L09_-_MCMC

# W on any connected a1 e1 1t a2 e2 2t a3 e3 3t a4 e4

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Unformatted text preview: c so it has 4 real eigenvalues λ1 = 1, λ2 = λ3 =⅓, λ4 = -⅓ (thanks Matlab) coincidence: λ2 = λ3 and λ4 = - λ2 (4-cylce is a special graph) non-coincidence (true for any lazy r.w. on an undirected connected graph): –  largest eigenvalue =1 –  all others have absolute value &lt;1 •  left eigenvector corresponding to λ1 = 1? •  e1= (¼, ¼, ¼, ¼) is a left eigenvector for λ1 •  Wow. Why would xt → e1 as t→∞? Proving xt → (¼, ¼, ¼, ¼) ? •  Recall A= ⅓ ⅓ 0 ⅓ ⅓0⅓ ⅓⅓0 ⅓⅓⅓ 0⅓⅓ 2 3 1 4 •  λ1 = 1, λ2 = λ3 =⅓, λ4 = -⅓ •  e1= (¼, ¼, ¼, ¼) •  Proof (see also notes): choose eigenvectors e2, e3, e4 for λ2, λ3, λ4, so that {e1, e2, e3, e4} is a basis of R4 (guaranteed by the spectral theorem, since A is symmetric) •  so x0= a1 e1 + a2 e2 + a3 e3+ a4 e4, for some a1, a2, a3 , a4 •  Now xt= x0 At = exact same proof for any = a1 e1 At + a2 e2 At + a3 e3 At + a4 e4 At lazy r.w. on any connected = a1 e1 λ1t + a2 e2 λ2t + a3 e3 λ3t + a4 e4 λ4t undirected graph. → a1 e1, as t→∞ •  since e1= (¼, ¼, ¼, ¼) is a distribution, it must be that a1=1 •  Hence xt→ (¼, ¼, ¼, ¼), as t→∞. Menu •  Random walks on graphs •  Markov Chains •  Examples: –  pagerank –  card-shuffling –  colorings General Setup •  Let G =(V, E) be a: –  (i) strongly-connected weighted directed graph (i.e. there is a path from every vertex to every vertex), –  with (ii) self-loop on every vertex (to avoid periodicities) •  The weight pij of every edge (i, j ) ∈ E represents a probability, namely: •  pij : probability of transitioning from i to j. ￿ •  Hence for all i: (i,j )∈E pij = 1. •  Transition matrix: A = [ pij ]. •  In the example on the left: A= .9 .075 .025 .15 .8 .05 .25 .25 .5 General Setup •  Transition matrix: A = [ pij ]. •  Still true that, if x0 is the distribution over vertices at time 0, then xt= x0 At. •  Claim: A...
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## This note was uploaded on 09/07/2013 for the course EECS 18.410 taught by Professor Erikdemaine during the Spring '13 term at MIT.

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