# 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