Multi-Step Coverage Modification Problem Solving

From Ground-Up Loss to Insurer Payment

A coverage problem may give a distribution for the loss but ask about a different quantity: the amount the insurer pays. The calculation is not complete until the contract terms have been applied in the correct order. A deductible removes an initial portion of a loss, a payment limit caps what remains, and coinsurance scales the resulting amount.

This tutorial uses a standard hypothetical contract with an ordinary deductible, a limit on the amount after the deductible but before coinsurance, and a coinsurance share. Losses and coverage amounts must be expressed in the same units. We do not apply any claim trend or change in price basis.

Definition: Let $X$ be a nonnegative ground-up loss, $d\geq 0$ an ordinary deductible, $u>0$ a limit on the insurer’s payment before coinsurance, and $c$ the insurer’s share, where $0<c\leq 1$. The insurer payment per loss is $$ Y^L=c\min\bigl((X-d)_+,u\bigr), \qquad (X-d)_+=\max(X-d,0). $$ Thus, $Y^L=0$ when $X\leq d$, increases with the loss between $d$ and $d+u$, and equals $cu$ when $X\geq d+u$.

The order in the formula matters. First subtract the deductible, then apply the limit $u$, and then multiply by $c$. In this convention, the maximum insurer payment is $cu$, not $u$. A contract with a stated maximum payment after coinsurance would require its terms to be translated carefully into this formula or another appropriate payment function.

How the Payment Distribution Is Related to the Loss Distribution

Write $F_X(x)=P(X\leq x)$ and $S_X(x)=P(X>x)$. Since the payment is a transformation of $X$, events involving $Y^L$ can be translated into thresholds for $X$. For any $y$ strictly between zero and the maximum payment $cu$, the payment is at most $y$ exactly when the loss is at most $d+y/c$.

F_{Y^L}(y)= \begin{cases} 0, & y<0,\\ F_X(d+y/c), & 0\leq y<cu,\\ 1, & y\geq cu. \end{cases}

This expression also identifies two important point masses. The payment is zero whenever $X\leq d$, and it reaches its cap whenever $X\geq d+u$. Therefore, including possible atoms at the boundary losses,

P(Y^L=0)=F_X(d), \qquad P(Y^L=cu)=P(X\geq d+u).

For a discrete or mixed loss distribution, these probabilities may be positive. For a continuous loss distribution, a single boundary value has probability zero, but the regions below and above the boundaries still determine the zero and capped payments. When calculating a strict probability such as $P(Y^L>y)$, preserve the strict inequality as you translate it to a loss threshold.

Expected Payment and Its Tail-Integral Justification

For a nonnegative random variable $W$, its expectation can be written as the integral of its survival probabilities: $E[W]=\int_0^\infty P(W>t)\,dt$, when the expectation exists. Here the payment cannot exceed $cu$. For $0\leq t<cu$, the event $Y^L>t$ is equivalent to $X>d+t/c$. Integrating those probabilities gives the expected insurer payment.

Theorem: Under the payment model above, with finite expected payment, $$ E[Y^L]=c\int_0^u S_X(d+t)\,dt. $$ Equivalently, $E[Y^L]=c\int_d^{d+u}S_X(x)\,dx$. The integral uses loss units, so if losses are in thousands of dollars, the resulting expectation is also in thousands of dollars.

To justify the formula, start from the tail-integral expression and substitute $t=cs$, so $dt=c\,ds$. For payment thresholds $0\leq t<cu$, $P(Y^L>t)=P(X>d+t/c)=S_X(d+t/c)$. Hence

E[Y^L]=\int_0^{cu}S_X(d+t/c)\,dt =c\int_0^u S_X(d+s)\,ds.

This derivation works for discrete, continuous, and mixed loss distributions. Values at individual integration endpoints do not change the integral. It can be especially efficient when the survival function is simpler to integrate than a payment distribution is to enumerate.

Write the payment rule.
Identify the deductible, the amount to which the limit applies, and whether the limit is before or after coinsurance.

Translate the requested event.
Use the piecewise payment rule to find the corresponding loss range or threshold. Check whether the question uses $>$, $\geq$, $=$, or a conditional event.

Calculate and check.
Use the loss probabilities or the tail integral, retain the original monetary units, and verify the answer against the possible payment range from zero to $cu$.

Worked Applications

Worked Example: A Uniform Loss with a Deductible, Limit, and Coinsurance

A hypothetical claim has $X\sim\operatorname{Uniform}(0,20)$, measured in thousands of dollars. The contract has $d=5$, $u=10$, and $c=0.5$. Find $P(Y^L>3)$ and $E[Y^L]$. Give the probability to four decimals and the expectation to three decimals in thousands of dollars.

Method. Apply the payment transformation $Y^L=0.5\min((X-5)_+,10)$, then use the loss threshold for the probability and either direct integration or the survival integral for the mean.

For the probability, $Y^L>3$ requires $0.5\min((X-5)_+,10)>3$. Dividing by $0.5$ gives $\min((X-5)_+,10)>6$. Since $6<10$, this is equivalent to $X>11$. The uniform loss has length 20, so

P(Y^L>3)=\frac{20-11}{20}=\frac{9}{20}=0.4500.

For the expectation, payments rise from zero to $0.5(10)=5$ as $X$ ranges from 5 to 15, and remain at 5 for losses from 15 to 20. The density of $X$ is $1/20$. Thus the contribution from the rising-payment region is

\frac{1}{20}\int_5^{15}0.5(x-5)\,dx =\frac{1}{20}\left[0.25(x-5)^2\right]_5^{15} =\frac{25}{20}=1.25.

The capped region contributes $5(5/20)=1.25$. Therefore $E[Y^L]=1.25+1.25=2.500$ thousand dollars. As a check, the payment is never above 5 thousand dollars, and the calculated mean is within that range.

Common mistake avoided: The mean is not obtained by treating the payment as linear over the entire interval from 5 to 20. It is capped once $X\geq15$; including the capped region gives $2.500$, not $1.875$, thousand dollars.

Worked Example: A Discrete Loss and a Conditional Payment Probability

For a hypothetical portfolio claim, $X$, in thousands of dollars, takes values $1,4,6,10$ with probabilities $0.20,0.30,0.30,0.20$, respectively. The contract has $d=3$, $u=5$, and $c=0.5$. Find $P(Y^L\geq1.5)$, $P(Y^L=2.5\mid Y^L>0)$, and $E[Y^L]$. Report probabilities to four decimals and the expectation in thousands of dollars.

Method. Calculate the payment at every possible loss first. This avoids assigning probability to payment amounts that the discrete loss model cannot produce.

\begin{array}{c|c|c} X & \min((X-3)_+,5) & Y^L=0.5\min((X-3)_+,5)\\ 1 & 0 & 0\\ 4 & 1 & 0.5\\ 6 & 3 & 1.5\\ 10 & 5 & 2.5 \end{array}

The event $Y^L\geq1.5$ occurs at losses 6 and 10, with probabilities 0.30 and 0.20. Therefore $P(Y^L\geq1.5)=0.30+0.20=0.5000$.

A positive payment occurs at losses 4, 6, and 10, so $P(Y^L>0)=0.30+0.30+0.20=0.80$. A payment of 2.5 occurs only at loss 10. Consequently,

P(Y^L=2.5\mid Y^L>0)=\frac{0.20}{0.80}=0.2500.

The expected payment is the probability-weighted sum of the payment values:

$$ E[Y^L]=0.20(0)+0.30(0.5)+0.30(1.5)+0.20(2.5) =0+0.15+0.45+0.50=1.10. $$

Thus the mean is $1.10$ thousand dollars. As a check, the largest possible payment is $cu=0.5(5)=2.5$, and the mean is below that cap.

Common mistake avoided: The conditional denominator is the probability of any positive payment, $0.80$, not the probability of a payment at least 1.5. The numerator is $0.20$, so the conditional probability is $0.2500$.

Worked Example: Exponential Losses and the Survival-Integral Mean

A hypothetical claim severity $X$, measured in thousands of dollars, has an exponential distribution with mean 8 thousand dollars. Thus its rate is $\lambda=1/8$ per thousand dollars and $S_X(x)=e^{-x/8}$ for $x\geq0$. A contract has $d=2$, $u=4$, and $c=0.75$. Find the probabilities of zero and maximum payment, $P(Y^L>1.5)$, and $E[Y^L]$. Give probabilities and the expectation in thousands of dollars to four decimal places.

Method. Translate each payment event into a loss threshold. Use the survival integral for the mean, retaining the loss unit of thousands of dollars.

The maximum payment is $cu=0.75(4)=3$ thousand dollars. A zero payment occurs for $X\leq2$, so

P(Y^L=0)=1-S_X(2)=1-e^{-2/8}=1-e^{-0.25}=0.2212.

The payment reaches its maximum when $X\geq d+u=6$. Therefore $P(Y^L=3)=S_X(6)=e^{-6/8}=e^{-0.75}=0.4724$.

For the requested threshold, $0.75\min((X-2)_+,4)>1.5$ is equivalent to $\min((X-2)_+,4)>2$, which is equivalent to $X>4$, because 2 is below the cap threshold 4. Hence $P(Y^L>1.5)=e^{-4/8}=e^{-0.5}=0.6065$.

Finally, apply the survival-integral formula:

\begin{aligned} E[Y^L] &=0.75\int_0^4 e^{-(2+t)/8}\,dt\\ &=0.75\left[-8e^{-(2+t)/8}\right]_0^4\\ &=6(e^{-0.25}-e^{-0.75})\\ &=6(0.7788008-0.4723666)\\ &=1.8386\text{ thousand dollars}. \end{aligned}

As a boundary check, the zero-payment probability is approximately 0.2212, the probability of a strictly partial payment is $e^{-0.25}-e^{-0.75}\approx0.3064$, and the capped-payment probability is 0.4724. These add to 1.0000, up to rounding. The mean is also between zero and the 3-thousand-dollar payment cap.

Common mistake avoided: The event $Y^L>1.5$ does not mean $X>1.5$. The deductible and the coinsurance share both affect the corresponding ground-up loss threshold.

Boundary and Unit Checks

Integrated problems often combine a familiar distribution with coverage terms whose boundaries are easy to misread. Before finalizing a result, check the payment at losses below, at, and above the deductible and the cap threshold. In this model, the payment is zero for $X\leq d$, equals $c(X-d)$ for $d<X<d+u$, and is $cu$ for $X\geq d+u$. This check catches a limit mistakenly applied to the ground-up loss or coinsurance applied in the wrong place.

Check the maximum. Under this convention, no payment can exceed $cu$.
Check the event boundary. For a continuous loss, changing a strict threshold to a non-strict one at a single point does not change its probability. For a discrete or mixed loss, it may.
Check conditional denominators. For a conditional payment probability, the denominator must represent exactly the conditioning event, such as $Y^L>0$.
Check the units. If $X,d,$ and $u$ are in thousands of dollars, the expectation calculated from the payment function is also in thousands of dollars.
Check the mean. Since $0\leq Y^L\leq cu$, its expectation must also lie between zero and $cu$.

These steps make coverage modification calculations more reliable: define the insurer payment first, translate the requested event second, and only then use the loss distribution. The approach applies whether the loss model is discrete, continuous, or mixed.

Check Your Understanding

For each hypothetical contract, determine the requested quantities using the payment rule in this tutorial. Monetary values are in thousands of dollars unless stated otherwise.

$X\sim\operatorname{Uniform}(0,12)$, $d=2$, $u=6$, and $c=0.5$. Find $P(Y^L=0)$ and $P(Y^L=cu)$. (Answer: $0.1667$; $0.3333$)
A discrete loss takes values $2,5,9$ with probabilities $0.25,0.50,0.25$. If $d=3$, $u=4$, and $c=0.5$, find $E[Y^L]$. (Answer: $1.00$ thousand dollars)
$X$ is exponential with mean 10, and the contract has $d=2$, $u=3$, and $c=0.8$. Find $P(Y^L>1.6)$. (Answer: $e^{-0.4}=0.6703$)
$X\sim\operatorname{Uniform}(0,30)$, $d=6$, $u=8$, and $c=0.5$. Find $E[Y^L]$ and the probability of maximum payment. (Answer: $2.6667$ thousand dollars; $0.5333$)
A discrete loss takes values $1,4,8,12$ with probabilities $0.10,0.20,0.40,0.30$. If $d=4$, $u=5$, and $c=0.4$, find $P(Y^L=2\mid Y^L>0)$. (Answer: $0.4286$)

What You'll Learn