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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,.