Ties and Weighted Cox Partial Likelihood

Why Tied Event Times Need a Convention

The Cox partial likelihood compares each person who experiences an event with the people still at risk at that time. When several people fail at the same recorded time, their ordering is unknown. The ordinary single-event contribution therefore does not specify one unique denominator for the whole tied group. Breslow and Efron are two standard ways to approximate that tied contribution.

In a weighted Cox model for a survival marginal structural model, both the event terms and the risk-set contributions use weights. The tie convention matters because it determines how much of the tied event mass is removed from the risk set across the denominators. It can therefore change the fitted coefficient, especially when ties are numerous, event weights vary, or a tied group accounts for a substantial share of the weighted risk set.

Weighted Risk-Set Notation

Consider one stratum and one tied event time $t$. Let $R(t)$ be the people at risk just before $t$, and let $D(t)\subseteq R(t)$ be the people who experience the event at $t$. Write $d=|D(t)|$ for the number of event cases, $w_i\geq 0$ for person $i$'s analysis weight, and $x_i(t)$ for the model covariate vector. The linear predictor is $\eta_i(t)=\beta^\mathsf{T}x_i(t)$. We assume finite covariate values, nonnegative finite weights, and a positive weighted risk-set denominator.

For $k=0,1$, define the weighted risk-set sums and tied-event sums by

S_0(t;\beta)=\sum_{j\in R(t)}w_j e^{\eta_j(t)},\qquad S_1(t;\beta)=\sum_{j\in R(t)}w_j e^{\eta_j(t)}x_j(t),

S_{0D}(t;\beta)=\sum_{i\in D(t)}w_i e^{\eta_i(t)},\qquad S_{1D}(t;\beta)=\sum_{i\in D(t)}w_i e^{\eta_i(t)}x_i(t), \qquad W_D=\sum_{i\in D(t)}w_i.

The weighted event numerator in the log partial likelihood is $\sum_{i\in D(t)}w_i\eta_i(t)$. In a stratified Cox model, these sums are formed within the relevant stratum; the tie formulas below apply separately at each tied time in each stratum.

Definition: The weighted Breslow contribution at a tied event time is $$ \ell_B(t;\beta)=\sum_{i\in D(t)}w_i\eta_i(t)-W_D\log S_0(t;\beta). $$ For this tutorial, the analytic-weight Efron contribution uses $d$ fractional removals, each removing $1/d$ of the tied events' weighted risk contribution: $$ \ell_E(t;\beta)=\sum_{i\in D(t)}w_i\eta_i(t) -\frac{W_D}{d}\sum_{r=0}^{d-1} \log\left\{S_0(t;\beta)-\frac{r}{d}S_{0D}(t;\beta)\right\}. $$ These formulas require $d\geq1$, and every logarithm argument must be positive.

The Efron definition above makes explicit one analytic-weight convention: it assigns the average event weight $W_D/d$ to each of $d$ denominator steps. For general noninteger analysis weights, there is not a single universal weighted extension of Efron. Software may implement a different convention, so the weight meaning and tie method need to be checked before comparing results.

Deriving the Two Score Contributions

The score is the derivative of the log partial likelihood with respect to $\beta$. For Breslow, differentiating the event term gives $\sum_{i\in D}w_i x_i(t)$, while differentiating $\log S_0$ gives $S_1/S_0$. Hence

U_B(t;\beta)=\sum_{i\in D(t)}w_i x_i(t) -W_D\frac{S_1(t;\beta)}{S_0(t;\beta)}.

For the defined Efron convention, set $S_0^{(r)}=S_0-(r/d)S_{0D}$ and $S_1^{(r)}=S_1-(r/d)S_{1D}$. Since the derivative of $S_0^{(r)}$ is $S_1^{(r)}$, differentiating each logarithm gives the Efron score.

U_E(t;\beta)=\sum_{i\in D(t)}w_i x_i(t) -\frac{W_D}{d}\sum_{r=0}^{d-1} \frac{S_1(t;\beta)-(r/d)S_{1D}(t;\beta)} {S_0(t;\beta)-(r/d)S_{0D}(t;\beta)}.

Result: In the Breslow approximation the risk-set log denominator has coefficient $W_D$ in the tied event group's contribution. In the analytic-weight Efron approximation, the tied-event risk contribution is progressively removed over $d$ denominators. Both methods retain the same weighted event numerator.

To verify the score derivation directly, note that $\nabla_\beta S_0^{(r)}=S_1^{(r)}$, so $\nabla_\beta\log S_0^{(r)}=S_1^{(r)}/S_0^{(r)}$. Multiplying by the coefficient $W_D/d$ and summing over $r$ gives the displayed Efron expression. The Breslow derivative follows from the same calculation with the denominator $S_0$ repeated with total coefficient $W_D$. $\square$

Worked Example: A Tied Risk Set at Zero Coefficients

At one time, suppose the risk set contains three people. Their covariates are $0,1,2$, and their weights are $1,2,1.5$, respectively. The people with covariates 1 and 2 fail together. Calculate both log-likelihood contributions and scores at $\beta=0$.

At $\beta=0$, every $e^{\beta x_i}=1$. Thus $S_0=1+2+1.5=4.5$ and $S_1=1(0)+2(1)+1.5(2)=5$. For the tied failures, $d=2$, $W_D=2+1.5=3.5$, $S_{0D}=3.5$, and $S_{1D}=2+3=5$. The weighted event numerator is zero because $\beta=0$.

Breslow likelihood. Its contribution is $-3.5\log(4.5)$, approximately $-5.264$. The score is $5-3.5(5/4.5)=5-3.8889=1.1111$.

Efron likelihood. The two denominators are $4.5$ and $4.5-(1/2)(3.5)=2.75$. Each is multiplied by $W_D/d=1.75$, giving $-1.75\{\log(4.5)+\log(2.75)\}$, approximately $-4.402$.

For the score, the two risk-set covariate means in the denominators are $5/4.5=1.1111$ and $(5-2.5)/2.75=2.5/2.75=0.9091$. Therefore $U_E=5-1.75(1.1111+0.9091)\approx5-3.5354=1.4646$. The calculations show that the methods agree on the weighted event numerator but not on the risk-set part.

Worked Example: One Event Gives the Same Contribution

Suppose just one person fails at time $t$, with weight $w_i=2$, covariate $x_i=1$, and linear predictor $\eta_i=\beta$. Other people may also be at risk, so let the full risk-set denominator be $S_0$.

Here $d=1$ and $W_D=w_i=2$. Breslow gives $2\beta-2\log S_0$. Efron has only $r=0$; its coefficient is $W_D/d=2$, and its denominator is $S_0-(0/1)S_{0D}=S_0$. It therefore gives $2\beta-2\log S_0$ as well. The event count is $d=1$, not $W_D$: a weight of 2 does not turn this one observed event into two event cases under the analytic-weight convention.

The equality follows in general for a single event: $d=1$, $W_D=w_i$, and the Efron sum contains only $r=0$. Both methods use denominator $S_0$ and the same weighted event numerator $w_i\eta_i$. $\square$

What the Efron Removal Represents

When all weights equal one, the familiar Efron idea is to imagine the tied events leaving the risk set in a sequence. At step $r$, a fraction $r/d$ of the tied events' total risk contribution has been removed. Breslow instead leaves the risk set unchanged for all $d$ denominator contributions. The weighted analytic version above preserves that fractional-removal idea using weighted risk contributions, while assigning total denominator coefficient $W_D$.

The denominators are well behaved under the stated conditions. Because $D(t)\subseteq R(t)$ and weights are nonnegative, $0\leq S_{0D}\leq S_0$. For $0\leq r\leq d-1$,

S_0-\frac{r}{d}S_{0D} \geq S_0-\frac{d-1}{d}S_0 =\frac{S_0}{d}>0.

Thus, if $S_0>0$, every Efron logarithm is defined. This also verifies that the fractional-removal construction cannot produce a negative denominator merely because some tied events have large weights. It does not, however, guarantee numerical stability: a very small positive denominator can still be sensitive to weight concentration.

Worked Example: Checking Event-Weight Concentration

Suppose a risk set has $S_0=10$. Two tied events contribute $S_{0D}=8$, and their weights sum to $W_D=5$. There are $d=2$ observed event cases. Find the Efron denominators and compare them with the Breslow denominator.

The Efron denominators are $10$ and $10-(1/2)(8)=6$, both positive. Each receives coefficient $W_D/d=2.5$, so the total denominator coefficient is $2(2.5)=5=W_D$. Breslow uses denominator $10$ with coefficient $5$. Efron therefore gives more weight to a smaller denominator in its second term, reflecting partial removal of the tied events' risk contribution.

The positivity check also follows from the bound: the smallest possible denominator at the last step is at least $S_0/d=10/2=5$; the actual last denominator, 6, satisfies that bound. The weights here are not equal, which is why the event count $d=2$ and the event-weight total $W_D=5$ must be kept distinct.

Frequency Weights, Diagnostics, and Limitations

A frequency weight has a different interpretation from an analytic or inverse-probability weight. If integer weights literally mean that an observation represents replicated individuals, expanding the data creates $M=W_D$ event copies in the tied group. A replication-based Efron calculation can then use $M$ removal steps, with denominators $S_0-(r/M)S_{0D}$ for $r=0,\ldots,M-1$, rather than the $d$ steps and coefficient $W_D/d$ defined here. When all tied event weights are one, $M=d$ and the expressions coincide. With unequal or noninteger weights, they need not. The analyst should identify whether weights represent replication or analytic weighting and use a tie method consistent with that interpretation.

Useful checks in an applied fit include the number and size of tied groups, the largest event weight relative to the rest of its group, the weighted event risk fraction $S_{0D}/S_0$, and whether software reports the intended tie convention. If $S_{0D}/S_0$ is close to one, later Efron denominators can be much smaller than $S_0$. Also check for zero-weight-only risk sets, nonfinite weights, and incorrect risk-set membership; a tie formula cannot repair an empty or misaligned risk set.

Neither approximation resolves why times are tied. Exact ties may arise from discrete observation schedules; ties can also result from rounding a more precise event time. If the recorded time precision changes, the tied groups and hence the fitted approximation may change. For large tied groups, substantial weight dispersion, or sensitivity of estimates to the convention, report the tie method and consider a sensitivity analysis. Breslow is simpler and often computationally convenient; Efron accounts for progressive removal and often better approximates the ordering of tied events. Neither is a substitute for correct weights, a suitable marginal structural Cox model, or appropriate uncertainty estimation.

Common pitfall: Do not replace the number of tied event cases $d$ with their total weight $W_D$ in the analytic-weight Efron formula. For one event of weight 2, $d=1$ but $W_D=2$. The Efron sum has one term, with coefficient $W_D/d=2$, and denominator $S_0$. Treating $d$ as 2 would incorrectly create a second removal step.

Check Your Understanding

Work each question, then compare with the solutions below.

A tied group has $d=3$, $W_D=6$, $S_0=12$, and $S_{0D}=6$. Give the three Efron denominators and their coefficients.
For one event with weight $0.7$, what are $d$, $W_D$, and the Efron denominator coefficient?
In a tied group, why must $S_{0D}\leq S_0$ under the stated assumptions?
Name one reason a frequency-weight Efron calculation may differ from the analytic-weight convention defined here.

Solutions. (1) The denominators are $12$, $12-(1/3)6=10$, and $12-(2/3)6=8$. Each coefficient is $W_D/d=6/3=2$. (2) There is one observed event, so $d=1$, $W_D=0.7$, and the one denominator coefficient is $W_D/d=0.7$; its denominator is $S_0$. (3) The event set is a subset of the risk set, and all terms $w_j e^{\eta_j}$ are nonnegative, so summing over the subset cannot exceed summing over the full set. (4) Integer frequency weights can represent replicated event copies, producing $W_D$ removal steps rather than $d$ fractional steps. $\square$