Stochastic Modeling

Download Mathematical aspects of mixing times in Markov chains by Ravi Montenegro, Prasad Tetali PDF

By Ravi Montenegro, Prasad Tetali

Offers an advent to the analytical elements of the idea of finite Markov chain blending instances and explains its advancements. This publication appears to be like at numerous theorems and derives them in easy methods, illustrated with examples. It contains spectral, logarithmic Sobolev concepts, the evolving set technique, and problems with nonreversibility.

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14. Given f : Ω → R then 1 ˆ y) (f (x) − f (y))2 π(x)P(x, EPˆ (f, f ) = 2 x,y∈Ω = 1 2 (f (x) − f (y))2 π(x)π(y) = Varπ (f ) . x,y∈Ω It follows that EP (f, f ) ≥ by definition of λ and A. 1 ˆ (f, f ) A EP ≥ 1 A Varπ (f ), and the result follows There is a related bound on the smallest eigenvalue of a reversible chain. 1 of [33]). 18. Consider a reversible Markov chain P on state space Ω, and a set of cycles γx of odd length from each vertex x ∈ Ω to itself. Then the smallest eigenvalue λn−1 of P satisfies the relation −1 1 1 + λn−1 ≥ 2 max π(x)|γxy |rx (a, b) , a,b: P(a,b)=0 π(a)P(a, b) x: (a,b)∈γx where rx (a, b) is the number of times edge (a, b) appears in path γx .

C1 Given that spectral profile is a fairly new tool, it has not been widely studied yet. However, mixing time methodologies that have been developed separately can sometimes be used to lower bound spectral profile, and still obtain the same mixing results. Hence this method subsumes many other results. For instance, in [34] the authors show that the log-Sobolev constant and a Nash inequality induce the following lower bounds: Λ(r) ≥ ρ log(1/r) 1−r and Λ(r) ≥ 1 1 − . 5) ρ T λ This is only a factor two weaker than that found with our more direct approach earlier.

The matrix M = √ √ −1 ( π) P( π) is a symmetric matrix because M (x, y) = π(x)P(x, y) π(y)P(y, x) π(x) P(x, y) = = = M (y, x) . π(y) π(x)π(y) π(x)π(y) It follows from the spectral theorem that since P is similar to a symmetric real matrix then it has a real valued eigenbasis. In this eigenbasis, suppose v is a left eigenvector w a right eigenvector, with corresponding eigenvalues λv and λw . Then, λv v w = (vP)w = v(Pw) = λw v w . 13) In particular, if λv = λw then v and w are orthogonal. A special case of this is the eigenvalue 1 with right eigenvector 1, as then if eigenvalue λi = 1 has left eigenvector vi then x vi (x) = v1 = 0.

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