only at the final step, to three decimal places in thousands.

Common Mistakes in Treaty Comparisons

A comparison is only meaningful if each option is evaluated on the same portfolio and on the same payment basis. In particular, do not mix per-claim recoveries with aggregate stop-loss payments, or apply a per-claim attachment to the annual aggregate. The treaty definition determines which random variable is paid.

Using the wrong compound variance formula. For the first treaty above, $E[Z]=0.6321206$ and $\operatorname{Var}(Z)\approx0.128906$. The incorrect calculation $10\operatorname{Var}(Z)\approx1.289$ omits claim-count variability. The correct aggregate variance is $10E[Z^2]=10(0.5284822)=5.284822$.
Confusing a limit with an unlimited layer. A finite layer pays at most $L$ per claim. The unlimited-layer formula $E[(X-a)_+]=\theta e^{-a/\theta}$ does not give the expected payment for a capped layer.
Calling expected recovery the premium. A premium is a contract price. Unless the problem states otherwise, calculate expected recovery and premium as separate quantities.
Interpreting an approximate percentile as exact. A normal approximation gives a convenient common comparison, but does not remove skewness or guarantee accurate tail probabilities.

Key takeaway: For each treaty, identify the per-claim insurer and reinsurer payments first. Then compare expected retained claims, expected recovery, any stated premium, and aggregate risk measures using one consistent frequency-severity model. State the assumptions and approximation behind every tail comparison.

Check Your Understanding

Use the stated assumptions in each question, and keep the payment basis consistent.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ claims per year and exponential severity with mean 2 thousand dollars. An unlimited per-claim layer attaches at 2 thousand dollars. Find annual expected recovery and annual expected retained claims. (Answer: expected recovery $2.943$ thousand dollars; expected retained claims $5.057$ thousand dollars.)
A hypothetical portfolio has Poisson mean 3 claims and exponential severity with mean 500 dollars. An unlimited per-claim layer attaches at 500 dollars. Find annual expected retained claims, in dollars. (Answer: $\$948.18$.)
For exponential severity with mean 1 thousand dollars, a Poisson claim count with mean 5, and an unlimited per-claim layer attaching at 1 thousand dollars, find the variance of annual aggregate retained claims. (Answer: $2.6424$ thousand-dollars squared.)
A hypothetical portfolio has Poisson mean 8 claims and exponential severity with mean 1 thousand dollars. A per-claim layer attaches at 0.5 thousand dollars and has limit 1.5 thousand dollars. Find the expected annual reinsurance recovery. (Answer: approximately $3.770$ thousand dollars.)

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