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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.

$Y_{j,k}$ is the number in group $k$ still observed and event-free immediately before event time $a_j$: its group-specific risk set size. Set $Y_j=Y_{j,0}+Y_{j,1}$. Under the common hazard function and conditional on exactly one event at that time, each at-risk individual has the same infinitesimal event probability. ThereforeThe indicator $g_{\pi (j)}$ records whether the event came from group 1, while $p(j,1)$ is its conditional expectation under equal hazards. Thus $z_j$ is the observed-minus-expected group-1 event count at that time. Under the null these are martingale differences, with conditional varianceThe log-rank statistic $z={\sum }_j{z}_j$ has null mean zero. Its predictable variance estimate is $V={\sum }_j{p}_j(1-p_j)$, so the equal-hazards test usesUse a two-sided large-sample log-rank test if no direction was prespecified; $V>0$ and adequate information are needed. The variance is the no-ties form. A positive $z$ indicates more group-1 events than expected. Independent censoring conditional on the model is required for these at-risk event intensities to describe the data.

Under the proportional hazards model, the total instantaneous group-$k$ rate is $Y_{j,k}{e}^{\beta k}{h}_0(a_j)$. Conditioning on the next event cancels the baseline hazard, givingKeep $z_j$ defined using its null probability $p(j,1)$, as in the question. Its conditional expectation under the alternative isIt is zero at $\beta =0$, as required, and has the sign of $\beta$ when both groups remain at risk. Replacing $p(j,1)$ inside the definition of $z_j$ with the alternative probability would give a different residual, whose mean would instead be zero under that alternative.

The history ${\mathcal{H}}_{t-}$ contains the information observed strictly before $t$: baseline characteristics, previous events and censoring, and therefore which individuals are currently eligible and under observation. Formally it is the pre-$t$ information in the relevant filtration. The at-risk process $Y_i(t)$ is 1 if individual $i$ is observed and event-free immediately before $t$, and 0 otherwise. Let $Y_{+}(t)={\sum }_i{Y}_i(t)$ and write $dH(t)=h(t)dt$ for the cumulative hazard increment.

The counting-process intensity in survival analysis givesThe common conditional hazard function is assumed to remain applicable after conditioning on the observed history, as under suitable independent censoring. Summing gives the conditional mean of the total event increment:On times with $Y_{+}(t)>0$, invert this relation to estimate the hazard increment:The last sum is for untied events. This is the Nelson–Aalen estimator; for ties replace 1 by the number of events at that time. No increment can be estimated once the risk set is empty.

Put $q_t=\mathrm{Pr}(d{N}_{+}(t)=1\mid {\mathcal{H}}_{t-})$. A binary event increment has conditional variance $q_t(1-q_t)$. Since $q_t=Y_{+}(t)h(t)dt+o(dt)$, its squared mean is of second order, givingTreating the predictable risk set size as known given the history, the Nelson–Aalen estimator increment therefore has first-order varianceSubstitute $d\hat{H}=d{N}_{+}/Y_{+}$ for $dH$ to obtain the estimated increment variance $d{N}_{+}/Y_{+}^2$. Summing these estimated predictable variances gives the usual Nelson–Aalen variance estimator:The martingale $d{N}_{+}-Y_{+}dH$ has orthogonal increments, which justifies accumulating the predictable variances of the estimation error. This is an estimated sampling variance on the observed at-risk range, not a claim of exact finite-sample unbiasedness after the risk set becomes empty. For multiple events at an event time the usual extension replaces the numerator 1 by the event count.

Multiply survival to $t$ by the conditional type-A event probability in the next infinitesimal interval. The unconditional type-A increment probability isEquivalently, its exact value over a positive interval of length $u$ is ${\int }_t^{t+u}\alpha e^{-(\alpha +\beta )v}dv$. The factor $e^{-(\alpha +\beta )t}$ is essential because a prior B event removes an individual from the relevant risk set.

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.

The risk sets at the six event times are $(7,6,5,3,2,1)$, each with one event: the censoring between $t_3$ and $t_4$ removes the seventh individual before the fourth event. Thus the survivor estimate after the fifth event isConditional on reaching $t_4$, the event mass at $t_4$ is $1/3$, that at $t_5$ is $(2/3)(1/2)$, and that at $t_6$ is $(2/3)(1/2)(1)$. Hence the conditional event probabilities are

Now treat the seventh observation as an actual event at $t_4$, not as a censoring at that time. The risk sets are $(7,6,5,4,2,1)$ and the event counts are $(1,1,1,2,1,1)$. The Kaplan–Meier estimator therefore givesConditional on reaching $t_4$, the two events at that time have total probability $2/4=1/2$. Surviving them has probability $1/2$, which is then divided equally between the remaining two event times. Thus

For fractional event imputation, let $q_j$ be the fractional event weight assigned to the seventh individual at $t_j$. Its event contribution at $t_j$ is $q_j$. It belongs to the risk set at that time precisely for allocations at $t_j$ or later, so its fractional risk set contribution is ${\sum }_{j^{\prime }\ge j}{q}_{j^{\prime }}$. This justifies both parts of the hint.

For $(q_4,q_5,q_6)=(1/2,1/4,1/4)$, the fractional counts at the last three times areEarlier Kaplan–Meier estimator factors still give survival $4/7$ immediately before $t_4$. ConsequentlyThe three values are ${\hat{F}}_5^0=1/7\simeq 0.1429$, ${\hat{F}}_5^1=5/28\simeq 0.1786$, and ${\hat{F}}_5^{\ast }=4/21\simeq 0.1905$, in strictly increasing order. Spreading the censored individual's mass beyond $t_4$ raises the survivor estimate from the immediate-event imputation and moves it toward the original Kaplan–Meier estimator.

The transformed event time is unit exponential, and the transformed censoring time remains a censoring time because $H$ is increasing. Under independent censoring and a correctly fitted model, the Kaplan–Meier estimator of the survivor function for the Cox–Snell residuals $\hat{H}(x_i)$ should approximate $e^{-u}$.

Thus a logarithmic survivor plot should follow a straight line through the origin with slope minus one:The empirical plot is a step function around that line. Systematic curvature suggests model misspecification; the late part is less stable when the risk set is small. The retained event indicators $v_i$ must be used rather than treating censored residuals as complete event times.

In the counting-process intensity in survival analysis, the counting process $N_i(t)$ jumps once if an observed event occurs; the at-risk process $Y_i(t)$ is one immediately before $t$ only while individual $i$ is still observed and event-free. Conditional on that information, the chance of an event in a short interval is $h_i(t)dt+o(dt)$ for an at-risk individual and zero otherwise. This is the meaning ofThe history ${\mathcal{H}}_{t-}$ is the filtration generated by the observed event and censoring histories, entry information, and available covariates strictly before $t$. In particular, it determines the current risk set but does not reveal future event times. The right-hand side is the infinitesimal compensator of a counting process; the equality is interpreted as an intensity statement to first order in $dt$.

Let $N(t)={\sum }_i{N}_i(t)$ and $Y(t)={\sum }_i{Y}_i(t)$ be the total observed counting process and size of the risk set. When the individual cumulative hazard functions are all $H_0$, summing the conditional intensity equations givesReplace the expected increment by its observed value and divide by the predictable risk set size. Thus the Nelson–Aalen estimator in integral form iswith the integrand defined as zero when $Y(u)=0$. This is an integral against the step counting process, so each event time contributes its event count divided by the risk set just before that time. The Nelson–Aalen estimator estimates integrated hazard over times with observable risk; it does not supply information after observation has ended.