Expected Limited Payments from Survival Functions

Why the Survival Function Gives an Expected Limited Payment

A policy limit caps the amount paid for a single loss. If the ground-up loss is $X$ and the payment limit is $u$, with no deductible or coinsurance, the payment is $Y=\min(X,u)$. Calculating its expectation from the possible payment amounts can be direct for a simple distribution, but a survival function often makes the calculation shorter and more transparent.

The key idea is to add up the chances that the payment exceeds each level from zero to the limit. This works for continuous, discrete, and mixed losses. In particular, it does not require a probability density, and it remains valid when the loss distribution has point masses.

Definition: Let $X$ be a nonnegative ground-up loss and define its survival function by $S_X(x)=P(X>x)$. For a finite limit $u\geq 0$, the limited payment is $Y=\min(X,u)$. Its expected value is

E[\min(X,u)]=\int_0^u S_X(x)\,dx.

The integral is with respect to the loss amount. It applies to any nonnegative loss distribution, including discrete and mixed distributions; no density is required.

The convention $S_X(x)=P(X>x)$ matters when there are atoms. For a continuous distribution, $S_X(x)=1-F_X(x)$ and the distinction between $P(X>x)$ and $P(X\geq x)$ has probability zero at each fixed $x$. For a discrete distribution, the two probabilities can differ at a support point. The integral formula uses the strict survival function as defined above.

Deriving the Tail-Integral Formula

The formula follows by viewing a nonnegative payment as the total length of the levels it exceeds. For each possible loss value, the limited payment is between zero and $u$. The indicator inside the following integral is one at levels below that payment and zero above it.

Write the payment as a level integral.
For every nonnegative value of $X$, $Y=\min(X,u)$ satisfies $Y=\int_0^u \mathbf{1}_{\{Y>t\}}\,dt$. The interval of levels $t$ that the payment exceeds has length exactly $Y$.

Take expectations.
The integrand is nonnegative, so the order of expectation and integration can be interchanged. Thus $E[Y]=\int_0^u P(Y>t)\,dt$.

Relate payment exceedance to loss exceedance.
For $0\leq t<u$, $\min(X,u)>t$ holds exactly when $X>t$. Therefore $P(Y>t)=S_X(t)$ throughout the integration interval, giving $E[Y]=\int_0^u S_X(t)\,dt$.

The value at the single endpoint $t=u$ does not affect an integral. The result therefore remains correct if $X$ has an atom at $u$. Also, because $Y\leq u$, its expectation is finite for every finite limit even if $E[X]$ is infinite.

Using the Formula for Continuous Losses

For a continuous loss with density $f_X$, the survival function is $S_X(x)=1-F_X(x)$. Substituting it into the tail integral gives a direct route to the expected payment. The density is useful for deriving the survival function, but the expected limited payment is calculated by integrating the survival function, not the density itself.

Worked Example: An Exponential Loss With a One-Thousand-Dollar Limit

A hypothetical insurer models a claim amount $X$, in dollars, as exponential with mean $1{,}000$. Thus the rate is $\lambda=1/1{,}000=0.001$ per dollar, and $S_X(x)=e^{-0.001x}$ for $x\geq0$. The policy pays $\min(X,1{,}000)$. Find the expected payment and the probability that the payment equals the limit.

Method. Integrate the ground-up survival function from zero to the payment limit. The exponential mean and rate are stated in dollars, so the limit must also be entered in dollars.

The expected payment is

\begin{aligned} E[\min(X,1{,}000)] &=\int_0^{1{,}000} e^{-0.001x}\,dx\\ &=\left[-1{,}000e^{-0.001x}\right]_0^{1{,}000}\\ &=1{,}000(1-e^{-1})\\ &=1{,}000(1-0.3678794)\\ &=632.1206\text{ dollars}. \end{aligned}

Rounded to the nearest cent, the expected payment is $632.12. The payment equals the limit exactly when $X\geq1{,}000$. This exponential distribution is continuous, so $P(X\geq1{,}000)=P(X>1{,}000)$:

P(Y=1{,}000)=P(X\geq1{,}000)=e^{-0.001(1{,}000)}=e^{-1}=0.3679.

The calculation is consistent with the model: about 36.79% of claims reach the cap, while the expected payment is lower than the $1,000 limit. A common error is to report $1,000 times the probability of reaching the limit as the full expected payment. That would count only claims at the cap and omit all payments below it; the survival integral includes both.

Discrete Losses and Stepwise Survival Functions

For a discrete loss, the survival function is a step function. Integrating it means adding the areas of the constant sections. An equivalent direct calculation is to multiply each possible limited payment by its probability. The tail-integral form is especially useful when the survival probabilities are easier to obtain than a full payment distribution.

When $X$ takes nonnegative integer values and $u$ is a nonnegative integer, the integral has a familiar sum form:

E[\min(X,u)]=\sum_{k=0}^{u-1}P(X>k).

On each interval from $k$ to $k+1$, the survival probability is constant and equals $P(X>k)$, which explains the sum. This particular formula requires integer-valued losses and an integer limit. For other discrete supports or a noninteger limit, use the integral or sum over the actual possible loss values.

Worked Example: A Discrete Loss With a Limit Between Possible Values

A hypothetical equipment insurer models a claim $X$, in dollars, as follows: $P(X=0)=0.10$, $P(X=500)=0.20$, $P(X=1{,}000)=0.30$, and $P(X=1{,}500)=0.40$. The payment is capped at $u=900$. Find the expected payment and the probability that the payment equals the cap.

Method. Since the loss is discrete and the cap is not a support point, identify the survival probability on each interval up to $900 and add the rectangular areas.

For $0\leq x<500$, the loss exceeds $x$ unless $X=0$, so $S_X(x)=0.20+0.30+0.40=0.90$. For $500\leq x<900$, only losses of $1,000 and $1,500 exceed $x$, so $S_X(x)=0.30+0.40=0.70$. Therefore,

\begin{aligned} E[\min(X,900)] &=\int_0^{500}0.90\,dx+\int_{500}^{900}0.70\,dx\\ &=(500)(0.90)+(900-500)(0.70)\\ &=450+280\\ &=730\text{ dollars}. \end{aligned}

For a direct check, the possible payments are $0, $500, $900, and $900. Thus $0(0.10)+500(0.20)+900(0.30)+900(0.40)=100+270+360=730$ dollars, matching the integral.

The payment equals $900 when $X\geq900$. In this model that means $X=1{,}000$ or $X=1{,}500$, so the probability is $0.30+0.40=0.70$. The answer is $730.00 expected payment and 0.7000 probability of payment at the limit.

A common mistake is to integrate only to $500 because $500 is the last loss value below the cap. The policy limit is $900, so the survival function remains 0.70 from $500 up to $900, and that additional area contributes $280.

A Second Continuous Model and a Useful Check

A uniform loss provides a simple check on both the integration and the interpretation. The survival function declines linearly, so its area to the limit can be calculated as a rectangle minus a triangle. The probability of reaching the limit provides a separate check on where the cap applies.

Worked Example: Uniform Losses and a Midpoint Limit

A hypothetical insurer models a property loss $X$, in dollars, as uniform on $[0,2{,}400]$. The payment limit is $1,200. Find the expected limited payment and the probability that the payment is capped.

Method. For $0\leq x\leq2{,}400$, the uniform distribution has $F_X(x)=x/2{,}400$, so $S_X(x)=1-x/2{,}400$. Integrate this expression only to $1,200.

\begin{aligned} E[\min(X,1{,}200)] &=\int_0^{1{,}200}\left(1-\frac{x}{2{,}400}\right)\,dx\\ &=\left[x-\frac{x^2}{4{,}800}\right]_0^{1{,}200}\\ &=1{,}200-\frac{1{,}440{,}000}{4{,}800}\\ &=1{,}200-300\\ &=900\text{ dollars}. \end{aligned}

The loss is continuous, and the cap is reached when $X\geq1{,}200$. The interval from $1,200 to $2,400 has length $1,200 out of the total $2,400, giving $P(X\geq1{,}200)=0.5000$. As a check on the expectation, losses below the cap contribute an average payment of $300 to the unconditional mean: $\int_0^{1{,}200}x(1/2{,}400)\,dx=300$. Losses at or above the cap contribute $1{,}200(0.5)=600$. Their sum is $900, as calculated by the survival integral.

The common error here is to use the mean of the ground-up loss, $1,200, as the expected payment. The cap reduces payments for all losses above $1,200, so the expected limited payment is $900, not $1,200.

Atoms, Interpretation, and Exam Checks

The integral method needs no special adjustment for an atom at zero, at an interior loss amount, or at the limit. A jump in the distribution produces a jump in the survival function, and the integral simply uses the appropriate survival level on each side. A jump at one point has no effect on the area of the integral at that exact point.

However, the probability of a payment equal to the limit must be handled with care. In general,

P(Y=u)=P(X\geq u),

not necessarily $S_X(u)=P(X>u)$. If there is positive probability that $X=u$, then that mass also produces a payment equal to $u$. For a continuous model the distinction vanishes, but for a discrete or mixed model it can matter.

The formula also gives useful boundary checks. If $u=0$, the payment and its expectation are zero. For a nonnegative loss and finite $u$, the expectation must satisfy $0\leq E[\min(X,u)]\leq u$. Increasing the limit cannot decrease the expected payment, because the payment for each possible loss cannot decrease. If $E[X]$ is finite, letting the limit grow without bound recovers the ground-up expectation through the full tail integral $E[X]=\int_0^\infty S_X(x)\,dx$.

Key takeaway: For a nonnegative loss with payment $\min(X,u)$, integrate the ground-up strict survival function from zero to the limit. This is a single method for continuous, discrete, and mixed losses. Check the units, the upper integration bound, and whether an atom at the limit affects a separate probability calculation.

Check Your Understanding

For each problem, use the stated loss model and limit. Amounts are in dollars unless specified otherwise.

A hypothetical device claim has an exponential distribution with mean $500 and payment limit $750. Find the expected limited payment and the probability that the payment equals the limit. (Answer: expected payment $388.43; probability 0.2231.)
A discrete loss has probabilities $P(X=0)=0.15$, $P(X=200)=0.25$, $P(X=700)=0.35$, and $P(X=1{,}100)=0.25$. With a $600 limit, find the expected payment and the probability of payment at the limit. (Answer: expected payment $410.00; probability 0.6000.)
A hypothetical loss is uniform on $[0,3{,}000]$, with a $900 payment limit. Calculate the expected limited payment. (Answer: $765.00.)
A loss takes values $0, $400, and $1,000 with respective probabilities 0.20, 0.50, and 0.30. The limit is $700. Calculate the expected payment using survival-function intervals. (Answer: $410.00.)
With probability 0.40 a hypothetical claim is zero; with probability 0.60 it is exponential with mean $1,000. A $500 limit applies. Find the expected payment. (Answer: $236.08.)

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