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5 Jun 2020: Theorem 1: For all , all , andany positive integer. Remarks:. ... of. Theorem 2: For all , all , and everypositive integer ,. Remark 17: Note that here, unlike the fixed-rate (or fixed-distortion) case, there is no need for time sharing: optimal

5 Jun 2020: finite functionV : X [1,], such that,. ‖Pn(x, ) π‖V B V (x)ρn, for all n 0, π-a.e. ... for all x X, A B.

5 Jun 2020: Case I: and. By Lemma 1, for all. Therefore,. so in this case. ... Case II:. Here, by Lemma 1, for all. so is achieved at , giving.

5 Jun 2020: 2For example, let B be a countable dense subset of A, and take E to be the col-lection of all open balls with rational radii centered at the points of ... Assume that , for all andsome. Then. B. Proof of Theorem 1.

10 Jan 2020: tβn. are joined by an edge if, after renumbering:. 2. (a) αi is isotopic to βi for all i > 1;. ... b) if Sα2,.,αn is a one-holed torus then i(α1,β1) = 1;.

9 Nov 2020: Proposition 4.4. If Def (1,r), as above, satisfies all the above properties, then so do the spacesDef(a,b), for 1 a b r. ... ya1)(fa1) for all fa1 Fa1. Here, F(a,b) D(a,b) is the. filtration corresponding to the point s and the element Ua,b.

2 May 2020: Pp(AB) > Pp(A)Pp(B),. for all increasing events A,B. Notation 2.8. Let = {0, 1}E with E finite. ... We add a boundary condition: either b = 0 and all edges outside Λ areclosed, or b = 1 and all edges outside Λ are open.

5 Jun 2020: with ψ( ) := R(x0, ): For all x X and all A B such thatR(x0,A) > 0, we have,. ... All of the approximations in this paper are obtained with respect to ||| |||v.

5 Jun 2020: Theorem B. [8]: Let X be a stationary ergodic source with first-order marginal distribution P ,and let Q be an arbitrary codebook distribution on  with Dav <. Then for all ... 2For example, let B be a countable dense subset of Â, and take E to be

5 Jun 2020: accurate and computationally more efficient than the corresponding baseline classifier. All the proofs. ... Although the statements of the more general theorems seem rather complicated, all the quantities in-.

5 Jun 2020: iv) For all n, VarĤn log22 n/n. Mutual information. Let. pnij =1n. ... sequences coinciding in all but the kth position,then for every string xn1. =

11 Nov 2020: are all hyperspecial, when b is basic and unramified. This reproves a result of Xiao–Zhu [XZ17], see Remark 1.4.3 below. ... Moreover the universal homeomorphism in (1) and the morphisms X1 Xswσ(s)(b)and X2 Xsw(b) in (2) are all equivariant for the

5 Jun 2020: In typical applications, X will often be a (Borelmeasurable) subset of Rd together with the collection B of all its (Borel) measurable subsets.(Precise definitions and detailed assumptions are given in ... k,where X0 π. Then Γ.G/ = K.G/, i.e., for all

5 Jun 2020: Upper Bound] For all n 2,pn. log pp. log n 2 log 2. ... B :=(. 2n 1n. )=. (2n 1)!n!(n 1)! ,. since it divides the numerator but not the denominator, and hence the product of all theseprimes also divides B.

5 Jun 2020: and we write R for R1.If for some σ -finite measure ψ on B, some θ > 0, and all functions s : X. ... To ensure that these arefinite for all x X and a suitable class of A B, we usually consider θ ’s in therange θ > |α|‖F ‖ (and as before, we

9 Nov 2020: 7 Let ρ : G GL(V ) be a representation of G of dimension d.(a) Compute the dimension of SnV and ΛnV for all n.(b) Let g G and let ... constituent of IndGKχ, then all |G : K| conjugates of χ are constituents of ResGKφ.

5 Jun 2020: Note thatand also (by Jensen’s inequality) , for all. Moreover, is twice differentiable in with. ... where the infimum is over all stationary probabilitymeasures on such that the -marginal of isand.

10 Mar 2020: and there are g1,. ,gn G, such that giAi = Bi for all i.If A C for some C B, we write A. ... Note that, if h : A B is a realisation of A B,then S h(S) for all S A.

22 Oct 2020: ii)Let F : Cc B be a functor such that F(w) is an isomorphism for all weakequivalences w, then f r g implies F(f) = F(g). ... f).Since RQ(A) = A for all A Ccf , the induced functor Θ : πCcf B is fully faithful.

22 Oct 2020: However, if G is residually split, thenfor [b] B(G,{µ}), all the straight representatives of [b] lie in the admissible set for µ. ... In particularthese are all open in X({µ},b)K and one cannot connect points in different strata.