Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 2

Let $N(t)$ count observed failures and let $Y(t)$ be the at-risk process, counting subjects under observation and still alive immediately before $t$. Under independent censoring, a short interval of length $dt$ has conditional expected failure count $Y(t)h(t)dt=Y(t)dH(t)$. The estimating equation obtained by replacing the count expectation by its observation is $d\hat{H}(t)=dN(t)/Y(t)$. Each distinct failure gives $\Delta N(a_m)=1$ and $Y(a_m)=r_m$, so the Nelson–Aalen estimator isThe same increment maximizes the local working likelihood for a Poisson distribution with mean $r_m\Delta H_m$: differentiating $\Delta N_m\log (\Delta H_m)-r_m\Delta H_m$ gives $\Delta H_m=\Delta N_m/r_m$. The counting-process derivation explains why the risk set, rather than the initial sample size, supplies the denominator.

Collect distinct observed failure times as $a_m$, with $d_m$ failures and $r_m$ individuals in the risk set just before $a_m$. The grouped Nelson–Aalen estimator isAll failures recorded at one time use the same pre-event risk set. When failures and censoring share a recorded time, the usual convention includes those censored at that time in the risk set and processes failures before removing censored individuals. A known different ordering should instead be honored. If ties arise solely through rounding of a continuous-time process, an alternative is to resolve the unobserved ordering: $d_m$ successive failures without intervening censoring would contribute ${\sum }_{s=0}^{d_m-1}(r_m-s{)}^{-1}$. That is not the grouped jump $d_m/r_m$; the difference represents a choice about what the recorded ties mean.

At a distinct event time $a_m$, estimate the conditional probability of surviving that time by $1-d_m/r_m$. Multiplication of these conditional survival probabilities gives the Kaplan–Meier estimator, and taking minus its logarithm gives an integrated hazard estimate:The conditional failure count has a binomial likelihood whose maximizing event fraction is $d_m/r_m$. Tied failures therefore enter directly as one count. Even ordering $d_m$ failures arbitrarily, with no intervening censoring, produces the same survivor function factor becauseThe failure-versus-censoring ordering convention still matters when both have the same recorded time. If $d_m=r_m$, the survivor function becomes zero and the logarithmic integrated hazard is infinite.

Put $u_m=d_m/r_m$. The Taylor series of the logarithm gives the small-jump comparison of cumulative hazard estimators:Consequently, if ${\max }_m{u}_m\le \epsilon <1$, the two integrated hazard estimates obeyThey are close when every event fraction is small. In particular, with distinct single failures and large risk sets, $u_m=1/r_m$ is small. With tied data, large risk sets alone are insufficient if a substantial fraction of the risk set fails together.

At the last observed failure time $a_{\ast }$, any individual still in the risk set must either fail there or be censored there or later. If there are no such censored observations, every remaining individual fails, so $d_{\ast }=r_{\ast }$. Thus the final Kaplan–Meier estimator factor is zero, while the final Nelson–Aalen estimator jump is one:There is no small-jump approximation at this endpoint. Without ties the final risk set has size one, so the same discrepancy follows from its last single failure.