 Solution

For subject $i$, write ${\delta }_i=1_{\{T_i\le C_i\}}$, and define the counting process and pre-event at-risk process byUnder independent censoring and a common hazard, each observed event-free subject has conditional event probability $h(t)dt$ in a short interval. Therefore $\mathbb{E}(dN(t)\mid \text{past})=R(t)dH(t)$ to first order. Equivalently the compensator of a counting process is ${\int }_0^{t}R(s)dH(s)$. Dividing the observed increments by current exposure gives the estimating equation $d\hat{H}=dN/R$, henceHere $t_j$ are distinct event times, $d_j$ the number of events at that time, and $r_j=R(t_j)$ the number at risk immediately before it. One can also obtain the same increment by maximizing the local multiplicative-intensity likelihood $a^{d_j}{e}^{-r_{j}a}$: differentiation gives $a=d_j/r_j$. This derives the Nelson–Aalen estimator, rather than merely naming it.

A censored observation contributes to the denominator while under observation, and then leaves the risk set; it contributes no event jump. For tied events, use the single increment $d_j/r_j$, not $d_j$ successive one-event increments with shrinking denominators. If censoring and events share a recorded time, a convention is needed; counting the event before removing same-time censored subjects uses the pre-time risk set above. Rounded or interval-censored data may require a model suited to their observation mechanism.

For the distinct ordered observations in the question, let ${\delta }_j$ indicate whether $X_j$ is an event. The risk set at $X_j$ contains precisely the $n-j+1$ subjects whose recorded time is at least $X_j$. ThusTherefore the sum of the fitted integrated hazards at all observation times equals the observed number of events, the event-count identity for Nelson–Aalen cumulative hazards. The grouped Nelson–Aalen event-count identity extends the same argument to tied observations with their multiplicities and the stated risk-set convention.