I am trying to understand the inner working of Hamiltonian Monte Carlo (HMC), but can't fully understand the part when we replace the deterministic time-integration with a Metropolis-Hasting proposal. I am reading the awesome introductory paper A Conceptual Introduction to Hamiltonian Monte Carlo by Michael Betancourt, so I will follow the same notation used therein.
Background
The general goal of Markov Chain Monte Carlo (MCMC) is to approximate the distribution $\pi(q)$ of a target variable $q$.
The idea of HMC is to introduce an auxiliary "momentum" variable $p$, in conjunction with the original variable $q$ that is modeled as the "position". The position-momentum pair forms an extended phase space and can be described by the Hamiltonian dynamics. The joint distribution $\pi(q, p)$ can be written in terms of microcanonical decomposition:
$\pi(q, p) = \pi(\theta_E | E) \hspace{2pt} \pi(E)$,
where $\theta_E$ represents the parameters $(q, p)$ on a given energy level $E$, also known as a typical set. See Fig. 21 and Fig. 22 of the paper for illustration.
The original HMC procedure consists of the following two alternating steps:
A stochastic step that performs random transition between energy levels, and
A deterministic step that performs time integration (usually implemented via leapfrog numerical integration) along a given energy level.
In the paper, it is argued that leapfrog (or symplectic integrator) has small errors that will introduce numerical bias. So, instead of treating it as a deterministic step, we should turn it into a Metropolis-Hasting (MH) proposal to make this step stochastic, and the resulting procedure will yield exact samples from the distribution.
The MH proposal will perform $L$ steps of leapfrog operations and then flip the momentum. The proposal will then be accepted with the following acceptance probability:
$a (q_L, -p_L | q_0, p_0) = min(1, \exp(H(q_0,p_0) - H(q_L,-p_L)))$
Questions
My questions are:
1) Why does this modification of turning the deterministic time-integration into MH proposal cancel the numerical bias so that the generated samples follow exactly the target distribution?
2) From the physics point of view, the energy is conserved on a given energy level. That's why we are able to use Hamilton's equations:
$\dfrac{dq}{dt} = \dfrac{\partial H}{\partial p}, \hspace{10pt} \dfrac{dp}{dt} = -\dfrac{\partial H}{\partial q}$.
In this sense, the energy should be constant everywhere on the typical set, hence $H(q_0, p_0)$ should be equal to $H(q_L, -p_L)$. Why is there a difference in energy that allows us to construct the acceptance probability?