Event-History Likelihoods From Intensities

From an Intensity to a Path Likelihood

An intensity describes the instantaneous rate of an event conditional on the history observed just before that time. A likelihood answers a different question: how compatible is an entire observed event history with a specified intensity model? The connection is direct. Each observed event contributes the intensity at its event time, while all time spent under observation contributes a factor that decreases with accumulated intensity.

The result is a likelihood for an observed path, conditional on the initial state and any covariate history treated as given. It does not say that the intensity itself is a probability, nor is the likelihood of a path generally the probability of that exact path. For continuous event times, the path likelihood is a density with respect to a reference measure on event times.

Definition: Let $N(t)$ be a simple counting process observed on $[0,\tau]$, with conditional intensity $\lambda(t\mid\mathcal H_{t-})$. Here $\mathcal H_{t-}$ denotes the history just before time $t$, and the intensity is predictable and nonnegative. If the observed path has ordered event times $0<t_1<\cdots<t_m\leq\tau$, its conditional likelihood density, up to factors not involving the event-process model, is $$ L=\left\{\prod_{j=1}^{m}\lambda(t_j\mid\mathcal H_{t_j-})\right\} \exp\left\{-\int_0^\tau \lambda(u\mid\mathcal H_{u-})\,du\right\}. $$ The product is the event contribution; the exponential is the accumulated-exposure contribution.

This expression applies when the intensity specifies the conditional event law, the process has no simultaneous jumps, and the integrated intensity is well-defined over the observation interval. The model must assign positive intensity at each observed event time, at least in the sense required for a positive density. The path is evaluated using its actual history: if the intensity changes after an event, its later values are evaluated after that event.

For multiple event types, write $\lambda_k(t\mid\mathcal H_{t-})$ for the type-$k$ intensity and $\lambda_\cdot(t)=\sum_k\lambda_k(t)$ for the total intensity. When at most one type can occur at a time, each event contributes the intensity for its observed type, and the accumulated term uses the total intensity. This describes a typed path; it does not assume the type-specific intensities are constant or independent.

L= \left\{\prod_{j=1}^{m}\lambda_{k_j}(t_j\mid\mathcal H_{t_j-})\right\} \exp\left\{-\int_0^\tau\sum_k\lambda_k(u\mid\mathcal H_{u-})\,du\right\}.

Why the Exposure Term Appears

The accumulated term reflects the requirement that no event occur during the intervals between the recorded events, including any remaining interval after the last event. To see this, consider a short interval of length $dt$ in which the history is fixed up to its start. By the definition of an intensity, the conditional probability of an event in that interval is approximately $\lambda(t)dt$, and the conditional probability of no event is approximately $1-\lambda(t)dt$. Multiplying the no-event contributions across small intervals gives the exponential survival factor.

Divide time into short intervals.
On an interval without an event, its conditional no-event contribution is approximately $1-\lambda(u)\,du$.

Accumulate the no-event contributions.
The product over intervals approaches $\exp\{-\int_0^\tau\lambda(u)\,du\}$, since $\log(1-\lambda(u)\,du)$ is approximately $-\lambda(u)\,du$.

Insert the observed jumps.
At each event time, the event-time density contributes $\lambda(t_j)$. Multiplying these terms by the no-event factor gives the path likelihood.

This is a justification through increasingly fine intervals: the event terms are densities in time, while the exponential accounts for the whole observed interval. The formula is conditional on the specified model and observation setup. If the distribution of covariates or the initial state is also part of the statistical model, its likelihood contribution must be included for a joint likelihood.

Worked Example: Constant Intensity Over Four Days

Suppose a hypothetical process has constant intensity $\lambda(t)=0.30$ events per day. One subject is observed for four days, with events at days 1 and 3 and no other events. Assume the constant-intensity model applies throughout follow-up and that observation is complete to day 4.

There are two event contributions, each $0.30$ per day. The integrated intensity over the four days is $0.30(4)=1.20$. Thus the path likelihood density is

L=(0.30)(0.30)\exp\{-0.30(4)\} =0.0900\exp(-1.20) =0.0900(0.3012) \approx 0.02711\ \text{per day}^2.

The units are per day squared because the path contains two continuous event times. The likelihood is a density, not the probability of observing events at exactly days 1 and 3. The exposure term includes all four days, not merely the gaps before the observed events.

Histories With Changing Intensities

An intensity need not be constant. It may vary with time or with the event history. In either case, the same path-likelihood structure applies: evaluate the intensity immediately before each observed event, then integrate the intensity along the observed path over the entire observation period. For a deterministic piecewise-constant intensity, the integral is the sum of intensity multiplied by time spent in each segment.

Worked Example: Piecewise-Constant Intensity

Consider a hypothetical event process observed from time 0 to time 5 days. Its intensity is $0.20$ events per day from day 0 up to day 2, then $0.50$ events per day from day 2 through day 5. One event occurs at day 3.5, with no other events. Assume this intensity is the correct conditional intensity on both intervals.

The event occurs in the second interval, so its event contribution is $0.50$ per day. Accumulated intensity is

\int_0^5\lambda(u)\,du =(0.20)(2)+(0.50)(3)=0.40+1.50=1.90.

Therefore,

L=0.50\exp(-1.90) =0.50(0.1496) \approx 0.07478\ \text{per day}.

The first two days contribute exposure even though they contain no event. Omitting that interval would overstate the likelihood because the observed path includes survival without an event through that time.

For a history-dependent intensity, the same calculation must follow the observed history segment by segment. For example, if an event changes the process state and hence changes its subsequent intensity, the post-event intensity is used only after that event. The integral is therefore not necessarily an integral of a fixed curve chosen independently of the path.

Worked Example: A Typed Event Path

Suppose a hypothetical subject can experience event type A or type B during three years of follow-up. Assume the conditional intensities are constant at $\lambda_A=0.20$ per year and $\lambda_B=0.10$ per year, with no simultaneous events possible. The observed path has one type-A event at year 1.5 and no other event by year 3.

The observed event contributes $\lambda_A=0.20$ per year. The total intensity during follow-up is $0.20+0.10=0.30$ per year, so the accumulated total intensity is $0.30(3)=0.90$. Hence the typed-path density is

L=0.20\exp(-0.90) =0.20(0.4066) \approx 0.08131\ \text{per year}.

The exponential accounts for remaining free of either event type when no event is observed. At the observed jump, the event contribution is the type-A intensity, not the total intensity. This calculation relies on the stated intensity model and complete observation through year 3.

Conditions, Diagnostics, and Alternatives

Before using a path likelihood, check that the observation scheme and intensity model match. In particular, the formula above presumes the process is observed over the stated interval, the path records all relevant jumps, and its conditional intensity is specified relative to the available history. If follow-up ends earlier for a reason related to the event process, or if some events are unrecorded, the observed path is not described by this expression without further modeling.

Useful diagnostics and failure signals include:

An impossible observed event. If the model intensity is zero at an observed event time, the event contribution is zero and the likelihood density for that path is zero under the model.
Incorrect exposure limits. Integrate over all time the subject is under observation, including intervals before the first event and after the last event. Use the actual observation interval, not automatically the full study calendar.
Wrong intensity after a jump. For a history-dependent model, use the intensity corresponding to the current state or history on each segment.
Missing event types in the no-event term. If any of several types could occur, the probability of no event of any type depends on their total intensity.
Unmodeled observation or covariate process. A conditional event-process likelihood does not automatically account for informative observation, or for a stochastic covariate process whose distribution is part of the target joint model.

A direct event-time calculation using conditional densities and probabilities of no event is an equivalent route to the same likelihood when it uses the same intensity assumptions. A discrete-time model instead assigns probabilities to events in time intervals; it can be suitable when data are genuinely recorded in intervals, but it is not automatically identical to a continuous-time intensity likelihood. Standard survival analysis gives a familiar single-event formulation; the intensity expression is useful for path data, recurrent events, or multiple event types because it keeps event contributions and exposure explicit. Detailed model-specific likelihood constructions require additional assumptions about the event process and observation design.

Common pitfall: Multiplying the intensity at each observed event and stopping there ignores the fact that the path also records periods with no event. The correction is to include $\exp\{-\int_0^\tau\lambda(u)\,du\}$, or the corresponding total-intensity term for multiple event types. Conversely, the exponential alone ignores the observed jumps. Both parts are necessary.

Key takeaway: A path likelihood has two complementary parts: intensities evaluated at observed jumps and an exponential penalty determined by integrated intensity over observed exposure. The event terms describe what happened; the accumulated term describes the time during which events could have happened but did not.

Check Your Understanding

Answer the questions, then compare with the solutions below.

A process has constant intensity $0.4$ per hour and is observed for 2 hours with no events. What is the path likelihood?
For a path with two events, why does its continuous-time likelihood density have units involving time?
Two event types have intensities $0.15$ and $0.25$ per week. What intensity determines the no-event contribution?
If an observed event occurs where the model intensity is zero, what does the model assign to that path?
In a piecewise-constant model, what time periods must be included in the integrated intensity?

Solutions. (1) With no event, $L=\exp\{-(0.4)(2)\}=\exp(-0.8)\approx0.4493$. (2) Each event-time intensity contributes inverse-time units, so two events give a density with inverse-time-squared units. (3) The total intensity is $0.15+0.25=0.40$ per week. (4) Its event contribution is zero, so the path likelihood density is zero under that model. (5) Include every observed interval, including time before the first event and after the last event, splitting intervals wherever the intensity changes.

Event-History Likelihoods From Intensities

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