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Right censoring occurs when a subject's exact failure time $T$ is unobserved but it is known that $T>C$, where $C$ is the censoring time. The observed pair is $U=\min (T,C)$ and $\Delta =1_{\{T\le C\}}$. For example, a living participant at study closure contributes a survival lower bound rather than a recorded death at closure. Right censoring differs from simply omitting the subject: time spent in the risk set remains informative.

Let $p_{rs}(t)=P(X(u+t)=s\mid X(u)=r)$ and $P(t)=e^{Qt}$, where time homogeneity removes dependence on $u$. A recorded state at the next clinic visit contributes a transition probability; an exactly observed death contributes a statistical probability density, not the probability of being dead at that time. If the last recorded living state is $r$, the mixed panel and exact-death likelihood factor after an interval $t$ isThis sums over the unobserved living state immediately before death.

Conditioning on the recorded initial states, the contribution of the three displayed patient histories isIn this irreversible illness-death model, $p_{21}=0$, $p_{11}(t)=e^{-(a+b)t}$ and $p_{22}(t)=e^{-ct}$, simplifying it toThe factor $c$ is essential: the death time is known exactly. Replacing the final statistical probability density by $p_{23}(34.6)$ would instead model interval observation of death and give a different likelihood.

The assumptions are independent patient histories with common rates; the Markov property; constant rates over calendar/follow-up time in this model; the stated absence of recovery and absorption at death; accurate state labels and death times; and an observation/follow-up mechanism that is noninformative for the latent process given the observed history. Clinic dates are conditioned on. The displayed living endpoints contribute only the shown observations, with noninformative right censoring if they are follow-up endpoints. Initial state probabilities are omitted by conditioning on them. Progression between visits can be unobserved, which is precisely why panel-observed multi-state likelihood uses the matrix exponential rather than assuming a transition occurs at a visit.

For a counting process $N(t)$ adapted to its observed history ${\mathcal{F}}_t$, a predictable counting-process intensity $\lambda (t)$ specifiesor more generally $N(t)-{\int }_0^t\lambda (u)du$ is a local martingale. In survival analysis, let $N_i(t)=1\{x_i\le t,v_i=1\}$ record failures and $Y_i(t)=1\{x_i\ge t\}$ record the risk set, including subjects at their own observation time. Under independent right censoring and a common hazard function $h$, the aggregate intensity is $Y(t)h(t)$, where $N={\sum }_i{N}_i$ and $Y={\sum }_i{Y}_i$.

Writing $H(t)={\int }_0^{t}h(u)du$, the conditional increment equation is $E(dN\mid {\mathcal{F}}_{t-})=YdH$. Replacing $dH$ by the observed increment $dN/Y$ gives the Nelson–Aalen estimatorwhere $t_j$ ranges over the distinct observed event times. Censored observations remove subjects from subsequent risk sets but do not produce hazard jumps.

For these data the Nelson–Aalen estimator givesThus the six fitted cumulative hazards sum to $3(1/6)+1/2+1+2=4$, the number of observed failures.

The event-count identity for Nelson–Aalen cumulative hazards follows by exchanging the finite sums. With everyone entering at time zero,This includes each failure subject in its own risk set. A delayed-entry dataset requires a different risk indicator and is not covered by this particular identity.

Under a constant hazard survival model, $\hat{H}(x_i)=\hat{\theta }{x}_i$. Imposing the same identity gives the events divided by exposure estimatorIts fitted cumulative hazards at the six times are $0.3,0.4,0.5,0.6,0.9,1.3$, again summing to four. It is also the maximum-likelihood estimate under independent right censoring, since the parameter-dependent likelihood is ${\theta }^d{e}^{-\theta {\sum }_i{x}_i}$. This is a sensible estimate if the exponential survival model is appropriate, but four failures give little precision. The event-count identity by itself does not validate constant hazard; the large final jump in the Nelson–Aalen estimator also reflects a risk set of one, rather than by itself proving an increasing hazard.

For a proper continuous event time, the survival function is $S(t)=e^{-H(t)}$. The assumed invertibility of the cumulative hazard function gives, for $u\ge 0$,ThereforeThis is the cumulative hazard probability transformation; it applies also conditionally on a subject's covariates, using that subject's correctly specified cumulative hazard function.

The fitted transformed times are Cox–Snell residuals. They retain their event/right censoring indicators, so a right-censored residual represents an exponential observation known only to exceed its displayed value. Under the fitted model and independent right censoring, calculate the Kaplan–Meier estimator of residual survival and compare it with $e^{-y}$, or calculate the residual Nelson–Aalen estimator and compare its cumulative hazard with the diagonal $H(y)=y$. Systematic departures reveal model inadequacy; sparse extreme residual risk sets and parameter estimation require caution. Treating all censored residuals as observed event times would invalidate this diagnostic.

Without right censoring, the residual mean should be approximately one. With right censoring, use the modified Cox–Snell residualFor a true unit-rate exponential distribution, the memoryless property gives $E(U\mid U>c)=c+1$. Thus an event keeps its known transformed time, while a censored observation is replaced by the conditional expected event time. Under independent right censoring, iterated expectation makes the mean of these adjusted values one when the true hazards are used, and approximately one when fitted hazards are used. These mean-imputed values do not themselves have an exponential distribution, so the survival-curve diagnostic should still use the original censored residual dataset. Moreover fitting equations can force the adjusted sample mean to one, making its mean alone a weak diagnostic.

For the proposed mixture of a finite right censoring time $c\ge 0$ and no right censoring, $P(C<U)=\pi e^{-c}$ andConsequentlyWhen right censoring has positive probability this choice is unique; if $\pi e^{-c}=0$, no correction is needed and any $k$ has the same effect. Its independence from $c$ is the content of exponential memorylessness: the expected extra lifetime after any right censoring time is one. Conditioning on an arbitrary independent right censoring time proves the same correction beyond this special two-point mixture.

For an event at $t_j$ from subject $i_j$, with covariate vector $z_i$, define the Schoenfeld functionThe second term is the hazard-weighted mean covariate in the risk set just before the event. The Schoenfeld residual is this function evaluated at the fitted coefficient, $r_j=s_j(\hat{\beta })$. Calculate one residual vector per event, using every at-risk subject, including those who will subsequently be censored. There is no ordinary event residual assigned at a right censoring time.

The Cox partial likelihood score function is ${\sum }_j{s}_j(\beta )$. At the true constant coefficient in a Cox proportional-hazards model, the conditional event subject is selected with weights proportional to $e^{{\beta }^T{z}_i}$, so each Schoenfeld function has conditional mean zero. If the coefficient varies with time, that centering changes. Plot residuals against event time or a transformation of it, smooth them, and investigate departures from zero. Scaled Schoenfeld residuals account for the risk-set covariate variance and can display departures in coefficient units; score function tests based on residual-time association provide a formal check. Risk-set composition affects unscaled residual variance, and the total residual score function can be zero by fitting even when a time trend is present.

Use the Aalen–Johansen estimator of the cumulative incidence function, updating the overall Kaplan–Meier estimator for either event:Right censoring changes subsequent risk sets, not the survival function by itself. Starting with the given estimates at $a_k$, the complete calculation isThe middle event is of the competing type: it decreases overall survival while leaving the disease cumulative incidence function unchanged at that instant. The event-free intervals leave every estimate and risk set unchanged.

At the final time, both the event subject and the subject censored at that time are in the just-before risk set, so its denominator is eight. This is the usual event-before-censoring convention for recorded ties. After the event and right censoring, six subjects remain at risk. ConsequentlyThe nine numbered source items are the inputs to this single calculation, not nine further questions.

The Kaplan–Meier estimator is $\hat{S}(t)={\prod }_{t_j\le t}(1-d_j/r_j)$, with event count $d_j$ and risk set size $r_j$. Its median is the first event time with estimated survival at most $1/2$. Since $\hat{S}(2)=0.5749$ and $\hat{S}(3)=0.4286$, the estimated median completion time is $3$ minutes.

The original last event exhausts the final risk set, so the fitted survival function drops to zero at $18$ and its conventional full area is finite. Integrate the right-continuous step function: its height is one before time one, and after each event the height is the newly computed survival value. There are three-minute gaps from $12$ to $15$ and from $15$ to $18$. ThusUsing the exact risk-set products rather than rounded printed survival values gives $4.187856$. This is not the sample mean of the observed event-or-censoring times; right censoring changes the estimated survival weights.

If the final observation at $18$ is instead censored, every earlier Kaplan–Meier estimator value is unchanged and no event factor is inserted at $18$. The median remains $3$ minutes, but the curve stays at $\hat{S}(15)\simeq 0.0107$ at the end of follow-up. By terminal censoring and survival-mean identifiability, a full unrestricted mean is no longer determined nonparametrically by these observations. Extending the final positive step forever would give an infinite integral, but that plotting convention does not establish an infinite population mean. Extra assumptions about the unobserved tail would be needed for a finite full-mean estimate.

The restricted mean survival time through $18$ is still $4.187856$ minutes in either version: the two curves differ only at or after the endpoint, which does not affect the area up to it. All these interpretations require independent censoring. Giving up may be related to how long a person would have taken to finish, so that assumption needs scientific justification in this study.

Throughout the survival questions, use the paper's notation $F(t)=\mathbb{P}(T>t)$ for the survivor function, not for a cumulative distribution function. To construct the Kaplan–Meier estimator, sort the distinct observed event times $a_j$. Let $r_j$ be the size of the risk set immediately before $a_j$, and $d_j$ the number of events at that time. The estimated conditional chance of surviving that event time is $1-d_j/r_j$. Multiplying these conditional probabilities givesA right censoring time reduces later risk sets but causes no jump itself. The usual interpretation as a population survivor function uses independent censoring.

Visits can be missed or rescheduled because of illness, clinic availability or patient choice; follow-up may end through dropout or right censoring. The correct panel-observed multi-state likelihood uses each observed interval length ${\Delta }_{ik}$:Thus the four aggregated one-year counts are generally no longer sufficient; records must retain actual durations. Numerical maximum likelihood estimation uses the corresponding transition probabilities, or a matrix exponential of the generator when a larger model is fitted. One must not treat a two-year interval as a single one-year transition or as two known transitions through an unobserved intermediate state.

This conditional calculation is valid when visit scheduling and loss to follow-up are ignorable given the modeled information. If symptomatic patients are systematically examined earlier or leave the study differently, observation itself may be informative; the visit or dropout process may require a joint statistical model. Simply replacing one year by the actual interval does not correct informative observation.