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Let $N(t)$ count observed failures and let $Y(t)$ be the at-risk process, counting subjects under observation and still alive immediately before $t$. Under independent censoring, a short interval of length $dt$ has conditional expected failure count $Y(t)h(t)dt=Y(t)dH(t)$. The estimating equation obtained by replacing the count expectation by its observation is $d\hat{H}(t)=dN(t)/Y(t)$. Each distinct failure gives $\Delta N(a_m)=1$ and $Y(a_m)=r_m$, so the Nelson–Aalen estimator isThe same increment maximizes the local working likelihood for a Poisson distribution with mean $r_m\Delta H_m$: differentiating $\Delta N_m\log (\Delta H_m)-r_m\Delta H_m$ gives $\Delta H_m=\Delta N_m/r_m$. The counting-process derivation explains why the risk set, rather than the initial sample size, supplies the denominator.

Put $u_m=d_m/r_m$. The Taylor series of the logarithm gives the small-jump comparison of cumulative hazard estimators:Consequently, if ${\max }_m{u}_m\le \epsilon <1$, the two integrated hazard estimates obeyThey are close when every event fraction is small. In particular, with distinct single failures and large risk sets, $u_m=1/r_m$ is small. With tied data, large risk sets alone are insufficient if a substantial fraction of the risk set fails together.

At the last observed failure time $a_{\ast }$, any individual still in the risk set must either fail there or be censored there or later. If there are no such censored observations, every remaining individual fails, so $d_{\ast }=r_{\ast }$. Thus the final Kaplan–Meier estimator factor is zero, while the final Nelson–Aalen estimator jump is one:There is no small-jump approximation at this endpoint. Without ties the final risk set has size one, so the same discrepancy follows from its last single failure.

The proposed hybrid recording rule updates the event times of students who leave but does not update the follow-up of those who remain. Thus the recorded end of observation depends on the subsequent outcome: this is outcome-dependent updating of survival follow-up.

In the four-student illustration, all four are known to be event-free at month six. Under the proposed coding, the three students without event reports are censored at month six, while the student whose event is subsequently reported remains in the risk set until month seven. The recorded risk set at that event therefore has size one, although the other three may still be studying. The Kaplan–Meier estimator factor would be $1-1/1=0$ and the Nelson–Aalen estimator jump would be $1/1=1$. This creates an artificial complete loss of estimated survival at that event.

If event notification is complete through the common eight-month cutoff, the absence of a notification establishes that the other three have not had the event by that cutoff. They should remain in the risk set at month seven. With only these four subjects, the corresponding factors are instead $1-1/4=3/4$ and $1/4$. Updating only failures while backdating nonfailures' censoring creates a severely biased risk set. If notifications are incomplete, absence of a report does not establish continued study; positive follow-up confirmation is then required.

Choose a common endpoint for the eight-month analysis and ascertain every student's status through that endpoint. With a complete event register, record each departure at its event time and right-censor every remaining student at the same administrative cutoff. Otherwise obtain a confirmation for all students at the cutoff, not just the most recent quarterly list, and verify intervening departure dates consistently.

A second coherent design is to analyse only up to the last common verified quarterly date, postponing a longer analysis until a further complete follow-up round is available. Extra event reports after that date cannot be combined with earlier censoring of all nonfailures as though the resulting risk set were independently censored. Genuine losses to follow-up should be recorded separately and their possible informative censoring assessed. Use a common observation rule for failures and nonfailures, independent of their subsequent outcome.

A censoring observation means an event time is not observed exactly: the information supplies a bound or interval instead. Under right censoring, one observes $X=\min (T,C)$ and the event indicator $\delta =1_{\{T\le C\}}$; if $\delta =0$, one knows only $T>C$. Left censoring gives an upper bound, while interval censoring locates the event between two observations.

Uninformative censoring means that, conditional on the covariates used in the analysis, the censoring mechanism supplies no additional information about the event time. Independence $T\perp C$ within those covariate strata is a sufficient formulation: subjects remaining in the observed risk set have the same future hazard as the corresponding event-free population. A fixed study cutoff unrelated to prognosis is an example of administrative censoring.

Informative censoring occurs when that condition fails. For example, students at high risk of leaving may also become harder to contact before their departure is formally recorded; censoring at loss of contact selectively removes high-risk students. Ordinary Kaplan–Meier estimator and Nelson–Aalen estimator calculations then use an unrepresentative risk set. The distinction determines whether ordinary survival estimates can be interpreted as population survival. Merely knowing that an observation is censored does not establish independence.

For subject $i$, write ${\delta }_i=1_{\{T_i\le C_i\}}$, and define the counting process and pre-event at-risk process byUnder independent censoring and a common hazard, each observed event-free subject has conditional event probability $h(t)dt$ in a short interval. Therefore $\mathbb{E}(dN(t)\mid \text{past})=R(t)dH(t)$ to first order. Equivalently the compensator of a counting process is ${\int }_0^{t}R(s)dH(s)$. Dividing the observed increments by current exposure gives the estimating equation $d\hat{H}=dN/R$, henceHere $t_j$ are distinct event times, $d_j$ the number of events at that time, and $r_j=R(t_j)$ the number at risk immediately before it. One can also obtain the same increment by maximizing the local multiplicative-intensity likelihood $a^{d_j}{e}^{-r_{j}a}$: differentiation gives $a=d_j/r_j$. This derives the Nelson–Aalen estimator, rather than merely naming it.

A censored observation contributes to the denominator while under observation, and then leaves the risk set; it contributes no event jump. For tied events, use the single increment $d_j/r_j$, not $d_j$ successive one-event increments with shrinking denominators. If censoring and events share a recorded time, a convention is needed; counting the event before removing same-time censored subjects uses the pre-time risk set above. Rounded or interval-censored data may require a model suited to their observation mechanism.

For the distinct ordered observations in the question, let ${\delta }_j$ indicate whether $X_j$ is an event. The risk set at $X_j$ contains precisely the $n-j+1$ subjects whose recorded time is at least $X_j$. ThusTherefore the sum of the fitted integrated hazards at all observation times equals the observed number of events, the event-count identity for Nelson–Aalen cumulative hazards. The grouped Nelson–Aalen event-count identity extends the same argument to tied observations with their multiplicities and the stated risk-set convention.

Use $F(t)=\mathrm{Pr}(T>t)$, a nonincreasing right-continuous survivor function. The empirical likelihood treats an exact event as a probability mass rather than maximizing an unrestricted probability density function at isolated data points. Under a noninformative observation mechanism, the event-time likelihood contributions areThe second, third and fourth rows account for right censoring, left censoring and interval censoring, respectively. A censoring mechanism with unknown dependence on $T$ cannot simply be dropped; these are the event-time factors under independent censoring or another ignorable mechanism.

There is a useful finite-dimensional reduction before optimization. Let $A_i$ be the set in which observation $i$ tells us its event time lies. Partition the time axis using all observed endpoints, keeping endpoint singletons when required. Within each cell, membership in every $A_i$ is constant. Assign cell masses $w_j\ge 0$, with ${\sum }_j{w}_j=1$, including a tail cell when necessary, and let $A_{ij}$ be the cell-membership indicator. ThenMass locations within a cell are not identifiable and need not be separate parameters. Repeated observation sets can be grouped and their factors raised to their counts. Cells with the same membership column can also be combined. If one column is componentwise dominated by another, transferring its mass to the dominating column weakly increases every factor; hence an unconstrained maximizing distribution can omit the dominated cell. This proves likelihood support reduction under censoring directly. The remaining log-likelihood is a concave function of the probability masses on a simplex, often far smaller than the original unrestricted distribution problem. Added constraints can restrict which transfers remain permissible.

For right censoring alone, take a grid of all distinct observation times $s_j$ and initially allow a discrete hazardLet $d_j$ be the number of events at $s_j$ and $n_j=\#\{i:x_i\ge s_j\}$ the risk set immediately before that time. For tied recorded times, use the convention that events precede censoring, so those censored at $s_j$ remain in this risk set. The mass at $s_j$ is $q_j{\prod }_{k<j}(1-q_k)$, and the probability of surviving past $s_j$ is ${\prod }_{k\le j}(1-q_k)$. Substitution into the exact-event and right-censoring factors yieldsThe exponent $n_j-d_j$ counts all observed individuals whose information requires survival beyond that jump. In particular, censoring observations affect the risk set even though they give no event factor.

Each factor can now be maximized separately. For $0<d_j<n_j$, its derivative is $d_j/q_j-(n_j-d_j)/(1-q_j)$, giving ${\hat{q}}_j=d_j/n_j$; the second derivative is negative. The same result holds at the boundaries: ${\hat{q}}_j=0$ if $d_j=0$, and ${\hat{q}}_j=1$ if $d_j=n_j$. The Kaplan–Meier estimator is thereforeOnly event times contribute nontrivial factors. Residual mass after the last observation may be placed anywhere later without changing the empirical likelihood. A terminal plateau is therefore not evidence of permanent survival; the tail is unobserved.

For a constrained Kaplan–Meier estimator with $F(t)=p$, first split the support cells at $t$. In the probability-mass representation the exact optimization problem isThe last sum excludes a singleton mass at $t$, because $F(t)=\mathrm{Pr}(T>t)$. This is a concave maximization with linear equality constraints, so it directly gives a constrained nonparametric maximum-likelihood estimator. It also handles a time lying between observed events or beyond all follow-up.

An equivalent hazard formulation adds $t$ to the grid, permits zero-event jumps when needed, and imposesFor an interior solution using only positive-event jumps before $t$, add $\lambda \{{\sum }_{s_j\le t}\log (1-q_j)-\log p\}$ to the log-likelihood. Differentiating givesThe multiplier is found by solving the scalar equationwith denominators and fitted hazards in their admissible ranges. On an admissible event-only interior branch, this product is increasing in $\lambda$, making a one-dimensional root search straightforward.

A constraint can require mass at a time with no observed event. The event-only multiplier equation must not be used blindly. To include boundary cases, put $u_j=-\log (1-q_j)\ge 0$. Before $t$, the constraint becomes ${\sum }_j{u}_j=-\log p$, and each contribution isFor $d_j>0$ its multiplier equation again gives $q_j=d_j/(n_j+\lambda )$. For $d_j=0$, its derivative is $-n_j$, so the maximum has $u_j=0$ when $\lambda >-n_j$, and may have $u_j>0$ when $\lambda =-n_j$. Such a jump supplies mass needed by the constraint; an unobserved tail with $n_j=0$ is included by the same rule. For example, if every observation is right-censored after $t$ and there are no events, the ordinary estimate has $F(t)=1$. To impose $p<1$, put mass $1-p$ at $t$ and the remaining $p$ after all censoring times. The maximized likelihood is $p^n$, although there are no event-time factors at all. The mass-constrained optimization above automatically accounts for this possibility and for any remaining tail nonuniqueness.

The log-rank test compares event counts with their conditional expectations under equal group hazard functions, using the individuals currently in each risk set. Assume the usual independent censoring conditions and independent strata. At pooled event time $t_j$, let $n_{Aj},n_{Bj}$ be the group numbers at risk, $n_j=n_{Aj}+n_{Bj}$, and let $d_{Aj},d_{Bj}$ be the event counts with total $d_j$. Under the null hypothesis, conditioning on the risk counts and total events gives the hypergeometric distributionThus the null expected A count and its conditional variance areWhen there is only one person at risk, the event allocation is deterministic and its variance is zero. The unstandardized log-rank statistic is $U={\sum }_j(d_{Aj}-E_j)$, with variance estimate $V={\sum }_j{V}_j$. Under standard large-sample conditions, $U/\sqrt{V}$ is approximately standard normal under the null; the test detects systematic group differences in event rates. It is particularly effective for proportional-hazards alternatives, while crossing effects can cancel in the score.

For a stratified log-rank statistic, construct these risk sets and observed-minus-expected differences separately within each stratum $s$, then sum:This allows different baseline hazard functions between strata and avoids comparing an event with individuals in another stratum. The null is equal A and B hazards within each stratum, together with a censoring mechanism that does not invalidate the within-risk-set comparison.

In a paired stratum, both individuals initially form the risk set. A first event while both are still observed gives one nontrivial event allocation. After either individual leaves, every subsequent event is certain to belong to the only remaining group and has observed count equal to expected count. Therefore each pair supplies at most one informative table. A pair supplies none if the first exit is censoring before either observed event, if neither has an event during their joint follow-up, or if both fail simultaneously so that the tied event allocation is fixed. An event-censoring tie needs a specified convention; with events processed before same-time censoring, the other individual remains at risk for that event and the pair is informative. The untied continuous-time case avoids this ambiguity.

If A has the informative first event, the table at that time isThe phrase in the second column refers only to this event time; it does not mean that B can never fail. The expected A count is $1\cdot 1/2=1/2$, so this table contributes $1-1/2=1/2$ to $U$, with variance $1/4$. An informative B-first pair contributes $-1/2$. All other tables contribute zero. Therefore the requested unnormalized score isA positive score means more observed early events on A. There is no need to standardize it to answer the question.

To obtain an exact conditional law, put $m=d_A+d_B$. Under the within-pair equal-hazard null, conditional on an event while both are at risk, either subject supplies it with probability $1/2$. Equivalently, before the pair's first exit, the cause-specific event intensities for A and B are equal; integration over the common period of joint observation gives equal probabilities for the two informative outcomes. This argument requires noninformative observation of the failure process, not just equality of unadjusted marginal survival curves. Different independent pairs then have independent fair signs when informative. Conditioning on the set of informative pairs gives $m$ independent Bernoulli outcomes with probability $1/2$; conditioning only on their count mixes identical binomial laws and gives the same result:This proves the paired log-rank reduction to a sign test. The conditional mass function is $\left(\genfrac{}{}{0ex}{}{m}{k}\right)2^{-m}$, $0\le k\le m$, so an exact two-sided p-value sums the masses for $k$ with $|2k-m|\ge |d_A-d_B|$. If $m=0$, there is no information for a treatment comparison. Failure-time distances beyond their ordering play no role in this paired score.