= Peskun ordering
{c}
{wiki}
Two <Markov kernels> $P,Q$ with the same invariant <probability distribution> $\pi$ satisfy Peskun ordering $P\succeq Q$ if $P(x,A)\geq Q(x,A)$ 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 $\mathcal E_P(f)\geq\mathcal E_Q(f)$ 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 $\sigma_P^2(f)=2\langle f,(I-P)^{-1}f\rangle_\pi-\langle f,f\rangle_\pi$ proves the variance comparison.<Optimum Monte-Carlo sampling using Markov chains by P. H. Peskun>[Biometrika 60 (1973), 607–612, asymptotic-variance ordering of transitions with a common stationary distribution]{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.
Back to article page
