Coinsurance as a Proportional Payment

Introduction

Coinsurance divides a covered loss between the insurer and the policyholder in fixed proportions. If the insurer pays a specified share of every dollar of loss, its payment is a scaled version of the original loss. That simple relationship lets us obtain the payment distribution and its moments directly from the severity model.

Here, coinsurance means a fixed insurer share applied proportionally to the entire loss. It does not mean a deductible, a payment cap, or a share applied only after some threshold. Those distinctions matter: with a pure proportional payment, the insurer’s payment changes in the same ratio as the loss.

Definition: Let $X$ be a nonnegative loss and let $c$, with $0\le c\le 1$, be the insurer’s fixed coinsurance share. The insurer’s payment per loss is $Y^L=cX$. The policyholder pays the remaining share $(1-c)X$. This definition assumes there is no deductible, payment limit, or other modification to the insurer’s share.

The share $c$ is fixed and does not vary with the size of the loss. For example, when $c=0.7$, the insurer pays 70% of a loss of $1,000 and also 70% of a loss of $10,000. The resulting payment is measured in the same units as $X$, such as dollars.

Distribution of the Proportional Payment

For $c>0$, the payment distribution follows by translating an event about $Y^L$ into the corresponding event about $X$. Since multiplication by a positive constant preserves order, $Y^L\le y$ exactly when $X\le y/c$.

F_{Y^L}(y)=P(Y^L\le y) =P(cX\le y) =F_X\!\left(\frac{y}{c}\right),\qquad c>0.

For a nonnegative loss, both $F_X(t)$ and $F_{Y^L}(t)$ are zero when $t<0$. The payment’s survival function likewise follows by rescaling the threshold:

S_{Y^L}(y)=P(Y^L>y) =S_X\!\left(\frac{y}{c}\right),\qquad c>0.

If $X$ has a density $f_X$, then the density of the payment is obtained by the change-of-variable rule. The factor $1/c$ is needed so that the density still integrates to one. If $X$ is discrete, each possible loss value $x$ instead maps to payment $cx$ with the same probability. In particular, any probability mass at zero remains at zero.

f_{Y^L}(y)=\frac{1}{c}f_X\!\left(\frac{y}{c}\right),\qquad c>0.

The case $c=0$ is not obtained by substituting zero into these formulas. The insurer pays nothing for every loss, so $Y^L=0$ with probability one. When $c=1$, the insurer pays the full loss and $Y^L=X$.

Worked Example: Rescaling an Exponential Loss Distribution

A hypothetical covered loss $X$ has an exponential distribution with mean $4,000 and rate $\lambda=1/4{,}000=0.00025$ per dollar. A contract pays 25% of every loss, with no deductible or limit. Find the payment distribution, the probability that payment is at most $1,500, the probability it exceeds $2,000, and its mean and variance.

Method. The payment is $Y^L=0.25X$. An exponential loss has survival function $S_X(x)=e^{-\lambda x}$ for $x\ge0$. Therefore, for $y\ge0$,

S_{Y^L}(y)=S_X\!\left(\frac{y}{0.25}\right) =\exp\!\left(-0.00025\frac{y}{0.25}\right) =e^{-0.001y}.

Thus $Y^L$ is exponential with rate $0.001$ per dollar and mean $1/0.001=$1{,}000$. Its CDF is $F_{Y^L}(y)=1-e^{-0.001y}$ for $y\ge0$.

At $1,500, the CDF is $1-e^{-1.5}=1-0.223130\ldots=0.7769$, to four decimal places. The probability of exceeding $2,000 is $e^{-0.001(2000)}=e^{-2}=0.1353$, to four decimal places.

An exponential variable with rate $0.001$ has variance $1/(0.001)^2=$1{,}000{,}000$ dollars squared. The results are consistent with scaling the original exponential mean and variance: the mean is $1,000 and the variance is $1,000,000 dollars squared.

Check. The payment’s mean is one quarter of $4,000, as required by a 25% share. Its probability of exceeding $2,000 is also the loss probability of exceeding $2000/0.25=$8{,}000$, namely $e^{-0.00025(8000)}=e^{-2}$.

Why the Mean and Variance Scale Differently

The same fixed-share definition gives the payment’s moments. For a finite loss mean and variance, multiplying the random variable by $c$ multiplies its mean by $c$ and its variance by $c^2$. The square in the variance arises because variance is based on squared deviations from the mean.

Theorem: If $Y^L=cX$ for a fixed constant $c$, and $E[X]$ and $\operatorname{Var}(X)$ are finite, then $$ E[Y^L]=cE[X],\qquad \operatorname{Var}(Y^L)=c^2\operatorname{Var}(X). $$ Consequently, $\operatorname{SD}(Y^L)=c\operatorname{SD}(X)$ for $c\ge0$.

To justify the mean formula, use linearity of expectation. For the variance, subtract the payment’s mean before squaring the deviation:

Write the payment and its mean.
$Y^L=cX$, so $E[Y^L]=E[cX]=cE[X]$.

Express the deviation from the payment mean.
$Y^L-E[Y^L]=cX-cE[X]=c(X-E[X])$.

Square and take the expectation.
$\operatorname{Var}(Y^L)=E[(Y^L-E[Y^L])^2]=E[c^2(X-E[X])^2]=c^2\operatorname{Var}(X)$.

The standard deviation is the square root of variance, so its multiplier is $c$, not $c^2$. These results require the insurer share to be fixed. If the share depends on the size of the loss, the payment is not generally a constant multiple of $X$, and these scaling formulas need not apply.

Worked Example: Transforming a Discrete Loss Model

A hypothetical loss $X$ takes values $0, $1,000, $3,000, and $6,000 with respective probabilities 0.10, 0.20, 0.40, and 0.30. The insurer pays 60% of each loss. Find the payment distribution, $P(Y^L\le $1{,}800)$, the mean, and the variance.

Method. Multiply each possible loss by $c=0.60$, while retaining its probability:

Loss $X$	Probability	Payment $Y^L=0.60X$
$0	0.10	$0
$1,000	0.20	$600
$3,000	0.40	$1,800
$6,000	0.30	$3,600

The event $Y^L\le $1{,}800$ includes the payments $0, $600, and $1,800. Therefore its probability is $0.10+0.20+0.40=0.70$.

First calculate the loss mean:

$$ E[X]=0(0.10)+1000(0.20)+3000(0.40)+6000(0.30) =0+200+1200+1800=$3{,}200. $$

Then $E[Y^L]=0.60(3200)=$1{,}920$. For the variance, calculate the second moment of the loss:

E[X^2]=0^2(0.10)+1000^2(0.20)+3000^2(0.40)+6000^2(0.30) =0+200{,}000+3{,}600{,}000+10{,}800{,}000 =14{,}600{,}000.

Thus $\operatorname{Var}(X)=14{,}600{,}000-3{,}200^2=14{,}600{,}000-10{,}240{,}000=4{,}360{,}000$ dollars squared. The insurer’s payment variance is $0.60^2(4{,}360{,}000)=0.36(4{,}360{,}000)=1{,}569{,}600$ dollars squared.

Check. The four payment probabilities still sum to $0.10+0.20+0.40+0.30=1$. Directly from the payment table, the mean is $600(0.20)+1800(0.40)+3600(0.30)=120+720+1080=$1{,}920$, agreeing with the scaling result.

Worked Example: Scaling Given Loss Moments

A hypothetical loss has mean $2,500 and variance 400,000 dollars squared. The insurer pays 40% of each loss, without a deductible or limit. Find the payment mean, variance, and standard deviation.

Method. Here $c=0.40$. The mean payment is $0.40(2500)=$1{,}000$. The variance payment is $0.40^2(400{,}000)=0.16(400{,}000)=64{,}000$ dollars squared.

The standard deviation is the square root of the payment variance:

\operatorname{SD}(Y^L)=\sqrt{64{,}000} =252.9822\ldots \approx $252.98.

The mean and standard deviation are reported to the nearest cent. Check. The loss standard deviation is $\sqrt{400{,}000}=632.4555\ldots$, and $0.40(632.4555\ldots)=252.9822\ldots$, the same payment standard deviation.

Common Exam Trap: Scaling Standard Deviation Like Variance

A frequent error is to multiply variance by $c$ instead of $c^2$, or to multiply the standard deviation by $c^2$. Both confuse the squared-deviation definition of variance with the original scale of the random variable.

For the preceding 40% coinsurance example, an incorrect calculation would be $\operatorname{Var}(Y^L)=0.40(400{,}000)=160{,}000$. This is wrong because the deviation $Y^L-E[Y^L]$ is $0.40$ times the loss deviation, and squaring it produces a factor of $0.40^2=0.16$. The correct variance is 64,000 dollars squared, and the correct standard deviation is $\sqrt{64{,}000}=$252.98$, not $253.00 when rounded to cents.

A second trap is to rescale the event in the wrong direction. For $c>0$, the event $Y^L\le y$ is $X\le y/c$, not $X\le cy$. A smaller insurer share means that a larger underlying loss is needed to reach the same payment threshold.

When the Scaling Results Apply

These formulas are useful when the policy changes only the insurer’s proportional share. They do not by themselves handle a deductible, a policy limit, or a coinsurance share that changes across loss amounts. In those cases, the payment is a different function of $X$, so its distribution and moments must account for that payment function.

Before applying the results, check that the problem’s wording means a fixed percentage of the whole loss is paid by the insurer. Then identify $c$ as the insurer’s share—not the policyholder’s share—and distinguish the requested measure: mean scales by $c$, variance by $c^2$, and standard deviation by $c$. These basic checks keep the distribution transformation and moment calculations consistent.

Key takeaway: For a fixed insurer share $c$, the payment per loss is $Y^L=cX$. For $c>0$, $F_{Y^L}(y)=F_X(y/c)$; when the relevant moments exist, $E[Y^L]=cE[X]$, $\operatorname{Var}(Y^L)=c^2\operatorname{Var}(X)$, and $\operatorname{SD}(Y^L)=c\operatorname{SD}(X)$.

Check Your Understanding

For each hypothetical contract, assume the insurer pays the stated fixed share of each nonnegative loss, with no deductible or limit.

A loss $X$ is exponential with mean $2,000, and the insurer share is $c=0.50$. Find $E[Y^L]$ and $P(Y^L\le $1{,}000)$. (Answer: mean $1,000; probability 0.6321.)
A loss takes values $0, $2,000, and $5,000 with probabilities 0.25, 0.50, and 0.25, respectively. If $c=0.40$, find $P(Y^L\ge $2{,}000)$. (Answer: 0.25.)
A loss is uniform on the interval from $0 to $6,000, and $c=0.30$. Find $E[Y^L]$ and $P(Y^L>$900)$. (Answer: mean $900; probability 0.5000.)
A loss has mean $12,000 and variance 9,000,000 dollars squared. If $c=0.75$, find the payment mean, variance, and standard deviation. (Answer: mean $9,000; variance 5,062,500 dollars squared; standard deviation $2,250.00.)
For a positive share $c=0.20$, suppose $F_X($6{,}000)=0.9000$. Find $P(Y^L\le $1{,}200)$. (Answer: 0.9000.)

What You'll Learn