Piecewise-Constant Intensities Across Observation Intervals

Why Calendar-Time Changes Matter

A panel study records a subject's state at scheduled visits, not the exact times of transitions between visits. If the transition process changes over calendar time—for example, because a treatment policy begins or the underlying risk changes—the probability of moving between observed states depends on when the observation interval crosses that change. Treating the entire interval as if it had one constant intensity can therefore misstate the transition probability.

The construction here uses a constant-intensity transition matrix within each calendar-time segment, then combines those matrices in time order. It supplies the transition probabilities needed for a panel-data likelihood; the observed states need not reveal how many transitions occurred between visits.

Definition: For a continuous-time Markov process on states \(1,\ldots,K\), the intensity matrix \(Q(t)\) has off-diagonal entries \(q_{rs}(t)\geq 0\), the instantaneous transition intensity from state \(r\) to state \(s\), and diagonal entries \(q_{rr}(t)=-\sum_{s\ne r}q_{rs}(t)\). A piecewise-constant intensity model specifies known calendar-time cut points \(c_0<c_1<\cdots<c_M\) and matrices \(Q_m\) such that \(Q(t)=Q_m\) within segment \(m\), for \(c_{m-1}\leq t<c_m\). Each row of \(Q_m\) sums to zero.

Use the row-vector convention: \(P_{rs}(a,b)\) is the probability of being in state \(s\) at time \(b\), given state \(r\) at time \(a\). Thus the row of a transition matrix corresponding to the starting state gives probabilities over destination states. The value of \(Q(t)\) at an isolated cut point does not affect these probabilities.

Constructing Probabilities Across Change Points

If \(Q_m\) is constant for a duration \(d\), the transition matrix over that segment is the matrix exponential \(\exp(Q_m d)\). If an observation interval \([a,b]\) crosses cut points, split it into consecutive pieces. Let their lengths be \(d_1,\ldots,d_L\), and let the applicable intensity matrices in chronological order be \(Q_1^*,\ldots,Q_L^*\). Then the full interval transition matrix is the product below.

P(a,b)=\exp(Q_1^*d_1)\exp(Q_2^*d_2)\cdots\exp(Q_L^*d_L).

The order follows from the Chapman–Kolmogorov property: the probability of reaching a destination is obtained by summing over the state at each segment boundary. With row-vector probabilities, the first segment's matrix is on the left. The durations are calendar-time lengths, and each segment matrix acts over the portion of the interval lying in its segment. For an interval that stays within one segment, this reduces to \(P(a,b)=\exp\{Q_m(b-a)\}\).

Justification: Insert the change times \(u_1,\ldots,u_{L-1}\) between \(a\) and \(b\). The Markov property gives \(P(a,b)=P(a,u_1)P(u_1,u_2)\cdots P(u_{L-1},b)\). On each subinterval the generator is constant, so its transition matrix is the exponential of that generator times the subinterval length. Substitution gives the ordered product. No assumption that the segment matrices commute is needed.

In general, matrix multiplication is not commutative: \(\exp(Q_1d_1)\exp(Q_2d_2)\) need not equal \(\exp(Q_2d_2)\exp(Q_1d_1)\). Therefore, changing calendar-time order can change the probability of a transition, even when the same two intensity regimes operate for the same total durations. Nor is the product generally equal to the exponential of a single matrix formed by averaging the generators.

Worked Examples

Worked Example: A One-Way Transition With a Change in Risk

Suppose a hypothetical process has two states: 0 (not yet experienced an event) and 1 (event experienced). State 1 is absorbing. From calendar time 0 to 2 years, the intensity from 0 to 1 is \(0.20\) per year; from year 2 to year 3, it is \(0.50\) per year. Find the probability of remaining in state 0 and the probability of reaching state 1 by year 3, starting in state 0.

Assume a continuous-time Markov process, the stated rates are constant within their respective calendar-time segments, and state 1 is absorbing. These assumptions specify the generator in each segment as \(Q_1=\begin{pmatrix}-0.20&0.20\\0&0\end{pmatrix}\) and \(Q_2=\begin{pmatrix}-0.50&0.50\\0&0\end{pmatrix}\). For this one-way model, survival in state 0 over duration \(d\) is \(\exp(-\lambda d)\), giving \(P_m(d)=\begin{pmatrix}\exp(-\lambda_m d)&1-\exp(-\lambda_m d)\\0&1\end{pmatrix}\).

The first segment lasts 2 years and the second lasts 1 year. Thus the probability of remaining in state 0 is \( \exp(-0.20\times2)\exp(-0.50\times1)=\exp(-0.90)=0.4066\). The probability of reaching state 1 is its complement, \(1-0.4066=0.5934\). The probability matrix is therefore approximately \(\begin{pmatrix}0.4066&0.5934\\0&1\end{pmatrix}\). The row sums equal 1, as required.

Worked Example: Why Reversing the Segments Changes the Result

Consider two hypothetical reversible states, A and B. During the first year, the intensity A to B is \(0.40\) per year and B to A is \(0.10\) per year. During the second year those intensities are \(0.10\) and \(0.40\) per year, respectively. Starting in A, calculate the probability of being in A at the end of the two years.

Assume the process is Markov and each generator remains constant during its specified year. For a two-state process with rates \(\alpha\) from A to B and \(\beta\) from B to A, the stationary probabilities are \(\pi_A=\beta/(\alpha+\beta)\) and \(\pi_B=\alpha/(\alpha+\beta)\). Over duration \(d\), the transition matrix is \( \begin{pmatrix}\pi_A+\pi_Be^{-(\alpha+\beta)d}&\pi_B(1-e^{-(\alpha+\beta)d})\\\pi_A(1-e^{-(\alpha+\beta)d})&\pi_B+\pi_Ae^{-(\alpha+\beta)d}\end{pmatrix}\).

For both years, \(e^{-0.5}=0.6065\), since the sum of the two rates is \(0.50\) per year. The first-year matrix is approximately \(P_1=\begin{pmatrix}0.6852&0.3148\\0.07869&0.9213\end{pmatrix}\); the second-year matrix is \(P_2=\begin{pmatrix}0.9213&0.07869\\0.3148&0.6852\end{pmatrix}\). The calendar-time order is first \(P_1\), then \(P_2\), so the full matrix is \(P_1P_2\). Its A-to-A entry is \(0.6852(0.9213)+0.3148(0.3148)\approx0.7304\). Thus the probability of being in A after two years, starting in A, is about 0.7304.

If the order were reversed, the corresponding entry would be \(0.9213(0.6852)+0.07869(0.07869)\approx0.6375\), not 0.7304. The same regimes and durations produce different results because the first regime changes the distribution of states entering the second.

Worked Example: A Panel-Data Likelihood Contribution

In a hypothetical study, subjects are in state 0 at time 0 and are assessed at year 3. The intensity from state 0 to absorbing state 1 is \(0.20\) per year for the first 2 years and \(0.50\) per year for the final year. Suppose 10 independent subjects are all observed in state 0 at baseline; at year 3, four are in state 0 and six are in state 1. Find the likelihood contribution for these observations, treating the baseline states as fixed.

Assume independent subjects, the same Markov transition model for all subjects, and the specified piecewise-constant rates. The first example gives \(P_{00}(0,3)=0.4066\) and \(P_{01}(0,3)=0.5934\). Because only the two endpoint states are observed, each subject contributes the probability of that observed endpoint; the likelihood is \(L=\{P_{00}(0,3)\}^4\{P_{01}(0,3)\}^6\).

Substituting the values gives \(L=(0.4066)^4(0.5934)^6\), approximately \(0.00119\). Equivalently, the log-likelihood contribution is \(4\log(0.4066)+6\log(0.5934)\approx-6.73\). This calculation does not assert that the six subjects in state 1 each made exactly one transition at a known time; it uses the probability of their observed endpoint under the model, which includes all paths compatible with that endpoint.

Assumptions, Checks, and Alternatives

The product construction is exact under the specified continuous-time Markov model. Its practical interpretation depends on several assumptions:

Markov property: conditional on the current state and the model's time scale, the future transition process does not depend on the earlier path. If duration in the current state or prior history affects risk, a state-only Markov model may be inadequate unless that information is incorporated into the state or model.
Calendar-time structure: cut points and the assignment of segments to calendar time are specified correctly. An individual observed over \([a,b]\) uses the portions of segments actually intersected by that interval.
Within-segment constancy: each intensity is constant throughout its segment. Abrupt changes are represented at the cut points, not gradually between them.
Valid generators: off-diagonal intensities are nonnegative and each row sums to zero. The resulting transition matrices should have nonnegative entries and rows that sum to one, apart from small numerical rounding error.

A useful basic check is to compute the segment matrix dimensions, verify that durations are nonnegative and sum to \(b-a\), and confirm that the final matrix has valid transition-probability rows. A result that depends on reversing the product order, or a row that does not sum to one beyond rounding tolerance, signals a calculation or convention error. These checks do not establish that the Markov or piecewise-constancy assumptions fit the data.

A single constant generator across the entire interval is simpler, but it forces the same intensities before and after every calendar-time change. The piecewise-constant model permits prespecified shifts while retaining a continuous-time transition mechanism within each segment. An alternative is to model interval transition probabilities directly, without deriving them from a continuous-time generator. That approach need not impose the same within-interval path structure, but it also does not provide the same intensity-based interpretation. The choice depends on the scientific question and the available observation schedule.

Common Mistakes and Limitations

Reversing the multiplication order. With row-vector probabilities, the earliest segment matrix comes first on the left. The order follows the chronological Chapman–Kolmogorov decomposition.
Averaging the rates and exponentiating. A duration-weighted average generator generally does not reproduce the ordered product when the generators differ. In particular, the matrices need not commute.
Ignoring how the observation interval meets cut points. Use each segment's actual overlap with \([a,b]\); do not assign the whole interval to the regime at its endpoint.
Interpreting endpoint data as exact transition histories. Panel observations usually identify endpoint states, not the number or timing of intervening transitions. Likelihood contributions must account for all paths through the transition matrix.

The model also has limitations. Choosing cut points does not guarantee that rates are truly constant within each segment, and sparse visits may provide little information about the separate intensities or the timing of changes. If a change point is estimated rather than prespecified, its uncertainty and identifiability require additional treatment; the formulas above alone do not resolve those issues. Further, the construction does not correct for an incorrectly specified state process or observation mechanism.

Key takeaway: For calendar-time piecewise-constant intensities, split each panel interval at every change point, exponentiate each segment's generator over its actual duration, and multiply the resulting matrices in chronological order.

Check Your Understanding

Use the stated assumptions in each question. Solutions follow the questions.

An interval crosses two constant-intensity segments, with the first lasting 1.5 years and the second 0.5 years. Write the transition matrix for the full interval using the row-vector convention.
Why can a duration-weighted average of two generators fail to give the same transition matrix as the chronological product?
For a one-way transition to an absorbing state, the intensity is \(0.10\) per year for 2 years and \(0.30\) per year for 1 year. What is the probability of no transition over the full interval?
A subject starts in state A and is observed in state B at the next visit. Does this observation establish that exactly one transition occurred? Explain.

Solutions. (1) If the generators in chronological order are \(Q_1\) and \(Q_2\), the matrix is \(\exp(1.5Q_1)\exp(0.5Q_2)\). (2) Matrix exponentials generally do not commute, so the ordered product can depend on regime order; a single averaged generator need not preserve that order-dependent effect. (3) The no-transition probability is \(\exp\{-(0.10\times2+0.30\times1)\}=\exp(-0.50)=0.6065\). (4) No. The endpoint observation identifies the final state, not the path or number of transitions between visits; its probability is obtained from the relevant transition-matrix entry.