Absorbing-State Probabilities from the Forward Equation

Introduction

In a continuous-time multiple-state model, an absorbing state represents an outcome that cannot be left once entered. Examples in hypothetical insurance models might include death or a terminal claim outcome. The probability of being in such a state by a specified time can be found by following the probability flow into it. When there are several absorbing outcomes, the calculations also show how that probability is divided among them.

The key is to use the forward equation with the correct state-index convention. The rate of change of a probability in an absorbing state is determined by probability arriving from every state with a direct transition into it—not merely by the starting state’s direct transition rate.

Definition: Let \(X_t\) be a time-homogeneous continuous-time Markov chain with finite state space and generator \(Q=(q_{ij})\), using the row convention. The transition probability \(p_{ij}(t)=P(X_t=j\mid X_0=i)\) is the probability of being in state \(j\) at time \(t\), given that the process started in state \(i\). A state \(a\) is absorbing if \(q_{aa}=0\) and \(q_{aj}=0\) for every \(j\ne a\). Its row of \(Q\) is therefore zero.

Forward Equations for Absorbing States

For a time-homogeneous model under the row convention, the forward equation is \(P'(t)=P(t)Q\), where \(P(t)=(p_{ij}(t))\). In particular, for a fixed initial state \(i\) and destination state \(a\),

\frac{d}{dt}p_{ia}(t)=\sum_k p_{ik}(t)q_{ka}.

If \(a\) is absorbing, then \(q_{aa}=0\). Also, no transition leaves \(a\), so its probability does not contribute to the rate of increase in \(p_{ia}(t)\). The derivative is the sum of the incoming flows from states with \(q_{ka}>0\). This may include several transient states, and the process may need to pass through one or more of them before reaching \(a\).

The equation can be justified by considering a short interval of length \(h\). Conditional on being in state \(k\) at time \(t\), the probability of a transition from \(k\) to \(a\) during that interval is \(q_{ka}h+o(h)\). Summing over the possible states \(k\) at time \(t\) gives the probability of a transition into \(a\), to first order in \(h\). Dividing by \(h\) and taking the limit yields the forward equation above.

If the model’s states consist of transient states and absorbing states, let \(T_A\) denote the first time the process enters any absorbing state. Because an absorbing state cannot be left, being in an absorbing state at time \(t\) is equivalent to having entered one by time \(t\). Thus, for an initial state \(i\),

P(T_A\le t\mid X_0=i)=\sum_{a\text{ absorbing}}p_{ia}(t).

This is a finite-time probability. It need not equal 1: the process may still be in a transient state at time \(t\), or may have a positive probability of never reaching an absorbing state. The sum of the absorbing-state probabilities should equal 1 minus the sum of the transient-state probabilities.

Worked Example: Two Absorbing Outcomes and an Intermediate State

A hypothetical contract model has transient states \(A\) and \(B\), and absorbing states \(D_1\) and \(D_2\). From \(A\), the process moves to \(B\) at rate 1 per year or to \(D_1\) at rate 1 per year. From \(B\), it moves to \(D_1\) at rate 1 per year or to \(D_2\) at rate 1 per year. The process starts in \(A\). Find the probabilities of being in each absorbing state after one year, and check the total probability of absorption.

Method. Write the forward equations for the transient-state probabilities first, then integrate each incoming flow into an absorbing state. The total exit rate from both \(A\) and \(B\) is 2 per year. The forward equations for the transient probabilities are \(p_{AA}'(t)=-2p_{AA}(t)\) and \(p_{AB}'(t)=p_{AA}(t)-2p_{AB}(t)\), with \(p_{AA}(0)=1\) and \(p_{AB}(0)=0\).

p_{AA}(t)=e^{-2t},\qquad p_{AB}(t)=te^{-2t}.

For example, the second expression follows by multiplying the equation for \(p_{AB}\) by the integrating factor \(e^{2t}\): \((e^{2t}p_{AB}(t))'=1\). Integrating from 0 to \(t\) gives \(e^{2t}p_{AB}(t)=t\).

There is an incoming transition to \(D_1\) from each transient state, at rate 1 per year. The only incoming transition to \(D_2\) is from \(B\), also at rate 1 per year. Therefore,

p_{AD_1}'(t)=p_{AA}(t)+p_{AB}(t),\qquad p_{AD_2}'(t)=p_{AB}(t).

Integrating from time 0, when both absorbing-state probabilities are 0, gives

p_{AD_1}(t)=\int_0^t (1+s)e^{-2s}\,ds =\frac{3-(3+2t)e^{-2t}}{4},

p_{AD_2}(t)=\int_0^t se^{-2s}\,ds =\frac{1-(1+2t)e^{-2t}}{4}.

At \(t=1\), \(e^{-2}=0.1353\) to four decimals. Substitution gives \(p_{AD_1}(1)=0.5808\) and \(p_{AD_2}(1)=0.1485\). Their sum is \(0.7293\) to four decimals. The transient probabilities are \(p_{AA}(1)=0.1353\) and \(p_{AB}(1)=0.1353\), summing to \(0.2707\). Before rounding, the absorbing probabilities sum to \(1-2e^{-2}=0.729329\), and the transient probabilities sum to \(2e^{-2}=0.270671\); together they equal 1. The rounded values also check to 1 at four decimals.

Common mistake avoided. Using only the direct rate from the starting state gives \(p_{AD_1}(t)=(1-e^{-2t})/2\), or \(0.4323\) at one year. That omits flow from \(B\) into \(D_1\). The forward equation correctly includes both incoming flows, \(p_{AA}(t)\) and \(p_{AB}(t)\).

Several Routes and Probability Checks

For a collection of absorbing states, each absorbing-state equation can be written separately by listing its incoming transitions. After solving for the transient-state probabilities, integrate those flows from the initial time to the required time. Adding the resulting absorbing-state probabilities gives the probability of absorption by that time. A separate check using the transient states can reveal a missing route, a wrong rate, or an arithmetic error.

Worked Example: A Transient State That Feeds One Absorbing Outcome

Consider a hypothetical model with transient states \(W\) and \(T\), and absorbing states \(D_1\) and \(D_2\). Starting in \(W\), the process moves to \(D_1\) at rate 2 per year and to \(T\) at rate 1 per year. From \(T\), it moves to \(D_2\) at rate 3 per year. Find the absorbing-state probabilities after one year.

The total exit rate from \(W\) is 3 per year. The forward equation for \(p_{WW}\) gives \(p_{WW}(t)=e^{-3t}\). The equation for \(p_{WT}\) is \(p_{WT}'(t)=p_{WW}(t)-3p_{WT}(t)\), with \(p_{WT}(0)=0\). Multiplication by \(e^{3t}\) and integration give \(p_{WT}(t)=te^{-3t}\).

The incoming flow to \(D_1\) is at rate 2 from \(W\); the incoming flow to \(D_2\) is at rate 3 from \(T\). Integrating each flow gives

p_{WD_1}(t)=2\int_0^t e^{-3s}\,ds =\frac{2}{3}(1-e^{-3t}),

p_{WD_2}(t)=3\int_0^t se^{-3s}\,ds =\frac{1-(1+3t)e^{-3t}}{3}.

At one year, \(e^{-3}=0.0498\) to four decimals. The absorbing probabilities are \(p_{WD_1}(1)=0.6335\) and \(p_{WD_2}(1)=0.2670\). Their unrounded sum is \(0.900426\), which rounds to \(0.9004\). The transient probabilities are \(p_{WW}(1)=0.0498\) and \(p_{WT}(1)=0.0498\), with unrounded sum \(0.099574\), or \(0.0996\) to four decimals. The unrounded absorbing and transient totals sum to 1. The separately rounded absorbing components add to \(0.9005\); the \(0.0001\) difference from the rounded total is due to rounding.

The calculation illustrates why an absorbing-state probability is not generally just the probability of a direct transition from the initial state. The probability of reaching \(D_2\) depends on being in \(T\) at the time the transition to \(D_2\) occurs.

Worked Example: Three Transient States and Two Absorbing Outcomes

A hypothetical model has transient states \(0,1,2\) and absorbing states \(D_1,D_2\). From state 0, the process moves to state 1 or to \(D_1\), each at rate 1 per year. From state 1, it moves to state 2 or to \(D_1\), each at rate 1 per year. From state 2, it moves to either absorbing state, each at rate 1 per year. The process starts in state 0. Find the probability of each absorbing outcome after one year.

Each transient state has total exit rate 2 per year. Solving the transient forward equations successively gives

p_{00}(t)=e^{-2t},\qquad p_{01}(t)=te^{-2t},\qquad p_{02}(t)=\frac{t^2}{2}e^{-2t}.

The first expression solves \(p_{00}'=-2p_{00}\). The next solves \(p_{01}'=p_{00}-2p_{01}\) with initial value 0. The third solves \(p_{02}'=p_{01}-2p_{02}\), also with initial value 0; multiplying by \(e^{2t}\) reduces these successive equations to integrating 1 and then \(t\).

There is an incoming transition to \(D_1\) from all three transient states, each at rate 1 per year. There is an incoming transition to \(D_2\) only from state 2, also at rate 1 per year. Hence

p_{0D_1}(t)=\int_0^t e^{-2s}\left(1+s+\frac{s^2}{2}\right)\,ds, \qquad p_{0D_2}(t)=\int_0^t \frac{s^2}{2}e^{-2s}\,ds.

Evaluating the integrals,

p_{0D_1}(t)= \frac{1-e^{-2t}}{2} +\frac{1-(1+2t)e^{-2t}}{4} +\frac{1-(1+2t+2t^2)e^{-2t}}{8},

p_{0D_2}(t)=\frac{1-(1+2t+2t^2)e^{-2t}}{8}.

At \(t=1\), the three terms in \(p_{0D_1}(1)\) are \(0.4323\), \(0.1485\), and \(0.0404\) to four decimals. Using unrounded values, \(p_{0D_1}(1)=0.621246\), or \(0.6212\) to four decimals, and \(p_{0D_2}(1)=0.040415\), or \(0.0404\). Their unrounded total is \(0.661662\), which rounds to \(0.6617\). The transient total is \(e^{-2}(1+1+1/2)=0.338338\), or \(0.3383\). The unrounded totals sum to 1; the separately rounded absorbing probabilities sum to \(0.6616\), a rounding difference of \(0.0001\).

Practical Checks and Exam Traps

A reliable calculation follows the direction of probability flow. For each absorbing state, list the states that can enter it directly, multiply each state-occupancy probability by its transition intensity, and integrate over time. If a route passes through another transient state first, its contribution appears through that state’s occupancy probability in the incoming-flow equation.

Do not use the absorbing state's row to calculate incoming flow. Its generator row is zero because it has no exits. Incoming probabilities are determined by the relevant entries in the absorbing state's column.
Do not omit indirect routes. A transition into an absorbing state from an intermediate transient state contributes even when the initial state has no direct transition to that outcome.
Distinguish finite-time absorption from eventual absorption. The sum of absorbing-state probabilities at time \(t\) is the probability of absorption by time \(t\), not automatically 1.
Keep the rate and time units consistent. If rates are per year, \(t\) in the equations must be measured in years.
Check the full state total. The absorbing-state total plus the transient-state total should equal 1, apart from rounding.

These calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In a multiple-state insurance model, the same flow accounting applies to hypothetical outcomes such as recovery, disability termination, or death, provided the states and transition intensities are specified consistently.

Check Your Understanding

For each problem, calculate the requested probabilities using the incoming flows to the absorbing states and check the total against the transient-state probability.

A process starts in transient state \(U\). It moves directly to absorbing states \(A\) and \(B\) at rates 2 and 1 per year, respectively. Find the probabilities of being in \(A\) and \(B\), and the probability of absorption, at \(t=(\ln 2)/3\) years. (Answer: \(p_{UA}=0.3333\), \(p_{UB}=0.1667\), total absorption \(=0.5000\).)
A process starts in \(U\). From \(U\), it moves to absorbing state \(A\) at rate 1 or to transient state \(V\) at rate 2. From \(V\), it moves to absorbing state \(B\) at rate 3. Find both absorbing-state probabilities after one year. (Answer: \(p_{UA}=0.3167\), \(p_{UB}=0.5339\), total absorption \(=0.8506\) to four decimals.)
A process starts in \(R\). From \(R\), it moves to absorbing state \(A\) at rate 1 or to transient state \(S\) at rate 2. From \(S\), it moves to \(A\) at rate 1 or absorbing state \(B\) at rate 2. Find the absorbing-state probabilities at \(t=1\), and verify the transient total. (Answer: \(p_{RA}=0.4947\), \(p_{RB}=0.3559\), transient total \(=0.1494\); total absorption \(=0.8506\).)
A process starts in transient state 0. It moves from 0 to transient state 1 at rate 2 and to absorbing state \(A\) at rate 1. It moves from 1 to absorbing state \(B\) at rate 2. Both transient states have total exit rate 3 per year. Write the state-occupancy probabilities and calculate the absorbing-state probabilities at \(t=1\). (Answer: \(p_{00}=e^{-3}\), \(p_{01}=2e^{-3}\), \(p_{0A}=0.3167\), \(p_{0B}=0.5339\); total absorption \(=0.8506\).)
In a model with transient states \(M,N\) and absorbing states \(C,D\), the only transitions into \(C\) are from \(M\) at rate 2 and \(N\) at rate 1. State \(D\) can be entered only from \(N\), at rate 3. Starting in \(M\), state the forward-flow equations for \(p_{MC}\) and \(p_{MD}\). (Answer: \(p_{MC}'(t)=2p_{MM}(t)+p_{MN}(t)\), \(p_{MD}'(t)=3p_{MN}(t)\), with both absorbing-state probabilities initially 0.)

What You'll Learn