Cause-Specific Rates And Absolute Event Risk

From an Instantaneous Rate to an Event Probability

A cause-specific hazard describes the instantaneous rate of a particular event among people who are still event-free. An absolute event risk answers a different question: what fraction of the original event-free population experiences that cause by a stated time? The distinction matters because a person can experience a competing event first and then no longer be at risk of experiencing the event of interest as a first event.

Consequently, knowing the cause-specific hazard for one cause does not, by itself, determine the probability of that cause’s event by a time horizon. To calculate that probability, we also need to know how quickly people leave the event-free state through all causes. This tutorial develops that relationship and shows how to use it for constant and time-varying rates.

Definition: Let \(T\) be the time to the first event, and let \(J\) denote its cause, with possible causes \(1,\ldots,K\). The event-free probability at time \(t\) is \(S(t)=P(T>t)\). The cumulative incidence function for cause \(k\) is \(F_k(t)=P(T\leq t,\ J=k)\), the probability that the first event is cause \(k\) by time \(t\). The cause-specific hazard \(\lambda_k(t)\) is the instantaneous event rate for cause \(k\) among individuals who remain event-free just before time \(t\).

Assume for now that there is an event-free population at time zero, event times have continuous distributions, and the cause-specific hazards describe that population over the period of interest. In a short interval from \(t\) to \(t+\Delta t\), the probability of cause \(k\) occurring, conditional on being event-free at \(t\), is approximately \(\lambda_k(t)\Delta t\). But only the fraction \(S(t)\) of the original population is still event-free and able to experience a first event at that time.

\text{Probability of cause }k\text{ in }[t,t+\Delta t) \text{ in the original population} \approx S(t)\lambda_k(t)\Delta t

That factor \(S(t)\) is the essential link between a cause-specific rate and an absolute probability. The rate is conditional on remaining event-free; the cumulative incidence counts events in the original population, including the fact that some people have already had a competing event.

Deriving the Cumulative Incidence Formula

The overall hazard among people still event-free is the sum of the cause-specific hazards. For continuous event times, this gives the event-free probability \(S(t)\) and the cumulative incidence for cause \(k\):

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

The first expression says that remaining event-free requires avoiding every cause. The second accumulates the chance of a cause-\(k\) event at each time, weighted by the probability of still being event-free at that time. These expressions presume that the hazards are defined for the population and time scale being analyzed; their integral must be well-defined.

Derivation: Over a short interval, the probability of any first event is approximately \(S(t)\sum_{j=1}^K\lambda_j(t)\Delta t\). Thus \(S(t+\Delta t)-S(t)\approx -S(t)\sum_j\lambda_j(t)\Delta t\). Dividing by \(\Delta t\) and taking the limit gives \(S'(t)=-S(t)\sum_j\lambda_j(t)\), with \(S(0)=1\). Solving this differential equation gives the expression for \(S(t)\). In the same interval, the increase in the probability of a cause-\(k\) first event is \(S(t)\lambda_k(t)\Delta t\). Therefore \(F_k'(t)=S(t)\lambda_k(t)\), and \(F_k(0)=0\); integrating gives the expression for \(F_k(t)\).

This derivation also explains why a single cause-specific hazard is insufficient: \(F_k(t)\) depends on \(S(u)\), and \(S(u)\) depends on the hazards for all causes. A cause-specific hazard can describe how rapidly cause \(k\) occurs among those currently event-free, but it does not tell us how many people remain event-free at later times.

Worked Example: Two Constant Hazards

Suppose, hypothetically, that two mutually exclusive first-event causes have constant cause-specific hazards over four years: \(\lambda_1=0.03\) per year and \(\lambda_2=0.01\) per year. Assume everyone is event-free at time zero, there are no other first-event causes, and the hazards apply throughout the four years. Find the probability of a cause-1 event by year 4.

The total hazard is \(0.03+0.01=0.04\) per year, so the event-free probability at time \(u\) is \(S(u)=e^{-0.04u}\). Integrating the cause-1 hazard weighted by this event-free probability gives:

\begin{aligned} F_1(4) &=\int_0^4 e^{-0.04u}(0.03)\,du\\ &=\frac{0.03}{0.04}\left(1-e^{-(0.04)(4)}\right)\\ &=0.75(1-e^{-0.16})\\ &\approx 0.1109. \end{aligned}

Thus the hypothetical absolute risk of a cause-1 first event by four years is about 11.09%. The calculation uses both hazards: cause 2 removes some individuals from the event-free group before they can have a cause-1 first event.

Constant Hazards: The General Formula

When all cause-specific hazards are constant, write \(\lambda_{\text{tot}}=\sum_{j=1}^{K}\lambda_j\). Substituting \(S(u)=e^{-\lambda_{\text{tot}}u}\) into the cumulative incidence integral gives a useful closed form:

F_k(t)= \frac{\lambda_k}{\lambda_{\text{tot}}} \left(1-e^{-\lambda_{\text{tot}}t}\right), \qquad \lambda_{\text{tot}}>0

The term \(1-e^{-\lambda_{\text{tot}}t}\) is the probability of any first event by time \(t\). Under constant hazards, the fraction \(\lambda_k/\lambda_{\text{tot}}\) allocates that probability to cause \(k\). This allocation is a consequence of the constant-hazard assumption; it is not a general rule for hazards that change over time.

If the other cause-specific hazards change while \(\lambda_k\) remains fixed, then \(\lambda_{\text{tot}}\), and hence the event-free probability, changes. The absolute risk \(F_k(t)\) can therefore change even though the cause-specific hazard for cause \(k\) does not.

Worked Example: Same Cause-Specific Rate, Different Absolute Risk

Consider two hypothetical populations followed for four years. In both, the constant cause-1 hazard is \(0.03\) per year. In population A, the cause-2 hazard is \(0.01\) per year; in population B, it is \(0.09\) per year. Assume in each population that these are the only first-event causes and that individuals are event-free at time zero.

For population A, the total hazard is \(0.04\) per year, so:

F_{1,A}(4) =\frac{0.03}{0.04}\left(1-e^{-(0.04)(4)}\right) =0.75(1-e^{-0.16}) \approx 0.1109.

For population B, the total hazard is \(0.12\) per year. Therefore:

F_{1,B}(4) =\frac{0.03}{0.12}\left(1-e^{-(0.12)(4)}\right) =0.25(1-e^{-0.48}) \approx 0.0953.

The cause-1 rate is identical, but its four-year absolute risk is about 11.09% in population A and 9.53% in population B. The larger cause-2 hazard in population B means fewer individuals remain event-free long enough to experience cause 1. The difference is a comparison of hypothetical risks under the stated hazard patterns, not evidence that changing cause 2 would causally change cause 1.

Time-Varying Hazards and Piecewise Calculations

With time-varying hazards, the risk of cause \(k\) still follows \(F_k(t)=\int_0^t S(u)\lambda_k(u)\,du\), but the constant-hazard shortcut generally does not apply. If hazards are constant within time intervals, calculate the contribution to the cumulative incidence separately in each interval. The contribution from a later interval must be multiplied by the probability of still being event-free at its start.

Worked Example: Piecewise-Constant Hazards

Suppose, hypothetically, that the hazards are constant in each of two periods. From year 0 to year 2, let \(\lambda_1=0.04\) and \(\lambda_2=0.06\) per year. From year 2 to year 5, let \(\lambda_1=0.01\) and \(\lambda_2=0.04\) per year. Assume these are the only first-event causes and the population is event-free at time zero. Calculate the cause-1 cumulative incidence by year 5.

During the first two years, the total hazard is \(0.10\) per year. The cause-1 contribution is:

F_1(2)=\frac{0.04}{0.10}\left(1-e^{-(0.10)(2)}\right) =0.4(1-e^{-0.2}) \approx 0.07251.

The probability of remaining event-free at year 2 is \(S(2)=e^{-0.2}\approx 0.8187\). During years 2 to 5, the total hazard is \(0.05\) per year. The cause-1 contribution in that interval, first calculated among those event-free at year 2 and then weighted by their share of the original population, is:

\begin{aligned} \text{Contribution from years 2 to 5} &=S(2)\frac{0.01}{0.05}\left(1-e^{-(0.05)(3)}\right)\\ &=e^{-0.2}(0.2)(1-e^{-0.15})\\ &\approx 0.02281. \end{aligned}

Adding the contributions gives \(F_1(5)\approx 0.07251+0.02281=0.09532\). The hypothetical probability of a cause-1 first event by year 5 is therefore about 9.532%. Omitting the factor \(S(2)\) would mistakenly treat the people still event-free at year 2 as if they were the entire original population.

What an Absolute Risk Calculation Requires

A cause-specific hazard model and an absolute-risk calculation answer related but distinct questions. The model describes cause-specific rates among those currently event-free. To obtain an absolute risk at a horizon, the hazards for all first-event causes must be combined to determine how the event-free population changes over time, and then the cause of interest must be accumulated over that changing population.

A standard alternative is to use one minus a Kaplan–Meier estimate after treating competing events as censored. That calculation does not generally estimate the observed probability of the event of interest by a time horizon: it removes competing events from the calculation as though those individuals could remain eligible for the event of interest. It may be useful for a different, hypothetical quantity, but it should not be reported as the real-world cumulative incidence when competing events occur.

The formula also relies on adequate information about the relevant hazards. In observed data, censoring is not itself a first-event cause, but it can prevent event times from being observed. Estimation then requires methods and assumptions that account for censoring, such as appropriate conditional independent-censoring assumptions. Sparse follow-up near the chosen horizon, changing event definitions, or hazards that are poorly represented by an assumed constant pattern can make calculated risks unstable or misleading.

Common Mistakes and Diagnostics

Calling a hazard a probability. A hazard has units of inverse time and is conditional on remaining event-free. A cumulative incidence is a probability by a stated time and has no rate units.
Using only the hazard of interest. This misses the effect of competing causes on \(S(t)\). Check that the event-free probability includes all causes capable of being the first event.
Multiplying a rate by the time horizon. The approximation \(\lambda_k t\) can be reasonable only in limited short-time, low-event settings. In general, it ignores both survival depletion and competing events; use the cumulative incidence integral or an appropriate special case.
Using the constant-hazard allocation for changing rates. The ratio \(\lambda_k/\lambda_{\text{tot}}\) is constant only when the hazards are constant. For time-varying hazards, integrate \(S(u)\lambda_k(u)\) over time.
Ignoring the population represented by the hazard. A conditional hazard model and the resulting risk apply to the population and covariate pattern for which the hazards are specified. Check that the modeled hazards, time origin, causes, and horizon match the risk being reported.

Key takeaway: The absolute probability of a cause-\(k\) event by time \(t\) is \(F_k(t)=\int_0^t S(u)\lambda_k(u)\,du\). The cause-specific hazard supplies the rate among those still event-free; the hazards of all competing causes help determine how many people remain able to experience that event.

Check Your Understanding

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

In one sentence, explain why \(\lambda_1(t)\) alone does not generally determine \(F_1(t)\).
Two constant cause-specific hazards are \(0.02\) and \(0.03\) per year. Using the constant-hazard formula, calculate the probability of a cause-1 event by two years.
Why must a piecewise-constant calculation for an interval after year 2 include the probability of remaining event-free through year 2?
What does one minus a Kaplan–Meier estimate generally fail to represent when competing events are treated as censored?

Solutions. (1) \(F_1(t)\) integrates \(\lambda_1(u)\) weighted by \(S(u)\), and \(S(u)\) depends on all cause-specific hazards. (2) The total hazard is \(0.05\) per year, so \(F_1(2)=(0.02/0.05)(1-e^{-0.05(2)})=0.4(1-e^{-0.1})\approx 0.03807\), or about 3.807%. (3) Only the fraction still event-free at year 2 can experience a first event later; the interval-specific risk must be weighted by that fraction to express it in the original population. (4) It generally does not represent the observed cumulative incidence of the event of interest; treating competing events as censored instead removes them as though they could remain eligible for that event.