Why Separate Hazard Predictions Must Be Combined
A prediction model for one cause describes that cause’s instantaneous event rate among people who are still event-free. It does not, by itself, give the probability of that cause occurring by a future horizon. To calculate that probability, predictions for all competing causes must be combined: an event from any cause removes a person from the event-free group and therefore changes the opportunity for every other cause to occur.
The target is the cumulative incidence for cause \(k\) by time \(t\), conditional on a specified prediction profile. It is the probability that cause \(k\) is the first event by time \(t\). The event-free probability is the probability that no event of any cause has occurred by that time. These quantities must fit together: the cause-specific cumulative incidences sum to the probability of any event, and the remaining probability is event-free.
The survival weighting is the essential combination step. A prediction for cause \(k\) cannot be converted to an absolute risk by integrating its hazard alone: that would act as if competing events did not occur. Nor can separate cause-specific risks be calculated as though each cause had its own event-free population. All causes share the same event-free probability.
Conditions for Combining Predictions
The formulas above assume that the cause-specific hazards describe the same target population or specified profile, use the same time origin and time scale, and cover every event type that removes a person from the event-free state. The cause labels must be mutually exclusive for a first event. Each predicted hazard must be nonnegative and defined over the prediction horizon. For the displayed integral formula, hazards are rates over continuous time; predictions with discrete event-time jumps require a jump or interval-based calculation instead.
These are conditions on the predictions and their interpretation, not an assumption that latent times to different causes are independent. Cause-specific hazards are defined among those who remain event-free. The combination uses those hazards and the shared event-free probability; it does not require specifying independent latent event times.
In practice, hazards may be available as functions of time or as piecewise-constant predictions on a time grid. The same logic applies in either case: first calculate the event-free probability using the hazards for all causes, then allocate the probability lost from the event-free group to each cause according to its hazard. The introductory material on Aalen-Johansen absolute risk predictions gives the corresponding discrete event-time recursion; the examples below focus on combining separately supplied cause-specific predictions.
Worked Example: Constant Cause-Specific Hazard Predictions
Suppose a hypothetical prediction profile has two constant cause-specific hazards over three years: \(\lambda_1=0.12\) per year and \(\lambda_2=0.08\) per year. Assume the hazards are valid and constant over this horizon, both causes are included, and there are no other event types that remove people from the event-free state. Find the two cumulative incidences and event-free probability at three years.
The combined hazard is \(0.12+0.08=0.20\) per year. Thus the probability of remaining event-free is
For constant hazards, integrating the cause-specific hazard against survival gives
For cause 1,
For cause 2,
The predictions are coherent: \(F_1(3)+F_2(3)=0.4512\), which agrees, up to rounding, with \(1-S(3)=1-0.5488=0.4512\). In context, the model predicts a 27.07% probability of cause 1 as the first event within three years for this profile, and an 18.05% probability for cause 2.
Derivation and a Piecewise Calculation
The combination follows from the change in event-free probability over a short time interval. Given survival to time \(t\), the probability of a cause-\(k\) event in a short interval of length \(dt\) is approximately \(\lambda_k(t\mid x)dt\). The total instantaneous probability of leaving the event-free state is therefore approximately \(\sum_j\lambda_j(t\mid x)dt\). In differential form,
Dividing by \(S(t\mid x)\) and integrating from 0 to \(t\), using \(S(0\mid x)=1\), gives the exponential expression for \(S(t\mid x)\). The probability lost from the event-free group during \(dt\) that is assigned to cause \(k\) is \(S(t\mid x)\lambda_k(t\mid x)dt\). Integrating that contribution over time gives \(F_k(t\mid x)\). This also shows why survival in the incidence integral must include every cause.
The formulas imply an important identity. Summing the cause-specific incidence integrals gives
The last equality follows because the integrand is \(-dS(u\mid x)/du\). Thus a combined prediction should satisfy \(S(t\mid x)+\sum_k F_k(t\mid x)=1\), apart from numerical rounding.
Worked Example: Piecewise-Constant Hazards
Consider hypothetical predictions for two causes over two years. During year 1, the hazards are \(\lambda_1=0.20\) and \(\lambda_2=0.10\) per year. During year 2, they are \(\lambda_1=0.10\) and \(\lambda_2=0.30\) per year. Assume these piecewise-constant hazards apply to the same profile, cover both causes, and use continuous time. Calculate both cumulative incidences and event-free probability at year 2.
During the first year, the total hazard is \(0.30\). Therefore \(S(1)=e^{-0.30}=0.740818\). The first-year incidences are
During year 2, the total hazard is \(0.40\). The probability of still being event-free at year 2 is
The probability remaining event-free at the start of year 2, \(0.740818\), must weight the incidences accrued during that year. The additional incidences are
Adding the first- and second-year contributions gives
Their sum is \(0.503415\), matching \(1-S(2)=1-0.496585=0.503415\) to rounding. The calculation also illustrates why applying the second-year hazards to the original population would overstate the second-year contribution: only the fraction event-free at the start of that year remains at risk.
Discrete-Time Predictions and Coherence Checks
Some models provide event probabilities over intervals rather than continuous-time hazard rates. Let \(q_{kj}\) be the probability of cause \(k\) during interval \(j\), conditional on being event-free at that interval’s start. For causes that are mutually exclusive within each interval, \(\sum_k q_{kj}\leq 1\). If \(S_{j-1}\) is the probability event-free at the start of interval \(j\), then the interval updates are
These are interval event probabilities, not continuous-time rates. Substituting them into a continuous-time exponential formula would mix incompatible quantities.
Worked Example: Combining Discrete Interval Probabilities
Suppose hypothetical predictions give two mutually exclusive cause probabilities in each of two intervals, conditional on being event-free at that interval’s start. In interval 1, \(q_{11}=0.10\) and \(q_{21}=0.20\). In interval 2, \(q_{12}=0.05\) and \(q_{22}=0.15\). Starting with \(S_0=1\) and zero cumulative incidence, calculate the risks after interval 2.
After interval 1, the event-free probability is \(S_1=1(1-0.10-0.20)=0.70\). The cumulative incidences are \(F_{11}=1(0.10)=0.10\) and \(F_{21}=1(0.20)=0.20\).
In interval 2, the event-free probability at its start weights both cause probabilities. Thus \(F_{12}=0.10+0.70(0.05)=0.135\) and \(F_{22}=0.20+0.70(0.15)=0.305\). The remaining event-free probability is \(S_2=0.70(1-0.05-0.15)=0.56\). The check is \(0.135+0.305+0.56=1.000\).
Useful checks for either representation include verifying that every predicted hazard or interval probability is nonnegative, that the event-free probability never increases, and that each cumulative incidence is nondecreasing. At each horizon, confirm that the event-free probability plus all cause-specific incidences equals one within numerical tolerance. Failure of this identity often indicates that a cause was omitted, a survival weight was not applied, or rates and probabilities were mixed.
The calculation does not establish that the input predictions are accurate. A model may produce internally coherent risks but still be poorly calibrated, rely on unsupported extrapolation, or omit an event type relevant to the target. Uncertainty in the separately fitted hazards also carries into every combined risk; treating the hazard predictions as known can understate uncertainty. If hazards are supplied on a grid, finer intervals may improve numerical integration when rates vary quickly, but cannot repair misspecified or incompatible predictions.
How This Combination Relates to Other Approaches
Aalen-Johansen absolute risk prediction combines cause-specific event contributions through the event-free probability, using estimated event-time increments. The formulas here express the same competing-event accounting for separately supplied hazard predictions, including smooth or piecewise hazard functions. By contrast, a direct cumulative-incidence model targets the cause-specific cumulative incidence itself rather than constructing it from all cause-specific hazards. Such approaches have different modeling assumptions; neither should be interpreted as making competing events disappear from the probability of the target event.
A common mistake is to calculate a “net risk” for each cause by integrating its own hazard as if it were the only possible event, then add those risks. For example, the cause 1-only risk in the constant-hazard example would be \(1-e^{-0.12(3)}=0.3023\), not the competing-risk incidence \(0.2707\). The former answers a different hypothetical question in which cause 2 does not remove anyone from risk. For absolute risk in the presence of both causes, the correct contribution is survival from all causes multiplied by the cause-specific hazard.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For constant hazards \(\lambda_1=0.06\) and \(\lambda_2=0.04\) per year, what is the event-free probability at two years?
- In the same setting, calculate the cumulative incidence for cause 1 at two years.
- Why is the cause 1 hazard alone insufficient to calculate its competing-risk cumulative incidence?
- In one discrete interval, the cause probabilities are \(q_1=0.12\) and \(q_2=0.18\). Starting event-free, what is the remaining event-free probability and the total event probability?
Solutions. (1) The total hazard is \(0.10\) per year, so \(S(2)=e^{-0.20}=0.8187\). (2) \(F_1(2)=\frac{0.06}{0.10}(1-e^{-0.20})=0.60(0.181269)=0.1088\). (3) Cause 2 can occur first and remove a person from the event-free group, so the cause 1 hazard must be weighted by survival from both causes. (4) The event-free probability is \(1-0.12-0.18=0.70\), and the total event probability is \(0.30\); these sum to one.