Why Compare the Two Definitions?
The cause-specific hazard and the subdistribution hazard both describe the instantaneous occurrence of a particular event type, but they use different denominators when other event types can occur. That difference matters only if there are competing events: if only one event cause is possible, the two denominators are the same. Establishing this special case makes clear what competing events change, and what they do not.
The definitions below concern the distribution of the event process, before considering how it might be estimated from censored observations. We use \(T\) for the time to the first event, \(J\) for its cause, \(S(t)=P(T>t)\) for the probability of remaining event-free through time \(t\), and \(F_k(t)=P(T\leq t,J=k)\) for the cumulative incidence of cause \(k\). For cause \(k\), let \(f_k(t)\) denote the density of its cumulative incidence, so that \(F_k(t)=\int_0^t f_k(u)\,du\) over the time range considered.
These density-based definitions require the relevant event-time and cause-specific distributions to be absolutely continuous with respect to time over the interval being considered. Separately, assume that simultaneous event types have probability zero, so an individual does not experience two competing first-event causes at the same instant. The causes are mutually exclusive labels for the first event. These assumptions make the densities and cumulative-incidence decomposition used below well-defined.
The Single-Cause Result
Suppose there is exactly one possible event cause, \(k\). At any finite time \(t\), everyone who has experienced an event by \(t\) has experienced cause \(k\). Thus \(F_k(t)=P(T\leq t)\), while \(S(t)=P(T>t)\). These are complementary events, so \(F_k(t)+S(t)=1\). This remains true if some individuals may never experience the event: at a finite time, they are included in \(P(T>t)\).
The result follows directly by substituting the complementary-probability identity into the subdistribution denominator:
This is an equality of definitions, not a claim that the two names refer to the same construction in every setting. With competing causes, individuals who have already experienced another cause are no longer event-free, but they have not experienced cause \(k\). They therefore contribute differently to the two denominators.
Worked Example: One Exponential Event Cause
Suppose a hypothetical population has one possible event cause, with event time following an exponential distribution at rate \(0.20\) per year. Find both hazards at \(t=2\) years.
For this model, \(S(t)=e^{-0.20t}\), and the event density is \(f_k(t)=0.20e^{-0.20t}\). At two years,
The cause-specific hazard is \(f_k(2)/S(2)=0.1341/0.6703=0.2000\) per year. The cumulative incidence is \(F_k(2)=1-e^{-0.4}=0.3297\), so the subdistribution hazard is
The matching values are expected: with one cause, \(1-F_k(2)\) is exactly the event-free probability \(S(2)\). The exponential model is used here for convenient arithmetic; the equality itself does not require an exponential distribution.
How Competing Causes Change the Denominator
Now suppose there are several possible first-event causes, indexed by \(j\), and write \(F_j(t)\) for the cumulative incidence of cause \(j\). By time \(t\), an individual has either remained event-free or experienced one of the causes. The outcomes are mutually exclusive, giving
For a focal cause \(k\), it follows that \(1-F_k(t)=S(t)+\sum_{j\ne k}F_j(t)\). The subdistribution denominator therefore includes both those still event-free and those who have experienced another cause. Using \(f_k(t)=\lambda_k(t)S(t)\), the relationship between the hazards is
When competing-event incidence is positive, the fraction multiplying \(\lambda_k(t)\) is less than one, provided \(S(t)>0\). Thus, if the focal cause has positive density at that time, its subdistribution hazard is smaller than its cause-specific hazard. The reason is the extra competing-event probability in the subdistribution denominator, not a change in the focal cause’s density.
Worked Example: Two Constant Cause-Specific Hazards
Consider a hypothetical population with two possible causes. The cause-specific hazard is \(0.20\) per year for cause \(k\) and \(0.10\) per year for cause \(\ell\), constant over time. Assume the first event is the only event counted and simultaneous causes have probability zero. Find both hazards for cause \(k\) at two years.
The event-free probability under these constant hazards is \(S(t)=e^{-(0.20+0.10)t}\). The cause-\(k\) density is \(f_k(t)=0.20S(t)\), and its cumulative incidence is the integral of this density:
The cause-specific hazard is \(0.1098/0.5488=0.2000\) per year. For comparison, the competing cause has \(F_\ell(2)=(0.10/0.30)(1-e^{-0.6})=0.1504\). Hence the subdistribution denominator is \(1-F_k(2)=S(2)+F_\ell(2)=0.6992\), and
The two hazards differ because the subdistribution denominator includes the \(0.1504\) probability of having experienced cause \(\ell\). The calculation also verifies that \(S(2)+F_k(2)+F_\ell(2)\) is approximately one, allowing for rounding.
When Can They Still Coincide?
Having multiple possible causes does not force the hazards to differ at every time. Their relationship shows the exact condition for equality, assuming both denominators are positive:
Since \(1-F_k(t)=S(t)+\sum_{j\ne k}F_j(t)\), equality holds if \(f_k(t)=0\), or if the cumulative incidence of every competing cause is zero at that time. The first case gives equality because both numerators are zero; it is generally not informative about a positive event rate. In the second case the denominators match. If cause \(k\) has positive density and some competing-event incidence has accumulated, the hazards differ.
Worked Example: A Competing Cause Has Not Yet Occurred
Consider a hypothetical process with cause-\(k\) hazard \(0.10\) per year from time zero. A second cause has hazard zero before year one and becomes possible after year one. Evaluate the two cause-\(k\) hazards at \(t=0.5\) years.
Before year one, cause \(k\) is the only possible event. Therefore \(S(0.5)=e^{-0.10(0.5)}=e^{-0.05}=0.9512\), and \(F_k(0.5)=1-e^{-0.05}=0.04877\), while \(F_\ell(0.5)=0\). The cause-\(k\) density is \(f_k(0.5)=0.10e^{-0.05}=0.09512\) per year. Consequently,
The hazards coincide because no probability has accumulated for the competing cause by that time. This does not imply they will continue to coincide after cause \(\ell\) becomes possible and its cumulative incidence becomes positive.
Checking the Argument and Avoiding Misinterpretation
When calculating or interpreting these quantities, check the definition of each denominator before simplifying. A useful diagnostic is to verify the probability decomposition \(S(t)+\sum_jF_j(t)=1\), then compare \(S(t)\) with \(1-F_k(t)\). If they differ, the difference must equal the accumulated incidence of causes other than \(k\). This check often reveals a denominator error immediately.
- Using \(S(t)\) in both definitions with multiple causes. That substitutes the cause-specific denominator into the subdistribution hazard. Use \(1-F_k(t)\) for the latter.
- Omitting the competing incidence from the complement. In a multiple-cause setting, \(1-F_k(t)\) is \(S(t)+\sum_{j\ne k}F_j(t)\), not just \(S(t)\).
- Reversing a cause-specific fraction. For constant hazards, \(F_k(t)\) is proportional to cause \(k\)'s rate: \(F_k(t)=\lambda_k(\,1-e^{-\sum_j\lambda_jt}\,)/\sum_j\lambda_j\). Check that each cause’s incidence increases with its own rate.
- Claiming equality merely because only one cause is of interest. Focusing the analysis on one cause does not remove other possible event types. Equality follows when there is only one possible cause, or under the specific time-dependent conditions above.
The two hazard definitions are useful for different questions. The cause-specific hazard uses the currently event-free population as its denominator. The subdistribution hazard uses a denominator tied to whether the focal cause has occurred, so prior competing events remain represented in it. With a single cause, these populations coincide; with competing causes, they generally do not. Neither definition eliminates the need to state the event process and assumptions clearly. In data analysis, censoring is a separate issue: the definitions above describe event distributions, while estimating them from censored data requires appropriate handling of censoring.
Check Your Understanding
Answer each question, then check the solutions that follow.
- With one possible cause, suppose \(S(t)=0.80\) and \(f_k(t)=0.12\) per year. Calculate both hazards and state why they match.
- At a particular time in a two-cause setting, \(S(t)=0.60\), \(F_k(t)=0.25\), and \(f_k(t)=0.09\) per year. Calculate the cause-specific and subdistribution hazards for cause \(k\).
- In a setting with competing causes, what condition on the cumulative incidence of the other causes makes the denominators equal?
- A student says, “Cause \(k\) is the only event type in my analysis, so its two hazards must match,” even though cause \(\ell\) can occur. Explain the error.
Solutions. (1) The cause-specific hazard is \(0.12/0.80=0.1500\) per year. Since there is one cause, \(1-F_k(t)=S(t)=0.80\), so the subdistribution hazard is also \(0.1500\) per year. (2) The cause-specific hazard is \(0.09/0.60=0.1500\) per year. The subdistribution denominator is \(1-0.25=0.75\), giving \(0.09/0.75=0.1200\) per year. The denominators differ because the other cause has accumulated incidence \(1-F_k-S=0.15\). (3) The denominators are equal when \(\sum_{j\ne k}F_j(t)=0\). (4) Naming one focal cause does not make it the only possible cause. If cause \(\ell\) can occur and has accumulated positive incidence, it contributes to \(1-F_k(t)\), but not to \(S(t)\); the hazards therefore generally differ when \(f_k(t)>0\).