Hamiltonian Monte Carlo and leapfrog correction — Examples

For $\mathrm{\mid }q\mathrm{\mid }\le 1$, $U(q)\le \sqrt{2}$, so monotonicity of the exponential gives $e^{-U(q)}\ge e^{-\sqrt{2}}$; [F8]–[F10] imply $Z_U\ge {\int }_{[-1,1]}{e}^{-U(q)}d{\lambda }_1(q)\ge e^{-\sqrt{2}}{\lambda }_1([-1,1])=2e^{-\sqrt{2}}>0.$ For all $q$, $U(q)\ge \mathrm{\mid }q\mathrm{\mid }$, hence $e^{-U(q)}\le e^{-\mathrm{\mid }q\mathrm{\mid }}$; by [F6] AC supplies the Countable Choice required by [F7], and [F9] gives $Z_U<\mathrm{\infty }$. Differentiating gives $U^{\mathrm{\prime }}(q)=\frac{q}{\sqrt{1+q^2}},U^{\mathrm{\prime }\mathrm{\prime }}(q)=\frac{1}{(1+q^2{)}^{3\mathrm{/}2}}\in (0,1]\mathrm{.}$ Applying [F5] to $U^{\mathrm{\prime }}$, which is continuous on each closed interval and differentiable on its interior, the bound $0<U^{\mathrm{\prime }\mathrm{\prime }}\le 1$ makes $U^{\mathrm{\prime }}$ globally $1$-Lipschitz.

For $z=(q,p)$ and $w=(\tilde{q},\tilde{p})$, $\mathrm{\parallel }F(z)-F(w){\mathrm{\parallel }}_2^2=\mathrm{\mid }p-\tilde{p}{\mathrm{\mid }}^2+\mathrm{\mid }{U}^{\mathrm{\prime }}(q)-U^{\mathrm{\prime }}(\tilde{q}){\mathrm{\mid }}^2\le \mathrm{\mid }p-\tilde{p}{\mathrm{\mid }}^2+\mathrm{\mid }q-\tilde{q}{\mathrm{\mid }}^2=\mathrm{\parallel }z-w{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus the Hamiltonian vector field is globally $1$-Lipschitz.

Apply [F2] to obtain the unique global flow. Since $U\in C^2$, [F4] and its Countable Choice hypothesis supplied by [F6] give energy and volume preservation; its sufficient-condition bound has $C=\max \{1,1\}=1$.

Applying [F3] with the same constant gives $\mathrm{\parallel }{\mathrm{\Phi }}_t(z)-{\mathrm{\Phi }}_t(w){\mathrm{\parallel }}_2\le e^{\mathrm{\mid }t\mathrm{\mid }}\mathrm{\parallel }z-w{\mathrm{\parallel }}_2$ for every pair of initial states and every real $t$. This is a flow-stability bound, not a discretization-error estimate.

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.

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

Substituting $ϵ=\sqrt{2}$ gives $A=\left(\begin{array}{cc}0& \sqrt{2}\\ -\sqrt{2}\mathrm{/}2& 0\end{array}\right)$ 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.

By [F2], acceptance is one and every returned position equals the input $q$. Starting at $q=0$ therefore gives law ${\delta }_0$ at every iteration. On $[1,2]$, the density [F3] is at least $(2\pi {)}^{-1\mathrm{/}2}{e}^{-2}>0$, so [F6] and [F7] give target mass at least $(2\pi {)}^{-1\mathrm{/}2}{e}^{-2}{\lambda }_1([1,2])=(2\pi {)}^{-1\mathrm{/}2}{e}^{-2}>0$. Thus the chain remains trapped at a point while the target is not a point mass.

The first half kick gives $p_{1\mathrm{/}2}=1-\frac{1}{2}{U}^{\mathrm{\prime }}(0)=1$. The drift gives $q_1=0+1\cdot p_{1\mathrm{/}2}=1$. The final half kick gives $p_1=1-\frac{1}{2}{U}^{\mathrm{\prime }}(1)=1\mathrm{/}2$, so the flipped proposal is $(q^{\ast },p^{\ast })=(1,-1\mathrm{/}2)$.

The initial energy is $H(0,1)=1\mathrm{/}2$, while the proposed energy is $H(1,-1\mathrm{/}2)=1\mathrm{/}2+1\mathrm{/}8=5\mathrm{/}8$. The discrete step therefore does not conserve $H$ at this point.

The acceptance probability is $\min \{1,e^{-5\mathrm{/}8+1\mathrm{/}2}\}=e^{-1\mathrm{/}8}<1$, since $-1\mathrm{/}8<0$ and [F3] gives $e^{-1\mathrm{/}8}<e^0=1$; rejection retains $q=0$ by [F2]. This finite calculation distinguishes energy error from the exact target-invariance statement proved for the full Metropolized kernel.