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Consider six subjects observed from time zero. Recorded survival data are $A:(1,1)$, $B:(2,0)$, $C:(3,1)$, $D:(3,1)$, $E:(3,0)$, and $F:(5,0)$, with each pair giving $(Y,\delta )$ in years. Thus $A,C,D$ have observed events, while $B,E,F$ are right-censored. The letters label subjects, not treatment groups. Here are all exit times, including rows without events:

At year one, $\hat{S}(1)=1(1-1/6)=5/6$. Removing $B$ at year two leaves the Kaplan–Meier estimator unchanged. Immediately before year three, $C,D,E,F$ are still in the risk set: the two events give $\hat{S}(3)=(5/6)(1-2/4)=5/12\approx 0.416667$. Only afterwards do $C,D,E$ leave, leaving $F$ alone. Removing $E$ before counting the tied events would incorrectly use $n=3$ and give $(5/6)(1-2/3)=5/18$. The event-first convention is part of the recorded-time interpretation; coarsely rounded timestamps do not reveal a hidden ordering within that time bin.

The exact Kaplan–Meier estimator is $1$ on $[0,1)$, $5/6$ on $[1,3)$, and $5/12$ on $[3,5]$. The three right censoring ticks are $(2,5/6)$, $(3,5/12)$, and $(5,5/12)$; at a shared event time the tick is on the post-event step. Beyond year five the risk set is empty, so the horizontal endpoint does not determine later survival. The Kaplan–Meier median survival time is three years, since this is the first step at or below $1/2$.

Now separate the tied events by changing only $D$'s recorded event time from year three to year four. At year three the risk set still has four subjects, but $d=1$, so $\hat{S}(3)=(5/6)(3/4)=5/8$. After the event and $E$'s right censoring, only $D,F$ remain. At year four $n=2,d=1$, giving $\hat{S}(4)=(5/8)(1/2)=5/16=0.3125$. The intervening right censoring prevents the factors from telescoping to the original tied-event factor. If $E$ instead remained observed past year four, the factors would be $(3/4)(2/3)=1/2$, giving the original $5/12$ after both events. This is Kaplan–Meier telescoping across a censor-free interval in numbers.

Incorrectly counting every right-censored exit as an event estimates the distribution of observation-ending times $Y$, rather than survival times $T$. For the original tied cohort, the spurious curve is $5/6$ at year one, $(5/6)(4/5)=2/3$ at year two, $(2/3)(1-3/4)=1/6$ at year three, and zero at year five. That zero asserts a failure for $F$, whose only known information is $T_F>5$.

The computation uses the risk set and event-before-censoring convention in Lu Tian's STAT331 notes, Unit 3, pages 2–6; this six-subject example supplies its own exact counts.

The survival estimates favor the new treatment at the two reported times, with absolute survival differences $0.199$ at two years and $0.157$ at five years. These point differences alone do not establish efficacy. Randomization supports a causal comparison of the treatment assignments, but sampling variance, censoring, follow-up, protocol adherence and the prespecified endpoint still matter. One should not infer a statistically established benefit merely from two larger estimates.

A Kaplan–Meier estimator iswhere $r_j$ is the risk set and $d_j$ the event count. Its uncertainty depends on the actual risk sets, not just the initial group size. The Greenwood formula estimatesWithout the missing survival tables one cannot calculate these standard errors or the uncertainty of the group difference. In particular, replacing the estimator variance by the variance of an uncensored binomial proportion would generally be wrong.

I would advise presenting both curves with numbers at risk and confidence intervals, checking independent censoring, and reporting a prespecified overall comparison plus an effect estimate and interval. For an unadjusted log-rank test, write $O_1={\sum }_j{d}_{1j}$ and $E_1={\sum }_j{d}_j{r}_{1j}/r_j$. Its hypergeometric null variance isTerms with no possible variation contribute zero. This compares the entire event history rather than selecting two arbitrary times, uses the available risk sets, and avoids choosing favorable time points after examining the curves. It is especially effective under proportional hazards. Crossing curves can make a log-rank comparison insensitive; a prespecified restricted mean survival time difference or another suitable time-varying analysis may then be more informative. An overall comparison does not make a clinically important prespecified time-point contrast invalid.

The Kaplan–Meier median survival time is the first event time at which $\hat{S}(t)\le 1/2$. Monotonicity of the reported estimates impliesIf the new-treatment curve never reaches one half during observed follow-up, its median is not estimable from that curve. The exact crossing times cannot be reconstructed from two survival probabilities: the standard curve could first cross at three or at four years and still have the same values at two and five years. The point estimates suggest benefit, but the evidence and exact medians require the missing event/censoring tables. Neither a conclusive agreement nor a conclusive rejection of benefit is justified from these numbers alone.