Put

S (x) = \sum_{i = 1}^{d} p (x, i) .

The normalizing identity

\int S (x) ν (d x) = \sum_{i} \int ν (d x_{- i}) \int μ (d x_{i} ∣ x_{- i}) = d

shows that

π (d x) = \frac{S ( x )}{d} ν (d x)

x

and

y

differ only in coordinate

i

, the transition density from

x

y

p (x, i) μ (y_{i} ∣ x_{- i}) / S (x)

. Therefore

π (x) K (x, y) = \frac{1}{d} ν (x) p (x, i) μ (y_{i} ∣ x_{- i}) = \frac{1}{d} ν (x_{- i}) μ (x_{i} ∣ x_{- i}) μ (y_{i} ∣ x_{- i}),

which is symmetric in

x_{i}, y_{i}

π

-reversible.

b

Solution

The summand intended in the question is

d, h (X (t)) / S (X (t))

. Under stationarity,

E_{π} [\frac{d , h ( X )}{S ( X )}] = \int \frac{d , h ( x )}{S ( x )} \frac{S ( x )}{d} ν (d x) = \int h d ν = H .

Moreover

S (x) \geq d c_{1}

, so the summand is bounded by

1/ c_{1}

. A stationary geometrically ergodic Markov chain satisfies the Markov-chain law of large numbers; hence

H_{n} = \frac{1}{n} \sum_{t = 1}^{n} \frac{d , h ( X ( t ))}{S ( X ( t ))} ⟶ H

almost surely, and therefore in probability. In particular,

Pr (∣ H_{n} - H ∣ > ϵ) \to 0.

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