Why Compare Risks at More Than One Horizon?
An absolute-risk prediction is tied to a time horizon and to the covariate profile for which it is calculated. Two profiles can have different predicted risks at one horizon and a smaller, larger, or even oppositely directed difference later. This can happen because their target-cause hazards evolve differently, because their competing-event hazards differ, or both. A single number therefore cannot describe the whole pattern over follow-up.
It is also important to distinguish a risk for one covariate profile from a risk averaged over a subgroup. The subgroup-average risk depends on which profiles are represented and how they are weighted. A fair comparison between groups often uses the same target-population weights for both, rather than allowing different covariate compositions to determine the average.
Risk Across Horizons: Definitions and Assumptions
With two causes, let \(\lambda_1(u\mid z)\) and \(\lambda_2(u\mid z)\) be the cause-specific hazards at time \(u\) for profile \(z\). The event-free probability through time \(u\) is \(S(u\mid z)\). Under the usual continuous-time competing-risks formulation, the cause-1 cumulative incidence is the probability of remaining event-free up to each time \(u\), multiplied by the instantaneous cause-1 hazard and accumulated over follow-up:
This formula accounts for the fact that a person who has already experienced either cause is no longer at risk of experiencing cause 1 later. The corresponding cause-2 cumulative incidence is obtained by replacing \(\lambda_1\) with \(\lambda_2\). In a two-cause model, the two cumulative incidences and the event-free probability sum to one at any time, when all event types are represented and the predictions are coherent.
For a subgroup with profiles \(z_1,\ldots,z_m\), fixed weights \(w_1,\ldots,w_m\) define the subgroup-average cause-1 risk as \(F_1^{\mathrm{avg}}(t)=\sum_{j=1}^m w_jF_1(t\mid z_j)\), where \(\sum_j w_j=1\). The weights describe the population over which the prediction is averaged. This is generally not the risk calculated by inserting average covariate values into a prediction model: risks can be nonlinear functions of covariates.
These calculations assume that the risk predictions are valid for the profiles and horizons being compared. In particular, the model must represent the competing causes appropriately, the predicted probabilities must be coherent, and the relevant follow-up must be supported by the data. When predictions are estimated from censored data, estimation also relies on appropriate assumptions about censoring, commonly conditional independence given the modeled information. A contrast in predicted risks is descriptive of the specified model and population; by itself, it does not establish a causal effect.
Worked Example: Profile Differences Can Change Direction
Worked Example: Profile Differences Can Change Direction
Suppose two hypothetical profiles, P and Q, have predictions over three successive one-year intervals. The entries below are conditional probabilities for someone event-free at the beginning of an interval: the probability of cause 1, the probability of cause 2, and the probability of neither event during that interval. Each row sums to one, and all entries are nonnegative.
| Profile | Interval | Cause 1 | Cause 2 | Neither |
|---|---|---|---|---|
| P | 0–1 year | 0.20 | 0.05 | 0.75 |
| P | 1–2 years | 0.05 | 0.05 | 0.90 |
| P | 2–3 years | 0.02 | 0.05 | 0.93 |
| Q | 0–1 year | 0.08 | 0.02 | 0.90 |
| Q | 1–2 years | 0.18 | 0.02 | 0.80 |
| Q | 2–3 years | 0.15 | 0.02 | 0.83 |
For these interval probabilities, the cause-1 cumulative incidence is the sum of each interval’s cause-1 probability multiplied by the probability of having remained event-free up to that interval. This recurrence is valid because the interval probabilities are conditional on being event-free at the interval’s start.
At one year, the risks are simply the first-interval probabilities: \(F_1(1\mid P)=0.20\) and \(F_1(1\mid Q)=0.08\). The difference P minus Q is \(0.20-0.08=0.12\), or 12 percentage points.
At two years, P’s event-free probability after year 1 is \(0.75\). Thus \(F_1(2\mid P)=0.20+(0.75)(0.05)=0.2375\). For Q, the event-free probability after year 1 is \(0.90\), so \(F_1(2\mid Q)=0.08+(0.90)(0.18)=0.2420\). The difference is \(0.2375-0.2420=-0.0045\), or -0.45 percentage points. The ordering has just reversed.
At three years, P’s probability of remaining event-free through two years is \((0.75)(0.90)=0.675\). Therefore \(F_1(3\mid P)=0.2375+(0.675)(0.02)=0.2510\). For Q, that probability is \((0.90)(0.80)=0.72\), giving \(F_1(3\mid Q)=0.2420+(0.72)(0.15)=0.3500\). The three-year difference is \(0.2510-0.3500=-0.0990\), or -9.90 percentage points.
The predictions are internally coherent: each interval has valid event probabilities, and each cumulative incidence increases with time. The example shows why the horizon must accompany a risk difference: P has higher cause-1 risk at one year, while Q has higher risk at two and three years. These are hypothetical model predictions, not evidence that profile membership causes the differences.
Subgroup Averages and Standardization
A subgroup-average comparison can differ from comparisons made at individual profiles if the groups have different profile compositions. To focus on risk differences rather than composition, calculate each group’s risks at the same set of profiles and average them using a common target-population distribution. This operation is often called standardization. The resulting average answers a specific question: what would the model-predicted average risk be if each group had the chosen target population’s profile distribution?
Worked Example: Comparing Standardized and Crude Averages
Consider two hypothetical groups, E and C, and two profiles, U and V. The table gives predicted cause-1 risks at one and three years. Assume the predictions are coherent and supported at both horizons. For a common target population, use weights \(w_U=0.25\) and \(w_V=0.75\).
| Group | Profile U, 1 year | Profile V, 1 year | Profile U, 3 years | Profile V, 3 years |
|---|---|---|---|---|
| E | 0.12 | 0.04 | 0.25 | 0.16 |
| C | 0.08 | 0.03 | 0.22 | 0.15 |
At one year, E’s standardized risk is \((0.25)(0.12)+(0.75)(0.04)=0.030+0.030=0.060\). C’s standardized risk is \((0.25)(0.08)+(0.75)(0.03)=0.020+0.0225=0.0425\). The standardized difference E minus C is \(0.060-0.0425=0.0175\), or 1.75 percentage points.
At three years, E’s standardized risk is \((0.25)(0.25)+(0.75)(0.16)=0.0625+0.1200=0.1825\). C’s standardized risk is \((0.25)(0.22)+(0.75)(0.15)=0.0550+0.1125=0.1675\). The difference is \(0.1825-0.1675=0.0150\), or 1.50 percentage points. The standardized difference is smaller at three years, although it remains positive.
Now suppose group E has 20% profile U and 80% V, while group C has 80% U and 20% V. Their one-year averages under those different compositions are E: \((0.20)(0.12)+(0.80)(0.04)=0.056\), and C: \((0.80)(0.08)+(0.20)(0.03)=0.070\). The difference is -0.014, or -1.40 percentage points, opposite in direction to the standardized contrast. This is not an arithmetic contradiction: the two averages refer to different profile distributions. The common weights make the comparison answer a different, composition-standardized question.
Why Competing Events Affect Risk
A cause-specific hazard describes the instantaneous rate of a particular event among people who are still event-free. It is not itself the probability of that event by a chosen horizon. Even if the cause-1 hazard is held fixed, a higher cause-2 hazard can reduce the number of people who remain available to experience cause 1. The cumulative incidence therefore depends on both cause-specific hazards.
Worked Example: Same Target Hazard, Different Competing Hazard
Suppose two hypothetical profiles have constant cause-1 hazard \(\lambda_1=0.10\) per year. Profile A has cause-2 hazard \(\lambda_2=0.20\) per year; profile B has \(\lambda_2=0.05\) per year. Assume constant hazards throughout follow-up, no other event types, and predictions are required only at horizons within the model’s supported follow-up.
For constant hazards with total hazard \(\lambda=\lambda_1+\lambda_2>0\), integrating the cause-1 hazard times event-free survival gives \(F_1(t)=\{\lambda_1/\lambda\}\{1-\exp(-\lambda t)\}\). For A, \(\lambda=0.10+0.20=0.30\) per year. At one year, \(F_1(1\mid A)=(0.10/0.30)\{1-\exp(-0.30)\}=0.08639\), or 8.639%. At three years, \(F_1(3\mid A)=(0.10/0.30)\{1-\exp(-0.90)\}=0.1978\), or 19.78%.
For B, \(\lambda=0.10+0.05=0.15\) per year. At one year, \(F_1(1\mid B)=(0.10/0.15)\{1-\exp(-0.15)\}=0.09286\), or 9.286%. At three years, \(F_1(3\mid B)=(0.10/0.15)\{1-\exp(-0.45)\}=0.2416\), or 24.16%.
B’s cause-1 risk exceeds A’s by 0.647 percentage points at one year and 4.377 percentage points at three years, despite the same cause-1 hazard. The different competing hazards alter event-free survival, and hence the accumulated probability of cause 1. The constant-hazard formula requires \(\lambda>0\); if both hazards are zero, then \(\lambda=0\) and the cause-1 probability is zero, not a ratio involving division by zero.
Calculation Checks, Alternatives, and Limitations
A useful derivation behind the constant-hazard calculation also clarifies its boundary case. When both hazards are constant over an interval of length \(d\), the cause-1 probability for someone event-free at the beginning is \(\int_0^d e^{-(\lambda_1+\lambda_2)u}\lambda_1\,du\). If \(\lambda=\lambda_1+\lambda_2>0\), integration yields \(\lambda_1\{1-e^{-\lambda d}\}/\lambda\). If \(\lambda=0\), nonnegative hazards imply \(\lambda_1=\lambda_2=0\), and the integral is zero. This separate case prevents an undefined division by zero.
Before interpreting a comparison, check that all profile-specific predictions use the same cause definition, time origin, and horizon convention. Confirm that interval event probabilities are nonnegative and sum to at most one, that cumulative incidence does not decrease over time, and that the cause-specific incidences plus event-free probability obey the appropriate probability bounds. Examine whether the data support each profile and horizon; a numerically precise extrapolation is not made reliable merely by displaying more digits. Uncertainty intervals should also be considered when available, especially when differences are small.
Profile-specific contrasts are useful when the question concerns particular covariate patterns. Standardized averages are preferable when comparing groups on a common target-population composition. Crude averages describe each group as actually composed, which may be the right descriptive estimand, but they mix risk differences with composition differences. A hazard ratio answers a different question from an absolute risk difference, and it does not substitute for horizon-specific cumulative incidence. Each summary should be chosen to match the question rather than treated as interchangeable.
The main limitations are that predictions depend on model specification, covariate values, follow-up support, and the chosen standardization weights. Averages may conceal meaningful profile variation, while individual-profile predictions may not represent a whole subgroup. Neither a standardized contrast nor a profile-specific difference is automatically causal; causal interpretation requires additional design and identification assumptions beyond the risk calculation itself.
Check Your Understanding
Answer each question, then compare with the short solutions.
- Two profiles have cause-1 risks of 0.10 and 0.07 at one year, and 0.18 and 0.22 at three years. What are the risk differences, first minus second, and what changes over follow-up?
- A subgroup has two profiles with risks 0.30 and 0.10 and target weights 0.40 and 0.60. Calculate its average risk.
- Why can two groups’ crude average risks be ordered differently from their standardized risks?
- If both cause-specific hazards are zero over an interval, what is the probability of cause 1 in that interval?
Solutions. (1) At one year the difference is \(0.10-0.07=0.03\), or 3 percentage points; at three years it is \(0.18-0.22=-0.04\), or -4 percentage points. The direction reverses. (2) The average is \((0.40)(0.30)+(0.60)(0.10)=0.12+0.06=0.18\), or 18%. (3) Crude averages use each group’s own profile weights, whereas standardized averages use the same target weights; differing compositions can change the contrast. (4) It is zero, since neither cause can occur under the stated hazards.