# Defining Hazard Rate at a Point Mass

The hazard rate function $h_T(t)$, also known as the force of mortality or the failure rate, is defined as the ratio of the density function and the survival function. That is, $\displaystyle h_T(t)=\frac{f_T(t)}{S_T(t)}$, where $T$ is the survival model of a life or a system being studied. In this definition, $T$ is usually taken as a continuous random variable with nonnegative real values as support. In this post we attempt to define the hazard rate at the places that are point masses (probability masses). This definition will cover discrete survival models as well as mixed survival models (i.e. models that are continuous in some interval and also have point masses). This post is in reponse to one comment posted by a reader. The comment is in response to the post The hazard rate function, an introduction
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If the suvival model $T$ is an exponential distribution, the hazard rate is constant. When the exponential survival model is censored on the right at some value of maximum lifetime, what is the hazard rate at the maximum? This is essentially the question posted by one reader of this blog. The following is the graph of the cdf $F_T(t)=1-e^{-0.25 t}$ censored at $t_{max}=5$.

We attempt to define the hazard at a probablity mass such as the one in Figure 1. The same definition woulod apply for any discrete probability model.

As indicated at the beginning of the post, the hazard rate function is defined as the following ratio:

$\displaystyle (1) \ \ \ \ \ h_T(t)=\frac{f_T(t)}{1-F_T(t)}=\frac{f_T(t)}{S_T(t)}$

where $f_T$, $F_T$ and $S_T$ are the density function, cumulative distribution function (cdf) and the survival function of a given survival model $T$. This definition is usually made at the points $T=t$ where it makes sense to take derivative of $F_T(t)$. The hazard rate thus defined 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 has survived up to time $t$.

Suppose that $T=t$ is a point mass (such as $T=5$ in Figure 1). The hazard rate at such points is defined by the same idea. We define the hazard rate at a point mass as the probability of failing at time $t$ given that the life has survived up to that time.

$\displaystyle (2) \ \ \ \ \ h_T(t)=\frac{P(T=t)}{P(T \ge t)}$

Note that both $(1)$ and $(2)$ are of the same general form (the ratio of density to suvival function) and have the same interpretation. However, $(2)$ is actually a conditional probability, while $(1)$ can only be a rate of failure. The hazard rate as in $(1)$ technically cannot be a probability since it can be greater than 1.

The hazard rate at $T=5$ in Figure 1 is 1.0. We can derive this using $(2)$, or we can think about the meaning of $(2)$. Note that the point mass in Figure 1 is the maximum lifetime. Any life reaches that point is considered a termination (perhaps the person drops out of the study). So given that the life reaches this maximum point, it is certain that the life fails at this point (hence the conditional probability as defined by $(2)$ is 1.0).

So if the point mass is at the last point of the time scale in the surviva model, the hazard rate is 1.0, representing that 100% of the survived lives die off. However, the hazard rate at a point mass at $T=t$ prior to the maximum point is less than 1.0 and is the size of the jump in the cdf at $T=t$ as a fraction of the probability of survival up to that point.

We close with a simple example illustrating the calculation of hazard rate for discrete survival model. Our example is the uniform model $T$ at $t=1,2,3,4,5$. The following is the graph of its cdf.

The following table defines the hazard rates.

$\displaystyle (3) \ \ \ \ \ \begin{bmatrix} \text{t}&\text{ }&P(T=t) &\text{ }&P(T \ge t) &\text{ }&h_T(t) \\\text{ }&\text{ }&\text{ } \\ 1&\text{ }&\displaystyle \frac{1}{5}&\text{ }& \displaystyle \frac{5}{5}&\text{ }& \displaystyle \frac{1}{5} \\\text{ }&\text{ }&\text{ } \\ 2&\text{ }& \displaystyle \frac{1}{5}&\text{ }& \displaystyle \frac{4}{5}&\text{ }& \displaystyle \frac{1}{4} \\\text{ }&\text{ }&\text{ } \\ 3&\text{ }& \displaystyle \frac{1}{5}&\text{ }& \displaystyle \frac{3}{5}&\text{ }& \displaystyle \frac{1}{3} \\\text{ }&\text{ }&\text{ } \\ 4&\text{ }& \displaystyle \frac{1}{5}&\text{ }& \displaystyle \frac{2}{5}&\text{ }& \displaystyle \frac{1}{2} \\\text{ }&\text{ }&\text{ } \\ 5&\text{ }& \displaystyle \frac{1}{5}&\text{ }& \displaystyle \frac{1}{5}&\text{ }&1 \end{bmatrix}$

The hazard rates in the above table are calculated using $(2)$. We would like to point out that the calculated hazard rates conform to the mortality pattern that is expected in a uniform model. Note that at the first point mass, one fifth of the lives die off. At the second point mass, one fourth of the survived die off and so on. Then at the last point mass, 100% of the survived die off.

# The hazard rate function, an introduction

The goal of this post is to introduce the concept of hazard rate function by modifying one of the postulates of the approximate Poisson process. The rate of changes in the modified process is the hazard rate function. When a “change” in the modified Poisson process means a termination of a system (be it manufactured or biological), the notion of the hazard rate function leads to the concept of survival models. We then discuss several important examples of survival probability models that are defined by the hazard rate function. These examples include the Weibull distribution, the Gompertz distribution and the model based on the Makeham’s law.

We consider an experiment in which the occurrences of a certain type of events are counted during a given time interval or on a given physical object. Suppose that we count the occurrences of events on the interval $(0,t)$. 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 $\lambda(t) \delta$ where $\delta$ is sufficiently small and $\lambda(t)$ is a nonnegative function of $t$.

For the lack of a better name, throughout this post, we call the above process the counting process (*). The approximate Poisson process is defined by conditions 1 and 2 and the condition that the $\lambda(t)$ in condition 3 is a constant function. Thus the process we describe here is a more general process than the Poisson process.

Though the counting process indicated here can model the number of changes occurred in a physical object or a physical interval, we focus on the time aspect by considering the counting process as models for the number of changes occurred in a time interval where a change means “termination” or ‘failure” of a system under consideration. In many applications (e.g. in actuarial science and reliability engineering), the interest is on the time until termination or failure. Thus, the distribution for the time until failure is called a survival model. 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. In actuarial science, the hazard rate function is known as the force of mortality.

Two random variables naturally arise from the counting process (*). 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 (or next) change.

Claim 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, $\displaystyle P[N_t=0]=e^{-\Lambda(t)}$.

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:

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

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 $(a)$, we have the following derivation:

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

$\displaystyle \frac{d}{dy} P[N_y=0]=-\lambda(y) P[N_y=0]$

$\displaystyle \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]}=-\lambda(y)$

$\displaystyle \int_0^{t} \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]} dy=-\int_0^{t} \lambda(y)dy$

Integrating the left hand side and using the boundary condition of $P[N_0=0]=1$, we have:

$\displaystyle ln P[N_t=0]=-\int_0^{t} \lambda(y)dy$

$\displaystyle P[N_t=0]=e^{-\int_0^{t} \lambda(y)dy}$

Claim 2
As discussed above, let $T$ be the length of the interval that is required to observe the first change in the counting process (*). Then the following are the distribution function, survival function and pdf of $T$:

• $\displaystyle F_T(t)=\displaystyle 1-e^{-\int_0^t \lambda(y) dy}$
• $\displaystyle S_T(t)=\displaystyle e^{-\int_0^t \lambda(y) dy}$
• $\displaystyle f_T(t)=\displaystyle \lambda(t) e^{-\int_0^t \lambda(y) dy}$

In Claim 1, we derive the probability $P[N_y=0]$ for the discrete variable $N_y$ derived from the counting process (*). We now consider the continuous random variable $T$. Note that $P[T > t]$ is the probability that the first change occurs after time $t$. This means there is no change within the interval $(0,t)$. 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.

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

• $\displaystyle \lambda(t)=\frac{f_T(t)}{1-F_T(t)}$
• $\displaystyle \lambda(t)=\frac{-S_T^{'}(t)}{S_T(t)}$

Remark
Based on the condition 3 in the counting process (*), the $\lambda(t)$ is the rate of change in the counting process. Note that $\lambda(t) \delta$ is the probability of a change (e.g. a failure or a termination) in a small time interval of length $\delta$. Thus the hazard rate function can be interpreted as the failure rate at time $t$ given that the life in question has survived to time $t$. Claim 3 shows that the hazard rate function is the ratio of the density function and the survival function of the time until failure variable $T$. Thus the hazard rate function $\lambda(t)$ is the conditional density of failure at 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 claim 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).

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 claim 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:

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

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

$\displaystyle \lambda(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1}$

$\displaystyle f_T(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}}$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}}$

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:

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

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

$\displaystyle \Lambda(t)=\int_0^t \alpha e^{\beta y} dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

$\displaystyle F_T(t)=\displaystyle 1-e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

$\displaystyle f_T(t)=\displaystyle \alpha e^{\beta t} \thinspace e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

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:

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

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

$\displaystyle \Lambda(t)=\int_0^t (\alpha e^{\beta y}+\mu) dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$

$\displaystyle F_T(t)=\displaystyle 1-e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$

$\displaystyle f_T(t)=\biggl( \alpha e^{\beta t}+\mu t \biggr) \thinspace e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$