Coinsurance After a Deductible

Introduction

A deductible and coinsurance can both leave part of a claim with the policyholder, but they apply in sequence. With coinsurance after an ordinary deductible, the policyholder first bears losses up to the deductible. Above it, the insurer pays a specified share of the excess and the policyholder bears the remaining share. Getting that order right is essential when calculating either an individual claim payment or its expected value.

Throughout this tutorial, $X$ denotes a nonnegative loss, $d\geq 0$ is an ordinary deductible, and $c\in[0,1]$ is the insurer’s share of the loss above the deductible. There is no policy limit. All insurance scenarios below are hypothetical.

Definition: The insurer payment per loss is $Y^L=c(X-d)_+$, where $(z)_+=\max(z,0)$. The policyholder retention is the part of the loss not paid by the insurer: $R=X-Y^L$. Thus $R=X$ when $X\leq d$, and $R=d+(1-c)(X-d)$ when $X>d$.

Payment and Retention Functions

The ordinary deductible determines the amount of loss that enters the coinsurance calculation. If the loss does not exceed $d$, there is no insurer payment. If it exceeds $d$, only the excess $X-d$ is shared. The insurer pays fraction $c$ of that excess; the policyholder pays the deductible plus fraction $1-c$ of the excess.

Y^L = \begin{cases} 0, & X\leq d,\\ c(X-d), & X>d, \end{cases} \qquad R = \begin{cases} X, & X\leq d,\\ d+(1-c)(X-d), & X>d. \end{cases}

The two functions must add to the original loss in every case. For $X\leq d$, $Y^L+R=0+X=X$. For $X>d$,

$$ Y^L+R=c(X-d)+d+(1-c)(X-d)=d+(X-d)=X. $$

This pointwise identity is a useful check on both formulas and any numerical calculation. The endpoint cases also behave as expected: if $c=0$, the insurer pays nothing; if $c=1$, the insurer pays all loss above the deductible and the policyholder retains $\min(X,d)$.

Expected Payment

For a nonnegative loss with finite mean, the expected insurer payment is $c$ times the expected excess loss above the deductible. Write $S_X(x)=P(X>x)$ for the survival function. The tail-integral identity gives a direct way to calculate the expected excess, whether the severity distribution is continuous, discrete, or mixed.

Theorem: If $E[X]<\infty$ and $d\geq0$, then $$ E[Y^L]=cE[(X-d)_+]=c\int_d^\infty S_X(x)\,dx. $$ If the distribution has a density $f_X$, the same expectation can also be calculated as $c\int_d^\infty (x-d)f_X(x)\,dx$. For a discrete loss, sum $c(x-d)_+$ times the probability at each possible loss value.

To see why the tail-integral form holds, take any nonnegative random variable $Z$ with finite mean. Its value can be expressed as the area under the indicator that it exceeds each level: $Z=\int_0^\infty \mathbf{1}_{\{Z>t\}}\,dt$. Taking expectations gives $E[Z]=\int_0^\infty P(Z>t)\,dt$. Set $Z=(X-d)_+$. For $t\geq0$, $Z>t$ exactly when $X>d+t$, so a change of variable $x=d+t$ yields the stated formula. This argument also explains why an atom at $X=d$ contributes zero payment.

When $E[X]$ is finite, expected retention follows from $R=X-Y^L$:

$$ E[R]=E[X]-E[Y^L]. $$

This relationship is often the quickest way to find expected retention. It does not replace checking the payment function: an incorrect payment formula will make the difference incorrect as well.

Worked Example: Exponential Loss With a Deductible

A hypothetical equipment policy has a loss $X\sim\operatorname{Exponential}(\text{mean }500)$, in dollars, so $S_X(x)=e^{-x/500}$ for $x\geq0$. The ordinary deductible is $d=\$200$, and the insurer pays $c=0.80$ of loss above it. Find the expected insurer payment and expected policyholder retention.

Method. Use the survival-function formula for the expected excess, then multiply by the insurer’s share.

Expected payment.

E[Y^L]=0.80\int_{200}^{\infty}e^{-x/500}\,dx =0.80\left[ -500e^{-x/500}\right]_{200}^{\infty} =0.80(500e^{-0.4}).

Using $e^{-0.4}=0.670320046$, the expected excess is $500(0.670320046)=335.160023$ dollars. The expected payment is $0.80(335.160023)=268.128018$ dollars, or \$268.13 to the nearest cent.

Expected retention. The mean loss is $E[X]=\$500$. Therefore $E[R]=500-268.128018=231.871982$, or \$231.87 to the nearest cent.

Check. Before rounding, $268.128018+231.871982=500$, the expected loss. The calculation applies coinsurance to the excess, not to the full loss.

Discrete Losses and the Deductible Boundary

For a discrete severity model, calculate the payment and retention at every possible loss value. A loss equal to the deductible has zero excess and therefore zero insurer payment, while the policyholder retains the entire loss. This boundary remains true even when the distribution places positive probability exactly at $d$.

Worked Example: Discrete Loss With an Outcome at the Deductible

A hypothetical small-business policy has losses $X$ of \$0, \$100, \$200, \$400, or \$700, with respective probabilities $0.10, 0.20, 0.15, 0.30,$ and $0.25$. The deductible is $d=\$200$, and the insurer pays $c=0.75$ of the excess. Calculate expected payment and expected retention.

Method. Apply $Y^L=0.75(X-200)_+$ at each possible loss, then take the probability-weighted average.

Loss $x$	Probability	Excess $(x-200)_+$	Payment $0.75(x-200)_+$	Retention $x-Y^L$
\$0	0.10	\$0	\$0	\$0
\$100	0.20	\$0	\$0	\$100
\$200	0.15	\$0	\$0	\$200
\$400	0.30	\$200	\$150	\$250
\$700	0.25	\$500	\$375	\$325

Expected payment. The only positive payment amounts are \$150 and \$375:

E[Y^L]=0.30(150)+0.25(375)=45+93.75=\$138.75.

Expected loss and retention.

E[X]=0.10(0)+0.20(100)+0.15(200)+0.30(400)+0.25(700) =0+20+30+120+175=\$345.

Thus $E[R]=345-138.75=\$206.25$. Directly from the retention column, the same result is $0.20(100)+0.15(200)+0.30(250)+0.25(325)=20+30+75+81.25=\$206.25$.

Check. The probabilities sum to $0.10+0.20+0.15+0.30+0.25=1$, and expected payment plus expected retention is $138.75+206.25=345$, the expected loss. The outcome at the deductible contributes \$200 of retention and no payment.

Using a Density: Uniform Losses

A bounded continuous model illustrates the same calculation without an exponential tail. Suppose $X$ is uniform on $[0,b]$. Its density is $1/b$ on that interval, and there is no loss above $b$. For $0\leq d\leq b$, the expected excess is the area under the excess-loss function weighted by the density:

E[(X-d)_+]=\frac{1}{b}\int_d^b(x-d)\,dx =\frac{(b-d)^2}{2b}.

Worked Example: Uniform Loss With Coinsurance Above the Deductible

A hypothetical event policy has $X$ uniform on $[0,\$1{,}000]$. Its deductible is $d=\$250$, and the insurer pays $c=0.60$ of the excess. Find expected payment and expected retention.

Method. Integrate the excess over the part of the uniform range above the deductible, then use the expected-loss identity for retention.

The density is $1/1000$, and the excess can be positive only from \$250 to \$1,000. Therefore,

E[(X-250)_+] =\frac{1}{1000}\int_{250}^{1000}(x-250)\,dx =\frac{1}{1000}\left[\frac{(x-250)^2}{2}\right]_{250}^{1000} =\frac{750^2}{2000} =\$281.25.

The expected insurer payment is $0.60(281.25)=\$168.75$. The mean of a uniform distribution on $[0,1000]$ is $E[X]=(0+1000)/2=\$500$, so expected retention is $500-168.75=\$331.25$.

Check. The expected amounts add to $168.75+331.25=500$, matching the mean loss. Losses below \$250 are fully retained; for losses above \$250, the insurer and policyholder divide only the excess in the ratio $0.60:0.40$.

Common Mistake: Applying Coinsurance to the Full Loss

A frequent error is to calculate insurer payment as $cX$ whenever $X>d$. That treats the deductible as if it did not exist. For example, in the discrete example, a \$400 loss has a \$200 excess. The correct payment is $0.75(400-200)=\$150$, not $0.75(400)=\$300$. The incorrect calculation overstates the insurer payment by \$150 and understates retention by the same amount.

Another tempting shortcut is to subtract the deductible from the expected loss and then multiply by $c$, without accounting for losses at or below the deductible. In general, $E[(X-d)_+]\neq E[X]-d$, because the positive-part operation must be applied to each loss outcome before averaging. The survival-integral formula or a direct expectation over the loss distribution handles this correctly.

Key takeaway: For coinsurance after an ordinary deductible, calculate the excess first, then apply the insurer share: $Y^L=c(X-d)_+$. The policyholder retains $R=X-Y^L$, so $Y^L+R=X$ for every loss. This framework is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical policy, calculate the requested amount. Keep exact values until the final rounding step.

A claim is \$500, the ordinary deductible is \$100, and the insurer pays 70% of the excess. Find the insurer payment and policyholder retention. (Answer: payment \$280; retention \$220)
A loss is \$0, \$200, or \$500 with probabilities $0.25, 0.25,$ and $0.50$, respectively. With $d=\$200$ and $c=0.80$, find expected payment and expected retention. (Answer: payment \$120; retention \$180)
A loss has an exponential distribution with mean \$400. For $d=\$100$ and $c=0.60$, find expected payment to the nearest cent. (Answer: \$186.91)
A loss is uniform on $[0,\$600]$. For $d=\$150$ and $c=0.40$, find expected payment and expected retention. (Answer: payment \$67.50; retention \$232.50)
A loss is \$0, \$100, \$300, or \$900 with probabilities $0.10, 0.20, 0.40,$ and $0.30$, respectively. For $d=\$200$ and $c=0.70$, calculate expected payment and expected retention, and verify their sum against expected loss. (Answer: payment \$175; retention \$235; expected loss \$410)

What You'll Learn