Alphabeta Math
RemarkRemark: Literature-sourcedProof: Not applicablePipeline-generatedaudited 2026-10-02
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Aperiodicity separates ordinary convergence from ergodic averages

Remark

The three convergence results of this page do not assume the same hypotheses, and the difference is exactly aperiodicity. This remark records the comparison; it proves nothing new and asserts no convergence statement beyond the three results it compares. The AC assumptions of those results are retained.

Why aperiodicity cannot simply be dropped

The ordinary-time conclusion of the first result is genuinely false without its aperiodicity hypothesis, and the companion page computes two obstructions with no aperiodicity and no randomness:

  • the deterministic two-state alternation with P(0,1)=P(1,0)=1 is irreducible on a finite state space, hence positive recurrent, with π=(1/2,1/2); from state 0 its law is δ0 at even times and δ1 at odd times, so ∥p(n)(0,⋅)−π∥TV=1/2 for every n and the sequence of laws does not converge at all;
  • the deterministic directed three-cycle is irreducible with uniform π=(1/3,1/3,1/3) and ∥p(n)(0,⋅)−π∥TV=2/3 for every n.

In both cases the failure has the same cause: the positive return times of a state are contained in a proper arithmetic progression dN with d≥2, so the n-step law keeps cycling through the residue classes of n modulo d instead of settling. The two ergodic-average results are unaffected, because averaging over all 0≤k<n samples every residue class with asymptotic frequency 1/d; for the two examples just named the Cesàro averages equal π for every n divisible by the period and converge to π in general, and the empirical frequencies of the visited states converge to 1/d, which is the stationary mass of each state.

Aperiodicity is needed for the ordinary-time convergence theorem from every deterministic start; it is not needed for the Cesàro or almost-sure ergodic averages. A stationary start is a different assertion: if the initial law is π, then L(Xn)=π, equivalently πp(n)=π, for every n≥0, even for a periodic chain (Invariant initial law makes a Markov chain stationary). This marginal identity does not assert p(n)(x,⋅)=π for a deterministic start x.

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Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources