Calculating the variance of insurance payment

Posted on January 16, 2019 Updated on March 4, 2019

The post supplements a three-part discussion on the mathematical models of insurance payments: part 1, part 2 and part 3. This post focuses on the calculation of the variance of insurance payments.

There are three practice problem sets for the 3-part discussion on the mathematical models of insurance payments – problem set 7, problem set 8 and problem set 9. Problems in these problem sets are on calculation of expected payments. We present several examples in this post on variance of insurance payment. A practice problem set will soon follow.

In contrast, the next post is a discussion on the insurance payment per payment.

Coverage with an Ordinary Deductible

To simplify the calculation, the only limit on benefits is the imposition of a deductible. Suppose that the loss amount is the random variable $X$ . The deductible is $d$ . Given that a loss has occurred, the insurance policy pays nothing if the loss is below $d$ and pays $X-d$ if the loss exceeds $d$ . The payment random variable is denoted by $Y_L$ or $(X-d)_+$ and is explicitly described as follows:

(1)…… $\displaystyle Y_L=(X-d)_+=\left\{ \begin{array}{ll} \displaystyle 0 &\ X \le d \\ \text{ } & \text{ } \\ \displaystyle X-d &\ X > d \end{array} \right.$

The subscript L in $Y_L$ is to denote that this variable is the payment per loss. This means that its mean, $E(Y_L)$ , is the average payment over all losses. A related payment variable is $Y_P$ which is defined as follows:

(2)…… $\displaystyle Y_P=X-d \ \lvert X > d$

The variable $Y_P$ is a truncated variable (any loss that is less than the deductible is not considered) and is also shifted (the payment is the loss less the deductible). As a result, $Y_P$ is a conditional distribution. It is conditional on the loss exceeding the deductible. The subscript P in $Y_P$ indicates that the payment variable is the payment per payment. This means that its mean, $E(Y_P)$ , is the average payment over all payments that are made, i.e. average payment over all losses that are eligible for a claim payment.

The focus of this post is on the calculation of $E(Y_L)$ (the average payment over all losses) and $Var(Y_L)$ (the variance of payment per loss). These two quantities are important in the actuarial pricing of insurance. If the policy were to pay each loss in full, the average amount paid would be $E(X)$ , the average of the loss distribution. Imposing a deductible, the average amount paid is $E(Y_L)$ , which is less than $E(X)$ . On the other hand, $Var(Y_L)$ , the variance of the payment per loss, is smaller than $Var(X)$ , the variance of the loss distribution. Thus imposing a deductible not only reduces the amount paid by the insurer, it also reduces the variability of the amount paid.

The calculation of $E(Y_L)$ and $Var(Y_L)$ can be done by using the pdf $f(x)$ of the original loss random variable $X$ .

(3)…… $\displaystyle E(Y_L)=\int_d^\infty (x-d) \ f(x) \ dx$

(4)…… $\displaystyle E(Y_L^2)=\int_d^\infty (x-d)^2 \ f(x) \ dx$

(5)…… $\displaystyle Var(Y_L)=E(Y_L^2)-E(Y_L)^2$

The above calculation assumes that the loss $X$ is a continuous random variable. If the loss is discrete, simply replace integrals by summation. The calculation in (3) and (4) can also be done by integrating the pdf of the payment variable $Y_L$ .

(6)…… $\displaystyle f_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ y=0 \\ \text{ } & \text{ } \\ \displaystyle f(y+d) &\ y > 0 \end{array} \right.$

(7)…… $\displaystyle E(Y_L)=\int_0^\infty y \ f_{Y_L}(y) \ dy$

(8)…… $\displaystyle E(Y_L^2)=\int_0^\infty y^2 \ f_{Y_L}(y) \ dy$

It will be helpful to also consider the pdf of the payment per payment variable $Y_P$ .

(9)…… $\displaystyle f_{Y_P}(y)=\frac{f(y+d)}{P[X > d]} \ \ \ \ \ \ \ y>0$

Three Approaches

We show that there are three different ways to calculate $E(Y_L)$ and $Var(Y_L)$ .

Using basic principle.
Considering $Y_L$ as a mixture.
Considering $Y_L$ as a compound distribution.

Using basic principle refers to using (3) and (4) or (7) and (8). The second approach is to treat $Y_L$ as a mixture of a point mass of 0 with weight $P(X \le d)$ and the payment per payment $Y_P$ with weight $P(X >d)$ . The third approach is to treat $Y_L$ as a compound distribution where the number of claims $N$ is a Bernoulli distribution with $p=P(X >d)$ and the severity is the payment $Y_P$ . We demonstrate these approaches with a series of examples.

Examples

Example 1
The random loss $X$ has an exponential distribution with mean 50. A coverage with a deductible of 25 is purchased to cover this loss. Calculate the mean and variance of the insurance payment per loss.

We demonstrate the calculation using the three approaches discussed above. The following gives the calculation based on basic principles.

$\displaystyle \begin{aligned} E(Y_L)&=\int_{25}^\infty (x-25) \ \frac{1}{50} \ e^{-x/50} \ dx \\&=\int_{0}^\infty \frac{1}{50} \ u \ e^{-u/50} \ e^{-1/2} \ du \\&=50 \ e^{-1/2} \int_{0}^\infty \frac{1}{50^2} \ u \ e^{-u/50} \ du \\&=50 \ e^{-1/2}=30.33 \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=\int_{25}^\infty (x-25)^2 \ \frac{1}{50} \ e^{-x/50} \ dx \\&=\int_{0}^\infty \frac{1}{50} \ u^2 \ e^{-u/50} \ e^{-1/2} \ du \\&=2 \cdot 50^2 \ e^{-1/2} \int_{0}^\infty \frac{1}{2} \ \frac{1}{50^3} \ u^2 \ e^{-u/50} \ du \\&=2 \cdot 50^2 \ e^{-1/2} \end{aligned}$

$\displaystyle Var(Y_L)=2 \cdot 50^2 \ e^{-1/2}-\biggl( 50 \ e^{-1/2} \biggr)^2=2112.954696$

In the above calculation, we perform a change of variable via $u=x-25$ . We now do the second approach. Note that the variable $Y_P=X-25 \lvert X >25$ also has an exponential distribution with mean 50 (this is due to the memoryless property of the exponential distribution). The point mass of 0 has weight $P(X \le 25)=1-e^{-1/2}$ and the variable $Y_P$ has weight $P(X > 25)=e^{-1/2}$ .

$\displaystyle \begin{aligned} E(Y_L)&=0 \cdot (1-e^{-1/2})+E(Y_P) \cdot e^{-1/2}=50 \ \cdot e^{-1/2} \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=0 \cdot (1-e^{-1/2})+E(Y_P^2) \cdot e^{-1/2} \\&=(50^2+50^2) \cdot e^{-1/2} =2 \ 50^2 \ \cdot e^{-1/2} \end{aligned}$

$\displaystyle Var(Y_L)=2 \cdot 50^2 \ e^{-1/2}-\biggl( 50 \ e^{-1/2} \biggr)^2=2112.954696$

In the third approach, the frequency variable $N$ is Bernoulli with $P(N=0)=1-e^{-1/2}$ and $P(N=1)=e^{-1/2}$ . The severity variable is $Y_P$ . The following calculates the compound variance.

$\displaystyle \begin{aligned} Var(Y_L)&=E(N) \cdot Var(Y_P)+Var(N) \cdot E(Y_P)^2 \\&=e^{-1/2} \cdot 50^2+e^{-1/2} (1-e^{-1/2}) \cdot 50^2 \\&=2 \cdot 50^2 \ e^{-1/2}-50^2 \ e^{-1} \\&=2112.954696 \end{aligned}$

Note that the average payment per loss is $E(Y_L)=30.33$ , a substantial reduction from the mean $E[X]=50$ if the policy pays each loss in full. The standard deviation of $Y_L$ is $\sqrt{2112.954696}=45.97$ , which is a reduction from 50, the standard deviation of original loss distribution. Clearly, imposing a deductible (or other limits on benefits) has the effect of reducing risk for the insurer.

When the loss distribution is exponential, approach 2 and approach 3 are quite easy to implement. This is because the payment per payment variable $Y_P$ has the same distribution as the original loss distribution. This happens only in this case. If the loss distribution is any other distribution, we must determine the distribution of $Y_P$ before carrying out the second or the third approach.

We now work two more examples that are not exponential distributions.

Example 2
The loss distribution is a uniform distribution on the interval $(0,100)$ . The insurance coverage has a deductible of 20. Calculate the mean and variance of the payment per loss.

The following gives the basic calculation.

$\displaystyle \begin{aligned} E(Y_L)&=\int_{20}^{100} (x-20) \ \frac{1}{100} \ dx \\&=\int_0^{80} \frac{1}{100} \ u \ du =32 \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=\int_{20}^{100} (x-20)^2 \ \frac{1}{100} \ dx \\&=\int_0^{80} \frac{1}{100} \ u^2 \ du =\frac{5120}{3} \end{aligned}$

$\displaystyle Var(Y_L)=\frac{5120}{3}-32^2=\frac{2048}{3}=682.67$

The mean and variance of the loss distribution are 50 and $\frac{100^2}{12}=833.33$ (if the coverage pays for each loss in full). By imposing a deductible of 20, the mean payment per loss is 32 and the variance of payment per loss is 682.67. The effect is a reduction of risk since part of the risk is shifted to the policyholder.

We now perform the calculation using the the other two approaches. Note that the payment per payment $Y_P=X-20 \lvert X > 20$ has a uniform distribution on the interval $(0,80)$ . The following calculates according to the second approach.

$\displaystyle \begin{aligned} E(Y_L)&=0 \cdot (0.2)+E[Y_P] \cdot 0.8=40 \ \cdot 0.8=32 \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=0 \cdot (0.2)+E[Y_P^2] \cdot 0.8=\biggl(\frac{80^2}{12}+40^2 \biggr) \ \cdot 0.8=\frac{5120}{3} \end{aligned}$

$\displaystyle Var(Y_L)=\frac{5120}{3}-32^2=\frac{2048}{3}=682.67$

For the third approach, the frequency $N$ is a Bernoulli variable with $p=0.8$ and the severity variable is $Y_P$ , which is uniform on $(0,80)$ .

$\displaystyle \begin{aligned} Var(Y_L)&=E(N) \cdot Var(Y_P)+Var(N) \cdot E(Y_P)^2 \\&=0.8 \cdot \frac{80^2}{12} +0.8 \cdot 0.2 \cdot 40^2 \\&=\frac{2048}{3} \\&=682.67 \end{aligned}$

Example 3
In this example, the loss distribution is a Pareto distribution with parameters $\alpha=3$ and $\theta=1000$ . The deductible of the coverage is 500. Calculate the mean and variance of the payment per loss.

Note that the payment per payment $Y_P=X-500 \lvert X > 500$ also has a Pareto distribution with parameters $\alpha=3$ and $\theta=1500$ . This information is useful for implementing the second and the third approach. First the calculation based on basic principles.

$\displaystyle \begin{aligned} E(Y_L)&=\int_{500}^{\infty} (x-500) \ \frac{3 \cdot 1000^3}{(x+1000)^4} \ dx \\&=\int_{0}^{\infty} u \ \frac{3 \cdot 1000^3}{(u+1500)^4} \ du \\&=\frac{1000^3}{1500^3} \ \int_{0}^{\infty} u \ \frac{3 \cdot 1500^3}{(u+1500)^4} \ du\\&=\frac{8}{27} \ \frac{1500}{2}\\&=\frac{2000}{9}=222.22 \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=\int_{500}^{\infty} (x-500)^2 \ \frac{3 \cdot 1000^3}{(x+1000)^4} \ dx \\&=\int_{0}^{\infty} u^2 \ \frac{3 \cdot 1000^3}{(u+1500)^4} \ du \\&=\frac{1000^3}{1500^3} \ \int_{0}^{\infty} u^2 \ \frac{3 \cdot 1500^3}{(u+1500)^4} \ du\\&=\frac{8}{27} \ \frac{2 \cdot 1500^2}{2 \cdot 1}\\&=\frac{2000000}{3} \end{aligned}$

$\displaystyle Var(Y_L)=\frac{2000000}{3}-\biggl(\frac{2000}{9} \biggr)^2=\frac{50000000}{81}=617283.95$

Now, the mixture approach (the second approach). Note that $P(X > 500)=\frac{8}{27}$ .

$\displaystyle \begin{aligned} E(Y_L)&=0 \cdot \biggl(1-\frac{8}{27} \biggr)+E(Y_P) \cdot \frac{8}{27}=\frac{1500}{2} \ \cdot \frac{8}{27}=\frac{2000}{9} \end{aligned}$

$\displaystyle \begin{aligned} E(Y_L^2)&=0 \cdot \biggl(1-\frac{8}{27} \biggr)+E(Y_P^2) \cdot \frac{8}{27}=\frac{2 \cdot 1500^2}{2 \cdot 1} \ \cdot \frac{8}{27}=\frac{2000000}{3} \end{aligned}$

$\displaystyle Var(Y_L)=\frac{2000000}{3}-\biggl(\frac{2000}{9} \biggr)^2=\frac{50000000}{81}=617283.95$

Now the third approach, which is to calculate the compound variance.

$\displaystyle \begin{aligned} Var(Y_L)&=E(N) \cdot Var(Y_P)+Var(N) \cdot E(Y_P)^2 \\&=\frac{8}{27} \cdot 1687500 +\frac{8}{27} \cdot \biggl(1-\frac{8}{27} \biggr) \cdot 750^2 \\&=\frac{50000000}{81} \\&=617283.95 \end{aligned}$

Remarks

For some loss distributions, the calculation of the variance of $Y_L$ , the payment per loss, can be difficult mathematically. The required integrals for the first approach may not have closed form. For the second and third approach to work, we need to have a handle on the payment per payment $Y_P$ . In many cases, the pdf of $Y_P$ is not easy to obtain or its mean and variance are hard to come by (or even do not exist). For these examples, we may have to find the variance numerically. The examples presented are some of the distributions that are tractable mathematically for all three approaches. These three examples are such that the second and third approaches represent shortcuts for find variance of $Y_L$ because $Y_P$ have a known form and requires minimal extra calculation. For other cases, it is possible that the second or third approach is doable but is not shortcut. In that case, any one of the approaches can be used.

In contrast, the next post is a discussion on the insurance payment per payment.

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Insurance payment per payment vs payment per loss « Practice Problems in Actuarial Modeling said:
January 21, 2019 at 1:53 pm

[…] on the insurance payment per payment and points out how the two payments are related. Another previous post is on how to calculate the variance of the insurance payment per […]

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