Barker's algorithm accepts a proposed move from to

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Computational Physics · Monte Carlo and lattice simulation of physical systems

Error, Chance and the Machine, Ch Markov-chain Monte Carlo

Problem

Barker's algorithm accepts a proposed move from

x

y

, drawn from a proposal density

q(y\mid x)

that satisfies

q(y\mid x)=q(x\mid y)

(a symmetric proposal), with probability

\alpha_B(x,y) = \frac{p(y)}{p(x)+p(y)},

rather than the Metropolis choice

\min(1,p(y)/p(x))

, where

p

is the (unnormalised) target density. (a) Show that the kernel

K(x,y) = q(y\mid x)\,\alpha_B(x,y)

(for

y\ne x

) satisfies detailed balance,

p(x)K(x,y)=p(y)K(y,x)

, for any symmetric proposal

q

. (b) A particle occupies one of two energy levels,

E_1=0

and

E_2=\Delta

, with Boltzmann weights

p(i)\propto e^{-\beta E_i}

. A symmetric proposal always proposes the other level,

q(2\mid1)=q(1\mid2)=1

. For

\beta\Delta=2

, use the unnormalised weights

p(1)=1

p(2)=e^{-\beta\Delta}

to compute the Barker acceptance probabilities

\alpha_B(1,2)

and

\alpha_B(2,1)

, and verify numerically that

p(1)K(1,2)=p(2)K(2,1)

to three significant figures. (c) For the same two-level system at

\beta\Delta = 2

, write the

2\times2

transition matrix of the Barker chain and of the Metropolis chain, and find the second eigenvalue

\lambda_2

of each. Compute the integrated autocorrelation time

\tau_{\rm int}

of the energy for each chain. Which chain gives the smaller error bar on the mean energy for a fixed run length, and why does the chain with the larger acceptances win?

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