16  Key Functions in Survival Analysis

16.1 The survival function

Let \(T > 0\) be a continuous random survival time with distribution function \(F(t) = \Prob(T \leq t)\) and density \(f(t) = F'(t)\).

Definition: Survival function

The survival function is \[ S(t) = \Prob(T > t) = 1 - F(t), \qquad t \geq 0. \]

\(S(t)\) is a non-increasing function with \(S(0) = 1\) and \(\lim_{t\to\infty} S(t) = 0\). For continuous \(T\), \[ S'(t) = -f(t) \qquad \Rightarrow \qquad f(t) = -S'(t). \]

16.2 The hazard function

Definition: Hazard function

The hazard function (or hazard rate) is \[ h(t) = \lim_{\delta t \to 0} \frac{\Prob(T < t + \delta t \mid T \geq t)}{\delta t}. \]

Intuitively, \(h(t)\,\delta t \approx \Prob(\text{event in } [t, t+\delta t) \mid \text{survived to } t)\) for small \(\delta t\).

Using conditional probability and the definition of the density: \[ h(t) = \frac{f(t)}{S(t)} = -\frac{S'(t)}{S(t)} = -\frac{d}{dt}\log S(t). \]

The hazard can take any non-negative value. Unlike the density \(f(t)\), it is not a probability density and need not integrate to 1.

16.3 The cumulative hazard

Definition: Cumulative hazard

The cumulative hazard function is \[ H(t) = \int_0^t h(s)\,ds. \]

Since \(h(t) = -d\log S(t)/dt\), integrating gives \[ H(t) = -\log S(t) \qquad \Longleftrightarrow \qquad S(t) = e^{-H(t)}. \]

This is one of the most important relationships in survival analysis. The three functions \(S(t)\), \(h(t)\), and \(H(t)\) each completely characterise the distribution of \(T\): knowing any one determines the others.

Example: Exponential distribution

For the exponential distribution with rate \(\lambda > 0\):

\[ f(t) = \lambda e^{-\lambda t}, \qquad S(t) = e^{-\lambda t}, \qquad h(t) = \lambda, \qquad H(t) = \lambda t. \]

The hazard is constant: the risk of death in the next instant does not depend on how long the individual has already survived. This is the memoryless property of the exponential distribution.

16.4 Relationship between the three functions

Function From \(S(t)\) From \(h(t)\)
\(S(t)\) — \(e^{-H(t)}\)
\(f(t)\) \(-S'(t)\) \(h(t)e^{-H(t)}\)
\(h(t)\) \(-S'(t)/S(t)\) —
\(H(t)\) \(-\log S(t)\) \(\int_0^t h(s)\,ds\)

16.5 Mean and median survival time

The mean survival time is \[ \E[T] = \int_0^\infty t\,f(t)\,dt = \int_0^\infty S(t)\,dt, \] where the second equality follows from integration by parts (under Assumption 2 of Chapter 15, \(\Prob(T < \infty) = 1\)).

The median survival time \(\mu_{1/2}\) satisfies \(S(\mu_{1/2}) = 0.5\), i.e., it is the time by which half the population has experienced the event. The median is often preferred over the mean in survival analysis because it is less sensitive to the (often unknown) tail behaviour of \(S(t)\).

For the exponential distribution: \(\E[T] = 1/\lambda\) and \(\mu_{1/2} = \log(2)/\lambda\).

More generally, the \(p\)th quantile satisfies \(S(t_p) = 1 - p\).

16.6 Empirical estimation (no censoring)

When data are uncensored, \(S(t)\) can be estimated by the empirical survival function \[ \widehat{S}(t) = \frac{\#\{i : t_i > t\}}{n}. \] This is an unbiased estimator: \(\E[\widehat{S}(t)] = S(t)\).

Figure 16.1: Empirical survival function for 10 uncensored survival times.
Example: Empirical estimation

For the 10 uncensored observations \(t = 3, 3.5, 4, 5, 6, 7, 10, 12, 19.5, 30\), the empirical survival function takes the value \(\widehat{S}(4.5) = 7/10\) because 7 of the 10 observations exceed 4.5. The estimated median is the smallest \(t\) such that \(\widehat{S}(t) < 0.5\), which is \(t = 7\).

In R:

Code 
library(survival)
t_obs <- c(3, 3.5, 4, 5, 6, 7, 10, 12, 19.5, 30)
model <- survfit(Surv(t_obs) ~ 1)
summary(model)
plot(model)

When data include censored observations, the empirical survival function is biased. The Kaplan–Meier estimator, introduced in Chapter 17, handles this correctly.

Exercises

16.1. Suppose a survival time \(T\) has hazard function \[h(t) = \frac{2}{t + 8}, \qquad t \geq 0.\]

  1. Find the cumulative hazard \(H(t)\).
  2. Hence write down the survival function \(S(t)\) and the probability density function \(f(t)\).

Recall \(H(t) = \int_0^t h(u)\,\mathrm{d}u\), \(S(t) = e^{-H(t)}\), and \(f(t) = h(t)S(t)\).

16.2. Continuing from Exercise 16.1 with \(S(t) = 64/(8+t)^2\).

  1. Find the mean survival time \(\E[T]\) using the formula \(\E[T] = \int_0^\infty S(t)\,\mathrm{d}t\).
  2. Find the median survival time \(t_{0.5}\), defined by \(S(t_{0.5}) = 0.5\).

For part (a), integrate \(64/(8+t)^2\) from 0 to \(\infty\). For part (b), solve \(64/(8+t_{0.5})^2 = 1/2\) for \(t_{0.5}\).