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If a counting process $N$ has predictable conditional intensity $\lambda$, subtracting its compensator of a counting process gives the associated martingale $M$. Conditional expected event increments and compensator increments agree, so their difference is conditionally centered. With the necessary integrability, $M$ is a mean-zero martingale; for a simple no-ties process its predictable quadratic variation is ${\int }_0^t\lambda (u)du$. A predictable weighted martingale integral has mean zero and variance determined by the square of its weights. In survival analysis, $\lambda =Yh$ produces the event-minus-expected representation used to derive the Nelson–Aalen estimator.

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.

With all subjects entering at time zero and no tied observation times, the Nelson–Aalen estimator satisfies ${\sum }_i\hat{H}(x_i)=d$. Exchanging sums makes each event increment $1/Y(t_j)$ appear in exactly $Y(t_j)$ subject hazards. The identity relies on the risk-set count equalling the number of observation times at least $t_j$, and generally needs modification for delayed entry.

Even with a common event hazard, the exact expectation of the Nelson–Aalen estimator is ${\int }_0^t\mathbb{P}(Y(u)>0)h(u)du$, because no increments can be observed after the risk set is empty. Replacing ratios by ratios of expectations gives an approximate target and does not prove finite-sample unbiasedness. For one uncensored exponential event time, the estimator has mean $1-e^{-\lambda t}$ rather than $\lambda t$.

Suppose everyone enters observation at zero and has one event or censoring time. At distinct recorded time $t_j$, let $n_j$ observations terminate, of which $d_j$ are events, and use the pre-time risk set size $r_j={\sum }_{k\ge j}{n}_k$. For the grouped Nelson–Aalen estimator, $\hat{H}(t_j)={\sum }_{k\le j}{d}_k/r_k$. Exchanging sums gives ${\sum }_j{n}_j\hat{H}(t_j)={\sum }_k(d_k/r_k){\sum }_{j\ge k}{n}_j={\sum }_k{d}_k$. The unweighted sum over distinct times generally does not have this property. Processing each observation separately after breaking ties also restores the corresponding unweighted identity on the resulting artificial time grid.

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 $H_0(t)$.

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.

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.

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.

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.

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 the standard right censoring setup with everyone entering observation at time zero. At each observed time $x_j$, the pre-exit risk set has size $r_j=\#\{i:x_i\ge x_j\}$. The Nelson–Aalen estimator of the cumulative hazard function isCensoring removes an individual from later risk sets but contributes no event increment.

Exchange the two finite sums:Each event increment appears in exactly as many subjects' fitted cumulative hazards as there were subjects at risk for that increment. This proves the event-count identity for Nelson–Aalen cumulative hazards. Common entry is relevant: with delayed entry, a subject does not accumulate the increments before entering, so an entry-time subtraction would be needed.

One method for ties is to group observations at each distinct recorded time. If $d_k$ events occur at ${\tau }_k$, use the pre-time risk set size $R_k$ including observations censored at that same recorded time, and take one jump $d_k/R_k$. This is the usual events-before-censoring convention within a recorded tie. Since $R_k=\#\{i:x_i\ge {\tau }_k\}$,Thus the grouped Nelson–Aalen event-count identity preserves the required identity at the recorded observation times. It is a sum over individuals, not an unweighted sum over distinct times.

A second method, suitable when ties reflect rounded times, is to break ties by small perturbations or an artificial ordering, then apply the untied estimator. The event-count identity holds on that new, untied time grid. It does not generally persist if all estimates are mapped back to the original common time. For example, two tied failures among two individuals give a grouped jump $2/2=1$, so the original-time sum is $2$. Sequentially breaking the failures gives jumps $1/2$ and $1$, whose total is $3/2$; assigning the completed jump to both original tied observations gives sum $3$. On the perturbed grid the two values are $1/2$ and $3/2$, whose sum is again $2$. This distinction states exactly what the second method does and does not preserve.

Another common grouped survival approach is the Kaplan–Meier estimator, from which one may estimate cumulative hazard by $-\log \hat{S}(t)$. Its jump is $-\log (1-d_k/R_k)$, not $d_k/R_k$, and it likewise does not have the Nelson–Aalen event-count identity. In particular it becomes infinite at a time when all remaining individuals fail.

Let $N(t)$ count observed events and $Y(t)$ count individuals still in the risk set just before $t$. Under a common hazard function $h(t)$ and independent censoring, the expected event count over a short interval, conditional on the observed past, isHere $H(t)={\int }_0^{t}h(u)du$ is the integrated hazard. Thus an estimating increment for $dH$ is $dN/Y$, on the range where the risk set is nonempty. Accumulating these increments gives the Nelson–Aalen estimatorThe observation times $t_j$ include both events and censorings, $r_j=Y(t_j)$, and ${\delta }_j$ is one for an event and zero for censoring. Without ties each event adds $1/r_j$; censoring changes later risk sets but adds nothing immediately. Equivalently, the local event fraction estimates the local hazard increment, and summing those estimates produces the integrated hazard. The counting-process intensity in survival analysis justifies this estimating equation; it does not imply exact finite-sample unbiasedness after the risk set can become empty.

Assume the usual cohort setup: all individuals enter at time zero, each contributes one event or censoring time, and every such observation is included. Order these distinct observation times as $t_1<\cdots <t_m$. The risk set size just before $t_k$ is then $r_k=m-k+1$. Write ${\delta }_k=1$ for an event and zero for censoring. Including the jump at an event time, the Nelson–Aalen estimator isExchange the two finite sums:Thus the sum equals the number of observed events. Each event increment appears once for each individual still at risk at that event, exactly cancelling its denominator. This is the event-count identity for Nelson–Aalen cumulative hazards. The cohort and entry convention matter: delayed entry would invalidate the simple equation $r_k=m-k+1$ and require a modified identity.

One method breaks a tied group into a sequence of distinct artificial times, by a specified or random ordering. Process tied events before tied censorings if their common recorded time means the censored individuals were still under observation when the events occurred. At each artificial event, add the reciprocal of the current risk set size and then remove that individual. This applies the untied Nelson–Aalen estimator to the resulting grid. For example, two tied failures among two individuals give successive increments $1/2$ and $1$.

On that expanded grid there is one observation time per individual. The cancellation argument in (b) is unchanged, so breaking ties preserves the identity when the sum includes every artificially separated observation. The chosen ordering can affect the estimated curve within a tied group and must not be presented as an observed event order.

A second method retains the recorded tied times. For a group with $d_j$ events and pre-time risk set size $r_j$, the grouped Nelson–Aalen estimator adds $d_j/r_j$ once. All events in the group use the same denominator; censorings in the group are removed afterwards. If two individuals fail at the same time and there are no other observations, the curve has one jump of $2/2=1$. Its sum over the one distinct time is $1$, whereas the event count is $2$. Hence the unweighted distinct-time identity does not generally survive grouping ties. On the artificially separated grid in the first method, the two post-observation values would instead be $1/2$ and $3/2$, whose sum is $2$.

There is nevertheless an exact weighted version for the grouped method. Let $n_j$ count all observations, event or censoring, at distinct time $t_j$. With everyone entering at zero, $r_k={\sum }_{j\ge k}{n}_j$. ThereforeThis grouped Nelson–Aalen event-count identity sums over individuals, with repeated values for tied times, rather than once per distinct time. Keeping that distinction resolves the apparent conflict between a valid grouped estimator and failure of the literal unweighted formula.