Edge Flows and Currents in Finite Chains

From Transition Probabilities to Edge Flows

A transition probability describes the next move conditional on the chain’s current state. In a stationary chain, it is also useful to ask how much probability mass travels along each directed transition per time step. This quantity is an edge flow: it combines the probability of being at the starting state with the probability of making the move. Comparing the flow in opposite directions separates a chain’s overall movement from its net directional current.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with transition matrix \(P=(p_{ij})\). The matrix is row-stochastic, so \(\sum_{j\in S}p_{ij}=1\) for each \(i\). Suppose the initial distribution is the row vector \(\pi\), and \(\pi P=\pi\); then the chain is stationary. Its filtration may be taken as \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\). In this tutorial, \(\pi\) is a stationary distribution, not necessarily a unique one.

The flow \(q(i,j)\) is the probability that the chain is at \(i\) at time \(n\) and at \(j\) at time \(n+1\), when \(X_n\) has distribution \(\pi\). Indeed, that joint probability is \(\mathbb{P}(X_n=i,X_{n+1}=j)=\pi_i p_{ij}\). The flow includes self-transitions: \(q(i,i)=\pi_i p_{ii}\). For any pair of distinct states, \(J(i,j)>0\) means more stationary probability mass moves from \(i\) to \(j\) than from \(j\) to \(i\). It does not mean that the transition probability \(p_{ij}\) alone exceeds \(p_{ji}\).

Conservation of Stationary Flow

Stationarity makes the total flow leaving a state match the total flow entering it. This is a local conservation statement: the mass at a state does not build up or drain away from one step to the next.

Proof. By the definition of \(q\) and the row sums of \(P\), the total outgoing flow from \(i\) is \(\sum_j q(i,j)=\sum_j\pi_i p_{ij}=\pi_i\sum_jp_{ij}=\pi_i\). The total incoming flow is \(\sum_j q(j,i)=\sum_j\pi_jp_{ji}=(\pi P)_i=\pi_i\), using stationarity. Subtracting the incoming total from the outgoing total gives \(\sum_jJ(i,j)=\sum_jq(i,j)-\sum_jq(j,i)=\pi_i-\pi_i=0\). The term \(J(i,i)\) is zero, so including self-transitions does not change this conclusion. \(\square\)

There is also a useful converse for constructing chains from flows. The positivity condition below ensures that every row of the transition matrix can be recovered by dividing by its state’s stationary mass.

Proof. Since \(\pi_i>0\), each ratio \(p_{ij}=q(i,j)/\pi_i\) is defined and nonnegative. For each \(i\), \(\sum_jp_{ij}=(\sum_jq(i,j))/\pi_i=\pi_i/\pi_i=1\), so \(P\) is row-stochastic. For each \(j\), \((\pi P)_j=\sum_i\pi_i p_{ij}=\sum_iq(i,j)=\pi_j\), where the last equality uses the column sum assumption. Hence \(\pi P=\pi\). \(\square\)

If a row sum is zero, this construction cannot define the transition probabilities out of that state by division. The theorem therefore assumes strictly positive \(\pi_i\) on all of \(S\). A balanced flow is not by itself a transition matrix; its row sums must first be used to normalize each row.

Detailed Balance and Symmetric Flows

The edge-flow formulation makes the connection to reversibility exact. Detailed balance requires that each directed transition be matched by an equal flow in the reverse direction. Equivalently, no pair of states has a net current.

Proof. By definition, detailed balance means \(\pi_i p_{ij}=\pi_jp_{ji}\) for all \(i,j\). Since \(q(i,j)=\pi_i p_{ij}\), this is exactly the equality \(q(i,j)=q(j,i)\). By the definition \(J(i,j)=q(i,j)-q(j,i)\), that equality holds exactly when \(J(i,j)=0\). This proves both implications, including \(i=j\), for which the current is always zero. \(\square\)

This equivalence concerns flows under a specified stationary distribution. A transition matrix can have unequal probabilities in opposite directions and still satisfy detailed balance if the stationary masses at the two states differ appropriately. Conversely, stationarity alone only guarantees the conservation of total flow at each state; it does not force each pair of opposing flows to match.

Building a Chain from Its Flows

The converse theorem gives a direct way to work from stationary flows to transition probabilities. It is especially useful when the desired stationary behavior is easier to specify as a balanced collection of flows. The next example performs the normalization explicitly and checks the resulting chain in both directions.

Common Pitfall: Comparing Probabilities Instead of Flows

A common error is to call \(p_{ij}-p_{ji}\) the net current. That difference ignores how often the chain is at each state. In the reversible example above, this incorrect calculation gives \(p_{12}-p_{21}=\tfrac12-\tfrac14=\tfrac14\), which might suggest a current from 1 to 2. But the stationary flows are \(q(1,2)=\tfrac14\cdot\tfrac12=\tfrac18\) and \(q(2,1)=\tfrac12\cdot\tfrac14=\tfrac18\). The correct net current is therefore \(J(1,2)=\tfrac18-\tfrac18=0\). Use transition-probability differences only when comparing the probabilities themselves; to measure stationary movement, multiply by the relevant stationary mass first.

Flows and currents serve different purposes. The nonnegative \(q(i,j)\) records stationary probability mass moving along a directed transition. The signed \(J(i,j)\) records the imbalance between opposite directions. Stationarity forces the total current out of each state to equal the total current in, but it does not require every pairwise current to vanish. Pairwise vanishing is precisely the additional symmetry required by detailed balance.