Predictions With More Than Two Causes

Why More Causes Require a Joint Calculation

When more than two types of event can occur first, the calculation follows the same principle as for two: the chance of a particular event depends both on its own cause-specific hazard and on remaining event-free up to that time. The calculation must include every cause when updating event-free probability. Otherwise, people who experience another cause may incorrectly remain at risk of the event being predicted.

This tutorial extends the model-based absolute-risk calculations from the earlier material on cumulative incidence, cause-specific hazards and time-varying covariate paths. The focus is a specified covariate profile or path, not an average over a population. All numerical data below are hypothetical.

Definition: Suppose there are $K$ mutually exclusive causes of a first event, indexed by $k=1,\ldots,K$. Let $\lambda_k(t\mid x(t))$ be the cause-specific hazard for cause $k$ at time $t$, conditional on being event-free just before $t$, for covariate path $x(t)$. Let $S(t\mid x)$ be the probability of remaining free of all causes through time $t$, and let $F_k(t\mid x)$ be the probability that cause $k$ has occurred first by time $t$. With an event-free starting state at time zero, the model-based quantities satisfy $$ S(t\mid x)=\exp\left\{-\int_0^t\sum_{j=1}^K\lambda_j(u\mid x(u))\,du\right\}, \qquad F_k(t\mid x)=\int_0^t S(u-\mid x)\lambda_k(u\mid x(u))\,du. $$

These expressions assume that the causes form a complete set of mutually exclusive first events for the prediction question, that the cause-specific hazards describe the event-free state, and that the specified covariate path and fitted hazards are appropriate over the prediction horizon. In continuous time, tied first events have probability zero under the usual hazard formulation. The cumulative incidence $F_k$ is an absolute probability of cause $k$ occurring first; it is not the “net risk” obtained by imagining other causes could not occur.

Exact Calculation Within a Constant-Hazard Interval

Suppose the cause-specific hazards are constant during an interval of length $\Delta$, with values $\lambda_1,\ldots,\lambda_K$. Write $\Lambda=\sum_{j=1}^K\lambda_j$ for the total hazard. If $S_{\mathrm{start}}$ is the event-free probability at the interval’s start, the conditional probability of any first event during the interval is $1-\exp(-\Lambda\Delta)$. When $\Lambda>0$, the share of that event probability assigned to cause $k$ is $\lambda_k/\Lambda$. Thus:

\Delta F_k = S_{\mathrm{start}}\frac{\lambda_k}{\Lambda} \left(1-\exp(-\Lambda\Delta)\right), \qquad S_{\mathrm{end}}=S_{\mathrm{start}}\exp(-\Lambda\Delta).

There is a necessary special case: if $\Lambda=0$, every cause-specific hazard is zero. No event occurs in that interval, so set every $\Delta F_k=0$ and $S_{\mathrm{end}}=S_{\mathrm{start}}$. The share formula is not used because division by zero is undefined.

Why the interval updates add up: When $\Lambda>0$, summing the cause-specific increments gives $$ \sum_{k=1}^K\Delta F_k = S_{\mathrm{start}} \frac{\sum_{k=1}^K\lambda_k}{\Lambda} \left(1-\exp(-\Lambda\Delta)\right) = S_{\mathrm{start}}\left(1-\exp(-\Lambda\Delta)\right). $$ Adding the event-free probability at the interval’s end gives $S_{\mathrm{start}}$, since $S_{\mathrm{start}}\exp(-\Lambda\Delta)+S_{\mathrm{start}}(1-\exp(-\Lambda\Delta))=S_{\mathrm{start}}$. If $\Lambda=0$, the same accounting holds because no cause increment occurs and event-free probability is unchanged.

This identity justifies a useful computation: carry forward the event-free probability and the running cumulative incidence for each cause, adding that cause’s interval increment at each step. After any number of intervals, the sum of all cause-specific cumulative incidences plus the event-free probability remains one, subject to rounding.

Worked Example: Three Causes With Constant Hazards

For a hypothetical covariate profile, suppose the cause-specific hazards are constant for one year: $\lambda_1=0.12$, $\lambda_2=0.08$, and $\lambda_3=0.05$ per year. The individual is event-free at time zero. Calculate the three one-year cumulative incidences and the event-free probability.

The total hazard is $\Lambda=0.12+0.08+0.05=0.25$ per year. The probability of any first event by one year is $1-\exp(-0.25)=0.2211992$. The cause shares are $0.12/0.25=0.48$, $0.08/0.25=0.32$, and $0.05/0.25=0.20$. Therefore:

\begin{aligned} F_1(1)&=1(0.48)(0.2211992)=0.1061756,\\ F_2(1)&=1(0.32)(0.2211992)=0.07078375,\\ F_3(1)&=1(0.20)(0.2211992)=0.04423984,\\ S(1)&=\exp(-0.25)=0.7788008. \end{aligned}

The three risks sum to $0.2211992$, the probability of any first event. Including the event-free probability gives $0.1061756+0.07078375+0.04423984+0.7788008=1.000000$, up to rounding. In this hypothetical model, the one-year probability of cause 2 occurring first is about 0.07078, or 7.078%.

Combining Intervals and Changes in the Cause Mix

The hazards need not have the same values across the prediction horizon. Divide time into intervals on which the hazards are constant, or are treated as constant for a numerical approximation. For each interval, use the hazards for that interval to calculate its event probability and cause shares. Multiply each cause’s share by the event-free probability at the interval’s start, then update that cause’s cumulative incidence and the event-free probability. This is the same logic when covariates follow a specified path: use the hazard values corresponding to the path in each interval.

Set initial values.
For an event-free start, set $S=1$ and $F_k=0$ for every cause.

Find the total interval hazard.
For each interval, add the $K$ cause-specific hazards. If the total is positive, calculate the conditional event probability and each cause’s share.

Update each cause and event-free probability.
Add the interval contribution to each running $F_k$, and multiply the starting $S$ by the interval’s conditional no-event probability. If all hazards are zero, leave all values unchanged.

Check the accounting.
At the horizon, compare $\sum_k F_k+S$ with one, allowing only for displayed rounding.

Worked Example: Four Causes Across Two Intervals

Consider a hypothetical two-year prediction with four causes and an event-free start. In year 1, the hazards per year are $0.10, 0.05, 0.03, 0.02$. In year 2, they are $0.04, 0.08, 0.04, 0.04$. Calculate the two-year risks and event-free probability.

In year 1, the total hazard is $0.20$, so the conditional event probability is $1-\exp(-0.20)=0.1812692$. The cause shares are $0.50,0.25,0.15,0.10$. Since $S_{\mathrm{start}}=1$, the year-1 increments are $0.09063462,0.04531731,0.02719039,0.01812692$. The event-free probability after year 1 is $\exp(-0.20)=0.8187308$.

In year 2, the total hazard is also $0.20$, and the cause shares are $0.20,0.40,0.20,0.20$. The conditional event probability is again $0.1812692$, but the starting event-free probability is now $0.8187308$. The year-2 increments are:

\begin{aligned} \Delta F_1&=0.8187308(0.20)(0.1812692)=0.02968214,\\ \Delta F_2&=0.8187308(0.40)(0.1812692)=0.05936428,\\ \Delta F_3&=0.8187308(0.20)(0.1812692)=0.02968214,\\ \Delta F_4&=0.8187308(0.20)(0.1812692)=0.02968214. \end{aligned}

Adding each interval’s increments gives $F_1(2)=0.1203168$, $F_2(2)=0.1046816$, $F_3(2)=0.05687253$, and $F_4(2)=0.04780907$. Event-free probability is $S(2)=\exp(-0.40)=0.6703200$. The risks sum to $0.3296800$, and adding $S(2)$ gives one to the shown precision. Although causes 1 and 3 have the same second-year hazard, their two-year cumulative incidences need not match: their first-year hazards differed.

Derivation and Assumptions Behind the Interval Formula

Within a constant-hazard interval, the total hazard $\Lambda$ is the hazard of any first event while the individual remains event-free. The event-free probability over an interval of length $v$, conditional on being event-free at its start, is therefore $\exp(-\Lambda v)$. The cause-$k$ contribution over an interval of length $\Delta$ is the integral of the event-free probability at elapsed time $v$ times the cause-$k$ hazard:

S_{\mathrm{start}}\int_0^\Delta \exp(-\Lambda v)\lambda_k\,dv = S_{\mathrm{start}}\frac{\lambda_k}{\Lambda} \left(1-\exp(-\Lambda\Delta)\right),\qquad \Lambda>0.

This establishes the interval formula rather than assuming that the cause share applies only at the interval’s end. It applies exactly when hazards are constant within the interval. With hazards that vary continuously, use the cumulative-hazard and cumulative-incidence integrals in the definition, evaluated analytically when possible or numerically when needed. Piecewise-constant calculations are exact for a piecewise-constant model; they are an approximation if used to represent smoothly varying hazards.

The calculation also relies on using one coherent set of hazards for the full set of possible first-event causes. Hazards must be nonnegative, expressed on the same time scale, and conditional on the same event-free state. If the causes omit a relevant first event, or if incompatible time origins or covariate histories are mixed, the predicted quantities may not describe the stated event process.

Common Pitfall, Diagnostics, and Alternatives

A common mistake is to calculate a separate “net risk” for each cause as though all competing events had been removed, then treat those values as the cause-specific probabilities in the observed competing-risk setting. For example, with constant hazards $0.20,0.10,0.10$ per year over one year, the separate net risks sum to

(1-\exp(-0.20))+2(1-\exp(-0.10)) =0.3715944.

But the actual probability of any first event, allowing all three causes to compete, is $1-\exp(-0.40)=0.3296800$, and event-free probability is $\exp(-0.40)=0.6703200$. The net-risk sum plus event-free probability is approximately $1.0419145$, not one. The correction is to calculate each cause’s contribution using event-free probability that reflects all cause-specific hazards, not just the hazard of that cause.

Useful checks are to confirm that all hazards are nonnegative and in the intended time units; that the cause list is complete for the question; that each interval uses the appropriate hazard values; and that all predicted $F_k$ values are nonnegative and nondecreasing over time. The sum $\sum_k F_k(t)+S(t)$ should equal one, apart from rounding or numerical integration error. A discrepancy larger than numerical tolerance suggests an implementation or input problem, such as omitting a cause from the total hazard, using cause shares that do not sum to one, or failing to carry forward the event-free probability.

Cause-specific hazard models provide a direct route to joint risk predictions because their hazards can be combined in the total-hazard calculation. Aalen–Johansen prediction uses the corresponding product-integral updates in a multi-state setting and is a nonparametric alternative under its conditions. Fine–Gray models directly model a cause-specific cumulative incidence through its subdistribution hazard; separate fits for several causes do not automatically guarantee a jointly coherent set of risks unless the modeling approach imposes that constraint. These approaches answer related prediction questions but are not interchangeable without attention to their estimands and assumptions.

Model-based results remain conditional on the chosen covariate profile or path, the fitted hazard models, the prediction horizon, and the assumed cause definitions. A probability sum check is necessary but cannot establish that the hazards are well estimated or the model fits the data. Uncertainty in estimated risks also remains even when the arithmetic is exactly coherent.

Worked Example: A Zero-Hazard Interval

For a hypothetical prediction with three causes, suppose all three cause-specific hazards are zero during the first half-year. During the next half-year they are constant at $0.06,0.03,0.01$ per year. Starting event-free, calculate the probability of each first event by one year.

For the first interval, $\Lambda=0$. Therefore every cause increment is zero and event-free probability stays at one; no cause-share division is attempted. In the second interval, $\Lambda=0.10$ per year and $\Delta=0.5$ years, so the conditional event probability is $1-\exp(-0.10\times0.5)=0.04877058$. The cause shares are $0.60,0.30,0.10$. Hence:

\begin{aligned} F_1(1)&=1(0.60)(0.04877058)=0.02926235,\\ F_2(1)&=1(0.30)(0.04877058)=0.01463117,\\ F_3(1)&=1(0.10)(0.04877058)=0.004877058,\\ S(1)&=\exp(-0.05)=0.9512294. \end{aligned}

The three risks sum to $0.04877058$; adding the event-free probability gives one to rounding. The zero-hazard interval contributes no risk, but it does not require a different prediction method.

Check Your Understanding

Work through each question, then compare your result with the solutions below.

Three causes have constant hazards $0.03,0.02,0.05$ per year over one year. Starting event-free, what is the probability of any first event?
For that same example, calculate the cause-2 cumulative incidence and event-free probability.
In an interval, all $K$ cause-specific hazards equal zero. What should happen to each cause’s cumulative incidence and to event-free probability?
Why is the conditional no-event probability in an interval determined by the sum of all cause-specific hazards?

Solutions. (1) The total hazard is $0.10$ per year, so the probability of any first event is $1-\exp(-0.10)=0.09516258$. (2) Cause 2’s share is $0.02/0.10=0.20$, so $F_2(1)=0.20(0.09516258)=0.01903252$. Event-free probability is $\exp(-0.10)=0.9048374$. (3) Each cause increment is zero, and event-free probability stays at its interval-start value; the undefined share formula is not used. (4) While event-free, any one of the causes can end the event-free state, so the total hazard is $\sum_{j=1}^K\lambda_j$. Using only one cause’s hazard would incorrectly leave other-cause events in the event-free group.