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For independently left-truncated and right-censored survival data, let $E_i$ and $Y_i$ be entry and exit times on the same clock. At each distinct event time $t_j$, the risk set is $R_j=\{i:E_i<t_j\le Y_i\}$, with size $r_j$, and $d_j$ events occur. The Kaplan–Meier estimator isBefore entry a subject contributes neither an event nor time at risk. Thus adding observed entrants can increase successive risk sets; pretending that all subjects were observed from time zero distorts the estimated hazard function. The estimator identifies the survival function over ages covered by observation; normalization from birth requires observation arbitrarily close to time zero or additional knowledge of earlier survival.

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.

For untied events at $a_j$ with predictable risk set size $Y_j$, the estimated predictable variation of the Nelson–Aalen estimator is ${\sum }_{a_j\le t}{Y}_j^{-2}$. It comes from the martingale event increment variance $YdH$ and the estimator increment $dN/Y$. It applies on the range with individuals at risk; it is not an exact finite-sample unbiasedness statement.

Suppose individual event hazards have the additive form $h_i=h_{0i}+h_1$, where $h_{0i}$ is known and $h_1$ is common. The aggregate counting-process intensity in survival analysis is $\lambda ={\sum }_i{Y}_i{h}_{0i}+Y{h}_1$, with $Y={\sum }_i{Y}_i$ the current risk set size. Writing $dN=\lambda dt+dM$ and dividing by $Y$ givesThe last term is a counting-process martingale integral. This establishes the cumulative common-hazard target on the observed at-risk time range. The known offset must be averaged over the current risk set, not the initial sample. Between event jumps, its subtraction can make the unconstrained estimate decrease. If all known hazards equal $h_0$ and $Y>0$ up to $t$, the estimate reduces to the Nelson–Aalen estimator minus $H_0(t)$.

If a cohort is periodically confirmed event-free but exact event reports update only those who fail, censoring the remaining subjects at their last scheduled confirmation creates informative censoring. Between confirmations the nominal risk set retains future observed failures and removes nonfailures. An apparent final failure may therefore empty the recorded risk set even though other cohort members remain event-free. Use a common verified cutoff or a complete event register with a common administrative censoring date.

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.

Under independent censoring, order the distinct observed failure times as $t_1<\cdots <t_m$. Let $r_j=\#\{i:x_i\ge t_j\}$ be the size of the risk set just before $t_j$, and let $d_j=\#\{i:x_i=t_j,v_i=1\}$ be the number of failures there. The estimated conditional probability of surviving that event time is $1-d_j/r_j$. Multiplying these conditional survival probabilities gives the Kaplan–Meier estimatorA censored observation leaves the risk set after its censoring time but does not create a downward survival jump. With tied failure and censoring times, the displayed risk-set convention includes those censored at the time while processing failures, then removes them.

Since $H=-\log S$, the Kaplan–Meier estimator of the cumulative hazard function isThis is not exactly the Nelson–Aalen estimator ${\sum }_{t_j\le t}{d}_j/r_j$, although the two are close when the individual fractions are small. If a jump exhausts the risk set, $\hat{S}$ becomes zero and ${\hat{H}}_{\mathrm{KM}}$ becomes infinite; transformed residuals beyond that point require a finite-tail modelling convention or restriction of follow-up.

Let $T_i$ be an event time and $C_i$ a potential censoring time. Under independent censoring, observe $Y_i=\min (T_i,C_i)$ and ${\delta }_i=1\{T_i\le C_i\}$. For an exponential distribution of rate $\theta$, an observed failure contributes $\theta e^{-\theta Y_i}$ to the survival likelihood, whereas a right-censored observation contributes $e^{-\theta Y_i}$. The censoring-law factors may be discarded when they are independent of $\theta$. Thus, writing $d={\sum }_i{\delta }_i$ and $V={\sum }_i{Y}_i$ for total observed person-time,For $d,V>0$ the unique maximum-likelihood estimate isCensored times belong in $V$: they record time during which failure was not observed. With $d=0$ and $V>0$, the likelihood decreases with $\theta$, so the supremum is at the boundary $\theta \downarrow 0$; there is no positive interior estimate. This is exponential-rate estimation from censored exposure.

Uninformative censoring means that the potential censoring time supplies no information about the latent failure time after conditioning on the modeled variables. A study ending at a fixed follow-up duration independent of prognosis is an example of administrative censoring. Informative censoring occurs, for example, when a seriously deteriorating patient withdraws because of that deterioration and the prognostic information is omitted from the model. Under informative censoring, those remaining in the risk set are selected by their future failure propensity, so the ordinary survival likelihood and its estimate need not target the true rate.

Using the reported durations and flags as though they were valid right-censored records givesTreating all durations as failures instead givesRelabelling has halved the event count without shortening any observed follow-up. The resulting estimate is therefore exactly half the complete-data estimate.

This is not a simulation of uninformative ordinary right-censoring. For a record declared censored at $Y_i$, ordinary right censoring requires $T_i>Y_i$; the construction instead took $Y_i$ to be the simulated failure time itself. Keeping a failure time and hiding only its flag does not produce the minimum of that time and an independent earlier censoring time. Even choosing labels at random does not fix this: independence of a flag from the recorded duration is different from independent censoring of a latent event time. Indeed the labelled durations still have mean one while the failure fraction is one half, so their ordinary censored estimate converges to $1/2$, not the generating rate one. Thus it is not a valid uninformative censoring scheme for the intended failure distribution; strictly, it has not implemented ordinary right-censoring of those simulated failure times at all. One could reproduce the randomly labelled record distribution with independent latent event and censoring times both of rate $1/2$, but that would be a different failure model, not the claimed rate-one model. Calling the flags random is therefore insufficient to justify the target distribution or its censored likelihood. This is the random relabelling is not independent right censoring issue.

A valid simulation draws independently $T\sim \mathrm{Exp}(1)$ and $C\sim \mathrm{Exp}(1)$, and records $(\min (T,C),1\{T\le C\})$. Sinceit gives independent censoring with the requested probability. Equivalently generate $T=-\log U$ and $C=-\log V$ from independent uniform random numbers. Another valid choice is fixed censoring at $\log 2$, since $\mathrm{Pr}(T>\log 2)=1/2$. These give a random censored fraction with expectation one half; they do not promise exactly 500 censored records. For the independent-exponential scheme, observed durations have rate two and mean one half, so $d/\sum Y_i$ correctly converges to one.

For the numerical competing risks update, the estimated probability of remaining event-free just before the tied time isThere were no intervening events, so the cumulative incidence functions did not jump between the two stated times. The Aalen–Johansen estimator uses the risk set immediately before the event time, including the subject censored at that same time under the usual events-before-censoring tie convention. Therefore the cause-$A$ increment is $0.474(2/20)$, andFor a consistency check, ${\hat{F}}_B(7.8)=0.241+0.474/20=0.2647$ and $\hat{S}(7.8)=0.474(1-3/20)=0.4029$; these sum to one. The censoring does not create an incidence jump and leaves $16$ subjects for subsequent follow-up. Dividing by $19$ would instead assume that the censored subject had already left before the event time, contrary to the stated immediately preceding risk set. This is a cumulative incidence update with tied censoring.

Collect distinct observed failure times as $a_m$, with $d_m$ failures and $r_m$ individuals in the risk set just before $a_m$. The grouped Nelson–Aalen estimator isAll failures recorded at one time use the same pre-event risk set. When failures and censoring share a recorded time, the usual convention includes those censored at that time in the risk set and processes failures before removing censored individuals. A known different ordering should instead be honored. If ties arise solely through rounding of a continuous-time process, an alternative is to resolve the unobserved ordering: $d_m$ successive failures without intervening censoring would contribute ${\sum }_{s=0}^{d_m-1}(r_m-s{)}^{-1}$. That is not the grouped jump $d_m/r_m$; the difference represents a choice about what the recorded ties mean.