Cumulative Incidence Under Competing Risks

Why Event Type Changes the Probability

In duration data, a person may experience one of several mutually exclusive first events. For example, a patient may experience a relapse, die without a recorded relapse, or remain event-free through the end of follow-up. Death prevents a later relapse from being observed as the first event. It is therefore not simply missing follow-up for the relapse outcome: it changes the probability that relapse can occur first.

The probability of a particular event type by time $t$ is called its cumulative incidence. It depends both on the hazard of that event and on the hazards of all competing events, because those other events remove people from the event-free population. This is why the cumulative incidence is not generally obtained by treating competing events as ordinary censoring in a Kaplan–Meier calculation.

Definition: Let $T$ be the time to the first event and $J\in\{1,\ldots,K\}$ its cause. For cause $k$, the cause-specific cumulative incidence function is $$ F_k(t)=P(T\leq t,\ J=k). $$ It is the probability that an individual experiences cause $k$ as the first event by time $t$. The overall event-free survival function is $S(t)=P(T>t)$; it concerns remaining free of every event type.

A cause-specific hazard describes the instantaneous rate of cause $k$ among individuals who have experienced no event yet. In continuous time, write it as $\lambda_k(t)$, with cumulative cause-specific hazard $\Lambda_k(t)=\int_0^t\lambda_k(u)\,du$. This hazard is conditional on still being event-free, not on being free of cause $k$ alone. For example, after a competing event, a person is no longer at risk of becoming a later first event of cause $k$.

From Hazards to Cumulative Incidence

Assume continuous event times and cause-specific hazards $\lambda_1(t),\ldots,\lambda_K(t)$. While someone remains event-free, the instantaneous rate of any first event is the sum of the cause-specific hazards. Thus the event-free survival function is $S(t)=\exp\{-\sum_{j=1}^K\Lambda_j(t)\}$. The cause-$k$ cumulative incidence accumulates the probability of being event-free just before time $u$, multiplied by the instantaneous cause-$k$ event rate at $u$:

F_k(t)=\int_0^t S(u)\lambda_k(u)\,du =\int_0^t \exp\left\{-\sum_{j=1}^K\Lambda_j(u)\right\}\lambda_k(u)\,du.

This formula makes the competing-risk mechanism explicit. A larger hazard for a competing cause lowers $S(u)$, leaving fewer people available to experience cause $k$ at later times. The cumulative incidence functions also account for all first-event types: in the absence of other ways to leave observation, $S(t)+\sum_{k=1}^K F_k(t)=1$.

Worked Example: Constant Cause-Specific Hazards

Suppose, hypothetically, that a population has two constant cause-specific hazards: $\lambda_1=0.04$ per year for relapse and $\lambda_2=0.06$ per year for a competing event. Calculate the probability of relapse as the first event within five years.

The event-free survival function is $S(u)=\exp[-(0.04+0.06)u]=\exp(-0.10u)$. Therefore,

F_1(5)=\int_0^5 e^{-0.10u}(0.04)\,du =\frac{0.04}{0.10}\left(1-e^{-0.10(5)}\right) =0.4(1-e^{-0.5}) \approx 0.1574.

The estimated probability is about 15.74% of experiencing relapse as the first event within five years. The competing cause contributes to the decay of event-free survival throughout the interval.

By contrast, $1-e^{-0.04(5)}\approx0.1813$. That number is the net risk associated with the relapse hazard alone, not the actual probability of relapse first when the competing hazard is present. It is larger than the cumulative incidence here.

A Discrete-Time Calculation

The same logic works when follow-up is divided into intervals and event probabilities are specified directly. Let $q_{kj}$ be the conditional probability of cause $k$ during interval $j$, among individuals event-free at that interval's start. Let $q_{\cdot j}=\sum_k q_{kj}$ be the probability of any event in that interval. The probability of remaining event-free through interval $j$, conditional on reaching its start, is $1-q_{\cdot j}$. Thus the cumulative incidence increment for cause $k$ in interval $j$ is the probability of reaching that interval event-free times $q_{kj}$.

F_k(j)=F_k(j-1)+S_{j-1}q_{kj}, \qquad S_j=S_{j-1}(1-q_{\cdot j}), \qquad S_0=1.

Worked Example: Two Discrete Follow-Up Intervals

Consider a hypothetical two-interval setting with two possible events. In interval 1, the conditional probabilities are $q_{11}=0.10$ and $q_{21}=0.20$. In interval 2, they are $q_{12}=0.15$ and $q_{22}=0.10$. Calculate the cumulative incidence of cause 1 through interval 2.

At the beginning, $S_0=1$. In interval 1, the cause-1 contribution is $1(0.10)=0.10$, and the probability of remaining event-free is $S_1=1(1-0.10-0.20)=0.70$.

Only the 70% who reached interval 2 event-free can experience a first event there. The cause-1 contribution in that interval is $0.70(0.15)=0.105$. Hence

$$ F_1(2)=0.10+0.105=0.205. $$

The cumulative probability of cause 1 by the end of interval 2 is 20.5%. Multiplying the conditional cause-1 probabilities without accounting for the competing events would not give this event probability.

Estimating Cumulative Incidence From Event-Time Data

For observed event times, a standard nonparametric estimate accumulates cause-specific event counts while using the estimated probability of still being event-free. At distinct event time $t_j$, let $Y_j$ be the number at risk just before that time, $d_{kj}$ the number of cause-$k$ events, and $d_{\cdot j}$ the total number of events of all causes. With right censoring, the usual product-limit estimate uses the event-free risk set and requires censoring to be independent of the event process for the target population (or independent conditional on variables used to define the analysis).

\widehat S(t_j)=\widehat S(t_j-)\left(1-\frac{d_{\cdot j}}{Y_j}\right), \qquad \widehat F_k(t)=\sum_{t_j\leq t}\widehat S(t_j-)\frac{d_{kj}}{Y_j}.

Here $\widehat S(t_j-)$ is the estimated probability of being event-free immediately before $t_j$. At a tied event time, all event types contribute to $d_{\cdot j}$ when updating event-free survival, while only the cause-$k$ count contributes to the cause-$k$ incidence increment. A competing event is not counted as censoring in this calculation.

Worked Example: Product-Limit Incidence With Tied Events

In a hypothetical cohort, 100 individuals are at risk at the first event time. At that time, 8 experience cause 1 and 12 experience cause 2. At the next event time, 80 remain at risk; 6 experience cause 1 and 10 experience cause 2. Assume there is no censoring before either event time. Calculate the estimated cause-1 cumulative incidence through the second event time.

Initially, $\widehat S=1$. The first cause-1 increment is $1(8/100)=0.08$. Since 20 events of either type occurred, event-free survival after this time is $1(1-20/100)=0.80$.

The second cause-1 increment is $0.80(6/80)=0.06$. Therefore,

\widehat F_1=0.08+0.06=0.14.

The estimated probability of cause 1 as the first event by the second event time is 14.0%. The risk set of 80 reflects removal of both types of first event, not just cause 1.

Why One Minus Kaplan–Meier Overstates Event Probability

If the target is cause 1, a tempting calculation is to treat every competing event as if it were ordinary right censoring, calculate Kaplan–Meier survival for cause 1, and report one minus that curve. This removes competing events from the risk set but does not reduce the estimated survival through the event-time factor for all causes. It therefore answers a different question: the net risk derived from the cause-1 hazard alone.

The distinction follows directly from the formulas. The cause-specific cumulative incidence uses event-free survival under all causes, $S(u)=\exp\{-\sum_j\Lambda_j(u)\}$, whereas the net risk is $1-\exp\{-\Lambda_k(t)\}$. Since $\sum_j\Lambda_j(u)\geq\Lambda_k(u)$, event-free survival under all causes is no larger than survival under cause $k$ alone. Integrating gives

F_k(t)=\int_0^t e^{-\sum_j\Lambda_j(u)}\,d\Lambda_k(u) \leq \int_0^t e^{-\Lambda_k(u)}\,d\Lambda_k(u) =1-e^{-\Lambda_k(t)}.

The inequality is strict when competing hazards remove people during a period in which cause $k$ can occur. In the discrete example, the net-risk calculation for cause 1 is $1-(1-0.10)(1-0.15)=0.235$, or 23.5%; the actual cumulative incidence was 20.5%. In the event-time example, censoring cause 2 and applying Kaplan–Meier to cause 1 gives $1-(1-8/100)(1-6/80)=0.149$, or 14.9%, rather than the 14.0% cumulative-incidence estimate.

Net risk can be useful as a distinct estimand, but it should not be reported as the observed probability of cause $k$ in the presence of competing events. Interpreting it as risk under an intervention that removes competing causes requires additional assumptions; the calculation alone does not establish that such an intervention is meaningful or would leave the cause-$k$ hazard unchanged.

Common Mistake and Practical Interpretation

Wrong reasoning: “A competing event prevents observing cause 1, so I will censor it and use one minus Kaplan–Meier as the probability of cause 1.”
Correction: A competing event prevents cause 1 from occurring as the first event. The actual cause-1 probability must account for the loss of event-free individuals to every cause; use the cumulative-incidence formula or its product-limit estimate.
Interpret the estimand in context: State the cause, time horizon, and population. For example, “The estimated probability of relapse as the first event within five years is 15.7%,” not simply “survival is 84.3%” if other event types matter.

Key takeaway: A cause-specific hazard is a rate among those still event-free; a cause-specific cumulative incidence is the probability of that cause occurring first by a given time. Its calculation weights the cause-specific hazard by survival from every event type.

Check Your Understanding

Use the definitions and calculations above to answer the following questions.

In your own words, distinguish the cause-specific hazard for cause $k$ from its cumulative incidence function.
Suppose a competing-cause hazard increases while the cause-$k$ hazard stays fixed. What happens to the event-free survival factor in the cause-$k$ incidence integral, and why?
For two intervals with $q_{11}=0.08$, $q_{21}=0.12$, and $q_{12}=0.10$, calculate the cause-1 cumulative incidence through interval 2.
In the tied-event product-limit estimate, why does $d_{\cdot j}$, rather than just $d_{kj}$, update event-free survival?
What probability does one minus a Kaplan–Meier curve represent when competing events are treated as censored, and why can it exceed the actual cumulative incidence?