Why Coinsurance Changes Payment Risk
Under proportional coinsurance, the insurer pays a fixed fraction of each covered loss. Reducing that fraction reduces every payment by the same proportion. It also changes the spread of the payment distribution, but not by that same proportion: variance scales with the square of the fraction. Getting this distinction right is central to calculating and interpreting the risk transferred to the insurer.
This tutorial considers a single covered loss \(X\) and a fixed insurer share. It does not include a deductible, policy limit, or interest discounting. Those features change the payment function and therefore require different calculations.
The insurer’s share \(c\) is a proportion, not a dollar amount. For example, \(c=0.7\) means the insurer pays 70% of every covered loss. If \(c=0\), the insurer pays nothing; if \(c=1\), the insurer pays the full loss. In every case, the model assumes the same share applies to every loss and does not depend on the size or occurrence of the loss.
Moments of a Proportional Payment
Write the loss mean as \(E[X]\) and its variance as \(\operatorname{Var}(X)\). The payment’s first raw moment is its expected payment. Its second raw moment is \(E[(Y^L)^2]\), which is not the same quantity as its variance. The distinction matters because variance subtracts the square of the mean from the second raw moment.
The second raw moment can also be expressed in terms of the loss mean and variance. Since \(E[X^2]=\operatorname{Var}(X)+(E[X])^2\), it follows that
Why the Scaling Formulas Hold
The result follows directly from the definition of the payment as a constant multiple of the loss. For the mean and second raw moment, take expectations of \(cX\) and \((cX)^2\). For the variance, use the definition of variance as the expected squared deviation from the mean.
Since \(Y^L=cX\), linearity of expectation gives \(E[Y^L]=E[cX]=cE[X]\).
Pointwise, \((Y^L)^2=(cX)^2=c^2X^2\). Taking expectations gives \(E[(Y^L)^2]=c^2E[X^2]\).
By definition, \(\operatorname{Var}(Y^L)=E[((Y^L)-E[Y^L])^2]\). Substituting \(Y^L=cX\) and \(E[Y^L]=cE[X]\) gives \(E[(c(X-E[X]))^2]=c^2\operatorname{Var}(X)\).
The last step explains why the variance factor is \(c^2\): deviations from the mean are multiplied by \(c\), and variance averages their squares. For \(c>0\), the standard deviation scales by \(c\), because it is the square root of variance. When \(c=0\), the payment and its variance are both zero.
These identities require a fixed share \(c\). If the share varies randomly, or depends on the loss amount, then the payment is not simply a constant multiple of \(X\); the formulas above do not generally apply. The moments also require \(E[X^2]\) to be finite for the variance formula to be finite.
Worked Example: Scaling a Discrete Loss Distribution
A hypothetical policy covers one claim. The loss \(X\) is \$0 with probability 0.2, \$1,000 with probability 0.5, and \$3,000 with probability 0.3. The insurer pays a fixed 80% of the loss, so \(c=0.8\). Find the expected payment, second raw moment, and variance. Assume no interest discounting or mortality contingency.
Method. Calculate the loss moments first, then scale them. The distribution is discrete with the stated three outcomes and probabilities.
Loss mean.
Loss second raw moment.
The loss variance is \(3{,}200{,}000-(1{,}400)^2=3{,}200{,}000-1{,}960{,}000=1{,}240{,}000\text{ dollars}^2\). The payment values are \(0\), \(\$800\), and \(\$2{,}400\), with the same respective probabilities.
Expected payment.
Second raw moment and variance.
Check. Subtract the squared payment mean from its second raw moment: \(2{,}048{,}000-(1{,}120)^2=2{,}048{,}000-1{,}254{,}400=793{,}600\text{ dollars}^2\). The result agrees. The common mistake to avoid is multiplying the loss variance by 0.8 instead of by \(0.8^2\).
Comparing Insurer Shares
For the same underlying loss distribution, changing \(c\) has predictable effects. The expected payment changes in direct proportion to \(c\), while the variance changes in proportion to \(c^2\). Thus, if the share is cut in half, the expected payment is halved and the variance is reduced to one-quarter. For a positive share, the coefficient of variation—the standard deviation divided by the mean—does not change under scaling, because both standard deviation and mean are multiplied by \(c\).
| Quantity | Payment under share \(c\) | Effect of reducing \(c\) |
|---|---|---|
| Expected payment | \(cE[X]\) | Changes in direct proportion to \(c\) |
| Second raw moment | \(c^2E[X^2]\) | Changes in proportion to \(c^2\) |
| Variance | \(c^2\operatorname{Var}(X)\) | Changes in proportion to \(c^2\) |
| Standard deviation | \(c\sqrt{\operatorname{Var}(X)}\), for \(c\geq 0\) | Changes in direct proportion to \(c\) |
Worked Example: Comparing Two Insurer Shares
A hypothetical repair-cost loss \(X\) follows an exponential distribution with mean \(\$2{,}000\). The insurer is considering either a 65% share or a 40% share. Calculate the payment mean and variance under each option. Assume no interest discounting or mortality contingency.
Model and loss moments. For an exponential distribution with mean \(\theta\), \(E[X]=\theta\) and \(\operatorname{Var}(X)=\theta^2\). Here \(\theta=\$2{,}000\), so \(E[X]=\$2{,}000\), \(\operatorname{Var}(X)=\$4{,}000{,}000\text{ dollars}^2\), and \(E[X^2]=4{,}000{,}000+2{,}000^2=8{,}000{,}000\text{ dollars}^2\).
At \(c=0.65\). The mean payment is \(0.65(2{,}000)=\$1{,}300\). The variance is \(0.65^2(4{,}000{,}000)=0.4225(4{,}000{,}000)=1{,}690{,}000\text{ dollars}^2\). The second raw moment is \(0.4225(8{,}000{,}000)=3{,}380{,}000\text{ dollars}^2\).
At \(c=0.40\). The mean payment is \(0.40(2{,}000)=\$800\). The variance is \(0.40^2(4{,}000{,}000)=0.16(4{,}000{,}000)=640{,}000\text{ dollars}^2\). The second raw moment is \(0.16(8{,}000{,}000)=1{,}280{,}000\text{ dollars}^2\).
Check. At the 65% share, \(3{,}380{,}000-(1{,}300)^2=3{,}380{,}000-1{,}690{,}000=1{,}690{,}000\). At the 40% share, \(1{,}280{,}000-(800)^2=1{,}280{,}000-640{,}000=640{,}000\). The variance ratio is \(640{,}000/1{,}690{,}000\), which also equals \((0.40/0.65)^2\). This comparison makes clear that payment variability is not reduced by the same percentage as the payment mean.
Worked Example: Moments from a Gamma Loss Model
A hypothetical equipment claim \(X\) follows a gamma distribution with shape \(k=2\) and scale \(\theta=\$1{,}500\). The insurer pays 40% of each loss. Find the payment mean, second raw moment, and variance. Assume no interest discounting or mortality contingency.
Method and loss moments. For a gamma distribution parameterized by shape \(k\) and scale \(\theta\), \(E[X]=k\theta\) and \(\operatorname{Var}(X)=k\theta^2\). Therefore,
The loss second raw moment is \(E[X^2]=4{,}500{,}000+(3{,}000)^2=4{,}500{,}000+9{,}000{,}000=13{,}500{,}000\text{ dollars}^2\).
Scale to the payment. With \(c=0.40\), the expected payment is \(0.40(3{,}000)=\$1{,}200\). The second raw moment is \(0.40^2(13{,}500{,}000)=0.16(13{,}500{,}000)=2{,}160{,}000\text{ dollars}^2\). The variance is \(0.16(4{,}500{,}000)=720{,}000\text{ dollars}^2\).
Check. \(2{,}160{,}000-(1{,}200)^2=2{,}160{,}000-1{,}440{,}000=720{,}000\text{ dollars}^2\). The check confirms that the scaled second raw moment and mean produce the same variance as direct scaling.
Common Exam Trap: Confusing Variance and Standard Deviation
A frequent error is to scale variance by \(c\), as though it were an expected payment. That is incorrect: variance measures squared deviations, so it scales by \(c^2\). Standard deviation, the square root of variance, does scale by \(c\) for a nonnegative share. A second common error is to report \(E[(Y^L)^2]\) as the variance without subtracting \((E[Y^L])^2\).
For example, if \(\operatorname{Var}(X)=900{,}000\text{ dollars}^2\) and \(c=0.5\), the incorrect calculation \(0.5(900{,}000)=450{,}000\) overstates the payment variance. The correct calculation is \(0.5^2(900{,}000)=225{,}000\text{ dollars}^2\). The standard deviation is then \(\sqrt{225{,}000}\), which is one-half of the loss standard deviation \(\sqrt{900{,}000}\).
Check Your Understanding
For each hypothetical single-claim model, use the fixed insurer share and the stated loss distribution. Assume no interest discounting or mortality contingency.
- The loss is \(\$500\) with probability 0.4 and \(\$1{,}500\) with probability 0.6. The insurer pays 50%. Find the payment mean and variance. (Answer: mean \(\$550\); variance \(60{,}000\text{ dollars}^2\).)
- A loss is exponentially distributed with mean \(\$1{,}200\), and the insurer pays 75%. Find the payment mean and variance. (Answer: mean \(\$900\); variance \(810{,}000\text{ dollars}^2\).)
- A loss follows a gamma distribution with shape \(k=3\) and scale \(\theta=\$400\). The insurer pays 50%. Find the payment mean and variance. (Answer: mean \(\$600\); variance \(120{,}000\text{ dollars}^2\).)
- A loss has variance \(640{,}000\text{ dollars}^2\) and mean \(\$2{,}500\). Compare payment means and variances when the insurer’s share is 30% versus 80%. (Answer: at 30%, mean \(\$750\) and variance \(57{,}600\text{ dollars}^2\); at 80%, mean \(\$2{,}000\) and variance \(409{,}600\text{ dollars}^2\).)
- A loss has mean \(\$1{,}800\) and variance \(360{,}000\text{ dollars}^2\). What fixed insurer share gives an expected payment of \(\$990\), and what is the resulting payment variance? (Answer: share \(c=0.55\); variance \(108{,}900\text{ dollars}^2\).)