Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 34 / 4

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 $YdH$.

Keep the counting process and at-risk process notation from the introductory derivation. Because the individual hazards add, the aggregate conditional intensity isThe known individual contributions must be averaged over the current risk set, rather than over the original cohort. DefineThen $dN(t)/Y(t)$ estimates $\{{\overline{h}}_0(t)+h_1(t)\}dt$. Subtract the known average contribution to obtain the offset-adjusted cumulative hazard estimatorTake the quotient to be zero when $Y=0$. If $M(t)=N(t)-{\int }_0^t\lambda (u)du$, direct substitution givesThis establishes the target and the mean-zero estimation noise, under the usual integrability conditions. It estimates $H_1(t)$ while there is a nonempty risk set; no data identify its continuation after follow-up has ended. With common known $h_{0i}=h_0$ and individuals at risk throughout $[0,t]$, it reduces to ${\hat{H}}_1(t)=\hat{H}(t)-H_0(t)$.

The estimate may decrease between events because the known offset is subtracted continuously. That is not an algebraic error: it is an unconstrained estimating-equation estimator, not automatically a nonnegative monotone cumulative hazard estimate. If such shape constraints are imposed, the fitting procedure must explicitly account for them. The working hazard model itself must satisfy $h_{0i}+h_1\ge 0$.

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.