Comparison of quadratic forms #
For a Markov kernel K which maps the probability vector p to q, that is,
∑ i, p i * K i j = q j for every j, the difference ⟨φ, K φ⟩_p - p(φ) q(φ) is a sum over pairs
of indices of p i * (K i j - q j) times -(φ i - φ j) ^ 2 / 2. If K i j ≥ q j for i ≠ j, the
difference is therefore nonpositive. The third parts of Lemmas 1 and 2 of the paper follow in this
way from their second parts.
Lemma 3 of Tierney (1998) is also proved for finite index sets: of two Markov kernels which are
reversible with respect to p, the one with the larger off-diagonal entries has the smaller
quadratic form.
For coefficients c i j, the sum ∑ i, ∑ j, c i j * (φ i - φ j) ^ 2 is
∑ i, (∑ j, c i j) * φ i ^ 2 + ∑ j, (∑ i, c i j) * φ j ^ 2 - 2 * ∑ i, ∑ j, c i j * φ i * φ j.
If K is a Markov kernel and ∑ i, p i * K i j = q j for every j, then
⟨φ, K φ⟩_p - p(φ) q(φ) = -(∑ i, ∑ j, p i * (K i j - q j) * (φ i - φ j) ^ 2) / 2.
If K is a Markov kernel, ∑ i, p i * K i j = q j for every j, and K i j ≥ q j for
i ≠ j, then ⟨φ, K φ⟩_p ≤ p(φ) q(φ).
For a Markov kernel K reversible with respect to p,
⟨φ, K φ⟩_p = ∑_i p_i φ(i) ^ 2 - (∑_{i, j} p_i K(i, j) (φ(i) - φ(j)) ^ 2) / 2.
Lemma 3 of Tierney (1998), for a finite index set: if the Markov kernels K₁ and K₂ are
reversible with respect to p and K₂ i j ≤ K₁ i j for i ≠ j, then
⟨φ, K₁ φ⟩_p ≤ ⟨φ, K₂ φ⟩_p.
For a Markov kernel K reversible with respect to a probability vector p,
p(φ) ^ 2 - ⟨φ, K φ⟩_p ≤ p(φ ^ 2) - p(φ) ^ 2: minus the lag-one autocovariance of φ is at most
its variance.
The entries of a Markov kernel are at most one.
The Gibbs kernel K(i, ·) = p is reversible with respect to p.
The identity kernel is reversible with respect to every vector.
For a probability vector p, the Gibbs kernel K(i, ·) = p is a Markov kernel.
For the identity kernel, ⟨φ, φ⟩_p = p(φ ^ 2).