I am confused about the equation that serves as the definition of the hazard rate. I get the idea of what the hazard rate is, but I just don't see how the equation expresses that intuition.

If $x$ is a random variable which represents the point of time of death of someone on a time interval $[0,T]$. Then the hazard rate is:


Where $F(x)$ represents the probability of death until time point $x\in[0,T]$,
$1-F(x)$ represents the probability of having survived up until time point $x\in[0,T]$,
and $f(x)$ is the probability of death at point $x$.

How does dividing $f(x)$ by the survival rate explain the intuition of the probability of instantaneous death in the next $\Delta t$? Shouldn't it just be $f(x)$, making the calculation of the hazard rate trivial?

  • $\begingroup$ You should appreciate that the definition of "hazard" differs according to what kind of survival model you are working with. E.g., the hazard in Weibull survival analysis is different than the baseline hazard of discrete time logit hazard models. $\endgroup$
    – Alexis
    Commented Jun 13, 2021 at 4:48

4 Answers 4


Let $X$ denote the time of death (or time of failure if you prefer a less morbid description). Suppose that $X$ is a continuous random variable whose density function $f(t)$ is nonzero only on $(0,\infty)$. Now, notice that it must be the case that $f(t)$ decays away to $0$ as $t \to \infty$ because if $f(t)$ does not decay away as stated, then $\displaystyle \int_{-\infty}^\infty f(t)\,\mathrm dt = 1$ cannot hold. Thus, your notion that $f(T)$ is the probability of death at time $T$ (actually, it is $f(T)\Delta t$ that is (approximately) the probability of death in the short interval $(T, T+\Delta t]$ of length $\Delta t$) leads to implausible and unbelievable conclusions such as

You are more likely to die within the next month when you are thirty years old than when you are ninety-eight years old.

whenever $f(t)$ is such that $f(30) > f(98)$.

The reason why $f(T)$ (or $f(T)\Delta t$) is the "wrong" probability to look at is that the value of $f(T)$ is of interest only to those who are alive at age $T$ (and still mentally alert enough to read stats.SE on a regular basis!) What ought to be looked at is the probability of a $T$-year old dying within the next month, that is,

\begin{align}P\{(X \in (T, T+\Delta t] \mid X \geq T\} &= \frac{P\{\left(X \in (T, T+\Delta t]\right) \cap \left(X\geq T\right)\}}{P\{X\geq T\}} & \\ \scriptstyle{ \text{ definition of conditional probability}}\\ &= \frac{P\{X \in (T, T+\Delta t]\}}{P\{X\geq T\}}\\ &= \frac{f(T)\Delta t}{1-F(T)} & \\ \scriptstyle{ \text{because }X\text{ is a continuous rv}} \end{align}

Choosing $\Delta t$ to be a fortnight, a week, a day, an hour, a minute, etc. we come to the conclusion that the (instantaneous) hazard rate for a $T$-year old is

$$h(T) = \frac{f(T)}{1-F(T)}$$

in the sense that the approximate probability of death in the next femtosecond $(\Delta t)$ of a $T$-year old is $\displaystyle \frac{f(T)\Delta t}{1-F(T)}.$

Note that in contrast to the density $f(t)$ integrating to $1$, the integral $\displaystyle \int_0^\infty h(t)\, \mathrm dt$ must diverge. This is because the CDF $F(t)$ is related to the hazard rate through

$$F(t) = 1 - \exp\left(-\int_0^t h(\tau)\, \mathrm d\tau\right)$$ and since $\lim_{t\to \infty}F(t) = 1$, it must be that $$\lim_{t\to \infty} \int_0^t h(\tau)\, \mathrm d\tau = \infty,$$ or stated more formally, the integral of the hazard rate must diverge: there is no potential divergence as a previous edit claimed.

Typical hazard rates are increasing functions of time, but constant hazard rates (exponential lifetimes) are possible. Both of these kinds of hazard rates obviously have divergent integrals. A less common scenario (for those who believe that things improve with age, like fine wine does) is a hazard rate that decreases with time but slowly enough that the integral diverges.

  • 1
    $\begingroup$ "Let X denote the time of death (or time of failure if you prefer a less morbid description". Time until recovery is even less morbid. $\endgroup$
    – ryu576
    Commented Aug 29, 2019 at 23:19

Imagine that you are interested in the incidence of (first) marriage for men. To look at the incidence of marriage at age 20, say, you would select a sample of people who are not married at that age and see if they get married within the next year (before they turn 21).

The you could get a rough estimate for $$ P(\mathrm{marry\,\, before\,\, 21}| \mathrm{not\,\, married \,\,at\,\, 20}) $$ as the proportion of individuals who got married from your sample of single 20 year olds, i.e. $$ \frac{N(\mathrm{married \,\,before \,\,21\,\, and \,\, not\,\,married\,\, at \,\, 20})}{N(\mathrm{not\,\, married\,\, at\,\, 20})} $$

So basically this is just using the definition of conditional probability, $$ P(X|Y) = \frac{P(X,Y)}{P(Y)}. $$ Now imagine we make the age unit smaller and smaller, up to days for example. I.e. what is the incidence of marriage at age of 7300 days? Then you would do the same, but survey all individuals of 7300 days and look who gets married before the end of the day. If $T$ is a random variable age at marriage, then we could write $$ P(T \leq 7301)| T \geq 7300) = \frac{P(T \in [7300, 7301))}{P(T \geq 7300)} $$ by the same logic as before.

The hazard would then be the instantaneous probability of marriage at age $t$, for a non-married individual. We can write this as $$ h(t) dt= P(T \in [t, t+dt) | T\geq t) = \frac{P(T \in[t, t+dt))}{P(T \geq t)} $$


$f(x)$ is not the probability of death, but the probability density; the expected number of times you die within the next unit of time if the probability density remained constant during that unit of time.

Notice there is a problem: your probability of dying when you already died before is rather problematic. So it makes more sense to compute the probability of dying conditional on having survived thus far. $1-F(t)$ it the probability of having survived until $t$, so dividing the probabilty density by that probability, will get us the expected number of times we will die within the next unit of time conditional on not having died before. That is the hazard rate.


"Death of a person is a tragedy, deaths of millions is statistics" - Joseph Stalin

Hazard rate is just a renormalization of the probability space that takes pallid impersonal statistics on input and converts it into your own chances to live another day.

Suppose you're an average young man in the Wild West. You decide to pursue a questionable career of a train robber.

Assume that the chance of an average guy surviving his first train robbery is $\frac{1}{2}$. After that you get slightly more experienced and for your second train robbery your chance of survival is $\frac{2}{3}$. Now, you're even more experienced and for the third stint the chance of survival is $\frac{3}{4}$.

So the night before your third robbery you might ask yourself, whether it is worth the risk of dying with 25% chance tomorrow, or should you rather give up on train robberies altogether and move on to start a career in finance?

The data you want to ask yourself this question is the chance of survival tomorrow, which is the Hazard rate.

Unfortunately, it's impossible to get the data about your odds in the real life. What you could do instead is take a look at the cumulative distribution function $F(t)$ of a train robber's life expectancy, or, rather its counterpart $S(t) = 1-F(t)$, called the survival function:

enter image description here

Probability mass function (which is a discrete-case analogue of the continuous probability density function) of dying at your third robbery $p(\xi = 3) = \frac{1}{8}$ . We can more or less reformulate this as a continuous problem $p(3 \leq \xi < 4) = F_\xi(3) - F_\xi(4) = f_\xi(3)dx$, where $\xi$ is a random variable indicating the number of robberies an average train robber survives, $dx=1$, $F_\xi(x)$ is cumulative probability function and $f_\xi(x)$ is probability density function.

So you see, probability density function/probability mass function answers a wrong question. It says that out of all repeat train robbers the fraction that dies at their third robbery is ($\frac{1}{8}$). But the question you want to ask is: if I go for my third robbery tomorrow, what are my chances to survive it, and the answer you want is $(\frac{3}{4})$.

Now, let's start formalizing this. For a discrete-time variable, Hazard function is your chance to die during your next robbery number $t$:

$\underbrace{S(t) - S(t+1)}_\text{fraction of train robbers who die at t} = \underbrace{\lambda(t)}_\text{hazard function at t} \cdot \underbrace{S(t)}_\text{fraction of survivors by t} \ cdot \delta t$

Thus, hazard function is defined as:

$\lambda(t) = \frac{-\delta S(t)}{\delta t \cdot S(t)}$

Or, in continuous-time case:

$\lambda(t) = \frac{-\partial S(t)}{\partial t \cdot S(t)} = \frac{f(t)}{S(t)}$

Cumulative hazard rate $\Lambda(t)$ is a funny thing. It essentially enumerates and sums up all the chances of death you escaped by the current moment. So, for instance, at your first train robbery you had a chance to die of $1/2$, at the second - $1/3$, at the third - $1/4$.

So by the time you start contemplating your fourth robbery, the "number of deaths" you deserved by now $\Lambda(t) = 1/2 + 1/3 + 1/4 = 1.083333$, so in a fair world you would have already been more than dead, exercising your luck so readily...


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