Tight bounds on sparse perturbations of Markov Chains Romain Hollanders Giacomo Como Jean-Charles Delvenne Raphaël Jungers MTNSβ2014 UCLouvain University of Lund PageRank is the average portion of time spent in a node During an infinite random walk PageRank is the average portion of time spent in a node During an infinite random walk PageRank is the average portion of time spent in a node During an infinite random walk PageRank : π = ππ π, ππ π = 1 How much can a few nodes affect the PageRank values ? PageRank : π = ππ π, ππ π = 1 How much can a few nodes affect the PageRank values ? π¦ PageRank : π = ππ π, ππ π = 1 How much can a few nodes affect the PageRank values ? π¦ PageRank : π = ππ π, ππ π = 1 How much can a few nodes affect the PageRank values ? π¦ ~π ~ ~ PageRank : π = π π, ππ~ π=1 How much can a few nodes affect a consensus ? Consensus : π₯π‘ = ππ₯π‘β1 How much can a few nodes affect a consensus ? the weight of each agent in the final decision Consensus : π = ππ π, ππ π = 1 How much can a few nodes affect a consensus ? π¦ ~π ~ ~ Consensus : π = π π, ππ~ π=1 How large can π β π be ? Weak bounds already exist They depend more on the size than the structure of the network / perturbation πβπ π β€ π π β π β π π Condition number of π Sensitive mainly to the magnitude of the perturbation But what if perturbations only affect a few rows of π ? Typically blows up when the network size grows Unless π β π βΉ π vanishes We need better, tighter bounds, adapted to local perturbations ! Como & Fagnani proposed a bound for the 1-norm a nice increasing function escape time from π² πβπ 1 ππ²βπ± β€π πβ ππ±βπ² mixing time hitting time from π± to π² Captures local perturbations Provides physical insight Difficult (impossible?) to extend to other norms No reason to believe that it is tight It is possible to compute the maximum of π β π β Exactly and in polynomial time 1. Compute π from π = ππ π, ππ π = 1 2. For all π£, compute min ππ£ β ππ£ π and max ππ£ β ππ£ π Ξπππ₯ π£ Ξπππ π£ ππ£ min ππ£ π Ξπππ π£ 3. max ππ£ π Ξπππ₯ π£ Return the largest Ξ encountered over all π£βs Finding π¦ππ± π π is easy π· π£ probability 1 π¦ 1 ππ£ = expected time between two visits of π£ Finding π¦π’π§ π π is easy too butβ¦ π· π£ probability 1 π¦ 1 ππ£ = expected time between two visits of π£ Finding π¦π’π§ π π if we fix the escape time from π¦ π· Let us add the constraint that ππ²βπ± = T Furthest away from π£ π£ π’ ?? probability 1 β π(π) probability π(π) π¦ 1 ππ£ = expected time between two visits of π£ A counter example π’β² π£ π€ π’ A counter example distance 3.33 from π£ π’β² π£ π€ π’ distance 4 from π£ A counter example distance 3.33 from π£ π’β² π£ π’ probability 1 β π(π) distance 4 from π£ To minimize π π the optimal solution is to go all-in to πβ² and not to π π€ probability π(π) βΉ We need to loop through every candidate βworst-nodeββ¦ The algorithm to compute the maximum of π β π β over all perturbations of the nodes of π², under the escape time constraint ππ²βπ± = T 1. Compute π from π = ππ π, ππ π = 1 2. For each π£, compute: Ξπππ₯ = max ππ£ β ππ£ π£ π 1 computation of ππ£ all nodes of π² go to π£ with probability 1 Ξπππ = min ππ£ β ππ£ π£ π π computations of ππ£ all nodes of π² go to some node π’ with probability 1 β π(π) and stay in π² with probability π(π) 3. Return the largest Ξπππ or Ξπππ₯ encountered π£ π£ Perspectives Improve the computation of Ξπππ π£ by identifying the βworst-nodeβ on the go, based on its distance to π£ Extend the approach to other norms especially the 1-norm Compare the results with Como & Fagnaniβs bound to establish its quality Thank you
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