# The hazard rate function

In this post, we introduce the hazard rate function using the notions of non-homogeneous Poisson process.

In a Poisson process, changes occur at a constant rate $\lambda$ per unit time. Suppose that we interpret the changes in a Poisson process from a mortality point of view, i.e. a change in the Poisson process mean a termination of a system, be it biological or manufactured, and this Poisson process counts the number of of terminations as they occur. Then the rate of change $\lambda$ is interpreted as a hazard rate (or failure rate or force of mortality). With a constant force of mortality, the time until the next change is exponentially distributed. In this post, we discuss the hazard rate function in a more general setting. The process that counts of the number of terminations will no longer have a constant hazard rate, and instead will have a hazard rate function $\lambda(t)$, a function of time $t$. Such a counting process is called a non-homogeneous Poisson process. We discuss the survival probability models (the time to the next termination) associated with a non-homogeneous Poisson process. We then discuss several important examples of survival probability models, including the Weibull distribution, the Gompertz distribution and the model based on the Makeham’s law. We aso comment briefly the connection between the hazard rate function and the tail weight of a distribution.

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The Poisson Process
We start with the three postulates of a Poisson process. Consider an experiment in which the occurrences of a certain type of events are counted during a given time interval. We call the occurrence of the type of events in question a change. We assume the following three conditions:

1. The numbers of changes occurring in nonoverlapping intervals are independent.
2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
3. The probability of exactly one change in the short interval $(t,t+\delta)$ is approximately $\lambda \delta$ where $\delta$ is sufficiently small and $\lambda$ is a positive constant.

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When we interpret the Poisson process in a mortality point of view, the constant $\lambda$ is a hazard rate (or force of mortality), which can be interpreted as the rate of failure at the next instant given that the life has survived to time $t$. With a constant force of mortality, the survival model (the time until the next termination) has an exponential distribution with mean $\frac{1}{\lambda}$. We wish to relax the constant force of mortality assumption by making $\lambda$ a function of $t$ instead. The remainder of this post is based on the non-homogeneous Poisson process defined below.

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The Non-Homogeneous Poisson Process
We modifiy condition 3 above by making $\lambda(t)$ a function of $t$. We have the following modified counting process.

1. The numbers of changes occurring in nonoverlapping intervals are independent.
2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
3. The probability of exactly one change in the short interval $(t,t+\delta)$ is approximately $\lambda(t) \delta$ where $\delta$ is sufficiently small and $\lambda(t)$ is a nonnegative function of $t$.

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We focus on the survival model aspect of such counting processes. Such process can be interpreted as models for the number of changes occurred in a time interval where a change means “termination” or ‘failure” of a system under consideration. The rate of change function $\lambda(t)$ indicated in condition 3 is called the hazard rate function. It is also called the failure rate function in reliability engineering and the force of mortality in life contingency theory.

Based on condition 3 in the non-homogeneous Poisson process, the hazard rate function $\lambda(t)$ can be interpreted as the rate of failure at the next instant given that the life has survived to time $t$.

Two random variables naturally arise from a non-homogeneous Poisson process are described here. One is the discrete variable $N_t$, defined as the number of changes in the time interval $(0,t)$. The other is the continuous random variable $T$, defined as the time until the occurrence of the first change. The probability distribution of $T$ is called a survival model. The following is the link between $N_t$ and $T$.

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\displaystyle \begin{aligned}(1) \ \ \ \ \ \ \ \ \ &P[T > t]=P[N_t=0] \end{aligned}

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Note that $P[T > t]$ is the probability that the next change occurs after time $t$. This means that there is no change within the interval $(0,t)$. We have the following theorems.

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Theorem 1.
Let $\displaystyle \Lambda(t)=\int_{0}^{t} \lambda(y) dy$. Then $e^{-\Lambda(t)}$ is the probability that there is no change in the interval $(0,t)$. That is, $\displaystyle P[N_t=0]=e^{-\Lambda(t)}$.

Proof. We are interested in finding the probability of zero changes in the interval $(0,y+\delta)$. By condition 1, the numbers of changes in the nonoverlapping intervals $(0,y)$ and $(y,y+\delta)$ are independent. Thus we have:

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$\displaystyle (2) \ \ \ \ \ \ \ \ P[N_{y+\delta}=0] \approx P[N_y=0] \times [1-\lambda(y) \delta]$

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Note that by condition 3, the probability of exactly one change in the small interval $(y,y+\delta)$ is $\lambda(y) \delta$. Thus $[1-\lambda(y) \delta]$ is the probability of no change in the interval $(y,y+\delta)$. Continuing with equation $(2)$, we have the following derivation:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\frac{P[N_{y+\delta}=0] - P[N_y=0]}{\delta} \approx -\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{d}{dy} P[N_y=0]=-\lambda(y) P[N_y=0] \\&\text{ } \\&\frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]}=-\lambda(y) \\&\text{ } \\&\int_0^{t} \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]} dy=-\int_0^{t} \lambda(y)dy \end{aligned}

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Evaluating the integral on the left hand side with the boundary condition of $P[N_0=0]=1$ produces the following results:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &ln P[N_t=0]=-\int_0^{t} \lambda(y)dy \\&\text{ } \\&P[N_t=0]=\displaystyle e^{\displaystyle -\int_0^{t} \lambda(y)dy} \end{aligned}

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Theorem 2
As discussed above, let $T$ be the length of the interval that is required to observe the first change. Then the following are the distribution function, survival function and pdf of $T$:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &F_T(t)=\displaystyle 1-e^{-\int_0^t \lambda(y) dy} \\&\text{ } \\&S_T(t)=\displaystyle e^{-\int_0^t \lambda(y) dy} \\&\text{ } \\&f_T(t)=\displaystyle \lambda(t) e^{-\int_0^t \lambda(y) dy} \end{aligned}

Proof. In Theorem 1, we derive the probability $P[N_y=0]$ for the discrete variable $N_y$ derived from the non-homogeneous Poisson process. We now consider the continuous random variable $T$, the time until the first change, which is related to $N_t$ by $(1)$. Thus $S_T(t)=P[T > t]=P[N_t=0]=e^{-\int_0^t \lambda(y) dy}$. The distribution function and density function can be derived accordingly.

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Theorem 3
The hazard rate function $\lambda(t)$ is equivalent to each of the following:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{f_T(t)}{1-F_T(t)} \\&\text{ } \\&\lambda(t)=\frac{-S_T^{'}(t)}{S_T(t)} \end{aligned}

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Remark
Theorem 1 and Theorem 2 show that in a non-homogeneous Poisson process as described above, the hazard rate function $\lambda(t)$ completely specifies the probability distribution of the survival model $T$ (the time until the first change) . Once the rate of change function $\lambda(t)$ is known in the non-homogeneous Poisson process, we can use it to generate the survival function $S_T(t)$. All of the examples of survival models given below are derived by assuming the functional form of the hazard rate function. The result in Theorem 2 holds even outside the context of a non-homogeneous Poisson process, that is, given the hazard rate function $\lambda(t)$, we can derive the three distributional items $S_T(t)$, $F_T(t)$, $f_T(t)$.

The ratio in Theorem 3 indicates that the probability distribution determines the hazard rate function. In fact, the ratio in Theorem 3 is the usual definition of the hazard rate function. That is, the hazard rate function can be defined as the ratio of the density and the survival function (one minus the cdf). With this definition, we can also recover the survival function. Whenever $\displaystyle \lambda(x)=\frac{f_X(x)}{1-F_X(x)}$, we can derive:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &S_X(x)=\displaystyle e^{-\int_0^t \lambda(y) dy} \end{aligned}

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As indicated above, the hazard rate function can be interpreted as the failure rate at time $t$ given that the life in question has survived to time $t$. It is the rate of failure at the next instant given that the life or system being studied has survived up to time $t$.

It is interesting to note that the function $\Lambda(t)=\int_0^t \lambda(y) dy$ defined in Theorem 1 is called the cumulative hazard rate function. Thus the cumulative hazard rate function is an alternative way of representing the hazard rate function (see the discussion on Weibull distribution below).

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Examples of Survival Models

–Exponential Distribution–
In many applications, especially those for biological organisms and mechanical systems that wear out over time, the hazard rate $\lambda(t)$ is an increasing function of $t$. In other words, the older the life in question (the larger the $t$), the higher chance of failure at the next instant. For humans, the probability of a 85 years old dying in the next year is clearly higher than for a 20 years old. In a Poisson process, the rate of change $\lambda(t)=\lambda$ indicated in condition 3 is a constant. As a result, the time $T$ until the first change derived in Theorem 2 has an exponential distribution with parameter $\lambda$. In terms of mortality study or reliability study of machines that wear out over time, this is not a realistic model. However, if the mortality or failure is caused by random external events, this could be an appropriate model.

–Weibull Distribution–
This distribution is an excellent model choice for describing the life of manufactured objects. It is defined by the following cumulative hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\biggl(\frac{t}{\beta}\biggr)^{\alpha} \end{aligned} where $\alpha > 0$ and $\beta>0$

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As a result, the hazard rate function, the density function and the survival function for the lifetime distribution are:

\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \\&\text{ } \\&f_T(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}} \end{aligned}

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The parameter $\alpha$ is the shape parameter and $\beta$ is the scale parameter. When $\alpha=1$, the hazard rate becomes a constant and the Weibull distribution becomes an exponential distribution.

When the parameter $\alpha<1$, the failure rate decreases over time. One interpretation is that most of the defective items fail early on in the life cycle. Once they they are removed from the population, failure rate decreases over time.

When the parameter $1<\alpha$, the failure rate increases with time. This is a good candidate for a model to describe the lifetime of machines or systems that wear out over time.

–The Gompertz Distribution–
The Gompertz law states that the force of mortality or failure rate increases exponentially over time. It describe human mortality quite accurately. The following is the hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t} \end{aligned} where $\alpha>0$ and $\beta>0$.

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The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t \alpha e^{\beta y} dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta} \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \\&\text{ } \\&f_T(t)=\displaystyle \alpha \ e^{\beta t} \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}\biggr)} \end{aligned}

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–Makeham’s Law–
The Makeham’s Law states that the force of mortality is the Gompertz failure rate plus an age-indpendent component that accounts for external causes of mortality. The following is the hazard rate function:

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\lambda(t)=\alpha e^{\beta t}+\mu \end{aligned} where $\alpha>0$, $\beta>0$ and $\mu>0$.

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The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.

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\displaystyle \begin{aligned}. \ \ \ \ \ \ \ \ \ &\Lambda(t)=\int_0^t (\alpha e^{\beta y}+\mu) dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t \\&\text{ } \\&S_T(t)=\displaystyle e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&F_T(t)=\displaystyle 1-e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \\&\text{ } \\&f_T(t)=\biggl( \alpha e^{\beta t}+\mu t \biggr) \ e^{\displaystyle -\biggl(\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t\biggr)} \end{aligned}

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The Tail Weight of a Distribution
The hazard rate function can provide information about the tail of a distribution. If the hazard rate function is decreasing, it is an indication that the distribution has a heavy tail, i.e., the distribution significantly puts more probability on larger values. Conversely, if the hazard rate function is increasing, it is an indication of a lighter tail. In an insurance context, heavy tailed distributions (e.g. the Pareto distribution) are suitable candidates for modeling large insurance losses (see this previous post or [1]).

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Reference

1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004

# The mean excess loss function

We take a closer look at the mean excess loss function. We start with the following example:

Example
Suppose that an entity is exposed to a random loss $X$. An insurance policy offers protection against this loss. Under this policy, payment is made to the insured entity subject to a deductible $d>0$, i.e. when a loss is less than $d$, no payment is made to the insured entity, and when the loss exceeds $d$, the insured entity is reimbursed for the amount of the loss in excess of the deductible $d$. Consider the following two questions:

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1. Of all the losses that are eligible to be reimbursed by the insurer, what is the average payment made by the insurer to the insured?
2. What is the average payment made by the insurer to the insured entity?

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The two questions look similar. The difference between the two questions is subtle but important. In the first question, the average is computed over all losses that are eligible for reimbursement (i.e., the loss exceeds the deductible). This is the average amount the insurer is expected to pay in the event that a payment in excess of the deductible is required to be made. So this average is a per payment average.

In the second question, the average is calculated over all losses (regardless of sizes). When the loss does not reach the deductible, the payment is considered zero and when the loss is in excess of the deductible, the payment is $X-d$. Thus the average is the average amount the insurer has to pay per loss. So the second question is about a per loss average.

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The Mean Excess Loss Function
The average in the first question is called the mean excess loss function. Suppose $X$ is the random loss and $d>0$. The mean excess loss variable is the conditional variable $X-d \ \lvert X>d$ and the mean excess loss function $e_X(d)$ is defined by:

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$\displaystyle (1) \ \ \ \ \ e_X(d)=E(X-d \lvert X>d)$

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In an insurance context, the mean excess loss function is the average payment in excess of a threshold given that the loss exceeds the threshold. In a mortality context, the mean excess loss function is called the mean residual life function and complete expectation of life and can be interpreted as the remaining time until death given that the life in question is alive at age $d$.

The mean excess loss function is computed by the following depending on whether the loss variable is continuous or discrete.

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$\displaystyle (2) \ \ \ \ \ e_X(d)=\frac{\int_d^\infty (x-d) \ f_X(x) \ dx}{S_X(d)}$

$\displaystyle (3) \ \ \ \ \ e_X(d)=\frac{\sum \limits_{x>d} (x-d) \ P(X=x)}{S_X(d)}$

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The mean excess loss function $e_X(d)$ is defined only when the integral or the sum converges. The following is an equivalent calculation of $e_X(d)$ that may be easier to use in some circumstances.

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$\displaystyle (4) \ \ \ \ \ e_X(d)=\frac{\int_d^\infty S_X(x) \ dx}{S_X(d)}$

$\displaystyle (5a) \ \ \ \ \ e_X(d)=\frac{\sum \limits_{x \ge d} S_X(x) }{S_X(d)}$

$\displaystyle (5b) \ \ \ \ \ e_X(d)=\frac{\biggl(\sum \limits_{x>d} S_X(x)\biggr)+(w+1-d) S_X(w) }{S_X(d)}$

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In both $(5a)$ and $(5b)$, we assume that the support of $X$ is the set of nonnegative integers. In $(5a)$, we assume that the deductible $d$ is a positive integer. In $(5b)$, the deductible $d$ is free to be any positive number and $w$ is the largest integer such that $w \le d$. The formulation $(4)$ is obtained by using integration by parts (also see theorem 3.1 in [1]). The formulations of $(5a)$ and $(5b)$ are a result of applying theorem 3.2 in [1].

The mean excess loss function provides information about the tail weight of a distribution, see the previous post The Pareto distribution. Also see Example 3 below.
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The Mean in Question 2
The average that we need to compute is the mean of the following random variable. Note that $(a)_+$ is the function that assigns the value of $a$ whenever $a>0$ and otherwise assigns the value of zero.

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$\displaystyle (6) \ \ \ \ \ (X-d)_+=\left\{\begin{matrix}0&\ X

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The mean $E((X-d)_+)$ is calculated over all losses. When the loss is less than the deductible $d$, the insurer has no obligation to make a payment to the insured and the payment is assumed to be zero in the calculation of $E[(X-d)_+]$. The following is how this expected value is calculated depending on whether the loss $X$ is continuous or discrete.

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$\displaystyle (7) \ \ \ \ \ E((X-d)_+)=\int_d^\infty (x-d) \ f_X(x) \ dx$

$\displaystyle (8) \ \ \ \ \ E((X-d)_+)=\sum \limits_{x>d} (x-d) \ P(X=x)$

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Based on the definitions, the following is how the two averages are related.

$\displaystyle E[(X-d)_+]=e_X(d) \ [1-F_X(d)] \ \ \ \text{or} \ \ \ E[(X-d)_+]=e_X(d) \ S_X(d)$

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The Limited Expected Value
For a given positive constant $u$, the limited loss variable is defined by

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$\displaystyle (9) \ \ \ \ \ X \wedge u=\left\{\begin{matrix}X&\ X

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The expected value $E(X \wedge u)$ is called the limited expected value. In an insurance application, the $u$ is a policy limit that sets a maximum on the benefit to be paid. The following is how the limited expected value is calculated depending on whether the loss $X$ is continuous or discrete.

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$\displaystyle (9) \ \ \ \ \ E(X \wedge u)=\int_{-\infty}^u x \ f_X(x) \ dx+u \ S_X(u)$

$\displaystyle (10) \ \ \ \ \ E(X \wedge u)=\biggl(\sum \limits_{x < u} x \ P(X=x)\biggr)+u \ S_X(u)$

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Interestingly, we have the following relation.

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

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The above statement indicates that purchasing a policy with a deductible $d$ and another policy with a policy maximum $d$ is equivalent to buying full coverage.

Another way to interpret $X \wedge d$ is that it is the amount of loss that is eliminated by having a deductible in the insurance policy. If the insurance policy pays the loss in full, then the insurance payment is $X$ and the expected amount the insurer is expected to pay is $E(X)$. By having a deductible provision in the policy, the insurer is now only liable for the amount $(X-d)_+$ and the amount the insurer is expected to pay per loss is $E[(X-d)_+]$. Consequently $E(X \wedge d)$ is the expected amount of the loss that is eliminated by the deductible provision in the policy. The following summarizes this observation.

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

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Example 1
Let the loss random variable $X$ be exponential with pdf $f(x)=\alpha e^{-\alpha x}$. We have $E(X)=\frac{1}{\alpha}$. Because of the no memory property of the exponential distribution, given that a loss exceeds the deductible, the mean payment is the same as the original mean. Thus $e_X(d)=\frac{1}{\alpha}$. Then the per loss average is:

$\displaystyle E[(X-d)_+]=e_X(d) \ S(d) = \frac{e^{-\alpha d}}{\alpha}$

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Thus, with a deductible provision in the policy, the insurer is expected to pay $\displaystyle \frac{e^{-\alpha d}}{\alpha}$ per loss instead of $\displaystyle \frac{1}{\alpha}$. Thus the expected amount of loss eliminated (from the insurer’s point of view) is $\displaystyle E(X \wedge d)=\frac{1-e^{-\alpha d}}{\alpha}$.

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Example 2
Suppose that the loss variable has a Gamma distribution where the scale parameter is $\alpha$ and the shape parameter is $n=2$. The pdf is $\displaystyle g(x)=\alpha^2 \ x \ e^{-\alpha x}$. The insurer’s expected payment without the deductible is $E(X)=\frac{2}{\alpha}$. The survival function $S(x)$ is:

$\displaystyle S(x)=e^{-\alpha x}(1+\alpha x)$

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For the losses that exceed the deductible, the insurer’s expected payment is:

\displaystyle \begin{aligned}e_X(d)&=\frac{\int_d^{\infty} S(x) \ dx}{S(d)}\\&=\frac{\int_d^{\infty} e^{-\alpha x}(1+\alpha x) \ dx}{e^{-\alpha d}(1+\alpha d)} \\&=\frac{\frac{e^{-\alpha d}}{\alpha}+d e^{-\alpha d}+\frac{e^{-\alpha d}}{\alpha}}{e^{-\alpha d}(1+\alpha d)} \\&=\frac{\frac{2}{\alpha}+d}{1+\alpha d} \end{aligned}

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Then the insurer’s expected payment per loss is $E[(X-d)_+]$:

\displaystyle \begin{aligned}E[(X-d)_+]&=e_X(d) \ S(d) \\&=\frac{\frac{2}{\alpha}+d}{1+\alpha d} \ \ e^{-\alpha d}(1+\alpha d) \\&=e^{-\alpha d} \ \biggl(\frac{2}{\alpha}+d\biggr) \end{aligned}

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With a deductible in the policy, the following is the expected amount of loss eliminated (from the insurer’s point of view).

\displaystyle \begin{aligned}E[X \wedge d]&=E(X)-E[(X-d)_+] \\&=\frac{2}{\alpha}-e^{-\alpha d} \ \biggl(\frac{2}{\alpha}+d\biggr) \\&=\frac{2}{\alpha}\biggl(1-e^{-\alpha d}\biggr)-d e^{-\alpha d} \end{aligned}

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Example 3
Suppose the loss variable $X$ has a Pareto distribution with the following pdf:

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

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If the insurance policy is to pay the full loss, then the insurer’s expected payment per loss is $\displaystyle E(X)=\frac{\alpha}{\beta-1}$ provided that the shape parameter $\beta$ is larger than one.

The mean excess loss function of the Pareto distribution has a linear form that is increasing (see the previous post The Pareto distribution). The following is the mean excess loss function:

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

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If the loss is modeled by such a distribution, this is an uninsurable risk! First of all, the higher the deductible, the larger the expected payment if such a large loss occurs. The expected payment for large losses is always the unmodified expected $E(X)$ plus a component that is increasing in $d$.

The increasing mean excess loss function is an indication that the Pareto distribution is a heavy tailed distribution. In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. On the other hand, a decreasing mean excess loss function indicates a light tailed distribution. The exponential distribution has a constant mean excess loss function and is considered a medium tailed distribution.

Reference

1. Bowers N. L., Gerber H. U., Hickman J. C., Jones D. A., Nesbit C. J. Actuarial Mathematics, First Edition., The Society of Actuaries, Itasca, Illinois, 1986
2. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004

# The Pareto distribution

This post takes a closer look at the Pareto distribution. A previous post demonstrates that the Pareto distribution is a mixture of exponential distributions with Gamma mixing weights. We now elaborate more on this point. Through looking at various properties of the Pareto distribution, we also demonstrate that the Pareto distribution is a heavy tailed distribution. In insurance applications, heavy-tailed distributions are essential tools for modeling extreme loss, especially for the more risky types of insurance such as medical malpractice insurance. In financial applications, the study of heavy-tailed distributions provides information about the potential for financial fiasco or financial ruin. The Pareto distribution is a great way to open up a discussion on heavy-tailed distribution.

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Update (11/12/2017). This blog post introduces a catalog of many other parametric severity models in addition to Pareto distribution. The link to the catalog is found in that blog post. To go there directly, this is the link.

Update (10/29/2017). This blog post has updated information on Pareto distribution. It also has links to more detailed contents on Pareto distribution in two companion blogs. These links are also given here: more detailed post on Pareto, Pareto Type I and Type II and practice problems on Pareto.

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The continuous random variable $X$ with positive support is said to have the Pareto distribution if its probability density function is given by

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

where $\alpha>0$ and $\beta>0$ are constant. The constant $\alpha$ is the scale parameter and $\beta$ is the shape parameter. The following lists several other distributional quantities of the Pareto distribution, which will be used in the discussion below.

$\displaystyle S_X(x)=\frac{\alpha^\beta}{(x+\alpha)^\beta}=\biggl(\frac{\alpha}{x+\alpha}\biggr)^\beta \ \ \ \ \ \ \ \ \ \text{survival function}$

$\displaystyle F_X(x)=1-\biggl(\frac{\alpha}{x+\alpha}\biggr)^\beta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{distribution function}$

$\displaystyle E(X)=\frac{\alpha}{\beta-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mean},\beta>1$

$\displaystyle E(X^2)=\frac{2 \alpha^2}{(\beta-1)(\beta-2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{second momemt},\beta>2$

$\displaystyle Var(X)=\frac{\alpha^2 \beta}{(\beta-1)^2(\beta-2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{variance},\beta>2$

$\displaystyle E(X^k)=\frac{k! \alpha^k}{(\beta-1) \cdots (\beta-k)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{higher moments},\beta>k, \text{ k positive integer}$

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The Pareto Distribution as a Mixture
The Pareto pdf indicated above can be obtained by mixing exponential distributions using Gamma distributions as weights. Suppose that $X$ follows an exponential distribution (conditional on a parameter value $\theta$). The following is the conditional pdf of $X$.

$\displaystyle f_{X \lvert \Theta}(x \lvert \theta)=\theta e^{-\theta x} \ \ \ x>0$

There is uncertainty in the parameter, which can be viewed as a random variable $\Theta$. Suppose that $\Theta$ follows a Gamma distribution with scale parameter $\alpha$ and shape parameter $\beta$. The following is the pdf of $\Theta$.

$\displaystyle f_{\Theta}(\theta)=\frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta} \ \ \ \theta>0$

The unconditional pdf of $X$ is the weighted average of the conditional pdfs with the Gamma pdf as weight.

\displaystyle \begin{aligned}f_X(x)&=\int_0^{\infty} f_{X \lvert \Theta}(x \lvert \theta) \ f_\Theta(\theta) \ d \theta \\&=\int_0^{\infty} \biggl[\theta \ e^{-\theta x}\biggr] \ \biggl[\frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^{\beta-1} \ e^{-\alpha \theta}\biggr] \ d \theta \\&=\int_0^{\infty} \frac{\alpha^\beta}{\Gamma(\beta)} \ \theta^\beta \ e^{-\theta(x+\alpha)} \ d \theta \\&=\frac{\alpha^\beta}{\Gamma(\beta)} \frac{\Gamma(\beta+1)}{(x+\alpha)^{\beta+1}} \int_0^{\infty} \frac{(x+\alpha)^{\beta+1}}{\Gamma(\beta+1)} \ \theta^{\beta+1-1} \ e^{-\theta(x+\alpha)} \ d \theta \\&=\frac{\beta \ \alpha^\beta}{(x+\alpha)^{\beta+1}} \end{aligned}

In the following discussion, $X$ will denote the Pareto distribution as defined above. As will be shown below, the exponential distribution is considered a light tailed distribution. Yet mixing exponentials produces the heavy tailed Pareto distribution. Mixture distributions tend to heavy tailed (see [1]). The Pareto distribution is a handy example.

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The Tail Weight of the Pareto Distribution
When a distribution significantly puts more probability on larger values, the distribution is said to be a heavy tailed distribution (or said to have a larger tail weight). According to [1], there are four ways to look for indication that a distribution is heavy tailed.

1. Existence of moments.
2. Speed of decay of the survival function to zero.
3. Hazard rate function.
4. Mean excess loss function.

Existence of moments
Note that the existence of the Pareto higher moments $E(X^k)$ is capped by the shape parameter $\beta$. In particular, the mean $E(X)=\frac{\alpha}{\beta-1}$ does not exist for $\beta \le 1$. If the Pareto distribution is to model a random loss, and if the mean is infinite (when $\beta=1$), the risk is uninsurable! On the other hand, when $\beta=2$, the Pareto variance does not exist. This shows that for a heavy tailed distribution, the variance may not be a good measure of risk.

For a given random variable $Z$, the existence of all moments $E(Z^k)$, for all positive integers $k$, indicates with a light (right) tail for the distribution of $Z$. The existence of positive moments exists only up to a certain value of a positive integer $k$ is an indication that the distribution has a heavy right tail. In contrast, the exponential distribution and the Gamma distribution are considered to have light tails since all moments exist.

The speed of decay of the survival function
The survival function $S_X(x)=P(X>x)$ captures the probability of the tail of a distribution. If a distribution whose survival function decays slowly to zero (equivalently the cdf goes slowly to one), it is another indication that the distribution is heavy tailed.

The following is a comparison of a Pareto survival function and an exponential survival function. The Pareto survival function has parameters ($\alpha=2$ and $\beta=2$). The two survival functions are set to have the same 75th percentile ($x=2$).

$\displaystyle \begin{pmatrix} \text{x}&\text{Pareto }S_X(x)&\text{Exponential }S_Y(x)&\displaystyle \frac{S_X(x)}{S_Y(x)} \\\text{ }&\text{ }&\text{ }&\text{ } \\{2}&0.25&0.25&1 \\{10}&0.027777778&0.000976563&28 \\{20}&0.008264463&9.54 \times 10^{-7}&8666 \\{30}&0.00390625&9.31 \times 10^{-10}&4194304 \\{40}&0.002267574&9.09 \times 10^{-13}&2.49 \times 10^{9} \\{60}&0.001040583&8.67 \times 10^{-19}&1.20 \times 10^{15} \\{80}&0.000594884&8.27 \times 10^{-25}&7.19 \times 10^{20} \\{100}&0.000384468&7.89 \times 10^{-31}&4.87 \times 10^{26} \\{120}&0.000268745&7.52 \times 10^{-37}&3.57 \times 10^{32} \\{140}&0.000198373&7.17 \times 10^{-43}&2.76 \times 10^{38} \\{160}&0.000152416&6.84 \times 10^{-49}&2.23 \times 10^{44} \\{180}&0.000120758&6.53 \times 10^{-55}&1.85 \times 10^{50} \end{pmatrix}$

Note that at the large values, the Pareto right tails retain much more probability. This is also confirmed by the ratio of the two survival functions, with the ratio approaching infinity. If a random loss is a heavy tailed phenomenon that is described by the above Pareto survival function ($\alpha=2$ and $\beta=2$), then the above exponential survival function is woefully inadequate as a model for this phenomenon even though it may be a good model for describing the loss up to the 75th percentile. It is the large right tail that is problematic (and catastrophic)!

Since the Pareto survival function and the exponential survival function have closed forms, We can also look at their ratio.

$\displaystyle \frac{\text{pareto survival}}{\text{exponential survival}}=\frac{\displaystyle \frac{\alpha^\beta}{(x+\alpha)^\beta}}{e^{-\lambda x}}=\frac{\alpha^\beta e^{\lambda x}}{(x+\alpha)^\beta} \longrightarrow \infty \ \text{ as } x \longrightarrow \infty$

In the above ratio, the numerator has an exponential function with a positive quantity in the exponent, while the denominator has a polynomial in $x$. This ratio goes to infinity as $x \rightarrow \infty$.

In general, whenever the ratio of two survival functions diverges to infinity, it is an indication that the distribution in the numerator of the ratio has a heavier tail. When the ratio goes to infinity, the survival function in the numerator is said to decay slowly to zero as compared to the denominator. We have the same conclusion in comparing the Pareto distribution and the Gamma distribution, that the Pareto is heavier in the tails. In comparing the tail weight, it is equivalent to consider the ratio of density functions (due to the L’Hopital’s rule).

$\displaystyle \lim_{x \rightarrow \infty} \frac{S_1(x)}{S_2(x)}=\lim_{x \rightarrow \infty} \frac{S_1^{'}(x)}{S_2^{'}(x)}=\lim_{x \rightarrow \infty} \frac{f_1(x)}{f_2(x)}$

The Hazard Rate Function
The hazard rate function $h_X(x)$ of a random variable $X$ is defined as the ratio of the density function and the survival function.

$\displaystyle h_X(x)=\frac{f_X(x)}{S_X(s)}$

The hazard rate is called the force of mortality in a life contingency context and can be interpreted as the rate that a person aged $x$ will die in the next instant. The hazard rate is called the failure rate in reliability theory and can be interpreted as the rate that a machine will fail at the next instant given that it has been functioning for $x$ units of time. The following is the hazard rate function of the Pareto distribution.

\displaystyle \begin{aligned}h_X(x)&=\frac{f_X(s)}{S_X(x)} \\&=\frac{\beta}{x+\alpha} \end{aligned}

The interesting point is that the Pareto hazard rate function is an decreasing function in $x$. Another indication of heavy tail weight is that the distribution has a decreasing hazard rate function. One key characteristic of hazard rate function is that it can generate the survival function.

$\displaystyle S_X(x)=e^{\displaystyle -\int_0^x h_X(t) \ dt}$

Thus if the hazard rate function is decreasing in $x$, then the survival function will decay more slowly to zero. To see this, let $H_X(x)=\int_0^x h_X(t) \ dt$, which is called the cumulative hazard rate function. As indicated above, the survival function can be generated by $e^{-H_X(x)}$. If $h_X(x)$ is decreasing in $x$, $H_X(x)$ is smaller than $H_Y(x)$ where $h_Y(x)$ is constant in $x$ or increasing in $x$. Consequently $e^{-H_X(x)}$ is decaying to zero much more slowly than $e^{-H_Y(x)}$.

In contrast, the exponential distribution has a constant hazard rate function, making it a medium tailed distribution. As explained above, any distribution having an increasing hazard rate function is a light tailed distribution.

The Mean Excess Loss Function
Suppose that a property owner is exposed to a random loss $Y$. The property owner buys an insurance policy with a deductible $d$ such that the insurer will pay a claim in the amount of $Y-d$ if a loss occurs with $Y>d$. The insuerer will pay nothing if the loss is below the deductible. Whenever a loss is above $d$, what is the average claim the insurer will have to pay? This is one way to look at mean excess loss function, which represents the expected excess loss over a threshold conditional on the event that the threshold has been exceeded.

Given a loss variable $Y$ and given a deductible $d>0$, the mean excess loss function is $e_Y(d)=E(Y-d \lvert X>d)$. For a continuous random variable, it is computed by

$\displaystyle e_Y(d)=\frac{\int_d^{\infty} (y-d) \ f_Y(y) \ dy}{S_Y(d)}$

Applying the technique of integration by parts produces the following formula:

$\displaystyle e_Y(d)=\frac{\int_d^{\infty} S_Y(y) \ dy}{S_Y(d)}$

It turns out that the mean excess loss function is one more way to examine the tail property of a distribution. The following is the mean excess loss function of the Pareto distribution:

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

Note that the Pareto mean excess loss function is a linear increasing function of the deductible $d$. This means that the larger 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. On the other hand, a decreasing mean excess loss function indicates a light tailed distribution. The exponential distribution has a constant mean excess loss function and is considered a medium tailed distribution.

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The Pareto distribution has many economic applications. Since it is a heavy tailed distribution, it is a good candidate for modeling income above a theoretical value and the distribution of insurance claims above a threshold value.

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Reference

1. Klugman S.A., Panjer H. H., Wilmot G. E. Loss Models, From Data to Decisions, Second Edition., Wiley-Interscience, a John Wiley & Sons, Inc., New York, 2004