Pareto Losses Under Deductibles and Limits

Why Pareto Losses Need a Moment Check

A Pareto model is useful for describing hypothetical losses with a long right tail: large losses are uncommon, but the probability of an extreme loss decreases more slowly than it would under many lighter-tailed models. A deductible removes the smaller losses from the insurer's payments, while a payment limit caps the largest payment. The resulting payment probabilities and expected amount depend on both the coverage terms and the Pareto parameters.

The shape parameter is especially important. Depending on its value, the ground-up loss may have a finite mean, a finite variance, both, or neither. A finite payment limit changes that conclusion for the insurer's payment: a capped payment is bounded, so all of its positive moments are finite.

Definition: In this tutorial, a ground-up loss $X$ follows the Pareto Type II distribution, also called the Lomax distribution, with scale parameter $\theta>0$ and shape parameter $\alpha>0$. Its support is $x\geq 0$, and $$ F_X(x)=1-\left(\frac{\theta}{\theta+x}\right)^\alpha,\qquad S_X(x)=P(X>x)=\left(\frac{\theta}{\theta+x}\right)^\alpha,\qquad x\geq 0. $$ The scale $\theta$ has the same units as the loss; the shape $\alpha$ controls how quickly the tail decreases.

This parameterization matters. A different Pareto convention can put the lower endpoint at a positive value or define the scale differently. Here, losses can be arbitrarily close to zero, and $S_X(0)=1$. All formulas below use this Type II convention.

Deductible-and-Limit Payment

Let $d\geq 0$ be an ordinary deductible and let $u>0$ be the maximum payment per loss, measured after the deductible. The insurer pays nothing when $X\leq d$, pays $X-d$ when $d<X<d+u$, and pays the full amount $u$ when $X\geq d+u$.

Definition: The payment per loss is $$ Y^L=\min\bigl((X-d)_+,u\bigr),\qquad (z)_+=\max(z,0). $$ For a continuous Pareto ground-up loss, $Y^L$ has a probability mass $F_X(d)$ at zero and a probability mass $S_X(d+u)$ at the cap $u$. Between zero and $u$, its payment tail is $$ P(Y^L>y)=S_X(d+y)=\left(\frac{\theta}{\theta+d+y}\right)^\alpha,\qquad 0\leq y<u. $$ For $y\geq u$, $P(Y^L>y)=0$, while $P(Y^L\geq u)=S_X(d+u)$.

The distinction between $P(Y^L>u)$ and $P(Y^L\geq u)$ is important. No payment can be strictly greater than the cap, but losses at or above $d+u$ produce a payment equal to the cap. At the lower boundary, $P(Y^L>0)=S_X(d)$: this is the probability that a loss exceeds the deductible.

Worked Example: Expected Payment and the Cap Probability

A hypothetical policy has a deductible of $\$5{,}000$ and a per-loss payment limit of $\$15{,}000$. Ground-up loss $X$ is Pareto Type II with $\theta=\$10{,}000$ and $\alpha=3$. Find the expected payment and the probability that the insurer pays the full limit.

Method. Use the Pareto payment-tail formula for the cap probability and integrate the survival function over the payment range for the expected payment. Work in thousands of dollars: $\theta=10$, $d=5$, and $u=15$.

For $0\leq y<u$, the payment tail is $P(Y^L>y)=(10/(15+y))^3$. In particular, reaching the cap requires $X\geq d+u=20$, so $$ P(Y^L=15)=S_X(20)=\left(\frac{10}{10+20}\right)^3 =\left(\frac{1}{3}\right)^3=\frac{1}{27}=0.0370. $$

For $\alpha>1$, integrating the payment tail gives $$ E[Y^L]=\frac{\theta^\alpha}{\alpha-1} \left[(\theta+d)^{1-\alpha}-(\theta+d+u)^{1-\alpha}\right]. $$ Substitute the parameters, retaining the thousands-of-dollars unit: $$ E[Y^L]=\frac{10^3}{2}\left(15^{-2}-30^{-2}\right) =500\left(\frac{1}{225}-\frac{1}{900}\right) =500\left(\frac{3}{900}\right) =1.6667. $$ Thus, the expected payment is $1.6667$ thousand dollars, or $\$1{,}666.67$ to the nearest cent.

Check. The probability of paying the cap is $1/27$, which is less than the probability of any positive payment, $S_X(5)=(10/15)^3=8/27$. The expected payment is also below the maximum possible payment of $\$15{,}000$. A common mistake is to use $S_X(u)$ for the cap probability; the loss must exceed the deductible and the payment limit, so the correct threshold is $d+u$.

Expected Payment and Its Derivation

For a nonnegative payment bounded above by $u$, its expected value can be written as the area under its payment-tail probabilities. In this setting, the identity follows directly by expressing each realized payment as a range of payment layers that it exceeds:

Y^L=\int_0^u \mathbf{1}_{\{Y^L>y\}}\,dy \quad\Longrightarrow\quad E[Y^L]=\int_0^u P(Y^L>y)\,dy =\int_0^u S_X(d+y)\,dy.

The indicator $\mathbf{1}_{\{Y^L>y\}}$ equals one for layers below the realized payment and zero for layers above it. Taking expectations converts that indicator to $P(Y^L>y)$. Substituting the Pareto survival function and integrating yields

E[Y^L]= \begin{cases} \displaystyle \frac{\theta^\alpha}{\alpha-1} \left[(\theta+d)^{1-\alpha} -(\theta+d+u)^{1-\alpha}\right], & \alpha\ne 1,\\[1.2em] \displaystyle \theta\ln\!\left(\frac{\theta+d+u}{\theta+d}\right), & \alpha=1. \end{cases}

The logarithmic case is not an exception to the coverage model. It arises because integrating $(\theta+d+y)^{-1}$ gives a logarithm. The formula also shows why a finite limit permits a finite expected payment even when the ground-up mean does not exist: the integration stops at $u$, rather than extending over an unbounded payment range.

Worked Example: Payment-Tail Probabilities and Conditional Probability

For a hypothetical loss portfolio, suppose $X$ is Pareto Type II with $\theta=\$8{,}000$ and $\alpha=2$. A policy has deductible $d=\$4{,}000$ and payment limit $u=\$12{,}000$. Find (a) the probability the payment exceeds $\$8{,}000$, (b) the probability of paying the full limit, and (c) the probability of exceeding $\$8{,}000$ given that the insurer makes a positive payment.

Method. Convert each payment threshold into the corresponding ground-up loss threshold. Use $P(Y^L>y)=S_X(d+y)$ below the cap and $P(Y^L= u)=S_X(d+u)$.

(a) The payment exceeds $\$8{,}000$ when $X>d+\$8{,}000=\$12{,}000$. Therefore $$ P(Y^L>8)=S_X(12) =\left(\frac{8}{8+12}\right)^2 =\frac{64}{400}=0.1600. $$ Here and below, amounts in the survival-function calculations are in thousands of dollars.

(b) The cap is paid when $X\geq d+u=16$, so $$ P(Y^L=12)=S_X(16) =\left(\frac{8}{8+16}\right)^2 =\frac{64}{576}=\frac{1}{9}=0.1111. $$

(c) A positive payment requires $X>d=4$, with probability $$ P(Y^L>0)=S_X(4)=\left(\frac{8}{12}\right)^2=\frac{4}{9}. $$ Since exceeding $\$8{,}000$ implies a positive payment, $$ P(Y^L>8\mid Y^L>0) =\frac{P(Y^L>8)}{P(Y^L>0)} =\frac{0.16}{4/9}=0.36. $$ The requested conditional probability is $0.3600$.

Check. The conditional probability $0.3600$ is greater than the unconditional probability $0.1600$, as expected after conditioning on a positive payment. The cap probability is below the probability of exceeding $\$8{,}000$, because every capped payment also exceeds that threshold. A common error is to use $S_X(8)$ in part (a), forgetting that the ground-up loss must first cover the deductible.

Moment Existence: Ground-Up Loss Versus Payment

For Pareto Type II losses, the ground-up mean exists only when $\alpha>1$, and the ground-up second moment exists only when $\alpha>2$. More generally, for $r>0$,

E[X^r]=\frac{\theta^r\Gamma(r+1)\Gamma(\alpha-r)}{\Gamma(\alpha)} \quad\text{when }r<\alpha; \qquad E[X^r]=\infty\quad\text{when }r\geq\alpha.

In particular, $E[X]=\theta/(\alpha-1)$ for $\alpha>1$, and $E[X^2]=2\theta^2/[(\alpha-1)(\alpha-2)]$ for $\alpha>2$. A deductible alone does not remove the heavy upper tail. For an uncapped payment $Y^L=(X-d)_+$, the $r$th moment is finite exactly when $r<\alpha$: subtracting a fixed deductible changes the payment at moderate losses but not its asymptotic tail. Thus, an uncapped deductible payment has a finite mean only for $\alpha>1$, and a finite variance only for $\alpha>2$.

With a finite limit $u$, however, $0\leq Y^L\leq u$. Consequently $E[(Y^L)^r]\leq u^r$ for every $r>0$, so all positive payment moments exist, whatever the positive value of $\alpha$. This does not make the ground-up loss moments finite; it makes the insurer's capped payment bounded.

Worked Example: A Finite Capped Mean with an Infinite Ground-Up Mean

A hypothetical policy covers Pareto Type II losses with $\theta=\$10{,}000$, $\alpha=1$, deductible $d=\$10{,}000$, and payment limit $u=\$30{,}000$. Find the expected payment and determine whether the ground-up mean and capped-payment variance exist.

Method. Use the $\alpha=1$ logarithmic expected-payment formula. Then apply the moment conditions to the ground-up loss and use boundedness for the capped payment.

Measure amounts in thousands of dollars. The relevant endpoints are $\theta+d=20$ and $\theta+d+u=50$. Therefore $$ E[Y^L]=10\ln\!\left(\frac{50}{20}\right) =10\ln(2.5) =9.162907\text{ thousand dollars}. $$ The expected payment is $\$9{,}162.91$ to the nearest cent.

Because $\alpha=1$ does not satisfy $\alpha>1$, the ground-up mean is infinite. But the capped payment lies between $\$0$ and $\$30{,}000$, so its second moment and variance are finite. This is consistent: the cap limits the insurer's payment even though the underlying loss has an infinite mean.

Check. The expected payment is below the $\$30{,}000$ maximum. The common mistake is to conclude that the expected capped payment must also be infinite because $E[X]=\infty$. The finite cap makes the payment bounded, and therefore its expectation is finite.

Exam Checks for Pareto Coverage

For a deductible-and-limit question, first identify the exact payment definition and units. Then locate the ground-up thresholds corresponding to a payment event: $d+y$ for exceeding a payment amount $y<u$, and $d+u$ for reaching the cap. For an expected payment, use the appropriate formula for $\alpha$, including the logarithmic case at $\alpha=1$. Finally, check whether the question concerns the ground-up loss, an uncapped payment, or a bounded capped payment before deciding whether a moment exists.

Key takeaway: Under Pareto Type II losses, payment tails are evaluated at the deductible-adjusted threshold $d+y$. A finite limit produces a bounded payment with all positive moments finite; without a limit, moment existence continues to depend on the Pareto shape parameter.

Check Your Understanding

Use the Pareto Type II convention defined above. Unless stated otherwise, monetary parameters in each question are in thousands of dollars.

A loss has $\theta=6$ and $\alpha=2$. The policy has $d=3$ and $u=9$. Find $P(Y^L>4)$. (Answer: $36/169=0.2130$)
A loss has $\theta=12$, $\alpha=3$, deductible $d=0$, and payment limit $u=12$. Find the expected payment in dollars. (Answer: $\$4{,}500.00$)
A loss has $\theta=5$, $\alpha=1.5$, and deductible $d=5$, with no payment limit. Determine whether the mean payment exists and, if it does, calculate it in thousands of dollars. (Answer: It exists and equals $5\sqrt{2}\approx 7.0711$ thousand dollars.)
A loss has $\theta=10$, $\alpha=1$, deductible $d=0$, and limit $u=10$. Calculate the expected payment in dollars. (Answer: $\$10{,}000\ln 2=\$6{,}931.47$)
An uncapped deductible payment is based on a Pareto Type II loss with $\theta=4$, $\alpha=1.8$, and $d=2$. State whether its mean and variance are finite. (Answer: The mean is finite; the variance is infinite.)

What You'll Learn