Search

23 Nov 2013: In both cases, each thickened portion of w isbackward extendable. so that the old wa becomes the new w0.) Since loop-erasure does notincrease distances along the walk among the vertices

28 Feb 2013: Let new fees be as follows. p1(θ) = p1(θ1) x(θ) E[. ... Then adopt new payments:. p1(θ1, θ2, θ3,. , θn) = p1(θ1, θ2, θ3,. ,

25 Apr 2013: In this paper we establish a new connection between mixing times and hittingtimes for reversible Markov chains (Theorem 1.1). ... 8 A new proof of tprod tL for reversible chains. Recall the definition tprod = maxx,A.

12 Sep 2013: 1987). Tilings and Patterns. Freeman, New York.MR0857454. [8] KENYON, R. (2004). ... Springer, New York.MR0894549. [12] KESTEN, H. (1987). Scaling relations for 2D-percolation.

7 Feb 2013: is about to quit —, we increase the prize tog(y), which becomes the new ‘prevailing prize’. ... is about to quit —, we increase the prize tog(y), which becomes the new ‘prevailing prize’. •

30 Oct 2013: It remains to show that I2 <. By defininga new equivalence relation on rectangles in Dk, i.e. ... Springer-Verlag, New York, 1995. [10] D. Khoshnevisan and Y. Xiao. Brownian motion and thermal capacity.

7 Mar 2013: By Lemma 4.4again, X(ω′) X(ω) is finite, so no new infinite components have beencreated, and thus X(ω′) has strictly fewer that n infinite components.Since the modification

3 Jul 2013: It implies criticality of such values, therebyproviding a new proof of the critical point of inhomogeneous systems. ... A great deal of rigorous mathematics exists for critical site percolation on thetriangular lattice, but surprisingly little for other

14 Nov 2013: K. D. Glazebrook, D. J. Hodge, C. Kirkbride, R. J. Minty, Stochastic scheduling: Ashort history of index policies and new approaches to index generation for dynamicresource allocation, J Scheduling., 2013. ... Dating strategy: should I contact a new

10 Oct 2013: It is easy to check thatG = {iJBi : J N}.The natural thing to do is to define a new random variable X′ = E[X|G] as follows. ... So for allA Fn we have. E[Xn1(A)] = E[X1(A)]. Hence if we now define a new probability measure Q̃(A) = E[X1(A)], then Q(A) =