Why an Observed Path Has a Likelihood
A multi-state observation records more than a starting state and an endpoint. It can record which transitions occurred, when they occurred, and how long the process remained in each state. Under an intensity model, these details contribute to the likelihood in two ways: each observed jump contributes its transition intensity, and each period without a jump contributes a term involving the total intensity of leaving the occupied state.
For example, if a person is observed in state A and then moves to B, the path gives evidence about both the A-to-B transition and the absence of any other exit from A before that jump. After arrival in B, observation without another transition also carries information: it indicates that no transition out of B occurred during that time. The endpoint alone would not retain all of this path information.
The formula below concerns a fully observed path over a fixed observation interval. It relies on the earlier material on event-history likelihoods from intensities, but states explicitly how the contributions combine for a multi-state process.
Notation and Assumptions
Assume the process starts in a known state at time 0 and is observed continuously through a fixed time \(\tau\). Its path is piecewise constant, has finitely many jumps on \([0,\tau]\), and has no simultaneous transitions. The intensities are predictable, nonnegative, and integrable over the observed interval. A transition that cannot occur has intensity zero. We condition on any observed covariate paths used to define the intensities; if relevant covariates or history are unobserved, a likelihood for the observed states alone may require integrating over them.
Write the observed transition times as \(0<t_1<\cdots<t_M\leq\tau\). At time \(t_m\), the path moves from \(X(t_m-)\) to \(X(t_m)\). Given the initial state, the likelihood contribution of the path, expressed as a density with respect to the ordered jump times and transition sequence, is
If the initial state is random and its distribution has unknown parameters, its probability must also be included. If it is fixed by design or conditioned upon, it contributes no parameter-dependent factor. A period ending at \(\tau\) with no further jump still contributes the integrated exit intensity through \(\tau\): the absence of a jump is part of the observed path.
Why Jumps and Sojourns Both Contribute
The expression follows from the conditional behavior over a short interval of length \(du\). While the process is in state \(j\), the conditional probability of a jump to a particular state \(k\) is \(\lambda_{jk}(u\mid\mathcal F_{u-})du+o(du)\). The conditional probability of no exit to any state is \(1-\lambda_j(u\mid\mathcal F_{u-})du+o(du)\). Multiplying these contributions across short intervals gives a transition-intensity factor at each observed jump and an accumulation of no-exit factors between jumps.
To see the no-exit term, first consider a setting in which the intensity along the no-exit path is specified by the time and the relevant history or covariate trajectory. Let \(S(u)\) be the conditional probability of no exit up to \(u\), given the full history at time \(s\) and that trajectory. Over a short interval, \(S(u+du)=S(u)\{1-\lambda_j(u\mid\mathcal F_{u-})du+o(du)\}\). Thus \(dS(u)/du=-\lambda_j(u\mid\mathcal F_{u-})S(u)\), whose solution is
The qualification matters: this is not generally a survival formula conditional only on \(X(s)=j\). Different histories that end in the same state can imply different future intensities. If a future covariate path is random and not conditioned upon, the conditional survival probability generally averages the exponential over that path; it need not equal an exponential formed from an averaged intensity.
Applying the short-interval factors over the full path gives the likelihood above. During each sojourn, the integrand sums all intensities out of the occupied state, not just the intensity for the destination eventually observed. At a jump, however, the factor is the intensity for the particular transition that occurred. If an absorbing state has no possible exits, its total exit intensity is zero, so remaining there adds no further survival penalty.
Worked Example: A Two-Jump Path With Constant Intensities
Suppose a hypothetical process starts in state 0 at time 0. It can move only from 0 to 1, at intensity \(0.30\) per year, and from 1 to 2, at intensity \(0.40\) per year; state 2 is absorbing. The observed path is \(0\to1\) at year 2, followed by \(1\to2\) at year 5, and observation ends at year 6.
The process spends 2 years in state 0, 3 years in state 1, and 1 year in absorbing state 2. The two observed jumps contribute \(0.30\) and \(0.40\). The integrated exit intensity is \(0.30(2)+0.40(3)+0(1)=1.8\). Therefore,
This is a path density for two specified jump times, not a probability that the process follows precisely those times. Its units reflect the two jump-time density contributions. The final year in state 2 is included in the path but contributes no exit term because state 2 is absorbing.
Time-Varying and History-Dependent Intensities
The same construction applies when an intensity changes with calendar time or with the observed history. The path determines which state is occupied during each interval, and the relevant intensity is evaluated along that path. For a time-varying intensity, each no-jump contribution is an integral, not the intensity at one representative time multiplied by the interval length, unless the intensity is constant on that interval.
Worked Example: Integrating Time-Varying Exit Intensities
In a hypothetical two-state process, the only transition from A is to B, with intensity \(\lambda_{AB}(t)=0.20+0.10t\) per year. From B, the only transition is to A, with intensity \(\lambda_{BA}(t)=0.10+0.20t\) per year. The process starts in A, moves to B at year 1.5, returns to A at year 3, and is observed through year 4.
In A from 0 to 1.5, the integrated exit intensity is \(\int_0^{1.5}(0.20+0.10t)\,dt=0.30+0.1125=0.4125\). The jump intensity at year 1.5 is \(0.20+0.10(1.5)=0.35\) per year. In B from 1.5 to 3, the integrated exit intensity is \(\int_{1.5}^{3}(0.10+0.20t)\,dt=0.15+0.675=0.825\). The B-to-A jump intensity at year 3 is \(0.10+0.20(3)=0.70\) per year. Finally, the process is back in A from year 3 to 4, contributing \(\int_3^4(0.20+0.10t)\,dt=0.20+0.35=0.55\).
The rates are evaluated at the actual jump times, while the no-jump contributions use the integrated rates over each sojourn. The calculation assumes these specified calendar-time intensities apply along the observed path.
For a Markov model whose transition intensities depend only on time and the current state, the intensity on a sojourn in state \(j\) is determined by \(j\) and time. In a more general model, it may also depend on the history or covariates. The likelihood still uses the intensity evaluated along the observed history, but interpreting it as a conditional path likelihood requires that the relevant history be observed or that the likelihood be formed jointly with, or marginalized over, any unobserved components.
Worked Example: Competing Exits and the Total Hazard
Consider a hypothetical process that starts in A. From A it can move to B with intensity \(\theta\), or to C with intensity \(2\theta\); both B and C are absorbing. The path moves from A to B at year 1 and is observed through year 2. Conditional on starting in A, the likelihood as a function of \(\theta\) is the A-to-B intensity at the jump multiplied by the no-exit contributions before the jump and while in B.
The total exit intensity from A is \(\theta+2\theta=3\theta\), even though the observed destination is B. At \(\theta=0.40\) per year, the A sojourn contributes \(\exp\{-3(0.40)(1)\}\). The remaining year in absorbing state B contributes 1. Hence
The event intensity is \(\theta\), not \(3\theta\), because the observed jump was specifically to B. The total \(3\theta\) belongs in the no-exit term because either possible exit would have ended the sojourn in A.
Interpreting, Checking, and Comparing Path Likelihoods
Taking logarithms separates the contributions additively:
Here \(\lambda_{X(u-)}\) denotes the total exit intensity from the state occupied immediately before \(u\). This form makes a basic check straightforward: every observed jump has its own transition-specific term, and every observed sojourn contributes the integrated total exit intensity for its occupied state.
| Quantity | Contribution to the path likelihood | Meaning |
|---|---|---|
| Observed transition \(j\to k\) at time \(t\) | \(\lambda_{jk}(t\mid\mathcal F_{t-})\) | Rate for the particular transition observed |
| Time spent in state \(j\) without a jump | \(\exp\{-\int \lambda_j(u\mid\mathcal F_{u-})\,du\}\) | Contribution from no exit to any other state |
| Observed endpoint with no further jump | Same integrated exit term through the observation endpoint | The process remained in its final state until observation ended |
An alternative is to work with transition probabilities between two times. Those probabilities describe the chance of being in a given state at the later time and, unless the full path is specified, incorporate all possible intermediate paths. A path likelihood instead uses the actual observed transition sequence and times. It is therefore the appropriate object when the path itself is observed and modeled; it is not interchangeable with a probability for the endpoint alone.
Useful checks include verifying that each transition is allowed by the model, each jump intensity is positive for a transition that occurred, and each sojourn integral uses the state occupied over that interval. Compare observed transitions and state-specific exposure with the model-implied intensities when assessing fit. If the path contains unobserved transitions, interval-censored transition times, or unobserved variables that affect intensity, the fully observed-path formula is not by itself the likelihood for the recorded data. Simultaneous transitions also violate the no-simultaneous-jumps assumption used here and require an appropriate observation and model formulation.
Common Mistakes and Limitations
- Using only the intensity for the observed destination in the no-jump term. That omits the possibility of leaving the state by another transition. Use the sum of all exit intensities during each sojourn.
- Using the total exit intensity as the jump factor. The jump factor is for the specific transition observed; the total intensity belongs in the integrated no-exit term.
- Replacing time-varying intensity by a single rate. Unless it is constant, integrate it over the full sojourn.
- Conditioning only on the current state when intensity depends on history. The exponential survival expression applies conditional on the relevant history and covariate trajectory, evaluated along the no-exit path. Averaging over different histories may not preserve that exponential form.
- Treating a path density as a path probability. Exact jump times are continuously distributed; the likelihood is a density relative to a specified reference measure.
The likelihood also depends on the observation scheme. If follow-up ends at a fixed administrative time, the observed no-jump period contributes through that time. If observation ends for a reason related to the transition process, the observation mechanism may need to be modeled or appropriately conditioned upon. The formula is not a substitute for handling incomplete or selective observation.
Check Your Understanding
Try each problem, then compare your reasoning with the solutions.
- A process starts in state 0. It has only a \(0\to1\) transition, at constant intensity \(0.40\) per year, and a \(1\to2\) transition, at constant intensity \(0.20\) per year. It moves from 0 to 1 at year 1.5 and is observed in state 1 through year 2. Write and calculate the path likelihood.
- In state A, the intensities to B and C are \(0.10\) and \(0.25\) per year. What total exit intensity belongs in the no-exit contribution for a sojourn in A? If a jump to B occurs, what intensity is its jump factor?
- Conditional on the full relevant history, a process remains in state \(j\) while its exit intensity over years 0 to 2 is \(0.10+0.20t\) per year. Find the probability of no exit over that interval under the specified intensity trajectory.
- Given state \(j\) at time \(s\), two equally likely histories imply constant exit rates 1 and 3 per year for the next year. Compare the conditional survival probability with the exponential formed using the average rate.
Solutions. (1) The jump factor is \(0.40\); the process spends 1.5 years in state 0 and 0.5 years in state 1, so \(L=0.40\exp\{-0.40(1.5)-0.20(0.5)\}=0.40e^{-0.70}\approx0.1986\) per year. (2) The total exit intensity is \(0.10+0.25=0.35\) per year; for an observed jump to B, the jump factor is \(0.10\) per year. (3) The integrated intensity is \(\int_0^2(0.10+0.20t)\,dt=0.20+0.40=0.60\), so the conditional no-exit probability is \(e^{-0.60}\approx0.5488\). (4) The conditional survival probability is \((e^{-1}+e^{-3})/2\approx0.2088\). Using the average rate gives \(e^{-2}\approx0.1353\), which differs because averaging the rates before exponentiating is not the same as averaging the history-specific survival probabilities.