Disconnection, random walks, and random interlacements

Abstract

We consider random interlacements on ${\mathbb {Z}}^d$, $d\ge 3$, when their vacant set is in a strongly percolative regime. We derive an asymptotic upper bound on the probability that the random interlacements disconnect a box of large side-length from the boundary of a larger homothetic box. As a corollary, we obtain an asymptotic upper bound on a similar quantity, where the random interlacements are replaced by the simple random walk. It is plausible, but open at the moment, that these asymptotic upper bounds match the asymptotic lower bounds obtained by Xinyi Li and the author in (Electron. J. Probab. 19(17):1–26, 2014), for random interlacements, and by Xinyi Li in (A lower bound on disconnection by simple random walk. arXiv:1412.3959, 2014), for the simple random walk. In any case, our bounds capture the principal exponential rate of decay of these probabilities, in any dimension $d \ge 3$.

Appendix

In this appendix we provide the proof of Lemma 2.3. The arguments are similar to p. 50 of [9]. The proof is included for the reader’s convenience.

Proof of Lemma 2.3

The uniqueness in the statement is immediate. Only the existence is at stake. We first observe that if $z \in \partial U$ does not belong to the boundary the connected component of U containing B(0, L), all three members of (2.21) vanish (and (2.21) holds with $\psi ^{A,U}_{y,z} = 0$). We can thus assume from now on that U is connected.

The first equality $P_z[H_A < \widetilde{H}_{\partial U}, X_{H_A} = y] = P_y[T_U < \widetilde{H}_A, X_{T_U} = z]$, with $y \in A$ and $z \in \partial U$, is obtained by time-reversal (one sums over the possible values in the discrete skeleton of the walk, of the entrance time in A, for the probability on the left, and of the exit time of U, for the probability on the right). It then suffices to show that when $K \ge c$, for $y \in A$, $z \in \partial U$, one has

$$\begin{aligned} P_y[T_U < \widetilde{H}_A, X_{T_U} = z] = e_A(y) \,P_0 [X_{T_U} = z] (1 + \psi ), \;\hbox {with }|\psi | \le c'/K. \end{aligned}$$

(A1)

One introduces the function on ${\mathbb {Z}}^d$

$$\begin{aligned} h(x) = P_x[X_{T_U} = z],\quad \hbox {for }x \in {\mathbb {Z}}^d. \end{aligned}$$

It is positive and harmonic in $U \supseteq B(0,KL)$. When $K \ge c$, it follows from the gradient control in Theorem 1.7.1, on p. 42 of [9], and the Harnack inequality in Theorem 1.7.2, on p. 42 of [9], that for $x \in D \cup \partial D$, where $D = B(0,2L)$,

(A2)

(using chaining in the last step).

Noting that $T_D$ happens before $T_U$, we find that by the strong Markov property

$$\begin{aligned} P_y[T_U < \widetilde{H}_A, X_{T_U} = z] = E_y\big [T_D < \widetilde{H}_A, P_{X_{T_D}} [T_U < H_A, X_{T_U} = z]\big ]. \end{aligned}$$

(A3)

Now, for $x \in \partial D$ we have

$$\begin{aligned}&P_x [T_U < H_A, X_{T_U} = z] = h(x) - P_x[H_A < {T_U}, X_{T_U} = z] \nonumber \\&\qquad = h(x) - P_x[H_A + T_D \circ \theta _{H_A} < T_U, X_{T_U} = z] \nonumber \\&\quad \mathop {=}\limits ^\mathrm{strong\;Markov} h(x) - E_x[H_A + T_D \circ \theta _{H_A} < T_U, h(X_{T_D} \circ \theta _{H_A})] \nonumber \\&\quad = h(x) - E_x[H_A < T_U, h(X_{T_D} \circ \theta _{H_A})]. \end{aligned}$$

(A4)

As a result of (A2) we see that

(A5)

In addition, it follows by (A4) that for $x \in \partial D$

$$\begin{aligned}&P_x[T_U < H_A, X_{T_U}= z] \nonumber \\&\quad = h(0) (P_x[T_U < H_A] + \varphi (x) - E_x[H_A < T_U, \varphi (X_{T_D} \circ \theta _{H_A})]). \end{aligned}$$

(A6)

Further, we note that for $x \in \partial D$

$$\begin{aligned} P_x[T_U < H_A] \ge P_x[H_{B(0,L)} = \infty ] \ge c, \end{aligned}$$

(A7)

and setting for $x \in \partial D$

$$\begin{aligned} \widetilde{ \varphi }(x) = \varphi (x) / P_x[T_U < H_A], \; \hbox {we have} \; \sup \limits _{\partial D} |\widetilde{\varphi }| \le c/K. \end{aligned}$$

(A8)

As a result, we obtain that for $x \in \partial D$

$$\begin{aligned} P_x[T_U < H_A, X_{T_U}= z] = h(0) \, P_x[T_U < H_A] \big (1 + \gamma (x)\big ), \end{aligned}$$

(A9)

where we have set

(A10)

Coming back to (A3) we obtain (with $X_{T_D}$ playing the role of x in (A9)) that

(A11)

where we note that when $e_{A,U}(y) = P_y [T_U < \widetilde{H}_A] > 0$, then $e_A(y) = P_y[\widetilde{H}_A = \infty ] > 0$ (because $e_{A,U}(y) \le e_{A,B(0,KL)}(y)$ and (2.16)), and the ratio above is understood as 1 if $e_A(y)$ vanishes. We thus see that the ratio in the last line of (A11) lies between 1 and $1 + \rho _{A,B(0,KL)} \le (1 - \frac{c}{K^{d-2}})^{-1}_+$, by (2.18) of Lemma 2.2 and (2.8).

Thus, looking at the last line of (A11), we see that when $K \ge c$,

$$\begin{aligned} P_y[T_U < \widetilde{H}_A, X_{T_U} = z] = e_A(y) \,P_0[X_{T_U} = z] (1 + \psi ) \; \hbox {with} \; |\psi | \le c'/K, \end{aligned}$$

i.e. (A1) holds and this proves Lemma 2.3. $\square$