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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.

A frailty model represents unobserved individual heterogeneity by a random effect in the hazard function. In a proportional frailty model, an individual with frailty random variable $U=u$ has hazard $u{h}_0(t)$. The normalization $\mathbb{E}U=1$ identifies the scale of the baseline hazard: without it, multiplying all frailties by a constant and dividing $h_0$ by that constant would leave all individual hazards unchanged.

Let $H_0(t)={\int }_0^t{h}_0(v)dv$. Conditional survival follows by integrating the hazard function:Taking the expectation over the frailty law yieldsThis is the Laplace transform of the frailty density evaluated at the cumulative hazard. More generally, the same calculation integrates against a probability measure, allowing atoms. It averages conditional survival rather than substituting $\mathbb{E}U$ into its exponent. Indeed, the frailty distribution among survivors is proportional to $e^{-u{H}_0(t)}g(u)$, giving population hazardSurvival progressively selects smaller frailties, so the population hazard need not be the baseline or even retain the individual's proportional structure.

For the cure model, first take $0\le \pi <1$. The necessary mean-one frailty law and baseline hazard areThen $\mathbb{E}U=\pi \cdot 0+(1-\pi )/(1-\pi )=1$. A cured person's conditional hazard is zero, while a susceptible person's conditional hazard is $h_0/(1-\pi )=h_{\ast }$. Taking $U$ to be zero or one while leaving $h_0=h_{\ast }$ would instead have mean frailty $1-\pi$ and violate the required normalization.

The frailty law is the measurewhere the $\delta$ terms are Dirac measures. This is not an ordinary Lebesgue density; the cure construction requires permitting a discrete frailty law. Its Laplace transform isWriting $H_{\ast }(t)={\int }_0^t{h}_{\ast }(v)dv$, we have $H_0=(1-\pi )H_{\ast }$, and thereforeIf $H_{\ast }(t)\to \infty$, the susceptible class eventually experiences the event and $S(t)\to \pi$. If $H_{\ast }(\infty )<\infty$, the limiting survival is larger than $\pi$ because some susceptible subjects also never experience the event. At $\pi =1$, everyone is cured: set $h_0\equiv 0$ and, for example, $U\equiv 1$ to retain mean one, giving $S\equiv 1$ without using the singular formula $1/(1-\pi )$.