Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 6

Letbe the Hamiltonian. The Hamiltonian Monte Carlo proposal is accepted with the Metropolis–Hastings acceptance probabilityThe leapfrog map is volume preserving and reversible after momentum reversal, so no proposal-density Jacobian appears.

For $U(x)=x^{T}x/2$, one leapfrog step isTherefore $L$ steps givewhich is a linear transformation.

For fixed $L$, expansion of the matrix in part (b) givesHence its largest eigenvalue is $1+O({\epsilon }^3)$. With $z=(x(0),p(0))$ and $z^{\prime }=M_{\epsilon }^{L}z$,For each fixed starting state, or uniformly on any bounded set of starting states, this impliesUsing $e^{-u}\ge 1-u$ for $u\ge 0$ in the acceptance formula givesfor sufficiently small $\epsilon$. The constant necessarily depends on a bound for $\Vert z\Vert$: a uniform pointwise constant over all of ${\mathbb{R}}^{2d}$ would be impossible because the energy error is quadratic in the starting state.