Three Probability Targets That Should Not Be Confused
At a landmark, a prediction question needs a clearly specified outcome and time horizon. “What is the probability of being ill in two years?” asks about state occupation at a fixed time. “What is the probability of having become ill by two years?” asks whether the person entered the illness state at any time by then. “What is the probability of eventually dying?” asks about entry into an absorbing death state over an unlimited horizon. These are different estimands, even when they describe the same multi-state process.
The distinction matters because a person can enter a transient state and then leave it before the prediction horizon. Such a person counts toward the probability of ever having entered that state, but not toward the probability of occupying it at the horizon. By contrast, if a state is absorbing, entering it means remaining there; the probability of occupying that absorbing state by a finite horizon is the probability of having entered it by that horizon.
For a transient state \(j\), occupation at time \(t\) implies that the process has entered \(j\) by time \(t\), so \(P(X(t)=j\mid X(0)=i)\leq P(\tau_j\leq t\mid X(0)=i)\). The inequality can be strict if some people leave \(j\) before the horizon. For an absorbing state \(a\), the equivalence \(X(t)=a\) if and only if \(\tau_a\leq t\) holds when the process begins outside \(a\), because there is no transition out of \(a\). If it begins in \(a\), it occupies \(a\) throughout, and the probability of occupying it is one.
Transition Probabilities at a Landmark
Write \(p_{ij}(0,t)=P(X(t)=j\mid X(0)=i)\) for the probability of occupying \(j\) at time \(t\), conditional on starting in \(i\) at the landmark. In a time-homogeneous continuous-time Markov model on a finite state space, the transition rates are constant over time, and the future transition distribution depends on the past only through the current state. The transition-probability matrix is \(P(0,t)=\exp(Qt)\), where \(Q\) is the generator matrix of transition rates. Its entry in row \(i\), column \(j\) is \(p_{ij}(0,t)\).
These assumptions are useful for deriving the examples below, but they are not automatic. If rates vary with time, a time-homogeneous generator is not appropriate. If future transitions depend on time already spent in a state or on earlier history beyond the current state, a Markov prediction based only on current state may also be inadequate. The target distinction remains meaningful in those settings, but the calculation must match the model.
For an absorbing state \(a\), its probability at a finite horizon is a cumulative incidence for that absorbing event when the process starts outside \(a\). The eventual absorption probability is the limit as the horizon increases without bound. These quantities need not be equal: some individuals may still be in transient states at a finite horizon, and when there are multiple absorbing states, the probability of entering one depends on the competing routes.
Worked Example: Illness Occupation Versus Ever Entering Illness
Consider a hypothetical illness-death model with healthy state 0, illness state 1, and absorbing death state \(d\). From state 0, the transition rate to illness is 0.2 per year and the rate to death is 0.1 per year. From illness, the rate to death is 0.4 per year. There are no other transitions. Assume a time-homogeneous Markov model and start at the landmark in state 0. Find the probabilities of being healthy, being ill, and being dead at two years, and compare the probability of being ill with the probability of having entered illness by then.
The total rate of leaving healthy state 0 is \(0.2+0.1=0.3\) per year, so the probability of still being healthy at time \(t\) is \(p_{00}(0,t)=e^{-0.3t}\). To occupy illness at time \(t\), the process must enter illness at some time \(u\) and then remain alive in illness until \(t\). Thus
At \(t=2\) years, the healthy-state probability is \(e^{-0.6}=0.5488116\), or 0.5488 to four significant figures. The illness-state probability is \(2(e^{-0.6}-e^{-0.8})=0.1989653\), or 0.1990.
Because death is the only other state, its probability is the remainder: \(1-e^{-0.6}-2(e^{-0.6}-e^{-0.8})=0.25222301995\), or 0.2522230 to seven decimal places and 0.2522 to four decimal places. The probabilities sum to one before rounding.
The probability of ever entering illness by two years is the probability of a transition from 0 to 1 before that time. The process must still be in state 0 at the transition time \(u\), so
Therefore, about 30.08% have entered illness by two years, while about 19.90% occupy illness at two years. The difference, \(0.3007922-0.1989653=0.1018269\), is the probability of having entered illness and then died before the two-year horizon. This interpretation follows from the model’s structure: after entering illness, the only possible exit is to death.
Finite-Horizon and Eventual Absorption
An absorbing state is a state that cannot be left once entered. For a fixed horizon \(t\), the probability of occupying that state is the probability of absorption by \(t\), not necessarily the probability of ever being absorbed. The eventual probability is obtained by taking a limit as \(t\) tends to infinity. This distinction is important when a prediction is intended to describe a specified clinical time window rather than an unbounded future.
Worked Example: Competing Absorbing States
Suppose a hypothetical process starts in transient state 0 and has two possible transitions: to absorbing death state \(d\) at rate 0.1 per year, or to absorbing transplant state \(r\) at rate 0.2 per year. Assume constant rates and no other transitions. Calculate the probabilities of remaining in state 0, entering each absorbing state by two years, and eventually entering each absorbing state.
The total rate of leaving state 0 is \(0.1+0.2=0.3\) per year. Therefore the probability of remaining in state 0 at two years is \(e^{-0.3(2)}=e^{-0.6}=0.5488\).
The probability of entering death by two years is the integral of the probability of still being in state 0 at time \(u\) times the death transition rate:
Similarly, the probability of entering the transplant state by two years is \(\frac{0.2}{0.3}(1-e^{-0.6})=0.3007922\). The three probabilities add to one: \(0.5488116+0.1503961+0.3007922=1.000000\), up to displayed precision.
For eventual absorption, the process leaves state 0 with probability one because its total exit rate is positive. The probability that its first transition is to death is the death rate divided by the total exit rate, \(0.1/0.3=1/3\). Thus the eventual death probability is 0.3333, and the eventual transplant probability is \(0.2/0.3=0.6667\). These are greater than their respective two-year probabilities because some transitions occur after two years. The eventual probabilities sum to one; the finite-horizon absorption probabilities do not, because 0.5488 remains in the transient state at two years.
The calculations also show why a competing event cannot be ignored when interpreting an absorption probability. A person who enters transplant cannot later enter death in this model. The probability of death is therefore determined by the transition process while the person remains in state 0, rather than by treating transplant as ordinary censoring and imagining that death could still occur afterward.
Derivation and Interpretation
The integral in the examples has a general interpretation. If a process starts in \(i\), and a transition from state \(k\) into state \(j\) occurs at rate \(q_{kj}(u)\), then the contribution to first entry into \(j\) at time \(u\) is the probability of being in a state that can enter \(j\), multiplied by its transition rate. With multiple possible predecessor states and no prior entry into \(j\), the first-entry probability is accumulated over all such routes. In contrast, state occupation at time \(t\) requires both entry and survival within the state or subsequent path so that the process is in \(j\) at exactly \(t\).
In the illness-death example, entry into illness can only occur from state 0. The probability of still being in state 0 at \(u\) is \(e^{-0.3u}\), and the instantaneous rate of moving from 0 to 1 is 0.2 per year. Multiplying gives the probability density of a first transition into illness at \(u\); integrating from 0 to \(t\) gives the probability of entering by \(t\). For occupation at \(t\), each such entry must additionally be followed by remaining in illness up to \(t\), which contributes the factor \(e^{-0.4(t-u)}\). Since this factor is less than one for \(u<t\), occupation is lower than ever-entry probability.
This argument relies on the specified transition structure and rates. If there are multiple routes into a state, all routes must be included. If a transient state can be entered repeatedly, the probability of ever entering it is not generally obtained by simply adding expected transition counts: individuals can contribute multiple entries, and probabilities cannot count the same individual repeatedly. State-occupation probabilities from the transition matrix remain probabilities for the state at the horizon; first-entry probabilities require a first-passage calculation or a model representation that tracks whether entry has occurred.
Choosing and Checking the Estimand
Before calculating a prediction, state the event in words and attach a time horizon. A useful check is to ask whether a person who experiences the event and then changes state should count. If yes, the target may be a first-entry probability. If the question concerns the state at a specific horizon, only those occupying it at that time count. For an absorbing state, these targets coincide at a finite horizon, but the eventual probability is still a different target from the probability by a fixed time.
| Question | Probability target | What counts |
|---|---|---|
| “What is the probability of being ill in two years?” | Transient-state occupation at two years | In illness at the specified horizon |
| “What is the probability of having become ill by two years?” | First entry into illness by two years | Entered illness at any time up to the horizon, even if later left |
| “What is the probability of death by two years?” | Finite-horizon absorption probability | Entered the absorbing death state by two years |
| “What is the probability of eventual death?” | Eventual absorption probability | Enters death over an unlimited time horizon |
In applications, model-based transition probabilities may be compared with nonparametric state-occupation estimates such as Aalen-Johansen estimates. The choice of method does not change the definition of the target: occupation at a horizon and entry by a horizon must not be interchanged. The earlier material on competing risks establishes the role of competing events in event probabilities; here the same principle applies within a multi-state description.
A common alternative is to use a net-risk calculation that removes competing events from consideration. That answers a different hypothetical question, generally requiring additional assumptions, rather than estimating the actual probability of occupying a state or entering an absorbing event in the presence of competing transitions. Likewise, a long but finite prediction horizon should not automatically be described as “eventual.” It is eventual only if the limit as the horizon tends to infinity is the target and the model supports that limit.
Diagnostics, Failure Modes, and Limits
- Check the state definition. If a state is labelled absorbing but the model allows a transition out of it, the equality between occupancy and entry probability does not apply.
- Check the horizon. “By two years” and “eventually” are different statements. Report the time horizon and units alongside the probability.
- Check the probability accounting. State-occupation probabilities over all states should sum to one when the state space is exhaustive. At a finite horizon, absorption probabilities across competing absorbing states can sum to less than one because some probability remains in transient states.
- Check the transition structure. Omitting a competing transition or a route into a state changes the target probability, not just its numerical estimate.
- Check the Markov assumption. If sojourn time or earlier history affects transition rates beyond the current state, a current-state-only Markov model can give misleading predictions.
Predictions also depend on the population and conditioning information represented at the landmark. A probability conditional on starting in state 0 is not automatically the probability for a population that includes people starting in several states; the latter requires averaging over the starting-state distribution. Moreover, an eventual absorption probability may be poorly determined by observed follow-up if the data contain little information at long times. A mathematically defined limit is not necessarily a reliable empirical prediction.
Check Your Understanding
Answer each question by identifying the probability target and, where requested, calculating it. Solutions follow.
- A state is transient. Can its probability of occupation at time \(t\) exceed its probability of ever being entered by time \(t\)? Explain.
- In the illness-death example, what is the probability of occupying death at two years, and why does it equal the probability of having entered death by then?
- In the two-competing-absorbing-state example, why do the two-year death and transplant probabilities not sum to one?
- For the two-competing-absorbing-state example, calculate the probability of remaining in state 0 for three years.
- A report gives the probability of being in a transient state at five years and calls it the probability of ever having experienced that state. Identify the error.
Solutions. (1) No. Occupation at \(t\) implies entry by \(t\), so the occupation event is a subset of the first-entry event. (2) It is 0.2522, calculated as \(1-e^{-0.6}-2(e^{-0.6}-e^{-0.8})=0.25222301995\), or 0.2522230 to seven decimal places. Death is absorbing, so occupying death at two years is equivalent to having entered death by then. (3) Some probability remains in transient state 0 at two years: \(e^{-0.6}=0.5488\). Thus the absorption probabilities sum to \(1-0.5488=0.4512\), not one. (4) The probability is \(e^{-0.3(3)}=e^{-0.9}=0.4066\). (5) The report has substituted a fixed-horizon occupation probability for a first-entry probability. Someone may have entered the transient state and left it before five years, so the first-entry probability can exceed occupation at five years.