Consider a non-negative random variable $T$ arising in the context of survival/reliability analysis (usually representing the time until some event occurs). It is well-known that the survival function is related to the hazard function by the equation:

$$S(t) = \exp \Bigg( - \int \limits_0^t \lambda(r) \ dr \Bigg).$$

In some recent comments to an answer on this site there was some discussion over the properties of the hazard function, and the assertion that its integral diverges to infinity. It was pointed out that a finite integral for the hazard function means there is some non-zero probability of infinite survival. To see this, we simply note that $\mathbb{P}(T = \infty) = \lim_{t \rightarrow \infty} S(t)$, so we have:

$$\begin{equation} \begin{aligned} \mathbb{P}(T = \infty) = 0 & & \iff & & \int \limits_0^\infty \lambda(t) \ dt = \infty, \\[12pt] \mathbb{P}(T = \infty) > 0 & & \iff & & \int \limits_0^\infty \lambda(t) \ dt < \infty. \\[6pt] \end{aligned} \end{equation}$$

There was some debate in the linked comments over whether this latter case is sensible or not, but it was hard to get a clear answer in a comment thread.

Question: Can the hazard rate integrate to a finite value? Even if this is mathematically allowable, is there any realistic case in survival/reliability analysis where this would actually occur?

  • $\begingroup$ Re "it was pointed out:" as I remarked in an earlier thread, that invokes an incorrect concept of the integral. When the chance of an infinite outcome is nonzero, the integral is infinite. $\endgroup$
    – whuber
    Oct 22, 2019 at 13:22
  • $\begingroup$ Sorry, but I'm not quite sure what you mean. My understanding is that a fininte hazard integral yields a non-zero probability of infinite time value. I realise that if you are not working in the extended reals then "infinity" becomes "does not exist", but even then you get divergence towards infinity. Is that what you are referring to, or am I missing something else? $\endgroup$
    – Ben
    Oct 22, 2019 at 21:32
  • $\begingroup$ "Infinity" and "does not exist" are not synonymous, Ben. Also, I'm not referring to divergence, but to arithmetic in the extended reals as described, for instance, in Rudin's book on real and complex analysis. $\endgroup$
    – whuber
    Oct 23, 2019 at 14:23
  • $\begingroup$ I don't see any mathematical or practical reason why a hazard should integrate to infinity. Practically, there are often populations where the survival can asymptotically not reach 0 - ex: cure models, where a subpopulation is cured of the disease. $\endgroup$ Oct 23, 2019 at 15:48
  • 1
    $\begingroup$ I don't see any problems in either place. $\endgroup$
    – whuber
    Oct 23, 2019 at 21:09

1 Answer 1


There are many realistic cases where this phenomenon arises, but I will show one main class of cases. Suppose we have a reliability problem with some object that will fail within a finite time. Let $T \geqslant 0$ be the (continuous) random failure time for that object, and let $\lambda$ be the hazard function for the failure time. Since the object is assumed to fail within a finite time, we must have:

$$\int \limits_0^\infty \lambda(t) dt = \infty.$$

Now, suppose we start observing this object at some time $t_0 > 0$, and we are unable to observe what state the object is in, but we can observe the failure event itself. Let $T_*$ be the remaining time until failure, from the time at which we start observing the object. Now, if $0 \leqslant T < t_0$ then the object has already failed before we start observing it, so we will never observe the failure event, and so $T_* = \infty$. This outcome occurs with probability $\mathbb{P}(T_*=\infty) = F_T(t_0)$.

Assuming that this event occurs with non-zero probability (i.e., there is some non-zero probability of failure prior to $t_0$), let's see what this means for the hazard function of $T_*$. For all $0 \leqslant t < \infty$ we have:

$$\begin{equation} \begin{aligned} F_{T_*}(t) &= F_{T}(t_0+t) - F_{T}(t_0), \\[6pt] f_{T_*}(t) &= f_{T}(t_0+t). \\[6pt] \end{aligned} \end{equation}$$

Thus, the hazard rate for the remaining failure time $T_*$ is:

$$\begin{equation} \begin{aligned} \lambda_*(t) &= \frac{f_{T_*}(t)}{1-F_{T_*}(t)} \\[6pt] &= \frac{f_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \\[6pt] &= \frac{1-F_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \cdot \frac{f_{T}(t_0+t)}{1-F_{T}(t+t_0)} \\[6pt] &= \frac{1-F_{T}(t_0+t)}{1-F_{T}(t_0+t) + F_{T}(t_0)} \cdot \lambda(t+t_0). \\[6pt] \\[6pt] \end{aligned} \end{equation}$$

We can see that this hazard function is composed of the initial hazard function for $T$, multiplied by an adjustment term. If there is a non-zero probability that the object has already failed prior to observation, then the adjustment term diminishes to zero as $t \rightarrow \infty$. In fact, it can be shown that:

$$\int \limits_0^\infty \lambda_*(t) dt = - \ln \mathbb{P}(T_*=\infty) < \infty.$$

We therefore see that, if there is a non-zero probability that the object has already failed prior to observation, then we have a random variable with a hazard rate that does not integrate to infinity. Note that this latter random variable arises as a simple variation on the initial problem, where we begin observing the object after a period of time, and we only observe the failure event, not the state of the object.

Example: Suppose we have constant hazard rate $\lambda(t) = 1$, so that $F_T(t) = 1-\exp(- t)$ and $\mathbb{P}(T_*=\infty) = 1-\exp(- t_0)$. Then the hazard function for $T_*$ is:

$$\lambda_*(t) = \frac{\exp(-(t_0+t))}{\exp(-(t_0+t)) + 1-\exp(-t_0)}.$$

You can plot this hazard function and see that it reduces quite rapidly towards zero. Integrating this function gives:

$$\begin{equation} \begin{aligned} \int \limits_0^\infty \lambda_*(t) \ dt &= \Bigg[ - \ln \big| \exp(-t) + \exp(t_0) - 1 \big| \Bigg]_{t=0}^{t \rightarrow \infty} \\[6pt] &= \Bigg[ (- \ln \big| \exp(t_0) - 1 \big|) - (- \ln \big| \exp(t_0) \big| ) \Bigg] \\[6pt] &= - \ln \bigg( \frac{ \exp(t_0)-1}{\exp(t_0)} \bigg) \\[6pt] &= - \ln ( 1-\exp(-t_0) ) \\[6pt] &= - \ln \mathbb{P}(T_*=\infty). \\[6pt] \end{aligned} \end{equation}$$


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