Cumulative Incidence As an Event Probability

Why an Event Rate Is Not an Event Probability

A clinical question often asks, “What is the probability that a patient experiences event A within two years?” In a competing-risks setting, the answer depends on what counts as an event. If a patient experiences event B first, event A may no longer be possible as the first event under study. The probability of event A must therefore account for the possibility that other causes happen earlier.

The cumulative incidence function (CIF) gives this probability directly. It describes the chance that a specified cause is the first event and occurs by a stated time. It is not itself an instantaneous rate, and it is not the probability of avoiding every event.

Definition: Let $T$ be the time to the first event of any of $K$ mutually exclusive event causes, and let $J$ be the cause of that first event. For cause $k$, the cumulative incidence function is $$ F_k(t)=P(T\leq t,\ J=k). $$ Thus, $F_k(t)$ is the probability that cause $k$ occurs first by time $t$. The event-free survival function is $$ S(t)=P(T>t), $$ the probability of having experienced none of the event causes by time $t$.

The phrase “cause $k$ occurs first” is essential. If cause $j$ occurs first, then a later occurrence of cause $k$ does not count toward $F_k(t)$ in this first-event definition. Likewise, a person who has not experienced any cause by time $t$ contributes to $S(t)$, not to any cause-specific CIF by that time.

The CIF is a population probability defined by the event process. It does not require independent potential event times for the different causes. In practice, causes must be defined consistently and classified as mutually exclusive first events. If follow-up is censored, that affects how the probability can be estimated from observed data, not what probability the CIF defines.

How the Probabilities Fit Together

If the $K$ causes cover all possible first events, then by time $t$ exactly one of two things has happened: a first event of one of the causes has occurred, or no event has yet occurred. Those outcomes are mutually exclusive. Therefore,

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

In particular, $1-S(t)$ is the probability of having experienced some event by time $t$, whereas $F_k(t)$ is the probability of having experienced the specified cause $k$ by that time. The two are equal only in the special case where cause $k$ is the only possible event cause.

This identity also provides a basic check: each CIF must be between 0 and 1, it cannot decrease as time increases, and the sum of all cause-specific CIFs cannot exceed 1. The remaining probability is event-free survival.

Worked Example: Reading a One-Year Cause Probability

Suppose a hypothetical population is followed for one year. The first event by one year is cause A for 12% of people, cause B for 8%, and no event occurs by one year for 80%. Find the one-year CIF for A, the one-year CIF for B, and event-free survival.

By definition, the probability of cause A as the first event by one year is $F_A(1)$. The stated proportion is 12%, so $$ F_A(1)=0.12. $$ Similarly, $$ F_B(1)=0.08,\qquad S(1)=0.80. $$ The probability identity checks: $$ F_A(1)+F_B(1)+S(1)=0.12+0.08+0.80=1.00. $$

Thus the one-year probability of cause A as the first event is 12%, while the probability of remaining free of either event through one year is 80%. The probability of either event by one year is $1-0.80=0.20$, not 12%.

Connecting Cumulative Incidence to Cause-Specific Hazards

A cause-specific hazard describes an instantaneous event rate among people who have not yet experienced any event. It is not a probability accumulated over time. Under continuous-time assumptions, the CIF combines that rate with the probability of still being event-free at each time.

Suppose the event times are continuous, so there is no probability mass at tied event times, and let $\lambda_k(u)$ be the cause-specific hazard for cause $k$ at time $u$. By its definition, over a short interval of length $du$, the conditional probability of cause $k$ occurring is approximately $\lambda_k(u)du$ among those still event-free at $u$. The unconditional probability of such an event in that interval is therefore approximately $S(u)\lambda_k(u)du$. Adding these contributions over time gives

F_k(t)=\int_0^t S(u)\lambda_k(u)\,du.

This relationship explains why integrating a cause-specific hazard alone does not generally give the CIF: the hazard applies only to those still event-free, represented by $S(u)$. The equality is written for continuous event times; with discrete event times or event-time probability masses, the corresponding calculation uses interval or jump probabilities rather than this continuous integral.

Worked Example: Discrete-Time Competing Event Probabilities

Consider a hypothetical group of 100 people, all event-free at time zero. During the first interval, each person has conditional probability 0.10 of cause A, 0.05 of cause B, and 0.85 of remaining event-free. During the second interval, among those still event-free at its start, the corresponding probabilities are 0.10, 0.20, and 0.70. Calculate the CIFs and event-free probability at the end of the second interval.

The first-interval probabilities are already unconditional probabilities for the original group: $$ F_A(1)=0.10,\qquad F_B(1)=0.05,\qquad S(1)=0.85. $$ For interval two, the conditional probabilities must be multiplied by the probability of reaching that interval event-free. Thus, the second-interval contribution to the cause A CIF is $0.85(0.10)=0.085$, and the contribution to cause B is $0.85(0.20)=0.17$. Hence $$ F_A(2)=0.10+0.85(0.10)=0.185, $$ $$ F_B(2)=0.05+0.85(0.20)=0.220. $$ The event-free probability at the end of interval two is $$ S(2)=0.85(0.70)=0.595. $$ The values add to $0.185+0.220+0.595=1.000$, as required. In percentage terms, 18.5% experience cause A first by the end of interval two, 22.0% experience cause B first, and 59.5% have experienced neither cause.

The multiplication by the probability of reaching an interval event-free is the discrete-time counterpart of the $S(u)$ factor in the continuous-time integral. A conditional cause probability among event-free people cannot be treated as if it applied to the entire original population at every interval.

Worked Example: Constant Cause-Specific Hazards

Suppose a hypothetical population has constant cause-specific hazards $\lambda_A=0.10$ per year and $\lambda_B=0.20$ per year over two years, with no other event causes. Under the continuous-time model, calculate $F_A(2)$, $F_B(2)$, and $S(2)$.

For constant hazards, the total hazard of any first event is $0.10+0.20=0.30$ per year. The event-free probability at time $t$ is therefore $S(t)=\exp(-0.30t)$. At two years, $$ S(2)=\exp[-(0.30)(2)]=\exp(-0.60)\approx 0.5488. $$ Using $F_k(t)=\int_0^t S(u)\lambda_k\,du$, the cause A CIF is $$ F_A(2)=\int_0^2 e^{-0.30u}(0.10)\,du =\frac{0.10}{0.30}\{1-e^{-(0.30)(2)}\} \approx \frac{1}{3}(0.4512)=0.1504. $$ For cause B, $$ F_B(2)=\int_0^2 e^{-0.30u}(0.20)\,du =\frac{0.20}{0.30}\{1-e^{-(0.30)(2)}\} \approx \frac{2}{3}(0.4512)=0.3008. $$ The check is $0.1504+0.3008+0.5488=1.0000$, up to rounding. The two-year probability of cause A as the first event is about 15.0%, not 30.0%. Cause B is more frequent in this model, and some people remain event-free throughout the two years.

For comparison, if one used the cause A hazard alone to calculate $1-\exp[-(0.10)(2)]$, the result would be approximately 0.1813. That calculation removes cause B from the event-free process. It is not the observed-population CIF for A when cause B is also possible; under the stated competing-risks model the CIF is about 0.1504.

Choosing the Right Quantity for the Question

The CIF is appropriate when the question concerns the absolute probability of a specified first event by a time horizon in the population where the competing events can occur. Other quantities answer different questions:

Event-free survival $S(t)$ answers the probability of having experienced no event of any cause by time $t$.
The cause-specific hazard $\lambda_k(t)$ describes an instantaneous rate among those still event-free at time $t$, rather than an accumulated probability.
A subdistribution hazard is another rate-based quantity connected to the CIF, with a different risk-set interpretation. It is not interchangeable with the CIF itself.

These quantities can tell different stories without contradicting one another. A cause-specific hazard may remain fixed while the CIF changes because the competing causes alter how many people remain event-free and therefore available to experience the focal cause. The CIF is often the most direct answer for counseling, planning, or describing absolute event burden at a specified horizon.

Common Mistakes and Checks

Confusing event-free survival with a cause probability. $S(t)$ is the chance of no event of any cause; $F_k(t)$ is the chance of the specified cause as the first event. Use $1-S(t)$ for the probability of any event, not for one particular cause.
Calling a hazard a probability. A hazard has units of inverse time, such as per year. A CIF is a probability and has no units. A hazard value cannot be read directly as the probability of an event by a future time.
Ignoring the event-free proportion when accumulating conditional risks. Later conditional probabilities apply only to people still event-free at that time. Multiply them by the probability of being event-free at the interval's start before adding their contributions to a CIF.
Assuming a competing event is merely a later observation of the focal event. In the first-event estimand, a different cause occurring first is a competing outcome, not a focal event that can be counted later.

When reporting a CIF, state the cause, the time horizon, and that it refers to the first event in the presence of the competing causes. Also check that the CIF is nondecreasing over time and that the sum across causes plus event-free survival equals one when the cause list is exhaustive. A mismatch can indicate omitted event categories, inconsistent definitions, or arithmetic errors.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

In one sentence, define $F_k(t)$. What does the event $J=k$ specify?
At a time horizon, $F_A(t)=0.18$, $F_B(t)=0.27$, and $S(t)=0.55$. What is the probability of any event by that time?
In a discrete interval, 70% of the original group is still event-free at the start, and the conditional probability of cause A in that interval is 0.12. What is the interval's contribution to the CIF for A?
Why is the integral of a cause-specific hazard alone generally not the CIF?
With constant hazards $\lambda_A=0.04$ and $\lambda_B=0.06$ per year, what is the one-year event-free probability, to three significant figures?

Solutions. (1) $F_k(t)=P(T\leq t,J=k)$, the probability that cause $k$ is the first event and occurs by $t$; $J=k$ identifies the cause of the first event. (2) Any event has probability $1-S(t)=1-0.55=0.45$, or 45%. (3) The contribution is $0.70(0.12)=0.084$, or 8.4% of the original group. (4) The cause-specific hazard applies only among people who remain event-free, so each time-specific contribution is weighted by the event-free probability $S(u)$. (5) The total hazard is $0.04+0.06=0.10$ per year, so $S(1)=\exp(-0.10)\approx 0.905$, or 90.5%.