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

Posted on Updated on

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 $X \wedge u$. $\text{ }$

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

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

The lognormal distribution with mean 177.682811 and variance 13680.72152 corresponds to $\mu=5$ and $\sigma=0.6$. The following gives the limited expected value. \displaystyle \begin{aligned} E[X \wedge 250]&=e^{5+0.6^2/2} \ \Phi \bigg(\frac{\log(250)-5-0.6^2}{0.6} \biggr)+250 \ \biggl[1-\Phi \bigg(\frac{\log(250)-5}{0.6} \biggr) \biggr] \\&=e^{5.18} \ \Phi (0.269)+250 \ [1-\Phi (0.869 )]\\&=e^{5.18} \ 0.6064+250(1-0.8072)=155.7969 \end{aligned}

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

## 7 thoughts on “Mathematical models for insurance payments – part 1 – policy limit”

[…] 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. […]

[…] Lognormal distribution and limited expectation […]

[…] 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. […]

[…] 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 […]

[…] 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 […]

[…] 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 […]

[…] 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 […]