Derive Payment Moments Under a Deductible and Limit

Why Payment Moments Need Careful Contract Translation

A deductible removes the first portion of a covered loss, while a finite policy limit caps what the insurer pays. The expected insurer payment therefore depends on severity in a particular interval: losses must exceed the deductible to produce a payment, but the insurer payment cannot exceed the limit. Survival-function integrals describe this interval directly and give both the expected payment and its second moment.

The distinction between a limit on insurer payment and a cap on ground-up loss is important. Throughout this tutorial, $u$ is the maximum insurer payment after the deductible. The formulas would need to be adjusted if a contract instead capped the loss before applying the deductible.

Definition: Let $X$ be a nonnegative ground-up loss, $d\geq 0$ an ordinary deductible, and $u>0$ a maximum insurer payment. The payment per loss is $$ Y^L=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ Equivalently, $Y^L=0$ when $X\leq d$, $Y^L=X-d$ when $d<X<d+u$, and $Y^L=u$ when $X\geq d+u$.

The largest possible payment is $u$, so every valid calculation must satisfy $0\leq E[Y^L]\leq u$ and $0\leq E[(Y^L)^2]\leq u^2$. These bounds are useful checks, but they do not replace careful interpretation of the contract.

Survival-Function Formulas

Write $S_X(x)=P(X>x)$ for the survival function of the ground-up loss. For $0\leq y<u$, a payment greater than $y$ occurs exactly when the ground-up loss exceeds the deductible plus $y$:

P(Y^L>y)=P(X>d+y)=S_X(d+y), \qquad 0\leq y<u.

For a nonnegative random variable $Z$, the tail-integral identity gives $E[Z]=\int_0^\infty P(Z>y)\,dy$. Applying it to the bounded payment $Y^L$, and then using the corresponding identity for the second moment, gives the main results.

Theorem: For a nonnegative severity $X$, ordinary deductible $d$, and finite maximum payment $u$, $$ E[Y^L]=\int_0^u S_X(d+y)\,dy =\int_d^{d+u} S_X(x)\,dx, $$ and $$ E[(Y^L)^2]=2\int_0^u yS_X(d+y)\,dy =2\int_d^{d+u}(x-d)S_X(x)\,dx. $$ These tail-integral formulas apply whether or not the severity distribution is continuous. The second moment is not the variance; when needed, $\operatorname{Var}(Y^L)=E[(Y^L)^2]-\{E[Y^L]\}^2$.

The first expression for each moment uses the payment scale $y$; the second uses the ground-up loss scale $x=d+y$. The second-moment integrand includes the factor $y$, or equivalently $x-d$. It is not correct to integrate $xS_X(x)$ for the payment second moment, because the insurer pays the amount above the deductible, not the full ground-up loss.

Why the Tail Integrals Hold

For any value $z$ between zero and $u$, its length can be written as the area under an indicator:

z=\int_0^u \mathbf{1}(z>y)\,dy, \qquad z^2=\int_0^u 2y\mathbf{1}(z>y)\,dy.

Substitute $z=Y^L$ and take expectations. Interchanging expectation and integration is valid here because the integrands are nonnegative and bounded over the finite interval. Thus $E[Y^L]=\int_0^u P(Y^L>y)\,dy$, and $E[(Y^L)^2]=2\int_0^u yP(Y^L>y)\,dy$. For $y<u$, the contract definition implies $Y^L>y$ exactly when $X>d+y$. Substituting this probability proves both formulas. Changing variables from $y$ to $x=d+y$ gives their alternative forms.

This argument also explains why the finite limit sets the integration range: payments cannot exceed $u$. With no finite payment limit, the upper endpoint becomes infinity, and the corresponding integrals require the relevant moments to be finite.

Worked Distributional Calculations

Worked Example: Exponential Severity with a Fixed Payment Limit

A hypothetical insurer models a single ground-up claim by an exponential distribution with rate $\lambda=0.01$ per dollar, so $S_X(x)=e^{-0.01x}$ for $x\geq0$. The contract has an ordinary deductible of $d=\$20$ and a maximum insurer payment of $u=\$50$. Calculate the expected payment and its second moment.

Method. Use the survival-function formulas over payment amounts from 0 to 50. The exponential parameter is a rate per dollar, and the payment limit applies after the deductible.

For the expected payment,

\begin{aligned} E[Y^L] &=\int_0^{50}e^{-0.01(20+y)}\,dy\\ &=e^{-0.2}\int_0^{50}e^{-0.01y}\,dy\\ &=e^{-0.2}\frac{1-e^{-0.5}}{0.01}\\ &=32.2145\text{ dollars}. \end{aligned}

For the second moment, integrate $2y$ times the payment survival probability:

\begin{aligned} E[(Y^L)^2] &=2e^{-0.2}\int_0^{50}y e^{-0.01y}\,dy\\ &=2e^{-0.2}\frac{1-e^{-0.5}(1+0.5)}{(0.01)^2}\\ &=1{,}477.0559\text{ dollars}^2. \end{aligned}

The integral identity used in the second line is $\int_0^u ye^{-\lambda y}\,dy=[1-e^{-\lambda u}(1+\lambda u)]/\lambda^2$, obtained by integration by parts. As a check, the mean is between 0 and the \$50 maximum payment, and the second moment is below $50^2=2{,}500$. The results are rounded to four decimal places. The calculation avoids treating the payment as the full claim amount above zero.

Worked Example: Uniform Severity with a Deductible and Limit

For a hypothetical equipment policy, suppose $X$ is uniform on $[0,100]$ dollars. The ordinary deductible is $d=\$20$, and the maximum insurer payment is $u=\$30$. Find the expected payment, second moment, and variance.

For $0\leq y\leq30$, the survival function at the deductible-plus-payment threshold is $S_X(20+y)=(100-(20+y))/100=(80-y)/100$. Therefore,

\begin{aligned} E[Y^L] &=\int_0^{30}\frac{80-y}{100}\,dy\\ &=\frac{1}{100}\left[80y-\frac{y^2}{2}\right]_0^{30}\\ &=\frac{2{,}400-450}{100} =19.50\text{ dollars}. \end{aligned}

For the second moment,

\begin{aligned} E[(Y^L)^2] &=2\int_0^{30}y\frac{80-y}{100}\,dy\\ &=\frac{2}{100}\left[40y^2-\frac{y^3}{3}\right]_0^{30}\\ &=\frac{2}{100}(36{,}000-9{,}000) =540.00\text{ dollars}^2. \end{aligned}

The variance is $540.00-(19.50)^2=540.00-380.25=159.75$ dollars squared. Both moment bounds hold: $19.50\leq30$ and $540.00\leq30^2=900$. The values are exact to the displayed cent for the mean and variance. The method accounts only for losses in the interval relevant to insurer payments; losses above the attachment point do not contribute their full ground-up amount.

Worked Example: Pareto-Type Severity with a Higher Deductible

A hypothetical portfolio uses a Lomax, or Pareto Type II, severity model with survival function $S_X(x)=(1+x/100)^{-3}$ for $x\geq0$, where amounts are in dollars. A policy has $d=\$100$ and maximum payment $u=\$100$. Calculate the two payment moments.

Use the loss-scale integral for the mean and set $z=1+x/100$. The integration endpoints $x=100$ and $x=200$ become $z=2$ and $z=3$, with $dx=100\,dz$:

\begin{aligned} E[Y^L] &=\int_{100}^{200}\left(1+\frac{x}{100}\right)^{-3}\,dx\\ &=100\int_2^3z^{-3}\,dz\\ &=50\left(\frac14-\frac19\right) =\frac{250}{36} =6.9444\text{ dollars}. \end{aligned}

For the second moment, the payment above the deductible is $x-100$. Under the same substitution, $x-100=100(z-2)$:

\begin{aligned} E[(Y^L)^2] &=2\int_{100}^{200}(x-100)\left(1+\frac{x}{100}\right)^{-3}\,dx\\ &=20{,}000\int_2^3(z-2)z^{-3}\,dz\\ &=20{,}000\left[-\frac1z+\frac1{z^2}\right]_2^3\\ &=20{,}000\left(\frac1{36}\right) =555.5556\text{ dollars}^2. \end{aligned}

The payment is capped at \$100, so its second moment cannot exceed \$10,000; the calculated value satisfies that bound. The results are rounded to four decimal places. This example illustrates that a high deductible can produce a modest mean even when the severity model has a long right tail: only the survival probabilities across the covered payment interval enter the calculation.

Common Mistake: Applying the Limit to the Wrong Amount

The most frequent contract-translation error is to apply the limit to ground-up loss and then subtract the deductible. For the convention used here, the correct calculation is $Y^L=\min((X-d)_+,u)$: first determine the amount above the deductible, then cap the insurer payment.

For example, let a ground-up loss be $X=\$90$, with $d=\$20$ and maximum payment $u=\$50$. The incorrect calculation is $\min(90,50)-20=30$ dollars. It applies the \$50 cap to the ground-up loss, rather than to the insurer payment. The correct payment is $\min(90-20,50)=\min(70,50)=\$50$.

A different contract might explicitly state that its limit caps the ground-up loss before the deductible. That is a different contract definition and can produce a different payment. Read the wording before choosing a formula; the symbol $u$ alone does not determine where a limit applies.

Key takeaway: When $u$ is the maximum insurer payment after an ordinary deductible, integrate $S_X(d+y)$ from $y=0$ to $u$ for the mean and $2yS_X(d+y)$ over the same interval for the second moment. Check the contract interpretation and verify the results against the payment bounds.

These moment calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. They are useful whenever a model requires the mean or second moment of a covered claim payment rather than of the underlying ground-up severity.

Check Your Understanding

For each problem, treat the stated limit as the maximum insurer payment after the deductible. Use the survival-integral formulas and check the result against the maximum possible payment.

A hypothetical loss has an exponential distribution with rate $0.02$ per dollar. The deductible is \$10 and the maximum payment is \$20. Find the expected payment, to four decimal places. (Answer: \$13.4960)
A hypothetical severity is uniform on $[0,80]$ dollars. With $d=\$15$ and $u=\$25$, find the expected payment and second moment. (Answer: \$16.4063 and \$377.6042 dollars squared)
A hypothetical severity has survival function $S_X(x)=(1+x/50)^{-2}$, $x\geq0$, with amounts in dollars. If $d=\$50$ and $u=\$50$, find both payment moments to four decimal places. (Answer: \$8.3333 and \$360.6592 dollars squared)
A hypothetical exponential severity has rate $0.01$ per dollar, with $d=0$ and $u=\$100$. Find the expected payment and second moment to four decimal places. (Answer: \$63.2121 and \$5,284.8224 dollars squared)
A hypothetical loss is uniform on $[0,100]$ dollars. Its deductible is \$20, and its stated \$60 limit is explicitly a maximum insurer payment after the deductible. Find the expected payment and second moment. (Answer: \$30.00 and \$1,440.00 dollars squared)

What You'll Learn