Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 4

$N(t)$ is the event counting process, while $Y(t)$ is the at-risk process. Their key relationship is that, conditionally on the observed past, the expected event increment is $Y(t)dH(t)$. It separates exposure to risk from event occurrence and naturally incorporates right censoring.

Before the observed endpoint, $Y+N=1$. If censoring occurs before the event, both processes are zero after the censoring time, so their sum is zero. With the inclusive endpoint conventions in the question, at an observed event time $t=T\le C$ both are one, so the sum is two at that single instant.

The cumulative hazard function is defined through the conditional mean incrementEquivalently, $dH(t)=\mathbb{E}\{dN(t)\mid Y(t)=1\}$ and $H(t)={\int }_0^{t}dH(u)$.

Aggregate the individual increments. At an event time $x_i$, their conditional expectation is the numberat risk times $dH(x_i)$. Replacing expectation by the observed event increment gives $d\hat{H}(x_i)=v_i/r_i$. Thus the estimator is the Nelson–Aalen estimator

The quantities $v_i-\tilde{H}(x_i)$ act as event-count residuals. Requiring their sum to vanish calibrates the fitted total number of events to the observed total, just as an intercept score equation calibrates fitted means. This is the aggregate zero-residual property of a martingale residual.

Put $D_j=\tilde{H}(x_j)-\tilde{H}(x_{j-1})$, with $\tilde{H}(x_0)=0$. ThenThe natural local calibration is thereforeTaking $\tilde{H}$ to be the right-continuous step function with these increments givesBecause $n-j+1$ is exactly the risk-set size $r_j$, this is the estimator from part c and automatically satisfies ${\sum }_i\{v_i-\tilde{H}(x_i)\}=0$.