Expected Payment Under a Policy Limit

Why a Policy Limit Changes the Expected Payment

A policy limit caps the insurer’s payment on a single claim, but it does not change the underlying ground-up loss distribution. For a nonnegative loss $X$ and a limit $u$, the insurer pays the smaller of the loss and the limit. Calculating the expected payment therefore requires accounting both for losses below the limit, which are paid in full, and for losses above it, which are capped.

The ground-up survival function makes this calculation particularly direct. Integrating the probability that the loss exceeds each level from zero to the limit gives the expected capped payment. This works for continuous, discrete, and mixed severities, including distributions with probability mass at or above the limit.

Definition: For a nonnegative ground-up loss $X$ and a nonnegative payment limit $u$, the insurer’s payment per loss under a limit alone is $Y^L=\min(X,u)$. The expected payment is $E[\min(X,u)]$. The survival function is $S_X(t)=P(X>t)$.

The Survival-Integral Formula

For any nonnegative random variable $X$ and finite limit $u$, the expected capped payment is

E[\min(X,u)]=\int_0^u S_X(t)\,dt.

This formula needs no assumption that $X$ has a density, and it does not require $E[X]$ to be finite. The cap $u$ is finite, so the payment is bounded between zero and $u$, and its expectation exists. In the integral, isolated points at which the survival function jumps do not affect the value.

Result: For a nonnegative loss $X$ and finite limit $u$, $E[\min(X,u)]=\int_0^u P(X>t)\,dt$. Use the ground-up survival function over the interval from zero to the payment limit.

A second expression can be useful when the severity distribution is given in terms of probabilities or conditional moments:

E[\min(X,u)] = E[X\,\mathbf{1}_{\{X\leq u\}}]+uP(X>u).

Losses at or below the limit contribute their full amount to the first term; losses strictly above it contribute $u$ to the second. A loss exactly equal to $u$ is included in the first term and is paid $u$. Thus, probability mass at the limit is handled correctly by this expression as well.

Why Integrating the Survival Function Works

For each realized loss $x\geq0$, the quantity $\min(x,u)$ is the length of the interval from zero up to $\min(x,u)$. Equivalently, at a level $t$ between zero and $u$, the indicator that the loss reaches above $t$ is one precisely when $x>t$. Therefore the capped payment can be written as an area:

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

Taking expectations and interchanging the expectation and integral is valid because the integrand is nonnegative. At each level $t$, the expected indicator is the probability $P(X>t)=S_X(t)$. This gives the result:

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

The argument also explains why the formula accommodates atoms. It sums the survival probabilities across payment levels; it does not require a density or treat a point probability as though it were a density value.

Worked Applications

Worked Example: Exponential Loss With a Fixed Limit

A hypothetical insurer models the ground-up amount of a single claim as exponential with mean $\theta=\$10{,}000$, so $S_X(t)=e^{-t/10{,}000}$ for $t\geq0$. A policy pays at most $u=\$15{,}000$ for the claim. Find the expected insurer payment to the nearest cent.

Model and assumptions. The loss is $X\sim\operatorname{Exponential}(\text{mean }\$10{,}000)$, with the stated survival function. Apply the limit directly to the ground-up loss: $Y^L=\min(X,\$15{,}000)$. No interest discounting is applied; the loss and payment are measured at the same valuation date. No mortality assumption is involved.

Method. Use the survival-integral formula:

E[Y^L] = \int_0^{15{,}000} e^{-t/10{,}000}\,dt.

An antiderivative is $-10{,}000e^{-t/10{,}000}$. Substituting the bounds gives

E[Y^L] = 10{,}000\left(1-e^{-15{,}000/10{,}000}\right) = 10{,}000(1-e^{-1.5}).

Using $e^{-1.5}=0.2231301601$, the result is $10{,}000(0.7768698399)=\$7{,}768.698399\ldots$, or $\$7{,}768.70$ to the nearest cent.

Check and trap avoided. The payment cannot exceed $\$15{,}000$, and it cannot exceed the uncapped exponential mean of $\$10{,}000$; $\$7{,}768.70$ satisfies both checks. A common wrong approach is to calculate the uncapped mean and report $\$10{,}000$ as the capped expected payment. The positive probability of a loss above $\$15{,}000$ means that some payments are reduced, so the capped expectation is lower.

Worked Example: Standard Pareto Type I With a Limit Above Its Minimum

For a hypothetical single-claim model, let $X$ have a standard Pareto Type I distribution with minimum parameter $\theta=\$4{,}000$ and shape parameter $\alpha=2$. A policy has a per-claim payment limit $u=\$8{,}000$. Calculate the expected payment to the nearest cent.

Model and assumptions. The standard Pareto Type I survival function is

S_X(t)= \begin{cases} 1, & 0\leq t<\theta,\\ (\theta/t)^\alpha, & t\geq\theta. \end{cases}

Here, $X\geq\$4{,}000$, $\theta=\$4{,}000$, and $\alpha=2$. No interest discounting is applied, and no mortality assumption is involved.

Method. Split the survival integral at the Pareto minimum, where the survival expression changes:

\begin{aligned} E[\min(X,u)] &=\int_0^{4{,}000}1\,dt +\int_{4{,}000}^{8{,}000}\left(\frac{4{,}000}{t}\right)^2dt\\ &=4{,}000+16{,}000{,}000\int_{4{,}000}^{8{,}000}t^{-2}\,dt\\ &=4{,}000+16{,}000{,}000 \left[-\frac{1}{t}\right]_{4{,}000}^{8{,}000}\\ &=4{,}000+16{,}000{,}000 \left(\frac{1}{4{,}000}-\frac{1}{8{,}000}\right)\\ &=4{,}000+2{,}000 =\$6{,}000. \end{aligned}

The expected payment is $\$6{,}000.00$ to the nearest cent. As a check, the uncapped mean exists because $\alpha>1$ and equals $\alpha\theta/(\alpha-1)=\$8{,}000$. The capped expectation is below that mean and below the $\$8{,}000$ maximum payment.

Trap avoided. For standard Pareto Type I, survival is one below $\theta$; the loss cannot be less than its minimum. Using $S_X(t)=(\theta/(\theta+t))^\alpha$ instead would use the Lomax (Pareto Type II) survival function, not the standard Pareto Type I model specified here, and would produce a different answer.

Worked Example: A Discrete Severity With Losses Above the Limit

A hypothetical portfolio assigns a single claim the following ground-up severity distribution: $X=\$2{,}000$ with probability $0.20$, $X=\$5{,}000$ with probability $0.50$, and $X=\$10{,}000$ with probability $0.30$. The per-claim payment limit is $u=\$6{,}000$. Find the expected insurer payment.

Model and assumptions. The severity is discrete, with the three probabilities specified above and no other possible losses. The payment is $Y^L=\min(X,\$6{,}000)$. No interest discounting is applied, and no mortality assumption is involved.

Method 1: payment outcomes. The three ground-up losses produce payments of $\$2{,}000$, $\$5{,}000$, and $\$6{,}000$, respectively. The third payment is capped even though the ground-up loss is $\$10{,}000$. Thus,

E[Y^L] = 0.20(2{,}000)+0.50(5{,}000)+0.30(6{,}000) = 400+2{,}500+1{,}800 = \$4{,}700.

Method 2: survival integral check. The survival probability is 1 from zero to $\$2{,}000$, 0.80 from $\$2{,}000$ to $\$5{,}000$, and 0.30 from $\$5{,}000$ to $\$6{,}000$. Therefore,

\begin{aligned} \int_0^{6{,}000}S_X(t)\,dt &=1(2{,}000)+0.80(3{,}000)+0.30(1{,}000)\\ &=2{,}000+2{,}400+300 =\$4{,}700. \end{aligned}

Both calculations give an expected payment of $\$4{,}700.00$ to the nearest cent. The payment distribution has probability 0.30 at the cap because all claims with $X=\$10{,}000$ are paid $\$6{,}000$. A common error is to use the ground-up value $\$10{,}000$ for that outcome when calculating the insurer’s payment.

Using the Formula and Checking the Result

For a continuous severity with a convenient survival function, integrate that function from zero to $u$. If the survival function changes form at a threshold, split the integral there, as in the Pareto example. For a discrete severity, either calculate each capped payment and weight it by its probability or integrate the piecewise-constant survival function. Both approaches calculate the same expected payment.

For standard Pareto Type I, the survival integral also gives a general limited-mean expression. When $u\geq\theta$, $\theta>0$, and $\alpha>0$, the survival function equals one up to $\theta$, then $(\theta/t)^\alpha$. Integrating the two regions gives, for $\alpha\neq1$,

E[\min(X,u)] = \theta+\theta^\alpha\frac{u^{1-\alpha}-\theta^{1-\alpha}}{1-\alpha} = \theta+\frac{\theta}{\alpha-1} \left[1-\left(\frac{\theta}{u}\right)^{\alpha-1}\right].

When $\alpha=1$, the integral of $1/t$ gives

E[\min(X,u)]=\theta+\theta\ln(u/\theta),\qquad u\geq\theta.

If $0\leq u<\theta$, then $X\geq\theta>u$ with probability one, so the payment is always $u$ and $E[\min(X,u)]=u$. These limited-mean formulas are valid for the stated standard Pareto Type I model. The limited expectation is finite for any $\alpha>0$ when $u$ is finite, even when the uncapped mean does not exist for $\alpha\leq1$.

Check	What to verify
Payment cap	$0\leq E[\min(X,u)]\leq u$.
Comparison with uncapped mean	If $E[X]$ exists, $E[\min(X,u)]\leq E[X]$, because $\min(X,u)\leq X$.
Distribution definition	Use the survival function for the specified model; standard Pareto Type I and Lomax (Pareto Type II) have different survival functions.
Discrete values above the limit	Each such loss contributes the limit $u$, not its ground-up amount.

The survival-integral result is useful because the capped payment depends only on survival probabilities up to the limit. It provides a direct route from a specified severity model to the expected insurer payment without first deriving the full payment distribution. When checking an answer, the cap is an immediate upper bound; an uncapped mean, if finite, provides another upper bound.

Key takeaway: For a nonnegative single loss subject only to a payment limit $u$, integrate the ground-up survival function from zero to $u$: $E[\min(X,u)]=\int_0^u S_X(t)\,dt$. Keep the specified severity model intact, especially when distinguishing standard Pareto Type I from Lomax.

Check Your Understanding

For each problem, calculate the expected payment under the stated single-claim limit. Unless stated otherwise, assume no interest discounting and no mortality model; losses and payments are measured at the same valuation date.

A hypothetical severity is exponential with mean $\$5{,}000$. The policy limit is $\$5{,}000$. Calculate the expected payment to the nearest cent. (Answer: $\$3{,}160.60$)
A standard Pareto Type I loss has minimum parameter $\theta=\$3{,}000$ and shape parameter $\alpha=3$. The payment limit is $\$6{,}000$. Calculate the expected payment to the nearest cent. (Answer: $\$4{,}125.00$)
A discrete loss is $\$1{,}000$ with probability 0.25, $\$4{,}000$ with probability 0.50, and $\$9{,}000$ with probability 0.25. Find the expected insurer payment with a $\$5{,}000$ limit. (Answer: $\$3{,}500.00$)
A hypothetical severity has survival function 1 for $0\leq t<\$2{,}000$, 0.60 for $\$2{,}000\leq t<\$5{,}000$, and 0.20 for $\$5{,}000\leq t<\$9{,}000$. The payment limit is $\$7{,}000$. Calculate the limited expected payment. (Answer: $\$4{,}200.00$)
A loss is uniform on the interval from $\$0$ to $\$12{,}000$, and the policy limit is $\$9{,}000$. Use the survival-integral formula to find the expected payment to the nearest cent. (Answer: $\$5{,}625.00$)

Check	What to verify
Payment cap	\(0\leq E[\min(X,u)]\leq u\).
Comparison with uncapped mean	If \(E[X]\) exists, \(E[\min(X,u)]\leq E[X]\), because \(\min(X,u)\leq X\).
Distribution definition	Use the survival function for the specified model; standard Pareto Type I and Lomax (Pareto Type II) have different survival functions.
Discrete values above the limit	Each such loss contributes the limit \(u\), not its ground-up amount.

What You'll Learn