From an Event-Time Table to State Probabilities
A multi-state data set records more than whether an event occurred: each observed transition has an origin state, a destination state, and a time. To estimate state-occupation probabilities, the calculation must respect which individuals were at risk for each transition at each event time. A pooled event proportion does not do this: it can ignore time at risk, censoring, and paths through intermediate states.
This workshop uses a three-state model with transitions \(1\to2\), \(1\to3\), and \(2\to3\); state 3 is absorbing. The aim is to calculate transition-intensity increments from a compact event-time table and combine them into a product-integral estimate. The calculations use row vectors: rows index the origin state and columns index the destination state.
These are increments of cumulative transition intensity. At an event-time jump they are dimensionless; for a continuous-time intensity, the corresponding rate has units of inverse time. The diagonal is defined so every row of \(d\widehat A(t)\) sums to zero. The update matrix \(I+d\widehat A(t)\) therefore has rows summing to one, provided the observed outgoing transition count from an origin state does not exceed its risk set.
Assumptions and Calculation Recipe
The Aalen–Johansen product-integral calculation requires correctly observed state transitions and risk sets. Here we assume individuals enter in a known state, transitions are observed at their event times, and right censoring is independent of future transitions conditional on the observed state history. We also assume the three-state process is Markov: at a given time, transition intensities depend on the current state and time, rather than on unrecorded earlier history. The resulting interpretation of transition probabilities relies on that model assumption. The introductory material on independent censoring in Aalen–Johansen estimation gives further context for the censoring condition.
For event times \(t_1<\cdots<t_K\), estimate each transition increment using its own origin-state risk set. Then form \(I+d\widehat A(t_k)\), and multiply these matrices in chronological order. With an initial row distribution \(\boldsymbol\pi(0)\), the estimated distribution at time \(t\) is
For a single initial state \(i\), \(\boldsymbol\pi(0)\) has a 1 in position \(i\) and zeros elsewhere. Censoring removes an individual from subsequent risk sets; it is not itself a transition to a new state. When several transitions occur at the same recorded time, count all transitions in the corresponding \(dN_{ij}(t)\) values and use one matrix increment for that time.
Worked Example: Auditing Risk Sets and Intensity Increments
Consider 12 hypothetical individuals, all initially in state 1, followed in days. Their observed exits are: two transitions \(1\to2\) at day 1; one censoring in state 1 at day 1.5; one \(1\to3\) and one \(2\to3\) transition at day 2; one censoring in state 1 at day 2.5; and two \(1\to2\) transitions at day 3. There are no other exits before day 3. The risk sets and counts are:
| Time (days) | \(Y_1(t)\) | \(Y_2(t)\) | \(Y_3(t)\) | Observed exits at time |
|---|---|---|---|---|
| 1 | 12 | 0 | 0 | Two \(1\to2\) |
| 1.5 | 10 | 2 | 0 | One censoring from state 1 |
| 2 | 9 | 2 | 0 | One \(1\to3\); one \(2\to3\) |
| 2.5 | 8 | 1 | 2 | One censoring from state 1 |
| 3 | 7 | 1 | 2 | Two \(1\to2\) |
At day 1, \(Y_1(1)=12\) and \(dN_{12}(1)=2\), so \(d\widehat A_{12}(1)=2/12=1/6\). At day 1.5, the censoring is not a transition; it reduces the state 1 risk set from 10 to 9 for day 2. At day 2, the two origins have different risk sets: \(d\widehat A_{13}(2)=1/9\) and \(d\widehat A_{23}(2)=1/2\). At day 2.5, censoring reduces the state 1 risk set from 8 to 7. Thus, at day 3, \(d\widehat A_{12}(3)=2/7\).
This audit also checks the counts against the paths: immediately before day 3, seven individuals remain under observation in state 1, one in state 2, and two in state 3. Applying the two day-3 transitions leaves five in state 1 and three in state 2; the two already in state 3 remain there.
Constructing the Matrix Updates
For the data above, the nonzero matrix increments occur at days 1, 2, and 3. Day 1 has two transitions out of state 1, so the diagonal increment for that row is \(-2/12=-1/6\). Day 2 has one transition out of each of states 1 and 2, giving diagonal increments \(-1/9\) and \(-1/2\). Day 3 has two transitions out of state 1, giving a diagonal increment \(-2/7\). The resulting matrices are:
Censoring times have no transition matrix increment. They matter because they change later risk sets, not because they move individuals between states. The corresponding update matrices are \(I+d\widehat A(t)\); for example, at day 2:
The first row says that, for an individual in state 1 just before day 2, the estimated update allocates \(1/9\) to state 3 and \(8/9\) to remaining in state 1. The second row allocates \(1/2\) to state 3 and \(1/2\) to remaining in state 2. These are the empirical jump updates at that time, not estimates of an entire multi-day path on their own.
Worked Example: Updating the Probability Distribution Through Day 2
Starting in state 1, the initial row distribution is \((1,0,0)\). At day 1 the update matrix is \(\begin{pmatrix}5/6&1/6&0\\0&1&0\\0&0&1\end{pmatrix}\). Multiplying gives \[ (1,0,0) \begin{pmatrix}5/6&1/6&0\\0&1&0\\0&0&1\end{pmatrix} =(5/6,1/6,0). \]
At day 2, use the matrix shown above. The updated state 1 probability is \((5/6)(8/9)=20/27\). The updated state 2 probability is \((1/6)(1/2)=1/12\). The state 3 probability receives contributions from both origins: \((5/6)(1/9)+(1/6)(1/2)=5/54+1/12=19/108\). Therefore,
The entries sum to one: \(20/27+1/12+19/108=80/108+9/108+19/108=1\). This is an important arithmetic check, but it does not by itself establish that the model or censoring assumptions are appropriate.
Final State-Occupation Estimates
The product-integral estimate at day 3 is found by applying the day-3 update to the day-2 distribution. The day-3 matrix is \(\begin{pmatrix}5/7&2/7&0\\0&1&0\\0&0&1\end{pmatrix}\). State 1 probability becomes \((20/27)(5/7)=100/189\). State 2 probability combines the transitions from state 1 and continued occupation of state 2: \((20/27)(2/7)+1/12=40/189+1/12=223/756\). State 3 probability remains \(19/108\), because state 3 is absorbing and there are no transitions into it after day 2 in this table.
Worked Example: Reporting the Day-3 Occupation Probabilities
The estimated distribution, starting with all individuals in state 1, is
Thus, under the stated model and censoring assumptions, the estimated probabilities of occupying states 1, 2, and 3 at day 3 are approximately 52.91%, 29.50%, and 17.59%, respectively. The calculation checks because \(400/756+223/756+133/756=1\). These are estimated state-occupation probabilities for the specified initial state and horizon; they are not proportions obtained by dividing the final observed state counts by 12, since censoring means not every individual is observed through day 3.
Diagnostics, Alternatives, and Limitations
Before reporting results, verify that risk sets are state-specific and counted immediately before each event time. Check that a transition is counted once, under its origin state, and that censoring only affects later risk sets. Each \(I+d\widehat A(t)\) should have rows that sum to one. The final product should preserve a probability distribution: its entries should be between zero and one and sum to one, apart from small rounding differences. A negative diagonal in the update, a row sum other than one, or a transition count larger than its risk set signals a data or calculation error.
The product integral is a nonparametric, event-time-based alternative to imposing constant transition intensities. It accommodates changing event-time increments but can be unstable when risk sets are small; for example, a single transition from a risk set of two gives an increment of \(1/2\). A constant-intensity model can smooth such sparse data, but it makes a stronger assumption about how intensities behave over time. A crude comparison of observed final counts is simpler, but does not account for censoring or transitions through intermediate states. The choice depends on the scientific question and the credibility of each method’s assumptions.
In practice, examine whether risk sets become sparse, whether event times or transition types may be misclassified, and whether censoring plausibly satisfies the stated condition. The Markov assumption is also substantive: if transition chances depend on elapsed time in a state or earlier history not represented by the current state, these state-only transition probabilities may not have the intended interpretation. The detailed consequences of that issue belong to models that explicitly incorporate duration or history.
Check Your Understanding
Use the three-state setup and event-time table above. Try each question before reading the short solutions.
- At day 2, why is the estimated \(2\to3\) increment \(1/2\), rather than \(1/9\)?
- What happens to the state 1 risk set between day 2 and day 3?
- Write the day-3 update matrix.
- Why does censoring at day 2.5 have no transition increment?
Solutions. (1) The transition originates in state 2, whose risk set just before day 2 is \(Y_2(2)=2\); hence \(dN_{23}(2)/Y_2(2)=1/2\). (2) It falls from 8 just after day 2 to 7 just before day 3 because one individual in state 1 is censored at day 2.5. (3) The matrix is \(\begin{pmatrix}5/7&2/7&0\\0&1&0\\0&0&1\end{pmatrix}\). (4) Censoring removes an individual from later observation but does not move that individual to another state, so it changes later risk sets without contributing to \(dN_{ij}(2.5)\).