Accelerating diffusions

Details Accelerating diffusions Accelerating diffusions

2005-05-12

arxiv.org/abs/math/0505245v1

Annals of Applied Probability 2005, Vol. 15, No. 2, 1433-1444

Mathematics

Probability

Published at http://dx.doi.org/10.1214/105051605000000025 in the Annals of Applied Probability (http://www.imstat.org/aap/) by

Scientific paper

10.1214/105051605000000025

Let U be a given function defined on R^d and \pi(x) be a density function proportional to \exp -U(x). The following diffusion X(t) is often used to sample from \pi(x), dX(t)=-\nabla U(X(t)) dt+\sqrt2 dW(t),\qquad X(0)=x_0. To accelerate the convergence, a family of diffusions with \pi(x) as their common equilibrium is considered, dX(t)=\bigl(-\nabla U(X(t))+C(X(t))\bigr) dt+\sqrt2 dW(t),\qquad X(0)=x_0. Let L_C be the corresponding infinitesimal generator. The spectral gap of L_C in L^2(\pi) (\lambda (C)), and the convergence exponent of X(t) to \pi in variational norm (\rho(C)), are used to describe the convergence rate, where \lambda(C)= Sup{real part of \mu\dvtx\mu is in the spectrum of L_C, \mu is not zero}, {-2.8cm}\rho(C) = Inf\biggl{\rho\dvtx\int | p(t,x,y) -\pi(y)| dy \le g(x) e^{\rho t}\biggr}.Roughly speaking, L_C is a perturbation of the self-adjoint L_0 by an antisymmetric operator C\cdot\nabla, where C is weighted divergence free. We prove that \lambda (C)\le \lambda (0) and equality holds only in some rare situations. Furthermore, \rho(C)\le \lambda (C) and equality holds for C=0. In other words, adding an extra drift, C(x), accelerates convergence. Related problems are also discussed.

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