By Olle Häggström

This article is perfect for complicated undergraduate or starting graduate scholars. the writer first develops the mandatory heritage in chance concept and Markov chains sooner than utilizing it to review a number of randomized algorithms with vital purposes in optimization and different difficulties in computing. The publication will charm not just to mathematicians, yet to scholars of computing device technology who will detect a lot precious fabric. This transparent and concise advent to the topic has quite a few routines that may support scholars to deepen their knowing.

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1, an M < ∞ such that (P M )i, j > 0 for all i, j ∈ {1, . . , k} . Set α = min{(P M )i, j : i ∈ {1, . . , k}} , and note that α > 0. We get that P(T ≤ M) ≥ P(X M = X M ) ≥ P(X M = s1 , X M = s1 ) = P(X M = s1 )P(X M = s1 ) k k P(X 0 = si , X M = s1 ) = i=1 P(X 0 = si , X M = s1 ) i=1 17 This is what characterizes the coupling method: to construct two or more processes on the same probability space, in order to draw conclusions about their respective distributions. 36 5 Stationary distributions k P(X 0 = si )P(X M = s1 | X 0 = si ) = i=1 k P(X 0 = si )P(X M = s1 | X 0 = si ) × i=1 k ≥ k P(X 0 = si ) α α i=1 P(X 0 = si ) = α2 i=1 so that P(T > M) ≤ 1 − α 2 .

3 (4) Random chess moves. (a) Consider a chessboard with a lone white king making random moves, meaning that at each move, he picks one of the possible squares to move to, uniformly at random. Is the corresponding Markov chain irreducible and/or aperiodic? (b) Same question, but with the king replaced by a bishop. (c) Same question, but instead with a knight. 4 (6) Oriented random walk on a torus. Let a and b be positive integers, and consider the Markov chain with state space {(x, y) : x ∈ {0, .

1, there exists an integer N < ∞ such that (P n )i,i > 0 for all i ∈ {1, . . , k} and all n ≥ N . Fix two states si , s j ∈ S. By the assumed irreducibility, we can find some n i, j such Irreducible and aperiodic Markov chains 27 that (P n i, j )i, j > 0. Let Mi, j = N + n i, j . For any m ≥ Mi, j , we have P(X m = s j | X 0 = si ) ≥ P(X m−ni, j = si , X m = s j | X 0 = si ) = P(X m−ni, j = si | X 0 = si )P(X m = s j | X m−ni, j = si ) >0 (23) (the first factor in the second line of (23) is positive because m − n i, j ≥ N , and the second is positive by the choice of n i, j ).