Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 216 / 3

Markov chain Monte Carlo asymptotic variance for a stationary Markov chain and $\psi \in L^2(\pi )$ iswhenever the limit and series exist. If a reversible Markov chain has positive $L_0^2(\pi )$ spectral gap $\gamma$, then the spectral theorem for normal operators on a separable Hilbert space gives

For $f,g\in L^2(\pi )$, detailed balance says that the measure $\pi (dx)K(x,dy)$ is invariant under exchanging $x$ and $y$. ThereforeThis is precisely the defining identity for a self-adjoint operator.

By the stationary distribution property, $X_1$ and $X_2$ have the same marginal distribution $\pi$. Expanding the square givesThis is the probabilistic representation of the Dirichlet form of a Markov chain.

Subtracting the expected value of $f$ does not alter either side, so suppose $\pi (f)=0$. Since reversibility makes $K$ a self-adjoint operator and stationarity makes it a contraction,The Discrete-time Poincaré inequality for a Markov kernel is therefore equivalent toApplying this inequality successively to $f,Kf,\dots ,K^{t-1}f$ yieldsConversely, the asserted variance contraction with $t=1$ rearranges to the Poincaré inequality. Hence the two statements are equivalent.

Part c gives the integral representationThe integrand vanishes on the diagonal $y=x$. Thus the assumed off-diagonal Peskun ordering impliesfor every $f\in L^2(\pi )$.

On the mean-zero subspace, the variational characterization of the spectral gap of a positive reversible kernel isIt follows immediately thatEquivalently, the energy inequality says $K⪯Q$ in the Löwner order on $L_0^2(\pi )$. Positivity permits the operator monotonicity of the square root and hence $K^{1/2}⪯Q^{1/2}$; the spectral representations in the question identify the top spectral values and give the same gap inequality.