Can I do HMC with the wrong Hamiltonian?

Question

I am a novice HMC user. I am reading Neal's chapter in the Handbook of MCMC. I think I can present the HMC algorithm as :

1. Sample a new momentum
2. Propose a new momentum and a new position using a reversible and volume preserving integrator such as the Leapfrog Integrator
3. Accept or reject the proposed momentum and position using a Metropolis step

If I understood well, the critical point in the second step of the HMC algorithm is that the proposal is volume preserving and reversible, but I am free to use another position energy function than the negated log-likelihood of the target. I am aware of the fact that this would be very inefficient because the acceptance rate of HMC depends of the use of Hamiltonian dynamics to visit high-density regions. Is my understanding correct or did I miss something ?

Answer 1

If I understood well, the critical point in the second step of the HMC algorithm is that the proposal is volume preserving and reversible, but I am free to use another position energy function than the negated log-likelihood of the target. I am aware of the fact that this would be very inefficient because the acceptance rate of HMC depends of the use of Hamiltonian dynamics to visit high-density regions. Is my understanding correct or did I miss something ?

You are correct that you could use this approach and could still use it to generate a Markov chain with the correct invariant measure. This will usually lead to a worse Markov chain, but may be cheaper if the `wrong Hamiltonian' is faster to compute. An example would be this paper, in which the author uses a cheap surrogate to the desired probability to generate approximate HMC trajectories quickly.