Hazard function

Remarks

At every point $t$ where $f$ is continuous, the hazard is the limit of a conditional probability per unit time: $$h(t) = \lim_{\Delta \downarrow 0} \frac{P(t \le T < t + \Delta \mid T \ge t)}{\Delta}.$$ Since $h = -S'/S = -(\log S)'$ there, one gets $S(t) = e^{-H(t)}$, so $h$ determines the distribution of $T$.

Example: for the exponential distribution with rate $\lambda$, $f(t) = \lambda e^{-\lambda t}$ and $S(t) = e^{-\lambda t}$, so $h(t) = \lambda$ is constant and $H(t) = \lambda t$. For the Weibull distribution with shape $k > 0$ and scale $\lambda > 0$, $S(t) = e^{-(t/\lambda)^k}$ and $h(t) = (k/\lambda)(t/\lambda)^{k-1}$, which increases when $k > 1$ and decreases when $k < 1$.

Non-example: if $T$ takes only the values $1$ and $2$, each with probability $1/2$, then $T$ has no density and the formula does not apply. For discrete times one uses instead the discrete hazard $P(T = t \mid T \ge t)$, here $1/2$ at $t = 1$ and $1$ at $t = 2$.