advanced35 min studyLesson 805 in Mathematics

Ergodicity, weak mixing, and mixing

Distinguish invariant events, Cesàro decorrelation, and pointwise decorrelation using exact stationary correlations.

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What you’ll understand

  • State the different correlation quantifiers for ergodicity, weak mixing, and strong mixing.
  • Explain the implication hierarchy without identifying periodic ergodicity with decorrelation.
  • Calculate a stationary two-state correlation from conditional transition probabilities.

The word mixing is stronger than having correct long-run averages. Ergodicity eliminates nontrivial invariant events. Weak mixing removes persistent correlations on average in time, while strong mixing requires correlations to vanish at each sufficiently large time. The difference is in the quantifiers and the location of the absolute value. An exact stationary Markov example shows how transition relaxation produces correlations, and a periodic example shows that ergodicity alone can coexist with permanent oscillation.

Ergodicity removes invariant components

On a probability-preserving system, ergodicity means that every event invariant modulo null sets has probability zero or one. An equivalent correlation statement says that, for every pair of measurable events, the Cesàro average of their overlap probabilities converges to the product of their probabilities. The deviations can have different signs and cancel in this average. This criterion follows by applying the mean ergodic theorem to an event indicator and pairing with the other indicator. It does not require the individual overlap probabilities to converge.

A single finite cycle with uniform probability is ergodic: an invariant subset must be a union of full cycles, leaving only the empty set or the whole space. Nevertheless any two singleton events have periodic overlap probabilities. This example is already enough to distinguish exploration in time averages from loss of temporal memory. Ergodicity is always relative to the chosen invariant measure; assigning positive mass to two disjoint cycles creates a nontrivial invariant event and destroys whole-space ergodicity.

cn(A,B)=μ(A∩T−nB)−μ(A)μ(B),T ergodic  ⟺  1N∑n=0N−1cn(A,B)→0for all A,Bc_n(A,B)=\mu(A\cap T^{-n}B)-\mu(A)\mu(B),\qquad T\text{ ergodic}\iff\frac1N\sum_{n=0}^{N-1}c_n(A,B)\to0\quad\text{for all }A,B
The average here contains signed deviations. The absolute-value version below is a strictly stronger requirement, so cancellation must not be silently removed.↔ Scroll wide equations horizontally.

Weak mixing averages absolute deviations

Weak mixing requires the Cesàro mean of the absolute correlation deviations to tend to zero for every pair of measurable events. Strong mixing requires the deviation itself to tend to zero for every pair. Boundedness of correlations gives the implication from strong mixing to weak mixing, and the ordinary triangle inequality gives the implication from weak mixing to ergodicity. Neither name should be inferred just from the existence of an invariant measure or the convergence of one selected observable's time average.

These notions have equivalent bounded-observable formulations obtained by approximating measurable functions by simple functions. For invertible probability systems, weak mixing is also equivalent to ergodicity of the product transformation on two independent copies, and to absence of nonconstant L2 eigenfunctions. These are useful structural tests, but they do not make weak mixing identical to strong mixing. Examples of weakly mixing systems that are not strongly mixing exist; constructing them takes more work than the simple periodic and independent examples used here.

weak mixing: 1N∑n=0N−1∣cn(A,B)∣→0;strong mixing: cn(A,B)→0for every measurable A,B\text{weak mixing}:\ \frac1N\sum_{n=0}^{N-1}|c_n(A,B)|\to0;\qquad \text{strong mixing}:\ c_n(A,B)\to0\qquad\text{for every measurable }A,B
The pair A,B is arbitrary in both definitions. Strong mixing implies weak mixing, which implies ergodicity; convergence for a single event pair proves none of these universal properties by itself.↔ Scroll wide equations horizontally.

Stationarity is needed when interpreting chain correlations

A stationary Markov chain on a two-sided sequence space defines an invertible probability-preserving shift. Its event correlations can be calculated from a transition matrix, but the stationary initial law must be included. A transition instrument starting from state zero gives conditional transition probabilities instead of a stationary joint law. Multiplying its probability of returning to zero by the stationary mass of zero produces the two-time stationary joint probability. Subtracting the squared stationary mass then gives the indicator correlation.

For an irreducible aperiodic finite chain, transition probabilities converge to the stationary law, and the corresponding stationary shift is strongly mixing. First verify separated cylinder events using transition convergence, then approximate general events by cylinders. A periodic irreducible chain can be stationary and its shift ergodic while its correlations oscillate. Thus the period obstruction is not a failure of stationarity or of long-run component exploration; it specifically obstructs the decorrelation limit required for mixing.

Compute a stationary two-state covariance

Let transitions 0→10\to1 and 1→01\to0 have probabilities p=1/4p=1/4 and q=1/2q=1/2. For the stationary chain shift, take A=B={X0=0}A=B=\{X_0=0\}. Find its time-n correlation and its value at n equal to two.

  1. The stationary probability of zero is π0=q/(p+q)=2/3\pi_0=q/(p+q)=2/3. The other transition eigenvalue is 1−p−q=1/41-p-q=1/4, so Pn(0,0)=2/3+(1/3)4−nP^n(0,0)=2/3+(1/3)4^{-n}.

  2. The stationary joint probability is π0Pn(0,0)\pi_0P^n(0,0). Subtracting π02\pi_0^2 gives cn(A,A)=(2/9)4−nc_n(A,A)=(2/9)4^{-n}, which tends to zero.

  3. At n equal to two, the conditional return probability is 2/3+1/48=11/162/3+1/48=11/16, the joint probability is 11/2411/24, and the correlation is 11/24−4/9=1/7211/24-4/9=1/72. These conditional, joint, and centered quantities are different numbers.

The result

The exact decaying correlation is consistent with strong mixing of this stationary finite-state shift. One computed event pair illustrates the general finite-chain theorem rather than proving its universal quantifier alone.

Explore the model

Interactive lab

Interactive lab · markov

Conditional relaxation versus stationary correlation

Pause and Reset, set both p=0.25,q=0.5p=0.25,q=0.5, and use two Steps to reach time one, hence iteration two. Predict state-zero probability 11/1611/16 from the instrument's fixed initial state zero; multiply by 2/32/3 on paper for the stationary joint law. Reset and set both p=1,q=1p=1,q=1, predicting alternating states with no convergence. The instrument gives exact distributions, not a sampled stationary trajectory. Control changes retain time; Reset restores both parameters and clock.

Conditional relaxation versus stationary correlationExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA conditional transition probability becomes a stationary correlation only after stationary weighting and centering.

Check your understanding

Which condition expresses weak mixing for every measurable event pair?

Try it yourself

Consider the stationary two-state chain with p=q=1p=q=1 and stationary probabilities one half each. For A={X0=0}A=\{X_0=0\}, compute cn(A,A)c_n(A,A) and its absolute Cesàro mean. Is the associated two-point shift weakly mixing?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

At even times the state returns to zero with certainty given a zero start; at odd times it never does.

Compare with the solution

The stationary joint probability is one half at even times and zero at odd times. Subtracting μ(A)2=1/4\mu(A)^2=1/4 gives deviations alternating 1/41/4 and −1/4-1/4. Their signed Cesàro mean tends to zero, but their absolute mean is always one quarter, so weak mixing fails. The two alternating sequences form a single two-cycle under the shift, making this stationary shift ergodic nevertheless.

Take it with you

  • Ergodicity controls invariant events and signed correlation averages.
  • Weak mixing averages absolute deviations; strong mixing requires pointwise decay.
  • Stationary joint laws differ from transition laws with a fixed starting state.

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