Absorption and the Generator
In a continuous-time Markov chain, a state can be absorbing: once entered, the process stays there forever. The generator identifies this behavior directly. Its diagonal entry records the negative of the total rate of leaving a state, so a zero exit rate means there is no possible departure. The zero-rate case also needs careful interpretation: the usual formula for the next-jump probabilities divides by the exit rate, and therefore does not apply at an absorbing state.
Throughout, let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on a finite or countable state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Write \(P_{ij}(t)=\mathbb{P}_i(X_t=j)\) for its transition probabilities and \(Q=(q_{ij})\) for its generator. For \(j\ne i\), \(q_{ij}\ge0\) is the instantaneous rate of a transition from \(i\) to \(j\); the rows of \(Q\) sum to zero, and \(q_i=-q_{ii}=\sum_{j\ne i}q_{ij}\) is the exit rate from \(i\). We use the natural filtration \(\mathcal{F}_t=\sigma(X_s:0\le s\le t)\) when conditioning on the history of the process.
The standard holding-time description of a CTMC says that, when \(q_i>0\), the holding time \(H_i\) at \(i\) is exponential with rate \(q_i\): \(\mathbb{P}_i(H_i>t)=e^{-q_i t}\). Conditional on a jump occurring, its destination \(j\ne i\) has probability \(q_{ij}/q_i\). These formulas describe the first jump after entering \(i\). When \(q_i=0\), the holding time is instead \(\infty\) almost surely, and there is no next-jump destination to sample.
Proof. First suppose \(i\) is absorbing. For each \(j\ne i\), \(P_{ij}(t)=0\) for every \(t\ge0\). By the definition of the generator’s off-diagonal entries, \(q_{ij}=\lim_{t\downarrow0}P_{ij}(t)/t=0\). Since \(q_i=\sum_{j\ne i}q_{ij}\), it follows that \(q_i=0\). Conversely, if \(q_i=0\), then the nonnegative rates \(q_{ij}\) sum to zero, so each \(q_{ij}=0\). In the CTMC construction from the generator, there is no event that takes the process out of \(i\); equivalently, its holding time has survival probability \(1\) at every finite time. Thus it remains at \(i\) forever and is absorbing. For \(q_i>0\), the holding-time and destination formulas follow from the exponential holding-time construction: the total rate of competing departures is \(\sum_{j\ne i}q_{ij}=q_i\), and the rate-\(q_{ij}\) contribution to the first-jump density at time \(t\) is \(q_{ij}e^{-q_i t}\). Integrating this density over \(t\ge0\) gives \(q_{ij}/q_i\); summing over destinations gives \(1\). \(\square\)
What the Jump-Chain Convention Means
When \(q_i>0\), the probabilities \(q_{ij}/q_i\) describe the state reached at the next actual jump. Recording the states visited at successive jump times gives the jump chain. If the CTMC reaches an absorbing state, there is no subsequent actual jump, so this record terminates there. A discrete-time chain used for bookkeeping is sometimes extended by setting its transition probability from an absorbing state to itself equal to \(1\). That self-loop is a convention for extending the discrete-time sequence; it does not represent an actual CTMC jump.
In particular, at \(q_i=0\), the expression \(q_{ij}/q_i\) is undefined, not a probability formula that can be evaluated by cancellation. The absorbing state’s behavior is already determined: the process waits there forever. Keeping the actual-jump description separate from any extended discrete-time convention prevents a fictitious jump from being mistaken for a physical event.
First-Jump Equations for Absorption
Let \(A\subseteq S\) be a nonempty set of absorbing states, and define the absorption time \(T_A=\inf\{t\ge0:X_t\in A\}\), with \(\inf\varnothing=\infty\). For a state \(i\notin A\) with \(q_i>0\), define \(r_{ij}=q_{ij}/q_i\) for \(j\ne i\). Conditioning on the first holding time and destination gives equations for both the probability of eventual absorption and the expected time to absorption.
Proof. Starting from \(i\notin A\), the first holding time \(H_i\) has mean \(1/q_i\), and the next state is \(j\ne i\) with probability \(r_{ij}\). The process cannot enter \(A\) before that first jump. Conditional on the destination \(j\), the Markov property at the jump gives the same absorption probability \(h_j\) and remaining expected time \(e_j\) as a process started at \(j\). Partitioning according to the first destination therefore gives \(h_i=\sum_{j\ne i}r_{ij}h_j\). For the time, the first holding time is always spent, and the subsequent time has conditional mean \(e_j\); taking expectations gives \(e_i=\mathbb{E}_i[H_i]+\sum_{j\ne i}r_{ij}e_j=1/q_i+\sum_{j\ne i}r_{ij}e_j\). If a path never reaches \(A\), its absorption time is infinite, which is included in the extended expectation. \(\square\)
These equations are a first-jump method, not a promise that every formal solution is the desired one. In particular, an absorption-probability system can have multiple solutions if absorption is not assured. Examples below verify the relevant finiteness or uniqueness conditions before using a solution.
Worked Example: Two Absorbing Destinations
Take \(S=\{0,1,2\}\), with states \(0\) and \(2\) absorbing. From state \(1\), the rates to \(0\) and \(2\) are \(2\) and \(1\), respectively. Find the probability of absorption at each destination and the expected time to absorption, starting from \(1\).
In the state order \(0,1,2\), the generator is $$ Q=\begin{bmatrix} 0&0&0\\ 2&-3&1\\ 0&0&0 \end{bmatrix}. $$ The row sums are \(0+0+0=0\), \(2-3+1=0\), and \(0+0+0=0\), as required. The absorbing rows have zero exit rate; at state \(1\), \(q_1=3\), so the next-jump probabilities are \(r_{10}=2/3\) and \(r_{12}=1/3\).
Absorption occurs at the first jump, which goes to exactly one of \(0\) and \(2\). Thus the absorption probabilities are \(2/3\) at \(0\) and \(1/3\) at \(2\), and their sum is \(2/3+1/3=1\). The time to absorption is the holding time at state \(1\), exponential with rate \(3\), so \(\mathbb{E}_1[T_{\{0,2\}}]=1/3\).
Check: The first-jump probability equation gives \(h_1=(2/3)\cdot1+(1/3)\cdot1=1\), the total probability of absorption. The expected-time equation gives \(e_1=1/3+(2/3)\cdot0+(1/3)\cdot0=1/3\), matching the exponential mean.
Worked Example: Absorption After Several Jumps
Let \(S=\{0,1,2\}\), with \(0\) absorbing. From state \(1\), the rates to \(0\) and \(2\) are \(1\) and \(2\); from state \(2\), the rates to \(0\) and \(1\) are \(3\) and \(1\). Starting from either transient state, calculate the probability and expected time of absorption in \(0\).
The generator, in the order \(0,1,2\), is $$ Q=\begin{bmatrix} 0&0&0\\ 1&-3&2\\ 3&1&-4 \end{bmatrix}. $$ The three row sums are \(0+0+0=0\), \(1-3+2=0\), and \(3+1-4=0\). The exit rates are \(q_1=3\) and \(q_2=4\). Therefore state \(1\) jumps to \(0\) with probability \(1/3\) and to \(2\) with probability \(2/3\); state \(2\) jumps to \(0\) with probability \(3/4\) and to \(1\) with probability \(1/4\).
Write \(h_i=\mathbb{P}_i(T_{\{0\}}<\infty)\). Conditioning on the first jump gives $$ h_1=\frac13+\frac23h_2,\qquad h_2=\frac34+\frac14h_1. $$ Substitute the second equation into the first: \(h_1=\frac13+\frac23(\frac34+\frac14h_1)=\frac13+\frac12+\frac16h_1=\frac56+\frac16h_1\). Hence \(\frac56h_1=\frac56\), so \(h_1=1\), and then \(h_2=\frac34+\frac14=\!1\).
For the expected times \(e_i=\mathbb{E}_i[T_{\{0\}}]\), the first holding-time means are \(1/3\) at state \(1\) and \(1/4\) at state \(2\). The equations are $$ e_1=\frac13+\frac23e_2,\qquad e_2=\frac14+\frac14e_1. $$ Substitution gives \(e_1=\frac13+\frac23(\frac14+\frac14e_1)=\frac13+\frac16+\frac16e_1=\frac12+\frac16e_1\). Thus \(\frac56e_1=\frac12\), so \(e_1=\frac35\). Then \(e_2=\frac14+\frac14\cdot\frac35=\frac14+\frac3{20}=\frac25\).
These expectations are finite: at each jump from states \(1\) or \(2\), the probability of absorption is at least \(1/3\). Thus the probability of surviving \(m\) jumps is at most \((2/3)^m\), giving a finite expected number of jumps. Each holding time has mean at most \(1/3\), so the expected total time is finite as well.
Check: Substitution gives \(h_1=\frac13+\frac23(1)=1\) and \(h_2=\frac34+\frac14(1)=1\). For the times, \(\frac13+\frac23\cdot\frac25=\frac13+\frac4{15}=\frac35=e_1\), and \(\frac14+\frac14\cdot\frac35=\frac25=e_2\).
Worked Example: Starting in an Absorbing State
Consider the one-state space \(S=\{a\}\), with generator \(Q=[0]\), and start the process at \(a\). Compute its holding time and its hitting time of \(A=\{a\}\).
The only generator row sums to \(0\), and \(q_a=-q_{aa}=0\). By the zero-exit-rate characterization, the process stays at \(a\) forever, so its holding time is \(\infty\) almost surely. Since the definition of \(T_A\) permits time \(t=0\), the process has already hit \(A\) at its initial time: \(T_{\{a\}}=0\) almost surely.
If one extends a discrete-time jump-chain bookkeeping rule to this state, the conventional assignment is a self-loop from \(a\) to \(a\). This does not alter either calculation: there are no actual jumps, and the hitting time remains zero. In particular, the next-jump formula would require \(q_{aa}/q_a=0/0\), which is undefined and should not be used.
Check: \(P_{aa}(t)=1\) for every \(t\ge0\), as required for an absorbing state, while \(T_{\{a\}}=0\) follows directly from the infimum over \(t\ge0\).
A Common Error and Its Correction
A common mistake is to treat the holding-time mean \(1/q_i\) and jump probabilities \(q_{ij}/q_i\) as formulas that apply at every state. At an absorbing state \(a\), this would give \(1/q_a=1/0\) and \(q_{aj}/q_a=0/0\). Neither is a valid finite holding-time mean or a valid destination probability. The correct interpretation is that the holding time is infinite and no next actual jump exists. If a discrete-time bookkeeping chain is desired, its self-loop at \(a\) is an added convention, not a rate-derived jump.
Absorbing states are useful as terminal outcomes in models of failure, recovery, extinction, or completion. Checking the generator row first resolves the basic question: if every off-diagonal rate in a row is zero, the state is absorbing; if the exit rate is positive, first-jump probabilities and holding-time means can be used to analyze what happens next. For absorption calculations, check the generator row sums, handle zero-rate states separately, and verify the solved equations by substitution.
Check Your Understanding
Use the definitions and first-jump equations to answer the following.
- A state \(i\) has \(q_{ij}=0\) for every \(j\ne i\). What is its exit rate, and what is its holding time?
- A state \(i\) has rates \(q_{i0}=2\) and \(q_{i1}=3\), with no other outgoing rates. Find its exit rate and the probabilities of the two possible next states.
- Why does \(q_{ij}/q_i\) not define a next-jump probability when \(q_i=0\)? What does the process do instead?
- For the generator in the second worked example, substitute \(e_1=3/5\) and \(e_2=2/5\) into both first-jump equations.
- Explain why a self-loop sometimes added to a discrete-time representation of an absorbing state is not an actual CTMC jump.