The exact-flow Hamiltonian Monte Carlo limit

Here Hamilton's equations are $\dot{q}=p$ and $\dot{p}=-q$. For $z=(q,p)$ and $w=(\tilde{q},\tilde{p})$, $\mathrm{\parallel }F(z)-F(w){\mathrm{\parallel }}_2^2=\mathrm{\mid }p-\tilde{p}{\mathrm{\mid }}^2+\mathrm{\mid }q-\tilde{q}{\mathrm{\mid }}^2=\mathrm{\parallel }z-w{\mathrm{\parallel }}_2^2$, so the field is globally $1$-Lipschitz. The curves $q(t)=q\cos t+p\sin t,p(t)=p\cos t-q\sin t$ satisfy these equations and the initial condition $(q(0),p(0))=(q,p)$. By [F4], this global solution is the unique exact flow through $(q,p)$; [F3] applies with the Countable Choice hypothesis supplied by [F5].

At $t=\pi \mathrm{/}2$, the exact endpoint is $(p,-q)$; flipping momentum gives $z^{\ast }=(p,q)$. Since $H(q,p)=(q^2+p^2)\mathrm{/}2$, $H(z^{\ast })=H(q,p)$ and the Metropolis acceptance probability is $1$.

The returned position is the refreshed $p$, which is independent of the old $q$ and has law $N(0,1)$ by [F2]. Thus every update returns an independent draw from the target. This exact-flow argument does not apply to a finite leapfrog discretization.