Why Conditional Probabilities Build Cumulative Incidence
A cumulative incidence is an absolute event probability: the chance that a particular cause has occurred by a specified time, in the presence of the other competing causes. To calculate it, consider the possible times at which that cause could first occur. At each time, its conditional event probability matters only for individuals who have remained event-free up to that time. Earlier events of any cause remove individuals from that event-free group.
This conditioning step is the reason the Aalen–Johansen cumulative incidence formula weights cause-specific event increments by all-cause event-free survival. The weight is not an adjustment added for convenience; it follows directly from decomposing the target probability according to when the event occurs.
The event-free survival probability includes freedom from all causes. If an individual experiences a competing event, that individual cannot later experience cause \(k\) as the first event. Thus, the probability of experiencing cause \(k\) at a particular time depends on both the probability of being event-free just beforehand and the chance of cause \(k\) occurring then, conditional on being event-free.
Decomposing the Probability by Event Time
Begin with a finite set of distinct event times \(u_1<u_2<\cdots<u_m\). Define \(q_k(u_r)\) as the conditional probability of experiencing cause \(k\) at time \(u_r\), given that the person is still event-free immediately before \(u_r\). Let \(q_{\cdot}(u_r)\) be the corresponding conditional probability of experiencing any cause at that time. Because there is only one first event, \(q_{\cdot}(u_r)=\sum_j q_j(u_r)\).
For an event of cause \(k\) to occur at \(u_r\), two things must happen: the person must have been event-free just before \(u_r\), and then cause \(k\) must occur. The multiplication rule for conditional probability gives \(P(T=u_r,J=k)=P(T\geq u_r)q_k(u_r)\). The events at different times are mutually exclusive, so adding these probabilities gives the cumulative incidence through any time \(t\).
Here \(S(u_r-)\) is the probability of having avoided every cause before \(u_r\). The product follows by applying the conditional probability of avoiding all causes at each preceding event time: conditional on being event-free just before \(v\), that probability is \(1-q_{\cdot}(v)\). Multiplying these successive conditional probabilities gives survival to just before \(u_r\).
This also explains why the cause-specific event probabilities are not simply added without weights. A conditional probability at a later time applies only to the fraction still event-free at that time. The earlier event-free survival factor converts it to a probability on the original population scale.
Worked Example: Conditioning Across Two Follow-Up Periods
In a hypothetical population, during the first follow-up period the conditional probabilities of a first event are 0.10 for cause A and 0.05 for cause B. Among those still event-free at the start of the second period, the corresponding probabilities are 0.20 and 0.10. Find each cause’s cumulative incidence through the end of the second period.
At the end of the first period, the probability of a cause A event is \(1(0.10)=0.10\), and the probability of a cause B event is \(1(0.05)=0.05\). The fraction still event-free is \(1-0.10-0.05=0.85\).
At the second period, the conditional probability of cause A is 0.20, but it applies to the 0.85 still event-free. Its contribution to the original population’s cause A risk is \(0.85(0.20)=0.17\). Therefore \(F_A=0.10+0.17=0.27\). Similarly, the second-period contribution for cause B is \(0.85(0.10)=0.085\), giving \(F_B=0.05+0.085=0.135\).
The remaining event-free probability is \(0.85(1-0.20-0.10)=0.595\). The check \(0.27+0.135+0.595=1.000\) confirms that the two cumulative incidences and the event-free probability account for the population.
From Conditional Probabilities to Hazard Rates
In continuous time, suppose the cause-specific hazards exist and are represented by rates \(\lambda_k(u)\). In a short interval of width \(du\), conditional on being event-free at its start, the probability of cause \(k\) occurring is approximately \(\lambda_k(u)du\); the probability of two or more events in that interval is of smaller order than \(du\). The total event rate is \(\lambda_{\cdot}(u)=\sum_j\lambda_j(u)\).
Over a short interval, event-free survival consequently changes by approximately \(-S(u)\lambda_{\cdot}(u)du\). The resulting survival function is \(S(u)=\exp\{-\int_0^u\lambda_{\cdot}(v)\,dv\}\). The conditional probability of cause \(k\) in the interval, multiplied by the probability of being event-free at its start, gives an unconditional probability increment of approximately \(S(u)\lambda_k(u)du\). Summing these increments across time becomes an integral:
This is the continuous-time version of the same conditioning argument. The survival factor accounts for all causes, while the rate inside the integral selects the cause of interest. The formula assumes that the event-time distribution is continuous over the interval under consideration; when event-time probability has jumps, the corresponding probability increments and survival just before the jump must be used instead.
Worked Example: Constant Cause-Specific Rates
Suppose, hypothetically, that two causes have constant cause-specific hazards over the first two years: \(\lambda_A=0.20\) per year and \(\lambda_B=0.30\) per year. Find the event-free probability and both cumulative incidences at two years.
The total rate is \(0.20+0.30=0.50\) per year, so event-free survival at two years is \(S(2)=\exp(-0.50\times2)=e^{-1}=0.3679\). The integrated cause A risk is \(F_A(2)=\int_0^2 e^{-0.50u}(0.20)\,du\).
For cause B, the calculation is \(F_B(2)=\frac{0.30}{0.50}(1-e^{-1})=0.3793\). Thus, by two years, the hypothetical probabilities are 0.2528 for a first event of cause A, 0.3793 for a first event of cause B, and 0.3679 of remaining event-free. Their sum is 1.0000, up to rounding. The rates have units of inverse years; the cumulative incidences are probabilities and therefore have no units.
How the Identity Leads to the Aalen–Johansen Estimator
For observed right-censored data, let \(Y(u)\) be the number at risk immediately before event time \(u\), and let \(d_k(u)\) be the number of observed events of cause \(k\) at that time. Under appropriate independent censoring, the observed fraction \(d_k(u)/Y(u)\) estimates the conditional probability of a cause \(k\) event at \(u\) among those event-free and observed just beforehand. For this simple presentation, assume distinct observed event times; the treatment of ties requires a separate convention.
Substitute the estimated conditional cause probabilities into the decomposition. The estimated conditional probability of any event at \(u\) is \(\sum_j d_j(u)/Y(u)\). The estimated probability of remaining event-free through that event time is therefore multiplied by \(1-\sum_jd_j(u)/Y(u)\). This gives the Aalen–Johansen estimator:
The estimator is thus a plug-in version of the conditional-probability identity: estimate the cause-specific conditional event increment, estimate the probability of reaching that time event-free, and multiply the two. Its validity relies on risk sets that represent those still event-free in the target population. In particular, censoring must not selectively remove people with different future event risks, conditional on the information accounted for in the analysis. The details of independent censoring and estimation with right censoring require careful attention in their own right.
Worked Example: Calculating Two Cause-Specific Estimates
Consider hypothetical data with three distinct event times. At time 2 years, \(Y=100\) and there are 10 cause A events. At time 5 years, \(Y=80\) and there are 9 cause B events. At time 8 years, \(Y=60\) and there are 8 cause A events. Assume there are no other causes and that the risk sets are appropriate under independent censoring. Calculate the estimates through 8 years.
At 2 years, the cause A conditional increment is \(10/100=0.10\). Thus \(\widehat{F}_A\) becomes 0.10, and \(\widehat{S}\) becomes \(1-10/100=0.90\).
At 5 years, the cause B increment is \(9/80=0.1125\), weighted by the event-free estimate just before that time, 0.90. Hence its contribution is \(0.90(0.1125)=0.10125\), and \(\widehat{F}_B=0.10125\). The updated event-free estimate is \(0.90(1-9/80)=0.90(0.8875)=0.79875\).
At 8 years, the cause A increment is \(8/60=0.1333\) (to four significant figures). Its contribution is \(0.79875(8/60)=0.1065\) (rounded). Therefore \(\widehat{F}_A=0.10+0.1065=0.2065\). The event-free estimate after time 8 is \(0.79875(1-8/60)=0.6923\).
The estimates sum to approximately \(0.2065+0.10125+0.69225=1.0000\), with the displayed values rounded. The cause A estimate is not just the sum \(10/100+8/60\): the later increment must be weighted by the estimated probability of still being event-free at 8 years.
Checks, Alternatives, and Common Errors
The decomposition provides useful checks on both reasoning and arithmetic. Each cumulative incidence should be nondecreasing and lie between 0 and 1. With all causes represented, the estimated cause-specific incidences plus the estimated event-free survival should sum to 1 at a given time, apart from rounding. A failure of this check can reveal an omitted cause, an incorrect risk set, or a survival factor that was updated incorrectly.
The main alternative way to express the population calculation is to integrate the event-free survival function against a cause-specific hazard rate. In continuous time this gives \(\int_0^t S(u)\lambda_k(u)\,du\), the same conditional-probability argument in rate form. With complete follow-up and no censoring, one can also count the observed first events of each cause by time \(t\) and divide by the cohort size; that direct proportion estimates the same probability. With censoring, a raw proportion among those still observed generally does not represent the original cohort probability, which is why event-time risk sets and survival weights are needed.
A common mistake is to add conditional cause-specific increments as if each applied to the full original population. For example, in the earlier two-period calculation, this would give \(0.10+0.20=0.30\) for cause A, instead of the correct \(0.10+0.85(0.20)=0.27\). The correction is to weight each conditional increment by the probability of reaching its event time without any competing event.
A second diagnostic is to inspect the risk-set construction and censoring assumptions. If people who are censored have systematically different subsequent risks from those remaining under observation, the observed event fractions may not estimate the target conditional probabilities. The formula cannot repair biased risk-set information. The resulting cumulative incidence describes event probabilities under the specified population and follow-up conditions; by itself, it does not establish a causal effect.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- At one event time, the probability of being event-free just beforehand is 0.70, and the conditional probability of cause A at that time is 0.08. What is cause A’s unconditional probability contribution at that time?
- At a time with \(Y=50\), there are 3 cause A events and 2 cause B events. What are the estimated conditional cause A increment and the estimated conditional probability of any event?
- In continuous time, two constant cause-specific hazards are 0.1 and 0.2 per year. What is event-free survival at one year?
- Why is it generally incorrect to add the cause A conditional increments across event times without survival weights?
Solutions. (1) The contribution is \(0.70(0.08)=0.056\). (2) The cause A increment is \(3/50=0.06\); the any-cause increment is \((3+2)/50=0.10\). (3) Total hazard is \(0.3\) per year, so \(S(1)=e^{-0.3}=0.7408\). (4) A later conditional increment applies only to those still event-free immediately beforehand, so it must be multiplied by that event-free probability to express its contribution on the original population scale.