By Yun Long, Asaf Nachmias, Weiyang Ning, Yuval Peres

ISBN-10: 1470409100

ISBN-13: 9781470409104

The Swendsen-Wang dynamics is a Markov chain popular through physicists to pattern from the Boltzmann-Gibbs distribution of the Ising version. Cooper, Dyer, Frieze and Rue proved that at the whole graph Kn the blending time of the chain is at so much O( O n) for all non-critical temperatures. during this paper the authors exhibit that the blending time is Q (1) in excessive temperatures, Q (log n) in low temperatures and Q (n 1/4) at criticality. in addition they offer an top certain of O(log n) for Swendsen-Wang dynamics for the q-state ferromagnetic Potts version on any tree of n vertices

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**Additional info for A Power Law of Order 1/4 for Critical Mean Field Swendsen-wang Dynamics**

**Example text**

Put T = δm, then for t ≤ T we have that E (Mt − Mt−1 )2 ≥ 1 − δ , hence the process 2 Mt∧T − (1 − δ)(t ∧ T ) , is a submartingale and optional stopping gives (1 − δ)E[τ ∧ T ] ≤ EMτ2∧T . 44) We now bound EMτ2∧T from above. We have Wτ ∧T = −Zδ m 1{τ ≤T } + WT 1{τ >T } . Thus, EWτ2∧T ≤ EZδ2 m + O(m2 )P(τ > T ) . Since after time δm/2 the process is subcritical with constant negative drift we have that P(τ > T ) decays exponentially in m. 17 now yields that EWτ2∧T = EZδ2 m + o(1) = O( −2 ). Next we estimate Ef 2 (τ ∧ T ).

3). 8) P Xt+1 − γ0 n ≥ (Xt − γ0 n) Xt − γ0 n ≥ n 4 ≥ 1 − O(n− 2 ). 9) P Xα log n − γ0 n ≥ n 4 ≥ (1 − O(n− 2 ))α log n = 1 − o(1) , 3 1 when α > 0 is small enough constant. 4 and √ the Markov’s inequality, we have P |Xπ − γ0 n| ≥ A n ≤ 14 for some constant A. 10) Xα log4 n − Xπ T V ≥ − o(1) ≥ , 4 4 which gives a lower bound on the mixing time of magnetization SW chain Xt . This concludes the proof since any lower bound of the mixing time of Xt implies the same lower bound of mixing time of σt . 6.

Recall that |C(v)| is distributed as the ﬁrst hitting time τ of Yt at 0. We put T = 2 m + and condition on Y −2 and on τ ≥ −2 . 33) P(τ ≥ 2 m + ) = P(τ ≥ −2 )E P(τ ≥ T | Y −2 −2 ,τ ≥ ) . 16 that P(τ ≥ −2 ) = O( ). The second term will give us the exponential in the assertion of the Lemma simply because YT has small probability of being positive at this time. Indeed, since the increments of Yt are stochastically bounded above by Bin(m − t, p) − 1 we have that for any small α > 0 E eα(Yt −Yt−1 ) | Yt−1 ≤ e−α [1 + p(eα − 1)]m−t ≤ e−α+(1+ )(α+α2 )(1−t/m) , since eα − 1 ≤ α + α2 for small enough α.