# 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.