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Transition Intensities and Probabilities · Tutorial 301 of 699

State-Occupation Probabilities At A Fixed Horizon

Calculate the probability of occupying each state at a chosen time from a specified starting state, and interpret it separately from an individual transition probability.

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What You'll Learn

  • Define fixed-horizon state-occupation probabilities conditional on a specified initial state.
  • Extract the appropriate row of an estimated transition-probability matrix.
  • Calculate state-occupation probabilities from ordered transition-matrix updates.
  • Decompose a fixed-horizon probability into paths through intermediate states.
  • Check that estimated probabilities are nonnegative and sum to one.
  • Distinguish occupancy at a horizon from a particular transition and from ever visiting a state.

What Is the Probability of Occupying a State at a Chosen Time?

In a multi-state model, a subject occupies exactly one state at a given time. A fixed-horizon question asks for the probabilities of being in each possible state at a specified time \(t\), given the subject’s state at an earlier time \(s\). For example, after starting in a disease-free state, what are the probabilities of still being disease-free, being ill, or having died by week 12? These probabilities form a distribution across states at that horizon.

The answer is not generally the probability of one particular transition. A subject could occupy a destination state at time \(t\) after following several different paths, and the subject could also have left that state again before \(t\). The fixed-horizon probability concerns the state occupied at \(t\), not the full history of transitions.

Definition: Let \(X(t)\) denote the state occupied at time \(t\), and let the state space be \(\{1,\ldots,K\}\). The transition-probability matrix from time \(s\) to time \(t\), for \(s\leq t\), is \(P(s,t)\), with entry \(p_{ij}(s,t)=P(X(t)=j\mid X(s)=i)\). For a subject known to start in state \(i\) at time \(s\), the fixed-horizon state-occupation probability for state \(j\) is \(p_{ij}(s,t)\). The probabilities across all destination states \(j\) are the entries of row \(i\) of \(P(s,t)\).

The term “state-occupation probability” emphasizes the state occupied at the chosen time. It does not mean the probability of ever entering that state, the probability of making a direct transition to it, or the proportion of time spent in it. Those are different quantities. Throughout, probabilities are conditional on the stated initial state unless an initial distribution is explicitly given.

From an Estimated Matrix to Occupation Probabilities

Suppose an estimator, such as the Aalen-Johansen estimator, provides an estimated transition-probability matrix \(\widehat{P}(s,t)\). For a known initial state \(i\), the estimated fixed-horizon occupation probabilities are obtained by selecting row \(i\):

$$ \widehat{p}_{ij}(s,t)=\widehat{P}_{ij}(s,t),\qquad j=1,\ldots,K. $$

This row-extraction rule is simple, but the interpretation depends on keeping the origin state and horizon clear. The entry \(\widehat{P}_{ij}(s,t)\) estimates the probability of occupying state \(j\) at time \(t\), conditional on occupying state \(i\) at time \(s\). It is not necessarily the probability of a direct \(i\)-to-\(j\) transition during the interval.

Worked Example: Reading a Fixed-Horizon Row

Suppose a hypothetical three-state analysis estimates the transition-probability matrix from baseline to week 12 as

$$ \widehat{P}(0,12)= \begin{pmatrix} 0.55 & 0.30 & 0.15\\ 0.10 & 0.60 & 0.30\\ 0 & 0 & 1 \end{pmatrix}, $$

where the states are 1 = well, 2 = ill, and 3 = dead. Find and interpret the week-12 state-occupation probabilities for a subject who starts well.

The initial state is state 1, so use the first row: \(\widehat{p}_{11}(0,12)=0.55\), \(\widehat{p}_{12}(0,12)=0.30\), and \(\widehat{p}_{13}(0,12)=0.15\). The estimated probabilities of being well, ill, and dead at week 12 are therefore 0.55, 0.30, and 0.15, respectively.

They sum to \(0.55+0.30+0.15=1.00\), as required for an exhaustive set of mutually exclusive states. In particular, 0.30 is the estimated probability of being ill at week 12 given a well state at baseline; it is not necessarily the probability of making a direct well-to-ill transition.

How Intermediate States Contribute

When transition probabilities are constructed across successive time intervals, matrix multiplication combines all possible intermediate states. Suppose \(s<u<t\). The general law-of-total-probability decomposition, conditional on both the intermediate state and the initial state, is

$$ P(X(t)=j\mid X(s)=i) = \sum_{k=1}^{K} P(X(t)=j\mid X(u)=k, X(s)=i) P(X(u)=k\mid X(s)=i). $$

This expression is valid without assuming that the future depends only on the state at time \(u\): it retains \(X(s)=i\) in the first conditional probability. Under a Markov model, that first factor reduces to \(P(X(t)=j\mid X(u)=k)\). In that setting, the decomposition becomes the Chapman-Kolmogorov multiplication rule. The qualification matters: dropping the initial-state condition in the first factor is not justified by the law of total probability alone.

For estimated transition matrices under a model where this multiplication rule applies, the calculation is \(\widehat{P}(s,t)=\widehat{P}(s,u)\widehat{P}(u,t)\). The order matters: rows index origin states and columns index destination states, so the first interval’s matrix is on the left. The entry in row \(i\), column \(j\) adds the contributions from every possible state \(k\) at the intermediate time.

Worked Example: Summing Paths Through an Intermediate State

Consider a hypothetical three-state process over two successive intervals. The estimated transition matrices are

$$ \widehat{P}(0,1)= \begin{pmatrix} 0.80 & 0.20 & 0\\ 0 & 0.75 & 0.25\\ 0 & 0 & 1 \end{pmatrix}, \qquad \widehat{P}(1,2)= \begin{pmatrix} 0.90 & 0.10 & 0\\ 0 & 0.60 & 0.40\\ 0 & 0 & 1 \end{pmatrix}. $$

Find the estimated state-occupation probabilities at time 2 for a subject in state 1 at time 0. Multiply the matrices in chronological order and take the first row. For state 1 at time 2, the only contribution is remaining in state 1 in both intervals: \(0.80(0.90)=0.72\).

For state 2 at time 2, there are two paths: move to state 2 during the first interval and remain there, or remain in state 1 first and move to state 2 in the second interval. Thus \(0.20(0.60)+0.80(0.10)=0.12+0.08=0.20\). For state 3 at time 2, the path is to enter state 2 during the first interval and then move to state 3: \(0.20(0.40)=0.08\).

The resulting row is \((0.72,0.20,0.08)\), which sums to 1. It describes state occupancy at time 2, including paths through state 2; it does not report only transitions made during the second interval.

Calculating Occupancy from Event-Time Updates

In discrete event-time computation, the estimated transition matrix to a horizon is obtained by multiplying the successive estimated update matrices in time order. Each update describes the transition probabilities at one event time, with rows summing to one. If the initial state is fixed at \(i\), the \(i\)th row of the resulting product gives the horizon-specific occupation probabilities. This calculation assumes the states form a complete set for the analysis and that the estimated updates are appropriate for the transition model and data.

Worked Example: A Three-State Product from Hypothetical Counts

Suppose a hypothetical study has states 0 = well, 1 = ill, and 2 = dead. At the first event time, 10 subjects are at risk in state 0; two move from state 0 to state 1 and one moves from state 0 to state 2. No transitions occur from states 1 or 2 at that time. At the second event time, four subjects are at risk in state 1 and one moves from state 1 to state 2. Treat these as the event-time transition updates and calculate the estimated state probabilities after the second update for a subject starting in state 0.

At the first event time, the estimated probabilities from state 0 are \(1-2/10-1/10=0.70\) of staying in state 0, \(2/10=0.20\) of moving to state 1, and \(1/10=0.10\) of moving to state 2. At the second event time, the estimated probability of staying in state 1 is \(1-1/4=0.75\), and the probability of moving to state 2 is \(1/4=0.25\). The updates are

$$ A_1= \begin{pmatrix} 0.70&0.20&0.10\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad A_2= \begin{pmatrix} 1&0&0\\ 0&0.75&0.25\\ 0&0&1 \end{pmatrix}. $$

The chronological product is \(A_1A_2\). Its row for initial state 0 is \[ (0.70,0.20,0.10)A_2 = (0.70,\;0.20(0.75),\;0.10+0.20(0.25)) = (0.70,0.15,0.15). \]

Thus the estimated probabilities after the second update are 0.70 in state 0, 0.15 in state 1, and 0.15 in state 2. The state-2 probability includes both the 0.10 that moved there at the first update and the \(0.20(0.25)=0.05\) that first entered state 1 and then moved to state 2. The row sums to \(0.70+0.15+0.15=1.00\).

Initial Distributions, Checks, and Interpretation

If subjects do not all start in the same state, let \(\boldsymbol{\pi}\) be their row vector of initial-state probabilities, where \(\pi_i=P(X(s)=i)\). The corresponding state distribution at time \(t\) is \(\boldsymbol{\pi}P(s,t)\), and an estimated distribution is \(\boldsymbol{\pi}\widehat{P}(s,t)\) when \(\boldsymbol{\pi}\) is specified. This is a weighted average of the origin-state rows, not a new transition probability conditional on one particular origin.

Worked Example: Combining Origin-State Rows

Use the estimated matrix from the first example and suppose a hypothetical population has initial-state distribution \(\boldsymbol{\pi}=(0.50,0.30,0.20)\). Calculate the estimated state distribution at week 12.

Multiply \(\boldsymbol{\pi}\) by the matrix. The probability for state 1 is \(0.50(0.55)+0.30(0.10)+0.20(0)=0.275+0.030=0.305\). For state 2 it is \(0.50(0.30)+0.30(0.60)+0.20(0)=0.150+0.180=0.330\). For state 3 it is \(0.50(0.15)+0.30(0.30)+0.20(1)=0.075+0.090+0.200=0.365\).

The resulting distribution is \((0.305,0.330,0.365)\), which sums to 1. The values describe the population’s state distribution at week 12 under the stated initial mix. They are not the conditional probabilities for a subject starting in any one specific state.

Several checks help identify errors. Each estimated matrix row should contain probabilities between 0 and 1 and sum to 1, apart from negligible numerical rounding. A product of row-stochastic transition matrices should also be row-stochastic. At a fixed horizon, the selected row must correspond to the stated initial state. A failure of these checks can indicate reversed multiplication order, an incorrect row, omitted destinations, or inconsistent update matrices. These checks verify arithmetic and probability accounting; they do not establish that the model or estimator is appropriate for the data.

The usual interpretation also depends on the model and estimation conditions supporting the transition-probability estimates. The states must be defined consistently, and the event-time updates and risk sets must represent the process being studied. If the assumptions needed for the chosen estimator fail, a row can still sum to one while its entries do not estimate the intended probabilities. Uncertainty in the estimated entries is a separate issue from calculating and interpreting the point estimates.

Common Pitfall: Occupancy Is Not a Direct Transition

A common mistake is to read \(\widehat{P}_{ij}(s,t)\) as the estimated probability of a direct \(i\)-to-\(j\) transition. The correct interpretation is the probability of being in state \(j\) at \(t\), conditional on being in state \(i\) at \(s\). As the second example shows, this can include paths through intermediate states. Another distinct quantity is the probability of ever visiting state \(j\) by time \(t\); a subject who visits \(j\) and leaves before \(t\) may count toward that probability but not toward occupancy of \(j\) at \(t\). When the question concerns a particular path, a direct transition, or ever entering a state, the corresponding probability must be defined and estimated for that event rather than inferred from a single fixed-horizon matrix entry.

A fixed-horizon summary is useful because it puts all possible states on a common probability scale at a clinically meaningful time. It can show how the distribution of patients across states changes over time and can support clear comparisons between starting states or specified initial populations. Its limitation is equally important: a single horizon does not describe when transitions occurred or how many transitions a subject experienced along the way.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. If the row for initial state 2 in \(\widehat{P}(s,t)\) is \((0.15,0.50,0.35)\), what is the estimated probability of occupying state 3 at time \(t\), given state 2 at time \(s\)?
  2. In the two-interval example, why does the probability of state 2 at time 2 include two terms?
  3. For the first update in the hypothetical-counts example, verify that the estimated transition probabilities from state 0 sum to one.
  4. Explain why a fixed-horizon probability of occupying state \(j\) is not generally the probability of ever visiting state \(j\) by that horizon.

Solutions. (1) It is the third entry of row 2, \(0.35\). (2) State 2 at time 2 can be reached either by being in state 2 at time 1 and remaining there, or by being in state 1 at time 1 and then moving to state 2; the two contributions are \(0.20(0.60)\) and \(0.80(0.10)\), totaling \(0.20\). (3) The probabilities are \(0.70\), \(0.20\), and \(0.10\), so \(0.70+0.20+0.10=1.00\). (4) Someone can visit state \(j\) before the horizon and leave it before the horizon. That person contributes to ever visiting \(j\), but is not occupying \(j\) at the horizon.

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