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:

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

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?

**Answer**

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.

**Attribution***Source : Link , Question Author : user246315 , Answer Author : Cliff AB*