From Cause-Specific Hazards to Absolute Risks
A cause-specific hazard describes the instantaneous rate of a particular event among people who are still event-free. Absolute risk over a time horizon is different: it is the probability that the event has occurred by that horizon, accounting for the possibility that another event happens first. This tutorial works through that conversion for two competing causes, starting with hazard increments over short time intervals.
The calculation requires the hazards for both causes. A cause 1 hazard by itself is not enough to determine cause 1 risk: cause 2 can remove people from the event-free state before they experience cause 1. The resulting risks are cumulative incidences, not probabilities calculated as if each cause could occur independently of the other.
For a prediction horizon divided into intervals, write \(\Delta\Lambda_{kj}\) for the increment in cause \(k\)'s cumulative hazard during interval \(j\). In this tutorial, each interval uses a piecewise-constant cause-specific hazard: within that interval, each cause-specific rate is constant. Thus \(\Delta\Lambda_{kj}=\lambda_{kj}\Delta t_j\), where \(\lambda_{kj}\) is the rate and \(\Delta t_j\) is the interval length. The total hazard increment is \(\Delta\Lambda_{\cdot j}=\Delta\Lambda_{1j}+\Delta\Lambda_{2j}\).
The piecewise-constant assumption is what lets us calculate the exact event probabilities within an interval from the hazard increments alone. If increments are supplied for wide intervals over which the rates change substantially, this calculation represents a piecewise-constant approximation rather than a recovery of the unknown within-interval hazard pattern.
The Interval Update
Suppose the probability of remaining event-free at the beginning of interval \(j\) is \(S_{j-1}\). Under constant cause-specific rates, the event-free probability at the end of the interval is \(S_{j-1}\exp(-\Delta\Lambda_{\cdot j})\). The probability of any event in that interval, conditional on being event-free at its start, is therefore \(1-\exp(-\Delta\Lambda_{\cdot j})\). The cause-specific hazards divide this total event probability in proportion to their rates.
Here \(\Delta F_{kj}\) is the interval's contribution to cause \(k\)'s cumulative incidence. If the total increment is zero, both cause-specific increments are zero; use \(S_j=S_{j-1}\) rather than evaluating a zero-over-zero ratio. With more than one interval, carry the end-of-interval event-free probability forward as the next interval's starting probability, and add each interval's cause-specific contribution to its cumulative incidence.
Why the Allocation Formula Works
Within one interval, let \(u\) denote elapsed time from its start, and let \(\lambda_{\cdot}=\lambda_1+\lambda_2\). For constant hazards, the event-free probability at elapsed time \(u\) is \(S_{j-1}\exp(-\lambda_{\cdot}u)\). The probability contribution from cause \(k\) over the interval is the integral of this event-free probability times the cause \(k\) hazard:
The last equality follows because each increment is its constant rate multiplied by the same interval length. This derivation also shows why the cause-specific increments must be used together: the survival term in the integral depends on the total hazard, while the cause-specific rate determines the share of events assigned to each cause.
Worked Calculations Across Intervals
Worked Example: Updating Both Risks Over Two Intervals
Suppose a hypothetical prediction begins with \(S_0=1\) and \(F_1(0)=F_2(0)=0\). Over interval 1, the cause-specific cumulative-hazard increments are \(\Delta\Lambda_{11}=0.05\) and \(\Delta\Lambda_{21}=0.10\). Over interval 2, they are \(\Delta\Lambda_{12}=0.08\) and \(\Delta\Lambda_{22}=0.02\). Assume the rates are constant within each interval and these are the only two possible first-event causes. Calculate both cumulative incidences and event-free probability at the end of interval 2.
Interval 1. The total increment is \(0.05+0.10=0.15\). The event-free probability after this interval is \(S_1=1\exp(-0.15)=0.860708\). The total event probability during the interval, given that everyone starts event-free, is \(1-\exp(-0.15)=0.139292\). The cause 1 share is \(0.05/0.15=1/3\), and the cause 2 share is \(0.10/0.15=2/3\). Therefore:
Interval 2. The total increment is \(0.08+0.02=0.10\), and the interval starts with the unrounded event-free probability \(S_1=\exp(-0.15)\). Its event-free probability at the end is \(S_2=\exp(-0.15)\exp(-0.10)=\exp(-0.25)=0.778801\). The total event probability conditional on being event-free at the start of interval 2 is \(1-\exp(-0.10)=0.095163\); cause 1 has 0.8 of the total hazard and cause 2 has 0.2. Thus:
Adding interval contributions gives \(F_1=0.0464307+0.0655258=0.11196\) and \(F_2=0.0928613+0.0163814=0.10924\). The final accounting check is \(0.11196+0.10924+0.778801\approx1.00000\). These are the predicted probabilities of experiencing each first-event cause by the end of interval 2 and of still being event-free at that time.
Worked Example: Converting Constant Rates Over a Fixed Horizon
For a hypothetical individual, suppose the cause-specific rates are constant for two years: \(\lambda_1=0.05\) per year and \(\lambda_2=0.10\) per year. There are no other first-event causes, and the individual is event-free at time zero. Find the two risks and event-free probability at two years.
The two-year cumulative-hazard increments are \(\Delta\Lambda_1=(0.05)(2)=0.10\) and \(\Delta\Lambda_2=(0.10)(2)=0.20\). Their total is 0.30. Therefore, the event-free probability is \(S(2)=e^{-0.30}=0.740818\). The total event probability is \(1-e^{-0.30}=0.259182\), divided between the causes in the ratio \(0.10:0.20\):
The probabilities sum to \(0.08639+0.17279+0.740818\approx1\). The rates have units of per year; the risks and event-free probability are dimensionless probabilities. This calculation does not assert that the two latent event times are independent. It uses the specified cause-specific hazards among people still event-free.
Worked Example: Continuing From a Partially Event-Free Cohort
At the start of a hypothetical interval, suppose \(S=0.92\), \(F_1=0.05\), and \(F_2=0.03\). The accounting is coherent because \(0.92+0.05+0.03=1\). During the next interval, the hazard increments are 0.04 for cause 1 and 0.06 for cause 2. Under the piecewise-constant assumption, calculate the updated probabilities.
The total hazard increment is 0.10. The probability of an event during this interval, conditional on being event-free at its start, is \(1-e^{-0.10}=0.0951626\). Only the starting event-free fraction is exposed to these first-event hazards, so the unconditional interval contributions are:
The updated cumulative incidences are \(F_1=0.05+0.03502=0.08502\) and \(F_2=0.03+0.05253=0.08253\). Their sum with the updated event-free probability is approximately \(0.08502+0.08253+0.83245=1.00000\). Multiplying by 0.92 is essential: the interval hazards apply only to people who have not already experienced either cause.
Checks, Alternatives, and Limitations
A reliable calculation should satisfy several checks before its risks are interpreted. First, each hazard increment must be nonnegative, and the two cause-specific increments must add to the total increment. Second, event-free probability must not increase, and neither cumulative incidence may decrease. Third, at each horizon, the two cumulative incidences plus event-free probability must equal one, allowing for rounding. A materially negative probability, a cumulative incidence that decreases, or a sum far from one signals an implementation or input problem.
The formula above is for cumulative-hazard increments under a continuous-time, piecewise-constant hazard model. It is not the same as using conditional event probabilities in a discrete-time model. If, instead, an interval supplies mutually exclusive conditional probabilities \(p_1\) and \(p_2\) of the two events among those event-free at the interval's start, the updates are \(\Delta F_k=S_{\mathrm{start}}p_k\) and \(S_{\mathrm{end}}=S_{\mathrm{start}}(1-p_1-p_2)\), provided \(p_1+p_2\leq1\). Such probabilities should not be treated as cumulative-hazard increments.
At observed event times, the Aalen–Johansen approach updates event-free survival and cause-specific cumulative incidence using observed hazard jumps. That is a standard alternative for estimating risks from event-time data. It is not interchangeable with the exact continuous-time interval integration here unless the input representation and assumptions match. In particular, do not substitute broad-interval hazard increments into a jump-update formula without checking what those increments represent.
The main limitations are substantive as well as computational. The calculation includes only causes represented in the hazards; omitting a possible first-event cause can make the remaining probabilities incoherent or the predicted risks misleading. The piecewise-constant assumption may be a poor approximation when hazards vary sharply within an interval. Finally, correct arithmetic does not establish that the hazards themselves are well estimated or transportable to the prediction population. Those are modeling and data-support questions, not resolved by the interval update.
Check Your Understanding
Try each calculation before reading the solutions. Assume the stated hazards are piecewise constant where hazard increments are given.
- Starting event-free, a single interval has cause-specific hazard increments 0.02 and 0.03. Find both interval contributions and the final event-free probability.
- A discrete-time interval starts with event-free probability 0.90. Its mutually exclusive conditional event probabilities are 0.04 for cause 1 and 0.06 for cause 2. Find the unconditional contributions and final event-free probability, and state why the continuous-time formula is not used.
- At a horizon, computed probabilities are \(F_1=0.12\), \(F_2=0.09\), and \(S=0.79\). Is the probability accounting coherent? What additional checks should be made across earlier horizons?
- Explain why the cause 1 risk cannot generally be calculated from cause 1's hazard increment alone when cause 2 is also possible.
Solutions. (1) The total increment is 0.05, so \(1-e^{-0.05}=0.0487706\). The cause shares are 0.4 and 0.6, giving \(\Delta F_1=0.4(0.0487706)=0.01951\), \(\Delta F_2=0.6(0.0487706)=0.02926\), and \(S=e^{-0.05}=0.95123\). They sum to one. (2) These are discrete conditional probabilities, so \(\Delta F_1=0.90(0.04)=0.036\), \(\Delta F_2=0.90(0.06)=0.054\), and \(S_{\mathrm{end}}=0.90(1-0.04-0.06)=0.81\). Their sum is 0.90, the starting event-free probability; the remaining 0.10 had already experienced an event. (3) Yes: \(0.12+0.09+0.79=1\). Also check that both cumulative incidences are nondecreasing, event-free probability is nonincreasing, and all inputs and calculations correspond to the intended causes and horizon. (4) Cause 2 can happen first and remove a person from the event-free group, changing the probability available to experience cause 1. Cause 1's interval contribution therefore depends on the total hazard and the starting event-free probability, not just its own increment.