Absolute Risk From Cause-Specific Cox Fits

From Cox Fits to Absolute Risk

A cause-specific Cox model estimates how covariates relate to the instantaneous rate of one event type among people who are still event-free. Its regression coefficient is not, by itself, an absolute probability. To predict the probability of a particular cause by a chosen horizon, we need the fitted covariate effects, estimated baseline hazards, and a way to combine all causes so that a competing event removes a person from the possibility of later failure.

The calculation is conditional on a covariate profile: it answers the risk question for a person with the specified covariate values, under the fitted models. It is not automatically an average risk for a population. The key practical sequence is to estimate a baseline cumulative hazard for each cause, multiply its increments by the appropriate Cox relative hazard, and combine the resulting increments across causes.

Definition: For cause $k$, the cause-specific hazard at time $t$, conditional on covariates $z$ and still being event-free immediately before $t$, is modeled as $$ \lambda_k(t\mid z)=\lambda_{0k}(t)\exp(\beta_k^\mathsf{T}z). $$ Here $\lambda_{0k}(t)$ is the cause-specific baseline hazard and $\beta_k$ is the coefficient vector for cause $k$. The cumulative incidence $F_k(t\mid z)$ is the probability of experiencing cause $k$ by time $t$, before any competing event.

Estimating and Scaling the Baseline Hazard

In a separate cause-specific Cox fit for each cause, the regression coefficient is estimated with the Cox partial likelihood. For the fit for cause $k$, an event of another cause ends the individual’s time at risk for cause $k$, just as censoring does in that cause-specific risk-set calculation. It does not mean that the competing event is unobserved or irrelevant to absolute risk: it is accounted for when the cause-specific hazards are combined.

Once the coefficient estimate $\hat{\beta}_k$ is fitted, a common estimate of the baseline cumulative hazard is the Breslow estimator. Let $t_j$ be a distinct event time for cause $k$, let $d_{kj}$ be the number of cause-$k$ events at that time, and let $R(t_j)$ be the risk set immediately before $t_j$. Then the estimated baseline cumulative-hazard increment is

\Delta\hat H_{0k}(t_j)= \frac{d_{kj}}{\displaystyle\sum_{i\in R(t_j)} \exp(\hat{\beta}_k^\mathsf{T}z_i)}.

The risk-set denominator is a covariate-weighted risk set, not simply the number of people at risk. For a profile $z^\ast$, the corresponding estimated cause-specific cumulative-hazard increment is

\Delta\hat H_k(t_j\mid z^\ast)= \exp(\hat{\beta}_k^\mathsf{T}z^\ast)\, \Delta\hat H_{0k}(t_j).

These are hazard increments, not event probabilities. Their interpretation depends on the fitted Cox model being suitable: the proportional-hazards relationship must be adequate over the prediction period, the covariates used for prediction must be measured and coded consistently with fitting, and censoring must be conditionally independent of the event process given the modeled covariates. In particular, censoring must not selectively remove people with different future risks after conditioning on those covariates.

Worked Example: Estimating and Scaling a Baseline Increment

Suppose a hypothetical cause-1 Cox fit has coefficient $\hat\beta_1=\log(1.8)$ for a binary covariate $z$. At time $t=1$, two cause-1 events occur. Immediately before that time, the weighted risk-set denominator is 12.5. Calculate the baseline increment and the increment for a person with $z=1$.

The Breslow baseline increment is the event count divided by the weighted risk-set denominator:

\Delta\hat H_{01}(1)=\frac{2}{12.5}=0.1600.

For $z=1$, the relative hazard multiplier is $\exp(\log(1.8)\times1)=1.8$. Therefore,

\Delta\hat H_1(1\mid z=1)=1.8(0.1600)=0.2880.

For comparison, the baseline profile $z=0$ has increment $0.1600$. The value $0.2880$ is not a 28.8% event probability; it is the profile-specific cumulative-hazard increment at this step. The assumptions to check include that the event count and risk set are correct, the fitted proportional-hazards model is being used on its intended covariate scale, and the profile value $z=1$ is coded as it was in fitting.

Combining Cause-Specific Increments

To obtain absolute risk, combine the cause-specific increments in time order. A useful calculation treats the increments over each interval as integrated hazards under constant cause-specific hazards within that interval. If the increments for all causes over an interval are $\Delta\hat H_k$, the total integrated hazard is $\Delta\hat H_{\text{tot}}=\sum_k\Delta\hat H_k$. The probability of any event over that interval, conditional on being event-free at its start, is $1-\exp(-\Delta\hat H_{\text{tot}})$. This probability is allocated among causes in proportion to their hazard increments.

The resulting cause-$k$ conditional event probability for the interval is

\hat q_k= \frac{\Delta\hat H_k}{\Delta\hat H_{\text{tot}}} \left\{1-\exp(-\Delta\hat H_{\text{tot}})\right\}.

When the total increment is zero, all cause-specific interval event probabilities are zero. Otherwise, the probabilities sum to the total event probability. If $\hat S_{j-1}$ is the predicted event-free probability at the start of interval $j$, the update is

\Delta\hat F_{kj}=\hat S_{j-1}\hat q_{kj},\qquad \hat F_{kj}=\hat F_{k,j-1}+\Delta\hat F_{kj},\qquad \hat S_j=\hat S_{j-1}\exp(-\Delta\hat H_{\text{tot},j}).

Initialize with $\hat S_0=1$ and $\hat F_{k0}=0$. The probabilities $\hat F_{kj}$ accumulate the probability of each cause up to the end of interval $j$; $\hat S_j$ is the probability of remaining event-free. This interval calculation is exact under constant cause-specific hazards within each interval. When fitted baseline cumulative hazards are represented by jumps, treating each increment as an interval integrated hazard gives a coherent numerical prediction convention; the convention should be kept consistent across causes and reported when it matters.

Justification: If cause-specific hazards are constant within an interval, then the total hazard is $\lambda_{\text{tot}}=\sum_k\lambda_k$. The probability of remaining event-free for interval length $a$ is $\exp(-\lambda_{\text{tot}}a)$. Integrating the cause-$k$ event density gives $$ \int_0^a \lambda_k\exp(-\lambda_{\text{tot}}u)\,du = \frac{\lambda_k}{\lambda_{\text{tot}}} \{1-\exp(-\lambda_{\text{tot}}a)\}. $$ Replacing each $\lambda_k a$ by its interval hazard increment yields the stated allocation rule. It follows that the interval probabilities add up: $\sum_k\hat q_k=1-\exp(-\Delta\hat H_{\text{tot}})$, so event-free probability plus all cause-specific cumulative incidences equals one after every update.

Worked Example: Two-Cause Risk Through Two Intervals

For a hypothetical covariate profile, suppose the fitted cause-specific increments are $0.10$ and $0.05$ in the first interval, and $0.20$ and $0.15$ in the second. Calculate both cumulative incidences and event-free probability through the second interval.

In interval 1, the total increment is $0.10+0.05=0.15$, so the probability of any event conditional on being event-free at the start is $1-\exp(-0.15)=0.1393$. The cause proportions are $0.10/0.15=2/3$ and $0.05/0.15=1/3$. Thus the conditional probabilities are $\hat q_{11}=0.09286$ and $\hat q_{21}=0.04643$. Starting from $\hat S_0=1$, the cumulative incidences are $0.09286$ and $0.04643$, and event-free probability is $\hat S_1=\exp(-0.15)=0.8607$.

In interval 2, the total increment is $0.20+0.15=0.35$, and the conditional probability of any event is $1-\exp(-0.35)=0.2953$. The cause proportions are $0.20/0.35=4/7$ and $0.15/0.35=3/7$. Therefore, $\hat q_{12}=0.1688$ and $\hat q_{22}=0.1266$. Multiply by the probability $0.8607$ of reaching this interval event-free:

\begin{aligned} \hat F_1&=0.09286+0.8607(0.1688)=0.2381,\\ \hat F_2&=0.04643+0.8607(0.1266)=0.1554,\\ \hat S_2&=\exp[-(0.15+0.35)]=0.6065. \end{aligned}

The check is $0.2381+0.1554+0.6065=1.0000$, up to rounding. These predictions apply to the specified profile and horizon, not to a population average. The arithmetic also assumes the increments are defined on the same intervals and use the same hazard-to-probability convention for both causes.

Checking the Prediction and Avoiding Misinterpretation

A basic numerical diagnostic is to verify, at each prediction time, that all predicted probabilities are between zero and one and that $\sum_k\hat F_k(t\mid z^\ast)+\hat S(t\mid z^\ast)=1$, apart from rounding. Check that cumulative incidence does not decrease, that the cause-specific increments are nonnegative, and that the risk sets and event counts used for each fitted model match the observed event coding. Large jumps deserve particular scrutiny: confirm the event time, number at risk, tie handling, and covariate coding rather than assuming the calculation is stable.

The calculation also relies on modeling choices. Proportional hazards may be implausible if covariate effects change over time; sparse risk sets at late times can make baseline increments unstable; and extrapolating beyond the observed follow-up requires assumptions not supplied by the fitted data alone. If covariates are missing, measured with error, or represented differently at prediction than at fitting, the resulting profile-specific hazards need not be meaningful. Uncertainty in both the coefficients and baseline hazards is not reflected in a single point prediction.

The Aalen-Johansen estimator provides a nonparametric alternative for estimating cumulative incidence in observed groups, without imposing cause-specific Cox proportional-hazards models. Cox fits can provide covariate-specific predictions and can borrow structure across profiles through their regression form, but they rely on model adequacy. The earlier material on Aalen-Johansen absolute risk predictions gives the nonparametric construction; neither method removes the need to define the target population, time horizon, and handling of censoring clearly.

Worked Example: Why a Cause-Specific Net Risk Is Not the Absolute Risk

Suppose a hypothetical profile has increments $0.24$ for the event of interest and $0.04$ for a competing event over one interval. What is the event-of-interest absolute risk over that interval, and how does it compare with treating the competing event as absent?

The total increment is $0.28$, so the interval probability of any event is $1-\exp(-0.28)=0.2442$. The proportion assigned to the event of interest is $0.24/0.28=6/7$. Its absolute risk is therefore

\hat q_1=\frac{0.24}{0.28}\{1-\exp(-0.28)\}=0.2093.

If the competing event were incorrectly removed from the calculation, the resulting net-risk quantity would be $1-\exp(-0.24)=0.2134$, which is larger. It answers a different question: risk under a calculation that excludes the competing cause, not the probability of the event of interest occurring first in the presence of the competing event. The competing event must remain in the total hazard when predicting actual cumulative incidence.

Common mistake: A Cox relative hazard such as $\exp(\hat\beta_k)=1.8$ is sometimes reported as if it meant “an 80% higher absolute risk.” It means an 80% higher cause-specific hazard under the proportional-hazards model, conditional on being event-free. Absolute risk additionally depends on the baseline hazards for all causes, the covariate profile, and the prediction horizon.

Check Your Understanding

Try each question, then compare your work with the solutions below.

A cause-specific Cox fit has $d=3$ events at a time when the weighted risk-set denominator is 15. What is the Breslow baseline hazard increment?
For a binary covariate, $\hat\beta=\log(2)$ and the baseline increment is $0.08$. Find the profile-specific increment for $z=1$.
Two causes have increments $0.06$ and $0.04$ in one interval. What is the conditional probability of any event during the interval?
Why is $1-\exp[-\Delta\hat H_1]$ generally not the cause-1 absolute event probability when another cause has a positive hazard increment?

Solutions. 1. The Breslow increment is $3/15=0.2000$. 2. The multiplier is $\exp(\log(2))=2$, so the profile increment is $2(0.08)=0.1600$. 3. The total increment is $0.06+0.04=0.10$, giving $1-\exp(-0.10)=0.09516$. 4. That expression excludes the competing cause from the total hazard. With both causes present, cause 1 receives only its proportional share of the probability of any event: $\{0.06/0.10\}\{1-\exp(-0.10)\}=0.05710$.

Absolute Risk From Cause-Specific Cox Fits

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