Limited Losses Under a Per-Claim Limit

Why a Per-Claim Limit Changes Expected Loss

A per-claim limit caps what the insurer pays on any one covered claim. Losses below the limit are paid in full, while losses above it are capped. To estimate the insurer's expected cost per claim, the relevant quantity is therefore the expected value of the capped payment, not the expected ground-up loss with a cap applied afterward.

This tutorial considers a covered, nonnegative ground-up claim severity $X$ and a per-claim limit $u$, with no deductible, coinsurance, or other payment adjustment. It develops two equivalent ways to calculate the expected payment: directly from the limited severity and by integrating the severity's survival function.

Definition: Under a per-claim limit $u$, with full payment from the first dollar and no other adjustment, the insurer payment per loss is $Y^L=\min(X,u)$. The limited expected value at limit $u$ is $\operatorname{LEV}_X(u)=E[\min(X,u)]$. It is the expected insurer payment per claim under these assumptions.

The limit applies separately to each claim. It is not an aggregate policy cap across multiple claims. Also, $X$ must represent the covered loss to which the limit applies. If a deductible or other payment feature is present, the insurer's payment function changes and this formula alone is not the correct expected payment.

Deriving the Survival-Function Formula

For a nonnegative random variable, the survival function is $S_X(t)=P(X>t)$. The limited expected value can be written as the area under this function up to the limit:

\operatorname{LEV}_X(u)=E[\min(X,u)]=\int_0^u S_X(t)\,dt

The identity follows by representing each capped payment as the length of an interval. For any fixed loss value $x\geq 0$, the indicator $1_{\{x>t\}}$ equals 1 for thresholds below $x$ and 0 above $x$. Its integral from 0 to $u$ is the length of the part of that interval below both $x$ and $u$, namely $\min(x,u)$.

\min(x,u)=\int_0^u 1_{\{x>t\}}\,dt

Taking expectations on both sides and interchanging expectation and integration gives the result. This interchange is valid because the integrand is nonnegative; it is also bounded over the finite interval from 0 to $u$. At any single threshold where $X$ has a point mass, using $P(X>t)$ rather than $P(X\geq t)$ does not change the integral: a single point has zero length.

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

This derivation works for both discrete and continuous severity distributions. If $X$ has a density $f_X$, an equivalent calculation is to average payments below the limit and add the capped payment for losses above it:

\operatorname{LEV}_X(u)=\int_0^u x f_X(x)\,dx+uS_X(u)

That density expression assumes a continuous distribution. If there are point masses, calculate the expectation using the actual probability mass at each value, or use the survival integral, which applies without a density. The survival integral also makes clear that the limited expected value is finite for a finite limit $u$, even if the uncapped severity has no finite mean.

Worked Applications

Worked Example: Exponential Severity With a Per-Claim Cap

A hypothetical insurer covers claims with severity $X$ measured in dollars. Assume $X$ has an exponential distribution with rate $\lambda=0.01$ per dollar, so $S_X(t)=e^{-0.01t}$ for $t\geq0$. The per-claim limit is $u=\$100$, with no deductible or other payment adjustment. Find the expected insurer payment per claim and the expected amount removed by the cap. No interest discounting or mortality assumption is involved; this is a per-claim severity calculation.

Step 1: Identify the method. Apply the survival-function formula for the limited expected value:

\operatorname{LEV}_X(100)=\int_0^{100}e^{-0.01t}\,dt

Step 2: Integrate and substitute the limit.

\begin{aligned} \operatorname{LEV}_X(100) &=\left[-100e^{-0.01t}\right]_0^{100}\\ &=100(1-e^{-1})\\ &=100(1-0.3678794)\\ &=63.2121. \end{aligned}

The expected insurer payment is $\$63.21$ per claim, rounded to the nearest cent. The uncapped exponential mean is $1/\lambda=\$100$, so the expected amount removed by the cap is $\$100-\$63.2121=\$36.7879$, or $\$36.79$ per claim. The check is consistent: the capped payment cannot exceed $\$100$, and its expected value is below the uncapped mean. The method avoids treating the cap as though it simply reduced every claim by the same amount.

Worked Example: Discrete Severity and a Limit Below Some Losses

For a hypothetical class of covered claims, severity $X$ has the following distribution: $\$20$ with probability 0.20, $\$60$ with probability 0.30, $\$100$ with probability 0.30, and $\$200$ with probability 0.20. The per-claim limit is $u=\$80$. Calculate the expected insurer payment both from the payment outcomes and from the survival function. No interest discounting or mortality assumption is involved; the stated discrete distribution is the severity model.

Step 1: Cap each possible payment. The insurer pays $\$20,\ \$60,\ \$80,$ and $\$80$, respectively. Thus:

\begin{aligned} E[\min(X,80)] &=0.20(20)+0.30(60)+0.30(80)+0.20(80)\\ &=4+18+24+16\\ &=62. \end{aligned}

Step 2: Check using the survival function. From 0 to 20, all four outcomes exceed the threshold, so $S_X(t)=1$. From 20 to 60, the outcomes 60, 100, and 200 exceed it, so $S_X(t)=0.80$. From 60 to 80, only 100 and 200 exceed it, so $S_X(t)=0.50$. Therefore:

\begin{aligned} \operatorname{LEV}_X(80) &=\int_0^{20}1\,dt+\int_{20}^{60}0.80\,dt+\int_{60}^{80}0.50\,dt\\ &=20+32+10\\ &=62. \end{aligned}

Both methods give an expected payment of $\$62.00$ per claim. As another check, the uncapped mean is $0.20(20)+0.30(60)+0.30(100)+0.20(200)=\$92$; the cap reduces the expectation by $\$30$. Integrating the survival function is especially useful here because it handles the discrete mass without requiring a density.

Worked Example: Pareto-Type Severity With a Specified Limit

Suppose a hypothetical covered-loss severity follows a Pareto Type II (Lomax) distribution with scale parameter $\theta=\$150$ and shape parameter $\alpha=3$. Its survival function is $S_X(t)=(150/(150+t))^3$ for $t\geq0$. The per-claim limit is $u=\$150$, with no deductible or other adjustment. Find the limited expected value and compare it with the uncapped mean. No interest discounting or mortality assumption is involved.

Step 1: Set up the survival integral.

\operatorname{LEV}_X(150)=\int_0^{150}\left(\frac{150}{150+t}\right)^3dt

Step 2: Integrate. Let $z=150+t$, so the integration limits for $z$ are 150 and 300:

\begin{aligned} \operatorname{LEV}_X(150) &=150^3\int_{150}^{300}z^{-3}\,dz\\ &=150^3\left[-\frac{1}{2z^2}\right]_{150}^{300}\\ &=\frac{150}{2}\left(1-\left(\frac{150}{300}\right)^2\right)\\ &=75(1-0.25)\\ &=56.25. \end{aligned}

The expected insurer payment is $\$56.25$ per claim. For this Pareto Type II distribution with $\alpha>1$, the uncapped mean is $\theta/(\alpha-1)=150/2=\$75$. Thus the expected reduction due to the cap is $\$18.75$. The result passes a basic check: the limited expected value is below both the limit of $\$150$ and the uncapped mean of $\$75$.

What the Limit Formula Does—and Does Not—Mean

A capped payment is bounded above by $u$, but it is generally not correct to calculate it by taking the smaller of the mean loss and the limit. In general, the expectation of a capped random variable is not the cap applied to the expectation:

E[\min(X,u)]\ne\min(E[X],u)

The distinction matters because the limit changes each claim payment before the average is taken. The following trap uses the discrete severity distribution from the earlier example to show the error explicitly.

Common mistake: Using $E[X]=\$92$ and calculating $\min(92,80)=\$80$ gives the wrong expected insurer payment. The correct calculation caps each outcome first: $0.20(20)+0.30(60)+0.30(80)+0.20(80)=\$62$. The incorrect calculation treats a single average loss as though it were one claim.

A per-claim cap also should not be confused with a deductible or an aggregate limit. Here, a loss below $u$ is paid in full, and every loss above $u$ produces payment $u$. If the policy has a deductible, the payment depends on both the loss and the deductible; if a limit applies across multiple claims, the payments can depend on the other claims in the period. In either case, the payment function must be specified before calculating its expectation.

Key takeaway: For a nonnegative covered loss $X$ subject only to a per-claim limit $u$, the expected insurer payment is $E[\min(X,u)]$, equivalently $\int_0^u S_X(t)\,dt$. Apply the cap to each loss before taking the expectation.

Check Your Understanding

For each problem, use the per-claim payment definition and the survival-function formula where appropriate. Assume no deductible, interest discounting, or mortality unless the problem states otherwise.

A hypothetical claim severity is $\$40$ with probability 0.25, $\$120$ with probability 0.50, and $\$300$ with probability 0.25. The per-claim limit is $\$100$. What is the expected insurer payment per claim? (Answer: $\$85.00$)
A hypothetical severity has an exponential distribution with rate $\lambda=0.02$ per dollar. For a $\$50$ per-claim limit, calculate the expected payment to the nearest cent. (Answer: $\$31.61$)
A hypothetical severity is uniformly distributed from $\$0$ to $\$240$. Find the expected insurer payment under a $\$120$ per-claim limit. (Answer: $\$90.00$)
A hypothetical severity follows a Pareto Type II distribution with $\theta=\$200$, $\alpha=2$, and $S_X(t)=(200/(200+t))^2$. Calculate the limited expected value at $u=\$200$. (Answer: $\$100.00$)
A hypothetical severity has survival function $S_X(t)=1$ for $0\leq t<20$, $S_X(t)=0.60$ for $20\leq t<60$, and $S_X(t)=0.25$ for $60\leq t<100$. Find the limited expected value at $u=\$100$. (Answer: $\$54.00$)

What You'll Learn