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A finite reversible two-state Markov kernel

Example

Let $X = {0, 1}$ with its full sigma-algebra. For $0 < a, b \leq 1$ , set $P = (\begin{array}{cc} 1 - a & a \\ b & 1 - b \end{array}), π (0) = \frac{b}{a + b}, π (1) = \frac{a}{a + b} .$ Then $P$ is reversible with respect to $π$ , and therefore $π$ is invariant.

Facts & Assumptions

Given: The two-state measurable space, parameters $0 < a, b \leq 1$ , and the displayed matrix and probability vector.

[A1]

$0 < a, b \leq 1$ .

[F1]

Reversibility means symmetry of the joint flow measure $Λ_{π, P} (C) = \int P (x, C_{x}) π (d x)$ on the product sigma-algebra. (General-state reversibility by symmetric flow measure)

[F2]

If a probability kernel is reversible with respect to a probability measure, that probability is invariant. (General-state reversibility implies invariance)

[F3]

A probability measure has total mass one. (Probability measures and probability spaces)

[F4]

A probability kernel has probability-measure rows and measurable event evaluations. (Measure kernel and probability kernel)

Verification

Given: $X = {0, 1}$ , $0 < a, b \leq 1$ , $P = (\begin{array}{cc} 1 - a & a \\ b & 1 - b \end{array})$ , and $π = (b / (a + b), a / (a + b))$ .

Proof technique: Verify stochasticity and normalization, then compare the four point masses of the joint flow with their coordinate transposes.

1.1A1F3F4givenalgebra

Every matrix entry is nonnegative and each row sums to one, so [F4] makes $P$ a probability kernel. Also $a + b > 0$ and $π (0) + π (1) = (a + b) / (a + b) = 1$ , with both entries positive, so $π$ is a probability measure by [F3].

2.1F1step 1.1givenalgebra

Each diagonal flow atom is unchanged by swap, and the off-diagonal atoms agree since $Λ_{π, P} ({0} \times {1}) = π (0) a = a b / (a + b) = π (1) b = Λ_{π, P} ({1} \times {0})$ . The four atoms generate the finite product sigma-algebra, so $Λ_{π, P} = Λ_{π, P}^{⊤}$ and $P$ is reversible by [F1].

3.1F2step 2.1∎

The general-state implication [F2] now gives $π P = π$ , so $π$ is invariant.

Remarks

The boundary values $a = 1$ or $b = 1$ are allowed: their zero diagonal flow atoms are still symmetric in step 2.1. The excluded case $a = b = 0$ would make $a + b$ vanish, while outside the stated range a row may fail to be stochastic. No aperiodicity or convergence claim is made.

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