Why average individual risks?
A risk prediction for one covariate profile answers a conditional question: what is the model-predicted probability of a particular event by a specified time for someone with these covariates? A cohort-level prediction asks a different question: what is the average of those probabilities over the covariate profiles in a defined cohort? When members differ, the cohort answer requires making predictions for those members and averaging them. A prediction for a person with average covariates is generally not the same quantity.
Here, the outcome of interest is occurrence of a specified event cause by a common horizon \(\tau\), in the presence of competing event causes. The prediction for each member must account for all causes, because a competing event prevents the event of interest from occurring first. Standardization describes the average model-predicted risk for a stated covariate distribution; it does not, by itself, establish a causal effect.
From cause-specific hazards to member-specific risk
For member \(i\), let \(\lambda_{ik}(t)\) be the cause-specific hazard at time \(t\) for cause \(k\). It is the instantaneous rate of cause \(k\) at \(t\), conditional on being event-free immediately before \(t\). Write \(\Lambda_{ik}(t)=\int_0^t\lambda_{ik}(u)\,du\) for the corresponding cumulative hazard. Under continuous-time hazards, the probability of remaining free of every event through time \(t\) is \(S_i(t)=\exp\{-\sum_k\Lambda_{ik}(t)\}\). The cause-\(k\) cumulative incidence is the probability that cause \(k\) occurs by \(t\) before another cause does:
Thus the other causes enter through survival \(S_i(u)\), even though the integrand for cause \(k\) uses its own hazard. In cause-specific Cox models, for example, each cause has its own baseline hazard and regression coefficients: \(\lambda_{ik}(t)=\lambda_{0k}(t)\exp(\beta_k^\mathsf{T}z_i)\). The fitted hazards are used together to obtain each member’s cause-specific cumulative incidence. That prediction step must be done before the cohort average is taken.
A standardized risk is meaningful only relative to a defined target cohort and horizon. The fitted models must adequately represent the cause-specific hazards over the period used; covariates used for prediction must be measured or otherwise specified for the target members; and the data must support prediction to \(\tau\). Estimation of the hazards also relies on appropriate handling of censoring, commonly requiring censoring to be independent of event time conditional on the modeled covariates. Misspecification, unsupported extrapolation, or an inappropriate target covariate distribution can undermine the average even when the arithmetic is correct.
Worked Example: Averaging Predictions for a Small Cohort
Suppose a fitted competing-risk model gives the following hypothetical predictions of cause A by three years for five members of a target cohort. Each prediction already accounts for the competing causes.
| Member | Predicted cause A risk by 3 years |
|---|---|
| 1 | 0.08 |
| 2 | 0.12 |
| 3 | 0.12 |
| 4 | 0.20 |
| 5 | 0.28 |
The target is the equal-weight cohort-average predicted risk at the common three-year horizon. Apply the standardization formula:
The model-predicted cohort-average risk of cause A by three years is 16.0%. This is the average of five conditional predictions, not a claim that exactly 16% of these five people will experience cause A. With only five members, the realized proportion would change in increments of 20%; the average predicted probability is a model-based expectation across the specified cohort.
Why averaging and predicting do not usually commute
The standardization calculation first predicts at each observed covariate profile and then averages. In contrast, a mean-covariate prediction first calculates the average covariate vector \(\bar z\) and then predicts for that single profile. In general, \(\frac{1}{n}\sum_i \hat F_{ik}(\tau)\neq \hat F_k(\tau\mid\bar z)\), because the risk function is nonlinear in the hazards and covariates. The mean-covariate prediction also describes a hypothetical covariate profile, which may not correspond to any actual member.
Worked Example: Cause-Specific Risks and the Mean-Covariate Prediction
Consider a hypothetical cohort of four members, with two having \(x=0\) and two having \(x=1\). Suppose cause-specific hazards are constant over time and, per year, are \(\lambda_1(x)=0.04\exp\{(\log 2)x\}\) for cause 1 and \(\lambda_2(x)=0.02\) for a competing cause. Take \(\tau=5\) years. Assume these hazards adequately describe both causes over this horizon and that the cohort’s covariate values are known.
With constant hazards, the cause-1 cumulative incidence is \(F_1(\tau\mid x)=\frac{\lambda_1(x)}{\lambda_1(x)+\lambda_2(x)}[1-\exp\{-(\lambda_1(x)+\lambda_2(x))\tau\}]\). For \(x=0\), the rates are 0.04 and 0.02 per year, so:
For \(x=1\), the cause-1 hazard is \(0.04e^{\log 2}=0.08\) per year. Therefore:
Average the four individual predictions, giving equal weight to the two members in each covariate group:
The cohort-average predicted risk is approximately 24.38% by five years. Now compare it with a prediction at the mean covariate, \(\bar x=(0+0+1+1)/4=0.5\). The hazards for this hypothetical profile are \(\lambda_1(0.5)=0.04e^{(\log 2)(0.5)}\approx0.05657\) and \(\lambda_2(0.5)=0.02\), both per year. Its predicted cause-1 risk is:
The difference is about 0.0088, or 0.88 percentage points. The 23.50% value is the prediction for a profile with \(x=0.5\); the 24.38% value averages predictions for the actual distribution, half \(x=0\) and half \(x=1\). The hazards and risk formula are nonlinear, so there is no requirement that the two summaries agree.
Target weights, checks, and alternatives
Equal weighting is appropriate when the target is the empirical cohort itself. If the goal is a different target population, its covariate composition must be represented by suitable weights or a target sample. For example, averaging predictions in a study sample does not automatically estimate an average for a population with a different distribution of risk factors. The weights should be nonnegative and sum to one; state clearly which population they represent.
Worked Example: Standardizing to a Different Group Distribution
Suppose a hypothetical prediction model gives cause B risks by two years of 0.10, 0.20, and 0.40 for three covariate groups. The study cohort contains 50, 30, and 20 members in those groups, respectively. The target population instead has group weights 0.20, 0.30, and 0.50. Assume each group prediction is applicable to the corresponding target members and the two-year horizon is supported by follow-up.
The study-cohort average would be \((50(0.10)+30(0.20)+20(0.40))/100=(5+6+8)/100=0.190\). For the stated target population, use its weights instead:
The target-standardized predicted risk is 28.0% by two years. The difference from 19.0% is due to the different group composition in these calculations; it is not evidence that any individual’s risk changed.
Several checks help reveal errors before a standardized estimate is interpreted:
- Confirm that every member’s prediction refers to the same cause and horizon, and that all relevant competing causes entered the risk calculation.
- Check that weights are nonnegative and sum to one. With equal weights, verify that the denominator is the number of target members, not the number with a particular event status.
- Check probability coherence: each predicted cumulative incidence is between zero and one, and the sum over causes at a horizon cannot exceed one. A cohort average should preserve these bounds.
- Inspect covariate distributions and the number of members with profiles unlike those represented in model fitting. Predictions for unsupported profiles or beyond observed follow-up may be extrapolations.
- Compare group-specific predictions and the overall average. An unexpected average can reveal a coding, weight, or prediction-horizon mismatch.
The main alternative, predicting at mean covariates, is compact but answers a conditional, hypothetical-profile question rather than the average-risk question for a heterogeneous cohort. A crude observed event proportion is another distinct quantity: it summarizes observed outcomes and may be affected by censoring and follow-up, whereas standardization averages model-predicted risks at a fixed horizon. Neither alternative should be substituted without stating the target estimand.
Standardization also has limitations. It inherits errors in the fitted cause-specific hazards and can be unstable when predictions depend heavily on sparse covariate patterns. A large target sample does not repair a poorly supported model or an incorrect target distribution. Uncertainty in fitted parameters and baseline hazards is not visible from the arithmetic average alone; inference must account for uncertainty in the predictions as well as the target-data or weight estimation when relevant.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Three members have predicted cause A risks by one year of 0.05, 0.15, and 0.25. What is their equal-weight standardized risk?
- Why must cause-specific predictions account for competing causes?
- A target population has group weights 0.25 and 0.75, with predicted risks 0.12 and 0.32. Calculate its weighted standardized risk.
- Does a mean-covariate prediction generally equal an average of individual risks? Explain briefly.
Solutions. (1) \((0.05+0.15+0.25)/3=0.15\), or 15.0%. (2) A competing event can occur first and prevents the event of interest from occurring later as the first event; it therefore contributes to the probability of remaining event-free in the cumulative-incidence calculation. (3) \(0.25(0.12)+0.75(0.32)=0.03+0.24=0.27\), or 27.0%. (4) No. The risk is generally a nonlinear function of covariates through the hazards and competing-event survival, so predict-then-average need not equal average-covariates-then-predict.