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A Gaussian harmonic leapfrog resonance

Example

For $U (q) = q^{2} / 2$ , $M = 1$ , $ϵ = \sqrt{2}$ , and $L = 4$ , the leapfrog trajectory returns exactly to its initial phase point before the momentum flip. The accepted position therefore never moves; starting at $q = 0$ is a trapped state although the standard-normal target assigns positive probability to $[1, 2]$ .

Facts & Assumptions

Given: AC, the one-dimensional HMC target $N (0, 1)$ , unit mass, step size $ϵ = \sqrt{2}$ , and $L = 4$ leapfrog steps.

[F1]

The one-step map is the composition of a half force kick, full drift, and half force kick. One-step leapfrog map as force and drift shears

[F2]

The HMC proposal flips momentum and accepts according to the Hamiltonian difference. Leapfrog Hamiltonian Monte Carlo

[F3]

Under AC, the standard-normal density is $(2 π)^{- 1 / 2} e^{- q^{2} / 2}$ . Standard normal and normal laws

[F4]

The HMC proposal is the momentum flip composed with the $L$ -step leapfrog map, and its correction uses the Hamiltonian difference. Leapfrog Hamiltonian Monte Carlo

[F5]

AC implies Countable Choice, the assumption needed for the Lebesgue box measure used below. AC supplies countable selections and prescribed serial paths

[F6]

The exponential is positive and strictly increasing, and under Countable Choice $λ_{1} ([1, 2]) = 1$ . The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The exponential function is strictly increasing, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

[F7]

The probability of a Borel set is the integral of its density; the nonnegative integral is monotone, and a constant $c \geq 0$ integrates over $E$ to $c λ (E)$ . Integral over a measurable subset, Monotonicity and nonnegative homogeneity of the nonnegative integral, The integral of a nonnegative simple function

Proof

technique · exact matrix multiplication

1.1F1algebra

For $U^{'} (q) = q$ and $M = 1$ , the half kick and drift matrices are $K_{ϵ / 2} = (\begin{array}{cc} 1 & 0 \\ - ϵ / 2 & 1 \end{array}), D_{ϵ} = (\begin{array}{cc} 1 & ϵ \\ 0 & 1 \end{array}) .$ Multiplication in the order from [F1] gives $A = K_{ϵ / 2} D_{ϵ} K_{ϵ / 2} = (\begin{array}{cc} 1 - ϵ^{2} / 2 & ϵ \\ - ϵ (1 - ϵ^{2} / 4) & 1 - ϵ^{2} / 2 \end{array}) .$

2.1F1F4step 1.1algebra

Substituting $ϵ = \sqrt{2}$ gives $A = (\begin{array}{cc} 0 & \sqrt{2} \\ - \sqrt{2} / 2 & 0 \end{array})$ and direct multiplication yields $A^{2} = - I_{2}$ , hence $A^{4} = I_{2}$ . Thus after four steps $(q_{4}, p_{4}) = (q, p)$ . By [F4] the flipped proposal is $(q, - p)$ , whose Hamiltonian equals the initial Hamiltonian because $K$ is even.

3.1F2F3F5F6F7step 2.1algebra∎

By [F2], acceptance is one and every returned position equals the input $q$ . Starting at $q = 0$ therefore gives law $δ_{0}$ at every iteration. On $[1, 2]$ , the density [F3] is at least $(2 π)^{- 1 / 2} e^{- 2} > 0$ , so [F6] and [F7] give target mass at least $(2 π)^{- 1 / 2} e^{- 2} λ_{1} ([1, 2]) = (2 π)^{- 1 / 2} e^{- 2} > 0$ . Thus the chain remains trapped at a point while the target is not a point mass.

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Sources

Radford M. Neal, MCMC Using Hamiltonian Dynamics, Handbook of Markov Chain Monte Carlo, Chapter 5 (standard reference, not scraped)
Michael Betancourt, A Conceptual Introduction to Hamiltonian Monte Carlo (standard reference, not scraped)