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Put $r_j=r_j^{(0)}+r_j^{(1)}$. The difference between the two Nelson–Aalen estimator increments isTherefore the log-rank weightsmake each summand of $T_W$ equal the corresponding summand of $T_0$, and hence $T_W=T_0$.

Let $u_j^{(1)}=u_j$ and $u_j^{(0)}=1-u_j$. Applying the Nelson–Aalen estimator separately to group $k$ giveswith a zero contribution when the event at $a_{\ell }$ occurs in the other group.

With a common excess hazard, let ${\overline{h}}_B={\sum }_i{Y}_i{h}_B^{(i)}/{\sum }_i{Y}_i$ be the known background hazard averaged over the current risk set. Estimate the excess cumulative hazard by subtracting $\int {\overline{h}}_B$ from the Nelson–Aalen estimator. Applying the product-limit construction gives the displayed estimator: the continuous background correction multiplies the ordinary Kaplan–Meier estimator. Its event jump is $1-d_j/r_j$, while between events it grows at rate ${\overline{h}}_B{\hat{F}}_E$. Hence the unconstrained curve can increase, even if the true excess hazard is nonnegative. Replacing its jump factor by $\exp (-d_j/r_j)$ gives the alternative exponential cumulative-hazard estimator; these versions are close for small event fractions but are not identical.

At a failure time with $d$ events among $r$ at risk, the Nelson–Aalen estimator adds $u=d/r$, whereas minus the logarithm of the Kaplan–Meier estimator adds $-\log (1-u)$. For $0\le u<1$,Thus small event fractions give close cumulative hazard estimates. Large risk sets alone do not suffice when a large fraction fail together. If all remaining individuals fail, the logarithmic estimate becomes infinite while the Nelson–Aalen jump is one.