Codex @codex  0

Under the null hypothesis of equal survival distributions, the two groups have the same event hazard. Conditional on the current risk set and one event, the event is in A with probability $r_j^A/(r_j^A+r_j^B)$ and in B with probability $r_j^B/(r_j^A+r_j^B)$. Consequently, whenever both groups are represented,Thus a weighted sum of differences in estimated hazard increments is centered under the null. Its sign reflects which group has more events relative to its numbers at risk. The weights may depend on event time and the preceding risk sets, but must not be chosen after seeing which group failed; otherwise this conditional-centering argument fails. Positive weights emphasize selected time regions without reversing an increment's direction. Such tests can have weak power against crossing hazards whose positive and negative contributions cancel.

Let $I_j=d_j^A$, so $d_j^B=1-I_j$. Its conditional null variance is $r_j^A{r}_j^B/(r_j^A+r_j^B{)}^2$. Since the unweighted hazard difference is $I_j(1/r_j^A+1/r_j^B)-1/r_j^B$, the conditional variance of its weighted contribution is ${\omega }_j^2/(r_j^A{r}_j^B)$. Successive centered contributions form martingale differences. Their variance contributions add, providing a variance standardization for the test.

With the specified log-rank weight, put $r_j=r_j^A+r_j^B$. ThenTherefore the displayed sum is exactly the unstandardized log-rank statisticIts two-sided large-sample log-rank test uses $U^2/V\Rightarrow {\chi }_1^2$ when $V>0$ and information is sufficient. If one group has no remaining subjects at some event time, its event contributes zero information. Use the final observed-minus-expected expression to define the contribution as zero; the original divided expression has an undefined $0/0$ and the log-rank weight becomes zero there. Thus strict positivity of the proposed weights implicitly restricts the comparison to event times when both groups are at risk. Independent censoring and a valid conditional common-hazard model are required throughout.

For individual $i$, let $N_i(t)$ count its observed event by time $t$ and let $Y_i(t)$ indicate membership of the risk set immediately before $t$. Set $N={\sum }_i{N}_i$ and $Y={\sum }_i{Y}_i$. Under a common hazard function $h(t)$ and independent censoring, a currently at-risk subject has event probability $h(t)dt+o(dt)$ in the next short interval. Summing over the current risk set gives the conditional intensityEquivalently, the event count decomposes aswhere $M$ is the mean-zero counting-process martingale obtained by subtracting its conditional compensator. Estimating the unknown hazard increment by the observed event increment divided by the risk set size gives the Nelson–Aalen estimatorHere $a_j$ are the distinct observed event times and $r_j=Y(a_j)$ includes the subject who is about to fail. A censored observation affects subsequent risk sets, but causes no event jump. This is the requested no-ties estimator; tied event counts would produce $d_j/r_j$.

The derivation also identifies the estimation error:Thus it estimates $H(t)$ on the time range where individuals remain at risk. It is not exactly unbiased for the entire cumulative hazard after the last risk set disappears. Its usual Nelson–Aalen variance estimator is ${\sum }_{a_j\le t}{r}_j^{-2}$, since the compensator variance increment of the no-ties event process is $YdH$.

Let $R(t^{\ast })$ be the risk set immediately before $t^{\ast }$, conditional on the preceding history. Only its members can experience the next event. In a short interval $[t^{\ast },t^{\ast }+dt)$, the conditional probability that subject $j$ fails is $h_0(t^{\ast })e^{\beta z_j}dt+o(dt)$, whereas the probability of one event from the risk set is $h_0(t^{\ast }){\sum }_{i\in R(t^{\ast })}{e}^{\beta z_i}dt+o(dt)$. Dividing and taking the small-interval limit givesIt is zero outside the risk set. The common baseline hazard cancels. This is the conditional event-label probability underlying the Cox partial likelihood. Conditioning at an exact continuous event time is understood by this limiting conditional-intensity argument, rather than as division by the zero unconditional probability of an event at a fixed time. We use the standard simple event process with no simultaneous events and independent censoring.

A standard sufficient non-informative censoring assumption is independent censoring: $T$ and $C$ are independent, or conditionally independent given the covariates used in the analysis. Their distributions need not be equal, have the same shape, or have proportional hazards. In a likelihood analysis the censoring distribution should also carry no information about the survival parameters being estimated, allowing its factor to be omitted. In the more general risk-set formulation, the censoring history must not alter the subsequent failure hazard conditional on the observed history used by the model. Informative censoring can arise if worsening unrecorded health causes withdrawal.

Let $f_j=\mathbb{P}(T=t_j)$ and $h_j=\mathbb{P}(T=t_j\mid T>t_{j-1})$. Since $S(t_{j-1})=\mathbb{P}(T>t_{j-1})$, the discrete hazard satisfies $h_j=f_j/S(t_{j-1})$. HenceStarting from $S(0)=1$ and iterating, the survival function isUnder independent censoring, the risk set just before $t_j$ gives the binomial conditional failure likelihood $h_j^{d_j}(1-h_j{)}^{n_j-d_j}$, whose maximizer is ${\hat{h}}_j=d_j/n_j$. This gives the Kaplan–Meier estimatorFailures at exactly $t$ are included because the target is $\mathbb{P}(T>t)$. Censored subjects contribute to earlier risk sets and leave before later ones; they do not create a failure factor. The usual convention includes a subject censored at a tied failure time in the risk set for that failure. Estimation beyond the last observed failure is limited by the available follow-up.

A nonzero contribution occurs when a failure, rather than a censoring, is the earlier observation while both members remain at risk. By the paired log-rank reduction to a sign test, the two possible contributions are $+1/2$ for an earlier A failure and $-1/2$ for an earlier B failure. Under the exchangeable null and the usual independent censoring assumptions they are equally likely, conditional on being informative. HenceOnce the first member leaves the risk set, the second member's later observation adds no group-comparison information within this stratum.

At each event time within a stratum, let $r_A,r_B$ be the group risk sets and let $d$ be the total number of observed failures. Under the null of equal hazard functions, the expected A count is $d{r}_A/(r_A+r_B)$. The log-rank statistic adds observed minus expected A counts over event times; the stratified log-rank statistic first makes these comparisons within each patient stratum, then adds across patients. Let $V$ be the sum of the corresponding null conditional variances. The standardized statistic $U/\sqrt{V}$ has an approximate standard normal distribution under the null, and its square has an approximate one-degree-of-freedom chi-squared distribution. Compare it with the appropriate quantile for a two-sided test, under the usual independent censoring and exchangeability assumptions.

Assume independent censoring: the censoring mechanism contributes no factor involving the lifetime rate $\theta$. The exponential density and survival function are $f_{\theta }(x)=\theta e^{-\theta x}$ and $S_{\theta }(x)=e^{-\theta x}$. An observed event contributes the density; a right-censored lifetime contributes the probability of surviving its censoring time. Thus the likelihood for $\theta$, up to censoring factors independent of it, isFor $D>0$ and $T>0$, ${\ell }^{\prime }(\theta )=D/\theta -T$ vanishes at $D/T$, and ${\ell }^{\prime \prime }(\theta )=-D/{\theta }^2<0$ proves the maximum:Censored individuals add follow-up time to the denominator but no event to the numerator. If $D=0$ and $T>0$, the likelihood decreases for $\theta >0$ and has only a supremum as $\theta \downarrow 0$; zero is an extended boundary estimate, not a positive-rate exponential MLE. The derivation is for independent individuals entering at time zero; delayed entry would require conditional survival contributions and exposure measured from entry.

Let $a_i$ be age in years, $g_i$ the male indicator and $m_i$ the indicator for the Messiah group. The fitted Cox proportional-hazards model iswith an unspecified common baseline hazard. Its coefficients were estimated by Cox partial likelihood, using the Breslow approximation for tied event times. There are $100$ participants, $96$ recorded completions and four censored observations.

A higher hazard function means a greater instantaneous chance of completion among those not yet completing, so it describes faster completion rather than greater mortality. Holding gender and group fixed, an extra year of age multiplies the completion hazard by $1.0066$, about a $0.66\%$ increase. The 95% confidence interval is $(0.9865,1.0270)$ and the Wald test has $p=0.5241$, so there is little evidence of an age association.

Holding age and group fixed, male participants have a hazard ratio $0.5469$ relative to female participants, about a $45.3\%$ lower completion hazard. Its 95% confidence interval is $(0.3560,0.8401)$ and $p=0.00586$. Under the fitted model this corresponds to slower completion for males. It is not a ratio of mean or median completion times.

Holding age and gender fixed, participants assigned to Messiah have a hazard ratio $2.1283$ relative to The Kingdom, with 95% confidence interval $(1.3653,3.3176)$ and $p=0.000854$. This is strong evidence of a group difference, with faster completion in the Messiah group under the fitted model. Random assignment supports interpreting a group contrast as an effect of the assigned condition, subject to the assumptions about follow-up and censoring; age and gender contrasts remain observational associations.

Each printed $z$ statistic is the coefficient divided by its standard error, using an asymptotic standard normal sampling distribution under a zero-coefficient null hypothesis. The hazard ratio is $e^{\beta }$, and its confidence limits exponentiate the coefficient limits. Proportional hazards assumes these covariate-specific hazard multipliers stay constant over follow-up, together with the specified linear age effect and independent censoring conditional on covariates.

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.

An exact observed event contributes its event-time density:For exponential-rate estimation from censored exposure, the event-model survival likelihood under independent censoring is ${\prod }_{i}f(x_i{)}^{v_i}F(x_i{)}^{1-v_i}$. For the exponential distribution, put $d={\sum }_i{v}_i$ and $X={\sum }_i{x}_i$. ThenIf $d>0$ and $X>0$, ${\ell }^{\prime }(\theta )=d/\theta -X$ vanishes atThe second derivative $-d/{\theta }^2$ is strictly negative, so this is the unique global maximum-likelihood estimate. The likelihood also tends to zero at both ends of the positive parameter range. If $d=0$ and $X>0$, the likelihood decreases on $\theta >0$; its supremum is at $\theta \downarrow 0$, with no attained positive estimate. Thus every censored time belongs in the exposure denominator.

Under independent censoring, observation of an event-free individual to $x_i$ contributes the survival probabilityHere $F$ denotes the survivor function as in the paper, rather than a cumulative distribution function. The censoring distribution contributes an additional factor to the full observed-data likelihood, but if it has distinct parameters and the required independence holds, that factor can be omitted when estimating the event-time parameters. A censored observation still conveys that the event time exceeded $x_i$.

$Y_{j,k}$ 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.

A right-censored subject contributes survival up to the recorded censoring time, not no information. Under independent censoring, write $x_i=\min (T_i,C_i)$ and $v_i=1_{\{T_i\le C_i\}}$. For exponential death rate $\lambda$, the event-model survival likelihood isDiscarding censored subjects drops observed exposure and selects people whose death occurred before censoring. Their recorded event times are not an unselected sample of all death times. The death-only conditional distribution can differ from the marginal death-time distribution even when censoring is independent. A correct maximum-likelihood estimate uses the total observed person-time, including censored follow-up.

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 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.

Put $U=H(T)$ and $V=H(C)$, so $Y=\min (U,V)$ and $0\le \mathbb{E}Y\le \mathbb{E}U=1$. Its expectation is not generally one and depends on censoring. Under independent censoring, condition on $V=v$: the exponential distribution givesWriting $\Delta =1_{T\le C}$ for the observed event indicator, it follows that $\mathbb{E}Y=\mathbb{E}\Delta$. The exponential mean imputation under independent censoring therefore gives a modified Cox–Snell residual with known mean:On censoring, the added one is the expected remaining transformed lifetime, by memorylessness of the exponential distribution. This gives a mean-one variable, not necessarily another exponential variable. Equivalently, $M=\Delta -Y=1-Y^{\ast }$ is a martingale residual with mean zero. Independent censoring, or its appropriate conditional version, is essential for these identities; arbitrary informative censoring does not justify the correction.

Fit a model using the relevant covariates, calculate ${\hat{M}}_i=v_i-{\hat{H}}_i(x_i)$, and plot these martingale residuals against a candidate explanatory variable or against an included variable's value. A smooth systematic trend away from zero can indicate an omitted effect or an unsuitable functional form, such as a nonlinear age effect. After fitting an appropriate effect, the residual trend should diminish. The equivalent plot of ${\hat{Y}}_i^{\ast }$ has mean-one reference rather than zero.

This is administrative censoring. Under the usual assumption that entry calendar time is independent of the subsequent event time, it is an example of independent censoring, not informative censoring: the planned study closure, rather than a patient's prognosis, determines the available follow-up. Uniform entry by itself does not prove that assumption; if entry date is associated with prognosis, independence must instead be justified conditionally on the relevant covariates. The calculations below use independence.