Practice Problem Set 13 – variance of insurance payment per loss

The practice problems in this post are to reinforce the discussion of the calculation of the variance of insurance payment per loss presented in this blog post.

Let $X$ be a loss random variable. An insurance coverage with an ordinary deductible of $d$ is purchased to cover the loss $X$. Of interest is the insurance payment per loss random variable $Y_{L}$. The variable $Y_{L}$ is 0 if the loss $X$ is below the deductible $d$ and the variable $Y_{L}$ is $X-d$ if the loss $X$ exceeds the deductible $d$. The previous post demonstrates three approaches for calculating the variance of the payment per loss $Y_{L}$.

The problems below are to calculate the variance of $Y_{L}$ in one of the three approaches:

1. using basic principle,
2. considering $Y_{L}$ as a mixture,
3. considering $Y_{L}$ as a compound distribution.

For extra practice, as much as possible, we encourage that the calculation for $Y_{L}$ is to be worked out in all three different methods.

 Practice Problem 13-A Losses follow a uniform distribution on the interval $(0, 5000)$. The insurance coverage has a deductible of 2,000. Determine the mean and the variance of the random loss amount (prior to the application of the deductible). Determine the mean and the variance of the insurance payment per loss.
 Practice Problem 13-B Losses follow a distribution with the following density function: $\displaystyle f(x)=\frac{1}{200} ( 20-x )$…………….$0 The insurance coverage has a deductible of 4. Determine the mean and the variance of the random loss amount (prior to the application of the deductible). Determine the mean and the variance of the insurance payment per loss.
 Practice Problem 13-C Losses $X$ follow a distribution with the following density function: $\displaystyle f(x)=\left\{ \begin{array}{ll} \displaystyle 0.008 &\ \ \ \ 0 The insurance coverage has a deductible of 40. Determine $E(X)$ and $Var(X)$, the mean and the variance of random loss amount prior to the application of the deductible. Using basic principle, determine $E(Y_L)$ and $Var(Y_L)$, the mean and the variance of the insurance payment per loss, respectively. Determine $E(Y_P)$ and $Var(Y_P)$, the mean and the variance of the insurance payment per payment, respectively. Note that $Y_P$ is the conditional random variable $X-40 \lvert X>40$. Using the information about $Y_P$ to derive $E(Y_L)$ and $Var(Y_L)$.
 Practice Problem 13-D Losses $X$ follow an exponential distribution with mean 20. The insurance coverage has a deductible of 10. Determine $E(X)$ and $Var(X)$, the mean and the variance of the random loss amount (prior to the application of the deductible). Determine $E(Y_L)$ and $Var(Y_L)$, the mean and the variance of the insurance payment per loss, respectively.
 Practice Problem 13-E The loss $X$ follows a Pareto distribution with parameters $\alpha=3$ and $\theta=500$. The deductible of the coverage is 100. Determine $E(X)$ and $Var(X)$, the mean and the variance of random loss amount prior to the application of the deductible. Using basic principle, determine $E(Y_L)$ and $Var(Y_L)$, the mean and the variance of the insurance payment per loss, respectively. Determine $E(Y_P)$ and $Var(Y_P)$, the mean and the variance of the insurance payment per payment, respectively. Note that $Y_P$ is the conditional random variable $X-100 \lvert X>100$. Using the information about $Y_P$ to derive $E(Y_L)$ and $Var(Y_L)$. Note. The Pareto distribution used in this problem is the Pareto Type II distribution. For more information, see this post.
 Practice Problem 13-F The random loss $X$ is modeled by a mixture of two exponential random variables with the first one having mean 10 (weight 80%) and the second one having mean 50 (weight 20%). The deductible of the coverage is 5. Determine $E(X)$ and $Var(X)$, the mean and the variance of random loss amount prior to the application of the deductible. Determine $E(Y_L)$ and $Var(Y_L)$, the mean and the variance of the insurance payment per loss, respectively. Note: use any of the three methods.

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Answers

……….
13-A
• $E(X)=2500$ and $Var(X)=\frac{6250000}{3}=2083333.333$
• $E(Y_L)=900$, $E(Y_L^2)=1800000$ and $Var(Y_L)=990000$
13-B
• $E(X)=\frac{20}{3}=6.667$, $E(X^2)=\frac{200}{3}$ and $Var(X)=\frac{200}{9}=22.2222$
• $E(Y_L)=\frac{256}{75}=3.41333$, $E(Y_L^2)=\frac{2048}{75}$ and $Var(Y_L)=\frac{88064}{5625}=15.655822$
13-C
• $E(X)=57.5$, $E(X^2)=\frac{12625}{3}$ and $Var(X)=\frac{2706.25}{3}=902.08333$
• $E(Y_L)=23.9$, $E(Y_L^2)=\frac{3113}{3}$ and $Var(Y_L)=\frac{1399.37}{3}=466.45667$
• $E(Y_P)=\frac{23.9}{0.68}$, $E(Y_P^2)=\frac{3113}{3 \times 0.68}$
13-D
• $E(X)=20$ and $Var(X)=400$
• $E(Y_L)=20 e^{-\frac{10}{20}}=12.1306$, $E(Y_L^2)=800 e^{-\frac{10}{20}}$ and $Var(Y_L)=800 e^{-\frac{10}{20}}-(20 e^{-\frac{10}{20}})^2=338.07275$
13-E
• $E(X)=250$, $E(X^2)=500^2$ and $Var(X)=187500$
• $E(Y_L)=\frac{3125}{18}=173.6111$, $E(Y_L^2)=\frac{625000}{3}=$ and $Var(Y_L)=\frac{57734375}{324}=178192.5154$
13-F
• $E(X)=18$, $E(X^2)=1160$ and $Var(X)=836$
• $E(Y_L)=8 e^{-\frac{5}{10}}+10 e^{-\frac{5}{50}}=13.9$, $E(Y_L^2)=160 e^{-\frac{5}{10}}+1000 e^{-\frac{5}{50}}$ and $Var(Y_L)=E(Y_L^2)-[E(Y_L)]^2=808.655$

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Insurance payment per payment vs payment per loss

This post picks up where this previous post leaves off. The previous post is the second part of a 3-part discussion on mathematical model of insurance payment. The previous post ends with pointing out the difference between payment per loss and payment per payment. This post focuses 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 loss.

Comparing the Two Payments

The insurance coverage being considered here is that losses are paid subject to an ordinary deductible $d$. If a loss is less than the deductible, the coverage pays nothing and if a loss exceeds the deductible, the coverage pays the loss less the deductible. This is called the insurance payment per loss variable and is 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 L in the subscript of $Y_L$ indicates that the variable is the payment per loss. The variable $Y_L$ includes the probability $P(X \le d)$ as well as the probability $P(X > d)$. Thus the average $E(Y_L)$ is the average over all losses, hence the name payment per loss.

The other payment variable of interest is the payment per payment called $Y_P$. This variable captures the losses that require a claim payment, i.e. all losses exceeding the deductible. For $Y_P$, losses below the deductible are not considered and are truncated.

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

The variable $Y_P$ is truncated and shifted. It is truncated below at $d$ and is shifted to the left by the amount $d$. The expected value $E(Y_P)$ is the average of all losses that require a payment, or the average of all payments that are made, hence the name payment per payment. The average $E(Y_P)$ is also called the mean excess loss function.

(3)……$\displaystyle e_X(d)=E(Y_P)=E(X-d \ \lvert X > d )$

When the $X$ is understood, we may use the notation $e(d)$ instead of $e_X(d)$. When the random variable $X$ is a distribution for insurance losses, the function $e_X(d)$ is interpreted as mean excess loss – the average portion of a loss that is in excess of the deductible. If $X$ represents a distribution of lifetime, then the expectation is called the mean residual life or complete expectation of life. It is then notated by $\overset{\circ}{e}_d$ or $\overset{\circ}{e}(d)$.

We now compare the probability density functions (pdfs) and the cumulative distribution functions (cdfs) of the two payment variables. Let $f(x)$ and $F(x)$ be the pdf and cdf of the loss random variable $X$, respectively.

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

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

Both pdfs are a function of the pdf $f(x)$ of the loss $X$. Assuming that the distribution of $X$ is a continuous distribution, the pdf $f_{Y_L}(y)$ is a mixed distribution (part discrete and part continuous). It has a point mass at $y=0$ with probability $F(d)=P(X \le d)$, representing the scenario that no payment is made for a small loss. The pdf $f_{Y_P}(y)$ is a continuous one if $X$ is continuous. It is the result of $f(x+d)$ divided by $P(X > d)$ since it is a conditional distribution (only losses exceeding the deductible are considered). The cdfs of the payment per loss $Y_L$ and the payment per payment $Y_P$ are given by the following:

(6)……$\displaystyle F_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ y<0 \\ \text{ } & \text{ } \\ \displaystyle F(y+d) &\ y \ge 0 \end{array} \right.$

(7)……$\displaystyle F_{Y_P}(y)=\frac{F(y+d)-F(d)}{P(X > d)} \ \ \ \ \ \ \ y>0$

The cdf $F_{Y_L}(y)$ has a jump at $y=0$, reflecting the scenario that there is no payment for any loss less than the deductible $d$. Note that the size of the jump is $F_{Y_L}(0)=F(d)$. The cdf $F_{Y_P}(y)$ has no jump at $y=0$ since losses below the deductible are not accounted for in the variable $Y_P$.

Once the pdfs are derived, the calculation of the mean and variance of each of the two variables is possible. Once again, assuming the loss distribution is a continuous one, the following integrals give the calculation. When the loss distribution is discrete, simply replace integrals with summations.

(8)……$\displaystyle E(Y_L)=E[(X-d)_+]=\int_0^\infty y \ f_{Y_L}(y) \ dy=\int_0^\infty y \ f(y+d) \ dy$

(9)……$\displaystyle E(Y_L^2)=E[(X-d)_+^2]=\int_0^\infty y^2 \ f_{Y_L}(y) \ dy=\int_0^\infty y^2 \ f(y+d) \ dy$

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

(11)……$\displaystyle e_X(d)=E(Y_P)=\int_0^\infty y \ f_{Y_P}(y) \ dy=\int_0^\infty y \ \frac{f(y+d)}{P[X > d]} \ dy$

(12)……$\displaystyle E(Y_P^2)=\int_0^\infty y^2 \ f_{Y_P}(y) \ dy=\int_0^\infty y^2 \ \frac{f(y+d)}{P[X > d]} \ dy$

(13)……$\displaystyle Var(Y_P)=E(Y_P^2)-E(Y_P)^2$

Comparing (8) and (11) and comparing (9) and (12) give the following relations.

(14)……$\displaystyle e_X(d)=\frac{E(Y_L)}{P(X > d)}$……or……$\displaystyle E(Y_L)=P(X > d) \ e_X(d)$

(15)……$\displaystyle E(Y_P^2)=\frac{E(Y_L^2)}{P(X > d)}$……or……$\displaystyle E(Y_L^2)=P(X > d) \ E(Y_P^2)$

The relations (14) and (15) show that only one set of calculation needs to be made. Once the first and second moments of one variable are known, the moments of the other variable are obtained by adjusting with $P[X > d]$, the probability of having a claim.

Demonstrating the Calculation

We work one example to illustrate the above calculation.

Example 1
The loss $X$ has the density function $f(x)=\frac{3}{500} \ x (10-x)$ where $0. The coverage has a deductible of 3. Compute the mean and variance of the insurance payment per loss. Use the per loss results to obtain the mean and variance of the payment per payment. Write out the cdf of $Y_L$ and the cdf of $Y_P$.

First, obtain the density function of the payment per loss.

$\displaystyle f_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle F(3)=\frac{108}{500} &\ y=0 \\ \text{ } & \text{ } \\ \displaystyle f(y+3)=\frac{3}{500} (21+4y-y^2) &\ 0

The following calculates the mean and variance.

\displaystyle \begin{aligned} E(Y_L)&=\int_0^7 y \ \frac{3}{500} \ (21+4y-y^2) \ dy \\&=\int_0^7 \frac{3}{500} \ (21 y+4y^2-y^3) \ dy \\&=\frac{4459}{2000}=2.2295 \end{aligned}

\displaystyle \begin{aligned} E(Y_L^2)&=\int_0^7 y^2 \ \frac{3}{500} \ (21+4y-y^2) \ dy \\&=\int_0^7 \frac{3}{500} \ (21 y^2+4y^3-y^4) \ dy \\&=\frac{21609}{2500}=8.6436 \end{aligned}

$\displaystyle Var(Y_L)=8.6436-2.2295^2=3.6729$

The probability of a claim is $P(X > 3)=\frac{392}{500}$. Dividing the results by this probability gives the following results.

$\displaystyle e_X(3)=\frac{E(Y_L)}{P(X > 3)}=\frac{4459}{2000} \ \frac{500}{392}=\frac{11.375}{4}=2.84375$

$\displaystyle E(Y_P^2)=\frac{E(Y_L^2)}{P(X > 3)}=\frac{21609}{2500} \ \frac{500}{392}=\frac{55.125}{5}=11.025$

$\displaystyle Var(Y_P)=11.025-2.84375^2=2.9381$

To contrast, $E(X)=5$, the average payment for loss if there is no deductible. By imposing a deductible of 3, on average the amount of 5 – 2.2295 = 2.7705 per loss is shifted to the insured. The average payment per payment 2.84375 is higher than the average payment per loss 2.2295 because the per payment calculation centers on the loss exceeding the deductible.

The cdf of the original loss $X$ is $F(x)=\frac{3}{500} \ (5x^2-\frac{1}{3} x^3)$ where $0 \le x <10$. The cdfs of $Y_L$ and $Y_P$ are given by:

$\displaystyle F_{Y_L}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ y<0 \\ \text{ } & \text{ } \\ \displaystyle \frac{3}{500} \ \biggl[5(y+3)^2-\frac{1}{3} (y+3)^3 \biggr] &\ 0 \le y \le 7 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ 7 < y < \infty \end{array} \right.$

\displaystyle \begin{aligned} F_{Y_P}(y)&=\frac{F(y+3)-F(3)}{P[X > 3]} \\&=\frac{\frac{3}{500} \ \biggl[5(y+3)^2-\frac{1}{3} (y+3)^3 \biggr]-\frac{108}{500}}{\frac{392}{500}} \\&=\frac{3 \ \biggl[5(y+3)^2-\frac{1}{3} (y+3)^3 \biggr]-108}{392} \ \ \ \ \ 0 \le y \le 7 \end{aligned}

With the cdf of $Y_P$ set up, we can determine the probabilities of claim payments. For example, if a claim has to be paid, what is the probability that the claim payment is between 3 and 5?

\displaystyle \begin{aligned} F_{Y_P}(5)-F_{Y_P}(3)&=\frac{3 \ \biggl[5 \cdot 8^2-\frac{1}{3} \cdot 8^3 \biggr]-108}{392}-\frac{3 \ \biggl[5 \cdot 6^2-\frac{1}{3} \cdot 6^3 \biggr]-108}{392} \\&=\frac{340}{392}-\frac{216}{392} =\frac{124}{392}=0.3163 \end{aligned}

Using Limited Expectation

The above discussion focuses on developing the probability distribution of the payment per loss $Y_L$ and the probability distribution of the payment per payment, as well as how the two distributions relate. Once the distribution of a payment variable is established, we can then evaluate various distributional quantities regarding the payment in question – e.g. mean and variance.

Another way to calculate $E(Y_L)=(X-d)_+$ is through the limited loss random variable $X \wedge u$.

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

It is clear that $X=(X-d)_+ + X \wedge d$. In words, buying a policy with a deductible of $d$ and another policy with a policy limit of $d$ equals full coverage. The expectation $E(X \wedge u)$ is called the limited expectation. Immediately we have the following relationship.

(17)……$\displaystyle E(X)=E((X-d)_+)+E(X \wedge d)$……or……$\displaystyle E(Y_L)=E((X-d)_+)=E(X)-E(X \wedge d )$

One advantage of the relationship (17) is that the limited expectation $E(X \wedge u )$ is provided in a table of distributions (table). For several distributions such as exponential, lognormal, Pareto and a few others, the formulas of $E(X \wedge u)$ are quite easy to implement. When the loss distribution is one of these distributions, (17) provides another way to compute the expected payment per loss. Combining (17) and (14) gives the following formula for the expected payment per payment (or mean excess loss).

(18)……$\displaystyle e_X(d)=\frac{E(X)-E(X \wedge d )}{P(X > d)}$

Franchise Deductible

The deductible in the above discussion is an ordinary deductible. The key characteristic of ordinary deductible is that the first $d$ dollars are not paid by the insurer. A franchise deductible works like an ordinary deductible except that when the loss exceeds the deductible, the policy pays the loss in full.

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

Instead of building a new set of formulas, we can simply relate the expected payment under the franchise deductible with the expected payment under the ordinary deductible.

(20)……expected payment per loss under franchise deductible = $\displaystyle E(Y_L)+d \ P(X > d)$

(21)……expected payment per payment under franchise deductible = $\displaystyle e_X(d)+d$

The payment under a franchise deductible is larger than the payment under an ordinary deductible. By how much? The deductible $d$. The addition of $d \ P[X > d]$ in (20) reflects the additional benefit when $X > d$. The expected payment per payment is already conditioned on $X > d$. Thus $d$ is unconditionally added to $e_X(d)$ in (21). The variance can be calculated using basic principle. Example 2 shows how.

To distinguish between the two deductibles, we adopt the convention that deductible means ordinary deductible. If franchise deductible is used, “franchise” would have to be explicitly stated.

More Examples

Example 2
Work Example 1 assuming a franchise deductible.

To keep things straight, we let $Y$ to denote the payment per loss and $Y_{*}$ to denote the payment per payment (the notation is applicable just for this example). Then $\displaystyle E(Y)$ and $\displaystyle E(Y_*)$ are greater than their counterparts for ordinary deductible according to (20) and (21).

$\displaystyle E(Y)=E(Y_L)+3 \ P(X > 3)=\frac{4459}{2000}+3 \cdot \frac{392}{5000}=\frac{9163}{2000}=4.5815$

$\displaystyle E(Y_*)=E(Y_P)+3=\frac{11.375}{4}+3=\frac{23.375}{4}=5.84375$

To find the variance, it is helpful to work with the pdfs.

$\displaystyle f_{Y}(y)=\left\{ \begin{array}{ll} \displaystyle F(3)=\frac{108}{500} &\ y=0 \\ \text{ } & \text{ } \\ \displaystyle f(y)=\frac{3}{500} (10y-y^2) &\ 3

For franchise deductible, there is no need to shift the pdf by the deductible. There is a point mass at $y=0$ like before. The pdf continues with the interval (3, 10). With this pdf, the following gives the calculation of the variance.

$\displaystyle E(Y^2)=\int_3^{10} y^2 \ \frac{3}{500} (10y-y^2) \ dy=\frac{290766}{10000}=29.0766$

$\displaystyle Var(Y)=29.0766-4.5815^2=8.0865$

We can leverage the per loss variance to obtain the per payment variance.

$\displaystyle E(Y_*^2)=\frac{E[Y^2]}{P[X > 3]}=29.0766 \cdot \frac{500}{392}=37.0875$

$\displaystyle Var(Y_*)=37.0875-5.84375^2=2.9381$

The cdfs are given by the following:

$\displaystyle F_{Y}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ y<0 \\ \text{ } & \text{ } \\ \displaystyle F(3)=\frac{108}{500} &\ 0 \le y < 3 \\ \text{ } & \text{ } \\ \displaystyle \frac{3}{500} \ \biggl[5 y^2-\frac{1}{3} y^3 \biggr] &\ 3 \le y \le 10 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ 10 < y < \infty \end{array} \right.$

$\displaystyle F_{Y_*}(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ y<3 \\ \text{ } & \text{ } \\ \displaystyle \frac{\frac{3}{500} \ \biggl[5 y^2-\frac{1}{3} y^3 \biggr]-\frac{108}{500}}{\frac{392}{500}}=\frac{3 \ \biggl[5 y^2-\frac{1}{3} y^3 \biggr]-108}{392} &\ 3 \le y \le 10 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ 10 < y < \infty \end{array} \right.$

If the insurer has to pay a claim, what is the probability that it is between 6 and 8?

$\displaystyle F_{Y_*}(8)-F_{Y_*}(6)=\frac{340}{392}-\frac{216}{392}=\frac{124}{392}=0.3163$

Note that this answer is the same one as in Example 1 since there is no shifting with a franchise deductible.

Example 3
The discussion is not complete without mentioning exponential distribution. Suppose that the pdf of the loss $X$ is $f(x)=\frac{1}{\theta} \ e^{-x/\theta}$ where $x >0$, i.e. the pdf for an exponential distribution with mean $\theta$. The deductible has an ordinary deductible of $d$. The expected payment per loss and the variance of payment per loss are:

$\displaystyle E(Y_L)= \theta \ e^{-d/\theta}$

$\displaystyle E(Y_L^2)= 2 \ \theta^2 \ e^{-d/\theta}$

$\displaystyle Var(Y_L)=2 \ \theta^2 \ e^{-d/\theta}-\theta^2 \ e^{-2d/\theta}$

The above results are obtained by using the pdf of the payment per loss $Y_L$, which can be obtained according to (4). The results for payment per payment are obtained by dividing by $P(X > d)=e^{-d/\theta}$.

$\displaystyle E(Y_P)= \theta$

$\displaystyle E(Y_P^2)= 2 \ \theta^2$

$\displaystyle Var(Y_P)=2 \ \theta^2 -\theta^2 =\theta^2$

The mean and variance of $Y_P$ are identical to the original exponential loss distribution. This is to be expected. Because the exponential distribution is memoryless, the payment variable $Y_P=X-d \ \lvert X > d$ is identical to the original loss $X$, an exponential distribution with mean $\theta$. For models of insurance payments with a deductible, knowing that the loss distribution is exponential simplifies the calculation greatly.

Example 4
This example illustrates how to obtain the mean payment per loss using the table approach – using limited expectation given in this table. Suppose that the loss distribution is a lognormal distribution with parameters $\mu=2$ and $\sigma=2$. The coverage has a deductible of 200. Compute the expected payment per loss $E(Y_L)$.

Using the formula for $E(X)$ and the formula for the limited expectation $E(X \wedge x)$ found in the table, the expected payment per loss is given by:

$\displaystyle F(200)=\Phi \biggl( \frac{\ln(200)-5}{2} \biggr)=\Phi(0.15)=0.5596$

$\displaystyle 1-F(200)=1-0.5596=0.4404$

$\displaystyle E(X)=e^{\mu+\frac{1}{2} \sigma^2}=e^7$

\displaystyle \begin{aligned} E(X \wedge 200)&=e^{\mu+\frac{1}{2} \sigma^2} \ \Phi \biggl( \frac{\ln(200)-\mu-\sigma^2}{\sigma} \biggr)+200 [1-F(200)] \\&=e^{7} \ \Phi \biggl( \frac{\ln(200)-5-2^2}{2} \biggr)+200 [0.4404] \\&=e^7 \ \Phi(-1.85)+88.08 \\&=e^7 \ (1-0.9678)+88.08 \\&=0.0322 \ e^7+88.08 \end{aligned}

\displaystyle \begin{aligned} E(Y_L)&=E(X)-E(X \wedge 200) \\&=e^7-(0.0322 \ e^7+88.08) \\&=0.9678 \ e^7-88.08=973.2415707 \end{aligned}

Example 5
This example continues Example 4. We now tackle the variance of the payment per loss. For the lognormal loss distribution, it is difficult to find the pdf of the payment per loss $Y_L$. In the table, there is no direct formula for $E[Y_L^2]$. However, there is a way to use the limited expectations $E(X \wedge d)$ and $E((X \wedge d)^2)$ to derive $E(Y_L^2)$. First, let’s break down the components that go into $E(Y_L^2)$.

\displaystyle \begin{aligned} E(Y_L^2)&=\int_d^\infty (x-d)^2 f(x) \ dx \\&=\int_d^\infty x^2 f(x) \ dx-2 \ d \int_d^\infty x f(x) \ dx+d^2 \ P(X > d) \end{aligned}

The $f(x)$ in above is the pdf of the lognormal distribution. Of course we are not to evaluate the above two integrals directly. However, they can be expressed using limited expectations as follows:

$\displaystyle \int_d^\infty x f(x) \ dx=E[X]-E[X \wedge d]+d \ P(X > d)$

$\displaystyle \int_d^\infty x^2 f(x) \ dx=E[X^2]-E[(X \wedge d)^2]+d^2 \ P(X > d)$

With respect to the lognormal parameters $\mu=2$ and $\sigma=2$ and the deductible 200, the limited expectation $E((X \wedge 200)^2)$ is given by the following. The first limited expectation is already calculated in Example 4.

$\displaystyle E(X^2)=e^{2 \ \mu+\frac{1}{2} \ 2^2 \ \sigma^2}=e^{18}$

\displaystyle \begin{aligned} E((X \wedge 200)^2)&=e^{2 \mu+\frac{1}{2} \ 2^2 \ \sigma^2} \ \Phi \biggl( \frac{\ln(200)-\mu-2 \ \sigma^2}{\sigma} \biggr)+200^2 [1-F(200)] \\&=e^{18} \ \Phi \biggl( \frac{\ln(200)-5-2 \cdot 2^2}{2} \biggr)+200^2 [0.4404] \\&=e^{18} \ \Phi(-3.85)+17616 \\&=e^{18} \ (1-0.9999)+17616 \\&=0.0001 \ e^{18}+17616 \end{aligned}

The calculation continues:

\displaystyle \begin{aligned}\int_{200}^\infty x f(x) \ dx&=E(X)-E(X \wedge 200)+200 \ P(X > 200) \\&=e^7-[0.0322 \ e^{7}+88.08]+88.08 \\&=0.9678 \ e^7 \end{aligned}

\displaystyle \begin{aligned}\int_{200}^\infty x^2 f(x) \ dx&=E(X^2)-E((X \wedge 200)^2)+200^2 \ P(X > 200) \\&=e^{18}-[0.0001 \ e^{18}+17616]+17616 \\&=0.9999 \ e^{18} \end{aligned}

\displaystyle \begin{aligned} E(Y_L^2)&=\int_{200}^\infty x^2 f(x) \ dx-2 \ 200 \int_{200}^\infty x f(x) \ dx+200^2 \ P(X > 200) \\&=0.9999 \ e^{18}-2 \cdot 200 \cdot 0.9678 \ e^7 +17616\\&=65246490.51 \end{aligned}

$\displaystyle Var(Y_L)=65246490.51-(0.9678 \ e^7-88.08)^2=64299291.36$

$\displaystyle \sigma_{Y_L}=\sqrt{64299291.36}=8018.683892$

Example 6
This example discusses the Pareto distribution as the loss distribution. The Pareto distribution is a two-parameter family of distributions. The pdf and cdf are given below, along with the mean and the variance.

$\displaystyle f(x)=\frac{\alpha \ \theta^\alpha}{(x+\theta)^{\alpha+1}} \ \ \ \ \ \ \ \ \ \ x>0$

$\displaystyle F(x)=1-\biggl(\frac{\theta}{x+\theta} \biggr)^\alpha \ \ \ \ \ x>0$

$\displaystyle E(X)=\frac{\theta}{\alpha-1}$

$\displaystyle E(X^2)=\frac{2 \ \theta^2}{(\alpha-1) \ (\alpha-2)}$

$\displaystyle Var(X)=E(X^2)-E(X)^2$

For the Pareto mean to exist, the parameter $\alpha$ must be greater than 1. For the variance to exist, the parameter $\alpha$ must be greater than 2. When the coverage has an ordinary deductible $d$, consider the insurance payment per loss $Y_L$ and the insurance payment per loss $Y_P$. In terms of these two payment variables, the Pareto distribution is very tractable. To find mean payment per loss, either start with the pdf of $Y_L$ or use the limited expectation $E(X \wedge d)$ from the table. Because the payment per payment is also a Pareto distribution, there are shortcuts for finding the variance of payment per loss (see Example 3 here).

When the loss $X$ is a Pareto distribution with parameters $\alpha$ and $\theta$, the payment per payment $Y_P=X-d \ \lvert X>d$ has a Pareto distribution with parameters $\alpha$ and $\theta+d$. Let’s examine the expected value of $Y_P$, or the mean excess loss.

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

Note that the mean excess loss $e_X(d)$ is an increasing function of $d$. This means that the higher the deductible, the larger the expected claim if such a large loss occurs! If a random loss is modeled by such a distribution, it is a catastrophic risk situation. In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. Thus the Pareto distribution is considered a heavy tailed distribution and is a suitable candidate for modeling situations that have potential extreme large losses.

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Calculating the variance of insurance payment

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. Practice problems are found in Practice Problem Set 13.

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 $Var(Y_L)$.

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.

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

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Practice Problem Set 9 – Expected Insurance Payment – Additional Problems

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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Problem Answer
9A
• $\displaystyle \frac{6800}{3}=2266.67$
9B
• $\displaystyle 1.15 e^{1.2/20.125}=1.22$
• About 22% increase
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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Practice Problem Set 8 – Expected Insurance Payment – Additional Problems

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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Problem Answer
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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Practice Problem Set 7 – Expected Insurance Payment

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 period. 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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Problem Answer
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
• $\displaystyle \frac{0.9441 e^8-722.4}{0.7224}= 2895.80$

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$\copyright$ 2017 – Dan Ma

Mathematical models for insurance payments – part 3 – other modifications

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

Mathematical models for insurance payments – part 2 – ordinary deductible

The post is the second post on models for insurance payments. The focus here is on the insurance payment when the loss is adjusted by a deductible. The first post is on insurance policy with a limit.

Ordinary Deductible

Suppose that the random variable $X$ is the size of the loss (if a loss occurs). Under an insurance policy with a 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 this policy provision, what is the expected amount of insurance payment for a given loss? The deductible described here is called an ordinary deductible. The insurance payment (per loss) under the presence of an ordinary deductible $d$ is denoted by $(X-d)_+$. The idea for this notation is that the insurance payment is $X-d$ as long as $X-d$ is not negative. The following describes the payment rule more explicitly.

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

The goal is to describe the probability model of the insurance payment $(X-d)_+$. First and foremost, we need to know how much it will cost the insurer on average to pay the insured, which is the mean $E[(X-d)_+]$. The variance of $(X-d)_+$ is also important as it is a measure of risk to the insurer. The mean $E[(X-d)_+]$ and the variance $Var[(X-d)_+]$ can be calculated by using the density function of $X$. At times it will be useful to know the the probability density function (PDF) and cumulative distribution function (CDF) of $(X-d)_+$ and other distributional quantities.

Moments

First, let’s focus on the moments of $(X-d)_+$. One way to obtain moments of $(X-d)_+$ is through the distribution of the unmodified loss $X$. 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. Then the moments of $(X-d)_+$ can be expressed using these distributional quantities of $X$.

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

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

$\displaystyle Var[(X-d)_+]=E[(X-d)_+^2]-E[(X-d)_+]^2$

The above calculation is based on first principle. When the loss $X$ is less than or equal to $d$, the insurance payment is zero. Note that the zero payment does not have to be explicitly stated in the above integrals. When the loss $X$ exceeds the deductible $d$, the payment is $X-d$, the amount of the loss in excess of $d$. Example 1 and Example 2 below demonstrate how this calculation is performed.

Another way to obtain the moments of $(X-d)_+$ is obtained by first finding the PDF of $(X-d)_+$. This is discussed in a section below.

Basic Example

Example 1
Suppose that the loss distribution has PDF $f(x)=1/5 (1- x/10) \ \ \ 0. Evaluate $E(X)$ and $E[(X-2)_+]$. What is the expected amount of the loss eliminated as a result of imposing the ordinary deductible of 2 (eliminated from the insurer’s perspective)?

\displaystyle \begin{aligned} E[(X-2)_+]&=\int_2^{10} (x-2) \ \frac{1}{5} (1- x/10) \ dx\\&=\int_2^{10} \frac{1}{5} (\frac{6}{5} x -x^2/10 -2) \ dx\\&=\frac{1}{5} \biggl[\frac{3 x^2}{5}-\frac{x^3}{30}-2x \biggr]_2^{10}=\frac{128}{75}=1.7067 \end{aligned}

From Example 1 in the previous post, $E(X)=10/3$. Thus when a loss occurs, the expected amount of the loss that is responsible by the insured is

$\displaystyle E(X)-E[(X-2)_+]=\frac{10}{3}-\frac{128}{75}=\frac{122}{75}=1.6267$

,which is also the expected amount of the loss eliminated (from the insurer’s perspective) as a result of imposing an ordinary deductible.

Example 2
Continue with Example 1. Calculate the variance of the insurance payment $(X-2)_+$.

\displaystyle \begin{aligned} E[(X-2)_+^2]&=\int_2^{10} (x-2)^2 \ \frac{1}{5} (1- x/10) \ dx \\ &=\int_2^{10} \frac{1}{5} \ (x^2-4x+4) \ (1- x/10) \ dx \\&=\int_2^{10} \frac{1}{5} (\frac{7}{5} x^2 -x^3/10-\frac{22}{5} +4) \ dx\\&=\frac{1}{5} \biggl[\frac{7 x^3}{15}-\frac{x^4}{40}-\frac{11x^2}{5} +4x \biggr]_2^{10}=\frac{512}{75} \end{aligned}

\displaystyle \begin{aligned} Var[(X-2)_+]&=E[(X-2)_+^2]-E[(X-2)_+]^2 \\ &=\frac{512}{75}-\biggl(\frac{128}{75} \biggr)^2 \\&=\frac{22016}{5625}=3.9140 \end{aligned}

From Example 2 in the previous post, $Var(X)=50/9=5.5556$. This shows that imposing a deductible that is significant enough has a variance reducing effect on the insurance payment. Had the deductible not present, there would be a greater fluctuation in the payment.

Connection with Policy Limit

The calculation in Example 1 and Example 2 are based on first principle. For many parametric families of distributions, the calculation for $E[(X-d)_+]$ can be done using formulas. To that end, consider summing the payment $(X-d)_+$ and the payment $X \wedge d$.

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

Summing up each case produces the original loss $X$. Thus buying a policy with an ordinary deductible $d$ along with a policy with a limit of $d$ equals to full coverage. From a calculation standpoint, $E[(X-d)_+]$ can be computed as follows:

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

The advantage of using the above idea is that the calculation of $E(X \wedge d)$ is routine for a large number of parametric families of distributions. For examples, most of the distributions in the table in this link have formulas for $E(X \wedge x)$ and $E[(X \wedge x)^k]$. Then subtracting $E(X \wedge x)$ into $E(X)$ would give $E[(X-d)_+]$. The following table gives the limited expectation $E(X \wedge x)$ for three distributions, taken from the table in the given link.

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: $\displaystyle E[(X \wedge x)^k]=\exp(k \mu+k^2 \sigma^2/2) \Phi \biggl(\frac{\log(x)-\mu-k \sigma^2}{\sigma} \biggr)+x^k \biggl[1-\Phi \biggl(\frac{\log(x)-\mu}{\sigma} \biggr) \biggr]$

Examples are given to demonstrate how these formulas are used.

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.

For the next loss that will occur, determine the expected amount that will be paid by the insurer.

The following shows the calculation.

$\displaystyle E(X)=\frac{\theta}{\alpha-1}=\frac{500}{2}=250$

\displaystyle \begin{aligned} E[X \wedge 1000]&=\frac{\theta}{\alpha-1} \biggl[1-\biggl(\frac{\theta}{100+\theta} \biggr)^{\alpha-1} \biggr] \\ &=\frac{500}{2} \biggl[1-\biggl(\frac{500}{600} \biggr)^{2} \biggr]=\frac{2750}{36}=76.389 \end{aligned}

$\displaystyle E[(X-1000)_+]=E(X)-E[X \wedge 1000]=250-\frac{2750}{36}=\frac{3125}{18}=173.611$

As a result of imposing a deductible of 1000, the payment made by the insurer per loss goes from 250 to 173.611. The reduction of payment of $E[X \wedge 100]$ = 76.389.

Example 4
You are given the following:

• Losses follow an exponential distribution with mean 2500.
• The coverage has an ordinary deductible of 1000.

Determine the expected amount per loss that is paid by the insurer.

The following shows the calculation.

$\displaystyle E(X)=2500$

$\displaystyle E(X \wedge 1000)=2500 (1-e^{-1000/2500})=2500 (1-e^{-0.4})=824.20$

\displaystyle \begin{aligned} E[(X-1000)_+]&=E(X)-E[X \wedge 1000] \\ &=2500-2500 (1-e^{-0.4}) =2500 e^{-0.4}=1675.80 \end{aligned}

Note that the reduction of payment by the insurer per loss is $\displaystyle E(X \wedge 1000)=824.20$, which is the amount per loss that has to be met by the insured.

One observation about exponential loss. When the loss $X$ is an exponential distribution, $E[(X-d)_+]$ can actually be directly calculated using first principle since the exponential distribution is very tractable mathematically.

Example 5
You are given the following:

• Losses follow a lognormal distribution with parameters $\mu=6.5$ and $\sigma=1.75$.
• The coverage has an ordinary deductible of 1000.

Determine the expected amount per loss that is paid by the insurer.

The following shows the calculation.

$\displaystyle E(X)=e^{\mu+(1/2) \sigma^2}=e^{6.5+0.5(1.75)^2}=e^{8.03125}=3075.583751$

\displaystyle \begin{aligned} E[X \wedge 1000]&=e^{6.5+0.5(1.75)^2} \ \Phi \biggl(\frac{\log(1000)-6.5-1.75^2}{1.75} \biggr)+1000 \biggl[1-\Phi \biggl(\frac{\log(1000)-6.5}{1.75} \biggr) \biggr] \\ &=e^{8.03125} \ \Phi (-1.52)+1000 [1-\Phi (0.23) ] \\ &=e^{8.03125} \ (1-0.9357)+1000 [1-0.5910 ]=606.7600352 \end{aligned}

\displaystyle \begin{aligned} E[(X-1000)_+]&=E(X)-E[X \wedge 1000] \\ &=e^{8.03125}-606.7600352=2468.82 \end{aligned}

PDF and CDF of Insurance Payment

The preceding section shows that $E[(X-d)_+]=E(X)-E(X \wedge d)$ is a great way to evaluate mean claim cost per loss when the loss distribution is from this list of distributions. For the distributions in this list, the limited expectation $E(X \wedge d)$ is readily available. To calculate the higher moments of $(X-d)_+$, it is helpful to know more about its distribution. To that end, we derive its PDF and CDF and the survival function.

Before deriving the PDF and CDF, note that the payment variable $(X-d)_+$ is a censored and shifted random variable. It is censored from below and is shifted to the left by the amount of the deductible $d$. As far as payment is concerned, any loss that is below the deductible is considered zero. In effect, any loss in the interval $(0,d)$ is recorded as zero. The insurance policy only pays for losses that are in excess of the deductible $d$. Hence the positive payments are derived by shifting losses in the interval $(d,\infty)$ to the left by $d$. The resulting distribution for the payment $(X-d)_+$ is a mixed random variable. It has a point mass at $y=0$ with probability $F_X(d)$ where $F_X(x)$ is the CDF of the loss $X$. On the interval $(0,\infty)$ or some appropriate interval $(0,M)$, the density curve of $(X-d)_+$ is continuous and is the resulting of shifting the density curve of $X$ to the left by the amount $d$. Let $Y=(X-d)_+$. Let $f_X(x)$ and $F_X(x)$ be the PDF and CDF of $X$, respectively. The following is the PDF of $Y$.

$\displaystyle f_Y(y)=\left\{ \begin{array}{ll} \displaystyle F_X(d) &\ \ \ \ y=0 \\ \text{ } & \text{ } \\ \displaystyle f_X(y+d) &\ \ \ \ y > 0 \end{array} \right.$

Note the point mass at $y=0$, which is the probability that the loss is less than or equal to the deductible. The curve $f_X(y+d)$ is the density curve $f_X(y)$ shifted to the left by the amount $d$. The following is the CDF and the survival function of $Y=(X-d)_+$.

$\displaystyle F_Y(y)=\left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ y<0 \\ \text{ } & \text{ } \\ \displaystyle F_X(y+d) &\ \ \ \ y \ge 0 \end{array} \right.$

$\displaystyle S_Y(y)=\left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ y<0 \\ \text{ } & \text{ } \\ \displaystyle S_X(y+d) &\ \ \ \ y \ge 0 \end{array} \right.$

Now that we have a description of the model for $Y=(X-d)_+$, moments and other distributional quantities can be derived. Consider the following two examples.

Example 6
You are given the following:

• Losses follow a uniform distribution on the interval $(0, 10)$.
• The coverage has an ordinary deductible of 4.

Determine the mean and variance of $Y=(X-4)_+$.

The following shows the PDF, CDF and survival function of the loss $X$.

$\displaystyle f_X(x)=\frac{1}{10} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

$\displaystyle F_X(x)=\frac{x}{10} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

$\displaystyle S_X(x)=1-\frac{x}{10} \ \ \ \ \ \ \ \ \ \ \ \ \ 0

The point mass is $F_X(4)=0.4$. The following is the PDF $f_Y(y)$.

$\displaystyle f_Y(y)=\left\{ \begin{array}{ll} \displaystyle 0.4 &\ \ \ \ y=0 \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{10} &\ \ \ \ 0

Note that the PDF $f_X(x)$ over the interval $(4,10)$ is shifted to the interval $(0, 6)$. The following gives the CDF and survival function of $Y$.

$\displaystyle F_Y(y)=\left\{ \begin{array}{ll} \displaystyle 0 & \ \ \ \ y < 0 \\ \text{ } & \text{ } \\ \displaystyle \frac{y+4}{10} & \ \ \ \ 0 \le y < 6 \\ \text{ } & \text{ } \\ \displaystyle 1 & \ \ \ \ y \ge 6 \end{array} \right.$

$\displaystyle S_Y(y)=\left\{ \begin{array}{ll} \displaystyle 1 & \ \ \ \ y < 0 \\ \text{ } & \text{ } \\ \displaystyle 1-\frac{y+4}{10}=\frac{6-y}{10} & \ \ \ \ 0 \le y < 6 \\ \text{ } & \text{ } \\ \displaystyle 0 & \ \ \ \ y \ge 6 \end{array} \right.$

The moments of $Y=(X-4)_+$ can be can be obtained by integrating $x^k$ with the PDF $f_Y(y)$.

$\displaystyle E(Y)=\int_0^6 \frac{1}{10} \ y \ dy=\frac{9}{5}=1.8$

$\displaystyle E(Y^2)=\int_0^6 \frac{1}{10} \ y^2 \ dy=\frac{36}{5}=7.2$

$Var(Y)=7.2-1.8^2=3.96$

Note that the variance of the unmodified loss is $Var(X)=100/12=8.33$. In this case, imposing a deductible of 4 reduces the variance by a little more than half.

Example 7
Consider the loss distribution and the coverage in Example 3. Determine the PDF, CDF and the survival function of $Y=(X-100)_+$. Then find the mean and variance of $Y=(X-100)_+$.

The loss $X$ has a Pareto distribution with parameters $\alpha=3$ and $\theta=500$. The following shows its PDF, CDF and survival function.

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

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

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

The point mass is $F_X(100)=91/216$. The following is the PDF $f_Y(y)$.

$\displaystyle f_Y(y)=\left\{ \begin{array}{ll} \displaystyle \frac{91}{216} &\ y=0 \\ \text{ } & \text{ } \\ \displaystyle \frac{3 \cdot 500^3}{(y+600)^4} &\ y > 0 \end{array} \right.= \left\{ \begin{array}{ll} \displaystyle \frac{91}{216} &\ y=0 \\ \text{ } & \text{ } \\ \displaystyle \biggl(\frac{500}{600} \biggr)^3 \ \frac{3 \cdot 600^3}{(y+600)^4} &\ y > 0 \end{array} \right.$

Note that the part for $y > 0$ is $S_X(100)$ times the Pareto density function for parameters $\alpha=3$ and $\theta=600$. Thus the moments of $Y=(X-1000)_+$ can be obtained by multiplying $S_X(1000)$ with the moments of this new Pareto distribution. Since the point mass is for $y=0$, we can ignore it.

$\displaystyle E(Y)=E[(X-100)_+]=\biggl(\frac{500}{600} \biggr)^3 \ \frac{600}{3-1}=\frac{3125}{18}=173.611$

$\displaystyle E(Y^2)=E[(X-1000)_+^2]=\biggl(\frac{125}{216} \biggr) \ \frac{600^2 \ 2}{(3-1)(3-2)}=\frac{625000}{3}$

$\displaystyle Var(Y)=E(Y^2)-E(Y)^2=\frac{625000}{3}-\biggl(\frac{3125}{18}\biggr)^2=178192.5154$

$\sigma_Y=\sqrt{178192.5154}=422.129$

To complete the example, the following gives the CDF and survival function.

$\displaystyle F_Y(y)=\left\{ \begin{array}{ll} \displaystyle 0 & \ \ \ \ y < 0 \\ \text{ } & \text{ } \\ \displaystyle 1-\biggl( \frac{500}{600+y} \biggr)^3 & \ \ \ \ 0 \le y < \infty \end{array} \right.$

$\displaystyle S_Y(y)=\left\{ \begin{array}{ll} \displaystyle 1 & \ \ \ \ y < 0 \\ \text{ } & \text{ } \\ \displaystyle \biggl( \frac{500}{600+y} \biggr)^3 & \ \ \ \ 0 \le y < \infty \end{array} \right.$

Insurance Payment Per Loss versus Per Payment

Note that the $E[(X-d)_+]$ is calculated over all losses (whether they are below or above the deductible). The expected value $E[(X-d)_+]$ reflects the probability $P(X \le d)$ that is assigned to the payment of zero (the point mass), and the probabilities that are applied to the payment $x-d$ over all $x$ in the interval $(d,\infty)$ (the integral). Thus $E[(X-d)_+]$ is the average payment per loss (or over all possible losses). Another average payment to consider is the average payment over all payments, i.e. over all losses larger than the deductible. This topic is continued in the next post.

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$\copyright$ 2017 – Dan Ma

Mathematical models for insurance payments – part 1 – policy limit

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}

Comments

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

Mathematical models for insurance payments – part 0 – introduction

Insurance losses depend on two random variables. The first one is the number of losses that will occur in a specified period in connection to an insured (or a group of insureds). This is commonly referred to as the loss frequency or claim frequency. Its probability distribution is called the loss (claim) frequency distribution. The second random variable is the amount of the loss (if a loss has occurred). The amount of loss is usually referred to as the severity and its probability distribution is called the severity distribution. Then putting these two distributions together will lead to the total loss distribution.

In the several posts that follow, the focus is on the severity distribution, i.e. we will focus on the size of the loss or size of the claim. Once the methodology for the severity is discussed in a fair amount of details, we will add claim frequency.

Given that a loss has occurred and that the amount is $X$, the insurance company will make a payment to the insured to cover the loss $X$. The insurance payment is a random quantity since $X$ is random. Due to the presence of policy provisions such as deductible and limit, the insurance payment is likely less than $X$. The focus is on determining the distributional quantities of the probability model of the insurance payment to the insured. First and foremost, we would like to calculate the mean and variance of the distribution of payment as well as probability density function and cumulative distribution function and other distributional quantities.

The subsequent post will show, from a mathematical standpoint, the random variable of the insurance payment is a truncated and/or censored variable of the loss $X$ due to policy provisions such as deductible and policy limit.

In practice, the mean and variance of the insurance payments are often estimated from historical observations rather than using a hypothesized distribution of losses. However, the parametric distribution approach is a good starting point of the discussion of estimation of insurance losses.

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$\copyright$ 2017 – Dan Ma