Expected Payments for Deductible-Limit Coverage

Why Integrate Over a Coverage Interval?

With an ordinary deductible and a per-loss limit, the insurer pays nothing on losses at or below the deductible, pays the amount above the deductible for intermediate losses, and pays no more than the limit for large losses. The expected payment can therefore be calculated directly from the ground-up loss distribution, without first constructing the full payment distribution.

The key is to use the survival function at each loss threshold between the deductible and the point where the policy reaches its limit. Integrating over that interval accumulates the probabilities that the payment exceeds successive amounts.

Definition: Let $X$ be a nonnegative ground-up loss, $d\geq 0$ an ordinary deductible, and $u\geq 0$ the maximum payment per loss. The insurer’s payment is $$ Y=\min\bigl((X-d)^+,u\bigr), \qquad (X-d)^+=\max(X-d,0). $$ The ground-up survival function is $S_X(x)=P(X>x)$. For a finite limit $u$, the expected payment is $$ E[Y]=\int_d^{d+u} S_X(x)\,dx. $$

The upper endpoint is $d+u$, not $u$: the insurer pays the full limit once the ground-up loss reaches the deductible plus the limit. This result applies to continuous, discrete, and mixed loss distributions, provided the stated survival function describes $X$. At isolated thresholds, strict versus non-strict inequalities do not change the integral.

Why the Survival-Function Formula Holds

A payment of amount $Y$ can be viewed as accumulating one dollar of payment for every payment threshold $t$ that it exceeds. More generally, the length of the interval of thresholds $t\in[0,u]$ below $Y$ is exactly $Y$. This gives a direct justification for the formula.

Write the payment as a threshold integral.
For each outcome, $Y=\int_0^u \mathbf{1}\{Y>t\}\,dt$. The indicator is one when the payment exceeds $t$, and zero otherwise.

Translate payment thresholds into loss thresholds.
For $0\leq t<u$, $Y>t$ exactly when $X>d+t$. A loss must exceed the deductible by more than $t$ for the payment to exceed $t$.

Take expectations and change variables.
Integrating the threshold probabilities gives $E[Y]=\int_0^u P(X>d+t)\,dt$. Substituting $x=d+t$ yields $E[Y]=\int_d^{d+u}S_X(x)\,dx$.

The expectation and integral can be interchanged because the indicator is nonnegative. This argument also explains why the result does not require a density: probabilities of exceeding thresholds are defined for discrete and mixed distributions as well as continuous ones.

Equivalent calculation: The same expected payment can be written $$ E[Y]=E\bigl[(X-d)\mathbf{1}\{d<X\leq d+u\}\bigr]+uP(X>d+u). $$ Losses in the intermediate interval contribute their amount above the deductible; losses beyond the interval contribute the full limit. This expression is exact even when $X$ has point masses.

Worked Examples

Worked Example: Uniform Ground-Up Loss

A hypothetical equipment policy covers a ground-up loss $X\sim\operatorname{Uniform}(0,1000)$ dollars, with deductible $d=\$100$ and per-loss limit $u=\$300$. Find the expected insurer payment. Also find the probability that the insurer pays the full limit.

Method. Use the ground-up survival function and integrate over $[d,d+u]$. For a uniform distribution on $[0,1000]$, the density is $1/1000$ on that interval, so for $0\leq x\leq1000$, $S_X(x)=1-x/1000$.

The integration interval ends at $d+u=100+300=400$. Thus

\begin{aligned} E[Y] &=\int_{100}^{400}\left(1-\frac{x}{1000}\right)\,dx\\ &=\left[x-\frac{x^2}{2000}\right]_{100}^{400}\\ &=\left(400-\frac{400^2}{2000}\right) -\left(100-\frac{100^2}{2000}\right)\\ &=(400-80)-(100-5)\\ &=320-95\\ &=\$225.00. \end{aligned}

The full limit is paid when $X\geq d+u=400$. The uniform distribution has no point mass at 400, so $P(Y=300)=P(X\geq400)=P(X>400)=1-400/1000=0.6000$.

Check. Losses from 100 to 400 generate payments from 0 to 300; losses above 400 generate a payment of 300. The integral gives a positive mean below the maximum payment, consistent with this structure. The common mistake avoided is integrating only from 100 to 300, which would confuse the payment limit with the ground-up loss threshold.

Worked Example: Exponential Losses and a Finite Limit

A hypothetical repair-cost model has $X\sim\operatorname{Exponential}(\lambda=0.002)$, where $\lambda$ is measured per dollar, so $S_X(x)=e^{-0.002x}$ for $x\geq0$. A contract has deductible $d=\$100$ and payment limit $u=\$200$. Calculate the expected payment, in dollars to the nearest cent.

Method. Integrate the exponential survival function from the deductible to the deductible-plus-limit. The interval is $[100,300]$, since $100+200=300$.

\begin{aligned} E[Y] &=\int_{100}^{300}e^{-0.002x}\,dx\\ &=\left[-\frac{1}{0.002}e^{-0.002x}\right]_{100}^{300}\\ &=\frac{e^{-0.2}-e^{-0.6}}{0.002}\\ &=\frac{0.8187308-0.5488116}{0.002}\\ &=\frac{0.2699192}{0.002}\\ &=134.9596\ldots\\ &=\$134.96. \end{aligned}

As a check, the probability of exceeding a payment threshold $t$ is $e^{-0.002(100+t)}$ for $0\leq t<200$. Integrating that probability from 0 to 200 gives $\int_0^{200}e^{-0.002(100+t)}dt =e^{-0.2}(1-e^{-0.4})/0.002=\$134.96$ after rounding, the same result.

The common mistake avoided is using the exponential mean, $1/0.002=\$500$, as the expected insurer payment. That is the mean ground-up loss, not the mean payment after the deductible and limit.

Worked Example: A Discrete Loss With a Payment Limit

A hypothetical portfolio has the following ground-up loss distribution: $P(X=0)=0.20$, $P(X=100)=0.15$, $P(X=250)=0.25$, $P(X=500)=0.25$, and $P(X=900)=0.15$, with amounts in dollars. The policy has deductible $d=\$150$ and limit $u=\$300$. Find the expected payment and the probability of a full-limit payment.

Method. First identify the integration interval: it is $[150,450]$. The survival function is $0.65$ from 150 up to 250, because the losses above those thresholds are 250, 500, and 900. It is $0.40$ from 250 up to 450, because only the losses 500 and 900 exceed those thresholds. Values at the individual endpoints do not affect the integral.

\begin{aligned} E[Y] &=\int_{150}^{250}0.65\,dx+\int_{250}^{450}0.40\,dx\\ &=(100)(0.65)+(200)(0.40)\\ &=65+80\\ &=\$145.00. \end{aligned}

A full-limit payment occurs when $X\geq d+u=450$. The possible losses meeting this condition are 500 and 900, so $P(Y=300)=0.25+0.15=0.4000$.

Check by direct payment calculation. The losses 0 and 100 produce payment 0; loss 250 produces $250-150=\$100$; losses 500 and 900 each produce the capped payment of $\$300$. Therefore $E[Y]=(0.25)(100)+(0.25+0.15)(300)=25+120=\$145.00$, agreeing with the integral. The common mistake avoided is treating every loss above the deductible as if it generated the full limit.

Interpreting the Result and Avoiding Traps

The integral adds up tail probabilities over payment thresholds, expressed as ground-up loss thresholds from $d$ to $d+u$. At a threshold $x$ in that interval, $S_X(x)$ is the probability that the ground-up loss is large enough to generate a payment exceeding $x-d$. This interpretation is useful for checking whether the limits and survival function have been aligned correctly.

Do not integrate from $d$ to $u$. The upper endpoint is $d+u$, because the insurer’s payment reaches $u$ when the ground-up loss reaches $d+u$.
Do not use one tail probability for the whole limit. The shortcut $uP(X>d)$ assumes every loss above the deductible produces the full limit. Many such losses may produce only a partial payment.
Use the ground-up survival function. The formula uses $S_X(x)=P(X>x)$, not a survival function for the modified payment.
Do not discard point masses. For discrete losses, calculate the survival function on each interval between loss values. Single boundary points have zero width in the integral, while the jump changes the survival probability on an interval.

As a quick bound, $0\leq Y\leq u$ implies $0\leq E[Y]\leq u$. The expected payment should also be no greater than the expected payment under the same deductible without a limit, whenever that expectation is finite. These checks can expose an incorrect interval or a payment calculation that exceeds the contract maximum.

Key takeaway: For an ordinary deductible $d$ and per-loss limit $u$, calculate the expected payment from the ground-up survival function over the interval $[d,d+u]$: $E[\min((X-d)^+,u)]=\int_d^{d+u}S_X(x)\,dx$. The formula works for continuous, discrete, and mixed losses.

Check Your Understanding

For each problem, identify the correct ground-up interval before calculating.

A hypothetical loss has survival function $S_X(x)=e^{-0.01x}$ for $x\geq0$. With deductible $d=20$ and limit $u=30$, find the expected payment to the nearest cent. (Answer: $\$21.22$)
A hypothetical loss is uniform on $[0,600]$ dollars. With deductible $d=\$120$ and limit $u=\$180$, calculate the expected payment. (Answer: $\$117.00$)
A discrete loss takes values $0,80,200,350$ dollars with respective probabilities $0.30,0.20,0.30,0.20$. For deductible $d=\$100$ and limit $u=\$150$, find the expected payment and the probability of a full-limit payment. (Answer: Expected payment $\$60.00$; full-limit probability $0.2000$)
A loss distribution has survival function $S_X(x)=1$ for $0\leq x<100$, $S_X(x)=0.8$ for $100\leq x<300$, $S_X(x)=0.3$ for $300\leq x<500$, and $S_X(x)=0$ for $x\geq500$. With deductible $d=150$ and limit $u=300$, calculate the expected payment. (Answer: $\$165.00$)

What You'll Learn