Cumulative hazard function ( $H (t)$ )

Suppose

T \geq 0

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) d u = - lo g S (t), S (t) = e^{- H (t)} .

(1)

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

lo g 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

- lo g S

+ \infty

. The normalization

S (0) = 1

also matters; an initial probability mass at zero would require an initial term

- lo g 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

P (a < T \leq b ∣ T > a) = 1 - e^{- [H (b) - H (a)]} .

(2)

For instance, a cumulative hazard increment of

2

gives event probability

1 - e^{- 2} \approx 0.864665

, rather than a probability of

2

. With

h (t) = 0.2

per year,

H (5) = 1

and

S (5) = e^{- 1} \approx 0.367879

. With the increasing hazard function

h (t) = 0.04 t

per year when

t

is expressed in years,

H (t) = 0.02 t^{2}

, so

H (5) = 0.5

gives

S (5) = e^{- 0.5} \approx 0.606531

. Its coefficient has units of inverse years squared.

Figure 1.
Constant and increasing hazard functions, their cumulative hazards, and their survival functions
. **Details**.

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 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

- lo g (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.

Table of contents

Mean accumulated hazard under independent censoring ( $E H (min (T, C)) = P (T \leq C)$ )

 0  0

Let

X = min (T, C)

V = 1_{{T \leq C}}

, and suppose

T

and

C

are independent, with continuous failure density

f = h S

. Then the Tonelli theorem gives

E H (X) = \int_{0}^{\infty} h (t) S (t) P (C \geq t) d t = P (T \leq C) = E V .

(1)

Thus

V - H (X)

has mean zero. The identity also holds conditionally on covariates when censoring is independent conditionally on them. For an estimated hazard, the mean error is controlled by

E ∣ H (X) - H (X) ∣

, not by pointwise fit alone.

// Target: survival-analysis.bigb

Mean accumulated hazard before fixed censoring ( $E H (min (T, c)) = P (T \leq c)$ )

 0  0

For a nonnegative continuous event time with survival function S, hazard function h and H the cumulative hazard, split the expectation at c. The censored contribution is S(c)H(c). The event contribution is

\int_{0}^{c} H (t) f (t) d t = 1 - S (c) - S (c) H (c)

by integration by parts, since

d H = h d t

and

f = h S

. Adding gives the displayed identity. Linearity extends it to the expected observed event count in a cohort, without requiring independence between subjects.