Backward Equations for Eventual Absorption Probabilities

Why Backward Equations Answer an Eventual-Absorption Question

In a continuous-time multiple-state model, a process may move through several transient states before reaching one of several absorbing states. If the question is which absorbing state will eventually be reached, the answer depends on the possible first transition from the starting state and on the same question in the state reached next. This recursive structure gives a system of backward equations.

These equations are useful when the desired quantities are eventual probabilities rather than transition probabilities at a specified time. They turn the continuous-time model into a finite system of linear equations. The intensities determine the relative chances of each first transition; the amount of time spent waiting does not affect which absorbing state is eventually reached in a time-homogeneous model.

Definition: Let $h_i^{(k)}$ be the probability that a process starting in transient state $i$ is eventually absorbed in absorbing state $k$. For an absorbing target $k$, its boundary value is $h_k^{(k)}=1$; for any other absorbing state $r$, $h_r^{(k)}=0$.

Assume a finite, time-homogeneous continuous-time Markov chain with constant transition intensities $q_{ij}\geq 0$ from state $i$ to a different state $j$. The total exit intensity from transient state $i$ is $\lambda_i=\sum_{j\ne i}q_{ij}$, which is assumed positive. The probability of a particular next state $j$, conditional on leaving $i$, is $q_{ij}/\lambda_i$. For the usual unique linear-system solution, assume that the process is eventually absorbed in one of the listed absorbing states with probability 1 from every transient state.

Deriving the Backward Equations

Fix an absorbing target $k$, and start in transient state $i$. The process first waits in $i$, then jumps to some other state. Conditional on that first jump being to $j$, its chance of eventually reaching $k$ is $h_j^{(k)}$, with the absorbing-state boundary values just defined. Conditioning on the first jump therefore gives

h_i^{(k)} =\sum_{j\ne i}\frac{q_{ij}}{\lambda_i}h_j^{(k)}.

Multiplying by $\lambda_i$ separates transitions to other transient states from transitions directly to absorbing states. If the transient states form a set $T$, the equations for $i\in T$ are

\lambda_i h_i^{(k)} =\sum_{\substack{j\in T\\j\ne i}}q_{ij}h_j^{(k)}+q_{ik}.

Here $q_{ik}$ is the intensity of a direct transition from $i$ to the target $k$. A direct transition to any other absorbing state contributes zero to this target’s equation because its boundary value is zero. If the target consists of a set of absorbing states, replace $q_{ik}$ by the sum of the intensities from $i$ to states in that target set.

Theorem: For each absorbing target $k$, solve one linear system for the transient-state probabilities $h_i^{(k)}$. In generator notation, let $Q_{TT}$ be the transient-to-transient block of the generator, with diagonal entries $q_{ii}=-\lambda_i$, and let $r_k$ have entries $q_{ik}$. Then $$ -Q_{TT}h^{(k)}=r_k. $$ When absorption in the listed absorbing states occurs with probability 1, this system has a unique solution.

The equation is called backward because it fixes the starting state $i$ and looks forward by conditioning on the first transition. It differs from a forward equation, which evolves probabilities across destination states as time passes. No interest rate or discount factor appears: an eventual-absorption probability concerns the destination, not the present value or timing of a payment.

Solving the Equations in Practice

For multiple absorbing targets, it is often efficient to solve for one target and then use the fact that the probabilities of all absorbing outcomes sum to 1. That shortcut is valid only if absorption in the listed states is certain. Otherwise, there may be a positive probability of never reaching any of them, and the probabilities for the listed targets need not sum to 1.

Set the absorbing boundary values.
For the target being calculated, assign 1 at that absorbing state and 0 at the other absorbing states.

Write each transient-state equation.
Use the total exit intensity on the left and the outgoing intensities times the corresponding eventual probabilities on the right.

Solve and check.
Substitute the solution into every equation, check that each probability lies in $[0,1]$, and verify the outcome probabilities sum to 1 when eventual absorption is certain.

Worked Example: Two Transient States and Two Absorbing Outcomes

A hypothetical policyholder moves between active state $U$ and disabled state $V$, and can be absorbed in recovery state $R$ or death state $D$. All intensities are constant and measured per year. From $U$, the intensities to $V,R,D$ are $2,1,1$, respectively. From $V$, the intensities to $U,R,D$ are $1,3,0$, respectively. Find the probabilities of eventual recovery and death from each transient starting state.

Method. First solve for eventual recovery. The total exit intensity is $4$ from each transient state. Recovery has boundary value 1 and death has boundary value 0, so the equations are

4h_U=2h_V+1,\qquad 4h_V=h_U+3.

From the second equation, $h_U=4h_V-3$. Substitution into the first gives

4(4h_V-3)=2h_V+1,\qquad 16h_V-12=2h_V+1,\qquad 14h_V=13.

Thus $h_V=13/14$, and $h_U=4(13/14)-3=52/14-42/14=10/14=5/7$. Since absorption must be in $R$ or $D$, the death probabilities are the complements: from $U$, $1-5/7=2/7$; from $V$, $1-13/14=1/14$.

Check. The probabilities sum to 1 from each starting state: $5/7+2/7=1$ and $13/14+1/14=1$. The recovery equations also hold: $4(5/7)=2(13/14)+1=26/14+1=20/7$, and $4(13/14)=5/7+3=52/14$.

Common mistake avoided. The intensity to recovery is not itself the recovery probability. For example, from $U$, the first-jump probability of moving to recovery is $1/4$; the remaining $3/4$ of paths first move to $V$ or $D$. The eventual probability must account for what happens after those jumps.

Worked Example: Three Transient States with Direct Target Transitions

A hypothetical long-term care model has transient states $1,2,3$ and absorbing states $A$ and $B$. Intensities per year are as follows: from state 1, the transitions to state 2, $A$, and $B$ have intensities $2,1,1$; from state 2, transitions to state 1, state 3, and $A$ have intensities $1,1,2$; from state 3, transitions to state 2 and $B$ have intensities $2,2$. Find the eventual absorption probabilities for both targets from each transient state.

Method. Every transient state has total exit intensity $4$. Write $a_i$ for the probability of eventual absorption in $A$ from state $i$. The backward equations are

4a_1=2a_2+1,\qquad 4a_2=a_1+a_3+2,\qquad 4a_3=2a_2.

The third equation gives $a_3=a_2/2$. Substituting this into the second gives $4a_2=a_1+a_2/2+2$, so $a_1=(7/2)a_2-2$. Substituting that expression into the first yields

4\left(\frac{7}{2}a_2-2\right)=2a_2+1,\qquad 14a_2-8=2a_2+1,\qquad 12a_2=9.

Therefore $a_2=3/4$, $a_1=(7/2)(3/4)-2=21/8-16/8=5/8$, and $a_3=(1/2)(3/4)=3/8$. The process is eventually absorbed in $A$ or $B$, so the probabilities of $B$ are the complements: $b_1=3/8$, $b_2=1/4$, and $b_3=5/8$.

Check. The pairs sum to 1 in each state. For $A$, the equations check directly: $4(5/8)=2(3/4)+1=5/2$; $4(3/4)=5/8+3/8+2=3$; and $4(3/8)=2(3/4)=3/2$.

Common mistake avoided. State 3 has no direct transition to $A$, but its eventual probability of $A$ is not zero: it can move to state 2, from which $A$ remains reachable. Its equation is $4a_3=2a_2$, not $a_3=0$.

Worked Example: Combining Starting-State Probabilities

Consider a hypothetical model with transient states $P,Q$ and absorbing outcomes $A,B$. From $P$, intensities per year to $Q,A,B$ are $3,2,1$; from $Q$, intensities to $P,B$ are $1,3$. An initial population is in $P$ with probability $1/3$ and in $Q$ with probability $2/3$. Find the overall probability of eventual absorption in each outcome.

Method. The total exit intensities are $6$ in $P$ and $4$ in $Q$. Write $a_P,a_Q$ for the probabilities of eventual absorption in $A$. The equations are $6a_P=3a_Q+2$ and $4a_Q=a_P$. Using $a_P=4a_Q$ in the first equation gives $24a_Q=3a_Q+2$, so $21a_Q=2$, $a_Q=2/21$, and $a_P=8/21$.

The initial mixture probability of $A$ is the probability-weighted average:

P(A)=\frac{1}{3}\frac{8}{21}+\frac{2}{3}\frac{2}{21} =\frac{8}{63}+\frac{4}{63} =\frac{12}{63} =\frac{4}{21}.

There are only two absorbing outcomes and absorption is certain, so the probability of $B$ is $1-4/21=17/21$. Alternatively, the state-specific probabilities are $b_P=13/21$ and $b_Q=19/21$; their mixture is $(1/3)(13/21)+(2/3)(19/21)=51/63=17/21$.

Check. The population-level probabilities sum to $4/21+17/21=1$. The mixture must weight starting-state probabilities, not transition intensities: the initial state is selected before the process evolves.

Conditions, Checks, and Exam Traps

The backward equations rely on the Markov property: conditional on the current state, the distribution of future transitions does not depend on the earlier path. With constant intensities, the next-state probabilities are the intensity proportions $q_{ij}/\lambda_i$. If the intensities vary with time, a constant system of algebraic equations does not generally describe eventual absorption from a fixed starting time; a time-dependent treatment is needed.

A common algebra error is to omit transitions that do not go directly to the target. They still contribute to the total exit intensity on the left and to the future-probability terms on the right. For example, if state $X$ has rates $3$ to $Y$, $1$ to target $A$, and $1$ to another absorbing state $B$, the correct equation for $A$ is $5h_X=3h_Y+1$. The incorrect equation $3h_X=3h_Y+1$ leaves out the $B$ transition from the total exit rate. The correction is to include every outgoing intensity in $\lambda_X=5$, while the direct contribution from $B$ remains zero because $h_B^{(A)}=0$.

Also distinguish an absorbing target’s boundary value from a transient-state probability. A process already in the target has probability 1 of eventual absorption there; that does not mean a process in a transient state has probability 1. After solving, test the equations themselves, not only whether the answers look plausible. For multiple absorbing outcomes, check that the probabilities sum to 1 only when eventual absorption among those outcomes is certain.

Key takeaway: For each target, assign its absorbing boundary values, condition on the first jump from every transient state, and solve the resulting linear system. Include all exits in each total exit intensity, then verify the solution and the probability totals under the stated absorption assumptions.

Check Your Understanding

For each hypothetical model, assume constant intensities in the stated time unit and eventual absorption in the listed outcomes. Set up the appropriate backward equations before checking the final answer.

A transient state $C$ has intensity $2$ per year to absorbing outcome $R$ and intensity $3$ per year to absorbing outcome $D$. What is the probability of eventual absorption in $R$ from $C$? (Answer: $2/5$.)
Transient states $X,Y$ lead to absorbing states $R,D$. From $X$, rates to $Y,R,D$ are $2,1,1$; from $Y$, rates to $X,R,D$ are $1,1,2$, all per year. Find the probabilities of $R$ from $X$ and $Y$, and of $D$ from each. (Answer: From $X$, $R=3/7$, $D=4/7$; from $Y$, $R=5/14$, $D=9/14$.)
Transient state $M$ has rates $1$ to $N$, $2$ to $A$, and $1$ to $B$; state $N$ has rates $3$ to $M$ and $1$ to $A$, all per year. Find the eventual probabilities of $A$ and $B$ from both transient states. (Answer: From $M$, $A=9/13$, $B=4/13$; from $N$, $A=10/13$, $B=3/13$.)
A population starts in transient state $P$ with probability $1/4$ and $Q$ with probability $3/4$. From $P$, rates per year to $Q,A,B$ are $3,2,1$; from $Q$, rates to $P,B$ are $1,3$. Find the population’s probability of eventual absorption in $A$ and in $B$. (Answer: $A=1/6$, $B=5/6$.)

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