Tied Event Times In Cox Regression

Why Tied Event Times Need Special Treatment

Cox partial likelihood is built around the ordering of failures: at each event time, the subject who fails is compared with everyone still at risk. If several subjects share an observed event time, the data do not specify an ordering among those failures. A tied time can arise because times are recorded only to a day or month, or because event times genuinely have limited precision. The standard single-failure denominator therefore does not say, by itself, how to form the contribution for that group.

Three standard approaches are the Breslow approximation, the Efron approximation, and the exact partial likelihood. They use the same Cox model and observed risk sets, but treat the tied failures differently. Their contributions can differ, and those differences can matter for estimating the regression coefficients when ties are frequent or large relative to the risk sets.

Definition: Suppose $d_j$ subjects fail at observed time $t_j$. Let $D_j$ be this set of failing subjects and $R_j$ the risk set immediately before $t_j$, including the failures. Under the Cox proportional-hazards model, write $w_i(\beta)=\exp(x_i^\top\beta)$, where $x_i$ is subject $i$'s covariate vector. Define $S_{R_j}(\beta)=\sum_{i\in R_j}w_i(\beta)$ and $S_{D_j}(\beta)=\sum_{i\in D_j}w_i(\beta)$. The numerator for the tied group is $\prod_{i\in D_j}w_i(\beta)$.

This tutorial assumes the Cox model $h_i(t\mid x_i)=h_0(t)\exp(x_i^\top\beta)$, with a common unspecified baseline hazard $h_0(t)$. As in ordinary Cox partial likelihood, subjects are independent conditional on their covariates, censoring is noninformative conditional on the model variables, and the risk sets correctly represent who remains under observation and at risk. The tie methods address how to construct a partial-likelihood contribution when observed failures share a time; they do not repair incorrect risk sets or informative censoring.

Three Contributions for One Tied Event Time

For a single failure, all three approaches reduce to the usual partial-likelihood factor: the failing subject's relative-risk weight divided by the sum of weights in the risk set. With $d_j$ tied failures, the methods differ in their denominator.

Breslow contribution: Raise the full risk-set sum to the number of failures. $$ L_{B,j}(\beta)= \frac{\prod_{i\in D_j}w_i(\beta)} {\left\{S_{R_j}(\beta)\right\}^{d_j}}. $$ Efron contribution: Use a sequence of denominators that removes increasing fractions of the tied failures' total weight. $$ L_{E,j}(\beta)= \frac{\prod_{i\in D_j}w_i(\beta)} {\prod_{\ell=0}^{d_j-1} \left\{S_{R_j}(\beta)-\frac{\ell}{d_j}S_{D_j}(\beta)\right\}}. $$ Exact partial-likelihood contribution: Compare the observed failure set with every possible subset of $d_j$ subjects from the risk set. $$ L_{X,j}(\beta)= \frac{\prod_{i\in D_j}w_i(\beta)} {\displaystyle\sum_{\substack{A\subseteq R_j\\|A|=d_j}} \prod_{i\in A}w_i(\beta)}. $$

The Breslow denominator treats the full risk set as available for each of the $d_j$ failures. Efron retains the observed tie-group weight but reduces the denominator in steps, by fractions of that weight. The exact denominator sums the relative-risk products for all size-$d_j$ subsets of the risk set, including the observed set. Each expression is a partial-likelihood contribution, not a statement that the observed failures have that probability under an unconditional model for event times.

For exact computation, the denominator is the elementary symmetric sum of order $d_j$ of the risk-set weights. It can be evaluated by enumerating subsets when the risk set is small, or by a recursion that builds these sums without listing every subset. The number of subsets in a direct enumeration is $\binom{|R_j|}{d_j}$, so exact calculation can become costly when both the risk set and the tied group are large.

Why the Efron Denominators Have Fractions

A useful way to understand the Efron approximation is to imagine resolving the tied group into $d_j$ sequential failures. The observed data do not tell us the order, so no particular ordering should be treated as known. After one tied failure, a sequential risk set would exclude that subject from later denominators. Efron approximates this removal by subtracting a fraction of the total tied-group weight at each step.

\text{Efron denominator at step }\ell = S_{R_j}(\beta)-\frac{\ell}{d_j}S_{D_j}(\beta), \qquad \ell=0,\ldots,d_j-1.

At $\ell=0$, nothing has yet been removed. At the final step, the denominator has subtracted $(d_j-1)/d_j$ of the tied-group weight. This is an approximation to sequential removal, not an assertion that a literal fraction of a person leaves the risk set. The exact method instead accounts for all possible failure subsets directly, while Breslow repeats the unadjusted sum.

Worked Example: A Large Risk Set With Three Tied Failures

Suppose a hypothetical risk set contains 100 subjects, and three fail at the same recorded time. Assume every subject has relative-risk weight 1 at the coefficient value being evaluated. Thus $S_R=100$, while only the three failures contribute to $S_D$, so $S_D=3$. The tied-group numerator is $1\cdot1\cdot1=1$.

Breslow. The denominator is $100^3=1{,}000{,}000$, so $L_B=1/1{,}000{,}000=0.000001000$.

Efron. The three factors are $100$, $100-(1/3)(3)=99$, and $100-(2/3)(3)=98$. Their product is $100\times99\times98=970{,}200$, so $L_E=1/970{,}200\approx0.000001031$.

Exact. Each subset of three subjects has weight product 1. There are $\binom{100}{3}=100\times99\times98/(3\times2\times1)=161{,}700$ such subsets. Therefore $L_X=1/161{,}700\approx0.000006184$.

These values show that the three contributions need not be numerically close just because the risk-set weights are equal at this coefficient value. However, that equality at one value of $\beta$ does not establish that the contribution is constant as a function of $\beta$. If covariates differ, weights can change differently as $\beta$ changes, and the event time can contain information about $\beta$. Only if the subjects have identical covariate vectors do their weights remain equal as $\beta$ varies.

Worked Comparison With Covariate-Dependent Weights

The following example makes the distinction between an evaluation at one coefficient value and a likelihood function explicit. Let the risk set contain four hypothetical subjects: two failures with $x=1$, and two nonfailures with $x=0$. Use a scalar coefficient $\beta$, so a subject with covariate $x$ has weight $\exp(x\beta)$. Write $a=\exp(\beta)$. Then the tied-group numerator is $a^2$, $S_R=2a+2$, and $S_D=2a$.

The three contributions as functions of $a$ are

L_B=\frac{a^2}{(2a+2)^2},\qquad L_E=\frac{a^2}{(2a+2)(a+2)},\qquad L_X=\frac{a^2}{a^2+4a+1}.

For the Efron denominator, the two factors are $S_R=2a+2$ and $S_R-\tfrac12S_D=2a+2-a=a+2$. For the exact denominator, the size-two subsets have total weight $a^2+4a+1$: one pair consists of both failures, four pairs contain one failure and one nonfailure, and one pair contains both nonfailures.

Worked Example: Comparing Methods at a Specified Coefficient

Use the four-subject risk set just described and evaluate the contributions at $\beta=\log 2$, so $a=\exp(\beta)=2$. The numerator is $a^2=4$, $S_R=2(2)+2=6$, and $S_D=2(2)=4$.

Breslow. Substitution gives $L_B=4/6^2=4/36=0.1111$.

Efron. The two denominator factors are $6$ and $6-(1/2)(4)=4$. Hence $L_E=4/(6\times4)=4/24=0.1667$.

Exact. The denominator is $a^2+4a+1=4+8+1=13$, so $L_X=4/13\approx0.3077$.

These are evaluations of three different likelihood contributions at the same coefficient value; they are not three estimates of $\beta$. To estimate coefficients, the contributions from all event times are combined into a partial likelihood and optimized. A change in one event-time contribution can affect the overall estimate if that contribution varies with $\beta$.

How Tie Frequency and Risk-Set Size Guide the Choice

When there are no ties, $d_j=1$, the methods coincide: the Breslow and Efron denominators each contain $S_{R_j}$, and the exact denominator sums the weights of all one-subject subsets, also $S_{R_j}$. Differences arise only for tied groups. Their practical importance depends not only on the number of tied failures, but also on how that number compares with the size and composition of the risk set.

When ties are sparse relative to risk-set size, Breslow and Efron often give similar contributions. Efron's stepwise adjustment is especially useful when ties are more numerous, while exact partial likelihood avoids the Breslow and Efron denominator approximations under its conditional formulation. But exactness is not free: if a risk set has many subjects and $d_j$ is neither very small nor very close to the risk-set size, $\binom{|R_j|}{d_j}$ may be enormous. Software may use an efficient recursion, but computational cost can still make the exact method impractical.

Worked Example: A Small Risk Set With Unequal Weights

Consider a hypothetical risk set of five subjects at $\beta=\log 2$. Let the two failures have covariates $x=1$ and $x=0$, giving weights 2 and 1. Let the three nonfailures have covariates $x=-1,0,1$, giving weights $0.5,1,2$. Thus the numerator is $2\times1=2$, $S_R=2+1+0.5+1+2=6.5$, and $S_D=2+1=3$.

Breslow. Since $d=2$, the denominator is $6.5^2=42.25$. Thus $L_B=2/42.25\approx0.04734$.

Efron. The denominator factors are $6.5$ and $6.5-(1/2)(3)=5$. The product is $32.5$, so $L_E=2/32.5\approx0.06154$.

Exact. Sum the products for all ten size-two subsets of weights $2,1,0.5,1,2$: $2+1+2+4+0.5+1+2+0.5+1+2=16$. Therefore $L_X=2/16=0.1250$.

Here the risk set is small enough that the exact denominator can be checked by direct enumeration. In larger risk sets, the relevant comparison is not just “exact versus approximate”: analysts also need to consider how many ties occur, how large each tied group is relative to its risk set, and whether their software implements the chosen method reliably.

Common Mistakes and Practical Reporting

Using the risk-set sum for the tied-group sum. $S_R$ includes every subject at risk; $S_D$ includes only the failures in the tie group. In the 100-subject example, $S_R=100$ but $S_D=3$.
Subtracting the wrong quantity in Efron's factors. The adjustment is $(\ell/d_j)S_D$, not $(\ell/d_j)S_R$. Check that the tied-group weight, not the full risk-set weight, is being reduced.
Inferring no coefficient information from equal weights at one value. Equal numerical weights at a particular $\beta$ do not show that the weights remain equal as $\beta$ changes. Examine the covariates and the likelihood's dependence on $\beta$.
Calling an exact contribution a probability of the observed failures in real time. It is a conditional partial-likelihood factor built from subsets of the risk set, not an unconditional event-time probability.
Choosing a method based only on the label “exact.” The exact denominator can be computationally expensive. Consider both the frequency and size of ties and the size of the risk sets, and report which tie method was used.

Key takeaway: Breslow repeats the full risk-set sum, Efron progressively subtracts fractions of tied-failure weight, and exact partial likelihood sums over all possible failure subsets of the observed size. The methods agree when there is one failure at a time, but can differ for ties; their practical comparison depends on tie frequency, risk-set composition, and computational feasibility.

Check Your Understanding

Answer each question using the definitions and calculations above.

A tie group has two failures. Which subjects contribute to $S_D$, and which contribute to $S_R$?
For $d_j=1$, explain why the Breslow, Efron, and exact denominators are identical.
A risk set has total weight $S_R=50$, and its four failures have total weight $S_D=8$. What are the four Efron denominator factors?
Why does equality of all risk-set weights at one evaluated coefficient value not prove that a tied event time contains no information about $\beta$?
What feature of the exact denominator can make direct computation difficult in a large risk set?