Tutorials › Competing Risks and Multi-State Models › Phase-Type Expansions For Non-Markov Sojourns

Semi-Markov and Non-Markov Models · Tutorial 329 of 699

Phase-Type Expansions For Non-Markov Sojourns

Learn how hidden transient phases turn a non-exponential sojourn into a Markov model, and how to calculate its duration and destination probabilities.

Advanced 11 min read

What You'll Learn

  • Define a phase-type waiting time using an initial phase distribution and a transient generator.
  • Derive its survival function, density, hazard, and moments.
  • Construct a Markov multi-state representation with phase-specific destination transitions.
  • Work through sequential phases, a mixture of exponential phases, and competing destinations.
  • Identify assumptions, diagnostics, and limitations of phase-type representations.

Why Represent a Sojourn With Hidden Phases?

A continuous-time Markov model has exponential holding times when the state is fully specified. If a clinically meaningful state instead has a non-exponential sojourn distribution, its duration contains information about how long the individual has already remained there. One way to retain a Markov description is to distinguish several unobserved transient phases within that state. Conditional on the current phase, the next transition has an exponential waiting time; the observed sojourn ends when the process leaves the collection of phases.

This construction is called a phase-type representation. It is useful when the target model needs a finite-state Markov formulation but the scientifically relevant state duration is not exponential. The phases are modeling states, not necessarily clinical events that are directly observed.

Definition: A phase-type random variable \(T_{\mathrm{abs}}\) is the time to absorption in a finite continuous-time Markov chain with \(m\) transient phases and one absorbing state. Its initial phase probabilities form the row vector \(\boldsymbol{\alpha}\), where \(\alpha_r\geq 0\) and \(\boldsymbol{\alpha}\mathbf{1}=1\). The \(m\times m\) transient subgenerator is \(S\): its off-diagonal entries are nonnegative, and its row sums are nonpositive. The vector of rates into the absorbing state is \(\mathbf{s}=-S\mathbf{1}\). We assume absorption occurs with probability one.

The initial distribution \(\boldsymbol{\alpha}\) specifies the phase at entry to the represented state. While the process is in phase \(r\), it may move to another transient phase according to the off-diagonal rates of \(S\), or be absorbed at rate \(s_r\). The diagonal entry \(S_{rr}\) is the negative total rate of leaving phase \(r\), including both transitions to other phases and absorption.

Sojourn-Time Distribution and Its Derivation

Let \(\mathbf{p}(t)\) be the row vector whose \(r\)th element is the probability that the process is in transient phase \(r\) at time \(t\), having started with distribution \(\boldsymbol{\alpha}\). The continuous-time Markov forward equation is \(\mathbf{p}'(t)=\mathbf{p}(t)S\), with \(\mathbf{p}(0)=\boldsymbol{\alpha}\). Its solution is \(\mathbf{p}(t)=\boldsymbol{\alpha}e^{St}\). The event \(T_{\mathrm{abs}}>t\) is exactly the event that the process is still in some transient phase, so summing the phase probabilities gives the survival function. Absorption at time \(t\) occurs at the total rate represented by \(\mathbf{s}\), giving the density.

$$ \begin{aligned} P(T_{\mathrm{abs}}>t) &= \boldsymbol{\alpha}e^{St}\mathbf{1},\\ f(t) &= \boldsymbol{\alpha}e^{St}\mathbf{s},\\ h(t) &= \frac{\boldsymbol{\alpha}e^{St}\mathbf{s}} {\boldsymbol{\alpha}e^{St}\mathbf{1}}, \qquad t\geq 0. \end{aligned} $$

Here \(h(t)\) is the instantaneous absorption hazard among processes that have not yet been absorbed. Unless the phase structure reduces to a single exponential holding time, this hazard will generally vary with elapsed time. The variation arises because the mix of phases among individuals still in the state can change over time.

If all eigenvalues of \(S\) have negative real parts, then \((-S)^{-1}\) exists and the absorption-time moments are \(\mathrm{E}(T_{\mathrm{abs}}^k)=k!\boldsymbol{\alpha}(-S)^{-k}\mathbf{1}\) for positive integers \(k\). In particular, the mean is \(\boldsymbol{\alpha}(-S)^{-1}\mathbf{1}\). This condition also ensures that the transient probability tends to zero, as required for eventual absorption.

Worked Example: Three Sequential Phases

Suppose a hypothetical sojourn lasts through three sequential hidden phases. Each phase has an exponential transition rate of \(0.5\) per month: phases 1 and 2 progress to the next phase, and phase 3 leads to absorption. Entry is always into phase 1. This is an Erlang distribution with shape 3 and rate \(0.5\) per month.

The phase generator and initial distribution are

$$ S= \begin{pmatrix} -0.5&0.5&0\\ 0&-0.5&0.5\\ 0&0&-0.5 \end{pmatrix}, \qquad \boldsymbol{\alpha}=(1,0,0). $$

The survival probability is the probability that fewer than three phase transitions have occurred by time \(t\). Equivalently, using the phase-type survival formula,

$$ P(T_{\mathrm{abs}}>t) =e^{-0.5t}\left(1+0.5t+\frac{(0.5t)^2}{2}\right). $$

At \(t=2\) months, this is \(e^{-1}(1+1+0.5)=2.5e^{-1}=0.9197\). The mean sojourn is the sum of three exponential phase means: \(3(1/0.5)=6\) months. The initial hazard is zero because absorption cannot occur immediately: the process must first pass through two intermediate phases. For example, the density at zero is zero, consistent with this delayed progression.

The assumptions are explicit in this construction: phase transitions have constant rates, the process starts in phase 1, and absorption occurs after the third phase. The resulting sojourn is not exponential even though the waiting time within each phase is exponential.

From Absorption to a Multi-State Model

To embed a phase-type sojourn in a multi-state process, replace a clinical state \(i\) by its transient phases \((i,1),\ldots,(i,m_i)\). Transitions among these phases follow the transient subgenerator for state \(i\). An exit from a phase represents a transition out of clinical state \(i\). If the destination is \(j\), the phase-specific exit rates must specify both how quickly the exit occurs and where the process goes.

For a state with possible destinations \(j=1,\ldots,J\), let \(\mathbf{t}_{ij}\) be the column vector of rates from each phase of \(i\) to destination \(j\). The within-state phase subgenerator \(S_i\) must satisfy \(\sum_j\mathbf{t}_{ij}=-S_i\mathbf{1}\). On entry to \(i\), assign a phase according to \(\boldsymbol{\alpha}_i\). If the destination state \(j\) itself has phases, an exit to \(j\) enters phase \(s\) with probability \(\alpha_{j,s}\). Thus the expanded-chain rate from \((i,r)\) to \((j,s)\) is \(t_{ij,r}\alpha_{j,s}\). If \(j\) is absorbing, the exit goes directly to that absorbing state.

This rate construction makes the expanded process Markov: its next-transition rates depend on the current phase, not on the time already spent in that phase or the preceding path. When phases are not observed, the clinical-state process generally has non-exponential holding times and is not itself Markov.

Worked Example: A Mixture of Two Exponential Sojourns

Consider a hypothetical state entered into a fast phase with probability \(0.75\) and a slow phase with probability \(0.25\). The fast phase exits at \(0.8\) per year, and the slow phase exits at \(0.2\) per year. There are no transitions between phases. This is a hyperexponential phase-type model.

The model has \(\boldsymbol{\alpha}=(0.75,0.25)\) and \(S=\operatorname{diag}(-0.8,-0.2)\). The survival function is the weighted sum of the phase-specific survivals:

$$ P(T_{\mathrm{abs}}>t) =0.75e^{-0.8t}+0.25e^{-0.2t}. $$

At one year, the survival probability is \(0.75e^{-0.8}+0.25e^{-0.2}=0.75(0.4493)+0.25(0.8187)=0.5417\). The mean is \(0.75(1/0.8)+0.25(1/0.2)=0.75(1.25)+0.25(5)=2.1875\) years.

For comparison, an exponential distribution with that same mean would have rate \(1/2.1875=0.4571\) per year and survival \(e^{-0.4571}=0.6330\) at one year. The phase mixture therefore has a different survival curve despite matching the mean. This construction assumes phase membership is selected at entry and remains fixed until exit; no unmodeled switching between these two phases is included.

Worked Example: Non-Exponential Sojourn With Two Exit Destinations

In a hypothetical model, a person enters phase 1 of a clinical state. From phase 1, the rates are \(0.4\) per month to phase 2, \(0.2\) per month to destination A, and \(0.1\) per month to destination B. From phase 2, the rates are \(0.1\) per month to A and \(0.3\) per month to B. There are no transitions back to phase 1.

The within-state subgenerator and destination exit-rate vectors are

$$ S= \begin{pmatrix} -0.7&0.4\\ 0&-0.4 \end{pmatrix}, \qquad \mathbf{t}_{A}= \begin{pmatrix}0.2\\0.1\end{pmatrix}, \qquad \mathbf{t}_{B}= \begin{pmatrix}0.1\\0.3\end{pmatrix}, \qquad \boldsymbol{\alpha}=(1,0). $$

The rates are internally consistent: \(\mathbf{t}_A+\mathbf{t}_B=(0.3,0.4)^\mathsf{T}=-S\mathbf{1}\). The joint density of exiting at time \(t\) to destination A is \(\boldsymbol{\alpha}e^{St}\mathbf{t}_A\); replacing \(\mathbf{t}_A\) by \(\mathbf{t}_B\) gives the corresponding density for B.

To calculate the destination probabilities, the expected time spent in each phase is needed. Phase 1 is occupied for \(1/0.7=1.4286\) months on average. The probability of moving from phase 1 to phase 2 is \(0.4/0.7\), so expected time in phase 2 is \((0.4/0.7)(1/0.4)=1.4286\) months. Multiplying these expected phase-occupancy times by the exit rates gives

$$ \begin{aligned} P(\text{exit to A})&=(1.4286)(0.2)+(1.4286)(0.1)=0.4286,\\ P(\text{exit to B})&=(1.4286)(0.1)+(1.4286)(0.3)=0.5714. \end{aligned} $$

The probabilities sum to 1, as they must because every path eventually exits to A or B. The expected time to either exit is \(1.4286+1.4286=2.8571\) months. The expanded Markov model has transitions from phase 1 to phase 2 at rate \(0.4\), and from each phase directly to A and B at the rates listed above.

Checking the Representation and Choosing an Approach

A phase-type construction imposes a specific structure, so its assumptions and fit should be checked rather than inferred from the fact that its transition rates are exponential. At minimum, verify that the initial probabilities are nonnegative and sum to 1, that off-diagonal entries of each transient subgenerator are nonnegative, and that the destination exit-rate vectors sum to the total rate of leaving the transient phases. Check also that absorption is certain. If the fitted model gives a nonzero probability of remaining in the phases forever, it does not define a proper finite sojourn distribution.

In practice, compare the implied survival curve and hazard with the intended or observed sojourn behavior, paying attention to early and late durations as well as the mean. A sharp mismatch can indicate too rigid a phase structure, an unsuitable initial phase distribution, or an omitted dependence on history or covariates. Different phase structures can also produce the same or very similar observable sojourn distributions, so the hidden phases may not be identifiable as unique scientific entities.

A direct semi-Markov specification describes the holding-time and destination distributions without requiring exponential phases. It can express the target duration law directly, but does not in general give a finite-state continuous-time Markov representation using only the clinical states. A phase-type representation instead supplies such a finite Markov construction, at the cost of choosing a phase structure and estimating its rates. Markovization by state augmentation is a broader modeling strategy: phase-type expansion is one specific way to use augmented states to represent a sojourn distribution.

Common pitfall: It is incorrect to say that a phase-type model gives an exponential sojourn just because each phase has an exponential holding time. The observed sojourn is the total time across a sequence or mixture of phases. Its survival is \(\boldsymbol{\alpha}e^{St}\mathbf{1}\), which is generally not a single exponential. The correction is to distinguish phase-level waiting times from the total time until absorption and assess the latter using its implied distribution.

Limitations and Practical Interpretation

The expanded chain is Markov only when the phase is included in the current state and the specified rates are adequate. If entry-phase probabilities or phase transitions depend on prior history, that dependence must be represented in the model rather than silently assumed away. Likewise, if an individual re-enters a clinical state, the model must specify the phase assignment on re-entry; using the same entry distribution each time is an assumption, not a consequence of the phase-type construction.

The hidden phases can be useful computational devices, but they should not automatically be interpreted as distinct biological stages. A good representation can approximate a desired sojourn law while remaining finite and Markov, yet it can require several phases and its parameters may be difficult to estimate from limited observations. The relevant check is whether the implied duration and destination behavior is suitable for the scientific question, not whether every phase has an independent clinical label.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. What two components determine a phase-type sojourn distribution?
  2. For a phase-type distribution, what does \(\boldsymbol{\alpha}e^{St}\mathbf{1}\) represent?
  3. A two-phase sequential model has rate \(2\) per day from phase 1 to phase 2 and rate \(2\) per day from phase 2 to absorption. What is its mean sojourn time?
  4. In the two-destination example, why must the destination exit-rate vectors sum to \(-S\mathbf{1}\)?
  5. Does an exponential waiting time in every hidden phase imply an exponential total sojourn? Explain.

Solutions. (1) The initial phase distribution \(\boldsymbol{\alpha}\) and the transient subgenerator \(S\), including its absorption rates. (2) The probability that absorption has not occurred by time \(t\). (3) Each phase has mean \(1/2\) day, so the total mean is \(1/2+1/2=1\) day. (4) Together, the destination-specific exits must account for all rates of leaving the transient phases; otherwise probability would be lost or an exit rate counted twice. (5) No. The total time is the time until absorption across the phase process, with survival \(\boldsymbol{\alpha}e^{St}\mathbf{1}\), which is generally not exponential.

Clinical Biostats