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21 Sep 2012: P{k Ŝn/2(A1) Ŝn/2(A2)}. over all complementary pairs A1, A2. R. J. ... for all τ (θ, 1]. (We take D(, t, , ) = 1 for t 0.).

15 Aug 2012: P(A B in C) = f(ρ β,λ β) for all ρ,λ,β. ... upward)arc with probability a, and by a leftward (respectively, downward) arc with prob-ability b; all such oriented arcs are placed independently of one another.

4 Sep 2012: J, such that (a) E[ f(X_1) | X_0=i ] f(i) for all i not in J, and (b) for each M>0 the set of states for which ... By the way, I would not expect an examiner to set a question in tripos that is as difficult to do by hand as doing all of (a) (b) and (c)

21 Nov 2012: Then for alltimes t, all starting states b and all sets Di Zdn we have. ... Since the graph is transitive, it follows that for all clusters i and all a,b.

6 Sep 2012: x. The unique value x maximizing the form is a stable point of thesystem to which all trajectories converge. ... a link between nodes b and c followed by routing adaptation to theextra capacity reduces the throughput of all connections.

21 Sep 2012: P{k Ŝn/2(A1) Ŝn/2(A2)}. over all complementary pairs A1, A2. R. J. ... for all τ (θ, 1]. (We take D(, t, , ) = 1 for t 0.).

3 Sep 2012: Otherwise we may impose an arbitrary uniform bound on ‖ f ‖and adapt to all B 1. ... Then, there exists a universal constant 0 < C < such thatfor all u > 0 and n N:.

15 Aug 2012: Thenthe critical exponent βJ(e) satisfies, in particular,. βJ(ef ) = βJ(eg) for all g K, (9). ... It is known that thecritical surface of the process is the set of all p satisfying φ(p) = 0 where.

19 Jan 2012: pt =1011 for all t. {c(0),c(1),c(2),c(3),c(4),. } = {0, 316,12,1,1,. }. Note that c(i)2 c(i 1)c(i 1) ... Adv Appl Probab 42(3):795–815. Bartroff J, Goldstein L, Samuel-Cahn E (2010b) The spend-it-all regionand small time results for the continuous

14 Mar 2012: k and V. 1m = B. 1m for all m. Continuing in the same way, i.e. ... uniformly in the ball B(0,R). Using Lemma 4.1 we deduce that, for all k,.

15 Aug 2012: An allocation α is said to ‘satisfy quota’ if αi {qi, qi} for all i. ... Letδ : {0, 1, 2,. } R be a given function satisfying b δ(b) b 1 for all b.

10 Jan 2012: THEOREM 1 (Low). Any confidence band Cn that is honest over P all withlevel α < 1 necessarily satisfies. ... Xn) that has asymptotic coverage for every fixed f P all;that is, Cn satisfies.

27 Jul 2012: E[(vol(Va(I,δ))). 2] 2E[vol(Va(I,δ))]2. (6.4). For all x define τx = inf{t I : Xt at B(x,δ)}. ... that for all ε sufficiently small. vol({Ψ > 0}) 12 log(1/ε).

7 Feb 2012: A Linear Programming Relaxation. Primal. maximize{zi}. iE. rizi. iS. ASi zi b(S) , for all S E,. ... Primal. maximize{zi}. iE. rizi. iS. ASi zi b(S) , for all S E,.

15 Aug 2012: written µ1 µ2, if µ1(f) µ2(f) for all increasing f : R. ... qφp,q(A) 0. pφp,q(A) for all increasing A. (2.7). The function γ(p,q) in Theorem 2.3(b) may be taken as any function which satisfiesthe

3 Aug 2012: constants. We then show that for all d 1 theMinkowski dimension of (B f)(A) is at least the maximum of the Minkowski dimension off(A) and that of B(A). ... The balls of radius ε centered at (B f)(ti) might not all be disjoint; the followinglemma

15 Aug 2012: 2.1) π(γ; α, β) = limr. πr(γ; α, β). for all triples (γ; α, β). ... Does there exist a constant A such that ηn An for all n?

1 Jun 2012: DX k f ‖ ‖ f ‖,. and C (M) is the intersection of all the spaces C k(M), k N. ... n. That uniformity over all densities between E22 J 1 and E22 J 2.

20 Apr 2012: P [u stays in B(0,R) for all s t] exp(ct/R2), (5). ... We write b = a j1and E = {at time b all cells are dense for scale j 1}.

15 Aug 2012: dTV(A, B) = supE. P(A E) P(B E). where the supremum is taken all events E. ... a) λn(α) cα as n ,(b) λn(α) c′α for all n, and(c) n5π2(αn.

21 Mar 2012: 10. D (discounted programming): 0 < β < 1, and |c(x, u)| < B for all x, u.N (negative programming): 0 < β 1, and c(x, u) 0 for all x, ... Lemma 5.5. Suppose all costs are bounded as follows. (a) K = supx.

15 Aug 2012: Proposition 3.2. For all p1,p2 satisfying 0 < p1 < p2 < 1, there exist strictlypositive numbers a = a(p1,p2) and b = b(p1,p2) such that, for any increasingcylinder event ... E B(n) 6=) exp{αk(p)nλk(n). } for all large n.We expect that the logarithmic

20 Apr 2012: qB2t1(x,x) qB2t(x,x) for all x B, t 0. (3.3). Letting B G this inequality extends to qt. ... Reff (y,B(x,r)c). Reff (x,B(x,r)c) C. for all x G, r 1.

9 Jan 2012: Proposition 4(a) extendsTheorem 1(b) to more general configurations than the all-1 configuration. ... a) all sites in [[1, n]]d1 {1} have color ;(b) all sites in [[1, n]]d1 {m} have color ; and(c) no site of color is adjacent in G to

26 Nov 2012: Xn B(0,R)} An,. and hence we deduce that a.s. for all sufficiently large n, the random walk at time n will stayoutside of the ball B(0,R). ... ij)dj=1, which are all strictly positive. Then for the matrix Mi.

15 Aug 2012: We declare the box S to be occupied if the two following conditions hold:(a) S contains only normal points and tunnels, and(b) M(x;y) occurs for all ... The equivalence class Cn(k) contains all normal pointslying in boxes B with 2 I.

15 Aug 2012: We define A B to be the set of all ω for which there exists a subset H of. ... 3.3). It is conjectured that such an inequality is valid for all events A and B, so longas A B is interpreted correctly.

15 Aug 2012: For a box , we write 0 for the subset of containing all congurations! ... by allocating state 1 to all edges in B lies in theevent A.If more than one such set B exists, we pick the earliest in some deterministic ordering of allsubsets

15 Aug 2012: i.Such a box B(a;b) may be turned into a graph by the addition of all relevant edgesfrom Ld. ... binary operation ,Pp(A B) Pp(A)Pp(B) for all increasing events A;B:The correct interpretation of A B turns out to be A and B occur disjointly'.

15 Aug 2012: µ(A B) µ(A)µ(B) for all increasing events A, B. imsart-aop ver. ... for all increasing A, B. Therefore, µρ is positively associated. Henceforth we restrict ourselves to positive density functions.

15 Aug 2012: for all e lying in B(n). Once this is shown, we sum over e to obtain by (4.9)that. ... We claim that there exists an integer M, chosen uni-formly for edges e in B(n) and for all large n, such that.

11 Apr 2012: a) f (y) > 0, and(b) for all x Zd1, either f (x) = 0 or the site (x, f (x)) is open. ... Let d = 3. By (42), for all a (0, 12 ) there exists ζ > 0.

23 Jul 2012: cSpringer-Verlag 2006. 2 Random-Cluster Measures [1.1]. negative of the sum of σxσy over all edges e = 〈x, y〉 of G. ... Theintersection of the TF over all finite sets F is called the tail σ-field and is denotedby T.

15 Aug 2012: 4 J. van den Berg, G. R. Grimmett, R. B. Schinazi(In this case, coexistence occurs for all suciently large values of. ) ... factors are strictly positive and do not depend on A, it suces toprove that P;(B j A) is uniformly (over all events A of the form

2 Jul 2012: It is an openquestion to prove OZ decay for the truncated function all the way down tothe critical point. ... 7. J. Balogh, B. Bollobás, H. Duminil-Copin, and R. Morris, The sharpthreshold for bootstrap percolation in all dimensions, Trans.

15 Aug 2012: P(S A B) P(S A)P(S B). for all pairs A, B of increasing events with the property that there exists E′ Esuch that A is defined in ... It clearlysuffices to verify (5.1) for all E containing E1, respectively E2.

20 Dec 2012: 6/where D.η, t, B, r/ denotes the maximum of P.X t/ over all r-concave random variables sup-ported on {0, 1=B, 2=B,. , ... In all our simulations, we used B = 50.Each scenario was run 500 times and, to determine the set Lq=p, in each scenario, we

15 Aug 2012: We saythat A is fully connected to B if, for all b B, there exists a A such that a b;in this case, we write A fc B. ... Pp(E,′ ) 1 β(ǫ) for all , ′ D,. whenever M M(ǫ).Proof.

15 Aug 2012: By making all edges within B open, we may obtain thatPp(I m 1) > 0. ... PERCOLATION 15(b) L = Rd, and 1t fW(t) [M;M]d eventually, a.s.,for all M > 0.Case (b) holds if and only if a typical time coordinate T

15 Aug 2012: We. write = Zd = {0, 1}Z. d. , and C(R) = B(R), the set of all configurations on B(R). ... percolation. Then Yy = 0 and Yx = 1 for all x P1y P2y N.

15 Aug 2012: Using Corollary (A.4), we deduce that1 Fn(;!) 1 FTn(!)() 1 eFn(;!)for all! ... PERCOLATION AND MINIMAL SPANNING TREES 17(B) for all y 2 eB, the point y is not r-occupied.Claim (A) follows from consideration of the set of points x for

15 Aug 2012: Let B0 (respectively, C0, F0) be the (decreasing)event that all edges in B (respectively, C, F) are closed. ... associated region is denoted by Λa,b and called a box-region. Write B forthe set of all box-regions of Ld.

15 Aug 2012: Let B(n) be thesubgraph of L consisting of all edges both of whose vertices lie in [n, n]3, andlet B(n) be the set of vertices V (B(n)) ... removing (respectively adding) all edges outside B(n).It is easily verified that.

15 Aug 2012: probability measure.(b) All extremal members of Rp,q are trivial on the tail σ-field T and lie in Wp,q.(c) All translation-invariant members of Wp,q lie ... B = limn. An =. n. An. exists. Furthermore A Am for all m, so that A B.

15 Aug 2012: We define B as the union of V together with all vertices x0 Z. ... The inequalities areimplied by (18) and (19).(b) Let δ DL, so that δ IL,M for all large M.

15 Aug 2012: 4.1) for all ρ,ρ′ R, ρ 2 ρ′ whenever ρ 1 ρ′. ... Thus we consider the set S of subsets Sof R2 such that:(a) (ρ,ρ) S for all ρ R,(b) (ρ1,ρ2) S whenever (ρ2,ρ1) S.The set R

15 Aug 2012: q) if p̂(J, β) = p̂(J′, β) for all β.(b) In addition, βc(J, q) > βc(J. ′, ... given by (2.4), where B = {σ(y) = 1 for all y Λ}, σΛ =(σΛ(x) : x Λ.

15 Aug 2012: By Zd, we mean the set of all d-vectors v = (v1,v2,. ... Note that. P(T t. DN) P(T t) 14η for all N,.

15 Aug 2012: the configuration σ is proportional to exp(βH(σ)), where β > 0and the ‘energy’ H(σ) is the negative of the sum of σxσy over all neighbouringpairs x,y of B. ... φ0p,q st φ st φ1p,q for all φ Wp,q. (c) For b = 0, 1, the measure φbp,q is

27 Jan 2012: Suppose we remove all closed edges, and consider the remainingopen subgraph of the lattice. ... T and all n n0, τ Hn and τ Vn possess crossings in L.