Probability Mass at a Policy Limit

Why a Policy Limit Creates Probability Mass

A policy limit caps the payment on a loss: once the loss reaches the cap, a larger loss does not produce a larger payment. Consequently, quite different ground-up losses can all produce the same payment. The payment distribution can therefore have a probability mass at the limit even when the ground-up loss has a continuous distribution and no point masses.

This tutorial considers a single ground-up loss $X$ and a fixed payment cap $u>0$, with no deductible or coinsurance. The payment is the lesser of the loss and the cap. The key task is to identify exactly which ground-up losses produce a payment equal to $u$. The same event-based method works whether the severity distribution is continuous, discrete, or mixed.

Definition: For a nonnegative ground-up loss $X$ subject to a fixed cap $u$, the payment per loss is $Y^L=\min(X,u)$. The payment reaches the cap exactly when $X\ge u$, so $$ P(Y^L=u)=P(X\ge u)=1-F_X(u-), $$ where $F_X(u-)=P(X<u)$ is the left limit of the ground-up loss distribution function at $u$. This formula applies to a direct payment cap; it does not by itself describe contracts in which other modifications are applied before the cap.

The left limit matters because the event being excluded is $X<u$, not $X\le u$. A loss exactly equal to the cap receives the full capped payment. In a discrete or mixed model, such a loss can have positive probability and must be included. If $X$ is continuous at $u$, the distinction between $P(X\ge u)$ and $P(X>u)$ does not affect the numerical answer, but the event $Y^L=u$ is still a tail event rather than the event $X=u$.

Deriving the Formula

The result follows directly from the payment rule. If $X<u$, then $Y^L=X<u$. If $X\ge u$, then $\min(X,u)=u$. Thus the events $\{Y^L=u\}$ and $\{X\ge u\}$ are identical. Taking the complement of $\{X\ge u\}$ gives $\{X<u\}$, whose probability is $F_X(u-)$.

\begin{aligned} \{Y^L=u\} &= \{X\ge u\},\\ P(Y^L=u) &= 1-P(X<u)\\ &=1-F_X(u-). \end{aligned}

The same result can be seen as the jump in the payment distribution at the cap. Since $Y^L\le u$, $F_{Y^L}(u)=1$. Also, $Y^L<u$ exactly when $X<u$, so $F_{Y^L}(u-)=F_X(u-)$. The jump is therefore

F_{Y^L}(u)-F_{Y^L}(u-)=1-F_X(u-)=P(Y^L=u).

This jump is the probability mass at the payment value $u$. It need not be a mass that already existed in the ground-up severity. Capping gathers the entire ground-up tail at and above $u$ into one payment amount.

Continuous Ground-Up Losses

For a continuous severity distribution, $P(X=u)=0$. Therefore, $P(X\ge u)=P(X>u)$, and the probability of a payment at the cap is the survival probability at $u$. For an exponential severity with rate $\lambda$, the survival function is $P(X>u)=e^{-\lambda u}$. This use of the survival function is valid for a continuous model; for a discrete or mixed model, check whether the ground-up distribution has an atom at the cap.

Worked Example: An Exponential Loss With a Dollar Cap

A hypothetical equipment policy pays $Y^L=\min(X,u)$, where the ground-up loss $X$ is exponential with rate $\lambda=1/10{,}000$ per dollar and the payment cap is $u=\$15{,}000$. Calculate the probability that the payment equals the cap. Give the probability to four decimal places.

Method. A payment of $\$15{,}000$ occurs for every loss of at least $\$15{,}000$, so calculate the ground-up survival probability at the cap.

The exponential survival function is $P(X> x)=e^{-\lambda x}$. Because the distribution is continuous, there is no probability mass at $X=u$, and $P(X\ge u)=P(X>u)$. The exponent is

\lambda u=\frac{1}{10{,}000}\times 15{,}000=1.5.

Therefore,

P(Y^L=15{,}000)=P(X\ge 15{,}000)=e^{-1.5}=0.2231301601\ldots

Rounded only at the final step to four decimal places, the probability is 0.2231. As a check, the probability of a loss below the cap is $1-e^{-1.5}=0.7768698399\ldots$, and $0.7768698399\ldots+0.2231301601\ldots=1$.

Common mistake avoided: $P(X=15{,}000)=0$ for this continuous model, but that is not the probability of a capped payment. Losses greater than $\$15{,}000$ also produce a payment of $\$15{,}000$.

Discrete Ground-Up Losses

For a discrete severity, calculate the probability of every loss amount at least as large as the cap and add those probabilities. In particular, include any probability at exactly the cap. Using $1-F_X(u-)$ is a reliable way to preserve that boundary case: $F_X(u-)$ contains only losses strictly below $u$.

Worked Example: A Discrete Loss With an Atom at the Cap

A hypothetical small-business policy has a cap of $u=\$5{,}000$. Its ground-up loss distribution is $P(X=\$0)=0.10$, $P(X=\$2{,}000)=0.25$, $P(X=\$5{,}000)=0.30$, and $P(X=\$9{,}000)=0.35$. Find the probability that the payment equals the cap, and compare it with the probability that the ground-up loss is strictly greater than the cap.

Method. Apply the event $Y^L=u$ if and only if $X\ge u$. The loss values meeting this condition are $\$5{,}000$ and $\$9{,}000$. Thus

P(Y^L=5{,}000)=P(X=5{,}000)+P(X=9{,}000)=0.30+0.35=0.65.

The left-limit check gives the same answer. The probability strictly below $\$5{,}000$ is $0.10+0.25=0.35$, so

1-F_X(5{,}000-)=1-0.35=0.65.

The probability that the ground-up loss is strictly greater than the cap is only $0.35$, from the $\$9{,}000$ outcome. The $\$5{,}000$ outcome also produces a payment at the cap, so the requested probability is 0.65. The four stated probabilities sum to $0.10+0.25+0.30+0.35=1.00$, as required.

Common mistake shown: Using $P(X>u)=0.35$ as the answer drops the $0.30$ probability at $X=u$. The correction is to use $P(X\ge u)$, which includes that atom.

Mixed Ground-Up Losses

A mixed severity can have both point masses and a continuous component. The event $X\ge u$ remains the same; the calculation simply combines whichever parts of the mixture can reach or exceed the cap. In particular, an atom exactly at $u$ contributes its full probability, while a continuous component contributes its tail probability above $u$.

Worked Example: Combining an Atom at the Cap and a Continuous Tail

Consider a hypothetical policy with cap $u=\$5{,}000$. The ground-up loss has this mixture distribution: with probability $0.20$, $X=\$0$; with probability $0.30$, $X=\$5{,}000$; and with probability $0.50$, $X=Z$, where $Z$ is exponential with mean $\$5{,}000$. Find the probability of a payment equal to the cap.

Method. Separate the mixture components and calculate the probability that each component produces a loss at least as large as $\$5{,}000$. The zero-loss component contributes zero. The point mass at $\$5{,}000$ contributes $0.30$, because equality reaches the cap. The exponential component has rate $\lambda=1/5{,}000$ per dollar, so its conditional probability of reaching the cap is

P(Z\ge 5{,}000)=e^{-(1/5{,}000)(5{,}000)}=e^{-1}=0.3678794412\ldots.

Weight that tail probability by the mixture probability $0.50$, then add the point mass at the cap:

\begin{aligned} P(Y^L=5{,}000) &=0.20(0)+0.30(1)+0.50(e^{-1})\\ &=0.30+0.1839397206\ldots\\ &=0.4839397206\ldots. \end{aligned}

Rounded only at the final step to four decimal places, the answer is 0.4839. To check using the left limit, the probability below the cap is $0.20+0.50(1-e^{-1})=0.5160602794\ldots$; its complement is $0.4839397206\ldots$. The atom at $\$5{,}000$ is correctly excluded from the probability below the cap.

Common mistake avoided: Applying the exponential tail formula to the entire mixture would ignore the separate point mass at the cap and the separate zero-loss component. Weight each component by its mixture probability.

Calculation Checklist and Exam Trap

Before calculating a limit-payment probability, identify the payment rule and the boundary event it implies. For a direct cap $Y^L=\min(X,u)$, the payment equals $u$ for $X\ge u$, including equality. The ground-up model determines how that tail probability is evaluated; it does not change the event itself.

Identify the cap event.
Write $Y^L=u$ if and only if $X\ge u$. Do not replace the weak inequality by a strict one when the severity may have an atom at $u$.

Choose the distribution calculation.
For a continuous model, use its survival probability at $u$. For a discrete model, sum the probabilities at all loss values at least $u$. For a mixture, calculate the contribution from each component and weight it by that component's probability.

Check the boundary and total probability.
Use $1-F_X(u-)$ as a check. Confirm that the probability below the cap and the probability at or above it add to one.

A frequent exam trap is to confuse a payment at the cap with a ground-up loss exactly equal to the cap. These are different events. A ground-up loss above the cap also produces a payment at the cap, so calculating only $P(X=u)$ is generally insufficient. For a continuous severity, that incorrect calculation is zero even when the cap-payment probability is substantial.

This distinction is useful whenever a question asks how often an insurer pays the full limit: the relevant probability is the ground-up tail at the limit, with the boundary included. It also explains why a capped payment distribution can have an atom at its maximum value without the original loss distribution having any atom at all. This treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Key takeaway: Under the direct cap $Y^L=\min(X,u)$, the probability of a payment equal to the limit is $P(X\ge u)=1-F_X(u-)$. The left limit handles atoms correctly, and the same event identity applies to continuous, discrete, and mixed ground-up severities.

Check Your Understanding

For each hypothetical severity model, find the probability that the capped payment equals the stated cap. Keep the boundary $X=u$ in mind.

A ground-up loss is uniform on $[0,\$12{,}000]$, and the payment cap is $\$9{,}000$. What is the probability of a payment at the cap? (Answer: 0.2500)
A discrete loss is $\$1{,}000$ with probability $0.20$, $\$4{,}000$ with probability $0.15$, $\$6{,}000$ with probability $0.25$, and $\$10{,}000$ with probability $0.40$. The cap is $\$4{,}000$. Find the cap-payment probability, including any loss exactly equal to the cap. (Answer: 0.8000)
A ground-up loss is exponential with mean $\$8{,}000$, and the cap is $\$6{,}000$. Calculate the cap-payment probability to four decimal places. (Answer: 0.4724)
A mixed loss is zero with probability $0.10$, equals $\$7{,}000$ with probability $0.25$, and otherwise follows an exponential distribution with mean $\$7{,}000$. The cap is $\$7{,}000$. Calculate the probability of a payment at the cap to four decimal places. (Answer: 0.4891)
A severity is a mixture: with probability $0.40$, it is uniform on $[0,\$20{,}000]$; with probability $0.60$, it is discrete, equaling $\$8{,}000$ with conditional probability $0.25$ and $\$12{,}000$ with conditional probability $0.75$. For a cap of $\$10{,}000$, combine the two components to find the probability of a payment at the cap. (Answer: 0.6500)

What You'll Learn