Payment Distributions Under Policy Limits

How a Policy Limit Changes a Loss

A policy limit caps the insurer’s payment for a covered loss. If the covered loss is below the limit, the payment equals the loss; if the loss reaches or exceeds the limit, the payment is capped. As a result, the payment distribution is not usually just the loss distribution cut off at the limit: all losses at or above the limit are combined into a single payment value.

This tutorial considers a nonnegative loss $X$ and a fixed policy limit $u>0$. There is no deductible or coinsurance in the payment considered here. The key feature is the transformation from loss $X$ to payment $Y=\min(X,u)$.

Definition: For a nonnegative loss $X$ subject to a policy limit $u>0$, the limited payment is $Y=\min(X,u)$. Thus, $Y=X$ when $X<u$, and $Y=u$ when $X\ge u$. Write $F_X(x)=P(X\le x)$ for the loss distribution function, and $F_X(x-)=P(X<x)$ for its left limit.

The distinction between $F_X(u)$ and $F_X(u-)$ matters when the loss can equal $u$. A loss exactly equal to the limit and a loss above the limit both produce payment $u$. The point mass at the cap must therefore include both possibilities.

The Distribution and the Point Mass at the Limit

For any payment value below the cap, the payment has not yet been altered by the limit. At and above the cap, however, the payment is always $u$. These facts give the full distribution function of $Y$.

Theorem: If $X\ge 0$ and $Y=\min(X,u)$, then $$ F_Y(y)=P(Y\le y)= \begin{cases} 0, & y<0,\\ F_X(y), & 0\le y<u,\\ 1, & y\ge u. \end{cases} $$ The payment has a point mass at the limit of $$ P(Y=u)=P(X\ge u)=1-F_X(u-). $$

The left limit in the point-mass formula ensures that the event $X=u$ is included. If $X$ is continuous at $u$, then $F_X(u-)=F_X(u)$, so the mass is $1-F_X(u)$. That simplification is not valid in general for a discrete or mixed loss distribution.

Why the Formula Holds

The distribution function follows directly from the events that define the payment. For $0\le y<u$, a payment is at most $y$ precisely when the original loss is at most $y$. For $y\ge u$, every payment is at most $y$, since no payment can exceed $u$. This establishes the piecewise formula.

For the point mass, the payment equals $u$ precisely when the loss is at least $u$. The complement of $X\ge u$ is $X<u$, whose probability is $F_X(u-)$. Therefore:

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

This reasoning also shows why it is useful to translate an event about payment back into an event about loss before calculating. Below the cap, the transformation leaves losses unchanged; at the cap, it gathers the entire upper tail into one payment value.

Calculating Probabilities in Payment Intervals

For an interval that stays strictly below the cap, the payment and loss have the same value. For example, if $0\le a<b<u$, then $a<Y\le b$ is exactly the event $a<X\le b$. The usual distribution-function subtraction applies, with the stated endpoint convention.

Interval rules: For $0\le a<b<u$, $$ P(a<Y\le b)=F_X(b)-F_X(a). $$ For $0\le a<u$, $$ P(a<Y\le u)=P(X>a)=1-F_X(a), $$ and $$ P(a<Y<u)=P(a<X<u)=F_X(u-)-F_X(a). $$ Use left limits when the event includes or excludes an endpoint at which $X$ may have a point mass.

The second rule includes the cap payment: $Y>a$ occurs exactly when $X>a$, and $Y\le u$ is always true. By contrast, $Y<u$ occurs exactly when $X<u$, so an interval ending strictly below $u$ excludes every loss at or above the limit. Endpoint symbols are consequential when the loss distribution has atoms.

Worked Example: An Exponential Loss With a Cap

A hypothetical policy pays $Y=\min(X,8)$, where $X$ has an exponential distribution with rate $\lambda=0.2$ per thousand dollars. Thus $F_X(x)=1-e^{-0.2x}$ for $x\ge0$, with amounts measured in thousands of dollars. Find (a) the probability of a payment equal to the limit, (b) $P(3<Y\le8)$, and (c) $P(3<Y<8)$. Give probabilities to four decimals.

Method. Translate each payment event into a loss event, then use the exponential survival probability $P(X>x)=e^{-0.2x}$. The exponential distribution is continuous, so the probability of a loss at any single exact value is zero.

(a) A payment of 8 occurs when $X\ge8$. By continuity, $P(X\ge8)=P(X>8)$:

P(Y=8)=e^{-0.2(8)}=e^{-1.6}=0.2018965\ldots\approx0.2019.

(b) Since $3<8$, the event $3<Y\le8$ is exactly $X>3$. Losses above 8 are included because they produce payment 8:

P(3<Y\le8)=P(X>3)=e^{-0.2(3)}=e^{-0.6}=0.5488116\ldots\approx0.5488.

P(3<Y<8)=P(X>3)-P(X\ge8) =e^{-0.6}-e^{-1.6} =0.3469151\ldots\approx0.3469.

Check. The events $Y\le3$, $3<Y<8$, and $Y=8$ partition all possible payments. Their probabilities are $1-e^{-0.6}=0.4512$, $0.3469$, and $0.2019$, respectively, which sum to $1.0000$. The common mistake is to treat $3<Y\le8$ as $3<X\le8$, which incorrectly drops losses above the limit.

Worked Example: A Discrete Loss With a Loss Exactly at the Limit

A hypothetical equipment policy has limit $u=12$ thousand dollars. The loss $X$, also measured in thousands of dollars, has the following distribution: $P(X=0)=0.10$, $P(X=5)=0.20$, $P(X=9)=0.25$, $P(X=12)=0.30$, and $P(X=20)=0.15$. Find the payment distribution, $P(Y=12)$, and $P(5<Y\le12)$.

Method. Map each possible loss value to $\min(X,12)$. Losses of 12 and 20 both map to payment 12; the loss exactly at the limit must not be omitted.

\begin{array}{c|ccccc} X & 0 & 5 & 9 & 12 & 20\\ \hline Y=\min(X,12) & 0 & 5 & 9 & 12 & 12 \end{array}

The payment probabilities are therefore $P(Y=0)=0.10$, $P(Y=5)=0.20$, $P(Y=9)=0.25$, and

$$ P(Y=12)=P(X=12)+P(X=20)=0.30+0.15=0.45. $$

For the requested interval, payments greater than 5 and at most 12 are 9 and 12. Hence:

P(5<Y\le12)=P(Y=9)+P(Y=12)=0.25+0.45=0.70.

Check. The payment probabilities sum to $0.10+0.20+0.25+0.45=1.00$. Also, $5<Y\le12$ corresponds to $X>5$, whose probability is $0.25+0.30+0.15=0.70$. A common error is to assign only $P(X>12)=0.15$ to the cap. That misses the $0.30$ probability at $X=12$; the correct cap mass is $P(X\ge12)=0.45$.

Worked Example: A Mixed Loss With an Interior Atom

For a hypothetical claim, $X=4$ thousand dollars with probability $0.25$. With probability $0.75$, $X$ is uniform on $[0,12]$ thousand dollars. The policy limit is $u=8$. Find $P(Y=8)$, $P(2\le Y<8)$, and $P(4\le Y\le8)$. Give probabilities to four decimals.

Method. Separate the point-mass component from the uniform component. The atom at 4 remains a payment of 4 because it is below the limit; only the uniform losses at or above 8 are mapped to the cap.

For the cap payment, the atom at 4 contributes nothing. Conditional on the uniform component, the probability of a loss of at least 8 is $(12-8)/12=4/12$. Therefore:

P(Y=8)=0.25(0)+0.75\left(\frac{4}{12}\right)=0.25.

For $2\le Y<8$, the atom at 4 contributes $0.25$. The uniform component contributes its probability of a loss in $[2,8)$, namely $(8-2)/12=6/12$, multiplied by its component probability:

P(2\le Y<8)=0.25+0.75\left(\frac{6}{12}\right) =0.25+0.375=0.6250.

For $4\le Y\le8$, the atom at 4 contributes $0.25$. The uniform component contributes losses in $[4,8)$, with probability $(8-4)/12=4/12$, and losses at or above 8, with probability $4/12$. Thus:

P(4\le Y\le8)=0.25+0.75\left(\frac{4}{12}+\frac{4}{12}\right) =0.25+0.50=0.7500.

Check. The payment below 8 has probability $0.25+0.75(8/12)=0.75$; adding the cap mass $0.25$ gives 1. The common mistake is to discard the point mass at 4 when calculating an interval that includes 4. A continuous component assigns zero probability to an exact value, but the full mixed distribution does not.

Common Endpoint Errors

When the loss distribution is continuous, changing a strict inequality to a non-strict one at a fixed interior point does not change a probability. That convenience can hide errors: it does not hold for discrete or mixed distributions, and the cap itself creates a payment point mass even if $X$ is continuous.

Using the wrong tail at the cap. The cap payment probability is $P(X\ge u)=1-F_X(u-)$, not always $1-F_X(u)$.
Dropping losses above the cap. For $a<u$, the event $a<Y\le u$ includes every loss $X>a$, including losses much larger than $u$.
Counting the cap when the interval excludes it. An event ending in $Y<u$ excludes all losses at least $u$.
Ignoring atoms inside an interval. If $X$ has a point mass at an included payment value below $u$, that mass remains in $Y$.

A reliable procedure is to identify whether the interval reaches the cap, translate its endpoint conditions to conditions on $X$, and then apply the distribution function with the correct strict or non-strict endpoints. As a final check, probabilities for a partition of all possible payments should sum to 1. This distributional view is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, first identify the loss event corresponding to the payment event. Probabilities are rounded to four decimals where needed.

A hypothetical loss is exponential with rate $0.25$ per thousand dollars and has policy limit $u=6$ thousand dollars. Find $P(Y=6)$, $P(2<Y<6)$, and $P(Y\le2)$. (Answer: $0.2231,\ 0.3834,\ 0.3935$, respectively.)
A discrete loss has probabilities $P(X=2)=0.15$, $P(X=7)=0.35$, $P(X=11)=0.20$, and $P(X=16)=0.30$. With limit $u=11$, give the payment distribution and find $P(7<Y\le11)$. (Answer: $P(Y=2)=0.15,\ P(Y=7)=0.35,\ P(Y=11)=0.50;\ P(7<Y\le11)=0.50$.)
A hypothetical loss is uniform on $[0,20]$ thousand dollars and the limit is $u=12$. Find $P(4\le Y<12)$, $P(Y=12)$, and $P(Y>4)$. (Answer: $0.4000,\ 0.4000,\ 0.8000$, respectively.)
A loss distribution satisfies $F_X(5)=0.40$, $F_X(5-)=0.30$, and $F_X(10-)=0.80$. With limit $u=10$, find $P(Y=10)$, $P(5\le Y<10)$, and $P(Y\le5)$. (Answer: $0.20,\ 0.50,\ 0.40$, respectively.)
A hypothetical loss equals 3 thousand dollars with probability $0.40$; otherwise it is exponential with rate $0.5$ per thousand dollars. With limit $u=4$, find $P(Y=4)$, $P(1<Y\le4)$, and $P(1<Y<4)$. (Answer: $0.0812,\ 0.7639,\ 0.6827$, respectively.)

What You'll Learn