Proof of Section 3 Clause Samples

Proof of Section 3. 3.1‌ Proof of Lemma 10. To begin with, note that Σ Σ i = w Ai 2 wi A w 2 A2 ! = , Σ ⩾ w wi w w Σ where in the first inequality we used the ▇▇▇▇▇▇’▇ inequality since wi/w > 0 and k i=1 wi/w = 1 . Let p be the probability of increasing a and q the probability of decreasing it. Using the above inequality, we get p = Σ Ai(Ai − 1) A2 A ⩾ i=1 n(n − 1)wi wn(n − 1) n(n − 1) n n − 1 n(n − 1) q = Σ ai A = aA . (3.14) Without loss of generality, assume that the current configuration is not in R1 , we must have that a < n(1 − ε) 1 and therefore, A ⩾ n − n(1 − ε) 1 . Thus, − A 1 1 −ε a 1 +w .