Coinsurance as a Payment Transformation

Introduction

A coinsurance clause assigns the insurer a specified fraction of an amount that would otherwise be payable. The important modeling question is therefore not merely how large the loss is, but what amount the coinsurance fraction is applied to. If the coverage terms first determine a payment amount and coinsurance then scales it, the resulting payment is a direct transformation of that base amount.

This tutorial derives the distribution and moments of that transformed payment. It treats the amount before coinsurance as already determined by the coverage being modeled; it does not assume that this amount must equal the ground-up loss.

Definition: Let $Y^L$ be the payment per loss under the stated coverage before applying coinsurance, and let $c$ be the insurer's coinsurance fraction, with $0\leq c\leq 1$. The insurer's payment after coinsurance is $$ Z=cY^L. $$ The model assumes coinsurance acts by multiplication on this otherwise payable amount, with no additional fixed payment or other modification at this step. The amount $Y^L$ is nonnegative.

The insured's share of this otherwise payable amount is $(1-c)Y^L$, if the contract assigns the remainder to the insured. Thus $c$ is the insurer's fraction, not the fraction retained by the insured. When $c=1$, the payment is unchanged; when $c=0$, the insurer pays zero.

Distribution of the Coinsurance Payment

For $c>0$, multiplying by $c$ preserves the order of possible payment amounts. An insurer payment at or below a threshold $z$ occurs exactly when the base payment is at or below $z/c$. This gives the distribution directly, whether the base payment is continuous, discrete, or mixed.

Theorem: If $Z=cY^L$ and $c>0$, then the cumulative distribution function of the insurer payment is $$ F_Z(z)=P(Z\leq z)=F_{Y^L}\left(\frac{z}{c}\right),\qquad -\infty<z<\infty. $$ For $z\geq 0$, the corresponding strict exceedance probability is $$ P(Z>z)=P\left(Y^L>\frac{z}{c}\right). $$ If $c=0$, then $Z=0$ with probability 1, so $F_Z(z)=0$ for $z<0$ and $F_Z(z)=1$ for $z\geq0$.

To justify the result for $c>0$, begin with the event defining the CDF. Since division by a positive number does not reverse an inequality,

\{Z\leq z\}=\{cY^L\leq z\} =\left\{Y^L\leq \frac{z}{c}\right\}.

Taking probabilities on both sides gives the stated formula. For the exceedance probability, the same reasoning gives $\{cY^L>z\}=\{Y^L>z/c\}$. The positivity of $c$ matters: dividing by zero is not defined, which is why $c=0$ must be handled separately.

For $0\leq a<b$ and $c>0$, the probability that the payment lies in an interval follows by subtracting CDF values:

P(a<Z\leq b) =F_Z(b)-F_Z(a) =F_{Y^L}\left(\frac{b}{c}\right)-F_{Y^L}\left(\frac{a}{c}\right).

The CDF formula is useful for distributions with jumps as well as for continuous distributions. The interval endpoints and inequality signs should be kept consistent when the base payment can take particular values with positive probability.

Moments and the Size of the Reduction

Scaling also gives a simple relationship between the moments of the base payment and those of the coinsurance payment. If the relevant moments exist, then for any positive integer $k$,

$$ E[Z^k]=E[(cY^L)^k]=c^kE[(Y^L)^k]. $$

In particular, the expected insurer payment is $cE[Y^L]$, and its variance is $c^2\operatorname{Var}(Y^L)$. The variance uses the square of the scaling factor; multiplying the variance by $c$ would be incorrect.

Result: For $Z=cY^L$ with finite first and second moments, $$ E[Z]=cE[Y^L],\qquad \operatorname{Var}(Z)=c^2\operatorname{Var}(Y^L). $$ These formulas also hold at $c=0$, since both the expectation and variance of the resulting zero payment are zero.

The expectation result follows from linearity of expectation: $E[cY^L]=cE[Y^L]$. For the variance, use the definition and pull out the constant:

\operatorname{Var}(Z) =E[(Z-E[Z])^2] =E[(cY^L-cE[Y^L])^2] =c^2E[(Y^L-E[Y^L])^2] =c^2\operatorname{Var}(Y^L).

The expected reduction from the base payment to the insurer's payment is $(1-c)E[Y^L]$, because $Y^L-Z=(1-c)Y^L$. This describes the expected amount shifted away from the insurer under the stated model; it does not by itself specify how any other contract terms allocate that amount.

Worked Applications

In each example, first identify the base payment distribution. Apply coinsurance to that payment—not automatically to the ground-up loss—and then use the distribution or moment relationship that answers the question.

Worked Example: Exponential Base Payments

A hypothetical coverage has a payment per loss $Y^L$ with an exponential distribution having mean $500. A coinsurance clause sets the insurer's fraction at $c=0.60$. Find the probability that the insurer pays more than $240, the probability that the payment is between $100 and $400, and the expected insurer payment.

Method and assumptions. For an exponential distribution with mean $500, the rate is $1/500$ per dollar. Thus, for $y\geq0$, $F_{Y^L}(y)=1-e^{-y/500}$. The coinsurance payment is $Z=0.60Y^L$.

Exceedance probability. Scale the threshold back to the base payment:

P(Z>240) =P\left(Y^L>\frac{240}{0.60}\right) =P(Y^L>400) =e^{-400/500} =e^{-0.8} \approx 0.4493.

Interval probability. The interval endpoints correspond to base payments $100/0.60=500/3$ and $400/0.60=2000/3$. Therefore,

\begin{aligned} P(100<Z\leq400) &=F_{Y^L}(2000/3)-F_{Y^L}(500/3)\\ &=e^{-1/3}-e^{-4/3}\\ &\approx 0.7165-0.2636\\ &=0.4529. \end{aligned}

Expected payment. The exponential base payment has mean $500, so

E[Z]=0.60(500)=\$300.00.

The answers are a probability of 0.4493, an interval probability of 0.4529, and an expected payment of $300.00. The method avoids the mistake of multiplying a threshold by $c$ when converting it to a threshold for $Y^L$; the correct conversion divides by $c$.

Worked Example: Uniform Base Payments

For a hypothetical set of covered losses, suppose the base payment $Y^L$ is uniformly distributed from $0 to $1,000. The insurer pays $c=0.75$ of that amount. Find the support and CDF of $Z$, the probability that $Z$ is between $300 and $600, and the mean and variance of $Z$.

Method and assumptions. A uniform distribution on $[0,1000]$ has CDF $F_{Y^L}(y)=y/1000$ for $0\leq y\leq1000$. Since $Z=0.75Y^L$, its smallest and largest possible values are $0.75(0)=0$ and $0.75(1000)=750$. For $0\leq z\leq750$,

F_Z(z)=F_{Y^L}(z/0.75)=\frac{z}{750}.

The full CDF is 0 below 0, $z/750$ on $[0,750]$, and 1 above 750. The requested interval probability is

P(300\leq Z\leq600)=\frac{600-300}{750}=0.4000.

For a uniform distribution on $[0,1000]$, the mean is $1000/2=\$500$, and the variance is $1000^2/12=83{,}333.333\ldots$ dollars squared. Scaling gives

E[Z]=0.75(500)=\$375.00,\qquad \operatorname{Var}(Z)=0.75^2\left(\frac{1000^2}{12}\right)=46{,}875\text{ dollars}^2.

The support, interval probability, mean, and variance are therefore $[0,\$750]$, 0.4000, $375.00, and 46,875 dollars squared. This calculation avoids treating the original support as the support after coinsurance: both endpoints must be scaled.

Worked Example: A Discrete Base Payment

A hypothetical plan has the following distribution for its payment per loss before coinsurance: $0 with probability 0.20, $400 with probability 0.50, and $1,000 with probability 0.30. The insurer's fraction is $c=0.60$. Find the distribution of its payment, $P(Z\leq300)$, the expected payment, and the variance.

Method and assumptions. Apply $Z=0.60Y^L$ to each possible base payment, keeping its probability attached to that outcome. The insurer-payment outcomes are $0, $240, and $600, with probabilities 0.20, 0.50, and 0.30, respectively.

Since both $0 and $240 are at or below $300,

P(Z\leq300)=0.20+0.50=0.7000.

The mean is

E[Z]=0.20(0)+0.50(240)+0.30(600) =0+120+180 =\$300.00.

For the variance, first compute the second moment:

E[Z^2]=0.20(0^2)+0.50(240^2)+0.30(600^2) =0+28{,}800+108{,}000 =136{,}800\text{ dollars}^2.

Then subtract the squared mean:

\operatorname{Var}(Z)=E[Z^2]-E[Z]^2 =136{,}800-300^2 =46{,}800\text{ dollars}^2.

The probability is 0.7000, the expected payment is $300.00, and the variance is 46,800 dollars squared. The check $E[Z]=0.60E[Y^L]=0.60[0.50(400)+0.30(1000)]=\$300.00$ confirms the direct calculation. The method avoids changing the outcome probabilities when scaling the payment amounts.

Common Mistakes and Modeling Checks

Coinsurance is simple algebraically, but errors often arise from applying that algebra to the wrong quantity or using the wrong direction when converting a threshold.

Scaling the ground-up loss instead of the base payment. Coinsurance here acts on $Y^L$, the amount already determined to be payable before coinsurance. If a hypothetical loss is $1,200 but its otherwise payable amount is $800, a 75% insurer fraction gives $0.75(800)=\$600$, not $0.75(1200)=\$900$.
Using the insured's share as the insurer's share. If $c=0.60$, the insurer pays 60% of the base amount, not the remaining 40%. For a base payment of $500, the insurer's payment is $300, while the remaining share is $200.
Multiplying instead of dividing a threshold. To find $P(Z>z)$, the equivalent base-payment threshold is $z/c$. For example, with $c=0.60$, an insurer payment above $240 requires a base payment above $400, not above $144.
Scaling variance by $c$. Since deviations from the mean are multiplied by $c$, squared deviations are multiplied by $c^2$. Thus $\operatorname{Var}(Z)=c^2\operatorname{Var}(Y^L)$.

Before calculating, check that the model specifies the amount coinsurance applies to, that $c$ is the insurer's fraction, and that any threshold has been converted in the correct direction. The CDF transformation is valid for $c>0$; the zero-fraction case is handled separately.

Key takeaway: Once the otherwise payable amount $Y^L$ is identified, coinsurance with fraction $c$ gives $Z=cY^L$. For $c>0$, evaluate its CDF at $z/c$; its mean scales by $c$, and its variance scales by $c^2$.

Check Your Understanding

In each problem, treat the stated payment amount as the base payment before coinsurance. Answers involving probabilities are rounded to four decimals.

A hypothetical covered loss has a base payment of $250, and the insurer's coinsurance fraction is 0.40. What is the insurer's payment? (Answer: $100.00)
The base payment is uniform on $0 to $400, and $c=0.50$. Find $P(Z\leq80)$ and $E[Z]$. (Answer: 0.4000 and $100.00)
The base payment is exponential with mean $250, and $c=0.80$. Find $P(Z>400)$ and $E[Z]$. (Answer: 0.1353 and $200.00)
A base payment is $0, $300, or $900 with probabilities 0.25, 0.50, or 0.25, respectively. If $c=2/3$, find $P(Z\leq200)$ and $E[Z]$. (Answer: 0.7500 and $250.00)
A base payment is zero with probability 0.20; otherwise it is uniform on $0 to $500. If $c=0.60$, find $P(Z>120)$ and $E[Z]$. (Answer: 0.4800 and $120.00)

What You'll Learn