Recovering Transition Intensities From Panel Observations

From Continuous-Time Rates to Visit-to-Visit Probabilities

Panel data record a subject’s state at scheduled visits, not every transition that may have occurred between visits. A transition probability over a visit interval therefore summarizes all possible paths during that interval. Recovering the rates that generated those probabilities is a separate question: it requires a continuous-time model, and a single interval may not contain enough information to determine the rates uniquely.

Assume a finite state space and a time-homogeneous continuous-time Markov chain. “Time-homogeneous” means that the instantaneous transition rate from one state to another does not depend on calendar time. The Markov assumption means that, conditional on the current state, the future evolution does not additionally depend on the earlier path. For now, suppose visits are separated by a known interval of length \(t\), and the state is observed without error at both endpoints.

Definition: The generator \(Q=(q_{ij})\) is the matrix of instantaneous transition rates. For \(i\ne j\), \(q_{ij}\ge 0\) is the rate of moving from state \(i\) to state \(j\). Each diagonal entry is \(q_{ii}=-\sum_{j\ne i}q_{ij}\), so every row of \(Q\) sums to zero. The transition matrix \(P(t)=(p_{ij}(t))\) has entries \(p_{ij}(t)=P(X(t)=j\mid X(0)=i)\).

We use a row-vector convention: probability distributions multiply transition matrices on the left. Under the assumptions above, the transition matrix over an interval of length \(t\) is

P(t)=\exp(tQ)=\sum_{m=0}^{\infty}\frac{(tQ)^m}{m!}.

This is a matrix exponential, not the entry-by-entry exponential of \(tQ\). The diagonal of \(P(t)\) includes paths with no transitions and paths that leave and return. An off-diagonal entry requires at least one transition from the starting state to a different ending state. More generally, it sums the probabilities of all paths that end in that different state, including paths with multiple transitions.

Why the Matrix Exponential Appears

The generator describes behavior over a very short interval. For a small positive \(h\), its definition gives \(P(h)=I+hQ+o(h)\), where \(I\) is the identity matrix and the remainder divided by \(h\) tends to zero as \(h\) tends to zero. The Markov property says that evolving for \(t+h\) is equivalent to evolving for \(t\), then for \(h\):

$$ P(t+h)=P(t)P(h). $$

Subtract \(P(t)\), divide by \(h\), and take the limit. This yields the forward differential equation \(P'(t)=P(t)Q\), with \(P(0)=I\). The unique solution is \(P(t)=\exp(tQ)\). Thus the exponential is not an approximation for the full visit interval: under the specified homogeneous Markov model, it is the exact relationship.

This argument also clarifies the role of assumptions. If rates vary over time, the same constant matrix \(Q\) does not describe the whole interval; if the Markov property fails, state alone may not summarize the information needed for future transitions. In either case, simply applying \(\exp(tQ)\) to a fitted constant generator may not represent the data-generating process.

Worked Example: Computing Visit Probabilities in a Two-State Model

Suppose a hypothetical process has states 0 and 1. The rate from 0 to 1 is \(a=0.4\) per year, and the rate from 1 to 0 is \(b=0.2\) per year. Assume these rates are constant and the process is Markov. Find the transition probabilities over two years.

The generator and the sum of the two rates are

Q= \begin{pmatrix} -0.4&0.4\\ 0.2&-0.2 \end{pmatrix}, \qquad k=a+b=0.6\text{ per year}.

For a two-state model, the probability of being in state 1 after time \(t\), starting in state 0, is \(\frac{a}{a+b}(1-e^{-(a+b)t})\). The reverse transition probability is \(\frac{b}{a+b}(1-e^{-(a+b)t})\). With \(t=2\) years, \(e^{-kt}=e^{-1.2}\approx0.3012\), so

P(2)= \begin{pmatrix} 1-\frac{0.4}{0.6}(1-0.3012)& \frac{0.4}{0.6}(1-0.3012)\\ \frac{0.2}{0.6}(1-0.3012)& 1-\frac{0.2}{0.6}(1-0.3012) \end{pmatrix} \approx \begin{pmatrix} 0.5341&0.4659\\ 0.2329&0.7671 \end{pmatrix}.

For example, a subject starting in state 0 has probability about 0.4659 of being in state 1 two years later. That probability is not the initial rate \(0.4\) multiplied by two years. It accounts for the possibility of moving from 0 to 1 and then back to 0, or moving again before the visit. The diagonal probabilities also include paths with no transition; the off-diagonal probabilities do not.

Recovering Rates in the Two-State Case

The two-state model provides a useful case in which the rate recovery problem has a simple solution. Write the observed transition matrix over an interval \(t>0\) as

P(t)= \begin{pmatrix} 1-p_{01}&p_{01}\\ p_{10}&1-p_{10} \end{pmatrix}.

For exact probabilities generated by a finite-rate, homogeneous two-state model, \(S=p_{01}+p_{10}\) satisfies \(0\le S<1\). The model gives \(S=1-e^{-(a+b)t}\). When \(S>0\), solving this equation and the ratio \(p_{01}:p_{10}=a:b\) recovers the rates:

a+b=-\frac{\log(1-S)}{t}, \qquad a=-\frac{\log(1-S)}{t}\frac{p_{01}}{S}, \qquad b=-\frac{\log(1-S)}{t}\frac{p_{10}}{S}.

If \(S=0\), the exact matrix is the identity and both rates are zero under this model. If \(S=1\), no finite pair of rates produces the matrix: the formula would require an infinite total rate. These boundary cases matter when assessing whether a proposed transition matrix is compatible with a finite-rate continuous-time model.

Worked Example: Recovering Two Rates From Panel Transition Counts

In hypothetical data, 100 subjects begin an interval of one year in state 0; 60 are observed in state 0 and 40 in state 1 at the next visit. Another 100 begin in state 1; 20 are observed in state 0 and 80 in state 1. Use the empirical transition proportions as plug-in probabilities to recover rates under a homogeneous two-state model.

The empirical matrix is \(\widehat P(1)=\begin{pmatrix}0.60&0.40\\0.20&0.80\end{pmatrix}\). Thus \(p_{01}=0.40\), \(p_{10}=0.20\), and \(S=0.60\). The total rate estimate is

\widehat{a+b}=-\frac{\log(1-0.60)}{1\text{ year}} =-\log(0.40)\text{ per year} \approx0.9163\text{ per year}.

The individual rate estimates are

\hat a=0.9163\frac{0.40}{0.60}\approx0.6109\text{ per year}, \qquad \hat b=0.9163\frac{0.20}{0.60}\approx0.3054\text{ per year}.

These plug-in values reproduce the empirical probabilities under the two-state model: the total rate is about \(0.9163\) per year, and its allocation between directions is in the ratio \(0.40:0.20\). They are not automatically maximum-likelihood estimates in every panel-data design; the full likelihood should reflect the actual observation schedule and any covariates or other modeling structure. Here, the calculation demonstrates inversion for one interval and one homogeneous two-state model.

Why One Interval May Not Identify the Generator

For a general state space, a tempting inversion is \(Q=t^{-1}\log P(t)\). This is not a complete recovery rule. A matrix can have multiple logarithms, and a matrix logarithm need not have nonnegative off-diagonal entries and zero row sums, as a valid generator must. Even when a valid generator exists, distinct valid generators can yield the same transition matrix at a particular interval.

The following three-state construction demonstrates exact nonuniqueness. Label the states 0, 1, and 2 in a cycle. Let \(r\) be the rate from each state to the next state in the cycle and \(s\) the rate to the other state. Each row then has diagonal entry \(-(r+s)\). The two nonzero eigenvalues of this circulant generator are

-\frac{3}{2}(r+s) \ \mathbin{\pm}\ i\frac{\sqrt{3}}{2}(r-s).

At interval length one, the exponential depends on the real part through the factor \(e^{-3(r+s)/2}\), and on the imaginary part through its angle \(\frac{\sqrt{3}}{2}(r-s)\). Changing that angle by \(2\pi\) leaves the exponential unchanged. This periodicity allows different rates to produce the same endpoint transition matrix.

Worked Example: Two Different Three-State Generators, One Transition Matrix

Compare the hypothetical rate pairs \((r,s)=(4,4)\) and \((r,s)=(7.628,0.3724)\), with the second pair defined more precisely by \(r+s=8\) and \(r-s=4\pi/\sqrt{3}\). Both have nonnegative rates. For the first pair, the nonzero eigenvalues are both \(-12\). For the second, the eigenvalues are \(-12\mathbin{\pm}i2\pi\). Over one time unit, both pairs therefore give eigenvalues \(e^{-12}\) on the two-dimensional subspace orthogonal to the constant vector; both give eigenvalue 1 on the constant vector.

Consequently, both exponentials equal the same matrix:

P(1)=\frac{1}{3} \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix} +e^{-12}\left(I-\frac{1}{3} \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\right).

Since \(e^{-12}\approx0.000006144\), the diagonal entries are \((1+2e^{-12})/3\approx0.3333\), and the off-diagonal entries are \((1-e^{-12})/3\approx0.3333\). The two generators are different, but their one-unit transition probabilities are identical. Thus, even an exact transition matrix at one interval need not identify the underlying intensities in a multistate model.

This is a structural limitation, not merely a problem of small sample size. More observations at the same interval can estimate that interval’s transition probabilities more precisely, but cannot distinguish generators that produce exactly the same matrix at that interval. Observations at additional interval lengths can provide further information, although whether they resolve ambiguity depends on the model and design.

Validity Checks, Alternatives, and Common Mistakes

A proposed recovery should be checked against both the data and the model. For an exact transition matrix from a finite-state, finite-rate continuous-time Markov chain, rows sum to one, entries are nonnegative, and the matrix is embeddable: at least one valid generator \(Q\) satisfies \(P(t)=\exp(tQ)\). Empirical transition proportions may contain zeros or sampling noise, so they need not themselves have the exact structure expected of a population transition matrix. A failed check can indicate sampling variation, an unsuitable model, numerical problems, or incompatible probabilities; it does not by itself identify which explanation is correct.

Worked Example: A Stochastic Matrix That Cannot Be a Finite-Rate Transition Matrix

Consider the hypothetical two-state matrix \(P=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) over a positive interval. Its rows sum to one and its entries are nonnegative, but it cannot equal \(\exp(tQ)\) for a finite-rate two-state generator. Starting in either state, the probability of no jump during the interval is \(e^{-\nu t}>0\), where \(\nu\) is the finite exit rate from that state. Therefore, the probability of still being in the starting state is at least \(e^{-\nu t}>0\), contradicting the zero diagonal of this matrix. The matrix is stochastic but not embeddable in this finite-rate continuous-time model. An empirical matrix with zero diagonal entries, however, may simply reflect a finite sample; it should not be treated as an exact population matrix without considering sampling uncertainty.

Two common alternatives answer different questions. A discrete-time transition model treats the visit-to-visit matrix itself as the model parameter and does not claim that it arose from constant continuous-time rates. A continuous-time model specifies a generator and uses its matrix exponential to obtain interval probabilities. The latter can describe unobserved transitions between visits, but requires stronger assumptions and may have identification limitations. Likelihood-based estimation uses the observed transition counts or panel histories together with \(P(t)=\exp(tQ)\); it can incorporate more of the data structure than separately taking a matrix logarithm, but it cannot overcome genuine nonidentifiability.

Common mistake: Treating \(p_{ij}(t)\) as \(q_{ij}t\). The correction is that \(q_{ij}t\) is only a first-order approximation over a very short interval; for a finite interval, \(\exp(tQ)\) includes paths with multiple transitions.
Common mistake: Saying an off-diagonal probability includes a path with zero transitions. Such a path remains in the starting state and contributes to a diagonal entry, not an off-diagonal one. In a two-state model, paths ending in the other state require an odd number of transitions.
Common mistake: Assuming every stochastic matrix has a unique generator. A stochastic matrix may have no valid generator, or may have multiple valid generators for a specified interval.

Key takeaway: Under a finite-state, time-homogeneous continuous-time Markov model, \(P(t)=\exp(tQ)\). Two-state rates can be recovered from an exact non-boundary transition matrix, but in multistate models one interval may not identify a unique generator, and a matrix logarithm must be checked for validity.

Check Your Understanding

Answer each question using the assumptions and definitions in this tutorial. Brief solutions follow the questions.

What does \(q_{ij}\) represent for \(i\ne j\), and what condition must each row of \(Q\) satisfy?
In a two-state model, an exact one-year transition matrix has \(p_{01}=0.30\) and \(p_{10}=0.20\). Calculate the implied rates \(a\) and \(b\).
Why can two different three-state generators give the same transition matrix at one interval?
Does a nonnegative matrix whose rows sum to one necessarily have a valid continuous-time generator? Explain.
Why is \(q_{ij}t\) generally not the exact probability of ending in state \(j\) after time \(t\)?

Solutions. (1) \(q_{ij}\) is the instantaneous rate from state \(i\) to state \(j\); \(q_{ii}=-\sum_{j\ne i}q_{ij}\), so each row sums to zero. (2) \(S=0.30+0.20=0.50\), so \(a+b=-\log(0.50)\approx0.6931\) per year. Thus \(a=0.6931(0.30/0.50)\approx0.4159\) per year and \(b=0.6931(0.20/0.50)\approx0.2773\) per year. (3) The matrix exponential can identify distinct valid generators at that interval; in the cyclic example, generator eigenvalue angles differing by \(2\pi\) have the same exponential. (4) No. It must also be embeddable as \(\exp(tQ)\) for a generator with nonnegative off-diagonal entries and zero row sums. (5) \(q_{ij}t\) is only a short-interval first-order approximation; over a finite interval, the endpoint probability accounts for paths involving multiple transitions.