Transient Survival in an Absorbing Two-State Chain

A One-Way Chain with an Absorbing State

In many models, a system remains in one condition until a single event occurs, after which it enters a state from which it cannot leave. Examples include a component that fails, a patient who reaches an absorbing endpoint, or a molecule that undergoes an irreversible reaction. The transient question is how likely the system is still in its initial state at time $t$; the closely related question is when absorption occurs.

Let $(X_t)_{t\ge0}$ be a continuous-time Markov chain on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, with state space $S=\{1,2\}$. State $1$ is transient and state $2$ is absorbing. The only possible jump is $1\to2$, at rate $\alpha$. Unless stated otherwise, assume $\alpha>0$, and let the initial state be $X_0=1$. We use the natural filtration $\mathcal{F}_t=\sigma(X_s:0\le s\le t)$. Define the absorption time by $T=\inf\{t\ge0:X_t=2\}$.

Definition: The generator of this chain is $$ Q=\begin{bmatrix}-\alpha&\alpha\\0&0\end{bmatrix}. $$ Its rows sum to zero. The rate $\alpha$ is the exit rate from state $1$; the zero second row records that state $2$ has exit rate zero and is absorbing.

Starting from state $1$, the holding time before the jump to state $2$ has exponential distribution with rate $\alpha$. Thus, for $t\ge0$, the survival probability is $\mathbb{P}_1(T>t)$. Since the only states are $1$ and $2$, and state $2$ cannot be left, the events $\{T>t\}$ and $\{X_t=1\}$ agree. This observation links the absorption-time law directly to the first row of the transition matrix.

Transition Probabilities and Survival

Write $P(t)=(p_{ij}(t))_{i,j\in S}$, where $p_{ij}(t)=\mathbb{P}_i(X_t=j)$. The forward Kolmogorov equation is $P'(t)=P(t)Q$, with $P(0)=I$. In this finite-state setting, the rate equation describes how probability flows: probability leaves state $1$ at rate $\alpha$, enters state $2$ at the same rate, and no probability leaves state $2$.

Theorem: For $\alpha>0$ and $t\ge0$, the transition matrix, survival probability from state $1$, and absorption-time distribution are $$ P(t)= \begin{bmatrix} e^{-\alpha t}&1-e^{-\alpha t}\\ 0&1 \end{bmatrix},\qquad \mathbb{P}_1(T>t)=e^{-\alpha t},\qquad \mathbb{P}_1(T\le t)=1-e^{-\alpha t}. $$ The absorption time has density $f_T(t)=\alpha e^{-\alpha t}$ for $t>0$, with no atom at $0$.

Proof. Let $u(t)=p_{11}(t)$. From $P'(t)=P(t)Q$, the first column gives $u'(t)=-\alpha u(t)$, and the initial condition is $u(0)=1$. Multiplying the differential equation by $e^{\alpha t}$ gives $\frac{d}{dt}(e^{\alpha t}u(t))=0$; hence $e^{\alpha t}u(t)=1$, so $u(t)=e^{-\alpha t}$. The other entry in the first row satisfies $p'_{12}(t)=\alpha p_{11}(t)$ and $p_{12}(0)=0$. Integrating gives $p_{12}(t)=\int_0^t\alpha e^{-\alpha s}\,ds=1-e^{-\alpha t}$. Starting in state $2$, the chain remains there, so the second row is $(0,1)$. Each row therefore sums to $1$, and all entries are nonnegative for $t\ge0$. Since $T>t$ exactly when the chain is still in state $1$, its survival probability is $p_{11}(t)$; taking the complement gives its distribution function. Differentiating $1-e^{-\alpha t}$ for $t>0$ gives the stated density. At $t=0$, the distribution function is $0$, so there is no atom at $0$. $\square$

This theorem distinguishes two probabilities that are easy to confuse: $e^{-\alpha t}$ is the probability of not yet being absorbed at time $t$, while $1-e^{-\alpha t}$ is the probability of having been absorbed by time $t$. The transition matrix includes both quantities in the first row. In row-vector notation, if the initial distribution is $\mu=(\mu_1,\mu_2)$, then the distribution at time $t$ is $\mu P(t)$.

Worked Example: Survival with Rate Two-Thirds

Suppose $\alpha=\tfrac23$, the chain starts in state $1$, and the time horizon is $t=\tfrac32$. First check the generator: its rows are $(-\tfrac23,\tfrac23)$ and $(0,0)$, whose sums are $0$. The rate is positive, so the theorem applies.

Substituting the specified rate and time gives $\alpha t=\tfrac23\cdot\tfrac32=1$. Therefore $$ P\left(\tfrac32\right)= \begin{bmatrix} e^{-1}&1-e^{-1}\\ 0&1 \end{bmatrix}. $$ The first row sums to $e^{-1}+1-e^{-1}=1$, and the second row sums to $0+1=1$.

The probability that absorption has not occurred by time $\tfrac32$ is $e^{-1}$. The probability of absorption by that time is $1-e^{-1}$. The density of the absorption time at $t=\tfrac32$ is $\tfrac23 e^{-1}$, because $f_T(t)=\alpha e^{-\alpha t}$. As a check, the distribution function is $1-e^{-2t/3}$; at $t=\tfrac32$, this is $1-e^{-(2/3)(3/2)}=1-e^{-1}$, matching the matrix entry $p_{12}(\tfrac32)$.

Reading the Absorption-Time Law

The distribution function and density answer different questions. For a specified time $t\ge0$, the distribution function gives the probability of absorption by that time. The density describes how probability is distributed over positive time intervals; for $0\le a<b$, the probability of absorption during $(a,b]$ is the integral of the density over that interval. In this model it equals $$ \mathbb{P}_1(a<T\le b)=\int_a^b\alpha e^{-\alpha s}\,ds=e^{-\alpha a}-e^{-\alpha b}. $$

The law also has a constant conditional failure rate. For $u,v\ge0$, the event $T>u$ has positive probability because $\alpha>0$. Using the survival formula, $$ \mathbb{P}_1(T>u+v\mid T>u) =\frac{\mathbb{P}_1(T>u+v)}{\mathbb{P}_1(T>u)} =\frac{e^{-\alpha(u+v)}}{e^{-\alpha u}} =e^{-\alpha v}. $$ Thus, conditional on still being in state $1$ at time $u$, the additional waiting time has the same survival function as the original holding time. This is the memoryless property of the exponential holding time, not a claim that the absorption time is independent of the chain.

Worked Example: Conditional Survival over a Further Interval

Let $\alpha=\tfrac32$, start in state $1$, and ask for the probability that the chain survives until time $\tfrac43$, given that it has survived until time $\tfrac23$. The generator rows are $(-\tfrac32,\tfrac32)$ and $(0,0)$, and each sums to zero. The conditioning event has positive probability, since $\mathbb{P}_1(T>\tfrac23)=e^{-1}>0$.

Apply the conditional survival formula directly: $$ \mathbb{P}_1\left(T>\tfrac43\mid T>\tfrac23\right) =\frac{e^{-(3/2)(4/3)}}{e^{-(3/2)(2/3)}} =\frac{e^{-2}}{e^{-1}} =e^{-1}. $$ Equivalently, the additional interval has length $\tfrac43-\tfrac23=\tfrac23$, so its survival probability is $e^{-(3/2)(2/3)}=e^{-1}$. The two calculations agree, checking the conditional result.

Arbitrary Initial Distributions and the Zero-Rate Case

If the initial distribution is $\mu=(\mu_1,\mu_2)$, where $\mu_1,\mu_2\ge0$ and $\mu_1+\mu_2=1$, then row-vector multiplication gives $$ \mu P(t)=\left(\mu_1e^{-\alpha t},\ \mu_2+\mu_1(1-e^{-\alpha t})\right). $$ The probability of still occupying the transient state at time $t$ is $\mu_1e^{-\alpha t}$. The absorption time $T=\inf\{t\ge0:X_t=2\}$ equals $0$ if the chain starts in state $2$. For $t\ge0$, its distribution function is therefore $$ \mathbb{P}(T\le t)=\mu_2+\mu_1(1-e^{-\alpha t})=1-\mu_1e^{-\alpha t}. $$ If $\mu_2>0$, this distribution has an atom of size $\mu_2$ at zero; for $t>0$, its density is $\mu_1\alpha e^{-\alpha t}$. This density integrates to $\mu_1$, the probability of starting in state $1$, while the remaining probability $\mu_2$ is already absorbed at time zero.

Worked Example: A Mixed Initial Distribution

Take $\alpha=2$, initial distribution $\mu=(\tfrac13,\tfrac23)$, and time $t=\tfrac12$. The generator has rows $(-2,2)$ and $(0,0)$, each summing to zero. Substitution gives $\alpha t=2\cdot\tfrac12=1$, so $$ P\left(\tfrac12\right)= \begin{bmatrix} e^{-1}&1-e^{-1}\\ 0&1 \end{bmatrix}. $$ Both rows sum to $1$.

Multiplying the row vector by this matrix gives $$ \mu P\left(\tfrac12\right) =\left(\tfrac13e^{-1},\ \tfrac13(1-e^{-1})+\tfrac23\right) =\left(\tfrac13e^{-1},\ 1-\tfrac13e^{-1}\right). $$ The entries sum to $\tfrac13e^{-1}+1-\tfrac13e^{-1}=1$. Thus the probability of being transient at time $\tfrac12$ is $\tfrac13e^{-1}$, not $e^{-1}$, because the chain starts in state $1$ only with probability $\tfrac13$.

For this initial distribution, $T=0$ with probability $\tfrac23$. For $t\ge0$, the distribution function is $1-\tfrac13e^{-2t}$, and for $t>0$, the density is $\tfrac23e^{-2t}$. At $t=\tfrac12$, the distribution function is $1-\tfrac13e^{-2(1/2)}=1-\tfrac13e^{-1}$, which matches the probability of occupying the absorbing state in the matrix calculation.

The positive-rate assumption matters. If $\alpha=0$, the generator is the zero matrix and both states are absorbing. Starting from state $1$, the chain never reaches state $2$: $P(t)=I$, $\mathbb{P}_1(T>t)=1$ for every finite $t$, and $\mathbb{P}_1(T\le t)=0$. In this degenerate case there is no finite absorption-time density, and the formula for positive $\alpha$ cannot be used to claim eventual absorption. The matrix formula itself has a continuous limit as $\alpha$ decreases to zero, but that limit describes no jump, not an exponential waiting time with a positive rate.

Common Calculation Pitfall

A frequent error is to put the survival probability in the probability of being absorbed. For example, with $\alpha=1$ and $t=1$, writing $p_{12}(1)=e^{-1}$ and $p_{11}(1)=1-e^{-1}$ reverses the two events. The corrected entries are $p_{11}(1)=e^{-1}$ for still being transient and $p_{12}(1)=1-e^{-1}$ for having been absorbed. The row-sum check catches the reversal only if the entries are not also arranged to sum to one; the event interpretation is essential as well. More generally, use the facts $p_{11}(t)=\mathbb{P}_1(T>t)$ and $p_{12}(t)=\mathbb{P}_1(T\le t)$, then check that the row entries sum to $1$.

The two-state model is useful because it isolates one rate and one absorbing endpoint. Its formulas provide a direct way to translate a hazard rate into survival probabilities, transition probabilities, and an absorption-time law. When a model has additional sequential states or competing transitions, the same quantities must be obtained from a larger system of rate equations; the single exponential formula here applies specifically because the only possible exit from the transient state leads directly to absorption.

Check Your Understanding

Assume a chain with state space $\{1,2\}$, state $2$ absorbing, and rate $\alpha>0$ from $1$ to $2$, unless a question specifies otherwise.

Write the generator and transition matrix at time $t$, and verify that both transition-matrix rows sum to $1$.
For a chain starting at state $1$, express the probability of absorption during $(a,b]$ in terms of $\alpha$, where $0\le a<b$.
If $\alpha=\tfrac45$, calculate the survival probability at $t=\tfrac54$ and the absorption probability by that time.
For initial distribution $\mu=(\tfrac25,\tfrac35)$, identify the probability mass of $T$ at zero and write its distribution function for $t\ge0$.
Explain what happens to the absorption-time distribution when $\alpha=0$ and the chain starts in state $1$.