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The survival estimates favor the new treatment at the two reported times, with absolute survival differences $0.199$ at two years and $0.157$ at five years. These point differences alone do not establish efficacy. Randomization supports a causal comparison of the treatment assignments, but sampling variance, censoring, follow-up, protocol adherence and the prespecified endpoint still matter. One should not infer a statistically established benefit merely from two larger estimates.

A Kaplan–Meier estimator iswhere $r_j$ is the risk set and $d_j$ the event count. Its uncertainty depends on the actual risk sets, not just the initial group size. The Greenwood formula estimatesWithout the missing survival tables one cannot calculate these standard errors or the uncertainty of the group difference. In particular, replacing the estimator variance by the variance of an uncensored binomial proportion would generally be wrong.

I would advise presenting both curves with numbers at risk and confidence intervals, checking independent censoring, and reporting a prespecified overall comparison plus an effect estimate and interval. For an unadjusted log-rank test, write $O_1={\sum }_j{d}_{1j}$ and $E_1={\sum }_j{d}_j{r}_{1j}/r_j$. Its hypergeometric null variance isTerms with no possible variation contribute zero. This compares the entire event history rather than selecting two arbitrary times, uses the available risk sets, and avoids choosing favorable time points after examining the curves. It is especially effective under proportional hazards. Crossing curves can make a log-rank comparison insensitive; a prespecified restricted mean survival time difference or another suitable time-varying analysis may then be more informative. An overall comparison does not make a clinically important prespecified time-point contrast invalid.

The Kaplan–Meier median survival time is the first event time at which $\hat{S}(t)\le 1/2$. Monotonicity of the reported estimates impliesIf the new-treatment curve never reaches one half during observed follow-up, its median is not estimable from that curve. The exact crossing times cannot be reconstructed from two survival probabilities: the standard curve could first cross at three or at four years and still have the same values at two and five years. The point estimates suggest benefit, but the evidence and exact medians require the missing event/censoring tables. Neither a conclusive agreement nor a conclusive rejection of benefit is justified from these numbers alone.

The log-rank test compares event counts with their conditional expectations under equal group hazard functions, using the individuals currently in each risk set. Assume the usual independent censoring conditions and independent strata. At pooled event time $t_j$, let $n_{Aj},n_{Bj}$ be the group numbers at risk, $n_j=n_{Aj}+n_{Bj}$, and let $d_{Aj},d_{Bj}$ be the event counts with total $d_j$. Under the null hypothesis, conditioning on the risk counts and total events gives the hypergeometric distributionThus the null expected A count and its conditional variance areWhen there is only one person at risk, the event allocation is deterministic and its variance is zero. The unstandardized log-rank statistic is $U={\sum }_j(d_{Aj}-E_j)$, with variance estimate $V={\sum }_j{V}_j$. Under standard large-sample conditions, $U/\sqrt{V}$ is approximately standard normal under the null; the test detects systematic group differences in event rates. It is particularly effective for proportional-hazards alternatives, while crossing effects can cancel in the score.

For a stratified log-rank statistic, construct these risk sets and observed-minus-expected differences separately within each stratum $s$, then sum:This allows different baseline hazard functions between strata and avoids comparing an event with individuals in another stratum. The null is equal A and B hazards within each stratum, together with a censoring mechanism that does not invalidate the within-risk-set comparison.

In a paired stratum, both individuals initially form the risk set. A first event while both are still observed gives one nontrivial event allocation. After either individual leaves, every subsequent event is certain to belong to the only remaining group and has observed count equal to expected count. Therefore each pair supplies at most one informative table. A pair supplies none if the first exit is censoring before either observed event, if neither has an event during their joint follow-up, or if both fail simultaneously so that the tied event allocation is fixed. An event-censoring tie needs a specified convention; with events processed before same-time censoring, the other individual remains at risk for that event and the pair is informative. The untied continuous-time case avoids this ambiguity.

If A has the informative first event, the table at that time isThe phrase in the second column refers only to this event time; it does not mean that B can never fail. The expected A count is $1\cdot 1/2=1/2$, so this table contributes $1-1/2=1/2$ to $U$, with variance $1/4$. An informative B-first pair contributes $-1/2$. All other tables contribute zero. Therefore the requested unnormalized score isA positive score means more observed early events on A. There is no need to standardize it to answer the question.

To obtain an exact conditional law, put $m=d_A+d_B$. Under the within-pair equal-hazard null, conditional on an event while both are at risk, either subject supplies it with probability $1/2$. Equivalently, before the pair's first exit, the cause-specific event intensities for A and B are equal; integration over the common period of joint observation gives equal probabilities for the two informative outcomes. This argument requires noninformative observation of the failure process, not just equality of unadjusted marginal survival curves. Different independent pairs then have independent fair signs when informative. Conditioning on the set of informative pairs gives $m$ independent Bernoulli outcomes with probability $1/2$; conditioning only on their count mixes identical binomial laws and gives the same result:This proves the paired log-rank reduction to a sign test. The conditional mass function is $\left(\genfrac{}{}{0ex}{}{m}{k}\right)2^{-m}$, $0\le k\le m$, so an exact two-sided p-value sums the masses for $k$ with $|2k-m|\ge |d_A-d_B|$. If $m=0$, there is no information for a treatment comparison. Failure-time distances beyond their ordering play no role in this paired score.

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.

At day 165 there are two A and two B individuals at risk. The event is in A, with expected A count $2/4=1/2$, so its score contribution is $1-1/2=1/2$ and its variance contribution is $1/4$. The remaining A individual is censored at 173 and leaves the risk set before either later event. The risk-set calculations are

ThusThe positive score indicates higher A event hazard in this late contribution. Although B has more observed events, its two events occur with no A individual at risk, so those times do not compare the groups. This is why raw event totals alone are insufficient for the log-rank test.

The log-rank test compares survival experience through the allocation of events within the successive risk sets. Its null is equal hazard functions between the groups over follow-up, with independent individuals and noninformative censoring within groups. At each distinct event time $t_j$, let $n_{Aj},n_{Bj}$ be the numbers at risk immediately before that time, and let $d_{Aj},d_{Bj}$ be event counts. Under the null, conditional on the total $d_j=d_{Aj}+d_{Bj}$, the expected number of A events is $e_{Aj}=d_j{n}_{Aj}/(n_{Aj}+n_{Bj})$.

The signed log-rank statistic is the observed-minus-expected scoreA positive score indicates relatively more A events than expected, hence a higher event hazard for A; B's score is its negative. Censoring times do not produce score terms, but remove people from all later risk sets.

At each event time the conditional null allocation is hypergeometric, as if the $d_j$ events were sampled without replacement from the risk set. Its variance gives the score-increment variance, with the finite-population correction for ties. Summing these conditional variances gives $V$, since distinct-time martingale score increments have zero cross covariance under the null. For sufficiently many informative events, $U_A/\sqrt{V}$ is approximately standard normal and $U_A^2/V$ approximately chi-squared with one degree of freedom. In the no-tie case, each variance contribution is $n_{Aj}{n}_{Bj}/(n_{Aj}+n_{Bj}{)}^2$. A time at which only one group remains at risk contributes no comparative information.

The numerical parts below report signed-score and variance contributions after day 160. They cannot be added as separate chi-squared statistics to the earlier follow-up: add the scores and variances first, and standardize once for the complete dataset. The log-rank test is particularly effective for proportional hazards alternatives; crossing hazards can produce cancelling score contributions.

$Y_{j,k}$ 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.

Summing the observed-minus-expected contributions gives the unstandardized log-rank statisticUnder the null it is centered at zero. A two-sided log-rank test compares its magnitude with the square root of its null variance.

Construct the two countries' period-specific risk sets with the same left truncation and right censoring rules, then compare their event counts by a log-rank test. This is a nonparametric comparison because it does not specify the shape of either country's hazard function.

The log-rank test compares observed failures in each group with the numbers expected under equal hazards, conditioning at every event time on the risk set and total failures. Its score sums $O_{1j}-Y_{1j}{d}_j/Y_j$ and is standardized by its null hypergeometric variance. It is most powerful for approximately proportional hazards. Strongly crossing survival curves can produce large positive and negative contributions that cancel, so very different distributions may yield a weak log-rank statistic.