= Kaplan–Meier six-subject cohort
{c}
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:
\Table[
|| Time $t$ (years)
|| <Risk set> just before $t$
|| At risk $n$
|| Events $d$
|| <Right censoring> count $c$
|| Survival factor $1-d/n$
|| <Kaplan–Meier estimator> $\widehat S(t)$
| $1$
| $A,B,C,D,E,F$
| $6$
| $1$
| $0$
| $5/6$
| $5/6$
| $2$
| $B,C,D,E,F$
| $5$
| $0$
| $1$
| $1$
| $5/6$
| $3$
| $C,D,E,F$
| $4$
| $2$
| $1$
| $1/2$
| $5/12$
| $5$
| $F$
| $1$
| $0$
| $1$
| $1$
| $5/12$
]
{title=Risk sets and exact Kaplan–Meier steps for the six-subject cohort}
At year one, $\widehat S(1)=1(1-1/6)=5/6$. 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 $\widehat S(3)=(5/6)(1-2/4)=5/12\approx0.416667$. 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 $\widehat S(3)=(5/6)(3/4)=5/8$. After the event and $E$'s <right censoring>, only $D,F$ remain. At year four $n=2,d=1$, giving $\widehat S(4)=(5/8)(1/2)=5/16=0.3125$. 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 $T_F>5$.
\Image[/kaplan-meier-censored-cohort.png]
{title=Six observed histories and the effect of ties and right censoring on Kaplan–Meier steps}
The computation uses the <risk set> and event-before-<censoring> convention in https://web.stanford.edu/~lutian/coursepdf/STAT331unit3.pdf[Lu Tian's STAT331 notes, Unit 3, pages 2–6]; this six-subject example supplies its own exact counts.
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