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A uniformly ergodic adaptive refresh family

Example

Let (X,X)=({0,1},2{0,1}) with uniform target π, let (Γ,G)=([1/2,1],B([1/2,1])), Π(x,⋅)=π(⋅), and for γ∈Γ set Pγ=(1−γ)I+γΠ. Use the deterministic adaptation Γn=12+12(n+1). The adaptive chain satisfies diminishing adaptation and its family is simultaneously uniformly ergodic, so its marginal law converges to π.

Facts & Assumptions

Given: The two-state space, uniform probability π, kernels Pγ, and deterministic sequence Γn in the Example.

[F1]

Measure kernel and probability kernel: a probability kernel has probability rows and measurable event evaluations.

[F2]

General-state reversibility by symmetric flow measure: a kernel is reversible when its joint flow measure under π is symmetric.

[F3]

General-state reversibility implies invariance: reversibility implies πP=π.

[F4]

Total variation distance for probability laws: total variation is the supremum over measurable events.

[F5]

Diminishing adaptation: Dn is the uniform-in-state total variation change between successive kernels.

[F6]

Adaptive MCMC marginal convergence under diminishing adaptation with uniform ergodicity or containment: diminishing adaptation and simultaneous uniform ergodicity imply marginal convergence.

Verification

Given: [F1]–[F6] and the finite state space and target in the Example.

1.1F1F2F3givenalgebra

The matrix of Pγ is (1−γ/2γ/2γ/21−γ/2). Its entries are continuous in γ, so the family is jointly measurable on the stated spaces; its rows are probability vectors by 0≤γ≤1 and [F1]. The matrix is symmetric, and π assigns equal mass to both states, so the four joint-flow atoms are symmetric under coordinate swap. Thus Pγ is reversible by [F2], and π is invariant by [F3].

1.2F4F5givenalgebra

The kernels satisfy Pγ′(x,⋅)−Pγ(x,⋅)=(γ′−γ)(π−δx), so Dn=∣Γn+1−Γn∣/2→0. This is diminishing adaptation by [F5].

2.1F4step 1.1algebra

Since Π2=Π and IΠ=ΠI=Π, induction gives Pγm=(1−γ)mI+(1−(1−γ)m)Π. For either state x, ∥δx−π∥TV=1/2, so ∥Pγm(x,⋅)−π∥TV=(1−γ)m/2≤2−m−1 uniformly in x and γ∈[1/2,1]. Therefore the family is simultaneously uniformly ergodic.

3.1F6step 2.1step 1.2∎

Steps 2.1 and 1.2 verify the two hypotheses in alternative 1 of [F6]. Hence ∥L(Xn)−π∥TV→0 for every initial state.

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