Multitype Contact Process on Z: Extinction and Interface by Daniel R. Valesin
By Daniel R. Valesin
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Is proved. 4. ) For all x, x (H I = ∞) = 0. Proof. ), note that x (τ1 = H I < ∞) = 1 − x (τ1 =1−0− = H I = ∞) − x (τ1 x (τ1 < ∞, τ1 < H I ) < ∞, |X τ1 | > L) ≥ 1 − C e−c L , which can be made positive by increasing L. ) Finally, note that x (H I = ∞) ≤ x (H I = ∞, τk < ∞ ∀k) + ∞ k=1 x (H I = ∞, τk = ∞). ) follows. 5. There exists C > 0 such that, for all x ∈ , x (H I > N) ≤ C|x| N . Proof. If x ∈ I, the left-hand side is zero. Assume that x ∈ / I. 9) for some C > 0. So the above sum is less than ∞ k+1 C|x| k+1 t k=0 i=1 C ′ |x| (1 − σ)k−2 ≤ t as required.
34. Springer-Verlag, New York-Heidelberg, 1976. xiii+408 MR0388547 [17] Sturm, Anja; Swart, Jan M. Tightness of voter model interfaces. Electron. Commun. Probab.
5) + (1 − εz )gz . 6) (Of course, if z ∈ I we must have bz1 = πz , bz2 = π). We will construct the process (X n ) coupled with other processes of interest. Let (X n , Zn ) be a Markov chain on × {0, 1} with transitions Q((x, i), ( y, j)) = ε x · b1x ( y − x) (1 − ε x ) · g x ( y − x) if j = 1; if j = 0. 7) We write x to represent any probability for this chain with X 0 = x, regardless of the law of Z0 . This abuse of notation is justified by the fact that Z0 has no influence on the distribution of the other variables of the chain, nor on the random variables to be defined below.