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17 Jun 2019: Assume p > q. Let Tj = inf{n > 1 : Xn = j} if this is finite, and Tj = otherwise. ... Let Ti = inf{n > 1 : Xn = i}. For each i 6= j letqij = P(Tj < Ti | X0 = i) and mij = E(Tj | X0 = i).

30 Aug 2019: Deduce that for every n 1,. Tn =d1j=1. cn(j)Tj. 2 Define the Weierstrass -function (z) associated to a lattice Λ.

19 Jun 2019: connected to nodes ti and tj with arcs of infinite capacity, and each nodeti is connected to a node s′ with an arc of capacity w wi 1 and w = wn.

1 May 2019: aposteriori). Specifically, we employ. ctot :=1. M 1. M. j=0. x(tj) x̃(tj), (2). ... tj := j tstopM. For the numerical study M = 29 = 512 has been used.

16 Jul 2019: a,b] X est définie par :. (c) = supa=t0<t1<<tk =b. k1j=0. dX (c (tj) ,c (tj1)). On dit que X est un espace de longueur (resp.

30 Aug 2019: j>0. bj (U (n))tj =n1j=0. (1 t2j1). END OF PAPER. Part III, Paper 16.

19 Jun 2019: Xj = min(Tj , cj ), for 1 j n. extending the notation of (b) in the natural way.

1 May 2019: ctot =1. M 1. Mj=0. f (tj) f̃ (tj), (4.1). where f denotes some function of the exact solution (e.g. ... M subintervals of equal length via. tj = j tstopM. 4.1 Performance of the new algorithm.

19 Jun 2019: Theeffects of recombination in the segment may be ignored. The time for which the samplehas j distinct ancestors is denoted by Tj , j = 2, 3,. ,

30 Aug 2019: The effects of recombination in the segment may be ignored.The time for which the sample has j distinct ancestors is denoted by Tj , j = 2, 3,. ,