Policy Limits as Payment Caps

Introduction

A policy limit caps the insurer’s payment for a covered loss. If the ground-up loss is smaller than the limit, the payment equals the loss; if the loss reaches or exceeds the limit, the payment cannot exceed the limit. This simple transformation changes the payment distribution: below the limit, it follows the ground-up distribution, while at the limit, different large losses all produce the same payment.

This tutorial models a limit applied directly to the payment, without a deductible or coinsurance. The ground-up loss is assumed nonnegative and continuous, and the limit $u$ is a fixed positive amount. There is no discounting or mortality assumption: the calculations concern one hypothetical loss and its undiscounted payment.

Definition: Let $X$ be a nonnegative continuous ground-up loss with CDF $F_X(x)=P(X\le x)$, survival function $S_X(x)=P(X>x)$, and density $f_X(x)$. Under a policy limit $u>0$, the payment per loss is $Y=\min(X,u)$. Thus $Y=X$ when $X<u$, and $Y=u$ when $X\ge u$.

Deriving the Payment Distribution

The key is to translate an event about the payment $Y$ into the corresponding event about the ground-up loss $X$. For any threshold below the limit, the cap has not yet changed whether the payment is at or below that threshold. At or above the limit, every possible payment is already at or below the threshold.

Consider a threshold below the limit.
For $0\le y<u$, $Y\le y$ holds exactly when $X\le y$. If $X>y$, either the payment remains above $y$, or it is capped at $u>y$; in both cases $Y\le y$ is false.

Use the ground-up CDF below the limit.
The event equivalence gives $F_Y(y)=P(Y\le y)=P(X\le y)=F_X(y)$ for $0\le y<u$.

Consider the limit and larger thresholds.
Since $Y\le u$ always, $F_Y(y)=1$ for $y\ge u$. For $y<0$, nonnegative losses imply $F_Y(y)=0$.

F_Y(y)= \begin{cases} 0, & y<0,\\ F_X(y), & 0\le y<u,\\ 1, & y\ge u. \end{cases}

The CDF is therefore unchanged below the cap, but it reaches 1 at the cap. Because the ground-up loss is continuous, $P(X=u)=0$. All losses $X\ge u$ generate the same payment $Y=u$, so the payment distribution has a point mass there:

P(Y=u)=P(X\ge u)=P(X>u)=S_X(u)=1-F_X(u).

For $0<y<u$, the payment distribution has density $f_Y(y)=f_X(y)$. There is no payment above $u$. The continuous part below the cap accounts for probability $F_X(u)$, and the remaining probability $S_X(u)$ is concentrated at $u$. In particular, the value of the CDF just below the cap is $F_X(u)$ by continuity, while $F_Y(u)=1$.

Theorem: For a continuous ground-up loss capped at $u$, probabilities entirely below the cap can be calculated from $F_X$. In particular, for $0\le a<b<u$, $P(a<Y\le b)=F_X(b)-F_X(a)$. At the cap, $P(Y=u)=S_X(u)$, and $P(Y\le u)=1$.

Worked Examples

Worked Example: Exponential Loss With a Limit

A hypothetical insurer covers one loss $X$ with an exponential distribution of rate $\lambda=0.1$ per dollar, so $F_X(x)=1-e^{-0.1x}$ for $x\ge0$. The policy limit is $u=\$15$. Find the probabilities that the payment is at most $\$4$, is strictly between $\$4$ and $\$10$, is strictly between $\$10$ and $\$15$, and equals $\$15$. No interest or mortality assumption applies. Round probabilities to four decimal places.

Method. Use the ground-up CDF for thresholds below $u$, and use the survival probability at $u$ for the capped payment. This avoids treating the cap as if it were an ordinary continuous value.

First, since $4<15$,

P(Y\le4)=F_X(4)=1-e^{-0.4}\approx0.3297.

For the interval strictly between $\$4$ and $\$10$, both endpoints are below the cap. The distinction between strict and non-strict endpoints does not affect the probability because the distribution is continuous in this range:

P(4<Y<10)=F_X(10)-F_X(4) =(1-e^{-1})-(1-e^{-0.4}) =e^{-0.4}-e^{-1} \approx0.3024.

Similarly, for a payment strictly between $\$10$ and $\$15$, the payment corresponds to a ground-up loss strictly between those values:

P(10<Y<15)=F_X(15)-F_X(10) =e^{-1}-e^{-1.5} \approx0.1447.

The payment equals the limit whenever the loss is at least $\$15$:

P(Y=15)=S_X(15)=e^{-1.5}\approx0.2231.

Check. These events partition the payment outcomes into $Y\le4$, $4<Y<10$, $10<Y<15$, and $Y=15$. Before rounding, their probabilities sum to $(1-e^{-0.4})+(e^{-0.4}-e^{-1})+(e^{-1}-e^{-1.5})+e^{-1.5}=1$.

Worked Example: Pareto Loss and a Limit Probability

A hypothetical equipment policy covers a ground-up loss $X$ with a Pareto Type I distribution, minimum loss $x_m=\$5$, and shape parameter $\alpha=2$. Thus $S_X(x)=(5/x)^2$ for $x\ge5$. The payment cap is $u=\$20$. Find $F_Y(10)$, $P(10<Y<20)$, and $P(Y=20)$. No interest or mortality assumption applies. Give probabilities to four decimal places.

Method. Below the cap, evaluate the Pareto CDF. At the cap, use the Pareto survival function. This specifically avoids using the ground-up CDF as the payment CDF at the limit.

At $\$10$, the ground-up CDF is

F_X(10)=1-\left(\frac{5}{10}\right)^2 =1-\frac14=\frac34=0.7500.

Since $10<20$, the payment CDF at $\$10$ equals this value: $F_Y(10)=0.7500$. The ground-up CDF at the limit is

F_X(20)=1-\left(\frac{5}{20}\right)^2 =1-\frac1{16}=\frac{15}{16}.

The probability of a payment strictly between $\$10$ and $\$20$ is the difference between the ground-up CDF values at those endpoints:

P(10<Y<20)=F_X(20)-F_X(10) =\frac{15}{16}-\frac34 =\frac{15-12}{16} =\frac3{16}=0.1875.

The payment equals the cap for every loss at least $\$20$:

P(Y=20)=S_X(20)=\left(\frac5{20}\right)^2 =\frac1{16}=0.0625.

Check and trap. The probabilities $P(Y\le10)=12/16$, $P(10<Y<20)=3/16$, and $P(Y=20)=1/16$ sum to $16/16=1$. A tempting but incorrect calculation is $F_Y(20)=F_X(20)=15/16$. The cap instead guarantees $Y\le20$, so the correct value is $F_Y(20)=1$; the missing $1/16$ is the probability mass at the limit.

Worked Example: Uniform Loss Capped Below Its Maximum

A hypothetical policy covers a loss $X$ uniformly distributed from $\$0$ to $\$30$, with density $f_X(x)=1/30$ on that interval. Its limit is $u=\$18$. Find $F_Y(6)$, $P(6<Y<18)$, and $P(Y=18)$, then check that the probabilities sum to one. No interest or mortality assumption applies. Round probabilities to four decimal places.

Method. Use the uniform CDF for values below the cap and the upper-tail probability for the payment at the cap.

For $0\le y\le30$, $F_X(y)=y/30$. Since $6<18$,

F_Y(6)=F_X(6)=\frac6{30}=0.2000.

A payment strictly between $\$6$ and $\$18$ occurs exactly when the loss is in that same interval. The interval has length $18-6=12$, so

P(6<Y<18)=\frac{18-6}{30} =\frac{12}{30}=0.4000.

Every loss from $\$18$ to $\$30$ produces a payment of $\$18$. The interval has length $30-18=12$, giving

P(Y=18)=P(X\ge18)=\frac{30-18}{30} =\frac{12}{30}=0.4000.

The check is $0.2000+0.4000+0.4000=1.0000$. The cap creates a point mass even though the ground-up loss itself has a continuous uniform distribution.

Common Mistakes and Interpretation

A cap does not truncate the distribution by discarding losses above $u$. Those losses remain in the model; their payments are all set equal to $u$. This is why the payment CDF jumps to 1 at the limit, and why calculations that concern the exact capped payment must account for the tail probability of $X$.

Using $F_X(u)$ as $F_Y(u)$. The correct value is $F_Y(u)=1$, because $Y\le u$ with certainty. The value $F_X(u)$ is the probability of a payment strictly below the cap when $X$ is continuous.
Forgetting the capped tail. $P(Y=u)$ is $S_X(u)$, not zero. Losses above the limit still generate payments.
Applying the cap to a threshold below $u$. For $y<u$, $F_Y(y)=F_X(y)$; the cap does not alter whether the payment is at or below $y$.
Confusing the loss with the payment. If $X>u$, the loss exceeds the limit but the payment remains $u$. An event about $Y$ must be translated using the payment transformation, not treated as an event about an uncapped loss without checking the boundary.

The distribution is a mixed distribution: it has a continuous component below $u$ and a point mass at $u$. This matters whenever probabilities are calculated from a payment model. The capped payment is not simply a continuous severity with a shorter range; it retains the ground-up distribution below the cap and collects the entire upper tail at one payment value.

Check Your Understanding

For each hypothetical policy, use the capped-payment distribution and pay particular attention to whether the requested value is below the limit or at the limit.

A ground-up loss is exponential with rate $\lambda=0.2$ per dollar and is capped at $u=\$8$. Find $F_Y(3)$ and $P(Y=8)$. (Answer: $F_Y(3)=1-e^{-0.6}\approx0.4512$; $P(Y=8)=e^{-1.6}\approx0.2019$.)
A loss is uniform on $[0,25]$ dollars and has a limit of $\$10$. Find $P(5<Y\le10)$, including the possibility that the payment equals the limit. (Answer: $0.8000$.)
A Pareto Type I loss has $x_m=\$4$, $\alpha=3$, and limit $u=\$12$. Find $F_Y(8)$ and $P(Y=12)$. (Answer: $F_Y(8)=0.8750$; $P(Y=12)=1/27\approx0.0370$.)
A ground-up loss is exponential with mean $\$5$, and the limit is $\$10$. Find the probability that the payment is strictly between $\$2$ and $\$10$. (Answer: $e^{-0.4}-e^{-2}\approx0.5350$.)
A loss is uniform on $[0,30]$ dollars and is capped at $\$20$. Find $P(8\le Y<20)$. (Answer: $12/30=0.4000$.)

What You'll Learn