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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.

P. H. Peskun authored the 1973 paper underlying Peskun ordering, a comparison of the efficiency of Markov chain Monte Carlo algorithms with the same stationary target.^[ref]

For finite reversible Markov chains, increasing every off-diagonal transition probability while preserving the stationary law cannot increase the asymptotic variance of time averages. The comparison formalizes why spending less time repeating the current state can help statistical estimation, provided the ordering's hypotheses hold. It compares the variance of estimators, not merely their one-step acceptance rates.^[ref]