Rate Balance Across State-Space Cuts

Stationary Flow Across a Cut

In a continuous-time Markov chain, probability moves between states at rates. For a set of states \(A\), transitions within \(A\) do not change the total probability assigned to \(A\); only transitions crossing between \(A\) and its complement can change it. Under stationarity, the total rate of probability flow into \(A\) must therefore equal the total rate flowing out. This set-level statement is called rate balance across a cut, and it provides a useful way to derive and check the equations for a stationary distribution.

Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Its generator is \(Q=(q_{ij})\): for distinct states \(i,j\), \(q_{ij}\ge0\) is the transition rate from \(i\) to \(j\), and \(q_{ii}=-\sum_{j\ne i}q_{ij}\), so every row sums to zero. Write \(P(t)=e^{tQ}\) for its transition matrix at time \(t\). A probability row vector \(\pi\) is stationary if, when \(X_0\) has distribution \(\pi\), \(X_t\) has distribution \(\pi\) for every \(t\ge0\). In finite state spaces all rates are bounded, and the matrix exponential gives the transition semigroup.

The factor \(\pi_i\) is essential: \(q_{ij}\) is a rate conditional on being in state \(i\), while \(\pi_iq_{ij}\) is the stationary rate of transitions from \(i\) to \(j\). A transition from one state of \(A\) to another state of \(A\) contributes to neither crossing flux.

Cut Balance and Global Balance

Proof. First suppose \(\pi\) is stationary. For any \(A\subseteq S\), let \(\mathbf{1}_A\) be the column vector whose \(i\)th entry is \(1\) when \(i\in A\) and \(0\) otherwise. The probability of being in \(A\) at time \(t\), when the initial distribution is \(\pi\), is \(\pi P(t)\mathbf{1}_A\). Stationarity makes this constant in \(t\), so its right derivative at \(0\) is zero. Since \(P(t)=e^{tQ}\), the derivative at \(0\) is \(\pi Q\mathbf{1}_A\). Expanding by rows gives \(\pi Q\mathbf{1}_A=\sum_i\pi_i\sum_{j\in A}q_{ij}\). For \(i\in A\), the inner sum is \(-\sum_{j\notin A}q_{ij}\), by the zero row sum of \(Q\). For \(i\notin A\), it is \(\sum_{j\in A}q_{ij}\). Thus the derivative equals incoming flux minus outgoing flux, which must be zero. This proves (i) implies (iii). Taking \(A=\{i\}\) in this identity gives \((\pi Q)_i=0\) for each \(i\), so (iii) implies (ii). Conversely, if \(\pi Q=0\), then \(\pi Q^k=0\) for every integer \(k\ge1\). The power series for the matrix exponential gives \(\pi P(t)=\pi e^{tQ}=\pi\), so \(\pi\) is stationary. This proves (ii) implies (i). Finally, if the singleton equations hold, they say precisely that each coordinate of \(\pi Q\) is zero, so (ii) holds and hence (iii) holds for every set. \(\square\)

The theorem makes cut balance a calculation tool as well as a check. Summing the singleton equations over states in \(A\) cancels all terms for transitions internal to \(A\), leaving exactly the incoming-equals-outgoing equation for the cut. Conversely, when a proposed distribution satisfies all singleton equations, it satisfies every cut equation. No irreducibility assumption is needed for these equivalences; if there is more than one stationary distribution, the equations may have more than one probability-vector solution.

Using Cuts to Solve for Stationary Probabilities

What Counts as Crossing the Cut?

A cut records only transitions whose starting and ending states lie on opposite sides of the chosen set. For the first example, the cut \(A=\{1,2\}\) has outgoing flux \(\pi_2\), because the rate from 2 to 3 is \(1\). Its incoming flux is \(3\pi_3\), because the rate from 3 to 1 is \(3\). The transition \(1\to2\) stays inside \(A\), so it contributes to neither side.

Common mistake. One might incorrectly write the incoming flux into \(\{1,2\}\) as \(2\pi_1+3\pi_3\), counting the rate \(1\to2\) as incoming. That transition is internal: both states belong to \(A\). The correct cut equation is \(\pi_2=3\pi_3\). The singleton equation for \(\{1\}\) is separately \(2\pi_1=3\pi_3\); together these imply \(\pi_2=2\pi_1\). The internal transition can appear in a singleton balance equation, but it cancels when balances are combined for a larger set.

For finite chains, cut balance is an exact reformulation of the stationary generator equations. It is especially convenient when a set has only a few crossing transitions, even if many transitions occur within the set. A useful calculation sequence is to list crossing transitions first, multiply each rate by the stationary probability of its source state, and only then equate the two fluxes. After solving, check normalization and substitute into the state balances; a cut calculation that omits a crossing rate or includes an internal transition will otherwise give a plausible-looking but incorrect equation.