Coinsurance on Deductible-Limit Payments

Introduction

A deductible determines when an insurer begins to pay, and a payment limit caps the amount eligible for payment. Coinsurance then scales that capped payment by the insurer’s share. Because these operations occur in a specified order, changing their order can produce a different distribution and expected payment.

This tutorial considers the payment $Y=c\min((X-d)^+,u)$, where $X$ is the ground-up loss. The focus is on calculating its distribution and expectation, while keeping the deductible, limit, and coinsurance sequence explicit.

Definition: Let $X\geq 0$ be a loss with cumulative distribution function $F_X(x)=P(X\leq x)$ and survival function $S_X(x)=P(X>x)$. For deductible $d\geq 0$, positive payment limit $u>0$, and coinsurance fraction $0<c\leq 1$, define $$ Y=c\min((X-d)^+,u), \qquad (X-d)^+=\max(X-d,0). $$ The deductible is applied first, the resulting payment is capped at $u$, and coinsurance scales that capped amount. Thus $Y$ cannot exceed $cu$.

The limit $u$ in this expression is a cap before coinsurance. Consequently, the maximum insurer payment is $cu$, not $u$. If $c=0$, the payment is zero for every loss; the formulas below assume $c>0$.

Distribution of the Payment

For a payment threshold below the maximum $cu$, the CDF of $Y$ can be calculated by converting that payment threshold into a threshold for $X$. This is useful when a problem asks for a payment probability or requires the distribution across several thresholds.

Theorem: For $0<c\leq 1$, $d\geq0$, and $u>0$, the CDF of $Y$ is $$ F_Y(y)= \begin{cases} 0, & y<0,\\ F_X(d+y/c), & 0\leq y<cu,\\ 1, & y\geq cu. \end{cases} $$

To see why the middle line holds, take $0\leq y<cu$, so that $y/c<u$. Then $$ Y\leq y \quad\Longleftrightarrow\quad \min((X-d)^+,u)\leq y/c \quad\Longleftrightarrow\quad X\leq d+y/c. $$ The last equivalence uses $y/c<u$: a capped amount is at most $y/c$ precisely when the amount above the deductible is at most $y/c$. Taking probabilities gives $F_Y(y)=F_X(d+y/c)$. At or above $cu$, every possible payment is at most $y$, so the CDF is 1. At $y=0$, the middle formula gives $F_Y(0)=F_X(d)$, as expected: a loss at or below the deductible produces no payment.

Expected Payment from the Survival Function

The payment is bounded by $cu$, so its expectation is finite even if the ground-up loss has no finite mean. The survival-function formula integrates the probability that the payment exceeds each threshold from zero to its maximum.

Theorem: Under the stated model, $$ E[Y]=c\int_d^{d+u} S_X(x)\,dx. $$

For justification, set $Z=\min((X-d)^+,u)$, so $Y=cZ$ and $0\leq Z\leq u$. The layer-cake identity for a nonnegative bounded random variable gives $$ E[Z]=\int_0^u P(Z>t)\,dt. $$ For $0\leq t<u$, the event $Z>t$ occurs exactly when $X>d+t$. Therefore $$ E[Z]=\int_0^u P(X>d+t)\,dt =\int_0^u S_X(d+t)\,dt =\int_d^{d+u} S_X(x)\,dx. $$ Multiplying by $c$ proves the formula. The integral only uses the survival function over the layer from the deductible $d$ to the attachment point $d+u$; losses above that point cannot increase the capped payment.

Identify the payment maximum.
Compute $cu$. This is the largest possible value of $Y$.

For a CDF threshold, undo coinsurance.
For $0\leq y<cu$, use the ground-up threshold $d+y/c$, not $d+y$.

For the mean, integrate across the covered layer.
Evaluate $c\int_d^{d+u}S_X(x)\,dx$, splitting the integral wherever the survival function changes form.

Check the endpoints and units.
The payment lies between 0 and $cu$, and its expected value must also lie in that range.

Worked Calculations

Worked Example: A Discrete Loss and a Capped Payment

A hypothetical repair policy has deductible $d=\$5$, payment limit $u=\$20$, and coinsurance fraction $c=0.50$. A claim loss $X$ takes values $\$5,\$15,\$25,\$35$ with probabilities $0.1000, 0.3000, 0.4000, 0.2000$, respectively. Find $E[Y]$ and $F_Y(\$5)$.

Method. Use the survival integral for the mean and the CDF threshold transformation for the requested payment probability. The payment maximum is $$ cu=0.50(\$20)=\$10. $$

The payment at each loss value is $$ \begin{array}{c|c|c} X & \min((X-5)^+,20) & Y=0.50\min((X-5)^+,20)\\ \$5 & \$0 & \$0\\ \$15 & \$10 & \$5\\ \$25 & \$20 & \$10\\ \$35 & \$20 & \$10 \end{array} $$ Thus the payment probabilities are $P(Y=0)=0.1000$, $P(Y=\$5)=0.3000$, and $P(Y=\$10)=0.6000$. In particular, $$ F_Y(\$5)=P(Y\leq \$5)=0.1000+0.3000=0.4000. $$ This also follows from the CDF formula: $d+y/c=5+5/0.50=15$, and $F_X(15)=0.4000$.

For the expected payment, the survival function equals $0.9000$ on $5\leq x<15$ and $0.6000$ on $15\leq x<25$. Therefore $$ E[Y]=0.50\left(\int_5^{15}0.9000\,dx+\int_{15}^{25}0.6000\,dx\right) =0.50\left(10(0.9000)+10(0.6000)\right) =0.50(15)=\$7.50. $$ As a check, the payment distribution gives $E[Y]=0.3000(\$5)+0.6000(\$10)=\$7.50$. The mean is below the maximum payment of $\$10$.

Common mistake avoided. Integrating only from $d$ to $d+u$ gives a pre-coinsurance mean of $\$15$; it is not yet the insurer’s mean payment. Multiplying by $c=0.50$ gives the correct $\$7.50$.

Worked Example: Exponential Losses and a Payment Threshold

For a hypothetical portfolio, let $X$ have an exponential distribution with rate $\lambda=0.1$ per dollar, so $S_X(x)=e^{-0.1x}$ for $x\geq0$. A contract has $d=\$5$, $u=\$15$, and $c=0.80$. Find $E[Y]$ and $F_Y(\$4)$.

Method. Substitute the exponential survival function into the limited-layer expectation formula. The maximum payment is $cu=0.80(\$15)=\$12$, so the threshold $\$4$ lies below the cap.

The expected payment is $$ \begin{aligned} E[Y] &=0.80\int_5^{20}e^{-0.1x}\,dx\\ &=0.80\left[-10e^{-0.1x}\right]_5^{20}\\ &=8(e^{-0.5}-e^{-2})\\ &=8(0.6065307-0.1353353)\\ &=8(0.4711954)=\$3.7696. \end{aligned} $$ Rounded to the nearest cent, $E[Y]=\$3.77$.

For the CDF at $\$4$, convert the payment threshold back to a loss threshold: $$ d+y/c=5+4/0.80=10. $$ Since $F_X(x)=1-e^{-0.1x}$ for $x\geq0$, $$ F_Y(\$4)=F_X(10)=1-e^{-1}=0.6321. $$ The results are plausible: the expected payment is below both the $\$12$ maximum and the $\$4$ CDF threshold, while a probability of $0.6321$ is between 0 and 1.

Common mistake avoided. Using $d+y=\$9$ as the loss threshold forgets to undo coinsurance. It would give $F_X(9)=1-e^{-0.9}=0.5934$, not the correct $F_X(10)=0.6321$. The correct threshold divides $y$ by $c$ before adding the deductible.

Worked Example: Checking the Probability at the Cap

A hypothetical equipment policy has $d=\$5$, $u=\$20$, and $c=1$. Loss $X$ equals $\$4$, $\$30$, or $\$40$ with probabilities $0.3000$, $0.3000$, or $0.4000$, respectively. Find the distribution of $Y$, its expectation, and the probability of a payment at the maximum.

Method. Apply the deductible and cap to each possible loss, then weight the resulting payments by their probabilities. The maximum payment is $cu=\$20$.

If $X=\$4$, it is below the deductible and $Y=\$0$. If $X=\$30$, the amount above the deductible is $\$25$, which is capped at $\$20$; hence $Y=\$20$. If $X=\$40$, the amount above the deductible is $\$35$, also capped at $\$20$; again $Y=\$20$. Therefore $$ P(Y=\$0)=0.3000,\qquad P(Y=\$20)=0.3000+0.4000=0.7000. $$ The expectation and cap probability are $$ E[Y]=0.3000(\$0)+0.3000(\$20)+0.4000(\$20)=\$14.00, \qquad P(Y=cu)=0.7000. $$ A check using the survival formula gives $\int_5^{25}S_X(x)\,dx=20(0.7000)=\$14.00$, since $S_X(x)=0.7000$ throughout that interval.

Common mistake avoided. The weighted sum $0.3000(20)+0.4000(20)+0.3000(0)$ equals $6+8+0=14$, not $16$. Recompute each product and include every outcome before reporting the expectation.

Interpreting the Results and Avoiding Traps

The CDF and expectation formulas answer different questions. The CDF translates one payment threshold into a ground-up loss threshold. The expectation averages payments over the whole covered layer, with the survival function accounting for the probability that a loss reaches each point in that layer. Both calculations depend on the same operation order: first subtract the deductible, then apply the payment cap, and finally multiply by $c$.

Do not cap at $u$ after coinsurance. The expression given here caps the eligible payment at $u$ and then scales it, so the payment maximum is $cu$.
Do not apply the deductible to $cX$. The deductible is applied to $X$; coinsurance scales the amount remaining after the deductible and limit.
Do not omit $c$ from the expectation. The integral gives the expected payment before coinsurance. Multiply the integral by $c$.
Do not integrate beyond the covered layer. Once $x$ reaches $d+u$, a larger loss cannot increase the payment.
Check the full CDF range. Below zero it is 0; at or above $cu$ it is 1. Within that range, use $F_X(d+y/c)$.

Key takeaway: For $Y=c\min((X-d)^+,u)$, use $F_Y(y)=F_X(d+y/c)$ for $0\leq y<cu$, and $E[Y]=c\int_d^{d+u}S_X(x)\,dx$. Keep the operation order and the distinction between the pre-coinsurance limit $u$ and the maximum payment $cu$ explicit.

Check Your Understanding

For each problem, identify the payment threshold or covered layer before calculating. Dollar answers are rounded to the nearest cent, and probabilities are given to four decimal places.

A hypothetical policy has $d=\$4$, $u=\$12$, and $c=0.50$. Loss $X$ equals $\$2,\$10,\$22$ with probabilities $0.2500,0.3500,0.4000$, respectively. Find $E[Y]$ and $F_Y(\$2)$. (Answer: $E[Y]=\$3.45$; $F_Y(\$2)=0.2500$)
Suppose $X$ is exponential with rate $0.125$ per dollar. For $d=\$4$, $u=\$12$, and $c=0.75$, find $E[Y]$ and $F_Y(\$3)$. (Answer: $E[Y]=\$2.83$; $F_Y(\$3)=0.6321$)
A loss $X$ equals $\$0,\$10,\$20,\$35$ with probabilities $0.2000,0.3000,0.2500,0.2500$. For $d=\$5$, $u=\$15$, and $c=0.80$, find $E[Y]$ and $P(Y=cu)$. (Answer: $E[Y]=\$7.20$; $P(Y=cu)=0.5000$)
Let $X$ be uniform on $[0,30]$ dollars. A policy has $d=\$6$, $u=\$12$, and $c=0.50$. Find $E[Y]$ and $F_Y(\$2)$. (Answer: $E[Y]=\$3.60$; $F_Y(\$2)=0.3333$)

What You'll Learn