Why a Zero Entry Matters
A state model is useful only when its allowed changes match the situation being represented. For example, a hypothetical benefit model might allow a person to move from active to disabled and later recover, but rule out a direct active-to-recovered jump. A zero in the intensity matrix records that restriction precisely. It does not, by itself, say that recovery can never occur: the person might reach recovery through an intermediate state.
The central distinction is between a direct transition, represented by one off-diagonal intensity, and a route through multiple states, represented by a sequence of transitions. A zero can forbid the first while leaving the second possible.
For a small time interval of length \(h\), the probability of a direct transition from \(i\) to \(j\), conditional on being in \(i\) at time \(t\), is \(q_{ij}(t)h+o(h)\). Thus, if the intensity is structurally zero, there is no direct \(i\to j\) jump. But if a path \(i\to k\to j\) is allowed, it may be possible to be in \(j\) after a longer interval. Such a path requires at least two jumps; it is not represented by \(q_{ij}(t)\).
Encoding Allowed and Forbidden Changes
Choose and label the states first. Then examine each ordered pair of distinct states. Assign a nonnegative intensity to a direct transition that the model allows, and assign zero to one it prohibits. The direction matters: allowing \(i\to j\) does not imply that \(j\to i\) is allowed. After assigning the off-diagonal entries, calculate each diagonal as the negative sum of the other entries in its row.
A zero should express a deliberate modeling restriction, not a missing calculation. For instance, a model may prohibit direct recovery from an active state but permit recovery after disability. That means the active-to-recovered entry is zero even though the active-to-disabled and disabled-to-recovered entries are positive. The state graph has a two-step route, not a direct edge.
Worked Example: A Prohibited Direct Recovery Transition
A hypothetical model has four states, in this order: active, disabled, recovered, and dead. At the time being modeled, the allowed direct changes are active to disabled at intensity \(0.10\), active to dead at \(0.02\), disabled to recovered at \(0.20\), disabled to dead at \(0.05\), recovered to active at \(0.30\), and recovered to dead at \(0.04\). Direct active-to-recovered transition is not allowed. No exit from dead is included in this model. Construct the intensity matrix and determine whether recovery is reachable from active.
Method. Enter each stated allowed transition in its row and column, use zero for the prohibited active-to-recovered transition, then set each diagonal to make its row sum to zero.
The active row has off-diagonal entries \(0.10,0,0.02\), whose sum is \(0.12\). Its diagonal is \(-0.12\). The disabled row has entries \(0,0.20,0.05\), summing to \(0.25\), so its diagonal is \(-0.25\). The recovered row has entries \(0.30,0,0.04\), summing to \(0.34\), so its diagonal is \(-0.34\). The dead row has no modeled exits, so all its entries are zero.
The active-to-recovered entry is \(q_{\text{active,recovered}}=0\), so there is no direct transition of that type. However, the entries for active to disabled and disabled to recovered are both positive. The route active \(\to\) disabled \(\to\) recovered is therefore allowed. Recovery is reachable from active by an indirect route, even though the direct change is forbidden.
Check. The first three row sums are \(-0.12+0.10+0+0.02=0\), \(0-0.25+0.20+0.05=0\), and \(0.30+0-0.34+0.04=0\). The final row also sums to zero. The common mistake this avoids is replacing “no direct recovery” with “recovery is impossible from active.”
Why a Structural Zero Does Not Erase an Indirect Route
The small-interval approximation explains the difference. If \(q_{ij}(t)=0\), the probability of a direct \(i\to j\) jump in an interval of length \(h\) has no term proportional to \(h\). A path through a distinct intermediate state \(k\), however, requires both an \(i\to k\) jump and a \(k\to j\) jump. For bounded intensities, the probability of two or more jumps in a sufficiently short interval is of smaller order than \(h\). It does not contribute to the instantaneous direct intensity, even though such a path can occur over a longer interval.
A simple calculation makes this concrete. Suppose the only transitions out of state 1 and state 2 are \(1\to2\) and \(2\to3\), each at constant intensity \(1\) per year. State 3 may have its own later transitions; those do not change the time at which it is first reached. Starting in state 1, reaching state 3 requires two successive waiting times. Each waiting time has an exponential distribution with rate \(1\). The probability of reaching state 3 by time \(t\) is the probability that their sum is at most \(t\):
In the integral, \(e^{-s}\) is the density of the first waiting time, and \(1-e^{-(t-s)}\) is the probability that the second waiting time ends within the remaining time. At \(t=1\), the probability is \(1-2e^{-1}\), approximately \(0.2642\). Yet the direct intensity from state 1 to state 3 is zero. This is not a contradiction: the positive probability comes from two jumps, not one direct jump.
This reasoning also identifies a limit on the conclusion. A zero intensity at one particular instant does not necessarily establish a structural prohibition for all times. To describe a forbidden transition in the model, specify that the intensity is zero throughout the relevant period. Conversely, a positive finite-interval probability between two states does not prove that a direct transition is allowed; an intermediate route may account for it.
Worked Example: Calculating Indirect Reachability
A hypothetical three-state process starts in state 1. Its only outgoing transition from state 1 is \(1\to2\), at a constant intensity of \(1\) per year. Its only outgoing transition from state 2 is \(2\to3\), also at a constant intensity of \(1\) per year. There is no direct \(1\to3\) transition. Find the probability that state 3 is reached by the end of one year.
Method. The person must complete two successive exponential waiting times, each with rate \(1\). Let \(S\) be the first waiting time. Its density is \(e^{-s}\) for \(s\ge0\). Given \(S=s\le1\), the probability the second waiting time is at most \(1-s\) is \(1-e^{-(1-s)}\). Therefore,
The direct intensity \(q_{13}\) is zero, but the probability of reaching state 3 within one year is positive because both allowed transitions can occur in sequence. Rounded to four decimal places, the answer is \(0.2642\). The common mistake avoided here is treating \(q_{13}=0\) as if it forced the finite-interval probability of reaching state 3 to be zero.
Check. The result is positive and less than one, as a probability must be. It is also consistent with the event requiring two waiting times to fit within one year; it is not the probability of a single transition with rate \(1\).
Constructing and Auditing Structural Zeros
When converting a verbal description into a matrix, make the allowed-transition decision before calculating the diagonal. This keeps two different tasks separate: the off-diagonal entries describe which direct changes can occur and at what rates, while the diagonal entries summarize total exit intensity from each row. A zero off-diagonal entry does not mean “the row has no exits”; it means that particular destination is not reached by a direct jump from that row.
Use the same order for both the rows and columns, and retain it when interpreting each matrix entry.
For each ordered pair of different states, enter the stated intensity if a direct change is allowed; otherwise enter zero.
Set each diagonal to the negative sum of the off-diagonal intensities in that row.
To decide whether one state can be reached from another, look for a sequence of allowed directed transitions, not just a single matrix entry.
Worked Example: Auditing a Proposed Transition Matrix
A hypothetical process has three states: enrolled, paused, and completed. A direct enrolled-to-completed change is prohibited; enrollment must first be paused. The allowed intensities are enrolled to paused \(0.12\), paused to enrolled \(0.30\), and paused to completed \(0.18\). There are no modeled exits from completed. A proposed matrix, in the stated state order, is
Check whether it is valid, and state whether completion is reachable from enrolled.
Method. Check the off-diagonal entries against the verbal rules, then check every row sum. The first row has the correct enrolled-to-paused intensity \(0.12\) and the required zero for direct enrolled-to-completed transition. Its diagonal is \(-0.12\), the negative of its total exit intensity \(0.12\). The second row has the allowed intensities \(0.30\) and \(0.18\), totaling \(0.48\), so its diagonal \(-0.48\) is correct. The final row has no modeled exits.
The row sums are \(-0.12+0.12+0=0\), \(0.30-0.48+0.18=0\), and \(0+0+0=0\). All off-diagonal entries are nonnegative, and the specified restriction is obeyed, so the proposed matrix is valid under these rules.
Completion is reachable from enrolled by the allowed route enrolled \(\to\) paused \(\to\) completed. The zero in row 1, column 3 prohibits only a direct enrolled-to-completed change. A common mistake would be to “fix” that zero by inserting a positive intensity there because completion is reachable; that would violate the stated restriction. The route is represented by the two existing positive entries, not by a direct edge.
Check. Both the structural restriction and the row-sum condition hold. Reachability by a sequence of transitions is consistent with a zero direct intensity.
Common Exam Traps
- Confusing direction. A zero for \(i\to j\) says nothing by itself about \(j\to i\). Check the ordered pair, not just the two state names.
- Replacing reachability with direct permission. A route through intermediate states can make a state reachable even when the direct intensity is zero.
- Reading a zero as a finite-interval statement. A structural zero rules out a direct jump. It need not rule out arrival by a sequence of jumps over a positive time interval.
- Forgetting the diagonal. The diagonal is not an additional transition to the same state. It is the negative sum of the row’s off-diagonal intensities.
- Overstating a time-specific zero. If an intensity is zero at one time, that alone does not establish that the transition is prohibited throughout the model period.
Check Your Understanding
For each problem, distinguish the direct transition rule from any route through intermediate states.
- A model has states \(A,B,C\), with allowed intensities \(A\to B=0.20\), \(B\to A=0.10\), and \(B\to C=0.30\). There are no modeled exits from \(C\), and direct \(A\to C\) is prohibited. Write the intensity matrix in the order \(A,B,C\). (Answer: \(\begin{pmatrix}-0.20&0.20&0\\0.10&-0.40&0.30\\0&0&0\end{pmatrix}\))
- In a three-state model, \(q_{12}=0.15\), \(q_{23}=0.25\), and \(q_{13}=0\). Is state 3 reachable from state 1 by a route of allowed direct transitions? Is a direct \(1\to3\) jump allowed? (Answer: Reachable by \(1\to2\to3\); a direct \(1\to3\) jump is not allowed.)
- A proposed row has off-diagonal entries \(0.08,0,0.12\), with the zero representing a prohibited direct transition. What must its diagonal entry be, and what is the row sum after it is inserted? (Answer: Diagonal \(-0.20\); row sum \(0\).)
- Starting in state 1, a process can move only \(1\to2\to3\), with each transition having constant intensity \(2\) per year. What is the probability of reaching state 3 within \(0.5\) years? (Answer: \(1-2e^{-1}\approx0.2642\).)
- A model states that direct \(X\to Y\) is forbidden, while \(X\to Z\) and \(Z\to Y\) are allowed. A candidate inserts a positive \(q_{XY}\) because \(Y\) is reachable from \(X\). Identify the error and the correct matrix-entry rule. (Answer: Reachability through \(Z\) does not authorize a direct edge; set \(q_{XY}=0\), retaining the allowed positive entries \(q_{XZ}\) and \(q_{ZY}\).)