Survival-Function Derivation of Deductible Payments

Why a Survival-Function Formula Helps

For an ordinary deductible $d$, the insurer pays the part of a covered loss $X$ that exceeds $d$. The payment per loss is $(X-d)_+=\max(X-d,0)$. The earlier material on limited expected value for deductible loss costs gives ways to calculate its expectation. A survival-function representation is especially useful when the tail probability $S_X(t)=P(X>t)$ is easier to describe than the density or probability mass function.

The representation also works when severity is not purely continuous. A severity distribution may include a continuous component, one or more point masses, or both. The key is to interpret $S_X(t)$ as a probability, not as a density: its value at each threshold gives the chance that the loss exceeds that threshold.

Definition: Let $X$ be a nonnegative loss severity with finite expected value, and let $d\geq 0$ be an ordinary deductible. The insurer's payment per loss is $Y^L=(X-d)_+$. The survival function is $S_X(t)=P(X>t)$. The expected payment is $$ E[Y^L]=E[(X-d)_+]=\int_d^\infty S_X(t)\,dt. $$ This formula applies to continuous, discrete, and mixed severity distributions. The integral is finite when the expected payment is finite; finite $E[X]$ is a sufficient condition.

Why the Integral Gives the Expected Payment

Fix one realized loss $X=x$. If $x\leq d$, the insurer pays zero, and there is no interval of thresholds above $d$ and below $x$. If $x>d$, the insurer pays $x-d$, which is exactly the length of the interval from $d$ to $x$. Equivalently, for every nonnegative loss value,

(x-d)_+=\int_d^\infty \mathbf{1}_{\{x>t\}}\,dt.

The indicator is one when the loss exceeds the threshold $t$ and zero otherwise. For a particular loss $x>d$, it is one for thresholds $d\leq t<x$, an interval of length $x-d$. Endpoint choices do not change that length.

Now take expectations. Since the integrand is nonnegative, Tonelli's theorem permits the expectation and integral to be interchanged, including when the distribution has atoms:

\begin{aligned} E[(X-d)_+] &=E\left[\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt\right]\\ &=\int_d^\infty E[\mathbf{1}_{\{X>t\}}]\,dt\\ &=\int_d^\infty P(X>t)\,dt\\ &=\int_d^\infty S_X(t)\,dt. \end{aligned}

The result uses the strict-tail convention $S_X(t)=P(X>t)$. One could instead use $P(X\geq t)$ inside this integral: the two functions can differ at atoms, but only at a set of threshold values with zero length, so their integrals are equal. Using the strict convention consistently makes the connection with the indicator $\mathbf{1}_{\{X>t\}}$ direct.

Important interpretation: The area under the survival curve from $d$ onward is the expected payment per loss. It is not the probability of a claim exceeding $d$, and it does not include claim frequency. At a threshold equal to a point mass, the survival function jumps; the jump itself has no width, but the changed tail probability affects the area over an interval of thresholds.

Applying the Formula to Continuous Severity

For a continuous distribution with density $f_X$, the survival function is $S_X(t)=\int_t^\infty f_X(x)\,dx$. Directly integrating the survival function can be simpler than integrating the payment function against the density, especially if a tail formula is already available. Both calculations must describe the same expected payment per loss.

Worked Example: A Uniform Loss and a High Deductible

Suppose a hypothetical covered loss has severity $X\sim\operatorname{Uniform}(0,10{,}000)$, measured in dollars, and the ordinary deductible is $d=\$2{,}500$. Assume the stated uniform distribution applies to each loss; there is no interest or mortality adjustment. Find the insurer's expected payment per loss using the survival-function integral.

Method: integrate the survival function. For $0\leq t\leq 10{,}000$, the probability that a uniform loss exceeds $t$ is the remaining interval length divided by the full interval length: $$ S_X(t)=\frac{10{,}000-t}{10{,}000}. $$ For $t>10{,}000$, the survival probability is zero. Since $d=2{,}500$, only the portion from $2{,}500$ to $10{,}000$ contributes: $$ \begin{aligned} E[(X-d)_+] &=\int_{2{,}500}^{10{,}000}\frac{10{,}000-t}{10{,}000}\,dt\\ &=\frac{1}{10{,}000} \left[10{,}000t-\frac{t^2}{2}\right]_{2{,}500}^{10{,}000}\\ &=\frac{1}{10{,}000} \left((50{,}000{,}000)-(21{,}875{,}000)\right)\\ &=2{,}812.5. \end{aligned} $$ The expected payment is **\$2,812.50 per loss**.

Check. A direct calculation gives $E[(X-2{,}500)_+]=\frac{1}{10{,}000}\int_{2{,}500}^{10{,}000}(x-2{,}500)\,dx$. This is the area of a triangle with base $7{,}500$ and height $7{,}500$, divided by $10{,}000$: $7{,}500^2/(2\cdot10{,}000)=2{,}812.5$, agreeing with the survival integral. The common mistake to avoid is integrating the survival curve from zero; that would give $E[X]$, not the payment above this deductible.

Atoms and Mixed Severity Distributions

A discrete severity distribution produces a step-shaped survival function. A point mass at $x_0$ contributes its probability to $S_X(t)$ for thresholds below $x_0$, but not for thresholds at or above $x_0$ under the strict-tail convention. Consequently, a payment amount at a point mass contributes the mass times its payment to the expected value, just as it would in a probability-weighted sum.

For a mixed distribution, the survival function includes both the tail from continuous portions and any point-mass probabilities that lie above the threshold. The same area formula remains valid; there is no need to assume a density exists everywhere.

Worked Example: A Point Mass and a Continuous Component

Consider a hypothetical severity model with the following mixture: with probability $0.30$, $X=\$4{,}000$; with probability $0.50$, $X$ is uniform on $[0,\$8{,}000]$; and with probability $0.20$, $X=0$. The components are mutually exclusive and exhaustive. The ordinary deductible is $d=\$3{,}000$. Find the expected insurer payment per loss. There is no interest or mortality adjustment.

Method: integrate the mixed survival function. Between $t=3{,}000$ and $t=4{,}000$, the point mass at $4{,}000$ is still above the threshold. The continuous component also contributes its tail: $$ S_X(t)=0.30+0.50\left(1-\frac{t}{8{,}000}\right), \qquad 3{,}000\leq t<4{,}000. $$ For $4{,}000\leq t\leq 8{,}000$, that point mass is no longer strictly above the threshold, so $$ S_X(t)=0.50\left(1-\frac{t}{8{,}000}\right). $$ Above $8{,}000$, the survival probability is zero. Thus, $$ \begin{aligned} E[(X-3{,}000)_+] &=\int_{3{,}000}^{4{,}000} \left[0.30+0.50\left(1-\frac{t}{8{,}000}\right)\right]dt\\ &\quad+\int_{4{,}000}^{8{,}000} 0.50\left(1-\frac{t}{8{,}000}\right)dt\\ &=300+0.50(562.5)+0.50(1{,}000)\\ &=300+281.25+500\\ &=1{,}081.25. \end{aligned} $$ The expected payment is **\$1,081.25 per loss**.

Check by mixture components. The point-mass component contributes $0.30(\$4{,}000-\$3{,}000)=\$300$. For the uniform component, the survival integral is $\int_{3{,}000}^{8{,}000}(1-t/8{,}000)\,dt=\$1{,}562.50$; its mixture-weighted contribution is $0.50(\$1{,}562.50)=\$781.25$. The zero-loss component contributes nothing. The total is $\$300+\$781.25=\$1{,}081.25$, as above.

The common mistake is to omit the point mass because it has no density. That would leave only the continuous contribution of $\$781.25$. A point mass does not need a density to affect the survival function or the expected payment.

Worked Example: A Discrete Loss at the Deductible

A hypothetical discrete severity $X$, in dollars, takes values $\$0,\$2{,}000,\$5{,}000,\$9{,}000$ with probabilities $0.25, 0.25, 0.30, 0.20$, respectively. The ordinary deductible is $d=\$5{,}000$. Find the expected payment per loss from the survival function. There is no interest or mortality adjustment.

Method: identify the survival intervals. For thresholds $5{,}000\leq t<9{,}000$, only the $\$9{,}000$ loss is strictly greater than $t$, so $S_X(t)=0.20$. For $t\geq9{,}000$, the survival probability is zero. Therefore, $$ \begin{aligned} E[(X-5{,}000)_+] &=\int_{5{,}000}^{9{,}000}0.20\,dt\\ &=0.20(9{,}000-5{,}000)\\ &=800. \end{aligned} $$ The expected payment is **\$800.00 per loss**.

Check. The loss at $\$5{,}000$ pays zero because it equals the deductible; the loss at $\$9{,}000$ pays $\$4{,}000$ with probability $0.20$; and the two smaller losses also pay zero. Thus the expected payment is $0.20(\$4{,}000)=\$800$. The atom at the deductible creates no payment, and including its probability in the tail for thresholds strictly above the deductible would be incorrect.

Common Interpretation and Calculation Traps

The survival integral is compact, but it is easy to misread what it measures. Keep the loss threshold $t$, the deductible $d$, and the probability $S_X(t)$ distinct. The integral begins at the deductible because losses below it do not generate insurer payment. Its integrand is a probability, so integrating over dollar-valued thresholds produces an expected dollar amount.

Using the wrong lower limit. $\int_0^\infty S_X(t)\,dt=E[X]$ for nonnegative $X$, the ground-up expected severity. The insurer's ordinary deductible payment instead uses $\int_d^\infty S_X(t)\,dt$.
Treating the tail as a density. The survival function is $P(X>t)$, not $f_X(t)$. It is integrated over thresholds; it is not multiplied by a loss amount inside this formula.
Dropping atoms. A point mass affects tail probabilities over a range of thresholds and must be included even though it has no density.
Counting a loss equal to the deductible as a payment. If $X=d$, then $(X-d)_+=0$. The value of the survival function at the single endpoint $t=d$ does not create a positive area by itself.
Calling the result an aggregate loss cost. The formula gives expected payment per loss. A claim frequency or exposure measure is needed to move from per-loss severity to an expected cost per exposure.

Key takeaway: For a nonnegative severity $X$ and an ordinary deductible $d\geq0$, insurer payment is the area under the severity survival function to the right of $d$: $E[(X-d)_+]=\int_d^\infty P(X>t)\,dt$. The derivation requires no continuous-density assumption, so it applies to discrete and mixed distributions as well.

Check Your Understanding

For each hypothetical severity model, calculate the expected insurer payment per loss using the survival-function integral. Unless stated otherwise, assume no interest or mortality adjustment.

$X$ is uniform on $[0,\$12{,}000]$ and the ordinary deductible is $\$4{,}000$. Find the expected payment. (Answer: \$2,666.67)
$X$ equals $\$0,\$2{,}500,\$7{,}000$ with probabilities $0.40,0.35,0.25$, respectively. The deductible is $\$3{,}000$. Find the expected payment, noting which outcomes exceed the deductible. (Answer: \$1,000.00)
With probability $0.40$, $X=\$6{,}000$; with probability $0.60$, $X$ is uniform on $[0,\$10{,}000]$. The deductible is $\$4{,}000$. Find the mixture expected payment, accounting for both components. (Answer: \$1,880.00)
A continuous severity has survival function $S_X(t)=(1-t/10{,}000)^2$ for $0\leq t\leq10{,}000$, and zero above $10{,}000$. The deductible is $\$2{,}000$. Calculate the expected payment to the nearest cent. (Answer: \$1,706.67)
With probability $0.20$, a loss is exactly $\$10{,}000$; with probability $0.80$, it is uniform on $[0,\$6{,}000]$. The deductible is $\$5{,}000$. Find the expected payment and explain why the atom at $\$10{,}000$ contributes. (Answer: \$1,066.67)

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