How Prediction Rankings Change Across Horizons

Why a Risk Ranking Depends on the Horizon

A predicted risk is incomplete without its time horizon. Two covariate profiles can have different predicted probabilities of experiencing a particular event by one year, and the ordering can change by five years. This is not a contradiction: cumulative incidence is built from both the target cause’s hazard and the chance of remaining free of every event that could occur first. A strong competing-event hazard can therefore reduce the target event’s later cumulative incidence, even when the target cause-specific hazard is high.

The question here is specifically about how the ordering of predicted risks changes with time. The profiles being compared must refer to the same target cause, prediction horizon, and estimand. When comparing subgroups, they should also be predictions for the same target population or clearly identified subgroup-specific predictions. Otherwise, a difference in ranking may reflect a change in the population being summarized as well as a change in time.

Definition: For cause \(k\), covariate profile \(x\), and horizon \(t\), the cause-\(k\) cumulative incidence \(F_k(t\mid x)\) is the probability, conditional on profile \(x\), that cause \(k\) occurs by time \(t\) and before any competing event. A horizon-specific ranking orders profiles by their \(F_k(t\mid x)\) values at that particular \(t\). A ranking reversal occurs when the ordering of two profiles differs at two horizons.

How Competing Events Shape the Ranking

Let \(\lambda_k(u\mid x)\) denote the cause-specific hazard for cause \(k\) at time \(u\), given profile \(x\). Let \(\Lambda_k(t\mid x)\) be its cumulative cause-specific hazard, and let \(S(t\mid x)\) be the probability of remaining free of every event through time \(t\). In a continuous-time competing-risks model with causes \(1,\ldots,K\) and absolutely continuous cause-specific cumulative hazards, the predicted cause-\(k\) cumulative incidence is

F_k(t\mid x)=\int_0^t S(u-\mid x)\,d\Lambda_k(u\mid x), \qquad S(t\mid x)=\exp\left\{-\sum_{j=1}^{K}\Lambda_j(t\mid x)\right\}.

The term \(S(u-\mid x)\) is the probability of being event-free just before time \(u\). Only people still event-free at that time can experience cause \(k\) then. Thus the cumulative incidence adds up cause-\(k\) event probability over time, weighted by the probability of still being at risk for any event. The cumulative hazards of competing causes affect \(S\), and hence affect \(F_k\), even though those causes are not the outcome being ranked.

For constant hazards of two causes, \(\lambda_1\) and \(\lambda_2\), the formula simplifies. Since \(S(u)=\exp\{-(\lambda_1+\lambda_2)u\}\), integration gives:

F_1(t)=\int_0^t e^{-(\lambda_1+\lambda_2)u}\lambda_1\,du =\frac{\lambda_1}{\lambda_1+\lambda_2} \left\{1-e^{-(\lambda_1+\lambda_2)t}\right\}.

The fraction \(\lambda_1/(\lambda_1+\lambda_2)\) is cause 1’s share of the total hazard under this constant-hazard model. The exponential term \(e^{-(\lambda_1+\lambda_2)t}\) is the probability of remaining event-free through time \(t\); one minus that term is the probability of experiencing any event by time \(t\). The formula makes clear why a larger cause-1 hazard alone does not guarantee a larger cause-1 cumulative incidence at every horizon.

Worked Example: A Ranking Reversal Under Constant Hazards

Suppose two hypothetical profiles, A and B, have constant hazards per year for cause 1 and a competing cause. Profile A has \(\lambda_1=0.12\) and \(\lambda_2=0.48\); profile B has \(\lambda_1=0.10\) and \(\lambda_2=0.05\). The model gives A the larger cause-1 hazard, but A also has a much larger competing hazard. Calculate the cause-1 risks at 0.5 and 5 years.

For A, the total hazard is \(0.12+0.48=0.60\) per year. At 0.5 years, \(F_1(0.5\mid A)=(0.12/0.60)\{1-e^{-0.60(0.5)}\}=0.2(1-e^{-0.30})\approx0.05184\).

For B, the total hazard is \(0.10+0.05=0.15\) per year. At the same horizon, \(F_1(0.5\mid B)=(0.10/0.15)\{1-e^{-0.15(0.5)}\}=(2/3)(1-e^{-0.075})\approx0.04817\). Thus A’s predicted cause-1 risk is slightly higher at 0.5 years.

At 5 years, A’s risk is \(F_1(5\mid A)=0.2(1-e^{-0.60(5)})=0.2(1-e^{-3})\approx0.1900\). B’s risk is \(F_1(5\mid B)=(2/3)(1-e^{-0.15(5)})=(2/3)(1-e^{-0.75})\approx0.3518\). The ranking has reversed: B has the higher cause-1 risk by 5 years. A’s cause-1 hazard is greater, but the large competing hazard leaves fewer people event-free long enough to experience cause 1. The constant-hazard formula has a smooth risk curve for each profile, so the ordering changes at a horizon between 0.5 and 5 years; numerical evaluation places the crossing between 0.85 and 0.86 years.

Changing Hazards Can Change the Ranking Later

With time-varying hazards, a profile’s risk accumulated before a given horizon remains part of its cumulative incidence. Later hazard patterns add risk only among those who have remained event-free up to that point. This means a late increase in a competing hazard can slow the subsequent accumulation of target-cause risk without undoing target events that have already occurred.

For piecewise-constant hazards on an interval from \(a\) to \(b\), with constant cause-specific hazards on that interval, the contribution to cause 1’s cumulative incidence is:

S(a\mid x)\frac{\lambda_1}{\sum_{j=1}^{K}\lambda_j} \left\{1-e^{-\left(\sum_{j=1}^{K}\lambda_j\right)(b-a)}\right\}.

This is the event-free probability at the start of the interval multiplied by the cause-1 share of the interval’s total hazard and the probability of any event during the interval.

Worked Example: A Competing Hazard Rises After Year One

Consider two hypothetical profiles. For A, during years 0 to 1, \(\lambda_1=0.12\) and \(\lambda_2=0.08\) per year; during years 1 to 3, \(\lambda_1=0.04\) and \(\lambda_2=0.36\). For B, the hazards remain \(\lambda_1=0.09\) and \(\lambda_2=0.01\) throughout. Calculate the cause-1 risks at years 1 and 3.

At year 1, A’s cause-1 risk is \(F_1(1\mid A)=(0.12/0.20)(1-e^{-0.20})=0.6(0.181269)\approx0.1088\). For B, it is \(F_1(1\mid B)=(0.09/0.10)(1-e^{-0.10})=0.9(0.095163)\approx0.08565\). A ranks higher at year 1.

From year 1 to year 3, A’s event-free probability at the start of the interval is \(e^{-0.20}\approx0.8187\). The additional cause-1 risk is \(e^{-0.20}(0.04/0.40)(1-e^{-0.40(2)})=0.818731(0.10)(1-e^{-0.80})\approx0.04509\). Therefore, \(F_1(3\mid A)\approx0.108761+0.045087=0.1538\).

For B, the year-3 risk is \(F_1(3\mid B)=(0.09/0.10)(1-e^{-0.10(3)})=0.9(1-e^{-0.30})\approx0.2333\). B ranks higher at year 3. A’s cause-1 hazard falls after year 1 while its competing hazard rises; both features limit its later cause-1 risk. B’s steady hazards lead to greater accumulated cause-1 incidence by year 3. All the calculations use the survival probability for each profile’s own hazards, rather than treating its cause-1 hazard in isolation.

Same Target Hazard, Different Predicted Risk

A further useful comparison holds the target cause-specific hazard fixed and varies only the competing hazard. The profiles then have the same instantaneous cause-1 rate among people still event-free, but their cause-1 cumulative incidences can still differ. This isolates the way competition affects the risk ranking.

Worked Example: Equal Cause-1 Hazards Do Not Imply Equal Risk

Suppose two hypothetical profiles have constant \(\lambda_1=0.08\) per year. Profile C has \(\lambda_2=0.02\), while profile D has \(\lambda_2=0.20\). At 4 years, C’s cause-1 risk is \(F_1(4\mid C)=(0.08/0.10)(1-e^{-0.10(4)})=0.8(1-e^{-0.40})\approx0.2637\).

For D, the total hazard is \(0.28\) per year, so \(F_1(4\mid D)=(0.08/0.28)(1-e^{-0.28(4)})=(2/7)(1-e^{-1.12})\approx0.1925\). Although the cause-1 hazards are identical, C has the higher 4-year cause-1 risk because D has the larger competing hazard. The calculation does not imply that changing the competing hazard would causally change anyone’s outcome; it compares predictions under the specified profiles and model.

How to Compare and Diagnose Rankings

A useful comparison begins by stating the target cause, profiles, prediction horizons, and risk estimand. Then calculate or read \(F_k(t\mid x)\) for each profile at each horizon. If the ranking changes, describe the horizons and the predicted probabilities rather than summarizing the profiles with one label such as “higher risk.” For example, “A is higher at 6 months, while B is higher at 3 years” is more informative than saying one profile is simply higher risk.

A concise display can make the horizon dependence explicit:

Profile	Cause-1 risk at 0.5 years	Cause-1 risk at 5 years	Ranking pattern
A	0.05184	0.1900	Higher early, lower later
B	0.04817	0.3518	Lower early, higher later

The table reports the same target cause for both profiles and makes clear that rankings are horizon-specific. When comparing subgroups, do not silently switch from subgroup-specific predictions to risks standardized to a common population: those answer different questions. Also distinguish a comparison of predicted absolute risks from a comparison of cause-specific hazard coefficients. The former incorporates the event-free probability under all causes; the latter describes the instantaneous target-cause rate among people who have not yet experienced any event.

Useful checks include confirming that all risks use the same time origin and units, that the cumulative incidence for each profile does not decrease as the horizon grows, and that the summed risks across causes do not exceed one. Inspect the underlying cause-specific hazard patterns and the event-free survival contribution when an ordering seems surprising. A large competing hazard, a hazard that changes over time, or a small difference between predicted risks can all produce or obscure a ranking reversal. If the reversal occurs where follow-up support is sparse, it may be sensitive to extrapolation; the computed ordering alone does not establish that it is stable.

Common mistake: “Profile A has the larger cause-1 hazard, so its cause-1 risk must be larger at every horizon.” This reasoning ignores that cumulative incidence is weighted by the probability of remaining free of every event. The correction is to compare \(F_k(t\mid x)\) at the stated horizons, using each profile’s hazards for all causes.

Rankings summarize predictions, not causal effects. They depend on the fitted models, the covariate profiles, the target cause, and the horizon. A profile can rank higher for one cause and lower for another, and two profiles can exchange rank as time passes. A single ordering is appropriate only when the horizon and target population are fixed and the ordering is supported across the horizons that matter for the decision.

Check Your Understanding

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

Under constant hazards \(\lambda_1=0.06\) and \(\lambda_2=0.04\) per year, calculate the cause-1 risk by 2 years.
Two profiles have the same cause-1 hazard, but one has a larger competing hazard. Which has the larger cause-1 risk at a positive horizon under constant hazards, and why?
Can a profile with a higher cause-1 hazard have a lower cause-1 cumulative incidence at a later horizon? State the mechanism.
Why is a cause-specific hazard ranking not automatically a cumulative-incidence ranking?

Solutions. (1) \(F_1(2)=(0.06/0.10)(1-e^{-0.10(2)})=0.6(1-e^{-0.20})\approx0.1088\), or about 10.88% by 2 years. (2) The profile with the smaller competing hazard has the larger cause-1 risk: with the target hazard held fixed, the smaller total hazard leaves more people event-free and able to experience cause 1. (3) Yes. A sufficiently larger competing hazard can reduce event-free survival enough that the profile’s later accumulated cause-1 incidence is lower, even if its cause-1 hazard is higher. (4) A cause-specific hazard concerns the instantaneous rate among those still event-free; cumulative incidence also depends on the event-free probability generated by all causes over time.