Why Expected Absorption Times Equal N Times One

Counting Transient Visits

The fundamental matrix records how often a chain visits each transient state before absorption. To obtain an expected absorption time from it, we need to connect that count to elapsed steps. The key observation is pathwise: before the first visit to an absorbing state, the chain spends exactly one time index at a transient state for each step it takes before absorption. Summing the indicators of those visits therefore counts the absorption time itself.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Assume its states are arranged in canonical absorbing-chain form: \(S=T\cup A\), where \(T\) is the set of transient states and \(A\) is the set of absorbing states. Once the chain enters \(A\), it stays there. Write \(Q=(q_{ij})_{i,j\in T}\) for the transient-to-transient block of its row-stochastic transition matrix. Thus \(Q\) is substochastic: its row sums are at most \(1\). Let \(T_A=\inf\{n\ge0:X_n\in A\}\), with \(\inf\varnothing=\infty\). We use the fundamental matrix \(N=\sum_{n=0}^{\infty}Q^n\), and assume its entries are finite, as in the finite absorbing-chain setting where the fundamental matrix exists.

The definition counts a visit at time \(0\). This matters: if the chain starts at a transient state and is absorbed after \(m\) transitions, it occupies a transient state at times \(0,1,\ldots,m-1\), which is \(m\) visits. If it starts in \(A\), then \(T_A=0\) and it has no transient visits.

Two Visit-Counting Results

Proof. Fix \(i,j\in T\). By the definition of \(V_j\) and the fact that its summands are nonnegative indicators, the monotone convergence theorem gives \(\mathbb{E}_i[V_j]=\sum_{n=0}^{\infty}\mathbb{P}_i(X_n=j)\). Because states in \(A\) are absorbing, a path that is at \(j\in T\) at time \(n\) cannot have visited \(A\) earlier. Hence the probability of being at \(j\) at time \(n\), starting from \(i\), is the \((i,j)\) entry of \(Q^n\): the matrix product sums the probabilities of all length-\(n\) paths that remain in \(T\) and end at \(j\). Therefore \(\mathbb{E}_i[V_j]=\sum_{n=0}^{\infty}(Q^n)_{ij}=N_{ij}\). For an initial transient distribution \(\alpha\), the time-\(n\) probabilities on \(T\) are the row vector \(\alpha Q^n\). Summing its \(j\)th component over \(n\) gives \(\sum_{n\ge0}(\alpha Q^n)_j=(\alpha N)_j\). \(\square\)

Proof. First compare \(T_A\) and \(V\) on each path. If \(T_A=m<\infty\), then \(X_0,\ldots,X_{m-1}\) lie in \(T\), and \(X_m\) and all later states lie in \(A\). Thus exactly \(m\) terms in the sum defining \(V\) equal \(1\), so \(V=m=T_A\). If \(T_A=\infty\), the chain never enters \(A\), so every \(X_n\) lies in \(T\); both \(V\) and \(T_A\) are then infinite. Consequently \(T_A=V\) pathwise, including when the chain starts in \(A\), for which both are zero.

Since all indicators are nonnegative, monotone convergence justifies exchanging expectation and the infinite sum: \(\mathbb{E}[T_A]=\mathbb{E}[V]=\sum_{n=0}^{\infty}\mathbb{E}[\mathbf{1}_{\{X_n\in T\}}]=\sum_{n=0}^{\infty}\mathbb{P}(X_n\in T)\). For a fixed starting state \(i\in T\), the probability at time \(n\) of being in each transient state is row \(i\) of \(Q^n\). Summing over transient states and then over \(n\) gives \(\mathbb{E}_i[T_A]=\sum_{n\ge0}\sum_{j\in T}(Q^n)_{ij}=\sum_{j\in T}N_{ij}\). For initial transient row vector \(\alpha\), the corresponding expression is \(\sum_{n\ge0}\alpha Q^n\mathbf{1}=\alpha N\mathbf{1}\). Any probability initially assigned to \(A\) contributes zero because the chain is already absorbed at time \(0\). Since \(N\) has finite entries and \(T\) is finite, these expressions are finite. In particular, \(T_A\) is finite almost surely: a nonnegative extended random variable that is infinite with positive probability has infinite expectation. \(\square\)

The exchange of expectation and summation is not justified by informally rearranging an infinite series. Here it is valid because the summands are nonnegative; monotone convergence applies even before finiteness is known. Finiteness of \(N\) then gives the stronger conclusion that the expected absorption time is finite. No uniform upper bound on the number of steps is required.

Worked Applications

What the Formula Does—and Does Not—Say

The multiplication by \(\mathbf{1}\) adds the expected visits across all transient states. The starting distribution determines which row, or weighted combination of rows, is used. Thus \(N\) alone is not the expected absorption time for every initial condition: its row sums give the state-specific expected times, and \(\alpha N\mathbf{1}\) gives the average for initial transient distribution \(\alpha\).

A common mistake is to add every entry of \(N\) and call the result the expected time from state \(1\). In the two-transient-state example, that incorrect calculation is \(2+\tfrac34+0+\tfrac32=\tfrac{17}{4}\). The expected time from state \(1\) is instead its row sum, \(2+\tfrac34=\tfrac{11}{4}\). The total \(\tfrac{17}{4}\) adds the visit counts from both possible starting states; it does not correspond to either one starting condition.

The finiteness condition is also substantive. For example, suppose a chain on \(\{1,2\}\) has \(P=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and the target is \(A=\{2\}\). Each row sums to \(1\), but from state \(1\) the target is never reached: \(T_A=\infty\). If state \(1\) were incorrectly treated as a transient block, then \(Q=[1]\), and the partial sums \(\sum_{n=0}^{m}Q^n=[m+1]\) grow without bound. The fundamental matrix is not finite, and there is no finite expected absorption time from state \(1\). In a correctly classified absorbing chain, state \(1\) here is not transient. The example shows why a finite fundamental matrix cannot be assumed when some paths can avoid absorption forever.

For an absorbing chain with finite \(N\), the visit-count argument gives a direct interpretation of every row of \(N\): its entries are expected transient-state visit counts, and their sum is the expected number of transitions before absorption. This interpretation also explains why the time formula uses \(N\mathbf{1}\), rather than a sum over columns independent of the starting state.