Deriving Expected Payments From the Severity Survival Function

Why Integrate the Survival Function?

A payment rule can be described as a function of the loss, but computing its expectation directly may require first finding the distribution of the payment. For many insurance and reinsurance contracts, there is a more direct route: express expected payment as an integral of the original loss distribution’s tail probabilities. This is especially useful when the survival function is simpler than the density or when the payment rule has a deductible, a cap, or an attachment point.

The key is to translate the contract wording into a payment map before integrating. A deductible determines where payments begin; a maximum payment determines where they stop increasing. Coinsurance scales the payment, but its position in the payment map matters. In particular, a cap on the insurer’s payment is not generally the same as a cap on the covered loss before coinsurance.

Definition: Let $X$ be a nonnegative loss, with distribution function $F_X(x)=P(X\leq x)$ and survival function $S_X(x)=P(X>x)=1-F_X(x)$. For a continuous loss, $S_X(x)$ is the probability that a loss exceeds $x$. The expected payment under a contract with payment map $g$ is $E[g(X)]$, when that expectation exists.

The identity $S_X(x)=P(X>x)$ is useful even when the loss distribution has point masses. At an individual endpoint, using $P(X>x)$ rather than $P(X\geq x)$ can change the tail probability. But changing the integrand at isolated points does not change an ordinary integral with respect to $x$. The survival-integral formulas below therefore also apply to nonnegative losses with atoms.

The Tail-Integral Identity

For any nonnegative random variable $Z$, its expectation can be obtained by adding up the probabilities that it exceeds each possible payment amount. To see why, fix an outcome where $Z=z$. The indicator $1_{\{Z>t\}}$ equals 1 for $0\leq t<z$ and 0 for $t\geq z$. Its integral over $t\geq 0$ is therefore $z$. Taking expectations gives the result:

Z=\int_0^\infty 1_{\{Z>t\}}\,dt \quad\Longrightarrow\quad E[Z]=\int_0^\infty P(Z>t)\,dt.

The exchange of expectation and integration is valid for this nonnegative integrand, even if the expectation is infinite. Applying the identity to a limited loss payment gives a particularly useful result. For $u\geq 0$, the payment $\min(X,u)$ exceeds $t$ exactly when $X>t$, for $0\leq t<u$. It cannot exceed $u$. Hence:

Theorem: For a nonnegative loss $X$ and a limit $u\geq 0$,

E[\min(X,u)]=\int_0^u S_X(t)\,dt.

This identity says that the expected limited payment is the area under the loss survival curve up to the payment limit. It does not require a density. If $X$ is continuous with density $f_X$, one could instead integrate $\min(x,u)f_X(x)$, but the tail integral often avoids splitting that calculation into several regions.

Deductibles, Payment Limits, and Coinsurance

A deductible $d\geq 0$ removes the first $d$ dollars of each loss. With no payment limit, the payment is $(X-d)^+=\max(X-d,0)$. Its tail-integral form is:

E[(X-d)^+]=\int_d^\infty S_X(x)\,dx.

To justify the lower endpoint, the payment exceeds an amount $t\geq 0$ precisely when $X>d+t$. Integrating this probability over $t\geq 0$, then changing variables to $x=d+t$, gives the displayed integral. If the contract also caps the insurer’s payment at $u$, the payment map becomes $\min((X-d)^+,u)$, so only tail probabilities from $d$ through $d+u$ contribute:

E[\min((X-d)^+,u)]=\int_d^{d+u}S_X(x)\,dx.

For coinsurance at rate $c$, where $0\leq c\leq 1$, the expected payment is multiplied by $c$ if the entire payment map is $c$ times the underlying covered layer. For example, $c\min((X-d)^+,u)$ has expectation $c\int_d^{d+u}S_X(x)\,dx$. By contrast, a maximum insurer payment $M$ applied after coinsurance gives $\min(c(X-d)^+,M)$. If $c>0$, that cap is reached when $X=d+M/c$, not when $X=d+M$.

An excess layer between attachment point $a$ and exhaustion point $b$, where $0\leq a<b$, pays $\min((X-a)^+,b-a)$. Its expected payment is the area under the survival function across that layer:

E[\min((X-a)^+,b-a)]=\int_a^b S_X(x)\,dx.

If a reinsurer pays a proportion $c$ of that layer, multiply this expected layer payment by $c$. The layer width is $b-a$, measured in dollars of loss; it is not automatically a maximum payment after coinsurance.

Worked Applications

Worked Example: Uniform Loss With a Deductible and Payment Limit

A hypothetical claim amount $X$ is uniform on $[0,1200]$ dollars. The insurer pays the amount above a $300 deductible, subject to a maximum insurer payment of $600 per claim. Find the expected payment.

Method. The payment map is $Y=\min((X-300)^+,600)$. Since the loss is uniform on $[0,1200]$, its survival function on that interval is $S_X(x)=1-x/1200$. The maximum payment is reached at loss $300+600=900$, so the tail integral runs from 300 to 900:

E[Y]=\int_{300}^{900}\left(1-\frac{x}{1200}\right)dx.

An antiderivative is $x-x^2/2400$. At the upper endpoint it is $900-900^2/2400=900-337.5=562.5$. At the lower endpoint it is $300-300^2/2400=300-37.5=262.5$. Thus:

E[Y]=562.5-262.5=\$300.00.

Check. The payment is never greater than $600, and losses above $900 all produce that maximum payment. The expectation of $300 is within the allowable payment range. A common mistake is to use 300 as the lower antiderivative value; the correct square term is $300^2/2400=37.5$, not 25. This example uses the full $600 payment layer.

Worked Example: Coinsurance With a Maximum Insurer Payment

A hypothetical loss $X$ has an exponential distribution with mean $800, so $S_X(x)=e^{-x/800}$ for $x\geq 0$. The insurer pays 75% of the amount above a $200 deductible, but its payment is capped at $450. Find the expected insurer payment, to the nearest cent.

Method. The map is $Y=\min(0.75(X-200)^+,450)$. The insurer reaches the $450 cap when $0.75(X-200)=450$, or $X=200+450/0.75=800$. The correct integral therefore uses the loss interval from 200 to 800 and includes the 75% coinsurance factor:

E[Y]=0.75\int_{200}^{800}e^{-x/800}\,dx =0.75(800)\left(e^{-200/800}-e^{-800/800}\right).

The exponential values are $e^{-0.25}=0.7788008$ and $e^{-1}=0.3678794$. Their difference is $0.4109214$. Multiplying by 600 gives $246.5528$, so the expected payment is $246.55, rounded to two decimal places.

Check. The payment cannot exceed $450, and the calculated expectation is below that cap. A common error is to treat the contract as 75% of a $450 loss layer, which would make the maximum payment only $337.50. Here the stated cap is instead $450 of insurer payment, so its exhaustion point must be found after applying coinsurance.

Worked Example: Expected Recovery on an Excess Layer

A hypothetical reinsurer pays 80% of the loss layer from $300 to $900. The underlying loss $X$ is exponential with mean $500, so $S_X(x)=e^{-x/500}$. Find the expected recovery per loss, to the nearest cent.

Method. The full layer payment is $\min((X-300)^+,600)$; the reinsurer receives 80% of it. Integrate the survival function across the attachment and exhaustion points, then apply the stated share:

E[\text{recovery}] =0.80\int_{300}^{900}e^{-x/500}\,dx =0.80(500)\left(e^{-300/500}-e^{-900/500}\right).

Here $e^{-0.6}=0.5488116$ and $e^{-1.8}=0.1652989$. Their difference is $0.3835127$. The expected payment for the full layer is $500(0.3835127)=191.7564$ dollars. The reinsurer’s 80% share is $0.80(191.7564)=153.4051$, or $153.41 to the nearest cent.

Check. The full layer is at most $600, so the reinsurer’s payment is at most $0.80(600)=\$480$. The expected recovery is below that maximum. The common mistake avoided here is integrating from zero: losses below the $300 attachment generate no layer payment.

Choosing the Correct Integral

The integration limits come from the payment map, not merely from the loss distribution. A deductible shifts the starting point of the integral. A cap on the covered loss layer sets the ending point at the deductible plus the layer width. A cap on the insurer’s payment must first be converted to the corresponding gross-loss threshold if coinsurance applies.

Payment rule	Expected payment	Where the integral stops
$(X-d)^+$, deductible only	$\int_d^\infty S_X(x)\,dx$	No finite stopping point unless the loss is bounded
$\min((X-d)^+,u)$	$\int_d^{d+u}S_X(x)\,dx$	At $d+u$
$c\min((X-a)^+,b-a)$	$c\int_a^b S_X(x)\,dx$	At the layer exhaustion point $b$

These expressions calculate expected payment for one loss. They do not, by themselves, calculate an expected total over a portfolio; that would also require a model for the number of losses and assumptions about how losses behave. Keeping the severity calculation separate makes it easier to combine it correctly with frequency information when needed.

Key takeaway: Write the payment map first. For a nonnegative loss, integrate $S_X(x)$ over exactly the loss amounts that can increase the payment, and apply any coinsurance factor to the payment rule it actually scales.

Check Your Understanding

For each problem, identify the applicable layer or payment cap before using the survival integral. Round dollar answers to the nearest cent.

A hypothetical loss is uniform on $[0,1000]$ dollars. The insurer pays all loss above a $250 deductible with no limit. Find the expected payment. (Answer: $281.25)
A hypothetical loss is exponential with mean $400. An insurer pays the loss in full up to a maximum insurer payment of $600. Find the expected payment. (Answer: $310.75)
A hypothetical loss is uniform on $[0,1500]$. The insurer pays 60% of the layer from $200 to $900. Find the expected payment. (Answer: $266.00)
A hypothetical loss is exponential with mean $1000. The insurer pays the amount above a $300 deductible, subject to a $500 maximum insurer payment. Find the expected payment. (Answer: $291.49)
A hypothetical loss is exponential with mean $600. A reinsurer pays 80% of the layer from $400 to $1000. Find the expected recovery per loss. (Answer: $155.78)

Payment rule	Expected payment	Where the integral stops
\((X-d)^+\), deductible only	\(\int_d^\infty S_X(x)\,dx\)	No finite stopping point unless the loss is bounded
\(\min((X-d)^+,u)\)	\(\int_d^{d+u}S_X(x)\,dx\)	At \(d+u\)
\(c\min((X-a)^+,b-a)\)	\(c\int_a^b S_X(x)\,dx\)	At the layer exhaustion point \(b\)

What You'll Learn