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Transition Intensities and Probabilities · Tutorial 294 of 699

Interpreting Diagonal Entries Of Transition Matrices

Interpret diagonal transition probabilities correctly, distinguish them from diagonal transition intensities, and identify what endpoint data can and cannot reveal about paths between observation times.

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What You'll Learn

  • Define a diagonal entry of a transition matrix as the probability of occupying the same state at two specified times.
  • Explain why a subject may leave a state and return before the second time.
  • Interpret a diagonal intensity as the negative total instantaneous rate of leaving a state.
  • Derive the short-interval relationship between a diagonal transition probability and its intensity.
  • Check transition and intensity matrices for basic probability and rate constraints.
  • Recognize when endpoint observations cannot establish whether a transition occurred in between.

Two Diagonals, Two Different Quantities

A diagonal entry can appear in both a transition-probability matrix and a transition-intensity matrix, but it has a different meaning in each. A diagonal entry of a transition-probability matrix is a probability about the states occupied at two endpoints. A diagonal entry of an intensity matrix is a rate-related quantity: it is the negative of the total instantaneous rate of leaving that state.

The distinction matters when interpreting multi-state estimates. A large diagonal transition probability says that many individuals starting in a state are also in that state at the later endpoint. It does not, by itself, say that those individuals remained there continuously. The two endpoints do not reveal the full path between them.

Definition: Let \(X(t)\) denote the state occupied at time \(t\). For states \(i\) and \(j\), the transition probability from time \(s\) to time \(t\) is \(P_{ij}(s,t)=P\{X(t)=j\mid X(s)=i\}\), when the conditioning event has positive probability or the model specifies the conditional probability. The diagonal entry \(P_{ii}(s,t)\) is the probability of being in state \(i\) at both endpoints, conditional on being in state \(i\) at time \(s\).

This definition concerns the state at \(s\) and the state at \(t\), not every state visited between them. In a Markov model, the transition probability is specified by the current state and the two times. Without the Markov assumption, the endpoint-conditional probability can still be defined, but it may depend on prior history or on how the group in state \(i\) at time \(s\) was selected.

A transition matrix is a matrix of these probabilities for the states in the model. If the state set is complete and the process is in exactly one of its states at time \(t\), each row sums to one: starting in state \(i\), the possible states at \(t\) exhaust the possibilities. Its diagonal entries are probabilities, so each is between zero and one. The earlier material on why product-integral estimates preserve probability concerns these probability-scale properties; it does not make a transition-probability diagonal an intensity.

Worked Example: Reading a Diagonal in an Illness-Death Model

Suppose a hypothetical model has three states: 1, disease-free; 2, ill; and 3, dead. For a specified interval from \(s\) to \(t\), its transition matrix is

$$ \mathbf{P}(s,t)= \begin{pmatrix} 0.72 & 0.20 & 0.08\\ 0.10 & 0.65 & 0.25\\ 0 & 0 & 1 \end{pmatrix}. $$

Each row sums to one, all entries are between zero and one, and the third row represents death as an absorbing state in this hypothetical model. These checks are consistent with a transition-probability matrix.

The entry \(P_{11}(s,t)=0.72\) means that, among individuals in the disease-free state at \(s\), the conditional probability of being disease-free again at \(t\) is 0.72. It does not establish that 72% experienced no illness during the interval. Some may have become ill and later returned to the disease-free state, if such transitions are allowed by the model.

Likewise, \(P_{22}(s,t)=0.65\) is the probability of being ill at both endpoints, conditional on being ill at \(s\). It is not a rate and does not have units of “per year.” In this model \(P_{33}(s,t)=1\) because death is specified as absorbing: someone already dead at \(s\) remains dead at \(t\).

What the Intensity-Matrix Diagonal Means

For a continuous-time multi-state process, the off-diagonal intensity \(q_{ij}(t)\), for \(j\ne i\), describes the instantaneous rate of moving from state \(i\) to state \(j\) at time \(t\). Intensities have units of inverse time, such as per year. The diagonal is defined to make the rows sum to zero:

$$ q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t). $$

Thus \(q_{ii}(t)\) is nonpositive. Its magnitude is the total instantaneous rate of leaving state \(i\), summing across the possible destinations. It is not a probability of staying in the state, and it is not a rate of making a transition from \(i\) back to \(i\). A move from a state to itself is not counted as an exit transition in this intensity representation.

Key distinction: \(P_{ii}(s,t)\) is a dimensionless endpoint probability in the range \([0,1]\). The intensity diagonal \(q_{ii}(t)\) is a nonpositive rate, with units of inverse time, equal to minus the sum of the off-diagonal exit rates from state \(i\).

The two quantities are related locally, but they are not interchangeable. Under regularity conditions for a continuous-time process—finite exit rates over the interval, a well-defined intensity, and negligible probability of multiple transitions to first order over a very short interval—the probability of an off-diagonal transition over a short interval of length \(\Delta\) is \(q_{ij}(t)\Delta+o(\Delta)\). The diagonal follows from the fact that each row of transition probabilities sums to one.

Why the Short-Interval Relationship Holds

Fix a state \(i\) and time \(t\). The probability of being in some state at \(t+\Delta\), given state \(i\) at \(t\), is one. For each \(j\ne i\), the definition of the transition intensity gives \(P_{ij}(t,t+\Delta)=q_{ij}(t)\Delta+o(\Delta)\) as \(\Delta\) decreases to zero. Summing these off-diagonal probabilities and subtracting from one yields the diagonal expansion.

$$ \begin{aligned} 1 &=P_{ii}(t,t+\Delta)+\sum_{j\ne i}P_{ij}(t,t+\Delta),\\ P_{ii}(t,t+\Delta) &=1-\sum_{j\ne i}\left\{q_{ij}(t)\Delta+o(\Delta)\right\},\\ &=1+q_{ii}(t)\Delta+o(\Delta). \end{aligned} $$

Equivalently, \(q_{ii}(t)\) is the limiting first-order change in the diagonal transition probability per unit time: \(q_{ii}(t)=\lim_{\Delta\downarrow0}\{P_{ii}(t,t+\Delta)-1\}/\Delta\). This local limit explains why the intensity diagonal is negative when the state has a positive exit rate. It does not say that \(P_{ii}(s,t)=1+q_{ii}(t)(t-s)\) for an arbitrary, non-small interval. Over longer intervals, rates can vary and paths can involve multiple transitions.

Worked Example: Interpreting an Intensity Diagonal Locally

Consider a hypothetical intensity matrix, measured per year, at a particular time:

$$ \mathbf{Q}(t)= \begin{pmatrix} -0.30 & 0.20 & 0.10\\ 0.05 & -0.25 & 0.20\\ 0 & 0 & 0 \end{pmatrix}. $$

For state 1, the two possible exit intensities are 0.20 per year to state 2 and 0.10 per year to state 3. Their sum is 0.30 per year, so \(q_{11}(t)=-(0.20+0.10)=-0.30\) per year. The row sums are zero, as required. State 3 has no exits in this example, so its diagonal intensity is zero.

For a short interval of \(\Delta=0.1\) year, the first-order approximation for remaining in state 1 at the endpoint is

$$ P_{11}(t,t+0.1)\approx 1+(-0.30)(0.1)=0.97. $$

The corresponding first-order probabilities of moving from state 1 to states 2 and 3 are approximately \(0.20(0.1)=0.02\) and \(0.10(0.1)=0.01\); together with 0.97, they sum to one. The value \(-0.30\) is not a 30% probability of leaving, nor is it the transition probability over the 0.1-year interval. The value 0.97 is only a first-order short-interval approximation; it is not asserted to be exact. This calculation uses nonnegative off-diagonal rates, a zero row sum, and a sufficiently short interval for the first-order approximation to be appropriate.

Endpoint Agreement Does Not Mean Continuous Residence

The event counted by \(P_{ii}(s,t)\) includes every possible path that starts in \(i\) at \(s\) and ends in \(i\) at \(t\). One such path is continuous residence in \(i\). Another is a departure followed by a return. If the model permits both, the endpoint probability combines them.

This can be justified by partitioning the event \(\{X(t)=i\}\), conditional on \(X(s)=i\), into mutually exclusive path histories. The probability of ending in \(i\) is the sum of the probabilities of all those histories. The diagonal entry does not label which history occurred; intermediate observations or additional assumptions are needed to distinguish them.

Worked Example: A Return Path Contributes to the Diagonal

In a hypothetical study, suppose individuals begin in state 1 and their interval histories fall into four mutually exclusive, exhaustive categories: 0.65 remain in state 1 throughout; 0.15 move from state 1 to state 2 and end in state 2; 0.10 move from state 1 to state 2 and return to state 1 before the endpoint; and 0.10 move from state 1 to state 3 and end there. The probabilities sum to \(0.65+0.15+0.10+0.10=1.00\), so the categories cover the hypothetical group.

The endpoint probability of being in state 1 is the sum of the categories that end in state 1:

$$ P_{11}(s,t)=0.65+0.10=0.75. $$

Thus 75% are in state 1 at both endpoints. Only 65% remained there throughout. The additional 10 percentage points come from people who left and returned. The example assumes that the listed histories are correctly classified and exhaustive; it illustrates why the endpoint diagonal alone cannot determine continuous residence.

Checks, Comparisons, and Failure Modes

Several basic checks help prevent a probability-rate mix-up. For a transition-probability matrix, check that entries are in \([0,1]\) and each row sums to one when the modeled states are exhaustive. For an intensity matrix, check that off-diagonal entries are nonnegative, diagonal entries are nonpositive, and each row sums to zero. Also check the units: probability entries have no time units, whereas intensities are rates.

QuantityMeaningRange or constraintUnits
\(P_{ii}(s,t)\)Probability of state \(i\) at both endpoints, conditional on state \(i\) at \(s\)Between 0 and 1None
\(q_{ii}(t)\)Negative total instantaneous rate of exiting state \(i\)Nonpositive; row sums of \(\mathbf{Q}(t)\) are zeroInverse time
\(P_{ij}(s,t)\), \(i\ne j\)Probability of being in state \(j\) at \(t\), given state \(i\) at \(s\)Between 0 and 1None
\(q_{ij}(t)\), \(i\ne j\)Instantaneous rate of transition from \(i\) to \(j\)NonnegativeInverse time

A useful diagnostic is whether an interpretation fits the sign and scale. A proposed diagonal transition probability of \(-0.30\) is impossible; a proposed intensity diagonal of \(0.97\) per year is inconsistent with the convention that the diagonal is nonpositive. Likewise, a transition matrix with a row sum below one may indicate omitted states, a modeling convention that does not represent all outcomes, or a computational problem. It should not automatically be treated as a valid probability matrix.

Endpoint probabilities should also be distinguished from claims about event-free survival or “never leaving” a state. Those claims concern the path over the whole interval and are generally not determined by the diagonal alone. When observations are intermittent, an individual observed in the same state at two visits may have made unobserved transitions between visits. The state definitions and observation schedule therefore matter to interpretation.

A common mistake is to read \(P_{ii}(s,t)\) as “the probability of no transition from state \(i\) during the interval.” That is generally wrong because departure and return can produce the same endpoint state. The correction is to say “the probability of being in state \(i\) at both endpoints”; a no-departure probability requires information or a model specifically about paths, not just this matrix entry.

A second mistake is to call \(q_{ii}(t)\) the probability of remaining in state \(i\). It is a negative rate, not a probability. Over a sufficiently short interval, the probability of remaining at the endpoint is approximately \(1+q_{ii}(t)\Delta\), subject to the local approximation and its assumptions. For longer intervals, using that linear expression as an exact probability can produce incorrect results.

Finally, a diagonal probability is conditional on the starting state. It is not automatically the fraction of an entire population in state \(i\) at time \(t\): that population fraction also depends on the distribution of starting states. If very few or no observations begin in state \(i\) at time \(s\), an empirical estimate of the corresponding row may be unstable or unavailable even though the theoretical quantity is well-defined under a model.

Key takeaway: Read \(P_{ii}(s,t)\) as an endpoint agreement probability, allowing for any intervening path that returns to \(i\). Read \(q_{ii}(t)\) as minus the total instantaneous exit rate from \(i\). Their short-interval relationship is an approximation arising from the intensity definition, not an equality between probabilities and rates.

Check Your Understanding

Answer the questions, then compare your reasoning with the solutions.

  1. If \(P_{22}(s,t)=0.68\), what event does this probability describe, and what does it not establish about the interval path?
  2. In a three-state model, the exit intensities from state 1 are \(0.12\) and \(0.08\) per month. Find \(q_{11}(t)\) and give its units.
  3. For a short interval of 0.2 month in Question 2, calculate the first-order approximation to the probability of remaining at the endpoint in state 1.
  4. Suppose 0.55 of a hypothetical group remain in state 1 throughout an interval, while 0.12 leave and return to state 1. What is \(P_{11}(s,t)\), assuming these are the only histories ending in state 1?
  5. Why can an estimated intensity matrix have a negative diagonal even though a transition-probability matrix cannot?

Solutions. (1) It is the conditional probability of being in state 2 at both endpoints, given state 2 at \(s\). It does not establish continuous residence; a leave-and-return path can also end in state 2. (2) \(q_{11}(t)=-(0.12+0.08)=-0.20\) per month. (3) \(1+(-0.20)(0.2)=0.96\), an approximate endpoint probability for this short interval. (4) \(P_{11}(s,t)=0.55+0.12=0.67\), because both listed histories end in state 1. (5) An intensity diagonal is the negative total exit rate and is constrained to be nonpositive; a transition-probability diagonal is an actual probability and must lie between zero and one.

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