Peskun ordering

Two Markov kernels $P,Q$ with the same invariant probability distribution $\pi$ satisfy Peskun ordering $P⪰Q$ if $P(x,A)\ge Q(x,A)$ whenever $x\notin A$. Thus $P$ moves at least as much probability off the current state.

For finite-state irreducible, aperiodic reversible Markov chains with common $\pi$, the ordering gives no larger Markov chain Monte Carlo asymptotic variance for $P$ than for $Q$, for every observable. The Dirichlet form of a Markov chain satisfies ${\mathcal{E}}_P(f)\ge {\mathcal{E}}_Q(f)$ because the off-diagonal transition weights multiply $(f(y)-f(x){)}^2$. On the mean-zero subspace, inversion reverses the order of the positive operators $I-P$ and $I-Q$. The formula ${\sigma }_P^2(f)=2⟨f,(I-P{)}^{-1}f{⟩}_{\pi }-⟨f,f{⟩}_{\pi }$ proves the variance comparison.^[ref]

The common invariant distribution and reversibility are substantive hypotheses. The comparison is per transition; a more expensive transition can still cost more per accurate estimate.