Absorbing States in Transition Matrices

Introduction

In a continuous-time state model, an absorbing state is one that cannot be left after it is entered. Recognizing such states from the transition intensities gives immediate information about the transition probability matrix: the row for an absorbing starting state is known at every duration, even when probabilities for other starting states require further calculation.

The distinction between a row and a column matters. A row describes the possible destination states from one starting state. Thus an absorbing state has a particularly simple row, but its column can still contain probabilities of arriving there from other states. This distinction is especially important when a model has more than one absorbing state.

Definition: In a finite-state continuous-time Markov model, state \(i\) is absorbing if, once entered, the process remains in \(i\) at all later times. For a time-homogeneous model, this means every outgoing intensity \(q_{ij}\) for \(j\ne i\) is zero. In the usual generator convention, \(q_{ii}=-\sum_{j\ne i}q_{ij}\), so an absorbing state has \(q_{ii}=0\) as well.

From Intensities to Absorbing Rows

Write \(p_{ij}(t)=P(X(t)=j\mid X(0)=i)\), where \(i\) is the starting state and \(j\) is the destination state after duration \(t\). The matrix \(P(t)=[p_{ij}(t)]\) uses starting states for rows and destination states for columns. A transition intensity \(q_{ij}\), for \(i\ne j\), describes the instantaneous rate of moving from \(i\) to \(j\); it is not itself a transition probability over a positive duration.

If state \(i\) is absorbing, starting in \(i\) means the process remains there with probability one. Consequently, \(p_{ii}(t)=1\) and \(p_{ij}(t)=0\) for every \(j\ne i\), for all \(t\ge 0\). In a time-inhomogeneous model, the corresponding condition is that all outgoing intensities from \(i\) remain zero at every subsequent time. An intensity that is zero only at the current instant is not enough to establish absorption.

p_{ij}(t)= \begin{cases} 1, & j=i,\\ 0, & j\ne i, \end{cases} \qquad \text{if }i\text{ is absorbing.}

This result determines an entire row of \(P(t)\). It does not determine every entry in the corresponding column. A transient state may lead into the absorbing state, so that column can have positive entries in rows for transient starting states.

Why the Absorbing Row Must Have This Form

The row follows directly from the definition of absorption. Conditional on starting in state \(i\), the process is in \(i\) at time zero. If it cannot leave \(i\), it is still in \(i\) after any duration \(t\). Therefore \(P(X(t)=i\mid X(0)=i)=1\), while the conditional probability of being in any distinct state \(j\) is zero. Since the entries in a transition probability row are these conditional probabilities, the row has a one in column \(i\) and zeros elsewhere.

This is consistent with the zero-duration transition matrix: at \(t=0\), every state is still its own destination, so \(P(0)\) is the identity matrix. Absorbing rows remain unchanged as duration increases; rows for other states need not.

Worked Example: One Absorbing State and a Two-State Matrix

A hypothetical model has states \(A\) (active) and \(D\) (dead). The only possible transition is from \(A\) to \(D\), with constant intensity \(0.25\) per year. Death is absorbing. Find the generator and the transition probability matrix after four years.

Method. First identify the absorbing state from the outgoing intensities. Then use the constant-rate no-transition probability for the active state and the row-sum condition.

The intensity from \(A\) to \(D\) is \(0.25\), so the diagonal generator entry in row \(A\) is \(-0.25\). There are no exits from \(D\). In the state order \(A,D\),

Q= \begin{pmatrix} -0.25 & 0.25\\ 0 & 0 \end{pmatrix}.

Starting in \(A\), the probability of no transition by time \(t\) is \(e^{-0.25t}\). At \(t=4\), this is \(e^{-1}=0.367879\ldots\), so \(p_{AA}(4)=0.3679\) to four decimal places. Since the only alternative destination is \(D\), \(p_{AD}(4)=1-0.367879\ldots=0.6321\). The absorbing row is \((0,1)\). Thus

P(4)= \begin{pmatrix} 0.3679 & 0.6321\\ 0 & 1 \end{pmatrix}.

Check. Each row sums to one, and the row for \(D\) is unchanged from the identity-matrix row. A common mistake is to use the generator row \((0,0)\) as the transition probability row for \(D\); the generator records rates, whereas the probability row is \((0,1)\).

Models With Multiple Absorbing States

Suppose there are several absorbing states. Each one contributes its own identity row: its row has a one in its own column and zeros in every other column. In the submatrix formed by the absorbing-state rows and columns, these entries form an identity matrix. However, the columns for absorbing states can receive probability from transient starting states.

For example, if a transient state can lead to either of two absorbing destinations, its row may have positive entries in both corresponding columns. Those entries depend on the rates and the duration. The absorbing rows alone do not determine them.

Worked Example: Two Competing Absorbing Destinations

A hypothetical claim state \(C\) has two possible transitions: to absorbing state \(D_1\) at constant intensity \(0.10\) per year, and to absorbing state \(D_2\) at constant intensity \(0.20\) per year. There are no other transitions. Construct the generator and the transition matrix after two years, using the state order \(C,D_1,D_2\).

Method. The total exit intensity from \(C\) is \(0.10+0.20=0.30\). The probability of remaining in \(C\) for two years is \(e^{-0.30(2)}=e^{-0.6}=0.5488116361\ldots\). The probability of leaving \(C\) by then is \(1-e^{-0.6}=0.4511883639\ldots\).

Conditional on leaving, the destination probabilities are proportional to the competing intensities: \(0.10/0.30=1/3\) for \(D_1\) and \(0.20/0.30=2/3\) for \(D_2\). Therefore

p_{CD_1}(2)=\frac{1}{3}(1-e^{-0.6})=0.1503961213\ldots, \qquad p_{CD_2}(2)=\frac{2}{3}(1-e^{-0.6})=0.3007922426\ldots.

The generator and transition matrix, rounded to four decimal places only in the final matrix, are

Q= \begin{pmatrix} -0.30 & 0.10 & 0.20\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}, \qquad P(2)= \begin{pmatrix} 0.5488 & 0.1504 & 0.3008\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}.

Check. The first row sums to \(0.5488+0.1504+0.3008=1.0000\), and each absorbing row sums to one. The absorbing-state submatrix, using rows and columns \(D_1,D_2\), is the \(2\) by \(2\) identity matrix.

Common mistake and correction. It is wrong to set the entire \(D_1\) column to zero because \(D_1\) is absorbing. That would incorrectly rule out arrivals from \(C\). Absorption means that the process cannot leave \(D_1\), not that it cannot enter \(D_1\). The correct \(D_1\) column has \(p_{CD_1}(2)=0.1504\) and \(p_{D_1D_1}(2)=1\).

Reading the Matrix Structure Carefully

When constructing or checking a transition matrix, separate facts forced by absorption from probabilities that still require calculation. If the state order is \(T_1,\ldots,T_r,A_1,\ldots,A_s\), where the \(T\) states are transient and the \(A\) states are absorbing, then the absorbing rows have the block form

P(t)= \begin{pmatrix} \text{transient-start rows}\\ 0\ \cdots\ 0\ \ I_s \end{pmatrix}.

The notation \(I_s\) denotes the \(s\) by \(s\) identity matrix. The zeros in the absorbing rows’ transient-state columns express that an absorbing state cannot be left. The entries in transient-start rows still need to be found from the model; they can include positive probabilities of reaching any absorbing state. Every full row must sum to one.

A state is not absorbing merely because its diagonal intensity is zero at a particular instant, or because no outgoing transition is listed in a partial description. Check all possible outgoing intensities under the model assumptions. For a time-varying model, check that they stay zero at all subsequent times. Also distinguish an absorbing state from a state that has no direct transition back to the starting state: a state can have outgoing transitions to other destinations and therefore is not absorbing.

Worked Example: A Sequential Model With an Absorbing Endpoint

A hypothetical process has states \(0,1,2\), with transitions \(0\to1\) at constant intensity \(0.50\) per year and \(1\to2\) at constant intensity \(0.25\) per year. State \(2\) has no exits. Find the generator and \(P(2)\), to four decimal places.

Method. Only state \(2\) has no outgoing intensity, so it is the sole absorbing state. The row for state \(2\) is \((0,0,1)\). For the other rows, use the one-way transition probabilities and the row-sum condition.

The generator, in order \(0,1,2\), is

Q= \begin{pmatrix} -0.50 & 0.50 & 0\\ 0 & -0.25 & 0.25\\ 0 & 0 & 0 \end{pmatrix}.

Starting in state \(0\), the probability of still being there after two years is \(p_{00}(2)=e^{-0.50(2)}=e^{-1}=0.3678794\ldots\). To be in state \(1\) at time \(2\), the process must move from \(0\) to \(1\) at some time \(s\) and then remain in state \(1\). Hence

\begin{aligned} p_{01}(2) &=\int_0^2 0.50e^{-0.50s}e^{-0.25(2-s)}\,ds\\ &=2(e^{-0.5}-e^{-1})\\ &=2(0.6065306597\ldots-0.3678794412\ldots)\\ &=0.4773024370\ldots. \end{aligned}

The remaining probability in row \(0\) is \(p_{02}(2)=1-p_{00}(2)-p_{01}(2)=0.1548181\ldots\). Starting in state \(1\), the probability of remaining there is \(p_{11}(2)=e^{-0.25(2)}=e^{-0.5}=0.6065307\ldots\), and \(p_{12}(2)=1-e^{-0.5}=0.3934693\ldots\). Therefore

P(2)= \begin{pmatrix} 0.3679 & 0.4773 & 0.1548\\ 0 & 0.6065 & 0.3935\\ 0 & 0 & 1 \end{pmatrix}.

Check. Before rounding, the first row sums to one by the definition of \(p_{02}(2)\); the second row sums to \(e^{-0.5}+1-e^{-0.5}=1\), and the third row sums to one. State \(1\) is not absorbing: it has an outgoing intensity of \(0.25\), even though it cannot return to state \(0\).

Exam Checks

A reliable construction starts with the state order, then checks outgoing intensities, then writes the forced absorbing rows. Only after those steps should the remaining transition probabilities be calculated. In particular, do not confuse a row of the generator with a row of the transition probability matrix, and do not infer a whole column from absorption alone.

Inspect outgoing intensities.
For each state, check whether every intensity to a different state is zero throughout the relevant future period.

Mark absorbing states.
Each qualifying state has no possible exit after entry; a zero rate at just one instant does not suffice.

Write their probability rows.
Place a one in each absorbing state’s own column and zeros in the rest of that row.

Check the remaining structure.
Allow transient states to reach absorbing states, and verify that every completed transition-probability row sums to one.

These checks are useful in multiple-state insurance models because death, maturity, or permanent claim closure may be represented as absorbing states, while other states can still move among themselves or into those endpoints. Correctly identifying the absorbing rows prevents structural errors before any valuation calculation uses the transition probabilities. This treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, state the absorbing state or states and use the specified state order. Answers are given for checking.

A two-state model has states \(W,Z\), with constant intensity \(0.08\) per year from \(W\) to \(Z\) and no exits from \(Z\). What is the absorbing row in the transition probability matrix? (Answer: \(Z\) is absorbing; in order \(W,Z\), its row is \((0,1)\).)
In state order \(H,C,D\), the only exits are \(H\to C\) at intensity \(0.12\), \(H\to D\) at intensity \(0.03\), and \(C\to H\) at intensity \(0.20\), all per year. State \(D\) has no exits. Which state is absorbing, and what is the \(D\) column’s entry in the \(D\) row? (Answer: Only \(D\) is absorbing; \(p_{DD}(t)=1\), while entries in the \(D\) column from \(H\) and \(C\) are not forced to be zero.)
In state order \(E,I,D,R\), the only exits are \(E\to I\) at intensity \(0.06\), \(E\to D\) at \(0.01\), \(I\to E\) at \(0.12\), and \(I\to D\) at \(0.08\), all per year. Neither \(D\) nor \(R\) has an exit. Give both absorbing rows. (Answer: \(D\) and \(R\) are absorbing; their rows are \((0,0,1,0)\) and \((0,0,0,1)\), respectively.)
A transient state \(C\) has two absorbing destinations \(A\) and \(B\), reached at constant intensities \(0.04\) and \(0.06\) per year. Starting in \(C\), find the row of transition probabilities after five years in order \(C,A,B\). (Answer: \((e^{-0.5},\,0.4(1-e^{-0.5}),\,0.6(1-e^{-0.5}))=(0.6065,0.1574,0.2361)\), to four decimal places.)
In a time-inhomogeneous model, \(Z\) has no outgoing intensity before time 3, but its intensity to absorbing state \(Y\) is \(0.20\) per year from time 3 onward. Starting in \(Z\) at time 0, is \(Z\) absorbing? Find \(p_{ZZ}(5)\) and \(p_{ZY}(5)\). (Answer: \(Z\) is not absorbing; \(p_{ZZ}(5)=e^{-0.20(2)}=0.6703\) and \(p_{ZY}(5)=1-e^{-0.4}=0.3297\).)

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