Why Define a Cause-Specific Hazard?
In competing-risks data, the first event can be one of several mutually exclusive causes. A cause-specific hazard describes the rate of occurrence of one chosen cause at a particular time, among subjects who have not yet experienced any event. It is a time-specific quantity: it can vary as time passes, even when the event and cause definitions remain fixed.
The denominator matters. The cause-specific hazard for cause \(k\) concerns subjects who are still event-free just before time \(t\), not everyone who began follow-up and not everyone who has ever experienced cause \(k\). The hazard is therefore a rate in the currently event-free population. It is not, by itself, the probability that a subject will experience cause \(k\) by time \(t\).
This definition uses a continuous-time setting in which event-time distributions have densities over the time range being considered. With continuous event times, \(P(T\ge t)\) and \(P(T>t)\) are equal. The hazard is defined only while there is a positive probability of remaining event-free. If all subjects must have experienced an event by some time, there is no event-free population beyond that time on which to define this rate.
What the Formula Says
The numerator \(f_k(t)\) describes how the probability of a first event from cause \(k\) is distributed over time. More precisely, it is the derivative of \(F_k(t)\). The denominator \(S(t)\) is the probability of having experienced no event by time \(t\). Dividing the cause-specific event-time density by the event-free probability puts the density on the scale of the event-free population.
The resulting units are the reciprocal of the time unit used. If time is measured in months, the hazard has units of per month. A value such as \(0.025\) per month is a rate, not a proportion of subjects and not a probability of \(0.025\) in each month. Turning a rate into an event probability requires considering how the rate behaves over time and how long subjects remain event-free; that is a different calculation.
A Short Justification of the Ratio
The ratio follows from the rule for conditional densities. For \(u\ge t\), the conditional density of a first event at time \(u\) from cause \(k\), given that the subject was event-free at time \(t\), is \(f_k(u)/S(t)\). The joint probability density contributes \(f_k(u)\), and conditioning on the event-free group divides by its probability \(S(t)\). At the time boundary \(u=t\), this density is \(f_k(t)/S(t)\), which is the cause-specific hazard at \(t\).
Equivalently, rearranging the definition gives a useful identity:
Thus, at a given time, the cause-\(k\) event-time density is the event-free probability multiplied by the cause-specific rate. This identity does not say that the hazard is a probability. It expresses the density of cause-\(k\) events in the full population in terms of the size of the event-free population and its instantaneous rate.
Worked Example: Computing a Hazard at 12 Months
Suppose a hypothetical continuous-time model for a study of a first event gives \(S(12)=0.72\). At 12 months, its cause-1 cumulative incidence function has derivative \(f_1(12)=0.018\) per month. Calculate and interpret the cause-1 hazard.
The formula is \(\lambda_1(t)=f_1(t)/S(t)\). Substituting the stated values gives $$ \lambda_1(12)=\frac{0.018\ \text{per month}}{0.72} =0.025\ \text{per month}. $$
At 12 months, the cause-1 instantaneous rate among subjects who remain event-free is \(0.025\) per month. The calculation is valid because the model has a continuous event-time distribution and \(S(12)=0.72>0\). The value does not mean that 2.5% of the original study population experiences cause 1 at month 12, or that the probability of cause 1 during the next month is exactly 2.5%.
Worked Examples with Cause-Specific Densities
Worked Example: Using a Cumulative Incidence Derivative
In a hypothetical model, at \(t=10\) months the cause-1 cumulative incidence is \(F_1(10)=0.25\), the cause-2 cumulative incidence is \(F_2(10)=0.15\), and event-free survival is \(S(10)=0.60\). The model is differentiable at 10 months and gives \(F'_1(10)=0.012\) per month. Find the cause-1 hazard at that time.
The hazard requires the derivative of the cause-1 cumulative incidence, not its value at 10 months. Here that derivative is \(f_1(10)=0.012\) per month. Therefore, $$ \lambda_1(10)=\frac{f_1(10)}{S(10)} =\frac{0.012}{0.60} =0.020\ \text{per month}. $$
The cause-1 hazard at 10 months is \(0.020\) per month among the event-free population at that time. The values \(F_1(10)=0.25\) and \(F_2(10)=0.15\) describe accumulated probabilities up to 10 months; neither is the density needed in the numerator. The calculation is defined because \(S(10)>0\).
Worked Example: Comparing the Same Cause at Two Times
A hypothetical fitted model reports the following values for cause 1. The event-free survival and cause-specific density are evaluated at each listed time.
| Time | \(S(t)\) | \(f_1(t)\) per month |
|---|---|---|
| 6 months | 0.80 | 0.016 |
| 18 months | 0.45 | 0.009 |
At 6 months, $$ \lambda_1(6)=\frac{0.016}{0.80}=0.020\ \text{per month}. $$ At 18 months, $$ \lambda_1(18)=\frac{0.009}{0.45}=0.020\ \text{per month}. $$
Although both the event-free probability and the cause-1 density are smaller at 18 months, their ratios are equal. The model therefore gives the same cause-1 instantaneous rate at these two particular times. This conclusion is limited to those times; two equal values do not establish that the hazard is constant between them. The definition applies at each time with \(S(t)>0\), as it does for both entries.
Assumptions, Data Conditions, and Diagnostics
The cause-specific hazard is a population quantity defined by the event-time and cause distributions. Its definition does not require a particular regression model. To use the formula as written, the relevant conditions are:
- A well-defined first event and cause. Each subject has a first-event time \(T\) and a cause label \(J\); the causes refer to mutually exclusive types of first event.
- A continuous-time density at the time of interest. \(F_k(t)\) must have a derivative \(f_k(t)\) at that time. The density-based definition does not directly describe a setting with point masses at event times.
- A nonempty event-free population in probability terms. \(S(t)>0\), so the denominator is positive.
These conditions concern the definition. In observed data, event times may be censored or recorded only at visits, and \(f_k(t)\) and \(S(t)\) are not generally known. Estimating the hazard then requires methods that account for the observation process and make appropriate assumptions about censoring and measurement. A formula applied to raw event counts without considering follow-up and censoring is not automatically an estimate of the population hazard.
Useful checks include confirming that the time units agree in the numerator and denominator, that the density is for the specified cause, and that the event-free survival is positive. If a fitted density or survival curve is irregular, the ratio can also be unstable; in particular, a small estimated \(S(t)\) can magnify uncertainty in both inputs. A reported hazard should therefore be accompanied by its time point, cause, units, and the population to which it refers.
How This Quantity Differs from Other Summaries
The cause-specific hazard is one way to summarize event occurrence, but it answers a different question from an accumulated probability. The cause-specific cumulative incidence \(F_k(t)\) is the probability of having experienced cause \(k\) as the first event by time \(t\). The hazard \(\lambda_k(t)\) is an instantaneous rate at time \(t\), conditional on being in the event-free population. The hazard is not a replacement for the cumulative incidence when the question concerns absolute event probability by a specified time.
The cause-specific hazard also differs from a rate calculated using the original cohort as a fixed denominator. Subjects who have already experienced any first event are no longer event-free, so they are not part of the population described by this hazard. The precise construction of observed risk sets and the estimation of cause-specific hazards from censored data require additional care; the population definition alone does not specify an estimator.
Common Mistakes
- Using a cumulative probability in the numerator. \(F_k(t)\) is an accumulated probability, while the numerator is its derivative \(f_k(t)\). A cumulative incidence value cannot be substituted for a density.
- Calling the hazard a probability. A hazard has units of inverse time. A value of \(0.020\) per month is not automatically a 2% probability over a month.
- Using the wrong denominator. The denominator is event-free survival at the time under study, not the number originally enrolled or the probability of having experienced cause \(k\).
- Comparing values without their time units. A rate per day and a rate per year are on different numerical scales. Convert them to a common time unit before comparing.
- Extrapolating from a few times. Equal hazards at two times do not prove a constant hazard over the intervening period.
Check Your Understanding
Answer each question using the definition, and include units where appropriate.
- At 8 months, a hypothetical model has \(S(8)=0.50\) and \(f_2(8)=0.015\) per month. What is the cause-2 hazard?
- Why is \(F_k(t)\) not the numerator in the definition of \(\lambda_k(t)\)?
- A model reports \(S(5)=0\). Can its cause-specific hazard at 5 months be calculated using the ratio in this tutorial? Explain.
- At one time point, a cause-specific hazard is \(0.03\) per week. Does this state that the probability of the cause-specific event in the next week is 3%? Why or why not?
- What additional condition is needed to use the density-based definition at a given time?
Solutions. (1) \(\lambda_2(8)=0.015/0.50=0.030\) per month. (2) \(F_k(t)\) is the accumulated probability of cause \(k\) by time \(t\); the numerator is its derivative, the event-time density \(f_k(t)\). (3) No. The denominator is zero, so the ratio is undefined; there is no event-free population in probability terms at that time. (4) No. The hazard is an instantaneous rate, not a one-week probability; a probability over a period requires accounting for the hazard over that period and the event-free status. (5) \(F_k\) must be differentiable at that time so that \(f_k(t)=F'_k(t)\) exists, and \(S(t)\) must be positive for the ratio to be defined.