Stationary Laws of Periodic Chains

Stationarity Does Not Require Convergence

A chain can have a stationary distribution even when its distribution at time \(n\) never settles down. A particle moving deterministically around a cycle is the simplest example: its location keeps changing, but a uniform initial location remains uniform at every time. This distinction matters whenever a chain is periodic. The stationary equations describe distributions unchanged by one transition; they do not, by themselves, say that a chain started from a particular state approaches such a distribution.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Its transition matrix \(P=(p_{ij})\) is row-stochastic: \(p_{ij}=\mathbb{P}(X_{n+1}=j\mid X_n=i)\), and \(\sum_{j\in S}p_{ij}=1\) for each \(i\). Distributions are row vectors, so if the law of \(X_0\) is \(\mu_0\), then the law at time \(n\) is \(\mu_n=\mu_0P^n\).

For a finite chain, existence of a stationary distribution is a standard finite-state result established in the earlier material on existence of stationary laws. Irreducibility is not needed for existence on a finite state space, but it is useful here because it gives a single cyclic structure and, as shown below, a unique stationary distribution. Periodicity does not prevent stationarity.

Deterministic Cycles

On a deterministic cycle, each state has exactly one successor, and the last state returns to the first. Starting at a fixed state makes the distribution rotate forever. In contrast, the uniform distribution assigns the same mass to every point of the cycle, so one transition merely permutes its entries.

The example illustrates the distinction precisely. If \(X_0\) has law \(\pi\), then the law of \(X_1\) is \(\pi P=\pi\); repeating this argument gives \(\pi P^n=\pi\) for every \(n\ge0\). A stationary law is therefore a choice of initial distribution that remains unchanged. It is not a claim about the long-run behavior from every initial state.

Cyclic Classes and Equal Stationary Mass

In an irreducible chain of period \(d\), the states can be divided into \(d\) nonempty cyclic classes \(C_0,\ldots,C_{d-1}\), with indices read modulo \(d\), so that every transition from a state in \(C_r\) goes to a state in \(C_{r+1}\). This is the standard cyclic decomposition theorem from periodicity theory. The classes explain the time pattern: if \(X_0\in C_r\), then \(X_n\in C_{r+n\ {\rm mod}\ d}\) with probability \(1\).

Proof. Write \(\pi(C_r)=\sum_{i\in C_r}\pi_i\). Stationarity says that the mass in \(C_{r+1}\) after one step is the mass that was in \(C_r\) before the step: transitions from \(C_r\) all enter \(C_{r+1}\), and transitions from any other class cannot enter \(C_{r+1}\), because transitions go only to the next class. Hence \(\pi(C_{r+1})=\pi(C_r)\) for every \(r\). The \(d\) class masses are equal, and their sum is \(\sum_{i\in S}\pi_i=1\), so each is \(1/d\). For the second assertion, start in \(C_r\) and consider the event \(X_n\in C_r\). Its probability is \(1\) when \(n\) is divisible by \(d\), and \(0\) otherwise. For \(d>1\), these probabilities do not converge. On a finite state space, convergence of the distributions would imply convergence of the probability of every subset, including \(C_r\), so the time-specific distributions cannot converge. \(\square\)

For a finite irreducible chain, the stationary distribution is unique. The finite-state existence result supplies at least one stationary distribution; the following argument proves uniqueness and also verifies that its coordinates are positive.

Proof. Let \(\pi\) be a stationary distribution. It has at least one positive coordinate because its coordinates sum to \(1\). If \(\pi_j=0\), stationarity gives \(0=\pi_j=\sum_i\pi_i p_{ij}\). All terms are nonnegative, so \(p_{ij}=0\) for every \(i\) with \(\pi_i>0\). Repeating this argument along paths shows that no state reachable from a positive-mass state can have positive incoming transition probability from the preceding state. Irreducibility supplies a path from a positive-mass state to \(j\), a contradiction. Thus \(\pi_j>0\) for every \(j\).

Now suppose \(\nu\) is another stationary distribution. Define \(r_i=\nu_i/\pi_i\), which is well-defined because \(\pi_i>0\). For each \(j\), stationarity of \(\nu\) and \(\pi\) gives \[ r_j=\sum_i \frac{\pi_i p_{ij}}{\pi_j}r_i. \] The coefficients \(\pi_i p_{ij}/\pi_j\) are nonnegative and sum to \(1\), since \(\sum_i\pi_i p_{ij}=\pi_j\). Choose \(j\) where \(r_j\) is maximal. A weighted average of values no larger than \(r_j\) equals \(r_j\), so every \(i\) with \(p_{ij}>0\) has \(r_i=r_j\). Applying the same argument backwards along paths ending at \(j\), irreducibility implies \(r_i=r_j\) for every state \(i\). Thus \(\nu=c\pi\) for a constant \(c\). Since both distributions sum to \(1\), \(c=1\) and \(\nu=\pi\). \(\square\)

Computing Stationary Laws in Periodic Chains

The stationary equations are ordinary balance equations: the mass assigned to a state must equal the total mass flowing into it in one step. Periodicity affects how the chain moves between classes, but it does not change how these equations are solved. The next examples show the equations and the cyclic structure together.

A Common Mistake: Confusing Invariance with a Limit

A common error is to average the states visited over a cycle and call that average the time-\(n\) distribution. For the four-state deterministic cycle started at state \(1\), the chain visits each state once in every four steps, so its long-run fraction of visits to each state is \(1/4\). But at time \(n=4k\), its distribution is the point mass at state \(1\), not the uniform distribution. The visit-frequency calculation describes an average across times, not the distribution at one specified time. In particular, it does not show convergence of \(\mu_0P^n\).

When computing a stationary law, check the balance equation \(\pi P=\pi\) and normalization \(\sum_i\pi_i=1\). When interpreting the result, separately check whether the chain is periodic. If it is irreducible with period \(d>1\), its cyclic classes force time-specific oscillation from any fixed starting state, even though the unique stationary law exists and assigns mass \(1/d\) to each class.