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A total-variation rate bounds strong-mixing coefficients

Statement

Assume AC. Let (Xk)k≥0 be a stationary time-homogeneous Markov chain with invariant probability π. Suppose there are a measurable M:X→[0,∞) with π(M)<∞ and r∈(0,1) such that ∥Pn(x,⋅)−π∥TV≤M(x)rnfor every x∈X, n≥1. Then the strong-mixing coefficient of the stationary process satisfies αX(n)≤π(M)rn(n≥1).

Facts & Assumptions

Given: AC, a stationary homogeneous Markov chain with kernel P and invariant law π, an integrable envelope M, and the displayed geometric total-variation bound.

[A1]

AC is the choice principle for the conditional classes used by the Markov-chain interfaces. (The Axiom of Choice)

[F1]

The strong-mixing coefficient is the supremum of ∣P(A∩B)−P(A)P(B)∣ over past and separated future events. (Strong, uniform, and maximal-correlation mixing coefficients)

[F2]

For every bounded product-measurable future-path functional H, the conditional expectation of its value on (Xn,Xn+1,…) is a measurable function of Xn. (Markov property for bounded future path functionals)

[F3]

For bounded measurable h, E[h(Xj+n)∣Fj]=Pnh(Xj). (Chapman-Kolmogorov equations)

[F4]

Total variation is ∥μ−ν∥TV=sup⁡A∣μ(A)−ν(A)∣. (Total variation distance for probability laws)

[F5]

A known measurable factor can be taken out of conditional expectation. (Taking out what is known)

Proof

1.1A1F2F3given

Fix n≥1, j≥0, A∈Fj, and a future event B∈σ(Xj+n,Xj+n+1,…). By [F2], for the path indicator defining B there is a measurable h:X→[0,1] with E[1B∣Fj+n]=h(Xj+n). Applying [F3] and the tower property gives E[1B∣Fj]=Pnh(Xj). Stationarity gives P(B)=π(h).

1.2F4construct

For probability measures μ,ν and measurable 0≤h≤1, the dyadic simple functions hm=2−m∑j=12m1{h≥j2−m} differ from h by at most 2−m. Their integrals differ by at most ∥μ−ν∥TV, since each summand is an indicator and [F4] bounds its measure difference. Letting m→∞ proves ∣μ(h)−ν(h)∣≤∥μ−ν∥TV.

2.1step 1.2given

Apply step 1.2 with μ=Pn(x,⋅) and ν=π. The given rate then yields ∣Pnh(x)−π(h)∣≤M(x)rn for every x.

3.1A1F1F2F3F5step 2.1given∎

Using [F5], [F2], and [F3], ∣P(A∩B)−P(A)P(B)∣=∣E[1A(Pnh(Xj)−π(h))]∣≤E∣Pnh(Xj)−π(h)∣≤π(M)rn. The first inequality uses 0≤1A≤1; the last follows from the preceding pointwise bound and stationarity of Xj. Since A,B,j were arbitrary, taking the supremum in [F1] proves the claim.

Source notes

The proof is local: the Markov property converts separated future events to a bounded function of the state, and the total-variation bound controls its expectation. The source's factor convention is matched to the repository definition of total variation as a supremum over events.

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