Alphabeta Math
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An invariant law integrates a drift function bounded on its small set

Statement

Let π be invariant for P, and suppose PV≤λV+b1C with V≥1, 0<λ<1, and sup⁡CV<∞. Then πV<∞.

Facts & Assumptions

Given: The invariant probability and drift inequality in the Statement.

The supremum is interpreted in the extended reals, with sup⁡∅=−∞; the empty-set case is handled separately below.

[F1]

Small-set minorization and geometric drift: V is finite-valued and PV≤λV+b1C pointwise.

[F2]

Invariant and stationary distribution for a Markov kernel: πP=π, including equality of integrals for bounded measurable functions.

[F3]

Measure kernel and probability kernel: P(V∧N) is measurable and bounded by N.

[F4]

Monotone convergence for the integral: ∫(V∧N)dπ↑∫Vdπ.

Proof

Given: The hypotheses and facts [F1]–[F4].

1.1F1F2given

If C=∅, then PV≤λV everywhere. For each integer N≥1, VN=V∧N satisfies PVN−VN≤−(1−λ)V on {V≤N} and PVN−VN≤0 on {V>N}. Invariance for the bounded function VN gives integral zero, so π({V≤N})=0. Since V is finite-valued, these sets increase to the whole space, contradicting that π is a probability. Thus no invariant probability exists in the empty-set case, and the implication is vacuous. Assume C≠∅ below.

1.2F1F3given

Set VN=V∧N. Since VN≤V and VN≤N, PVN≤min⁡(PV,N). On C, PVN−VN≤λsup⁡CV+b. On Cc∩{V≤N} it is at most −(1−λ)V, and on Cc∩{V>N} it is at most zero.

2.1F2step 1.2

Invariance for bounded VN gives ∫(PVN−VN)dπ=0. Using the three pointwise bounds from step 1.2, (1−λ)∫Cc∩{V≤N}V dπ≤(λsup⁡CV+b)π(C)≤λsup⁡CV+b.

3.1F4step 2.1∎

The domains Cc∩{V≤N} increase to Cc; monotone convergence bounds ∫CcVdπ by the right side divided by 1−λ. The integral over C is at most sup⁡CV. Hence πV<∞.

Depends on

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