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Equivalent criteria for recurrence and transience

Statement

Assume AC (The Axiom of Choice). Let X=(Xn,Fn)n≥0 be a time-homogeneous Markov chain on an at most countable state space E, with transition matrix p. Fix x∈E and use Px and Ex for the specified law with X0=x almost surely. For the visit count Nx=∑n≥01{Xn=x}, which includes the initial visit, the following are equivalent:

x is recurrent⟺Px(Nx=∞)=1⟺∑n=0∞p(n)(x,x)=∞.

If x is transient and rx:=Px(Tx+<∞)<1, then for every integer k≥1,

Px(Nx=k)=(1−rx)rxk−1,ExNx=∑n=0∞p(n)(x,x)=11−rx<∞.

Facts & Assumptions

Given: AC, a countable-state Markov chain, and a fixed state x∈E.

[A1]

AC is assumed for the specified chain law and the finite-dimensional and strong-Markov conditional-expectation interfaces used below. (The Axiom of Choice)

[F1]

Under the fixed initial state, X0=x almost surely, and Px,Ex denote that specified law and expectation. (Initial distribution of a Markov chain)

[F2]

Nx=∑n≥01{Xn=x} counts the time-zero visit. (Hitting, return, and visit times)

[F3]

Tx+=inf⁡{n≥1:Xn=x}, so a return must occur at a strictly positive time. (Hitting, return, and visit times)

[F4]

The successive returns are R0=0 and Rj=inf⁡{n>Rj−1:Xn=x} after a finite preceding return, with later returns set to +∞ after an infinite one. (Hitting, return, and visit times)

[F5]

x is recurrent exactly when rx=Px(Tx+<∞)=1, and is transient exactly when rx<1. (Recurrent and transient states)

[F6]

p(n)(x,y)=Kn(x,{y}) and p(0)(x,y)=1{x=y}. (Transition matrices and n-step probabilities)

[F7]

The finite-dimensional law at the single time n, with initial law δx and test 1{x}, gives Px(Xn=x)=p(n)(x,x). (Finite-dimensional laws of a Markov chain)

[F8]

At each finite return time Rj, the post-return path has the Px law in the conditional sense: for bounded measurable future-path H, Ex[Zj,H∣FRj]=1{Rj<∞}Ex[H(X0,X1,…)] a.s. (Renewal decomposition at successive returns)

[F9]

Each coordinate Xn is a measurable random element; hence its singleton event {Xn=x} is measurable. (Stochastic processes and their finite-dimensional distributions)

[F10]

If nonnegative measurable Ym↑Y, then ExYm↑ExY, allowing +∞. (Monotone convergence for the integral)

[F11]

For decreasing measurable events Ak in the probability measure Px, Px(⋂kAk)=lim⁡kPx(Ak) because Px(A1)≤1<∞. (Continuity from above when one set has finite measure)

Proof

technique · identify visit tails with finite successive returns, iterate the return-time Markov identity, then apply monotone convergence to both the time-indexed and tail-indexed visit counts
1.1F1F2F4given

For each k≥1, pathwise {Nx≥k}={Rk−1<∞}. Since X0=x [F1], the first visit is already counted, and at least k visits are exactly k−1 further finite returns. In particular, both events are certain for k=1 since R0=0. Also {Nx=∞}=⋂k≥1{Nx≥k}.

1.2A1F1F2F6F7F9F10given

The finite partial visit counts Vm:=∑n=0m1{Xn=x} are nonnegative measurable by [F9] and increase pointwise to Nx [F2]. Monotone convergence [F10] therefore gives, with extended values allowed, ExNx=lim⁡m→∞∑n=0mPx(Xn=x)=∑n=0∞p(n)(x,x), where the last equality uses the finite-dimensional law [F7] under AC [A1]. The n=0 term is 1 on each side by [F1] and [F6]; no initial visit is lost.

2.1A1F3F4F8F9step 1.1given

Define the bounded future-path functional Hx(ω)=1{∃m≥1:ωm=x}. Its event is a countable union of coordinate-cylinder events, hence measurable by [F9], and ExHx=rx by the definition of Tx+ [F3]. On {Rj<∞}, Zj,Hx=1{Rj+1<∞}; on {Rj=∞} it is zero. Applying the AC-based strong-Markov identity [F8] and taking expectations gives Px(Rj+1<∞)=rxPx(Rj<∞). Starting from Px(R0<∞)=1, induction yields Px(Rj<∞)=rxj. By step 1.1, Px(Nx≥k)=rxk−1 for every k≥1.

3.1F4F5F11F12step 2.1cases-exhaustivecases

The events {Nx≥k} decrease to {Nx=∞}. Continuity from above [F11] and step 2.1 give Px(Nx=∞)=lim⁡k→∞rxk−1. If x is recurrent, [F5] gives rx=1, so this limit is 1. If x is transient, [F5] gives 0≤rx<1, and [F12] makes the limit 0. These two cases exhaust all states, so x is recurrent if and only if Px(Nx=∞)=1.

3.2F2F4F5F10F13step 2.1step 1.2cases-exhaustive

For M≥1, let WM:=∑k=1M1{Nx≥k}. These are nonnegative measurable variables, increase pointwise to Nx, and [F10] together with step 2.1 gives ExNx=∑k=1∞Px(Nx≥k)=∑j=0∞rxj. If x is recurrent then [F5] gives rx=1 and this sum is +∞. If x is transient then [F5] gives rx<1 and [F13] gives ExNx=1/(1−rx)<∞. In view of step 1.2, the Green series diverges exactly in the recurrent case. This proves both directions of the recurrence/Green-series equivalence without subtracting extended values.

4.1F13F14step 2.1step 3.1step 3.2

In the transient case, for every k≥1 the nested tail events satisfy Px(Nx=k)=Px(Nx≥k)−Px(Nx≥k+1)=(1−rx)rxk−1 by step 2.1 and finite subtraction of probabilities in [0,1]. Since Px(Nx=∞)=0 by step 3.1, these masses account for all outcomes; their sum is (1−rx)∑j≥0rxj=1 by [F13]. When rx=0, the convention 00=1 [F14] gives Px(Nx=1)=1 and Px(Nx=k)=0 for k>1, as expected when no positive return occurs. The mean formula is the transient case of step 3.2.

5.1A1F1F2F3F4F5F6F8step 1.1step 2.1step 3.1step 1.2step 3.2step 4.1given∎

If E=∅, there is no x and the theorem is vacuous. For a one-state absorbing chain, rx=1, Nx=∞ almost surely, and p(n)(x,x)=1 for every n, agreeing with both recurrence criteria. If a deterministic chain started at x leaves and never returns, then rx=0, Nx=1 almost surely, and p(n)(x,x)=0 for n≥1, agreeing with the transient formulas. If instead a deterministic cycle returns after a fixed positive period d, then rx=1 and p(md)(x,x)=1 for all m≥0, so both the infinite-visit probability and Green series are infinite as asserted. The n=0 Green term and k=1 tail were treated explicitly in steps 1.1 and 1.2–3.2. AC [A1] is used for the specified chain law and the conditional strong-Markov identity [F8]; the return-series arithmetic itself uses no choice. Both stated equivalences have been proved in both directions in steps 3.1 and 3.2.

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