 Kaplan–Meier six-subject cohort

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.