Deriving Limited Expected Loss from the Survival Function

Why the Survival Function Gives Limited Expected Loss

A policy limit caps the insurer’s payment on an individual loss. To price that capped payment, the quantity of interest is the expected value of the capped severity—not the average severity among claims that happen to fall below the limit. A survival function provides a direct way to calculate this expected payment because it records, at every loss threshold, the probability that severity exceeds that threshold.

This tutorial derives that relationship and applies it to specified severity distributions. It is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. All examples use undiscounted losses: interest and mortality assumptions do not enter these calculations.

Definition: Let $X$ be a nonnegative ground-up loss and let $u\geq 0$ be a per-loss limit. The limited loss is $\min(X,u)$, and its expected value is the limited expected loss, $E[\min(X,u)]$. The survival function is $S_X(x)=P(X>x)$. For nonnegative $X$, the limited expected loss is $$ E[\min(X,u)]=\int_0^u S_X(x)\,dx. $$

The formula applies whether the severity distribution is continuous, discrete, or a mixture, and it does not require a density. The limit must be nonnegative and the loss must be nonnegative. A finite limit ensures that $0\leq\min(X,u)\leq u$, so the limited expected loss is finite even if the uncapped mean $E[X]$ is infinite.

Deriving the Survival-Integral Identity

For any fixed nonnegative loss value $X$, the quantity $\min(X,u)$ is the length of the interval from zero up to the smaller of $X$ and $u$. Equivalently, it is the accumulated length of all thresholds $x$ below both $X$ and $u$:

\min(X,u)=\int_0^u \mathbf{1}_{\{X>x\}}\,dx.

Here $\mathbf{1}_{\{X>x\}}$ equals one when $X>x$ and zero otherwise. If, for example, $X=40$ and $u=100$, the indicator is one for thresholds below 40 and zero above 40, so the integral equals 40. If $X=140$ and $u=100$, it is one throughout the integration interval, so the integral equals 100.

Taking expectations gives the result. Because the indicator is nonnegative, the order of expectation and integration can be interchanged by Tonelli’s theorem:

\begin{aligned} E[\min(X,u)] &=E\left[\int_0^u \mathbf{1}_{\{X>x\}}\,dx\right]\\ &=\int_0^u E[\mathbf{1}_{\{X>x\}}]\,dx\\ &=\int_0^u P(X>x)\,dx\\ &=\int_0^u S_X(x)\,dx. \end{aligned}

The probability in the integrand is a strict exceedance probability. Whether an atom occurs exactly at a threshold does not change the value of the integral: changing the integrand at an individual point does not change an ordinary integral. This is why the identity works for discrete losses as well as continuous losses.

How to Apply the Formula

When a survival curve is given, first identify the loss units and the limit in those units. Then integrate the curve from zero to the limit, splitting the integral at any points where its formula changes. The following process keeps the calculation tied to the insurance payment being valued.

Confirm the loss basis and cap.
Use the stated ground-up severity and the per-loss limit $u$. Do not replace the cap with a deductible or an aggregate limit.

Write the survival function over the integration range.
Check whether it is piecewise and whether the distribution has an atom, such as a probability mass at zero.

Integrate from zero to the limit.
Evaluate $\int_0^u S_X(x)\,dx$, splitting at any breakpoints and retaining the stated units.

Check and report.
The result must be between zero and $u$. If $E[X]$ is finite, it must also be no greater than $E[X]$. Round only the final monetary amount.

Worked Example: A Discrete Severity Distribution With a Cap

A hypothetical equipment policy has ground-up claim severity $X$, measured in thousands of dollars, with probabilities $P(X=0)=0.20$, $P(X=100)=0.40$, and $P(X=300)=0.40$. Calculate limited expected loss at a per-claim limit of $u=250$ thousand dollars. State the method and check the result. This is an undiscounted severity calculation; no interest or mortality assumption is involved.

Method. Integrate the survival function up to the limit. The specified distribution puts 20% probability at zero, so $P(X>0)=0.80$. For thresholds from 0 to 100, only losses of 100 or 300 exceed the threshold, giving survival 0.80. For thresholds from 100 to 250, only losses of 300 exceed the threshold, giving survival 0.40.

\begin{aligned} E[\min(X,250)] &=\int_0^{100}0.80\,dx+\int_{100}^{250}0.40\,dx\\ &=0.80(100)+0.40(250-100)\\ &=80+60\\ &=140\text{ thousand dollars}. \end{aligned}

The limited expected loss is therefore $140,000.00. As a direct check using the payment distribution, the capped payment is 0 with probability 0.20, 100 with probability 0.40, and 250 with probability 0.40; its mean is $0.20(0)+0.40(100)+0.40(250)=140$ thousand dollars. The answer is below the 250-thousand-dollar cap. A common mistake is to assign survival 0.40 over the full interval from zero to 250, overlooking that losses of 100 also exceed thresholds below 100.

Worked Example: A Heavy-Tailed Severity With an Infinite Mean

For a hypothetical liability coverage, let severity $X$ be measured in thousands of dollars. Its survival function is $S_X(x)=0.9(1+x/200)^{-1}$ for $x\geq0$; the remaining probability, 0.10, is at zero. Find limited expected loss at $u=600$ thousand dollars, and determine whether the uncapped mean is finite. This calculation is undiscounted, with no interest or mortality assumption.

Method. Integrate the specified survival curve to the cap. Use the substitution $z=1+x/200$, so $dx=200\,dz$. When $x=0$, $z=1$; when $x=600$, $z=4$.

\begin{aligned} E[\min(X,600)] &=\int_0^{600}0.9(1+x/200)^{-1}\,dx\\ &=0.9(200)\int_1^4\frac{1}{z}\,dz\\ &=180\ln(4)\\ &=249.5329850016\text{ thousand dollars}. \end{aligned}

Converting to dollars gives $249.5329850016\times 1{,}000=\$249{,}532.9850016$, which rounds to $249,532.99 to the nearest cent. For the uncapped mean, the same identity requires integration over the full nonnegative range. The survival curve behaves like a positive constant divided by $x$ for large $x$, and its integral diverges. More directly,

\int_0^\infty 0.9(1+x/200)^{-1}\,dx =180\int_1^\infty \frac{1}{z}\,dz =\infty.

Thus the uncapped mean is infinite, while the limited expected loss at 600 thousand dollars is finite. The limit bounds every payment by 600 thousand dollars. A common mistake is to multiply the limit by survival at the limit, obtaining $600(0.225)=135$ thousand dollars. That values only a rectangle of height $S_X(600)$; the correct formula integrates the varying survival probabilities over the entire range from zero to 600.

Worked Example: A Lomax Severity With a Finite Uncapped Mean

A hypothetical property severity $X$, in thousands of dollars, follows a Lomax distribution with scale 100 and shape 3. Its survival function is $S_X(x)=(1+x/100)^{-3}$ for $x\geq0$. Calculate limited expected loss at a limit of 200 thousand dollars, and compare it with the uncapped mean. The losses are undiscounted; no interest or mortality assumption is required.

Method. Integrate the survival function to 200. With $z=1+x/100$, $dx=100\,dz$, and the integration bounds become 1 and 3.

\begin{aligned} E[\min(X,200)] &=\int_0^{200}(1+x/100)^{-3}\,dx\\ &=100\int_1^3 z^{-3}\,dz\\ &=100\left[-\frac{1}{2}z^{-2}\right]_1^3\\ &=50\left(1-\frac{1}{9}\right)\\ &=\frac{400}{9}=44.444444\ldots\text{ thousand dollars}. \end{aligned}

Rounded to the nearest cent, the limited expected loss is $44,444.44. The uncapped mean is the integral of the survival function over all nonnegative thresholds:

E[X]=100\int_1^\infty z^{-3}\,dz =100\left[-\frac{1}{2}z^{-2}\right]_1^\infty =50\text{ thousand dollars}.

The capped mean is less than the uncapped mean, as it must be. A further check is that the expected amount above the cap is $50-44.444444\ldots=5.555555\ldots$ thousand dollars. This equals the integral of the survival curve beyond 200, $100\int_3^\infty z^{-3}\,dz=50/9$ thousand dollars. The calculation avoids the mistake of treating a finite uncapped mean as the limited mean; the limit removes some expected payment even when the uncapped mean exists.

Interpretation and Common Exam Traps

The survival-integral identity says that each threshold contributes its probability of being exceeded to the expected capped payment. At low thresholds, many claims may exceed the threshold; at high thresholds, fewer do. The area under that declining probability curve up to the limit is the expected payment under the cap.

Several checks help catch errors:

Respect the cap: $0\leq E[\min(X,u)]\leq u$. A result above the limit cannot be correct.
Compare with the uncapped mean when it is finite: $E[\min(X,u)]\leq E[X]$. Raising the cap cannot reduce limited expected loss.
Do not confuse a capped mean with a conditional mean: $E[\min(X,u)]$ is not $E[X\mid X\leq u]$. The latter excludes losses above the limit rather than capping their payments.
Integrate the whole curve: $uS_X(u)$ is generally not the limited expected loss. It uses the survival probability at just one threshold instead of all thresholds up to $u$.
Track units: If $X$ and $u$ are measured in thousands of dollars, the integral is also in thousands. Convert to dollars only after evaluating it.

The identity is useful when a survival function is specified directly, when the distribution has atoms, and when calculating a capped mean is easier than working from a density. It also makes the effect of the limit transparent: the survival probabilities between zero and the limit determine expected capped payment. When the cap changes, only the area added or removed over the affected threshold range changes.

Key takeaway: For nonnegative severity $X$ and a per-loss limit $u$, calculate limited expected loss as $\int_0^u S_X(x)\,dx$. Integrate the survival curve over the full interval, preserve the stated loss units, and check the answer against the cap.

Check Your Understanding

For each problem, use the survival-integral identity. Amounts stated in thousands are in thousands of dollars; round final monetary answers to the nearest cent.

A hypothetical severity is uniform on 0 to 400 thousand dollars. Calculate limited expected loss at a 100-thousand-dollar limit. (Answer: $87,500.00)
A hypothetical claim severity takes values 0, 50, and 250 thousand dollars with probabilities 0.25, 0.50, and 0.25, respectively. Find limited expected loss at a 100-thousand-dollar limit. (Answer: $50,000.00)
A hypothetical severity has exponential distribution with mean 80 thousand dollars, so $S_X(x)=e^{-x/80}$ for $x\geq0$. Calculate limited expected loss at 160 thousand dollars. (Answer: $69,173.18)
A hypothetical severity has survival function $S_X(x)=(1+x/100)^{-2}$ for $x\geq0$, with $x$ measured in thousands of dollars. Find limited expected loss at 300 thousand dollars and state the uncapped mean. (Answer: Limited expected loss is $75,000.00; uncapped mean is $100,000.00.)
A hypothetical severity, measured in thousands of dollars, has survival function $S_X(x)=0.7(1+x/250)^{-1}$ for $x\geq0$, with the remaining probability at zero. Find limited expected loss at 500 thousand dollars and determine whether the uncapped mean is finite. (Answer: Limited expected loss is $192,257.15; the uncapped mean is infinite.)

What You'll Learn