In the mathematical study of stochastic processes, a Harris chain is a Markov chain satisfying an additional property.
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A Markov chain {Xn} on state space Ω with stochastic kernel K is a Harris chain[1] if there exist A, B ⊆ Ω, ϵ > 0, and probability measure ρ with ρ(B) = 1 such that
In essence, this technical definition can be rephrased as follows: given two points x1 and x2 in A, then there is at least an ϵ chance that they can be moved together to the same point at the next time step.
Another way to say it is that suppose that x and y are in A. Then at the next time step I first flip a Bernoulli with parameter ϵ. If it comes up one, I move the points to a point chosen using ρ. If it comes up zero, the points move independently, with x moving according to P(Xn+1 ∈ C|Xn = x) = K(x, C) − ερ(C) and y moving according to P(Yn+1 ∈ C|Yn = y) = K(y, C) − ερ(C).
Given a countable set S and a pair (A′, B′ ) satisfying (1) and (2) in the above definition, we can without loss of generality take B′ to be a single point b. Upon setting A = {b}, pick c such that K(b, c) > 0 and set B = {c}. Then, (1) and (2) hold with A and B as singletons.
Let {Xn}, Xn ∈ Rd be a Markov Chain with a kernel that is absolutely continuous with respect to Lebesgue measure:
such that K(x, y) is a continuous function.
Pick (x0, y0) such that K(x0, y0 ) > 0, and let A and B be open sets containing x0 and y0 respectively that are sufficiently small so that K(x, y) ≥ ε > 0 on A × B. Letting ρ(C) = |B ∩ C|/|B| where |B| is the Lebesgue measure of B, we have that (2) in the above definition holds. If (1) holds, then {Xn} is a Harris chain.
In the following, R := inf {n ≥ 1 : Xn ∈ A}; i.e. R is the first time after time 0 that the process enters region A.
Definition: If for all L(X0), P(R < ∞|X0 ∈ A) = 1, then the Harris chain is called recurrent.
Definition: A recurrent Harris chain Xn is aperiodic if ∃N, such that ∀n ≥ N, ∀L(X0), P(Xn ∈ A|X0 ∈ A) > 0.
Theorem: Let Xn be an aperiodic recurrent Harris chain with stationary distribution π. If P(R < ∞|X0 = x) then as n → ∞, distTV (L(Xn|X0 = x), π) → 0.