# Insurance Loss Models

### Calculating the variance of insurance payment

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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.

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 $Var[Y_L]$.

1. Using basic principle.
2. Considering $Y_L$ as a mixture.
3. 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.

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### Practice Problem Set 9 – Expected Insurance Payment – Additional Problems

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are additional practice problems in addition to Practice Problem Set 7 and Practice Problem Set 8.

 Practice Problem 9A Losses follow a distribution that is a mixture of two equally weighted Pareto distributions, one with parameters $\alpha=2$ and $\theta=2000$ and the other with with parameters $\alpha=2$ and $\theta=4000$. An insurance coverage for these losses has an ordinary deductible of 1000. Calculate the expected payment per loss.
 Practice Problem 9B Losses, prior to any deductible being applied, follow an exponential distribution with mean 17.5. An insurance coverage has a deductible of 8. Inflation of 15% impacts all claims uniformly from the current year to next year. Determine the percentage change in the expected claim cost per loss from the current year to next year.
 Practice Problem 9C Losses follow a distribution that has the following density function. $\displaystyle f(x)=\left\{ \begin{array}{ll} \displaystyle 0.15 &\ \ \ \ 0 An insurance policy is purchased to cover these losses. The policy has a deductible of 3. Calculate the expected insurance payment per payment.
 Practice Problem 9D Losses follow a distribution with the following density function. $\displaystyle f(x)=\frac{1250}{(x+1250)^2} \ \ \ \ \ \ \ \ \ x>0$ An insurance coverage pays losses up to a maximum of 100,000. Determine the average payment per loss.
 Practice Problem 9E You are given the following information. Losses, prior to any application of a deductible, follow a Pareto distribution with $\alpha=3$ and $\theta=5000$. An insurance coverage is purchased to cover these losses. If the size of a loss is between 5,000 and 15,000, the coverage pays for the loss in excess of 5,000. Otherwise, the coverage pays nothing. Determine the average insurance payment per loss.
 Practice Problem 9F You are given the following information. An insurance coverage is purchased to cover a certain type of liability losses. The coverage has a deductible of 1000. Other than the deductible, there are no other coverage modifications. If the deductible is an ordinary deductible, the expected insurance payment per loss is 1,215. If the deductible is a franchise deductible, the expected insurance payment per loss is 1,820. Determine the proportion of the losses that exceed 1,000.
 Practice Problem 9G Losses follow a uniform distribution on the interval $(0, 1000)$. The insurance coverage has a deductible of 250. Determine the variance of the insurance payment per loss.
 Practice Problem 9H Losses follow an exponential distribution with mean 500. An insurance coverage that is designed to cover these losses has a deductible of 1,000. Determine the coefficient of variation of the insurance payment per loss.
 Practice Problem 9I Losses are modeled by an exponential distribution with mean 3,000. An insurance policy covers these losses according to the following provisions. The insured pays 100% of the loss up to 1,000. For the loss amount between 1,000 and 10,000, the insurance pays 80%. The loss amount above 10,000 is paid by the insured until the insured has paid 10,000 in total. For the remaining part of the loss, the insurance pays 90%. Determine the expected insurance payment per loss.
 Practice Problem 9J You are given the following information. The underlying loss distribution for a block of insurance policies is a Pareto distribution with $\alpha=2$ and $\theta=5000$. In the next calendar year, all claims in this block of policies are expected to be impacted uniformly by an inflation rate of 25%. In the next calendar year, the insurance company plans to purchase an excess-of-loss reinsurance policy that caps the insurer’s loss at 10,000 per claim. Determine the insurance company’s expected claim cost per claim after the effective date of the reinsurance policy.

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9A
• $\displaystyle \frac{6800}{3}=2266.67$
9B
• $\displaystyle 1.15 e^{1.2/20.125}=1.22$
9C
• $\displaystyle \frac{10}{3}=3.33$
9D
• 5493.061443
9E
• 312.5
9F
• 0.605
9G
• 61523.4375
9H
• $\sqrt{2 e^2-1}=3.77887956$
9I
• 1642.795495
9J
• $\displaystyle \frac{50000}{13}=3846.15$

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$\copyright$ 2017 – Dan Ma

### Practice Problem Set 8 – Expected Insurance Payment – Additional Problems

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are additional practice problems in addition to Practice Problem Set 7. Another problem set on expected insurance payment: Practice Problem Set 9.

 Practice Problem 8A Losses follow a uniform distribution on the interval $(0,50000)$. An insurance policy has an 80% coinsurance and an ordinary deductible of 10,000. The coinsurance is applied after the deductible so that a positive payment is made on the loss amount above 10,000. Determine the expected payment per loss.
 Practice Problem 8B Losses follow a uniform distribution on the interval $(0,50000)$. An insurance policy has an 80% coinsurance and an ordinary deductible of 10,000. The coinsurance is applied after the deductible so that a positive payment is made on the loss amount above 10,000. In addition to the deductible and coinsurance, the coverage has a policy limit of 24,000 (i.e. the maximum covered loss is 40,000). Determine the expected payment per loss.
 Practice Problem 8C Losses in the current year follow a uniform distribution on the interval $(0,50000)$. Further suppose that inflation of 25% impacts all losses uniformly from the current year to the next year. Losses in the next year are paid according to the following provisions: Coverage has an ordinary deductible of 10,000. Coverage has an 80% coinsurance. The coinsurance is applied after the deductible. The coverage has a policy limit of 24,000. Determine the expected payment per loss.
 Practice Problem 8D Liability claim sizes follow a Pareto distribution with shape parameter $\alpha=1.2$ and scale parameter $\theta=10000$. Suppose that the insurance coverage has a franchise deductible of 20,000 per loss. Given that a loss exceeds the deductible, determine the expected insurance payment.
 Practice Problem 8E Losses in the current year follow a Pareto distribution with parameters $\alpha=3$ and $\theta=5000$. Inflation of 10% is expected to impact these losses in the next year. The coverage for next year’s losses has an ordinary deductible of 1,000. Determine the expected amount per loss in the next year that will be paid by the insurance coverage.
 Practice Problem 8F Losses in the current year follow a Pareto distribution with parameters $\alpha=3$ and $\theta=5000$. Inflation of 10% is expected to impact these losses in the next year. The coverage for next year’s losses has a franchise deductible of 1,000. Determine the expected amount per loss in the next year that will be paid by the insurance coverage.
 Practice Problem 8G Losses follow a distribution that is a mixture of two equally weighted exponential distributions, one with mean 6 and the other with mean 12. An insurance coverage for these losses has an ordinary deductible of 2. Calculate the expected payment per loss.
 Practice Problem 8H Losses follow a distribution that is a mixture of two equally weighted exponential distributions, one with mean 6 and the other with mean 12. An insurance coverage for these losses has a franchise deductible of 2. Calculate the expected payment per loss.
 Practice Problem 8I You are given the following information. Losses follow a distribution with the following cumulative distribution function. $\displaystyle F(x)=1-\frac{1}{3} e^{-2x}-\frac{1}{3} e^{-x}-\frac{1}{3} e^{-x/2} \ \ \ \ x>0$ For each loss, the insurance coverage pays 80% of the portion of the loss that exceeds a deductible of 1. Determine the average payment per loss.
 Practice Problem 8J You are given the following information. Losses follow a lognormal distribution with $\mu=3$ and $\sigma=1.2$. An insurance coverage has a deductible of 10. Determine the percentage change in the expected claim cost per loss when losses are uniformly impacted by a 20% inflation.

All normal probabilities are obtained by using the normal distribution table found here.

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8A
• 12,800
8B
• 12,000
8C
• 14,400
8D
• 170,000
8E
• 1968.9349
8F
• 2574.761038
8G
• $\displaystyle 3 e^{-1/3}+6 e^{-1/6}=7.2285$
8H
• $\displaystyle 4 e^{-1/3}+7 e^{-1/6}=8.7915$
8I
• $\displaystyle 0.8 \biggl(\frac{1}{6} e^{-2}+\frac{1}{3} e^{-1}+\frac{2}{3} e^{-1/2} \biggr)=0.43963$
8J
• Claim Cost before inflation: 32.52697933.
• Claim Cost after inflation: 40.51721002.
• 24.56% change.

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Dan Ma actuarial

$\copyright$ 2017 – Dan Ma

### Practice Problem Set 7 – Expected Insurance Payment

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are basic problems on calculating average insurance payment (per loss or per payment).

Additional problem set: Practice Problem Set 8 and Practice Problem Set 9.

 Practice Problem 7A Losses follow a uniform distribution on the interval $(0,50000)$. An insurance policy has an ordinary deductible of 10,000. Determine the expected payment per loss. Suppose that in addition to the deductible of 10,000, the coverage has a policy limit of 30,000 (i.e. the maximum covered loss is 40,000). Determine the expected payment per loss.
 Practice Problem 7B Losses for the current year follow a uniform distribution on the interval $(0,50000)$. Further suppose that inflation of 25% impacts all losses uniformly from the current year to the next year. Determine the expect insurance payment per loss for the next year assuming that the policy has a deductible of 10,000. Determine the expect insurance payment per loss for the next year assuming that the policy has a deductible of 10,000 and a policy limit of 30,000 (maximum covered loss = 40,000).
 Practice Problem 7C Losses follow an exponential distribution with mean 5,000. An insurance policy covers losses subject to a franchise deductible of 2,000. Determine the expected insurance payment per loss.
 Practice Problem 7D Liability claim sizes follow a Pareto distribution with shape parameter $\alpha=1.2$ and scale parameter $\theta=10000$. Suppose that the insurance coverage pays claims subject to an ordinary deductible of 20,000 per loss. Given that a loss exceeds the deductible, determine the expected insurance payment.
 Practice Problem 7E Losses follow a lognormal distribution with $\mu=7.5$ and $\sigma=1$. For losses below 1,000, no payment is made. For losses exceeding 1,000, the amount in excess of the deductible is paid by the insurer. Determine the expected insurance payment per loss.
 Practice Problem 7F Losses in the current exposure period follow a lognormal distribution with $\mu=7.5$ and $\sigma=1$. Losses in the next exposure period are expected to experience 12% inflation over the current year. Determine the expected insurance payment per loss if the insurance contract has an ordinary deductible of 1,000.
 Practice Problem 7G Losses follow an exponential distribution with mean 2,500. An insurance contract will pay the amount of each claim in excess of a deductible of 750. Determine the standard deviation of the insurance payment for one claim such that a claim includes the possibility that the amount paid is zero.
 Practice Problem 7H Liability losses for auto insurance policies follow a Pareto distribution with $\alpha=3$ and $\theta=5000$. These insurance policies have an ordinary deductible of 1,250. Determine the expected payment made by these insurance policies per loss.
 Practice Problem 7I Liability losses for auto insurance policies follow a Pareto distribution with $\alpha=3$ and $\theta=5000$. These insurance policies make no payment for any loss below 1,250. For any loss greater than 1,250, the insurance policies pay the loss amount in excess of 1,250 up to a limit of 5,000. Determine the expected payment made by these insurance policies per loss.
 Practice Problem 7J Losses follow a lognormal distribution with $\mu=7.5$ and $\sigma=1$. For losses below 1,000, no payment is made. For losses exceeding 1,000, the amount in excess of the deductible is paid by the insurer. Determine the average insurance payment for all the losses that exceed 1,000.

All normal probabilities are obtained by using the normal distribution table found here.

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7A
• 16,000
• 15,000
7B
• 22,050
• 18,000
7C
• 4,692.24
7D
• 150,000
7E
• $0.9441 e^8-722.4 = 2091.92$
7F
• $1.071168 e^8-761.1 = 2432.01$
7G
• 2414.571397
7H
• 1,600
7I
• 1,106.17284
7J
• $\displaystyle \frac{0.9441 e^8-722.4}{0.7224}= 2895.80$

Daniel Ma actuarial

Dan Ma actuarial

$\copyright$ 2017 – Dan Ma

### Mathematical models for insurance payments – part 3 – other modifications

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This post is a continuation of the discussion on models of insurance payments initiated in two previous posts. Part 1 focuses on the models of insurance payments in which the insurance policy imposes a policy limit. Part 2 continues the discussion by introducing models in which the insurance policy imposes an ordinary deductible. In each of these two previous posts, the insurance coverage has only one coverage modification. A more interesting and more realistic scenario would be insurance coverage that contains a combination of several coverage modifications. This post is to examine the effects on the insurance payments as a result of having some or all of these coverage modifications – policy limit, ordinary deductible, franchise deductible and inflation. Additional topics: expected payment per loss versus expected payment per payment and loss elimination ratio.

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Ordinary Deductible and Policy Limit

The previous two posts discuss the expectations $E[X \wedge u]$ and $E[(X-d)_+]$. The first is the expected insurance payment when the coverage has a policy limit $u$. The second is the expected insurance payment when the coverage has an ordinary deductible $d$. They are the expected values of the following two variables.

$\displaystyle X \wedge u=\left\{ \begin{array}{ll} \displaystyle X &\ X \le u \\ \text{ } & \text{ } \\ \displaystyle u &\ X > u \end{array} \right.$

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

It is easy to verify that $X=(X-d)_+ + X \wedge d$. Buying a coverage with an ordinary deductible $d$ and another coverage with policy limit $d$ equals full coverage. Thus we have the following relation.

$(1) \ \ \ \ E[(X-d)_+]=E[X]-E[X \wedge d]$

The limited expectation $E[X \wedge d]$ has expressions in closed form in some cases or has expressions in terms of familiar functions (e.g. gamma function) in other cases. Thus the expectation $E[(X-d)_+]$ can be computed by knowing the original expected value $E[X]$ and the limited expectation $E[X \wedge d]$.

Inflation

Suppose that losses (or claims) in the next period are expected to increase uniformly by $100r$%. For example, $r=0.10$ means 10%. What would be the effect of inflation on the expectations $E[X \wedge u]$ and $E[(X-d)_+]$?

First, the effect on the limited expectation $E[X \wedge u]$. As usual, $X$ is the random loss and $u$ is the policy limit. With inflation rate $r$, the loss variable for the next period would be $(1+r)X$. One approach is to derive the distribution for the inflated loss variable and use the new distribution to calculate $E[(1+r)X \wedge u]$. Another approach is to express it in terms of the limited expectation of the pre-inflated loss $X$. The following is the expectation $E[(1+r)X \wedge u]$, assuming that there is no change in the policy limit.

$\displaystyle (2) \ \ \ \ E[(1+r) X \wedge u]=(1+r) \ E \biggl[X \wedge \frac{u}{1+r} \biggr]$

Relation (2) relates the limited expectation of the inflated variable to the limited expectation of the pre-inflated loss $X$. It says that the limited expectation of the inflated variable $(1+r) X$ is obtained by inflating the limited expectation of the pre-inflated loss but at a smaller policy limit $\frac{u}{1+r}$.

The following is the expectation $E[(1+r) (X-d)_+]=E[(1+r) X]-E[(1+r) X \wedge d]$.

\displaystyle \begin{aligned} (3) \ \ \ \ E[((1+r) X-d)_+]&=E[(1+r) X]-E[(1+r) X \wedge d] \\&=(1+r) E[X]-(1+r) \ E \biggl[X \wedge \frac{d}{1+r} \biggr] \\&=(1+r) \biggl( E[X]- \ E \biggl[X \wedge \frac{d}{1+r} \biggr] \biggr) \\&=(1+r) E \biggl[ \biggl(X-\frac{d}{1+r} \biggr)_+ \biggr] \end{aligned}

Similarly, (3) expresses the expected payment on the inflated loss in terms of the expected payment of the pre-inflated loss. It says that when the loss is inflated, the expected payment per loss is obtained by inflating the expected payment on the pre-inflated loss but at a smaller deductible.

Insurance Payment Per Loss versus Per Payment

The previous post (Part 2) shows how to evaluate the average amount paid to the insured when the coverage has an ordinary deductible. The average payment discussed in Part 2 is the average per loss (over all losses). As a simple illustration, let’s say the amounts of losses in a given period for an insured are 7, 4, 33 and 17 subject to an ordinary deductible of 5. Then the insurance payments are: 2, 0, 28 and 12. The average payment per loss would be (2+0+28+12)/4 = 10.5. If we only count the losses that require a payment, the average is (2+28+12)/3 = 14. Thus the average payment per payment is greater than the average payment per loss since only the losses exceeding the deductible are counted in the average payment per payment. In the calculation discussed here, the average payment per payment is obtained by dividing the average payment per loss by the probability that the loss exceeds the deductible. Note that 10.5/0.75 = 14.

Suppose that the random variable $X$ is the size of the loss (if a loss occurs). Under an insurance policy with an ordinary deductible $d$, the first $d$ dollars of a loss is responsible by the insured and the amount of the loss in excess of $d$ is paid by the insurer. Under such an arrangement, a certain number of losses are not paid by the insurer, precisely those losses that are less than or equal to $d$. If we only count the losses that are paid by the insurer, the payment amount is the conditional random variable $X-d \lvert X>d$. The expected value of this conditional random variable is denoted by $e_X(d)$ or $e(d)$ if the loss $X$ is understood.

Given that $P(X>d)>0$, the variable $X-d \lvert X>d$ is called the excess loss variable. Its expected value $e_X(d)$ is called the mean excess loss function. Other names are mean residual life and complete expectation of life (when the context is that of a mortality study).

For the discussion in this post and other posts in the same series, we use $Y_P$ to denote $X-d \lvert X>d$. The P stands for payment so that its expected value would be average insurance payment per payment, i.e. the expected amount paid given that the loss exceeds the deductible. When the random variable $X$ is the age at death, $e_X(x)$ would be the expected remaining time until death given that the life has survived to age $x$.

The expected value $e_X(d)$ is thus the expected payment per payment (or expected cost per payment) under an ordinary deductible. In contrast, the expected value $E[(X-d)_+]$ is the expected payment per loss (or expected cost per loss), which is discussed in the previous post. The two expected values are related. The calculation of one will give the other. The following compares the two calculation. Let $f_X(x)$ and $F_X(x)$ be the PDF and CDF of $X$, respectively.

$\displaystyle (4) \ \ \ \ E[(X-d)_+]=\int_d^\infty (x-d) \ f_X(x) \ dx$

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$\displaystyle (5) \ \ \ \ e_X(d)=\int_d^\infty (x-d) \ \frac{f_X(x)}{S_X(d)} \ dx=\frac{E[(X-d)_+]}{S_X(d)}$

Note that $e_X(d)$ is calculated using a conditional density function. As a result, $e_X(d)$ can be obtained by dividing the expected payment per loss divided by the probability that there is a payment. Thus the expected payment per payment is the expected payment per loss divided by the probability that there is a payment. This is described in the following relation.

$\displaystyle (6) \ \ \ \ e_X(d)=\frac{E[(X-d)_+]}{S_X(d)}=\frac{E[X]-E[X \wedge d]}{S_X(d)}$

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Distributions of Insurance Payment Variables

Relation (1) and Relation (4) are used for the calculation of expected insurance payment when the coverage has an ordinary deductible. For deriving other information about insurance payment in the presence of an ordinary deductible, it is helpful to know the distributions of the insurance payment (per loss and per payment).

Let $Y_L=(X-d)_+$ (payment per loss) and let $Y_P=X-d \lvert X>d$ (payment per payment). We now discuss the distribution for $Y_P$. First, the following gives the PDF, CDF and the survival function of $Y_L$.

Payment Per Loss
PDF $\displaystyle f_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle F_X(d) &\ \ \ \ y=0 \\ \text{ } & \text{ } \\ \displaystyle f_X(y+d) &\ \ \ \ y > 0 \end{array} \right.$
CDF $\displaystyle F_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ y<0 \\ \text{ } & \text{ } \\ \displaystyle F_X(y+d) &\ \ \ \ y \ge 0 \end{array} \right.$
Survival Function $\displaystyle S_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ y<0 \\ \text{ } & \text{ } \\ \displaystyle S_X(y+d) &\ \ \ \ y \ge 0 \end{array} \right.$

Note that the above functions have a point mass at $y=0$ to account for the losses that are not paid. For $Y_P=X-d \lvert X>d$, there is no point mass at $y=0$. We only need to consider $y>0$. Normalizing the function $f_X(y+d)$ would give the PDF of $Y_P=X-d \lvert X>d$. Thus the following gives the PDF, the survival function and the CDF of $Y_P=X-d \lvert X>d$.

Payment Per Payment
PDF $\displaystyle f_{Y_P}(y)=\frac{f_X(y+d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0$
CDF $\displaystyle F_{Y_P}(y)=\frac{F_X(y+d)-F_X(d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0$
Survival Function $\displaystyle S_{Y_P}(y)=\frac{S_X(y+d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0$

The following calculates the two averages using the respective PDFs, as a result deriving the same relationship between the two expected values.

$\displaystyle (7) \ \ \ \ E(Y_L)=E[(X-d)_+]=\int_0^\infty y \ f_X(y+d) \ dy$

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$\displaystyle (8) \ \ \ \ E(Y_P)=e_X(d)=\int_0^\infty y \ \frac{f_X(y+d)}{S_X(d)} \ dy$

Relation (1) and Relation (7) are identical. The former is calculated using the distribution of the original loss $X$ and the latter is calculated using the distribution of the payment per loss variable. Similarly, compare Relation (5) and Relation (8). The former is computed from the distribution of the unmodified loss $X$ and the latter is computed using the distribution of the payment per payment variable.

Note that Relation (7) and Relation (8) also lead to Relation (6), which relates the expected payment $E[(X-d)_+]$ and the expected payment $e_X(d)$.

With the PDFs and CDFs for the payment per loss and the payment per payment developed, other distributional quantities can be derived, e.g. hazard rate, variance, skewness and kurtosis.

Loss Elimination Ratio

When the insurance coverage has an ordinary deductible, a natural question is (from the insurance company’s perspective), what is the impact of the deductible on the payment that is made to the insured? More specifically, on average what is the reduction of the payment? The loss elimination ratio is the proportion of the expected loss that is not paid to the insured by the insurer. For example, if the expected loss before application of the deductible is 45 (of which 36 is expected to be paid by the insurer), the loss elimination ratio is 9/45 = 0.20 (20%). In this example, the insurer has reduced its obligation by 20%. More formally, the loss elimination ratio (LER) is defined as:

$\displaystyle LER=\frac{E(X)-E[(X-d)_+]}{E(X)}=\frac{E[X \wedge d]}{E(X)}$

LER is the ratio of the expected reduction in payment as a result of imposing the ordinary deductible to the expected payment without the deductible. Though it is possible to define LER as the ratio of the reduction in expected payment as a result of a coverage modification to the expected payment without the modification for a modification other than an ordinary deductible, we do not attempt to further generalize this concept.

Policy with a Limit and a Deductible

In part 1, expected insurance payment under a policy limit is developed. In part 2, expected insurance payment under a policy with an ordinary deductible is developed. We now combine both provisions in the same insurance policy. First, let’s define two terms. A policy limit is the maximum amount that will be paid by a policy. For example, if the policy limit is 10,000, the policy will paid at most 10,000 per loss. If the actual loss is 15,000, then the policy will paid 10,000 and the insured will have to be responsible for the remaining 5,000. On the other hand, maximum covered loss is the level above which no loss will be paid. For example, Suppose that the policy covers up to 10,000 per loss subject to a deductible of 1,000. If the actual loss is 20,000, then the maximum covered loss is 10,000 with the policy limit being 9,000. This is because the policy only covers the first 10,000 with the first 1,000 paid by the insured.

Suppose that an insurance policy has an ordinary deductible $d$, a maximum covered loss $u$ with $d and no other coverage modifications. Any loss below $d$ is not paid by the insurer. For any loss exceeding $d$, the insurer pays the loss amount in excess of $d$ up to the maximum covered loss $u$, with the policy limit being $u-d$. The following describes the payment rule more explicitly.

$\displaystyle (X \wedge u)-(X \wedge d)=\left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ X \le d \\ \text{ } & \text{ } \\ \displaystyle X-d &\ \ \ \ d < X \le u \\ \text{ } & \text{ } \\ \displaystyle u-d &\ \ \ \ X > u \end{array} \right.$

Under such a policy, the maximum amount paid (policy limit) is $u-d$, which is the maximum covered loss minus the deductible. The policy limit is reached when $X > u$. When $d < X \le u$, the payment $X \wedge u$ is $X$ and the payment $X \wedge d$ is $d$. Then the expected payment per loss under such a policy is:

$\displaystyle (9) \ \ \ \ E[X \wedge u]-E[X \wedge d]$

There is not special notation for the expected payment. In words, it is the limited expected value at $u$ (the maximum covered loss) minus the limited expected value at the ordinary deductible $d$. The higher moments can be derived by evaluating $E(Y^k)$ using the PDF of the loss $X$ where $Y=(X \wedge u)-(X \wedge d)$.

The payment $Y=(X \wedge u)-(X \wedge d)$ is on a per loss basis. It is also possible to consider payment or payment by removing the point mass at zero. The expected payment per payment is obtained by dividing (9) by the probability of a positive payment.

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Franchise Deductible

An alternative to the ordinary deductible is the franchise deductible. It works like an ordinary deductible except that when the loss exceeds the deductible, the policy pays the loss in full. The following gives the payment rule.

$\displaystyle Y=\left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ X \le d \\ \text{ } & \text{ } \\ \displaystyle X &\ \ \ \ X > d \end{array} \right.$

Note that when the loss exceeds the deductible ($X > d$), the policy with a franchise deductible pays more than a policy with the same ordinary deductible. By how much? By the amount $d$. Thus the expected payment per loss under a franchise deductible is $E[(X-d)_+]+d \ S_X(d)$. The addition of $d \cdot S_X(d)$ reflects the additional benefit when $X > d$. Instead of deriving calculation specific to franchise deductible, we can derive the payments under franchise deductible by adding the additional benefit appropriately.

$\displaystyle (10) \ \ \ \ E[(X-d)_+]+d \ S_X(d)$

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Benefit Tables

The previous two posts and this post discuss coverage with two types of deductible (ordinary and franchise) as well as coverage that may include a limit. In order to keep the expected payments straight, the following table organizes the different combinations of coverage options discussed up to this point.

A = Ordinary Deductible Only
B = Franchise Deductible Only
C = Ordinary Deductible + Limit
D = Franchise Deductible + Limit

Expected Cost Per Loss Expected Cost Per Payment
A $\displaystyle E[(X-d)_+]=E(X)-E(X \wedge d)$ $\displaystyle \frac{E(X)-E(X \wedge d)}{S_X(d)}=e_X(d)$
B $\displaystyle E(X)-E(X \wedge d)+d \ S_X(d)$ $\displaystyle \frac{E(X)-E(X \wedge d)}{S_X(d)}+d=e_X(d)+d$
C $\displaystyle E(X \wedge u)-E(X \wedge d )$ $\displaystyle \frac{E(X \wedge u)-E(X \wedge d)}{S_X(d)}$
D $\displaystyle E(X \wedge u)-E(X \wedge d )+d \ S_X(d)$ $\displaystyle \frac{E(X \wedge u)-E(X \wedge d)}{S_X(d)}+d$

Row A shows the expected payment in a coverage with an ordinary deductible for both per loss and per payment. The expected per loss payment in Row A is the subject of the previous post and the expected payment per payment is discussed earlier in this post. Row B is for expected payments in a coverage with a franchise deductible. Note that the coverage with a franchise deductible pays more than a coverage with an identical ordinary deductible. Thus Row B is Row A plus the added benefit, which is the deductible $d$.

Row C is for the coverage of an ordinary deductible with a policy limit. The per loss expected payment is $\displaystyle E(X \wedge u)-E(X \wedge d )$. The per payment expected value is a conditional one, conditional on the loss greater than the deductible. Thus the per payment expected value is the per loss expected value divided by $S_X(d)$.

Note that Row D is Row C plus the amount of $d$, the additional benefit as a result of having a franchise deductible instead of an ordinary deductible.

Rather than memorizing these formulas, focus on the general structure of the table. For example, understand the payments involving ordinary deductible (Row A and Row C). Then the payments for franchise deductible are obtained by adding an appropriate additional benefit.

The following table shows the expected payments under the influence of claim cost inflation. The maximum covered loss $u$ and the deductible $d$ are identical to the above benefit table. The losses are subject to a $(100r)$% inflation in the next exposure period. Note that $u^*$ and $d^*$ are the modified maximum covered loss and deductible that will make the formulas work.

E = Ordinary Deductible Only
F = Franchise Deductible Only
G = Ordinary Deductible + Limit
H = Franchise Deductible + Limit

$\displaystyle u^*=\frac{u}{1+r}$

$\displaystyle d^*=\frac{d}{1+r}$

Expected Cost Per Loss (with Inflation) Expected Cost Per Payment (with Inflation)
E $\displaystyle (1+r) \biggl \{ E(X)-E [X \wedge d^* ] \bigg \}$ $\displaystyle (1+r) \biggl \{ \frac{\displaystyle E(X)-E [X \wedge d^* ]}{\displaystyle S_{X} (d^*)} \biggr \}$
F $\displaystyle (1+r) \biggl \{ E(X)-E [X \wedge d^* ]+d^* \ S_{X}(d^* ) \bigg \}$ $\displaystyle (1+r) \biggl \{ \frac{\displaystyle E(X)-E [X \wedge d^* ]}{\displaystyle S_{X} (d^* )}+d^* \biggr \}$
G $\displaystyle (1+r) \biggl \{ E [X \wedge u^* ]-E [X \wedge d^* ] \bigg \}$ $\displaystyle (1+r) \biggl \{ \frac{\displaystyle E [X \wedge u^* ]-E [X \wedge d^*]}{\displaystyle S_{X} (d^* )} \biggr \}$
H $\displaystyle (1+r) \biggl \{ E [X \wedge u^* ]-E [X \wedge d^* ]+d^* \ S_{X} (d^* ) \bigg \}$ $\displaystyle (1+r) \biggl \{ \frac{\displaystyle E[X \wedge u^* ]-E [X \wedge d^* ]}{\displaystyle S_{X} (d^* )}+d^* \biggr \}$

There is really no need to mathematically derive the formulas for the second insurance benefit table (Rows E through H). Recall the effect of inflation on the expected payments $E[X \wedge u]$ and $E[ (X-d)_+]$ (see Relation (2) and Relation (3)). Then apply the inflation effect on the first insurance benefit table. Note that the second table is obtained by inflating the first table by $1+r$ but with smaller upper limit $u^*$ and deductible $d^*$. Also note the relation between Row E and Row F (same relation between Row A and Row B) and the relation between Row G and Row H (same relation between Row C and Row D).

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Calculation

One approach of modeling insurance payments is to assume that the unmodified insurance losses are from a catalog of parametric distributions. For certain parametric distributions, the limited expectations $E[X \wedge d]$ have convenient forms. Examples are: exponential distribution, Pareto distribution and lognormal distribution. Other distributions do not close form for $E[X \wedge d]$ but the limited expectation can be expressed as a function that can be numerically calculated. One example is the gamma function.

The table in this link is an inventory of distributions that may be useful in modeling insurance losses. Included in the table are the limited expectations for various distributions. The following table shows the limited expectations for exponential, Pareto and lognormal.

Limited Expectation
Exponential $E[X \wedge x]=\theta (1-e^{-x/\theta})$
Pareto $\displaystyle E[X \wedge x]=\frac{\theta}{\alpha-1} \biggl[1-\biggl(\frac{\theta}{x+\theta} \biggr)^{\alpha-1} \biggr]$, $\alpha \ne 1$
Lognormal $\displaystyle E[(X \wedge x)^k]=\exp(k \mu+k^2 \sigma^2/2) \Phi \biggl(\frac{\log(x)-\mu-k \sigma^2}{\sigma} \biggr)+x^k \biggl[1-\Phi \biggl(\frac{\log(x)-\mu}{\sigma} \biggr) \biggr]$

See the table in this link for the limited expectations of the distributed not shown in the above table. Once $E[X \wedge d]$ is computed or estimated, the expected payment for various types of coverage provisions can be derived.

Examples

The first example demonstrates the four categories of expected payments in the table in the preceding section.

Example 1
Suppose that the loss distribution is described by the PDF $\displaystyle f_X(x)=\frac{1}{5000} (100-x)$ with support $0. Walk through all the expected payments discussed in the above table (Rows A through D) assuming $d=12$ and $u=60$.

First, the basic calculation.

$\displaystyle E(X)=\int_0^{100} x \ \frac{1}{5000} (100-x) \ dx=\frac{100}{3}=33.3333$

$\displaystyle F_X(x)=\frac{1}{5000} (100x- \frac{1}{2} x^2); \ \ \ 0

$F_X(12)=0.2256$

$S_X(12)=1-0.2256=0.7744$

$F_X(60)=0.84$

$S_X(60)=1-0.84=0.16$

$\displaystyle E(X \wedge 12)=\int_0^{12} x \ \frac{1}{5000} (100-x) \ dx +12 \cdot S_X(12)=\frac{6636}{625}$

$\displaystyle E(X \wedge 60)=\int_0^{60} x \ \frac{1}{5000} (100-x) \ dx +12 \cdot S_X(60)=\frac{12864}{625}$

Four categories of expected payments are derived and are shown in the following table.

A = Ordinary Deductible Only
B = Franchise Deductible Only
C = Ordinary Deductible + Limit
D = Franchise Deductible + Limit

Expected Cost Per Loss Expected Cost Per Payment
A $\displaystyle \frac{100}{3}-\frac{6636}{625}=\frac{42592}{1875}=22.7157$ $\displaystyle e_X(d)=\frac{42592}{1875} \frac{1}{0.7744}=\frac{10648}{363}=29.3333$
B $\displaystyle \frac{42592}{1875}+12 \cdot 0.7744=\frac{60016}{1875}=32.0085$ $\displaystyle \frac{10648}{363}+12=\frac{15004}{363}=41.3333$
C $\displaystyle \frac{12864}{625}-\frac{6636}{625}=\frac{6228}{625}=9.9648$ $\displaystyle \frac{6228}{625} \frac{1}{0.7744}=\frac{6228}{484}=12.8678$
D $\displaystyle \frac{6228}{625}+12 \cdot 0.7744=\frac{12036}{625}=19.2576$ $\displaystyle \frac{6228}{484}+12=\frac{12036}{484}=24.8678$

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Example 2
Suppose that a coverage has an ordinary deductible and suppose that the CDF of the insurance payment per loss is given by:

$\displaystyle G(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ y<0 \\ \text{ } & \text{ } \\ \displaystyle 1-e^{-\frac{y+20}{100}} &\ \ \ \ y \ge 0 \end{array} \right.$

Determine the expected value and the variance of the insurance cost per loss.

Note that there is a jump in the CDF at $y=0$. Thus $y=0$ is a point mass. The following is the PDF of the insurance payment.

$\displaystyle g(y)=\left\{ \begin{array}{ll} \displaystyle 1-e^{-1/5} &\ \ \ \ y=0 \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{100} e^{-\frac{y+20}{100}}=e^{-1/5} \ \frac{1}{100} e^{-y/100} &\ \ \ \ y > 0 \end{array} \right.$

It is clear that $g(y)=f(y+20)$ where $f(y)$ is the PDF of the exponential distribution with mean 100. Thus $g(y)$ is the PDF of the insurance payment per loss when the coverage has an ordinary deductible of 20 where the loss has exponential distribution with mean 100. We can use $g(y)$ to calculate the mean and variance of $Y$. Using $g(y)$, the mean and variance are:

\displaystyle \begin{aligned} E(Y)&=\int_0^\infty y \ e^{-1/5} \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ \int_0^\infty y \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ (100)=81.873 \end{aligned}

\displaystyle \begin{aligned} E(Y^2)&=\int_0^\infty y^2 \ e^{-1/5} \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ \int_0^\infty y^2 \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ 2(100)^2 \end{aligned}

$\displaystyle Var(Y)=e^{-1/5} \ 2(100)^2-\biggl( e^{-1/5} \ (100) \biggr)^2=9671.414601$

Note that the calculation for $E(Y)$ and $E(Y^2)$ is done by multiplying $e^{-1/5}$ by mean and second moment of the exponential distribution with mean 100.

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Example 3
You are given the following:

• Losses follow a Pareto distribution with parameters $\alpha=3$ and $\theta=500$.
• The coverage has an ordinary deductible of 100.

Determine the PDF and CDF of the payment to the insured per payment. Determine the expected cost per payment.

Note that the setting of this example is identical to Example 3 in this previous post. Since this only concerns the insurance when the loss exceeds the deductible, the insurance payment is the conditional distribution $Y=X-100 \lvert X>100$ where $X$ is the Pareto loss distribution.

The following gives the PDF and CDF and other information of the loss $X$.

$\displaystyle f_X(x)=\frac{3 \ 500^3}{(x+500)^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ x> 0$

$\displaystyle F_X(x)=1-\biggl(\frac{500}{x+500} \biggr)^3 \ \ \ \ \ \ x> 0$

$\displaystyle S_X(x)=\biggl(\frac{500}{x+500} \biggr)^3 \ \ \ \ \ \ \ \ \ \ \ \ x> 0$

$\displaystyle S_X(100)=\biggl(\frac{500}{100+500} \biggr)^3=\biggl(\frac{500}{600} \biggr)^3=\frac{125}{216}$

The following shows the PDF and CDF for the payment per payment variable $Y$.

$\displaystyle f_Y(y)=\frac{f_X(y+100)}{S_X(100)}=\frac{3 \ 500^3}{(y+100+500)^4} \ \biggl(\frac{600}{500} \biggr)^3=\frac{3 \cdot 600^3}{(y+600)^4} \ \ \ \ \ \ \ \ \ y> 0$

$\displaystyle S_Y(y)=\frac{S_X(y+100)}{S_X(100)}=\biggl(\frac{500}{y+100+500} \biggr)^3 \ \biggl(\frac{600}{500} \biggr)^3=\biggl(\frac{600}{y+600} \biggr)^3 \ \ \ \ \ \ y> 0$

$\displaystyle F_Y(y)=1-S_Y(y)=1-\biggl(\frac{600}{y+600} \biggr)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y> 0$

Note that these PDF and CDF are for a Pareto distribution with $\alpha=3$ and $\theta=600$. Thus the expected payment per payment and the variance are:

$\displaystyle E(Y)=\frac{600}{3-1}=300$

$\displaystyle E(Y^2)=\frac{600^2 \cdot 2}{(3-1) (3-2)}=600^2$

$Var(Y)=600^2 - 300^2=270000$

Example 4
Suppose that losses $X$ follow a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta$. Suppose that an insurance coverage pays each loss subject to an ordinary deductible $d$. Derive a formula for the expected payment per payment $e_X(d)$.

Let $Y=X-d \lvert X>d$ be the payment per payment. In Example 3, we see that the CDF and PDF of $Y$ are also Pareto but with shape parameter $\alpha$ (same as for $X$) and scale parameter $\theta+d$ (the original scale parameter plus the deductible). Thus the following gives the expected payment per payment.

$\displaystyle e_X(d)=\frac{\theta+d}{\alpha-1}=\frac{\theta}{\alpha-1}+\frac{1}{\alpha-1} d$

Note that $e_X(d)$ in this case is an increasing linear function of the deductible $d$. The higher the deductible, the larger the expected payment. This is a clear sign that the Pareto distribution is a heavy tailed distribution.

Example 5
Suppose that losses follow a lognormal distribution with parameters $\mu=5$ and $\sigma=0.6$. An insurance coverage has an ordinary deductible of 100. Compute the expected payment per loss. Compute the expected payment per loss if the deductible is a franchise deductible.

The following calculates $E(X)$ and $E[X \wedge 100]$.

$E(X)=e^{5+0.6^2/2}=e^{5.18}=177.682811$

\displaystyle \begin{aligned} E[X \wedge 100]&=e^{5+0.6^2/2} \ \Phi \biggl(\frac{\log(100)-5-0.6^2}{0.6} \biggr)+100 \ \biggl[1-\Phi \bigg(\frac{\log(100)-5}{0.6} \biggr) \biggr] \\&=e^{5.18} \ \Phi (-1.26)+100 \ [1-\Phi (-0.66 )]\\&=e^{5.18} \ 0.1038+100(0.7454)=e^{5.18} \ 0.1038+74.54=92.98347578 \end{aligned}

As a result, the expected payment per loss is $E(X)-E[X \wedge 100]=84.69933522=84.70$. If the deductible of 100 is a franchise deductible, the expected payment per loss is obtained by adding an additional insurance payment, which is $d \ S(d)$. The following is the added payment.

\displaystyle \begin{aligned} \text{added payment}&=100 \ \biggl[1-\Phi \bigg(\frac{\log(100)-5}{0.6} \biggr) \biggr] \\&=100 \ \biggl[1-\Phi (-0.66 ) \biggr] \\&=100 (0.7454) \\&=74.54 \end{aligned}

Thus the expected payment per loss with a franchise deductible of 250 is 84.70 + 74.54 = 159.24.

Practice Problems

The following practice problem sets are to reinforce the concepts discussed here.

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models of insurance payment

ordinary deductible, franchise deductible

maximum covered loss

Daniel Ma Math

Daniel Ma Mathematics

Dan Ma Actuarial

$\copyright$ 2017 – Dan Ma

### Practice Problem Set 5 – Exercises for Severity Models

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This problem set has exercises to reinforce the various parametric continuous probability models discussed in the companion blog on actuarial modeling. Links are given below for the models involved.

This blog post in Topics in Actuarial Modeling has a catalog for continuous models.

 Practice Problem 5A Claim amounts for collision damages to insured cars are mutually independent random variables with common density probability density function $\displaystyle f(x)=\left\{ \begin{array}{ll} \displaystyle \frac{1}{1500} \ e^{-x/1500} &\ X > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \text{otherwise} \end{array} \right.$ For three claims that are expected to be made, calculate the expected value of the largest of the three claims.
 Practice Problem 5B The lifetime of an electronic device is modeled using the random variable $Y=20 X^{3}$ where $X$ is an exponential random variable with mean 0.5. Determine the variance of $Y$.
 Practice Problem 5C The lifetime (in years) of an electronic device is $2X +Y$ where $X$ and $Y$ are independent exponentially distributed random variables with mean 3.5. Determine the probability density function of the lifetime of the electronic device.
 Practice Problem 5D The time (in years) until the failure of a machine that is brand new is modeled by a Weibull distribution with shape parameter 1.5 and scale parameter 4. Calculate the 95th percentile of times to failure of the machines that are 2-year old.
 Practice Problem 5E The size of a bodily injury claim for an auto insurance policy follows a Pareto Type II Lomax distribution with shape parameter 2.28 and scale parameter 200. Calculate the proportion of claims that are within one-fourth standard deviations of the mean claim size.
 Practice Problem 5F Suppose that the size of a claim has the following density function. $\displaystyle f(x)=\left\{ \begin{array}{ll} \displaystyle \frac{1}{2.5 \ x \ \sqrt{2 \pi}} \ e^{- z^2/2} &\ x > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \text{otherwise} \end{array} \right.$ where $\displaystyle z=\frac{\text{ln}(x)-1.1}{2.5}$. A coverage pays claims subject to an ordinary deductible of 20. Determine the expected amount paid by the coverage per claim.
 Practice Problem 5G An actuary determines that sizes of claims from a large portfolio of insureds are exponentially distributed. For about 60% of the claims, the claim sizes are modeled by an exponential distribution with mean 1.2. For about 30% of the claims, the claim sizes are modeled by an exponential distribution with mean 2.8. For the remaining 10% of the claims, the claim sizes are considered high claim sizes and are modeled by an exponential distribution with mean 7.5. Determine the variance of the size of a claim that is randomly selected from this portfolio.
 Practice Problem 5H Losses are modeled by a loglosistic distribution with shape parameter $\gamma=2$ and scale parameter $\theta=10$. When a loss occurs, an insurance policy reimburses the loss in excess of a deductible of 5. Determine the 75th percentile of the insurance company reimbursements over all losses.
 Practice Problem 5I Losses are modeled by a loglosistic distribution with shape parameter $\gamma=2$ and scale parameter $\theta=10$. When a loss occurs, an insurance policy reimburses the loss in excess of a deductible of 5. Determine the 75th percentile of the insurance company reimbursements over all payments.
 Practice Problem 5J Claim sizes for a certain class of auto accidents are modeled by a uniform distribution on the interval $(0, 10)$. Five accidents are randomly selected. Determine the expected value of the median of the five accident claims.

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Problem Links for the relevant distributions
5A Exponential distribution
5B Exponential distribution
5C Hypoexponential distribution
5D Weibull distribution
5E Pareto distribution
5F Lognormal distribution and limited expectation
5G Hyperexponential distribution
5H Loglogistic distribution
5I Loglogistic distribution
5J Order statistics

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5A 2,750
5B 4,275
5C $\displaystyle \frac{2}{7} \ e^{-x/7}-\frac{1}{3.5} \ e^{-x/3.5}$
5D 8.954227
5E 0.4867
5F 61.106
5G 12.3459
5H 12.32
5I 15
5J 5

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

$\copyright$ 2017 – Dan Ma

### Mathematical models for insurance payments – part 1 – policy limit

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Suppose an individual (or an entity) faces a random loss and the loss is modeled by the random variable $X$. The individual purchases an insurance coverage to cover this loss. If the policy pays the loss in full (if a loss occurs), then the expected insurance payment is mean loss $E(X)$. Usually the expected insurance payment is less than $E(X)$ due to the presence of policy provisions such as a deductible and/or limit. This post is the first post in a series of posts in discussing the probability models of insurance payments.

Policy Limit

In this post, we assume that the random loss $X$ is a continuous random variable taking on the positive real numbers or numbers from an interval such as $(0,M)$. The methodology can be adjusted to handle discrete loss variable (mostly replacing integrals with summation).

Suppose that the random variable $X$ is the size of the loss (if a loss occurs). Under an insurance policy with a limit $u$, the insurer pays the loss in full if the loss $X$ is less than the limit. Furthermore if the loss is $u$ or greater, the insurance payment is capped at $u$. Under this policy provision, what is the expected amount of insurance payment if there is a loss? The insurance payment in the presence of a limit $u$ is denoted by $X \wedge u$. This is called the limited loss random variable.

$\displaystyle X \wedge u=\left\{ \begin{array}{ll} \displaystyle X &\ X \le u \\ \text{ } & \text{ } \\ \displaystyle u &\ X > u \end{array} \right.$

First, let’s consider the mean $E(X \wedge u)$ (called limited expectation). Let $f(x)$, $F(x)$ and $S(x)=1-F(x)$ be the probability density function (PDF), cumulative distribution function (CDF) and the survival function of the random loss, respectively.

\displaystyle \begin{aligned} E(X \wedge u)&=\int_0^u x \ f(x) \ dx+ u \ S(u) \\&=\int_0^u S(x) \ dx \end{aligned}

The first integral is based on the definition of the limited loss variable, summing up the payments in the interval $(0,u)$ and in the interval $(0, \infty)$. The second integral is an alternative way to evaluate the first integral (see here for more explanation). The following gives the second moment, which can be used in evaluating the variance of insurance payment.

\displaystyle \begin{aligned} E[(X \wedge u)^2]&=\int_0^u x^2 \ f(x) \ dx+ u^2 \ S(u) \\&=\int_0^u 2x \ S(x) \ dx \end{aligned}

$Var(X \wedge u)=E[(X \wedge u)^2]-E(X \wedge u)^2$

Once again, the second moment is also expressed using the alternative calculation. The following two examples demonstrate the evaluation of the first two moments of $X \wedge u$.

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Basic Example

Example 1
Suppose that the loss distribution has PDF $f(x)=1/5 (1- x/10) \ \ \ 0. Evaluate $E(X)$ and $E(X \wedge 4)$

The CDF of the loss $X$ is $F(x)=1/5 (x -x^2/20)$.

$\displaystyle E(X)=\int_0^{10} x \ 1/5 (1- x/10) \ dx=\frac{1}{5} \biggl[\frac{x^2}{2}-\frac{x^3}{30} \biggr]_0^{10}=\frac{10}{3}=3.33$

\displaystyle \begin{aligned} E(X \wedge 4)&=\int_0^4 x \ 1/5 (1- x/10) \ dx +4 \ S(4)\\&=\int_0^4 1/5 (x -x^2/10) \ dx+4 \ S(4)\\&=\frac{1}{5} \biggl[\frac{x^2}{2}-\frac{x^3}{30} \biggr]_0^4+4 \ \frac{9}{25} \\&=\frac{196}{75}=2.6133 \end{aligned}

\displaystyle \begin{aligned} E(X \wedge 4)&=\int_0^4 \biggl[1-\frac{1}{5} \biggl(x-\frac{x^2}{20} \biggr) \biggr] \ dx \\&=\int_0^4 \biggl[1-\frac{x}{5}+\frac{x^2}{100} \biggr] \ dx \\&=x-\frac{x^2}{10}+\frac{x^3}{300} \biggr|_0^4 \\&=\frac{196}{75}=2.6133 \end{aligned}

The limited expected value is calculated in two ways to demonstrate that the two approaches are equivalent. Without the policy limit, the expected loss is 3.3333. With the benefit payment capped at 4, the expected amount paid by the insurance policy is 2.6133 (per loss). The difference is 3.3333 – 2.6133 = 0.72, which is the expected loss responsible by the insured.

Example 2
Continue with Example 1. Calculate the variance of the insurance payment $X \wedge 4$.

\displaystyle \begin{aligned} E[(X \wedge 4)^2]&=\int_0^4 x^2 \ 1/5 (1- x/10) \ dx +16 \ S(4)\\&=\int_0^4 1/5 (x^2 -x^3/10) \ dx+16 \ \frac{9}{25} \\&=\frac{1}{5} \biggl[\frac{x^3}{3}-\frac{x^4}{40} \biggr]_0^4+16 \ \frac{9}{25} \\&=\frac{656}{75} \end{aligned}

$\displaystyle Var(X \wedge 4)=\frac{656}{75}- \biggl(\frac{196}{75} \biggr)^2=\frac{10784}{5625}=1.9172$.

To compare, the variance of $X$ is $\displaystyle Var(X)=\frac{50}{9}=5.5556$ as calculated below.

\displaystyle \begin{aligned} E[X^2]&=\int_0^{10} x^2 \ 1/5 (1- x/10) \ dx \\&=\int_0^{10} 1/5 (x^2 -x^3/10) \ dx \\&=\frac{1}{5} \biggl[\frac{x^3}{3}-\frac{x^4}{40} \biggr]_0^{10}=\frac{50}{3} \end{aligned}

$\displaystyle Var(X)=\frac{50}{3}- \biggl(\frac{10}{3} \biggr)^2=\frac{50}{9}=5.5556$

The example demonstrates that policy provisions such as limit has a variance reducing effect. Had the insurer been liable to pay for the entire loss amount, there would be a greater fluctuation in the payment.

A Censored Variable

Mathematically, the random variable $X \wedge u$ is an upper censored random variable. Any realized value of $X$ above the limit is known only known as the limit $u$ (as far as payment is concerned). On the other hand, applying a limit on the loss $X$ turns the continuous random variable $X$ into a mixed random variable. In the interval $(0,u)$, $X \wedge u$ is continuous while there is a point mass at the point $X=u$ with probability mass $P[(X \wedge u)=u]=S(u)$.

The policy provisions such as deductible and limit have the effect of turning the unmodified loss variable $X$ into censored or truncated variables. As a result, these variables are mixed random variables. The subsequent posts will further demonstrate this point.

Parametric Distributions

There is a vast inventory of parametric distributions that are potential candidates for models of random losses. For example, the table in this link is an inventory of distributions that may be useful in modeling insurance losses. Included in the table are the limited expectations for various distributions. The following table shows three of the distributions. They are listed here primarily because the calculation is tractable and is better suited for demonstration of the concept.

Limited Expectation
• Exponential: $E[X \wedge x]=\theta (1-e^{-x/\theta})$
• Pareto: $\displaystyle E[X \wedge x]=\frac{\theta}{\alpha-1} \biggl[1-\biggl(\frac{\theta}{x+\theta} \biggr)^{\alpha-1} \biggr]$, $\alpha \ne 1$
• Lognormal: $\displaystyle E[(X \wedge x)^k]=\exp(k \mu+k^2 \sigma^2/2) \Phi \biggl(\frac{\log(x)-\mu-k \sigma^2}{\sigma} \biggr)+x^k \biggl[1-\Phi \biggl(\frac{\log(x)-\mu}{\sigma} \biggr) \biggr]$

For formulas for the higher moments, refer to the table. This table will be used throughout the discussion (in subsequent blog posts) on estimating insurance payments. Here’s some blog posts on these three distributions – exponential, Pareto and lognormal.

The following example shows how these formulas are used.

Example 3
Evaluate $E(X \wedge u)$ for the following loss distributions.

• Exponential: mean 1000, $u$ = 2000.
• Pareto: mean 5, variance 75, $u$ = 10.
• Lognormal: mean 177.682811, variance 13680.72152, $u$ = 250.

The exponential distribution with mean 1000 has parameter $\theta=1000$ (the scale parameter). The following gives the limited expected value.

$E(X \wedge 2000)=1000 (1-e^{-2000/1000})=864.6647$

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The Pareto distribution with mean 5 and variance 75 translates to the parameters $\alpha=3$ and $\theta=10$. The following gives the limited expected value.

$\displaystyle E(X \wedge 10)=\frac{10}{3-1} \biggl[1-\biggl(\frac{10}{10+10} \biggr)^{3-1} \biggr]=3.75$.

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The lognormal distribution with mean 177.682811 and variance 13680.72152 corresponds to $\mu=5$ and $\sigma=0.6$. The following gives the limited expected value.

\displaystyle \begin{aligned} E[X \wedge 250]&=e^{5+0.6^2/2} \ \Phi \bigg(\frac{\log(250)-5-0.6^2}{0.6} \biggr)+250 \ \biggl[1-\Phi \bigg(\frac{\log(250)-5}{0.6} \biggr) \biggr] \\&=e^{5.18} \ \Phi (0.269)+250 \ [1-\Phi (0.869 )]\\&=e^{5.18} \ 0.6064+250(1-0.8072)=155.7969 \end{aligned}

The discussion here is just the beginning. The limited expected value $E(X \wedge u)$ is for a simple policy provision, having just one modification of loss. It is a building block for other payments under more complicated insurance payment rules.
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$\copyright$ 2017 – Dan Ma