Introduction
A transition probability answers a question such as: if a person is in a particular state now, what is the probability of being in another state after a specified interval? In a time-homogeneous model, the answer depends on the length of that interval, not on the calendar time at which it begins. This distinction matters whenever a transition calculation is split across multiple intervals or when probabilities from different observation periods are compared.
The key is to keep two pieces of information separate: the start and end times, and the elapsed duration between them. A model may be homogeneous even though transitions between different states have different rates. Homogeneity concerns whether the transition law changes as calendar time passes.
Definition: Time Homogeneity
The Markov assumption means that, conditional on the state at time \(s\), the future state probabilities do not require the earlier path of the process. Time homogeneity adds a separate condition: shifting the observation interval along the time axis without changing its length leaves the transition probabilities unchanged.
For example, the intervals \([2,5]\) and \([10,13]\) both have duration 3. A homogeneous model assigns them the same transition matrix. The interval \([2,6]\), however, has duration 4 and need not have the same matrix. In a time-inhomogeneous model, even two intervals of the same duration can have different transition probabilities.
Why Equal Durations Give the Same Transition Law
The assertion is built into the definition of homogeneity, but it also explains why transition matrices in a homogeneous model are indexed by duration alone. If \(P(u)\) denotes the transition matrix over duration \(u\), then any interval \([s,t]\) can be represented by \(P(t-s)\). In particular, intervals of equal duration have identical entries, not merely similar-looking rows.
There is also a useful connection with the Chapman-Kolmogorov relationship. In a row-vector convention, the transition from time \(s\) to \(t\), followed by the transition from \(t\) to \(r\), gives the transition from \(s\) to \(r\). Under homogeneity, the two durations are \(t-s\) and \(r-t\), so the relationship becomes a duration-based matrix identity.
The order of multiplication follows the row-vector convention: the first interval's matrix is on the left and the next interval's matrix is on the right. The identity does not say that \(P(u+v)=P(u)\) or that probabilities remain constant over time. It says that the transition over a combined duration can be calculated by multiplying the matrices for its component durations.
In a continuous-time Markov model with constant transition intensities, the rates may depend on the current state and destination, but not on calendar time. The earlier material on the matrix exponential gives the corresponding duration-based form \(P(u)=\exp(Qu)\), where \(Q\) is the constant intensity matrix. This formula is a standard way to calculate the transition matrix; the defining test for homogeneity remains whether shifting the interval without changing its duration changes the transition probabilities.
Worked Example: Comparing Equal Intervals in a Two-State Model
A hypothetical contract holder moves between active state \(A\) and disabled state \(D\). The model is a time-homogeneous continuous-time Markov chain. The transition intensity from \(A\) to \(D\) is \(0.5\) per year, and the intensity from \(D\) to \(A\) is also \(0.5\) per year. Find the transition matrix over \(\ln 2\) years, and compare the intervals \([0,\ln 2]\) and \([4,4+\ln 2]\).
Method. For a two-state model with equal transition intensities \(\lambda\) in both directions, the probability of being in the starting state after duration \(u\) is \((1+e^{-2\lambda u})/2\); the probability of being in the other state is \((1-e^{-2\lambda u})/2\). The two entries in each row sum to 1.
Here \(2\lambda u=2(0.5)(\ln 2)=\ln 2\), so \(e^{-2\lambda u}=e^{-\ln 2}=1/2\). Therefore, the probability of remaining in the starting state is \((1+1/2)/2=3/4\), and the probability of being in the other state is \((1-1/2)/2=1/4\). In the state order \(A,D\), the matrix is
Both specified intervals have duration \(\ln 2\) years. Thus each has this same transition matrix. In particular, starting in \(A\) at time 4, the probability of being in \(D\) at time \(4+\ln 2\) is \(0.25\), just as it is for a start in \(A\) at time 0. The row sums are \(0.75+0.25=1\) in both cases, as required.
Common mistake avoided: using the calendar endpoint \(4+\ln 2\) as the duration for the second interval. The duration is the difference between endpoint and start, \((4+\ln 2)-4=\ln 2\), not the endpoint itself.
Worked Example: A One-Way Transition at Different Start Times
In a hypothetical two-state model, state \(N\) changes to absorbing state \(E\) at a constant intensity of \(0.2\) per year. There are no transitions out of \(E\). Calculate the transition probabilities over three years for an interval beginning at year 1, and state the probabilities for an interval of the same length beginning at year 8.
Method. While the process remains in \(N\), the constant transition intensity gives an exponential holding time with rate \(0.2\) per year. The probability of no transition during three years is \(e^{-0.2(3)}=e^{-0.6}=0.5488\), to four decimal places. The probability of transition to \(E\) by the end of the interval is the complement: \(1-0.5488=0.4512\), using unrounded values before rounding.
Since \(E\) is absorbing, the transition matrix over three years, in the order \(N,E\), is
The interval beginning at year 1 ends at year 4; the interval beginning at year 8 ends at year 11. Both have duration 3, so both use \(P(3)\). For a starting state \(N\), the probabilities of being in \(N\) and \(E\) are \(0.5488\) and \(0.4512\), respectively, in either interval. The row totals are \(0.5488+0.4512=1\) and \(0+1=1\).
Common mistake avoided: treating the year-11 endpoint as an 11-year exposure when the question concerns only the interval from year 8 to year 11. The relevant exposure is three years.
What Changes When Rates Depend on Calendar Time?
A model is time-inhomogeneous when its transition law can change with calendar time. For instance, a transition intensity might be lower during one period and higher in a later period, even when the process is in the same state. Then the probability of no transition over an interval depends on the intensity encountered across that interval, not just on its length.
For a one-way transition with calendar-time intensity \(\lambda(t)\), the probability of no transition between times \(s\) and \(t\) is \(\exp(-\int_s^t \lambda(r)\,dr)\). The integral records the cumulative intensity over the particular interval. This formula is being used here only to illustrate the distinction; when \(\lambda\) varies with calendar time, equal interval lengths do not generally produce equal integrals.
Worked Example: Equal Durations With a Calendar-Time Change
Consider a hypothetical one-way transition from state \(N\) to absorbing state \(E\). Its intensity is \(0.1\) per year before calendar year 5 and \(0.4\) per year from year 5 onward. Find the probability of transition during \([4.5,5.5]\) and during \([5,6]\), given state \(N\) at the start of each interval. Both intervals are one year long.
Method. Calculate the cumulative intensity over each interval, then use the exponential of its negative for the probability of no transition. The transition probability is one minus that survival probability.
For \([4.5,5.5]\), the first half-year has intensity \(0.1\) and the second half-year has intensity \(0.4\). The cumulative intensity is \(0.1(0.5)+0.4(0.5)=0.05+0.20=0.25\). Therefore, the probability of no transition is \(e^{-0.25}=0.7788\), and the probability of transition is \(1-e^{-0.25}=0.2212\), to four decimal places.
For \([5,6]\), the intensity is \(0.4\) for the full year. The cumulative intensity is \(0.4(1)=0.4\). The probability of no transition is \(e^{-0.4}=0.6703\), and the probability of transition is \(1-e^{-0.4}=0.3297\), to four decimal places. Both intervals have duration one year, but their transition probabilities differ because they encounter different calendar-time intensities.
Check. In each case, the no-transition and transition probabilities sum to 1: \(0.7788+0.2212=1.0000\) and \(0.6703+0.3297=1.0000\). The difference between the answers is therefore due to the change in intensity, not a missing probability outcome.
Common mistake avoided: assigning the same probability to these intervals merely because they have equal duration. That shortcut is valid for a homogeneous model, not for the stated calendar-time-varying intensity.
How to Apply the Distinction
Before using a transition matrix, establish whether its time argument means an elapsed duration or a calendar time. For a homogeneous model, write \(P(u)\) for the matrix over duration \(u\). If a problem gives start and end times, calculate \(u=t-s\) first. For a time-inhomogeneous model, retain both times as \(P(s,t)\); replacing them by their difference alone can discard information about the rates encountered.
Check whether the transition probabilities or intensities are stipulated to be unchanged by calendar-time shifts. Constant rates in a continuous-time Markov model support the homogeneous assumption.
For an interval from \(s\) to \(t\), use \(u=t-s\). Do not substitute the calendar endpoint for the duration.
For a homogeneous model, use \(P(u)\). If rates vary with calendar time, keep the start and end times visible as \(P(s,t)\).
A common exam trap is to confuse homogeneity with an absence of state changes. A homogeneous model can have substantial movement among states; its defining feature is that the transition law for a given duration is unchanged when the whole interval is shifted. Another trap is to conclude that all equal-duration intervals have the same probabilities without checking the homogeneity assumption.
Check Your Understanding
For each problem, identify whether elapsed duration alone is sufficient and calculate the requested quantities.
- A two-state homogeneous model has intensity \(0.25\) per year in each direction. Find the probability of remaining in the starting state over two years, and state whether it changes when the interval begins at year 6 instead of year 0. (Answer: \(0.6839\) in either interval.)
- A homogeneous one-way transition has intensity \(0.15\) per year, with the destination absorbing. Find the probabilities of remaining in the initial state and reaching the absorbing state over four years. (Answer: \(0.5488\) and \(0.4512\), respectively.)
- A homogeneous three-state model has transitions \(0\to1\) at intensity \(0.2\), \(0\to2\) at intensity \(0.1\), and \(1\to2\) at intensity \(0.3\), all per year; state 2 is absorbing. Starting in state 0, find the probabilities of states 0, 1, and 2 after one year. (Answer: \(0.7408,\ 0.1482,\ 0.1110\), respectively.)
- A one-way transition has calendar-time intensity \(0.05\) per year before year 8 and \(0.25\) per year from year 8 onward. Starting in the initial state, compare the transition probabilities over \([7,9]\) and \([9,11]\). (Answer: \(0.2592\) over \([7,9]\) and \(0.3935\) over \([9,11]\).)
- A one-way transition has intensity \(0.1\) per year before year 5 and \(0.3\) per year from year 5 onward. Find the probability of no transition over \([4,6]\). (Answer: \(e^{-0.4}=0.6703\).)