Why Unobserved Transitions Matter
At a panel visit, a subject’s state is observed; between visits, it may not be. If the state is \(A\) at both ends of an interval, that observation does not by itself establish that the subject remained in \(A\). The subject may have left \(A\) and returned before the next visit. A likelihood that counts only paths consistent with “no change” can therefore omit possible histories and misrepresent the probability of the observed data.
The key calculation is to add the probabilities of all latent state sequences compatible with the observed endpoints. The transition-matrix likelihood for panel data supplies the endpoint probability; splitting an interval at an unobserved time makes the hidden-state summation explicit. This tutorial focuses on that summation, not on estimating a particular model’s transition parameters.
The state process must be Markov at the split times used for the calculation: conditional on the current state (and any modeled covariates or history), its future transition distribution cannot depend further on its earlier path. The state space and transition probabilities must also correctly represent the allowed transitions. If the visit schedule is informative about the unobserved path, its probability may need to be modeled too; the calculations below treat the schedule as fixed or noninformative for the state process.
Summing Over Hidden States
Suppose an observation interval from \(s\) to \(t\) is split at an unobserved time \(u\). Write \(P^{(1)}=P(s,u)\) and \(P^{(2)}=P(u,t)\). The state at \(u\) could be any state \(k\) in \(S\). Conditional on that state, the Markov property gives the probability of the two-part path as \(P^{(1)}_{ik}P^{(2)}_{kj}\). Because the intermediate state is not observed, add these mutually exclusive possibilities:
This is matrix multiplication. With more split times, sum over every sequence of hidden states, multiplying the transition probabilities along that sequence:
A term in this sum is the probability associated with one state sequence on the chosen time grid. Importantly, a matrix entry for a subinterval may itself aggregate paths within that subinterval. Thus, the grid describes which intermediate states are explicitly summed over; it does not necessarily enumerate every transition time.
Worked Example: A Same-State Observation With a Leave-and-Return Path
Consider two states, \(A\) and \(B\), and a hypothetical interval split into two equal parts. Suppose the transition-probability matrix for each part is
where rows give the starting state and columns give the ending state. A subject is observed in \(A\) at the beginning and \(A\) at the end. The hidden state at the split could be \(A\) or \(B\), so
The first term, \(0.64\), is the grid path \(A\to A\to A\). The second, \(0.06\), is \(A\to B\to A\): the subject leaves the observed state and returns by the end. The total endpoint likelihood contribution is \(0.70\), not \(0.64\). The calculation is conditional on the stated transition matrices and Markov assumption; it does not claim that the observed endpoints reveal which path occurred.
Return Paths in a Larger State Space
With more than two states, several distinct hidden states may provide routes back to the observed state. Each compatible route contributes its own product of transition probabilities. Keeping these terms separate during calculation can make it easier to verify that no allowed path has been omitted.
Worked Example: Summing Two Different Return Routes
Suppose a hypothetical three-state process has states \(0,1,2\), and an interval is split into two parts. Use the transition matrices
Each row sums to 1, as required for a transition-probability matrix. A subject observed in state \(0\) at both visits may have been in state \(0\), \(1\), or \(2\) at the split. Therefore,
The contributions are \(0.560\) for the grid path \(0\to0\to0\), \(0.050\) for \(0\to1\to0\), and \(0.015\) for \(0\to2\to0\). Both return routes belong in the likelihood because either is compatible with the observed endpoints. The remaining endpoint probability is not evidence that any particular route occurred; the intermediate state is unobserved.
From Interval Contributions to a Subject Likelihood
For observations \(x_0,x_1,\ldots,x_m\) at times \(t_0<t_1<\cdots<t_m\), the likelihood contribution for one subject under a Markov model is the product of the transition probabilities across consecutive visits. At each interval, its endpoint probability already sums over compatible hidden paths in that interval. Conditional independence across subjects then gives the full likelihood as the product of their subject-level contributions.
Here \(n\) indexes subjects, \(r\) indexes their observed intervals, and \(\theta\) denotes model parameters. This product uses observed states only. It does not require selecting a single hidden path for each subject. If a model instead includes a latent state at a split time, summing over that state recovers the same endpoint contribution.
Worked Example: A Subject With Two Panel Intervals
Use the two-state matrix \(K\) from the first example for each half-interval. In a hypothetical panel, one subject is observed in state \(A\) at time 0, in \(B\) at time 2, and in \(B\) at time 4. Each observed interval contains two half-intervals. The first contribution, from \(A\) to \(B\), is
The two terms correspond to \(A\to A\to B\) and \(A\to B\to B\). The second contribution, from \(B\) to \(B\), is
The \(0.06\) term is the return path \(B\to A\to B\). Multiplying the observed-interval contributions gives the subject-level likelihood:
Thus, under the specified model, the probability of this observed sequence is \(0.165\). The calculation does not assign the subject a known intermediate history; it sums over the alternatives for each interval.
Derivation, Checks, and Failure Modes
The hidden-state sum follows directly from conditioning on the state at the split. For the event \(X(s)=i\) and \(X(t)=j\), the events \(\{X(u)=k\}\), \(k\in S\), partition all possible states at \(u\). By the Markov property, for each such \(k\),
Adding over the disjoint possibilities yields
Repeated conditioning gives the corresponding sum over state sequences for any number of splits. This derivation depends on having a state process for which the Markov property holds at the split times; an omitted variable that carries relevant information about the future can invalidate the factorization.
Useful checks include verifying that every transition matrix has nonnegative entries and rows summing to 1, and that each endpoint probability lies between 0 and 1. If all endpoint probabilities are computed, the resulting transition matrix should also have rows summing to 1. For a same-state endpoint, check explicitly whether the model permits departures and returns. If it does, confirm that all allowed hidden-state routes have been included.
- Dropping return paths. Keeping only the “stay in the state” term confuses equal observed endpoints with an unchanged latent history. Sum over all compatible hidden states instead.
- Using a forbidden route. A path through a state that cannot be reached under the specified model must have probability zero, not an invented positive contribution. Check the model’s transition structure.
- Multiplying instead of summing over alternatives. Hidden states at a split are mutually exclusive possibilities, so their path probabilities are added. Probabilities across consecutive observed intervals are multiplied.
- Assuming the state alone is sufficient. If the future depends on prior history beyond the current state, the Markov factorization may fail unless the state definition is expanded or the dependence is modeled.
An alternative is to use the interval transition matrix directly, without explicitly listing hidden paths; its relevant entry is already the required endpoint probability. Explicit path summation is useful for understanding or checking that entry. Choosing a single “most likely” path is not equivalent to summing: it discards the probability mass of other compatible paths. In irreversible models, some return routes are impossible, but same-state endpoint probabilities still need to reflect every route the model does allow. The main limitation is that observed endpoints alone generally do not reveal the path or identify every transition probability; the likelihood calculation accounts for uncertainty in paths but cannot create information that was not observed.
Check Your Understanding
For each question, use the stated transition probabilities and show how hidden states enter the calculation. Solutions follow each question.
- For two states \(A,B\), let \(K=\begin{pmatrix}0.90&0.10\\0.40&0.60\end{pmatrix}\) apply over each of two subintervals. Find the probability of observing \(A\) at both endpoints, and identify the return-path contribution.
- Why are probabilities for different hidden states at a split added, while contributions from consecutive observed intervals are multiplied?
- In a three-state model, suppose a route \(0\to2\to0\) is forbidden. What should its contribution to \(P_{00}\) be?
- What assumption justifies factoring a two-part path probability into the product of its transition probabilities?
Solutions. (1) The hidden midpoint is \(A\) or \(B\), so \(P_{AA}=(0.90)(0.90)+(0.10)(0.40)=0.81+0.04=0.85\). The return path \(A\to B\to A\) contributes \(0.04\). (2) Hidden states are mutually exclusive alternatives, so their probabilities are summed; successive observed-interval contributions form a joint sequence probability and are multiplied under the Markov factorization. (3) Its contribution is zero because the model assigns that route probability zero. (4) The Markov property at the split time: conditional on the state there, the future transition distribution does not depend further on the earlier path.