Competing Events as Occupied Absorbing States

From Competing Events to Occupied States

After a landmark, a person may remain free of a terminal event or experience one of several mutually exclusive terminal events. For example, after a treatment landmark, the possible outcomes by a prediction horizon might be relapse, death before relapse, or neither event. The probability of each terminal event by that horizon is not only a cumulative incidence: in a multi-state representation, it is also the probability that the person occupies the corresponding absorbing state at that time.

This interpretation depends on the endpoint being terminal. Once a person enters a terminal-event state, that person stays in it for the prediction period. The discussion below concerns a person who is in the event-free state at landmark time \(s\), and probabilities conditional on being in that state and on the specified landmark information. The earlier material on landmark state-occupation prediction provides the broader prediction framework; here the focus is the connection between competing terminal events and absorbing-state probabilities.

Definition: Suppose that at landmark time \(s\) a person is event-free, and that the possible terminal event types are \(1,\ldots,K\). Let \(X(t)\) denote the state occupied at time \(t\), with state \(0\) denoting event-free and states \(1,\ldots,K\) denoting the respective terminal events. If each event state is absorbing, then the cause-\(k\) cumulative incidence by horizon \(t\geq s\) is \(F_k(t\mid s)=P\{X(t)=k\mid X(s)=0,\text{ landmark information}\}\). It is the probability that event \(k\) has occurred by \(t\), in the presence of the other competing events.

The phrase “in the presence of” matters. If event \(j\) occurs first, the person is no longer at risk of experiencing a different terminal event as the first event. The cause-\(k\) probability therefore depends both on the cause-\(k\) event rate and on how often other events remove people from the event-free state.

Hazard Formulation and Derivation

Let \(u\) denote elapsed time since the landmark. For a person still event-free at elapsed time \(u\), let \(\lambda_k(u)\) be the cause-specific hazard for terminal event \(k\): the instantaneous event rate for type \(k\), conditional on remaining event-free just before \(u\). Assume that the hazards are defined on the prediction interval, that events are mutually exclusive first events, and that there are no simultaneous jumps of different event types. Let \(\lambda_{\cdot}(u)=\sum_{k=1}^K\lambda_k(u)\) be the total hazard of leaving the event-free state.

Write \(S(u)\) for the probability of still occupying the event-free state at elapsed time \(u\), conditional on being event-free at the landmark. Under the hazard model, \(S(0)=1\) and \(S(u)=\exp\{-\int_0^u\lambda_{\cdot}(v)\,dv\}\). The probability of a cause-\(k\) event in a very short interval near \(u\) is approximately the probability of still being event-free, \(S(u)\), multiplied by its cause-specific hazard and the interval width. Thus the instantaneous rate of accumulation of cause-\(k\) probability is \(S(u)\lambda_k(u)\).

\frac{dF_k(u)}{du}=S(u)\lambda_k(u), \qquad F_k(t)=\int_0^t S(u)\lambda_k(u)\,du.

This is a direct consequence of the definition of the cause-specific hazard. More formally, over an interval of width \(du\), the probability of a cause-\(k\) event is \(S(u)\lambda_k(u)du+o(du)\). Summing these increments from the landmark to the horizon and taking the interval widths to zero gives the integral. In a finite-state multi-state model, the same result follows by multiplying the probability of occupying state 0 just before \(u\) by the transition intensity from state 0 to absorbing state \(k\).

Because the event states are absorbing, “having experienced cause \(k\) by time \(t\)” and “occupying state \(k\) at time \(t\)” are the same event. The event-free probability and all cause-specific cumulative incidences therefore partition the population:

S(t)+\sum_{k=1}^K F_k(t)=1.

This identity is a useful arithmetic check, but it also describes the model: by the horizon, each person is either event-free or in exactly one of the terminal-event states.

Worked Calculations

Worked Example: Two Constant Competing Hazards

Suppose a hypothetical landmark prediction starts with individuals event-free. Over the next three years, assume constant cause-specific hazards of \(0.12\) per year for event 1 and \(0.08\) per year for event 2. Assume these are the only possible terminal events, the hazards apply throughout the three-year period, and the two events cannot occur simultaneously as first events. Find the state-occupation probabilities at three years.

The total event-free exit hazard is \(0.12+0.08=0.20\) per year. Therefore the event-free probability at three years is \(S(3)=\exp(-0.20\times3)=\exp(-0.60)=0.5488\).

With constant hazards, the cause-\(k\) probability is the fraction \(\lambda_k/(\lambda_1+\lambda_2)\) of the total event probability \(1-S(3)\). For event 1:

F_1(3)=\frac{0.12}{0.20}\{1-\exp(-0.20\times3)\} =0.60(1-0.5488)=0.2707.

For event 2:

F_2(3)=\frac{0.08}{0.20}\{1-\exp(-0.20\times3)\} =0.40(1-0.5488)=0.1805.

Thus, by three years, the model assigns probability \(0.2707\) to occupying event state 1, \(0.1805\) to occupying event state 2, and \(0.5488\) to remaining event-free. The check is \(0.2707+0.1805+0.5488=1.0000\), up to rounding. Both event probabilities are actual competing-risk probabilities, not risks calculated as if the other event could not occur.

Worked Example: Piecewise-Constant Hazards

Consider a hypothetical model with cause-specific hazards, in events per year, of \((0.10,0.20)\) from year 0 to year 1 and \((0.05,0.15)\) from year 1 to year 3. Start event-free at the landmark. Assume within each interval the two hazards are constant, the intervals cover the full prediction period, and the terminal events are mutually exclusive and absorbing. Calculate the probabilities at year 3.

During the first year, the total hazard is \(0.30\) per year. The event-free probability at year 1 is \(S(1)=\exp(-0.30)=0.7408\). The cause-1 probability accumulated during that year is \((0.10/0.30)(1-\exp(-0.30))=0.08639\); the cause-2 probability is \((0.20/0.30)(1-\exp(-0.30))=0.1728\).

From year 1 to year 3, the total hazard is \(0.20\) per year over two years. Conditional on still being event-free at year 1, the probability of remaining event-free through year 3 is \(\exp(-0.20\times2)=\exp(-0.40)=0.6703\). The unconditional event-free probability at year 3 is therefore \(0.7408\times0.6703=0.4966\).

The additional cause-1 probability over the second interval is the probability of reaching year 1 event-free, multiplied by the conditional event probability in that interval and the cause-1 share of its total hazard: \(0.7408\times(0.05/0.20)\times(1-\exp(-0.40))=0.06105\). Similarly, the additional cause-2 probability is \(0.7408\times(0.15/0.20)\times(1-\exp(-0.40))=0.1832\). Adding each interval’s contribution gives \(F_1(3)=0.08639+0.06105=0.1474\) and \(F_2(3)=0.1728+0.1832=0.3560\).

The year-3 probabilities are therefore \(0.1474\) in state 1, \(0.3560\) in state 2, and \(0.4966\) event-free. Their sum is \(1.0000\), allowing for rounding. A common calculation error is to apply the second-interval hazards to the original cohort without multiplying by \(0.7408\); that would count people who had already entered an absorbing event state as still at risk.

Worked Example: Updating Probabilities in Discrete Intervals

Suppose a hypothetical discrete-time model gives, among individuals still event-free at the start of each interval, probabilities of \(0.10\) for event 1 and \(0.20\) for event 2 in the first interval. In the second interval, the corresponding conditional probabilities are \(0.20\) and \(0.10\). The remaining conditional probability in each interval is no event. Assume these probabilities are valid for the specified landmark group and that events are mutually exclusive, terminal events. Find the state probabilities after two intervals.

Initially everyone is event-free. After the first interval, the probabilities are \(0.10\) in event state 1, \(0.20\) in event state 2, and \(0.70\) event-free. Only the \(0.70\) still event-free can experience a first terminal event during the second interval. The second-interval additions are therefore \(0.70\times0.20=0.14\) for event 1 and \(0.70\times0.10=0.07\) for event 2. The event-free probability after two intervals is \(0.70\times(1-0.20-0.10)=0.49\).

The final probabilities are \(F_1=0.10+0.14=0.24\), \(F_2=0.20+0.07=0.27\), and \(S=0.49\). They sum to \(0.24+0.27+0.49=1.00\). The calculation illustrates the same occupation logic as the continuous-time model, using interval probabilities rather than hazards.

Assumptions, Estimation, and Diagnostics

The interpretation as absorbing-state occupation is exact when the event types are defined as mutually exclusive first terminal events and remaining event-free is the only alternative state. If a person can recover, relapse, and later die, then a first-event competing-risks description does not represent that full process; the states and transitions must instead encode the relevant sequence. Similarly, if an event state is not actually terminal for the question, treating it as absorbing changes the estimand.

For a model-based prediction, the cause-specific hazards or interval probabilities must describe the intended landmark population and prediction period. If the model conditions on landmark information, the interpretation is conditional on that information; it is not automatically a probability for a different target population. If probabilities are estimated from censored follow-up, the censoring mechanism must also be handled appropriately. In particular, treating a competing event as ordinary censoring does not estimate the observed cumulative incidence of another cause. Estimation from censored data requires suitable assumptions about censoring, such as independent censoring conditional on the information used in the analysis, along with adequate follow-up support for the requested horizon.

Several checks help identify a calculation or model that is going wrong:

Probability accounting: all cause-specific occupation probabilities and the event-free probability should be nonnegative and sum to one, apart from rounding.
Monotonicity: each cumulative incidence can stay constant or increase with time; it cannot decrease. Event-free probability cannot increase under a model with nonnegative exit hazards.
Competing-event contribution: increasing another cause’s hazard, while holding the cause of interest’s hazard fixed, generally reduces the time available for the event of interest and hence its cumulative incidence over a fixed horizon.
Scale and horizon: hazards must have units reciprocal to time, and the hazard must be integrated over the elapsed time from the landmark, not over an unrelated time origin.
Support: a fitted prediction far beyond observed follow-up may be driven by extrapolation, even if its probabilities sum correctly.

The main alternative calculation that causes confusion is the “net” risk obtained by considering one cause alone and removing competing events. For constant cause-1 hazard \(0.12\) per year over three years, that calculation is \(1-\exp(-0.12\times3)=0.3023\). It exceeds the actual cause-1 cumulative incidence \(0.2707\) in the first example because it acts as if event 2 could not remove anyone from the event-free state. The two quantities answer different questions. A cause-specific hazard describes the instantaneous rate among those still event-free; it is not, by itself, the probability of that cause by a future horizon.

For data-based state-occupation estimates, the Aalen-Johansen estimator combines estimated transition probabilities while retaining competing event types as distinct states. That is the appropriate multi-state counterpart for these absorbing terminal states. A product-limit calculation that censors other event types instead targets a different quantity. Neither approach eliminates the need to define the landmark population, prediction horizon, event states, and censoring assumptions clearly.

Key takeaway: In a competing-risks model with terminal absorbing event states, cause-\(k\) cumulative incidence by a horizon is exactly the probability of occupying state \(k\) at that horizon. It is accumulated only while individuals remain event-free, so the hazards of all competing causes matter.

Check Your Understanding

Answer each question, then compare with the solutions below.

Under constant hazards \(\lambda_1=0.06\) and \(\lambda_2=0.04\) per year, what are the event-free and two cause-specific occupation probabilities after two years?
In a discrete interval, the conditional probabilities are \(0.15\) for event 1 and \(0.25\) for event 2. What is the probability of remaining event-free through that interval?
Why does the probability of a cause-1 event by a horizon generally differ from \(1-\exp\{-\int_0^t\lambda_1(u)\,du\}\) when competing causes are present?
After one interval, \(0.60\) of a group remains event-free. In the next interval, conditional probabilities are \(0.10\) for event 1, \(0.20\) for event 2, and \(0.70\) for no event. What are the second-interval additions to the two cumulative incidences and the final event-free probability?
Name two conditions needed to interpret event-state probabilities as competing terminal-event cumulative incidences.

Solutions. (1) The total hazard is \(0.10\) per year, so \(S(2)=\exp(-0.20)=0.8187\). The total event probability is \(1-0.8187=0.1813\). The cause shares are \(0.06/0.10=0.60\) and \(0.04/0.10=0.40\), giving \(F_1(2)=0.1088\) and \(F_2(2)=0.07251\). These add to \(1.0000\) with the event-free probability. (2) The event-free probability is \(1-0.15-0.25=0.60\). (3) The cause-specific cumulative incidence accumulates at rate \(S(u)\lambda_1(u)\), where \(S(u)\) reflects removal by every cause. The expression \(1-\exp\{-\int_0^t\lambda_1(u)\,du\}\) excludes that competing removal and is a net-risk calculation. (4) The additions are \(0.60\times0.10=0.06\) and \(0.60\times0.20=0.12\). The final event-free probability is \(0.60\times0.70=0.42\). (5) For example, event types must be mutually exclusive first terminal events, and the corresponding event states must be absorbing for the period of interest. For estimation with censoring, appropriate censoring assumptions are also required.

Competing Events as Occupied Absorbing States

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