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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.

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.

Assign the seventh individual's event weights $(1/3,1/3,1/3)$, the conditional probabilities obtained from the original Kaplan–Meier estimator. The last three fractional counts areTheir conditional event fractions are $1/3$, $1/2$, and $1$, exactly as in the original censored-data calculation. Hence the fractional allocation reproduces the original estimate:This illustrates self-consistency of Kaplan–Meier estimation: distributing the unobserved event according to the current conditional fitted event distribution leaves the fitted distribution unchanged. The first imputation in part (b) was not such a self-consistent allocation.

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.

If the final individual experiences an event at $t_1$, the Kaplan–Meier estimator acquires the factor $1-1/1=0$. ThusThe conventional fitted curve is zero thereafter. Its area gives the finite plug-in estimateThe area is a sum of rectangles: if $a_0=0<a_1<\cdots <a_m=t_1$ are the event times, it equals ${\sum }_{j=0}^{m-1}\hat{F}(a_j)(a_{j+1}-a_j)$. Censoring times within a constant segment do not alter its area.

Both terminal estimates have poor reliability because the last risk set has size one: one observed outcome determines whether the entire remaining fitted tail stays positive or collapses to zero. The zero tail is an empirical endpoint convention, not evidence that every future member of the population must fail by $t_1$. With earlier right censoring, the displayed finite area can be especially sensitive to this last event and does not remove uncertainty about the population tail. In the final-censoring case the full mean lacks a determined tail area; in the final-event case the conventional full fitted mean is finite and calculable, but should be reported with this limitation. A restricted mean survival time at a prespecified cutoff within reliable follow-up often avoids the unstable terminal extrapolation.

Let $a_1<a_2<\cdots$ be the distinct observed event times, $r_j$ the risk set size immediately before $a_j$, and $d_j$ the events there. Under independent censoring, the Kaplan–Meier estimator isIt starts at one, jumps down at events, and is unchanged by a pure censoring time; right censoring removes people from subsequent risk sets. If events and censorings coincide, the convention here processes events before censoring, so those censored at the recorded time are included just before the event. The estimated median is the first time the curve reaches or falls below $1/2$. If it never does during observation, the median is not estimable from the observed curve without tail assumptions.

With one remaining individual censored at $t_1$, no event factor is introduced. HenceThis is the terminal observed step value. With an empty risk set after $t_1$, continued horizontal plotting is a convention and supplies no information about the true later survivor function. The tail rests on one person and is imprecise; this is terminal censoring and survival-mean identifiability. If ${\varphi }_0>0$, the unrestricted mean cannot be obtained nonparametrically from this censored tail: extending the last positive step to infinity would give an infinite area, which does not establish an infinite population mean. A restricted mean survival time up to a suitably supported finite cutoff remains estimable.

The estimate remains $(r-d)/r$. Individuals right-censored at $t_R$ are observed throughout the preceding interval, so they contribute to each risk set until the end. They must not be removed from the denominator at $t_L$.

If events and right censoring share the recorded time $t_R$, use the usual Kaplan–Meier estimator convention: count events first, using a risk set that still includes those censored at that time, and then remove the censored observations. With no earlier censoring, the product of event factors over the interval telescopes to $(r-d)/r$.

An individual is right-censored at $t$ when follow-up establishes that the event has not occurred through $t$, but does not reveal the subsequent event time. The observed information is $T>t$ for the continuous-time convention.

Knowing only the total number censored inside an interval does not specify the risk sets at the event times. For example, start with two individuals, one event and one censoring. If censoring precedes the event, the Kaplan–Meier estimator contributes $1-1/1=0$; if the event precedes censoring, it contributes $1-1/2=1/2$. The totals $r=2,d=1,c=1$ are identical.

Therefore the timing of right censoring relative to the events is needed. Independent censoring is also required for these changing risk sets to yield a valid estimate of the underlying survivor function.

Let $d_j$ be the number of events at potential time $a_j$ and let $r_j$ be the size of the risk set immediately before that time, including observations censored at $a_j$. Define the discrete conditional event probabilityUnder independent censoring, and ignoring the separate censoring distribution, the survival likelihood factorizes asThis follows by multiplying a survival factor for each individual at risk who does not fail at $a_j$, and an event factor for each failure. The log factor $d_j\log q_j+(r_j-d_j)\log (1-q_j)$ is maximized at ${\hat{q}}_j=d_j/r_j$, with the corresponding boundary values when $d_j=0$ or $d_j=r_j$.

The survival function is the product of successive conditional survival probabilities. ThereforeThis is the Kaplan–Meier estimator. A product over no event times is 1. If everyone remaining experiences an event, the product becomes zero and stays zero; subsequent empty risk sets supply no additional factor.

The set must contain every observed event time. Otherwise a discrete survival likelihood would assign probability zero to at least one exact event observation.

The set can include additional times without observed events. Include time zero when events at zero are present, and allow remaining tail probability beyond the fitted range. Estimation of a conditional event mass requires a nonempty risk set; once follow-up ends, later probability masses are not identified by the data. The usual Kaplan–Meier estimator retains the final estimated survival probability as its tail convention.

A potential event time is an allowed support point for the discrete event-time distribution used to derive the Kaplan–Meier estimator. A positive probability mass may be assigned there before fitting, even if no event was observed at that time.

This is a device for nonparametric maximum likelihood estimation from exact event times and right censoring; it does not assert that the underlying biological event time is genuinely discrete. Its estimated mass can be zero at an unobserved potential time.

At an added potential event time with no event and a nonempty risk set, $d_j=0$, so maximum likelihood estimation gives ${\hat{q}}_j=0$. Its factor in the Kaplan–Meier estimator is therefore $1-d_j/r_j=1$.

Adding or deleting such times also leaves the risk sets at actual event times unchanged: those depend on the observed event and censoring histories. Consequently the fitted survivor function depends only on observed event times and their risk sets. Beyond the last observed follow-up, the tail is not identified; the standard flat continuation is a convention, rather than an additional consequence of likelihood maximization.

Choose the potential right-censoring times to include every observed right censoring time up to $t$, and also the endpoints 0 and $t$. Then no individual is censored strictly inside $(c_{k-1},c_k)$. Use an explicit endpoint convention: let $r_k$ count individuals still event-free after any events at $c_k$, but before censoring at $c_k$. If $d_k$ events occur in $(c_{k-1},c_k]$, the number at the beginning of that interval, after censoring at its left endpoint, is $r_k+d_k$.

The no-censoring interval estimate from the preceding parts is therefore $r_k/(r_k+d_k)$. The possible event mass at zero contributes $r_0/(r_0+d_0)$. Multiplying yieldsFor a continuous event-time distribution, normally $d_0=0$. Within interval $k$, put $b_k=r_k+d_k$ and let its successive event counts be $e_1,\dots ,e_m$. Its Kaplan–Meier estimator factors telescope:The same calculation at zero provesThe paper's phrase “at risk at $c_k$” does not specify its side of an event at that endpoint. If it denotes the conventional pre-event risk set size $r_k^{-}$ instead, and $e_k$ events occur exactly at $c_k$, substitute $r_k=r_k^{-}-e_k$. The interval factor is then $(r_k^{-}-e_k)/(r_k^{-}+d_k-e_k)$; at zero it is $(r_0^{-}-d_0)/r_0^{-}$. This keeps the events-before-censoring convention consistent. After an empty risk set, stop the product; if the fitted survival has reached zero, retain zero rather than creating a $0/0$ factor.

Place discrete hazards $q_j$ at the ordered distinct event times $t_j$. If $d_j$ events occur among $r_j$ individuals at risk immediately before $t_j$, the nonparametric empirical likelihood isEquivalently, an observed event at $x$ contributes the probability mass at $x$, while a right-censored observation contributes the survivor probability beyond its censoring time. Maximization gives ${\hat{q}}_j=d_j/r_j$ and the Kaplan–Meier estimator

Left truncation means an individual is observed only conditional on surviving beyond an entry time $L$. Ignoring it overrepresents long survivors and creates survivorship bias. A subject with event or censoring time $X>L$ contributes its usual likelihood divided by $S(L)$; in risk-set form, that individual enters each $r_j$ only for event times satisfying $L<t_j\le X$.

The corrected Kaplan–Meier estimator is $2/3$ immediately after $t=2$ andimmediately after $t=7$. It first crosses one half there, so

At each distinct event time $t_j$, let $d_j$ events occur among $r_j$ people in the risk set. The Kaplan–Meier estimator iscensorings remove people from later risk sets but create no factor.

Left truncation, or delayed entry, means that an individual is observed only after surviving to an entry time. A registry assembled from patients alive when a clinic opens is a practical example. Adapt Kaplan–Meier by admitting each person to the risk set only at their entry time.

Period survival analysis is useful when recent prognosis is desired but complete long-term follow-up of a recent diagnosis cohort is unavailable. For a chosen calendar year, intersect every patient's observed follow-up with that year. Express the surviving pieces on the time-since-diagnosis scale, treat the beginning of the calendar window as delayed entry and its end as right censoring, and apply Kaplan–Meier with those entry and exit times.

Take the 2017 calendar window to be $[1,13]$ on the supplied month scale. Intersecting each follow-up interval with this window and translating to time since diagnosis gives:

Patient 2 died before the period and patient 10 entered after it, so neither contributes. At duration 6, patients 4, 5, 6, 7, and 8 are at risk, giving factor $1-1/5=4/5$. At duration 15, patients 1 and 3 are at risk, giving factor $1-1/2=1/2$. Therefore the period Kaplan–Meier estimator isover the range supported by the period data.

At duration seven there are six at risk and two deaths, at duration thirteen there are four at risk and one death, at duration twenty-six there are three at risk and one death, and at duration twenty-seven there are two at risk and one death. The Kaplan–Meier estimator is thereforeThe censoring of D at duration thirty-seven causes no multiplicative drop.