Why Ties Need Special Handling in Fine–Gray Models
A Fine–Gray model describes the subdistribution hazard for a specified target cause. Its defining risk set retains people who have already experienced a competing event, because those events remain relevant to the target cause’s cumulative incidence. Consequently, tied target failures are handled within a risk set that differs from the ordinary risk set in a cause-specific Cox model. The earlier material on Fine–Gray regression establishes that risk-set structure; here the focus is how ties change the partial-likelihood calculation.
Ties occur when two or more target events have the same recorded time. They may be genuinely simultaneous, or they may result from recording times only to the nearest day, month, or other interval. The partial likelihood for distinct event times does not directly specify how to order multiple failures at one recorded time. Breslow and Efron approximations resolve this in different ways; an exact conditional calculation is another possibility in an ordinary unweighted risk set.
For the basic formulas below, assume there is no censoring before the tied time, so the risk-set weights are all one. The formulas also show the common weighted-risk-set structure by including \(q_j(t)\); software may apply censoring weights through a weighted estimating equation, so the precise implementation should be checked. At a tied time, the target failures are counted once each in the event contribution. The key question is how the denominator accounts for the \(d_k(t)\) failures.
Two Common Approximations
Let \(D=D_k(t)\), \(d=d_k(t)\), and define the tied-event risk score total \(E^{(0)}(t)=\sum_{i\in D}q_i(t)e^{\eta_i}\). Under the Breslow approximation, all tied failures use the full risk-set denominator. Under Efron’s approximation, the tied failures are treated as if they leave the risk set gradually, in equal fractions. The resulting contributions to the log partial likelihood are:
For \(r=0\), Efron’s first denominator is the full risk score. The next denominator subtracts \(1/d\) of the tied failures’ risk score, and each subsequent one subtracts another \(1/d\). The subtraction is based on risk scores, not merely on the number of tied events. When all individuals have the same covariate score and unit weights, this corresponds to subtracting tied failures fractionally from the risk-set size.
Worked Example: Two Tied Failures With Equal Risk Scores
Suppose there are five individuals in the Fine–Gray risk set at time \(t\), including two tied target-cause failures. There is no censoring, all covariate scores are zero, and all weights are one. Thus \(S^{(0)}(t)=5\), \(E^{(0)}(t)=2\), \(d=2\), and the numerator contribution \(\sum_{i\in D}\eta_i\) is zero.
Breslow. Both failures use denominator 5:
Efron. There are two denominators, for \(r=0\) and \(r=1\):
The risk set has five members and there are two tied failures, so there are two denominator terms in either approximation. Efron’s second denominator is 4 because it removes half of the tied failures’ total risk score, which is \(2/2=1\).
How the Efron Denominators Arise
At distinct event times, a partial-likelihood factor compares the event’s risk score with the sum of risk scores in the current risk set. With \(d\) tied target failures, a direct ordering would require deciding which failure occurred first, second, and so on, even though the observed data do not distinguish their order. Breslow’s choice is to use the unchanged risk set for all \(d\) contributions. Efron’s choice approximates the gradual removal of the tied failures’ combined risk score.
The Efron construction can be justified by dividing the tied-event risk score equally across the \(d\) contributions. Before contribution \(r+1\), for \(r=0,\ldots,d-1\), remove \(r/d\) of the tied-event score from the full risk score. This gives \(S^{(0)}(t)-(r/d)E^{(0)}(t)\). Each tied failure contributes its own \(\eta_i\) to the numerator, so summing the log numerator terms gives \(\sum_{i\in D}\eta_i\). Taking the logarithm of each denominator and subtracting it gives the Efron expression above. At \(r=d\), all tied-event score would have been removed, but there is no \((d+1)\)st contribution; this is why the sum ends at \(d-1\).
The same reasoning applies at each distinct target-event time. Contributions from times with no target failure do not enter the target-cause partial likelihood. If a competing event occurs at the tied time, it is not an additional target failure in \(D\). The Fine–Gray risk set is defined just before the time, so the treatment of event status at that instant must be consistent across the data and software.
Worked Example: Comparing Breslow and Efron With Covariates
At one tied time, suppose four individuals are in the risk set, with unit weights. Two have target events: individual A has \(x=1\), and B has \(x=0\). The other two are in the risk set but do not have a target event at this time: C has \(x=0\), and D has \(x=1\). Consider the partial-likelihood contribution evaluated at \(\beta=\log(2)=0.6931\), so \(e^{\beta x}=2\) for \(x=1\) and 1 for \(x=0\).
The full risk score and tied-event risk score are
There are \(d=2\) tied failures. Breslow’s contribution is
For Efron, the denominators are \(6\) and \(6-(1/2)(3)=4.5\). Therefore,
These are contributions evaluated at one selected coefficient value, not fitted coefficient estimates or a test of a hypothesis. To fit the model, software sums the contributions over target-event times and optimizes the resulting objective (or its weighted estimating-equation counterpart).
Exact Conditional Handling and Its Limits
For an ordinary unweighted risk set, exact conditional handling uses a denominator that sums over all possible subsets of \(d\) individuals from the risk set. Each possible subset is treated as a candidate tied failure set. The observed tied set contributes its own product of risk scores in the numerator. This construction avoids approximating the tied event set by a sequential ordering.
Worked Example: Exact Contribution When All Scores Are Equal
Use a risk set of five individuals with \(\eta_j=0\) for everyone, and suppose two have tied target events. The observed tied set has numerator \(\exp(0+0)=1\). There are \(\binom{5}{2}=10\) possible two-person subsets in the denominator, and every subset has score product 1. Thus,
For the same data, the Breslow contribution is \(-2\log(5)=-3.2189\), and the Efron contribution is \(-\log(5)-\log(4)=-2.9957\). The values differ because the three methods construct the tied-event denominator differently. The exact result is not a correction factor to be applied to either approximation.
The exact formula is a conditional tied-set likelihood for an ordinary risk set; it is not automatically a universal gold standard for every Fine–Gray implementation. Fine–Gray fitting commonly uses censoring adjustments, which can produce weighted risk sets. A weighted partial-likelihood estimating procedure is not generally the same as the ordinary subset likelihood above, and software may not offer an exact option for its weighted formulation. Do not assume that selecting “exact” in a general Cox interface necessarily gives an exact Fine–Gray calculation.
Assumptions, Diagnostics, and Software Choices
Tie handling is one part of fitting the Fine–Gray model, not a remedy for failures of the model’s other assumptions. Interpretation of the subdistribution hazard ratio relies on an appropriate proportional subdistribution hazards specification over the modeled time range. With censoring adjustment, the censoring mechanism must be adequately handled; in particular, the relevant censoring weights require the censoring distribution to be estimable and not to assign effectively zero probability of remaining uncensored in regions where the analysis needs support. These assumptions concern the model and its risk-set construction, not the choice between Breslow and Efron.
Before fitting, check how many ties occur, how large the tied sets are relative to their risk sets, and whether recorded time precision is coarse. A large number of ties can make the choice consequential. Also check that target events have been distinguished correctly from competing events and censoring, and that event times and status codes were not inadvertently rounded or recoded. Where censoring weights are used, inspect whether the software reports extreme weights or convergence problems; those can indicate unstable risk-set contributions rather than a tie-method issue alone.
The main alternatives have different practical trade-offs:
| Method | Calculation | Practical consideration |
|---|---|---|
| Breslow | Uses the full risk-set denominator for each tied target failure. | Simple and widely available; may be a rough approximation when ties are substantial. |
| Efron | Fractionally removes tied-event risk scores across the \(d\) contributions. | Often a useful approximation with tied failures; confirm how the Fine–Gray software implements it. |
| Exact conditional | Sums over all candidate tied subsets of size \(d\). | Can be computationally demanding and may not match or be available for a censoring-weighted Fine–Gray procedure. |
Software interfaces differ: they may offer only one tie method, use different defaults, or expose a tie option through a Cox-model fitting routine after transforming the data. Record the method, software, and version used. If ties are numerous, compare estimates and standard errors under supported alternatives as a sensitivity analysis. A notable change is a reason to report the dependence on tie handling and investigate event-time resolution; it is not, by itself, evidence that one setting is correct.
Common Pitfall
Another mistake is to treat Efron as though it removes a fixed number of people from the risk set. It removes a fraction of the tied failures’ total risk score. When covariate scores or weights differ, the risk-score subtraction is not simply the tied-event count divided by the risk-set count.
Check Your Understanding
Answer the questions, then compare your work with the brief solutions.
- A Fine–Gray risk set has six members, of whom three have tied target events. How many denominator terms appear under Breslow and Efron?
- Suppose \(S^{(0)}(t)=8\), \(E^{(0)}(t)=3\), \(d=3\), and all tied event scores are zero. Write the Efron denominators.
- Why does a competing event at the tied time not add one to \(d\)?
- In one sentence, distinguish Breslow from Efron tie handling.
- Why should an analyst check whether an exact option is valid for the chosen Fine–Gray software?
Solutions. (1) Three terms under each method, because \(d=3\). (2) For \(r=0,1,2\), the denominators are \(8\), \(8-(1/3)3=7\), and \(8-(2/3)3=6\). (3) \(d\) counts target-cause failures only; a competing event is not a target event, although the Fine–Gray risk-set definition accounts for competing events. (4) Breslow uses the full risk score for every tied failure, whereas Efron subtracts successive fractions of the tied failures’ risk score. (5) Fine–Gray censoring adjustments can make the calculation a weighted procedure for which the ordinary exact subset likelihood may be unavailable or inapplicable.