= Cumulative hazard function
{title2=$H(t)$}
= Integrated hazard function
{synonym}
= Integrated hazard
{synonym}
= Cumulative hazard
{synonym}
Suppose $T\geq0$ has an <absolutely continuous probability distribution>, with <survival function> $S(0)=1$. On an <interval> where $S(t)>0$, its <cumulative hazard> is
$$
H(t)=\int_0^t h(u)\,du=-\log S(t),\qquad
S(t)=e^{-H(t)}.
$$
Here $h$ is the <hazard function>; its <integral> is dimensionless because the units of rate and time cancel. The identity follows by integrating the <derivative> of $\log S$, so mere <continuity> of the <survival function> is not enough: the <absolute continuity> assumption supplies a <probability density function>. If $S$ reaches zero, interpret $-\log S$ as $+\infty$. The normalization $S(0)=1$ also matters; an initial <probability mass> at zero would require an initial term $-\log S(0)$.
Accumulating <hazard functions> adds on the $H$ scale and multiplies survival factors on the $S$ scale. For $S(a)>0$ and $b>a$,
$$
\mathbb P(a<T\leq b\mid T>a)=1-e^{-[H(b)-H(a)]}.
$$
For instance, a <cumulative hazard> increment of $2$ gives event <probability> $1-e^{-2}\approx0.864665$, rather than a <probability> of $2$. With $h(t)=0.2$ per year, $H(5)=1$ and $S(5)=e^{-1}\approx0.367879$. With the increasing <hazard function> $h(t)=0.04t$ per year when $t$ is expressed in years, $H(t)=0.02t^2$, so $H(5)=0.5$ gives $S(5)=e^{-0.5}\approx0.606531$. Its coefficient has units of inverse years squared.
\Image[/survival-hazard-relationship.png]
{title=Constant and increasing hazard functions, their cumulative hazards, and their survival functions}
The constant <hazard function> produces an <exponential distribution>; the increasing example is a <Weibull distribution>. Their <survival functions> both decrease, although one <hazard function> stays constant and the other increases. At year ten both have $H(10)=2$ and $S(10)=e^{-2}$ despite different earlier histories. The accumulation identity is described in https://web.stanford.edu/~lutian/coursepdf/unit1.pdf[Lu Tian's Stanford notes, Section 1.1].
For a <discrete hazard> $q_j$ at each possible event time $t_j$, the exact counterpart is $S(t)=\prod_{t_j\leq t}(1-q_j)$. Taking its negative <logarithm> sums $-\log(1-q_j)$, rather than $q_j$. Therefore exponentiating a <Nelson–Aalen estimator> does not give exactly the <Kaplan–Meier estimator>; <small-jump comparison of cumulative hazard estimators> quantifies the difference.
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