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Detailed balance implies invariance

Statement

Let $p$ be a transition matrix on a countable state space $E$ (Transition matrices and n-step probabilities), and let $μ : E \to [0, + \infty)$ be a state measure with $μ (x) < + \infty$ for every $x \in E$ that satisfies detailed balance for $p$ (Reversible measure and detailed balance). Write

$(μ p) (y) : = \sum_{x \in E} μ (x) p (x, y) (y \in E),$

the measure-matrix product, a sum of nonnegative terms in $[0, + \infty]$ . Then

$μ p = μ, that is (μ p) (y) = μ (y) for every y \in E .$

In particular the conclusion holds for a reversible state measure of infinite total mass $\sum_{x \in E} μ (x) = + \infty$ ; and if the mass is one, $μ$ is an invariant (equivalently stationary) probability distribution for $p$ in the sense of Invariant and stationary distribution for a Markov kernel.

Facts & Assumptions

Given: A countable state space $E$ , a transition matrix $p$ on $E$ , and a state measure $μ$ satisfying detailed balance for $p$ .

[F1]

The transition entries satisfy $p (x, y) \geq 0$ and $\sum_{y \in E} p (x, y) = 1$ for every $x \in E$ ; the matrix is the countable form of a probability kernel and no choice principle enters its definition. (Transition matrices and n-step probabilities)

[F2]

A state measure is a function $μ : E \to [0, + \infty)$ with $μ (x) < + \infty$ for every $x$ , and it satisfies detailed balance for $p$ when $μ (x) p (x, y) = μ (y) p (y, x)$ for all $x, y \in E$ ; every product $μ (x) p (x, y)$ is then a well-defined element of $[0, + \infty)$ and no subtraction of infinite quantities occurs. (Reversible measure and detailed balance)

[F3]

On a countable state space, a probability measure $π$ is invariant for $p$ exactly when $π (y) = \sum_{x \in E} π (x) p (x, y)$ for every $y \in E$ . (Invariant and stationary distribution for a Markov kernel)

Proof

Given: A countable state space $E$ , a transition matrix $p$ on $E$ , and a state measure $μ$ satisfying detailed balance for $p$ .

Proof technique: direct termwise comparison of two nonnegative series, with the finite row-sum normalization.

1.1F1F2given

Fix $y \in E$ . Every term $μ (x) p (x, y)$ of the series defining $(μ p) (y)$ is a product of a finite nonnegative number and an element of $[0, 1]$ , hence lies in $[0, + \infty)$ ; so $(μ p) (y) \in [0, + \infty]$ is a well-defined nonnegative extended series.

1.2F2given

For each fixed $x \in E$ the detailed balance identity gives $μ (x) p (x, y) = μ (y) p (y, x)$ ; since the two families of nonnegative terms indexed by $x$ are equal term by term, the series they generate have the same value in $[0, + \infty]$ , that is $\sum_{x \in E} μ (x) p (x, y) = \sum_{x \in E} μ (y) p (y, x)$ . No rearrangement or interchange of summation is used.

2.1F1step 1.2given

The common factor $μ (y)$ is a fixed element of $[0, + \infty)$ , so it may be factored out of the nonnegative series: $\sum_{x \in E} μ (y) p (y, x) = μ (y) \sum_{x \in E} p (y, x) = μ (y) \cdot 1 = μ (y)$ , where the row sum is one by [F1]. This step is valid also when $μ (y) = 0$ , in which case both sides vanish.

3.1F2F3step 1.2step 2.1given∎

Combining steps 1.1–2.1, $(μ p) (y) = μ (y)$ for the arbitrary $y \in E$ , hence $μ p = μ$ . If additionally $\sum_{x \in E} μ (x) = 1$ , then [F3] identifies this identity as invariance of the probability distribution $μ$ . All quantities appearing are nonnegative, no difference of infinities is formed, and neither step selects an object, so the argument uses no choice principle and does not require $μ$ to have finite total mass or to be nonzero.

Depends on

Used by

An invariant law need not be reversible Counterexample
Random walk on a finite undirected graph is reversible Example
Stationary law of a finite birth-and-death chain Example

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Sources

Durrett, Probability: Theory and Examples, fifth edition, §5.5, stationary measures and reversibility (standard reference, not scraped)
Levin–Peres–Wilmer, Markov Chains and Mixing Times, second edition, §1.4 and Appendix C.1 (standard reference, not scraped)