Three-State Competing Exits and an Absorbing State

Introduction

A model can have only one state in which a person remains active and still require careful probability calculations. If that state has two possible exits, the chance of reaching either destination depends on both exit intensities: the first exit prevents the other exit from occurring. This is a continuous-time competing-exits model.

The key distinction is between the probability of reaching a particular absorbing state by time $t$, and the probability of that destination given that an exit has occurred by time $t$. With constant intensities, the destination probabilities have a simple closed form. The same formula also gives a useful check on the interpretation of each intensity.

Definition: Consider a time-homogeneous continuous-time Markov chain with states $0,1,2$. State $0$ is transient, and states $1$ and $2$ are absorbing. From state $0$, the transition intensity to state $1$ is $\alpha$, and the transition intensity to state $2$ is $\beta$, where $\alpha,\beta\geq 0$ and $\lambda=\alpha+\beta>0$. Once the process enters state $1$ or state $2$, it remains there.

Model and Closed-Form Probabilities

The total intensity of leaving state $0$ is $\lambda=\alpha+\beta$, not either exit intensity alone. Using the row convention for the generator, the intensity matrix is

Q= \begin{pmatrix} -\lambda & \alpha & \beta\\ 0&0&0\\ 0&0&0 \end{pmatrix}.

Let $p_{0j}(t)$ denote the probability of being in state $j$ at time $t$, given that the process started in state $0$ at time $0$. Because states $1$ and $2$ are absorbing, being in either destination at time $t$ means that the corresponding exit has occurred by time $t$.

Theorem: For $t\geq 0$, $$ p_{00}(t)=e^{-\lambda t},\qquad p_{01}(t)=\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right),\qquad p_{02}(t)=\frac{\beta}{\lambda}\left(1-e^{-\lambda t}\right). $$ Consequently, the probability of any exit by time $t$ is $1-e^{-\lambda t}$. Conditional on an exit having occurred by time $t$, the destination probabilities are $\alpha/\lambda$ for state $1$ and $\beta/\lambda$ for state $2$, provided $t>0$.

The conditional destination proportions do not depend on $t$. This happens because the two cause-specific intensities are constant: at every moment the process remains in state $0$, the relative rates of the two possible exits are the same.

Why the Formula Holds

The time to the first exit from state $0$ has an exponential distribution with rate $\lambda$. Thus the probability of still being in state $0$ at time $t$ is $e^{-\lambda t}$. To reach state $1$ at an instant near time $s$, the process must first survive in state $0$ until $s$, then exit to state $1$ during the next small time interval. The corresponding probability flow per unit time is $\alpha e^{-\lambda s}$.

Survive in the transient state.
The probability of no exit before time $s$ is $e^{-\lambda s}$, since the total exit intensity is $\lambda$.

Apply the cause-specific intensity.
Conditional on still being in state $0$, the probability of an exit to state $1$ in a short interval of length $ds$ is $\alpha\,ds+o(ds)$. The probability flow to state $1$ is therefore $\alpha e^{-\lambda s}\,ds$.

Accumulate flow over time.
Integrating the probability flow from time $0$ to $t$ gives the probability of reaching state $1$ by time $t$.

p_{01}(t)=\int_0^t \alpha e^{-\lambda s}\,ds =\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right).

The same argument with intensity $\beta$ gives $p_{02}(t)$. Adding the two destination probabilities gives $\frac{\alpha+\beta}{\lambda}(1-e^{-\lambda t})=1-e^{-\lambda t}$, because $\alpha+\beta=\lambda$. Together with $p_{00}(t)=e^{-\lambda t}$, the probabilities sum to one.

Worked Applications

Worked Example: Two Hypothetical Benefit Destinations

A hypothetical benefit-status model begins in state $0$. It moves to absorbing destination $1$ at constant intensity $\alpha=0.20$ per year and to absorbing destination $2$ at constant intensity $\beta=0.30$ per year. Find the probabilities of each state after two years, and the probability of destination $1$ conditional on an exit having occurred by then. Assume a time-homogeneous Markov model, with no return transitions.

Method. Add the competing intensities to obtain the total exit intensity: $\lambda=0.20+0.30=0.50$ per year. The two-year survival probability in state $0$ is $e^{-0.50(2)}=e^{-1}=0.367879$.

The probability of any exit by two years is $1-0.367879=0.632121$. The fraction of exits going to destination $1$ is $\alpha/\lambda=0.20/0.50=0.4$, and the fraction going to destination $2$ is $\beta/\lambda=0.30/0.50=0.6$. Therefore,

p_{01}(2)=0.4(0.632121)=0.252848,\qquad p_{02}(2)=0.6(0.632121)=0.379273.

The conditional probability of destination $1$, given an exit by year 2, is $0.252848/0.632121=0.4$. To four decimal places, the state probabilities are $p_{00}(2)=0.3679$, $p_{01}(2)=0.2528$, and $p_{02}(2)=0.3793$. Their rounded values sum to 1.0000.

Check and trap avoided. The probability of any exit is $0.252848+0.379273=0.632121$, as required. Using $\alpha$ alone in the survival exponent would incorrectly treat the exit to destination $2$ as if it did not end the time in state $0$.

Worked Example: Annual Intensities and a Longer Term

In a hypothetical long-term care model, an individual starts in a healthy transient state. The constant intensity of moving to an absorbing care state is $0.12$ per year, and the intensity of moving to an absorbing exit state is $0.08$ per year. Calculate the probabilities of being in each state after five years. State all rates per year and assume that there are no transitions out of either absorbing state.

Method. The total exit intensity is $\lambda=0.12+0.08=0.20$ per year. Since the time is measured in years, the exponent is $-\lambda t=-(0.20)(5)=-1$. Hence the probability of remaining healthy is $e^{-1}=0.367879$, and the probability of any exit is $1-e^{-1}=0.632121$.

The care-state share of exits is $0.12/0.20=0.6$; the exit-state share is $0.08/0.20=0.4$. Thus

p_{\text{care}}(5)=0.6(0.632121)=0.379273,\qquad p_{\text{exit}}(5)=0.4(0.632121)=0.252848.

To four decimal places, the probabilities are $0.3679$, $0.3793$, and $0.2528$, respectively. They sum to 1.0000 before rounding. The probability of being in the care state is not $0.6$: $0.6$ is the conditional destination share among those who have exited, while $0.379273$ also accounts for the chance of having no exit by year 5.

Check and trap avoided. The time and intensities use the same unit, years. Substituting five months or using $0.12$ alone in the exponent would change the model rather than merely change the calculation.

Worked Example: Recovering Destination Probabilities from Survival

A hypothetical state model has two absorbing destinations with constant intensities in the ratio $\alpha:\beta=2:1$. The probability of remaining in the initial state after three years is $0.25$. Find both intensities and the probabilities of occupying each state after three years. Assume time is measured in years and the model is time-homogeneous.

Method. The probability of remaining in state $0$ is $e^{-3\lambda}$. Therefore, $e^{-3\lambda}=0.25$, so $-3\lambda=\ln(0.25)=-\ln 4$ and $\lambda=\ln 4/3\approx0.462098$ per year. The ratio $2:1$ means the intensities are two-thirds and one-third of the total: $\alpha=(2/3)(0.462098)\approx0.308065$ and $\beta=(1/3)(0.462098)\approx0.154033$ per year.

The total exit probability is $1-0.25=0.75$. The destination shares are $2/3$ and $1/3$, so $p_{01}(3)=(2/3)(0.75)=0.50$ and $p_{02}(3)=(1/3)(0.75)=0.25$. Thus, to four decimal places, the state probabilities are $0.2500$, $0.5000$, and $0.2500$.

Check and trap avoided. The three probabilities sum to one. The observed survival probability identifies the total intensity, while the intensity ratio identifies the split between causes. The ratio alone cannot determine the total intensity.

Interpreting and Checking the Results

The formulas separate the calculation into two parts: the chance that an exit occurs, and the allocation of that exit between destinations. This separation is useful when checking calculations and when explaining results in context.

Quantity	Expression	Interpretation
Remain in state $0$	$e^{-\lambda t}$	No exit has occurred by time $t$
Any exit by time $t$	$1-e^{-\lambda t}$	The process is in one of the absorbing states
Reach state $1$ by time $t$	$(\alpha/\lambda)(1-e^{-\lambda t})$	An exit occurs by $t$ and its destination is state $1$
Reach state $2$ by time $t$	$(\beta/\lambda)(1-e^{-\lambda t})$	An exit occurs by $t$ and its destination is state $2$

Several checks follow directly. At $t=0$, the probabilities are $1,0,0$. As $t$ increases without bound, $p_{00}(t)$ tends to zero, while the destination probabilities tend to $\alpha/\lambda$ and $\beta/\lambda$. At every finite $t$, all probabilities are nonnegative and $p_{00}(t)+p_{01}(t)+p_{02}(t)=1$.

A frequent exam error is to report $\alpha/\lambda$ as the probability of reaching state $1$ by a specified finite time. That ratio is the conditional destination share, and also the eventual destination probability when $\lambda>0$. The finite-time probability must include the probability that an exit has occurred:

\underbrace{\frac{\alpha}{\lambda}}_{\text{destination share}} \quad\ne\quad \underbrace{\frac{\alpha}{\lambda}(1-e^{-\lambda t})}_{\text{destination probability by time }t} \qquad (0<t<\infty).

For example, with $\alpha=0.20$, $\beta=0.30$, and $t=2$, using the share $0.4$ as the destination probability overstates the correct value $0.252848$, because it ignores the $0.367879$ probability of no exit. If one intensity is zero, its destination probability is zero. If both intensities are zero, $\lambda=0$ and the ratio formulas do not apply; the process remains in state $0$ with probability one.

Key takeaway: Add the competing intensities to obtain the total exit rate. Multiply the probability of any exit by the relevant intensity's share of that total to obtain the finite-time probability of each absorbing destination.

Check Your Understanding

For each problem, assume a time-homogeneous model with one transient state and two absorbing destinations, and give probabilities to four decimal places unless stated otherwise.

Exit intensities are $0.40$ and $0.10$ per year. Find the probabilities of remaining in the transient state and reaching each destination after two years. (Answer: $0.3679,\ 0.5057,\ 0.1264$, respectively.)
A hypothetical model has destination intensities $0.06$ and $0.14$ per year. Find the three state probabilities after four years. (Answer: $0.4493,\ 0.1652,\ 0.3855$, respectively.)
Two absorbing destinations have intensities $0.10$ and $0.15$ per year. Find the probabilities of remaining in the transient state and reaching each destination after $t=\ln 2/0.25$ years. (Answer: $0.5000,\ 0.2000,\ 0.3000$, respectively.)
The probability of remaining in the initial state after six years is $0.20$, and the two constant exit intensities are in the ratio $3:2$. Find the probabilities of reaching the two destinations by year 6. (Answer: $0.4800$ and $0.3200$, respectively.)
With constant intensities $0.09$ and $0.21$ per year, find the probability that the first exit occurs between year 1 and year 3 and is to the destination with intensity $0.09$. (Answer: $0.1003$.)

Quantity	Expression	Interpretation
Remain in state \(0\)	\(e^{-\lambda t}\)	No exit has occurred by time \(t\)
Any exit by time \(t\)	\(1-e^{-\lambda t}\)	The process is in one of the absorbing states
Reach state \(1\) by time \(t\)	\((\alpha/\lambda)(1-e^{-\lambda t})\)	An exit occurs by \(t\) and its destination is state \(1\)
Reach state \(2\) by time \(t\)	\((\beta/\lambda)(1-e^{-\lambda t})\)	An exit occurs by \(t\) and its destination is state \(2\)

What You'll Learn