# Mean Excess Loss

### Insurance payment per payment vs payment per loss

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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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