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When cohorts share one time-since-entry survival distribution, their observations can be pooled using the Kaplan–Meier estimator, even when later cohorts have shorter administrative follow-up. A short-follow-up cohort still improves estimation of early survival, which is a factor in survival at later times. Pooling requires comparable survival across cohorts and independent censoring; calendar-period or cohort effects can invalidate that assumption.

The Kaplan–Meier estimator constrained to have survivor function value $p\in (0,1)$ at $t$ maximizes the right-censoring log-likelihood ${\sum }_j[d_j\log q_j+(n_j-d_j)\log (1-q_j)]$ subject to the displayed constraint. Here $q_j$ is the discrete hazard, $d_j$ the event count and $n_j$ the risk set size. In an interior event-time solution, a Lagrange multiplier gives $q_j=d_j/(n_j+\lambda )$ before the constrained time, with $q_j=d_j/n_j$ afterwards. The multiplier is chosen to satisfy the product constraint. It is essential to allow a zero-event jump at the constrained time or another relevant cell boundary: a constraint can require mass outside observed event times. A general implementation instead maximizes the concave probability-mass log-likelihood with total mass one and mass above $t$ equal to $p$, allowing all relevant cells and a terminal tail. This also handles unobserved tails and boundary solutions without imposing unjustified event-only support.

A Cox–Snell residual is the fitted cumulative hazard function evaluated at an observed event or censoring time. Under a correct model, complete transformed event times are unit exponential; censored transformed times retain their event indicators. A logarithmic Kaplan–Meier estimator survivor plot should be close to a line of slope minus one.

For a correct fitted survival law, complete Cox–Snell residuals are approximately unit exponential. Residuals computed at observed censored times retain their censoring indicators. Fit a Kaplan–Meier estimator to the transformed times and indicators and compare it with $e^{-r}$, or compare their Nelson–Aalen estimator cumulative hazard with $r$. Treating censored transformed times as complete exponential observations gives a distorted Q–Q check.

Suppose $T\ge 0$ has an absolutely continuous probability distribution, with survival function $S(0)=1$. On an interval where $S(t)>0$, its cumulative hazard isHere $h$ is the hazard function; its integral is dimensionless because the units of rate and time cancel. The identity follows by integrating the derivative of $\log S$, so mere continuity of the survival function is not enough: the absolute continuity assumption supplies a probability density function. If $S$ reaches zero, interpret $-\log S$ as $+\infty$. The normalization $S(0)=1$ also matters; an initial probability mass at zero would require an initial term $-\log S(0)$.

Accumulating hazard functions adds on the $H$ scale and multiplies survival factors on the $S$ scale. For $S(a)>0$ and $b>a$,For instance, a cumulative hazard increment of $2$ gives event probability $1-e^{-2}\approx 0.864665$, rather than a probability of $2$. With $h(t)=0.2$ per year, $H(5)=1$ and $S(5)=e^{-1}\approx 0.367879$. With the increasing hazard function $h(t)=0.04t$ per year when $t$ is expressed in years, $H(t)=0.02t^2$, so $H(5)=0.5$ gives $S(5)=e^{-0.5}\approx 0.606531$. Its coefficient has units of inverse years squared.

The constant hazard function produces an exponential distribution; the increasing example is a Weibull distribution. Their survival functions both decrease, although one hazard function stays constant and the other increases. At year ten both have $H(10)=2$ and $S(10)=e^{-2}$ despite different earlier histories. The accumulation identity is described in Lu Tian's Stanford notes, Section 1.1.

For a discrete hazard $q_j$ at each possible event time $t_j$, the exact counterpart is $S(t)={\prod }_{t_j\le t}(1-q_j)$. Taking its negative logarithm sums $-\log (1-q_j)$, rather than $q_j$. Therefore exponentiating a Nelson–Aalen estimator does not give exactly the Kaplan–Meier estimator; small-jump comparison of cumulative hazard estimators quantifies the difference.

At ordered possible failure times $t_j$, the discrete hazard is $h_j=\mathbb{P}(T=t_j\mid T>t_{j-1})$. Its survival function satisfies $S(t_j)=S(t_{j-1})(1-h_j)$, and hence $S(t)={\prod }_{t_j\le t}(1-h_j)$. Estimating $h_j$ by the observed number of failures divided by the risk set size under independent censoring yields the Kaplan–Meier estimator. Unlike a continuous-time hazard rate, a discrete hazard is a probability between zero and one.

If a censored individual is allocated event weights $q_j$ over later event times, its event-count contribution at time $t_j$ is $q_j$ and its risk set contribution is ${\sum }_{k\ge j}{q}_k$. These fractional counts can be used in the Kaplan–Meier estimator.

For the Kaplan–Meier estimator, the estimated variance isThe formula follows by summing approximate variances of the logarithms of the conditional survival factors and applying the delta method to their product. Its square root is the estimated standard error. The same expression uses the actual delayed-entry risk sets for the Kaplan–Meier estimator with delayed entry. When $0<\hat{S}(t)<1$ and every contributing $r_j>d_j$, an approximate log-scale confidence interval is $\exp \{\log \hat{S}(t)\pm z_{1-\alpha /2}\widehat{\mathrm{se}}(\hat{S}(t))/\hat{S}(t)\}$, with the upper endpoint capped at one. This interval can be inaccurate with very small risk sets.

A simple sufficient independent censoring assumption is that the potential observation-ending time $C$ and survival time $T$ are independent random variables. More generally, appropriate conditional independence given the covariates used in the analysis can suffice for analysis within those conditions. The requirement concerns the potential times, rather than independence of the observed duration $Y=\min (T,C)$ and event indicator $\delta$: those recorded quantities ordinarily depend on each other.

For an ordinary pooled Kaplan–Meier estimator, the observed risk set must represent the future event experience of everyone still event-free at each time. Conditional independence given a covariate does not alone justify ignoring that covariate in a pooled calculation. Also, there must be a positive probability of remaining observed through the horizons being estimated. If follow-up ends at a common closing time, it supplies no observations to identify the later survival function. Independent subjects with a common survival distribution and independent censoring form the basic model in Lu Tian's STAT331 notes, Unit 3, pages 1–4.

For example, subjects who stop attending because their condition is deteriorating can produce informative censoring. A predetermined common closing date can instead give administrative censoring, provided calendar entry and other relevant conditions do not select future survival times. The reason for stopping observation must be justified from study design and context; the recorded event indicators alone cannot verify independent censoring.

Informative censoring occurs when censoring carries information about the future event time beyond the variables conditioned on in the analysis. It can invalidate ordinary Kaplan–Meier estimator and survival-model residual interpretations.

At distinct observed event times $t_1<t_2<\cdots$, let $n_j$ be the risk set size immediately before $t_j$, and let $d_j$ count events at $t_j$. The Kaplan–Meier estimator of the survival function isThe empty product is one. Each factor estimates a conditional survival probability among those still observed and event-free; their product joins the successive steps. This is nonparametric statistics: no specified shape for the survival distribution is imposed. Use independent censoring and the appropriate risk sets to interpret the result as an estimate for the intended population.

The Kaplan–Meier estimator is a right-continuous function with downward jumps at event times. Right censoring changes later denominators without causing a jump. A tick on the curve marks right censoring, rather than a failure. For a tied group of $d_j$ events, use the one factor $(n_j-d_j)/n_j$; subjects right-censored at that recorded time remain in $n_j$ for this factor. The fraction $d_j/n_j$ estimates a discrete hazard, not a continuous hazard function in inverse-time units. With no censoring, the factors telescope to the fraction of the original cohort whose observed survival times exceed $t$, the complement of its empirical distribution function.

Edward L. Kaplan and Paul Meier published ^[ref]. Their paper addresses lifetime estimation when observation ends before failure and explicitly calls attention to the assumption of independence between lifetime and potential observation loss. The date 1958 is the original publication year; the publisher's later online-posting date is not the discovery date.

In his ^[ref], Kaplan traces Meier's interest to cancer-duration data in 1952 and his own to the lifetimes of vacuum tubes in undersea telephone-cable equipment a year later. The journal encouraged them to combine their manuscripts. The shared mathematical question was how to estimate a survival function when some lifetimes are still running or disappear from observation.

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.

The median survival estimate is the first event time at which the Kaplan–Meier estimator reaches or falls below one half. It is not identified beyond the observed follow-up if the curve never reaches that level. A few reported survival probabilities bound the crossing time but do not determine an exact median. Confidence intervals can be constructed by inverting confidence bands for the survivor function.

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 a fixed cohort with $r$ individuals at risk just after $a$, no censoring or entry inside $(a,b)$, and $d$ events in $(a,b]$, the Kaplan–Meier estimator factors telescope to the displayed ratio. A tied group of $d_j$ events contributes $(r_j-d_j)/r_j$, the same product as any sequential ordering without intervening censoring. Events at $b$ are counted before censoring at the same time under the usual convention.

For cause-specific hazard ${\lambda }_k$, the net survival is $S_k^{\mathrm{net}}(t)=\exp [-{\int }_0^t{\lambda }_k(u)du]$. Censoring all competing events in a cause-specific Kaplan–Meier estimator targets this quantity under appropriate observation-censoring assumptions. Its complement generally exceeds the actual cumulative incidence function ${\int }_0^{t}S(u){\lambda }_k(u)du$, because all-cause survival $S(u)$ also includes the other hazards. Identifying net survival with survival after a hypothetical elimination of competing events requires extra causal assumptions.

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.