Deriving Cumulative Incidence From Cause-Specific Hazards in Competing Risks

From a Cause-Specific Hazard to Absolute Risk

A cause-specific hazard describes the instantaneous rate of one event among people who remain event-free. Absolute risk asks a different question: what is the probability that this event has occurred by a specified time, accounting for the possibility that another cause happens first? The link between these quantities is an accumulation over time. At each time, the event-free population supplies the people who can still experience any first event; the cause-specific hazard determines the rate at which that population experiences the cause of interest.

The derivation below makes that link precise. It assumes a continuous-time event distribution with densities. Under that assumption, the cause-specific cumulative incidence is an integral of event-free survival multiplied by the cause-specific hazard. The survival factor is essential: a hazard increment at a later time applies only to individuals who have not already experienced any competing event.

Definition: Let $T$ be the time to the first event of any cause, and let $J$ identify its cause, with $J\in\{1,\ldots,K\}$. The event-free survival function is $S(t)=P(T>t)$. The cause-$k$ cumulative incidence function is $F_k(t)=P(T\leq t,\ J=k)$. When the relevant event-time distributions have densities, the cause-specific hazard for cause $k$ is $$ \lambda_k(t)=\lim_{h\downarrow 0}\frac{P(t\leq T<t+h,\ J=k\mid T\geq t)}{h}. $$ It has units of inverse time and describes the instantaneous rate of cause $k$ among people still event-free at time $t$.

For a continuous event-time distribution, whether the risk set is written as $T\geq t$ or $T>t$ does not change this definition at an individual time. In the derivation, assume there is no event mass at baseline, the event-time distribution is absolutely continuous over the time interval considered, and the cause-specific hazards are integrable there. Absolute continuity means the event probabilities can be represented by densities with respect to time. Merely having no point masses is not enough: an atomless distribution can still have a singular-continuous component that does not have a density.

Deriving the Integral

Write $f_k(t)$ for the density of experiencing the first event at time $t$ from cause $k$. Among individuals event-free at time $t$, the probability of cause $k$ in a short interval of length $h$ is approximately $\lambda_k(t)h$. The proportion of the original population still event-free is $S(t)$. Consequently, the population probability of a cause-$k$ event in that interval is approximately $S(t)\lambda_k(t)h$. Taking the limit gives the density identity $f_k(t)=S(t)\lambda_k(t)$.

The total event density is the sum of the cause-specific densities because a first event has exactly one cause. Thus $S'(t)=-\sum_{j=1}^{K} f_j(t)=-S(t)\sum_{j=1}^{K}\lambda_j(t)$. Solving this differential equation with $S(0)=1$ gives the event-free survival representation below. Integrating the cause-specific density from baseline to $t$ then yields the cumulative-incidence formula.

\begin{aligned} f_k(u)&=S(u)\lambda_k(u),\\ S'(u)&=-S(u)\sum_{j=1}^{K}\lambda_j(u),\\ S(u)&=\exp\!\left\{-\int_0^u\sum_{j=1}^{K}\lambda_j(v)\,dv\right\},\\ F_k(t)&=\int_0^t f_k(u)\,du =\int_0^t S(u)\lambda_k(u)\,du. \end{aligned}

This derivation also explains the integrand’s interpretation. Over a short interval $du$ at time $u$, $S(u)\lambda_k(u)\,du$ is the probability, in the original population, of still being event-free just before that interval and then experiencing cause $k$ within it. Summing those contributions over time gives the probability of cause $k$ by $t$. In continuous time, the distinction between survival just before and at $u$ vanishes under the stated assumptions.

Theorem: If first-event times have an absolutely continuous distribution, with integrable cause-specific hazards $\lambda_1,\ldots,\lambda_K$, then $$ F_k(t)=\int_0^t \exp\!\left\{-\int_0^u\sum_{j=1}^{K}\lambda_j(v)\,dv\right\} \lambda_k(u)\,du. $$ Each cause-specific risk accumulates its hazard only while the individual remains event-free from every cause.

Worked Calculations

Worked Example: Constant Hazards Over Five Years

Suppose, hypothetically, that two causes have constant cause-specific hazards $\lambda_1=0.04$ per year and $\lambda_2=0.06$ per year. Find each cause’s cumulative incidence by five years. The hazards are nonnegative and constant, hence integrable; assuming the hypothetical event-time distribution is generated by these continuous hazards gives an absolutely continuous distribution and satisfies the derivation’s assumptions.

The total hazard is $0.04+0.06=0.10$ per year, so event-free survival at time $u$ is $S(u)=e^{-0.10u}$. For cause 1,

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 probability of a first event from cause 1 by five years is about $15.74\%$. For cause 2,

F_2(5)=\frac{0.06}{0.10}(1-e^{-0.5}) =0.6(0.393469) \approx 0.2361.

The remaining event-free probability is $S(5)=e^{-0.5}\approx0.6065$. The checks $0.1574+0.2361+0.6065\approx1$ and $F_1(5),F_2(5)\geq0$ are consistent with the possible outcomes by five years.

Worked Example: Hazards That Change at Two Years

Consider a hypothetical population with two causes. From year 0 through year 2, let $\lambda_1=0.03$ and $\lambda_2=0.07$ per year. From year 2 through year 5, let $\lambda_1=0.08$ and $\lambda_2=0.02$ per year. These nonnegative piecewise-constant hazards are integrable, and the example assumes continuous event times generated by them. Calculate $F_1(5)$.

In the first interval the total hazard is $0.10$ per year. Its contribution to cause 1 is

\int_0^2 e^{-0.10u}(0.03)\,du =\frac{0.03}{0.10}(1-e^{-0.20}) \approx 0.0543808.

At year 2, the event-free probability is $S(2)=e^{-0.20}\approx0.818731$. In the second interval the total hazard is again $0.10$ per year, so the cause-1 contribution is

\int_2^5 e^{-0.20}e^{-0.10(u-2)}(0.08)\,du =e^{-0.20}\frac{0.08}{0.10}(1-e^{-0.30}) \approx 0.169760.

Adding the unrounded interval contributions gives $F_1(5)\approx0.224141$, or about $22.41\%$. Notice that the second interval’s contribution is multiplied by $S(2)$. It is not the risk that would accrue if everyone were still event-free at year 2.

Worked Example: Recomputing Risk When Both Hazards Change

In a separate hypothetical example, let $\lambda_1=0.02$ and $\lambda_2=0.03$ per year from year 0 to year 2, and let $\lambda_1=0.06$ and $\lambda_2=0.04$ per year from year 2 to year 4. Assume continuous event times generated by these nonnegative, integrable hazards. Find both cumulative incidences by year 4 and verify the event-free probability.

During years 0–2, the total hazard is $0.05$ per year; during years 2–4 it is $0.10$ per year. Therefore $S(2)=e^{-0.05(2)}=e^{-0.1}$, and $S(4)=e^{-0.1}e^{-0.10(2)}=e^{-0.3}$. For cause 1, the first and second interval contributions are

\begin{aligned} F_1(4) &=\frac{0.02}{0.05}(1-e^{-0.1}) +e^{-0.1}\frac{0.06}{0.10}(1-e^{-0.2})\\ &\approx 0.0380650+0.0984115 =0.136477. \end{aligned}

For cause 2, the corresponding calculation is

\begin{aligned} F_2(4) &=\frac{0.03}{0.05}(1-e^{-0.1}) +e^{-0.1}\frac{0.04}{0.10}(1-e^{-0.2})\\ &\approx 0.0570975+0.0656077 =0.122705. \end{aligned}

Finally $S(4)=e^{-0.3}\approx0.740818$, and $F_1(4)+F_2(4)+S(4)\approx1.000000$. The calculation respects both changes in the hazards and the changing event-free population.

Interpretation, Comparisons, and Diagnostics

A cause-specific hazard is a rate among those currently event-free; it is not itself an absolute probability. The cumulative incidence is an absolute probability that incorporates the competing causes through $S(u)$. For that reason, the cumulative incidence for a cause depends not only on its own hazard, but also on the hazards of the other causes. Increasing a competing cause’s hazard can reduce the event-free probability available for the cause of interest, even if the latter’s cause-specific hazard is unchanged.

This distinction separates the cumulative incidence from net risk. Net risk is obtained by considering the cause-specific hazard as if competing events did not remove individuals from the possibility of the event of interest. It is not generally the real-world probability of that event in the presence of competing causes. The Aalen–Johansen estimate is a standard nonparametric way to estimate cumulative incidence from observed competing-risk data; the integral here gives the population relationship from which such risk calculations can be understood. Estimation under censoring is a separate issue from this identity.

Quantity	What it describes	Role in this derivation
Cause-specific hazard $\lambda_k(u)$	Instantaneous rate of cause $k$ among individuals still event-free	Rate for cause $k$ at time $u$
Event-free survival $S(u)$	Probability of no cause having occurred by $u$	Fraction of the original population still able to experience a first event
Cumulative incidence $F_k(t)$	Probability of a first event from cause $k$ by $t$	Integral of $S(u)\lambda_k(u)$ over $0$ to $t$

Several checks help identify errors in a calculation. Hazards must be expressed in inverse units matching the time scale, and the integral must produce a dimensionless probability. Each contribution must be nonnegative. Across all causes, the cumulative incidences plus event-free survival must equal one under the stated setup:

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

A result that exceeds one, a negative cause-specific risk, or a sum inconsistent with one points to an error in the hazard totals, interval boundaries, survival factors, or arithmetic. The identity also makes clear why using the wrong survival factor is consequential: multiplying $\lambda_k(u)$ only by survival from cause $k$ would fail to remove individuals who already experienced a different first event.

Assumptions and Common Pitfalls

The ordinary time integral requires absolute continuity of the first-event distribution over the interval and integrable cause-specific hazard densities. If there are event-time atoms, such as a positive probability of events at a scheduled time, a density-only calculation does not capture those jumps. Likewise, an atomless but singular-continuous distribution need not be represented by an ordinary hazard density. Such settings require a measure-based treatment rather than the density derivation used here.

Ignoring competing causes. Integrating $\lambda_k(u)$ alone treats the hazard as if the original population remained available throughout. The correction is to weight it by event-free survival $S(u)$.
Using the wrong survival function. Survival from cause $k$ alone does not represent being free of every first event. Use survival from all causes combined.
Confusing rate with risk. A hazard has units such as per year; $F_k(t)$ is a probability by time $t$. Do not report a hazard value as a cumulative probability.
Applying an interval’s hazard to the original population. For a later interval, multiply its contribution by survival up to that interval. In the piecewise example, the factor $e^{-0.20}$ at year 2 accounts for events that already occurred.
Assuming no ties guarantees the formula. Absence of point masses alone does not ensure an event-time density. Check the absolute-continuity assumption rather than inferring it from the lack of ties.

Key takeaway: Under absolute continuity, cause-$k$ absolute risk by time $t$ is the time-accumulated product of event-free survival and the cause-$k$ hazard: $F_k(t)=\int_0^t S(u)\lambda_k(u)\,du$. Other causes matter because they reduce $S(u)$.

Check Your Understanding

Answer each question, then compare your work with the solutions.

For constant hazards $\lambda_1=0.03$ and $\lambda_2=0.07$ per year, calculate $F_1(4)$.
At a particular time, suppose $S(u)=0.80$ and $\lambda_1(u)=0.05$ per year. What is the cause-1 event-density contribution at that time, and what are its units?
Why does the ordinary hazard-density integral require more than the absence of point masses?
State the probability check relating event-free survival and the cause-specific cumulative incidences.

Solutions. (1) The total hazard is $0.10$ per year, so $F_1(4)=\frac{0.03}{0.10}(1-e^{-0.10(4)})=0.3(1-e^{-0.4})\approx0.098904$, or $9.890\%$. (2) The contribution per unit time is $S(u)\lambda_1(u)=0.80(0.05)=0.040$ per year. It is a density of event probability over time, not a probability by itself. (3) An atomless distribution may have a singular-continuous component without a density with respect to time, so absolute continuity is needed for the ordinary density-based integral. (4) $S(t)+\sum_{k=1}^{K}F_k(t)=1$: by time $t$, the individual is either event-free or has experienced a first event from exactly one cause.

Quantity	What it describes	Role in this derivation
Cause-specific hazard \(\lambda_k(u)\)	Instantaneous rate of cause \(k\) among individuals still event-free	Rate for cause \(k\) at time \(u\)
Event-free survival \(S(u)\)	Probability of no cause having occurred by \(u\)	Fraction of the original population still able to experience a first event
Cumulative incidence \(F_k(t)\)	Probability of a first event from cause \(k\) by \(t\)	Integral of \(S(u)\lambda_k(u)\) over \(0\) to \(t\)