Absorbing States in an Intensity Matrix

Introduction

In a multiple-state model, some states allow movement to other states, while others represent an endpoint from which the process cannot move. A state is absorbing when entering it rules out every subsequent transition to a different state. Recognizing this feature determines the entire row for that state in the intensity matrix.

The key point is that an absorbing-state row is not merely a row whose entries add to zero. Every off-diagonal entry must be zero, and the diagonal entry must then also be zero. This distinction matters when checking whether a proposed intensity matrix is valid.

Definition: In a continuous-time state model, a state \(r\) is absorbing if, once the process enters \(r\), it cannot leave for any other state. Using the intensity-matrix convention in which rows identify the current state and columns identify the destination state, every off-diagonal intensity in row \(r\) is zero. The diagonal entry is also zero, so the complete row consists only of zeros.

What the All-Zero Row Represents

Write \(q_{ij}\) for the intensity of a transition from state \(i\) to a different state \(j\), and write \(q_{ii}\) for the diagonal entry in row \(i\). The off-diagonal entries are transition intensities, so they must be nonnegative. For each row, the diagonal entry is the negative sum of that row’s off-diagonal entries:

q_{ii}=-\sum_{j\ne i}q_{ij}.

If \(r\) is absorbing, it has no permitted outgoing transitions. Therefore \(q_{rj}=0\) for every \(j\ne r\). Substitution into the diagonal formula gives \(q_{rr}=0\). Thus the row is zero in every column, including its diagonal.

This does not mean that there is an outgoing transition from \(r\) back to itself with intensity zero. The diagonal is a balancing entry in the matrix convention; it is not a self-transition intensity. An absorbing state has no outgoing transition intensities at all.

Key result: In an intensity matrix with rows as origins and columns as destinations, the row for an absorbing state is an all-zero row. A zero row expresses the absence of every outgoing transition, and the diagonal is zero because it balances the sum of the off-diagonal entries.

Worked Examples

Worked Example: Constructing a Terminal-State Row

A hypothetical income-protection model has states \(E\) (eligible for coverage), \(C\) (claim being paid), and \(D\) (dead), in that order. For this model, death is permanent. While in state \(E\), the transition intensities per year are \(q_{EC}=0.12\) and \(q_{ED}=0.03\). While in state \(C\), the only possible transition is to \(D\), with intensity \(q_{CD}=0.08\). Construct the intensity matrix and verify the row for \(D\). The stated rates are constant for this example.

Method. List the states in the required order, put each given transition intensity in its origin row and destination column, and calculate each diagonal as the negative sum of the other entries in its row.

In row \(E\), the total outgoing intensity is \(0.12+0.03=0.15\), so \(q_{EE}=-0.15\). In row \(C\), the total outgoing intensity is \(0.08\), so \(q_{CC}=-0.08\). Since death is permanent, there are no outgoing transitions from \(D\); every entry in row \(D\) is zero.

Q= \begin{pmatrix} -0.15 & 0.12 & 0.03\\ 0 & -0.08 & 0.08\\ 0 & 0 & 0 \end{pmatrix}.

Check. The row sums are \(-0.15+0.12+0.03=0\), \(0-0.08+0.08=0\), and \(0+0+0=0\). All off-diagonal entries are nonnegative. Row \(D\) is all zeros, as required for an absorbing state.

Common mistake avoided. Do not put a negative value on the diagonal of row \(D\) simply because other rows have negative diagonals. The diagonal balances outgoing intensities; there are none in an absorbing state.

Worked Example: Correcting an Invalid Absorbing-State Row

A hypothetical model has states \(V\) (covered and active), \(C\) (a permanent terminal claim status), and \(D\) (dead), in that order. A transition from \(V\) to \(C\) has intensity \(0.04\) per year, and a transition from \(V\) to \(D\) has intensity \(0.01\) per year. Both \(C\) and \(D\) are specified as absorbing. A draft matrix places the entries \((-0.04,0.04,0)\) in row \(C\). Identify the error and give the correct matrix.

Method. First use the row and column labels to locate each transition. Then check that every off-diagonal entry is nonnegative and that each diagonal balances the other entries in its row.

The transition \(V\to C\) belongs in row \(V\), column \(C\), so it contributes \(0.04\) to \(q_{VC}\). It does not belong in row \(C\). In the draft row \(C=(-0.04,0.04,0)\), the entry \(q_{CV}=-0.04\) is an invalid negative off-diagonal intensity, and the diagonal \(q_{CC}=0.04\) is positive. Although the entries in that row sum to zero, the row is not valid.

For row \(V\), the total outgoing intensity is \(0.04+0.01=0.05\), so \(q_{VV}=-0.05\). Both \(C\) and \(D\) are absorbing, so their rows consist of zeros. The corrected matrix is:

Q= \begin{pmatrix} -0.05 & 0.04 & 0.01\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}.

Check. The row sums are \(-0.05+0.04+0.01=0\), \(0+0+0=0\), and \(0+0+0=0\). The off-diagonal intensities are all nonnegative. Each specified transition is in its correct origin row, and both absorbing-state rows are all zeros.

Common mistake avoided. A row sum of zero does not by itself establish that a row is valid. The draft’s row summed to zero, but its negative off-diagonal and positive diagonal were both invalid.

Worked Example: Finding All Absorbing States in a Model

A hypothetical long-term benefit model uses states \(A\) (active), \(I\) (disabled), \(S\) (surrendered), and \(D\) (dead), in that order. Surrender and death are permanent endpoints under the model. The transition intensities per year are \(A\to I:0.10\), \(A\to S:0.02\), \(A\to D:0.01\), \(I\to A:0.05\), and \(I\to D:0.08\). Construct the matrix, identify every absorbing state, and verify the rows.

Method. Place each given intensity in the row of its starting state and the column of its destination state. Use the stated endpoint assumptions to fill the \(S\) and \(D\) rows, then calculate the remaining diagonals.

From \(A\), the total outgoing intensity is \(0.10+0.02+0.01=0.13\), so \(q_{AA}=-0.13\). From \(I\), it is \(0.05+0.08=0.13\), so \(q_{II}=-0.13\). No transition out of \(S\) or \(D\) is allowed, so both rows are zero. The matrix is:

Q= \begin{pmatrix} -0.13 & 0.10 & 0.02 & 0.01\\ 0.05 & -0.13 & 0 & 0.08\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}.

Check. The four row sums are \(-0.13+0.10+0.02+0.01=0\), \(0.05-0.13+0+0.08=0\), \(0+0+0+0=0\), and \(0+0+0+0=0\). All off-diagonal entries are nonnegative. The absorbing states are \(S\) and \(D\), because the scenario specifies that each is permanent and each has an all-zero row.

Common mistake avoided. A state is not absorbing just because it sounds like an endpoint in ordinary language. Here, the model’s specification makes \(S\) and \(D\) permanent. The row of any other state must be based on its stated transitions, not on a label alone.

Why the Diagonal Is Zero

The zero diagonal in an absorbing-state row follows directly from the intensity-matrix row convention. For a valid intensity matrix, the off-diagonal entries in row \(r\) describe every possible transition from \(r\) to another state. The diagonal is defined to make the row sum zero. If state \(r\) is absorbing, each of those off-diagonal entries is zero, so their sum is zero and the diagonal must also be zero.

q_{rj}=0\quad(j\ne r) \quad\Longrightarrow\quad q_{rr}=-\sum_{j\ne r}q_{rj}=0.

Under the model assumption that state \(r\) has no possible exit at any time, the process remains in \(r\) after it enters. In particular, the model assigns no positive probability to an exit during any finite interval. If a model allows an exit later, then the state is not absorbing, even if an exit intensity happens to be zero at one particular instant. The all-zero-row conclusion applies when there are no outgoing transitions in the model, not merely when one observed rate is momentarily zero.

Common Checks and Exam Traps

When checking a proposed absorbing-state row, apply both the model definition and the matrix rules. The following checks are useful:

Check the model’s transitions. An absorbing state has no permitted transition to any different state.
Check every off-diagonal entry. Each must be zero in an absorbing-state row. In any valid intensity matrix, off-diagonal entries must also be nonnegative.
Check the diagonal. It is zero when the off-diagonal entries are all zero; it is not an outgoing self-transition intensity.
Check the row sum, but do not stop there. Every valid row sums to zero, but a row can sum to zero and still have invalid entries.
Check the state order. A transition from \(i\) to \(j\) goes in row \(i\), column \(j\). Reversing those indices can incorrectly place an incoming transition in the absorbing state’s row.

The invalid row in the second example illustrates the main trap: \((-0.04,0.04,0)\) sums to zero but is not a valid row. Its negative off-diagonal entry violates the intensity rule, and its positive diagonal does not balance a set of nonnegative outgoing intensities. If the state is absorbing, the correction is not to adjust that row’s diagonal; the entire row must be zero.

Key takeaway: For an absorbing state, write a zero in every column of its row. This represents the absence of all outgoing transitions; the diagonal is zero because it balances a sum of zero off-diagonal intensities.

Check Your Understanding

For each problem, use the stated state order and check the model description as well as the row sums.

A hypothetical model has states \(W\) (waiting) and \(T\) (a permanent terminal status), in that order. The only transition is \(W\to T\) at intensity \(0.06\) per year. A draft gives row \(T=(0,0.06)\). Identify the error and state the correct row. (Answer: The \(W\to T\) intensity belongs in row \(W\), column \(T\); the absorbing row \(T\) is \((0,0)\).)
In state order \(A,I,D\), a hypothetical model has \(A\to I\) intensity \(0.09\), \(A\to D\) intensity \(0.01\), and \(I\to D\) intensity \(0.04\), all per year. State \(D\) is permanent. Find the total outgoing intensity from \(A\), its diagonal entry, and row \(D\). (Answer: The total outgoing intensity from \(A\) is \(0.10\) per year, \(q_{AA}=-0.10\), and row \(D=(0,0,0)\).)
A proposed row for an absorbing state \(M\), in state order \(A,M,D\), is \((0,-0.03,0.03)\). It sums to zero. Explain why it is invalid and give the correct row if \(M\) is absorbing. (Answer: It has a negative off-diagonal entry and a positive diagonal; row \(M\) must be \((0,0,0)\).)
A hypothetical model has states \(A,B,T\), in that order. The allowed transitions are \(A\to B\) at intensity \(0.07\), \(A\to T\) at intensity \(0.03\), \(B\to A\) at intensity \(0.02\), and \(B\to T\) at intensity \(0.05\), all per year. State \(T\) is permanent. Construct and check the intensity matrix. (Answer: \(Q=\begin{pmatrix}-0.10&0.07&0.03\\0.02&-0.07&0.05\\0&0&0\end{pmatrix}\); each row sums to zero, all off-diagonal entries are nonnegative, and \(T\) is absorbing.)
In a proposed matrix, row \(R\), for a state described as non-absorbing, is \((-0.09,0.06,0.03)\). The row sum is zero. What can be concluded about whether this row describes an absorbing state, and why? (Answer: It is not an absorbing-state row because it has positive off-diagonal entries \(0.06\) and \(0.03\); the zero row sum alone does not make a state absorbing.)

What You'll Learn