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The compensator of a counting process $N(t)$ is a predictable increasing process $\Lambda (t)$ such that $N(t)-\Lambda (t)$ is a local martingale. For an observed event process with at-risk process $Y(t)$ and hazard function $h(t)$, it is $\Lambda (t)={\int }_0^{t}Y(u)h(u)du$ under independent censoring. It is cumulative intensity, rather than an event probability.

An observed event counting process has conditional increment $\mathbb{E}(d{N}_i(t)\mid {\mathcal{H}}_{t-})=Y_i(t)h_i(t)dt$, where $Y_i$ is the at-risk process. The filtration records the past observed events, censoring, and covariates; the integral of $Y_i{h}_i$ is the compensator of a counting process.

If a counting process $N$ has predictable conditional intensity $\lambda$, subtracting its compensator of a counting process gives the associated martingale $M$. Conditional expected event increments and compensator increments agree, so their difference is conditionally centered. With the necessary integrability, $M$ is a mean-zero martingale; for a simple no-ties process its predictable quadratic variation is ${\int }_0^t\lambda (u)du$. A predictable weighted martingale integral has mean zero and variance determined by the square of its weights. In survival analysis, $\lambda =Yh$ produces the event-minus-expected representation used to derive the Nelson–Aalen estimator.

The entry count $N_s(t)$ counts jumps of a continuous-time Markov chain into state $s$ by time $t$, excluding initial occupation. It is a counting process. For homogeneous transition intensities, its mean has derivative ${\sum }_{i\ne s}{p}_{ri}(t)q_{is}$ conditional on initial state $r$, so its expected value is the sum of incoming rates times the expected occupation times of a continuous-time Markov chain. This density is an expected count rate, not necessarily a normalized probability density function.

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.

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 $h_{0i}+h_1\ge 0$.

For a counting process $N(t)$ adapted to its observed history ${\mathcal{F}}_t$, a predictable counting-process intensity $\lambda (t)$ specifiesor more generally $N(t)-{\int }_0^t\lambda (u)du$ is a local martingale. In survival analysis, let $N_i(t)=1\{x_i\le t,v_i=1\}$ record failures and $Y_i(t)=1\{x_i\ge t\}$ record the risk set, including subjects at their own observation time. Under independent right censoring and a common hazard function $h$, the aggregate intensity is $Y(t)h(t)$, where $N={\sum }_i{N}_i$ and $Y={\sum }_i{Y}_i$.

Writing $H(t)={\int }_0^{t}h(u)du$, the conditional increment equation is $E(dN\mid {\mathcal{F}}_{t-})=YdH$. Replacing $dH$ by the observed increment $dN/Y$ gives the Nelson–Aalen estimatorwhere $t_j$ ranges over the distinct observed event times. Censored observations remove subjects from subsequent risk sets but do not produce hazard jumps.

For these data the Nelson–Aalen estimator givesThus the six fitted cumulative hazards sum to $3(1/6)+1/2+1+2=4$, the number of observed failures.

The event-count identity for Nelson–Aalen cumulative hazards follows by exchanging the finite sums. With everyone entering at time zero,This includes each failure subject in its own risk set. A delayed-entry dataset requires a different risk indicator and is not covered by this particular identity.

Under a constant hazard survival model, $\hat{H}(x_i)=\hat{\theta }{x}_i$. Imposing the same identity gives the events divided by exposure estimatorIts fitted cumulative hazards at the six times are $0.3,0.4,0.5,0.6,0.9,1.3$, again summing to four. It is also the maximum-likelihood estimate under independent right censoring, since the parameter-dependent likelihood is ${\theta }^d{e}^{-\theta {\sum }_i{x}_i}$. This is a sensible estimate if the exponential survival model is appropriate, but four failures give little precision. The event-count identity by itself does not validate constant hazard; the large final jump in the Nelson–Aalen estimator also reflects a risk set of one, rather than by itself proving an increasing hazard.

In the counting-process intensity in survival analysis, the counting process $N_i(t)$ jumps once if an observed event occurs; the at-risk process $Y_i(t)$ is one immediately before $t$ only while individual $i$ is still observed and event-free. Conditional on that information, the chance of an event in a short interval is $h_i(t)dt+o(dt)$ for an at-risk individual and zero otherwise. This is the meaning ofThe history ${\mathcal{H}}_{t-}$ is the filtration generated by the observed event and censoring histories, entry information, and available covariates strictly before $t$. In particular, it determines the current risk set but does not reveal future event times. The right-hand side is the infinitesimal compensator of a counting process; the equality is interpreted as an intensity statement to first order in $dt$.

Apply the tower property of conditional expectation to the intensity equation. Since $h_i(t)$ is a specified individual hazard function,As in the conditional equation, this describes the first-order expected increment of the observed counting process.

Let $N(t)={\sum }_i{N}_i(t)$ and $Y(t)={\sum }_i{Y}_i(t)$ be the total observed counting process and size of the risk set. When the individual cumulative hazard functions are all $H_0$, summing the conditional intensity equations givesReplace the expected increment by its observed value and divide by the predictable risk set size. Thus the Nelson–Aalen estimator in integral form iswith the integrand defined as zero when $Y(u)=0$. This is an integral against the step counting process, so each event time contributes its event count divided by the risk set just before that time. The Nelson–Aalen estimator estimates integrated hazard over times with observable risk; it does not supply information after observation has ended.

Let $N(u)$ count all jumps of the continuous-time Markov chain. Its Markov-chain entry count into state $s$ isThusThis is a Lebesgue-Stieltjes integral against a counting process, so each jump contributes one. The initial occupation of state $s$ is not counted as an entry. An ordinary time integral of an indicator random variable asserting a jump at exactly time $u$ would be zero, since individual jump times have zero Lebesgue measure. The requested integral must therefore use this counting-measure interpretation. Assume non-explosion and finite expected jump counts on bounded intervals, as holds for a finite-state continuous-time Markov chain.

Define the event counting process and at-risk process byUnder a common hazard function and independent censoring, the conditional event rate is $Y(u)h(u)$. Equivalently, the counting-process compensator satisfies $\mathbb{E}[dN(u)\mid {\mathcal{F}}_{u-}]=Y(u)dH(u)$. Estimate a cumulative hazard increment by the observed event increment divided by the current risk set:With no ties, $dN$ has mass $v_i$ at $x_i$, and $Y(x_i)={\sum }_{j:x_j\ge x_i}1$. Thus the Nelson–Aalen estimator isOnly event times contribute, so no empty risk set is divided into an event count. Right-censored observations contribute time to the risk sets, but no event increment.

In the Cox proportional-hazards model, replace the number at risk in the Nelson–Aalen estimator by the sum of relative hazards:The total event counting process has conditional rate $W_{\beta }(u)h_0(u)$, so the estimated baseline hazard increment is $dN(u)/W_{\hat{\beta }}(u)$. Integrating gives the Breslow estimatorWhen $\hat{\beta }=0$, all relative hazards are 1 and this reduces to the Nelson–Aalen estimator. The denominator is positive at each observed event time.

$N(t)$ 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.

Let $Y(t)$ be the at-risk process and $N(t)$ the counting process for observed events. Over a short interval, the multiplicative-intensity model giveswhere $h$ is the hazard function and $H$ the cumulative hazard function. Solving this relation for the infinitesimal hazard increment suggests $d\hat{H}(t)=dN(t)/Y(t)$. Summing over distinct event times gives the Nelson–Aalen estimatorwhere $d_j$ events occur among $r_j$ individuals at risk. Here there are no ties, so $d_j=1$.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

For a bounded Borel set $W\subset {\mathbb{R}}^d$, the window count $N_W$ of a locally finite point process is a finite nonnegative integer random variable. It counts multiplicities, and the convention for window boundaries matters. On the time half-line, $N(t)=\Pi ((0,t])$ is a counting process and $N(t)-N(s)=\Pi ((s,t])$. For pairwise disjoint observation windows $W_1,\dots ,W_m$, the measure identityis valid for every realization. Additivity of counts supplies no independence assumption.