Interpreting Stochastic Intensities

What an Intensity Says

An event rate is easy to misread when its value changes with time or with what has happened so far. A stochastic intensity makes that dependence explicit: it describes the event rate conditional on the history available just before the time in question. It is not, in general, a probability density for event times in the population.

Let $N(t)$ count events up to time $t$, and let $\mathcal F_{t-}$ denote the observed history immediately before $t$. For a simple counting process, an event produces a jump of size one, and simultaneous jumps are excluded. We assume the process is locally finite, so only finitely many events occur in any bounded time interval. Its intensity $\lambda(t\mid\mathcal F_{t-})$ describes the conditional first-order event rate given that history.

Definition: A stochastic intensity $\lambda(t\mid\mathcal F_{t-})$ is a nonnegative, history-dependent rate such that, over a sufficiently short interval of length $h$, $$ P\{N(t+h)-N(t)=1\mid\mathcal F_{t-}\} =\lambda(t\mid\mathcal F_{t-})h+o(h), $$ and the conditional probability of two or more events is $o(h)$, as $h$ tends to zero. The intensity is predictable: its value at time $t$ is determined by information available just before $t$. It has units of events per unit time.

The $o(h)$ term means a remainder that becomes negligible relative to $h$ as $h$ shrinks. Thus, an intensity of $0.20$ per year does not mean that the event probability in every year is $0.20$. It means that, conditional on the current history, the probability of an event in a very short interval of length $h$ years is approximately $0.20h$. For finite intervals, the probability also depends on how the rate and event history may change during the interval.

History, Risk Sets, and Conditional Hazards

For a single terminal event, let $T$ be the event time and define $Y(t)=I(T\geq t)$, the indicator that the individual remains at risk just before $t$. If the event-time hazard conditional on covariates $Z$ and the observed history is $\lambda_0(t\mid Z,\mathcal F_{t-})$, the counting-process intensity is

\lambda(t\mid\mathcal F_{t-}) =Y(t)\lambda_0(t\mid Z,\mathcal F_{t-}).

While the individual is at risk, the intensity equals the conditional hazard. After the event, $Y(t)=0$, so the intensity for another occurrence of that terminal event is zero. In recurrent-event settings, by contrast, the count may continue to increase and the rate after an event can depend on the updated history. In either setting, the intensity is conditional on the history; it is not a stand-alone description of the distribution of $T$.

Worked Example: A Rate That Changes With Event History

Suppose a hypothetical recurrent-event process has intensity $\lambda(t)=0.05+0.10N(t-)$ per year, where $N(t-)$ is the number of previous events just before $t$. At a time when two events have already occurred, interpret the intensity and estimate the probability of another event in the next $0.01$ year.

With $N(t-)=2$, the rate is $$ \lambda(t)=0.05+0.10(2)=0.25\ \text{events per year}. $$ Conditional on the history with two previous events, the short-interval probability is $$ P\{\text{one event in }[t,t+0.01)\mid\mathcal F_{t-}\} \approx 0.25(0.01)=0.0025. $$

The approximation is $0.0025$, or about $0.25\%$, for that $0.01$-year interval. It is a local approximation, not an exact finite-interval probability: if another event occurs, $N$ changes and so does the intensity. The rate is conditional on the observed event count, and its units are events per year.

Intensity Is Not Event-Time Density

A density $f_T(t)$ describes the unconditional distribution of an event time: for a small interval, $P(t\leq T<t+h)$ is approximately $f_T(t)h$. The intensity instead describes a rate conditional on a particular history. The two can coincide in simple settings, but they generally answer different questions. In particular, a density is weighted by how much probability remains at risk at time $t$, whereas an at-risk conditional rate is not.

A useful illustration is a population with unobserved rate differences. Suppose half the individuals have an exponential event time with rate $0.5$ per year and half have rate $2$ per year. Conditional on an individual’s group, the intensity while the individual remains event-free is that group’s rate. But among individuals still event-free at time $t$, the groups are no longer equally represented: individuals in the higher-rate group tend to have had the event earlier.

Worked Example: Mixture Density and the Event-Free Population Rate

Let $G$ be the unobserved group, with $P(G=1)=P(G=2)=0.5$. Conditional on group $g$, let $T$ be exponential with rate $\lambda_1=0.5$ per year or $\lambda_2=2$ per year, respectively. Find the population survival probability, density, and hazard at $t=1$ year. Then explain why the density is not the intensity for a known group.

For an exponential time with rate $\lambda_g$, the conditional survival function is $e^{-\lambda_g t}$ and the conditional density is $\lambda_g e^{-\lambda_g t}$. Averaging over the two equally likely groups gives $$ S_T(t)=0.5e^{-0.5t}+0.5e^{-2t}, \qquad f_T(t)=0.5(0.5)e^{-0.5t}+0.5(2)e^{-2t}. $$

At $t=1$ year, the survival probability is $$ S_T(1)=0.5(0.6065306597)+0.5(0.1353352832) =0.3032653299+0.0676676416 =0.3709329715\approx 0.3709. $$ The population density is $$ f_T(1)=0.25(0.6065306597)+1(0.1353352832) =0.1516326649+0.1353352832 =0.2869679481\approx 0.2870\ \text{per year}. $$

The population hazard among those still event-free is the density divided by survival: $$ \frac{f_T(1)}{S_T(1)} =\frac{0.2869679481}{0.3709329715} \approx 0.7736383 \approx 0.7736\ \text{per year}. $$ This is neither the density $0.2870$ per year nor either group-specific rate. It is the event-free population’s conditional rate at year 1. Given the group, the intensity is $0.5$ per year for group 1 or $2$ per year for group 2. When group is unobserved and only event-free status is known, the conditional rate is the survival-weighted mixture, $f_T(1)/S_T(1)$. The faster-rate group has become less common among those still event-free.

Why the Density-Hazard Relation Holds

For an absolutely continuous event-time distribution, survival and density satisfy $f_T(t)=-S_T'(t)$. The hazard among those still event-free is defined as the instantaneous event rate conditional on survival to $t$. Since the probability of being event-free at $t$ is $S_T(t)$, dividing the unconditional density by this probability gives that conditional rate:

\lambda_{\mathrm{pop}}(t) =\lim_{h\downarrow0} \frac{P(t\leq T<t+h\mid T\geq t)}{h} =\frac{f_T(t)}{S_T(t)}.

The relation holds almost everywhere where $S_T(t)>0$, with $f_T(t)=-S_T'(t)$ almost everywhere. In the mixture example, this population hazard is an average of group rates using the group proportions among survivors at that time, not the original 50/50 proportions. The derivation also shows why a density and a conditional rate have different interpretations: the density concerns probability per unit time in the original population, while the hazard conditions on remaining event-free.

Separate Intensities for Competing Transitions

When multiple event types can occur, let $N_j(t)$ count transitions of type $j$, and let $\lambda_j(t\mid\mathcal F_{t-})$ be its intensity. Each rate describes the instantaneous occurrence rate for that particular type given the same history. If at most one transition can occur at a time, the total intensity for any transition is the sum of the type-specific intensities. This is a local statement; it does not say the type-specific event-time densities are the intensities.

Worked Example: Interpreting Competing Transition Rates

A hypothetical individual is in state 0 and can move either to state 1 or to state 2. At the current history, the respective transition intensities are constant over the next $0.1$ year at $0.08$ and $0.12$ per year. Compute the total rate, the short-interval approximation, and the exact probabilities under the stated constant-rate assumption.

The total instantaneous rate is $$ \lambda_{\mathrm{total}}=0.08+0.12=0.20\ \text{per year}. $$ The first-order approximation for any transition during $0.1$ year is $0.20(0.1)=0.020$, or $2.0\%$. Because the rates are assumed constant and the individual remains in state 0 until a transition, the exact probability of any transition by $0.1$ year is $$ 1-e^{-(0.20)(0.1)}=1-e^{-0.02}\approx 0.01980. $$

For a constant-rate competing transition model, the probability of a transition of type $j$ by time $h$ is $(\lambda_j/\lambda_{\mathrm{total}})(1-e^{-\lambda_{\mathrm{total}}h})$. Thus the exact probabilities are $$ P(0\to1)=\frac{0.08}{0.20}(1-e^{-0.02})\approx 0.007921, \qquad P(0\to2)=\frac{0.12}{0.20}(1-e^{-0.02})\approx 0.01188. $$ They sum to $0.01980$, the exact probability of any transition. The intensities $0.08$ and $0.12$ are rates conditional on the current history; the finite-interval probabilities account for the possibility of remaining in state 0 throughout the interval.

Assumptions, Diagnostics, and Common Pitfalls

The conditional-rate interpretation relies on the event process being well-defined over the observation window, the history being specified correctly, and the intensity being predictable and locally integrable. The small-interval interpretation also uses the simple-process condition that the probability of multiple events in an interval of length $h$ is negligible relative to $h$. For a terminal event, the risk indicator must be zero after the event; for transition-specific rates, the individual must be in a state from which that transition is possible.

In practice, check that the proposed rate uses only information available before the event time. A covariate measured after an event cannot be used as though it were known just before that event. Check that the stated units are inverse time, that the rate is zero when the event is impossible or the individual is no longer at risk, and that distinct event types have been identified clearly. If rates are reported as probabilities over a long interval without accounting for changes in risk or history, the approximation $\lambda h$ may be poor.

Mistake: calling an intensity a density. A density is unconditional probability per unit time; an intensity is a conditional instantaneous rate given history. The mixture example shows the numerical and interpretive difference.
Mistake: treating a stochastic rate as fixed for everyone. A rate can vary across histories or people. State what is being conditioned on before interpreting its value.
Mistake: multiplying a rate by a long time interval and calling the result an exact probability. The product is a short-interval approximation. Changes in rate and survival through the interval matter for an exact probability.

The event-time hazard is a useful description when the question concerns the rate among individuals still event-free. A stochastic intensity generalizes that conditional-rate idea to history-dependent counting processes and multiple event types. An unconditional event-time density instead describes the population distribution of event times. Choosing between them is not a matter of terminology: it determines which population and conditioning set an estimate describes.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If an intensity is $0.30$ per month at the current history, what is the first-order conditional probability of an event in the next $0.02$ month?
Why can the population hazard in the two-group mixture change over time even though each group’s rate is constant?
Two competing transition intensities are $0.04$ and $0.06$ per year. What is the total instantaneous intensity, and what is the first-order probability of any transition in $0.01$ year?
In one sentence, distinguish an event-time density from a conditional intensity.

Solutions. (1) The short-interval approximation is $0.30(0.02)=0.006$, or $0.6\%$, conditional on the stated history. (2) The proportion of each group among event-free individuals changes as events occur; higher-rate individuals leave the event-free group sooner. (3) The total intensity is $0.04+0.06=0.10$ per year, giving the first-order approximation $0.10(0.01)=0.001$, or $0.1\%$. (4) A density gives unconditional event-time probability per unit time, whereas an intensity gives an instantaneous rate conditional on the history immediately before the time considered.