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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.