Transient and Absorbing Blocks in Matrix Solutions

Why Separate Transient and Absorbing States?

In a multiple-state model, an absorbing state cannot be left once it is entered. Examples in a hypothetical insurance model might include a terminal claim state or a completed-benefit state. The other states may be left as the process evolves; call these states transient when the model assumes they are eventually exited. Separating the two kinds of states makes it possible to calculate both how much probability remains in the transient states and how that probability has accumulated in each absorbing state.

This partition is especially useful when there is more than one possible absorbing outcome. It keeps the calculation organized: the transient block determines the probability of still being in each nonabsorbing state, while the rates from transient states into absorbing states determine the accumulating absorption probabilities. The method below uses the row-vector convention for transition matrices and the forward equation.

Definition: Suppose a finite-state, time-homogeneous continuous-time Markov chain has transient states listed first and absorbing states listed last. Its generator has the block form $$ Q= \begin{pmatrix} S & R\\ 0 & 0 \end{pmatrix}. $$ Here $S$ contains rates between transient states, including the negative diagonal entries; $R$ contains rates from transient states to absorbing states. The zero blocks express that no absorbing state can be left. Each row of $Q$ sums to zero, so each row of $S$ plus the corresponding row of $R$ sums to zero.

We use $P(t)$ for the transition probability matrix over an interval of length $t$. In the row-vector convention, $P'(t)=P(t)Q$, and $P(0)=I$. Partition its rows and columns in the same order as $Q$: $$ P(t)= \begin{pmatrix} P_{TT}(t) & P_{TA}(t)\\ P_{AT}(t) & P_{AA}(t) \end{pmatrix}. $$ The first subscript describes the starting state group and the second the ending state group. Thus, $P_{TA}(t)$ contains probabilities of starting transient and ending absorbing.

Block Forward Equations

Substitute the two block matrices into $P'(t)=P(t)Q$ and multiply by blocks. The resulting equations are

\begin{aligned} P_{TT}'(t)&=P_{TT}(t)S, & P_{TA}'(t)&=P_{TT}(t)R,\\ P_{AT}'(t)&=P_{AT}(t)S, & P_{AA}'(t)&=P_{AT}(t)R. \end{aligned}

The initial conditions follow from $P(0)=I$: $P_{TT}(0)=I$, $P_{TA}(0)=0$, $P_{AT}(0)=0$, and $P_{AA}(0)=I$, with identity matrices of the appropriate sizes. Since $P_{AT}(0)=0$, its homogeneous equation gives $P_{AT}(t)=0$; then $P_{AA}'(t)=0$, so $P_{AA}(t)=I$. An absorbing state stays where it is.

The upper blocks give the occupancy and absorption calculations of interest. Solving the first equation and integrating the second gives

P_{TT}(t)=e^{St}, \qquad P_{TA}(t)=\int_0^t e^{Ss}R\,ds.

The integral uses elapsed time $s$ inside the interval from 0 to $t$. It is the accumulated flow into absorbing states: at time $s$, the transient-to-transient probabilities are $e^{Ss}$, and the instantaneous rates from those states into absorbing states are $R$.

Result: If the initial probability row vector is $(\alpha,\beta)$, where $\alpha$ assigns probabilities to transient states and $\beta$ to absorbing states, then $$ p_T(t)=\alpha e^{St}, \qquad p_A(t)=\beta+\alpha\int_0^t e^{Ss}R\,ds. $$ The term $\beta$ remains in its original absorbing states; the integral adds probability absorbed by time $t$. If $S$ is invertible, the integral can also be written $S^{-1}(e^{St}-I)R$. The integral form remains the direct result of the block equations and avoids needing this inverse form.

Justification: Why the Integral Gives Absorption

The equation $P_{TA}'(t)=P_{TT}(t)R$ says that the rate of increase of each transient-to-absorbing transition probability is the current transient probability multiplied by the corresponding rate into the absorbing state. Integrating from 0 to $t$ accumulates that rate over the whole interval. Because $P_{TA}(0)=0$, the result is exactly the probability of having reached each absorbing state by time $t$, starting from each transient state.

For an initial distribution $(\alpha,\beta)$, this argument gives $p_A(t)=\beta+\alpha P_{TA}(t)$. Similarly, $p_T(t)=\alpha P_{TT}(t)$: probability initially in an absorbing state cannot return to a transient state. These formulas rely on a time-homogeneous generator with absorbing states as described above. If intensities vary with time, the constant matrix-exponential solution $e^{St}$ does not generally apply; the corresponding time-dependent forward equations must be used.

Worked Calculations

Worked Example: A Two-Stage Process with One Absorbing State

Consider a hypothetical benefit process with transient states 1 and 2 and absorbing state $D$. From state 1, the rate of moving to state 2 is 1 per year and the rate of moving to $D$ is 1 per year. From state 2, the rate of moving to $D$ is 1 per year. The process starts in state 1. Calculate the probabilities of occupying each state at $t=\ln 2$ years.

Method. List transient states before the absorbing state and partition the generator. The total exit rate from state 1 is $1+1=2$; from state 2 it is 1. Thus

Q= \begin{pmatrix} -2&1&1\\ 0&-1&1\\ 0&0&0 \end{pmatrix}, \qquad S= \begin{pmatrix}-2&1\\0&-1\end{pmatrix}, \qquad R= \begin{pmatrix}1\\1\end{pmatrix}.

For initial state 1, the transient occupancy probabilities solve $p_T'(t)=p_T(t)S$ with $p_T(0)=(1,0)$. The first state has no inflow, so $p_1(t)=e^{-2t}$. The second-state equation is $p_2'(t)=p_1(t)-p_2(t)$, $p_2(0)=0$. Multiplying by the integrating factor $e^t$ gives $(e^t p_2(t))'=e^{-t}$, and therefore $p_2(t)=e^{-t}-e^{-2t}$.

At $t=\ln 2$, $e^{-t}=1/2$ and $e^{-2t}=1/4$. Hence $p_1=1/4$ and $p_2=1/2-1/4=1/4$. There is only one absorbing state, so its probability is $1-p_1-p_2=1-1/4-1/4=1/2$.

Check. The probabilities sum to $1/4+1/4+1/2=1$. The method avoids treating the rate from state 1 to state 2 as an absorption rate: state 2 is transient, not absorbing.

Worked Example: Competing Absorbing Outcomes

A hypothetical contract model has one transient state $T$ and two absorbing outcomes, $A$ and $B$. From $T$, the rates to $A$ and $B$ are 1 and 3 per year, respectively. Starting in $T$, find the transient and absorbing probabilities after 0.5 years. Give probabilities to four decimal places.

Method. The total rate out of $T$ is $1+3=4$, so the generator blocks are $S=(-4)$ and $R=(1\ \ 3)$. The transient probability is $e^{-4t}$. At $t=0.5$, this is $e^{-2}=0.1353353$.

The absorption probabilities are the integrated rates into the two outcomes:

\begin{aligned} P(T\to A\text{ by }0.5) &=\int_0^{0.5} e^{-4s}\,ds =\frac{1-e^{-2}}{4} =0.2161662,\\ P(T\to B\text{ by }0.5) &=\int_0^{0.5} 3e^{-4s}\,ds =\frac{3(1-e^{-2})}{4} =0.6484985. \end{aligned}

Rounded to four decimal places, the probabilities are $0.1353$ in $T$, $0.2162$ in $A$, and $0.6485$ in $B$. Before rounding, they sum to $e^{-2}+(1-e^{-2})/4+3(1-e^{-2})/4=1$.

Check. The two absorbing probabilities are in the ratio $1:3$, matching the two exit rates, and their sum is $1-e^{-2}$, the probability of having left $T$. The method avoids assigning all exits to a single absorbing destination.

Worked Example: Two Transient States and Two Absorbing States

In a hypothetical care model, the transient states are 1 and 2, and the absorbing outcomes are $A$ and $B$. From state 1, the rates to state 2 and $A$ are each 1 per year. From state 2, the rates to $A$ and $B$ are each 1 per year. Starting in state 1, calculate all four state probabilities after 1 year. Give probabilities to four decimal places.

Method. Each transient state has total exit rate 2. In the stated ordering,

S= \begin{pmatrix}-2&1\\0&-2\end{pmatrix}, \qquad R= \begin{pmatrix}1&0\\1&1\end{pmatrix}.

Since $S=-2I+N$, where $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $N^2=0$, its exponential is $e^{St}=e^{-2t}(I+tN)$. Thus

P_{TT}(t)=e^{-2t} \begin{pmatrix}1&t\\0&1\end{pmatrix}.

Starting from state 1, the transient probabilities at $t=1$ are $p_1=e^{-2}$ and $p_2=e^{-2}$. For absorption, the instantaneous flow into $A$ is $p_1(s)+p_2(s)=e^{-2s}(1+s)$; the flow into $B$ is $p_2(s)=se^{-2s}$. Integrating gives

\begin{aligned} p_A(1) &=\int_0^1 e^{-2s}(1+s)\,ds =\frac{1-e^{-2}}{2}+\frac{1-3e^{-2}}{4} =\frac34-\frac54e^{-2} =0.5808309,\\ p_B(1) &=\int_0^1 se^{-2s}\,ds =\frac{1-3e^{-2}}{4} =0.1484985. \end{aligned}

Using $e^{-2}=0.1353353$, the four probabilities rounded to four decimal places are $p_1=0.1353$, $p_2=0.1353$, $p_A=0.5808$, and $p_B=0.1485$.

Check. Before rounding, $p_1+p_2+p_A+p_B=2e^{-2}+3/4-5e^{-2}/4+(1-3e^{-2})/4=1$. This check confirms that both transient occupancy and both absorption flows have been included.

Common Matrix and Probability Traps

Reversing the block order. With the row-vector convention, $P'=PQ$, so $P_{TA}'=P_{TT}R$, not $RP_{TT}$. The dimensions and interpretation must agree: $P_{TT}$ maps starting transient states to transient states, and $R$ then gives rates into absorbing states.
Putting absorbing states in $S$. The lower-right zero block reflects that absorbing states cannot be left. An absorbing state is not a transient state with a small exit rate.
Using rates as probabilities. An entry of $R$ is an instantaneous rate, not a probability over a finite interval. Finite-time absorption requires integrating transient occupancy multiplied by the relevant rates.
Forgetting probability already absorbed at time zero. For an initial distribution with mass $\beta$ in absorbing states, the absorbing probabilities include $\beta$ as well as the later accumulated flow.
Confusing event-by-time probabilities with probabilities of absorption at an exact time. $P_{TA}(t)$ is the probability of being in each absorbing state at time $t$, which here equals the probability of having reached it by then. The integrand describes the instantaneous rate of accumulation; it is not itself the finite-time probability.

Key takeaway: For a time-homogeneous generator partitioned as $Q=\begin{pmatrix}S&R\\0&0\end{pmatrix}$, transient occupancy is governed by $e^{St}$, and absorption is the accumulated transient-to-absorbing flow $\int_0^t e^{Ss}R\,ds$. Check the state ordering, matrix dimensions, and total probability.

Check Your Understanding

For each problem, use the stated state order and row-vector convention. Answers are rounded only where indicated.

A single transient state has rates 0.5 and 1.5 per year to absorbing states $A$ and $B$. Starting in the transient state, find all three probabilities at $t=\ln 2$ years. (Answer: transient $0.2500$, $A$ $0.1875$, $B$ $0.5625$.)
There are transient states 1 and 2 and one absorbing state. The rates are 1 from state 1 to state 2, 2 from state 1 to absorption, and 2 from state 2 to absorption, all per year. Starting in state 1, calculate the two transient probabilities and the absorption probability at $t=\ln 2$ years. (Answer: $p_1=0.1250$, $p_2=0.1250$, absorption $=0.7500$.)
A transient state has rates 2 and 3 per year to absorbing states $A$ and $B$. Starting in the transient state, find the probabilities of being transient, in $A$, and in $B$ at $t=(\ln 5)/5$ years. (Answer: transient $0.2000$, $A$ $0.3200$, $B$ $0.4800$.)
There are transient states 1 and 2 and absorbing states $A$ and $B$. From state 1, the rates to state 2 and $A$ are 2 and 1 per year. From state 2, the rate to $A$ is 3 per year. Starting in state 1, calculate all state probabilities at $t=1$ year. (Answer: $p_1=e^{-3}=0.0498$, $p_2=2e^{-3}=0.0996$, $p_A=1-3e^{-3}=0.8506$, $p_B=0$.)

What You'll Learn