Random Walks in Cooling Random Environments

Abstract

We propose a model of a one-dimensional random walk in dynamic random environment that interpolates between two classical settings: (I) the random environment is sampled at time zero only; (II) the random environment is resampled at every unit of time. In our model the random environment is resampled along an increasing sequence of deterministic times. We consider the annealed version of the model, and look at three growth regimes for the resampling times: (R1) linear; (R2) polynomial; (R3) exponential. We prove weak laws of large numbers and central limit theorems. We list some open problems and conjecture the presence of a crossover for the scaling behaviour in regimes (R2) and (R3).

The research in this paper was supported through ERC Advanced Grant VARIS–267356 and NWO Gravitation Grant NETWORKS-024.002.003. The authors are grateful to David Stahl for his input at an early stage of the project, and to Zhan Shi for help with the argument in Appendix C. Thanks also to Yuki Chino and Conrado da Costa for comments on a draft of the paper.

Appendices

A Toeplitz Lemma

In this appendix we prove Lemma 1.

Proof

Estimate

$$\begin{aligned} \left| \sum _{k \in \mathbb {N}} \gamma _{k,n} z_k + \gamma _n z_{\bar{T}^n} - z^*\right| \le \sum _{k \in \mathbb {N}} \gamma _{k,n} \left| z_k-z^*\right| + \gamma _n \left| z_{\bar{T}^n} - z^*\right| . \end{aligned}$$

(28)

Hence it suffices to shows that, for n large enough, the right-hand side is smaller than an arbitrary $\epsilon >0$. To this end, pick $\epsilon _1>0$ (which will be fixed at the end), and choose $N_0=N_0(\epsilon _1)$ such that $\left| z_k-z^*\right| < \epsilon _1$ for $k>N_0$. Note further that, in view of (22), we can choose $N_1=N_1(\epsilon _1)$ such that, for $n > N_1$,

$$\begin{aligned} \sum _{k=1}^{N_0} \gamma _{k,n} \left| z_k-z^*\right| < \epsilon _1, \end{aligned}$$

(29)

and

$$\begin{aligned} \gamma _n = \frac{\bar{T}^n}{n}> \epsilon _1\quad \Longrightarrow \quad \bar{T}^n > N_0. \end{aligned}$$

(30)

Write the right-hand side of (28) as

$$\begin{aligned} \sum _{k=1}^{N_0} \gamma _{k,n} \left| z_k-z^*\right| + \sum _{k>N_0} \gamma _{k,n} \left| z_k-z^*\right| +\gamma _n \left| z_{\bar{T}^n} - z^*\right| . \end{aligned}$$

(31)

The first two terms in (31) are bounded from above by $\epsilon _1$ for $n>\max \{N_0,N_1\}$. Indeed, for the first term this is due to (29), while for the second term it is true because $\sum _{k \in \mathbb {N}} \gamma _{k,n} \le 1$ and $\left| z_k-z^*\right| <\epsilon _1$ for $k>N_0$. For the third term we note that either $\gamma _n= {\bar{T}^n}/{n}>\epsilon _1$, in which case (30) together with $\gamma _n \le 1$ guarantees that $\gamma _n \left| z_{\bar{T}^n} - z^*\right| <\epsilon _1$, or $\gamma _n\le \epsilon _1$, in which case $\gamma _n\left| z_{\bar{T}^n} - z^*\right| <K\epsilon _1$ with $K = \sup _{k \in \mathbb {N}} \left| z_k-z^*\right| < \infty$. We conclude that (31) is bounded from above by $(2+\max \{1,K\})\epsilon _1$. Now we let $\epsilon _1$ be such that $(2+\max \{1,K\})\epsilon _1 <\epsilon$, to get the claim.    $\square$

B Central Limit Theorem

Lemma 2

(Lindeberg and Lyapunov condition, Petrov [9, Theorem 22]). Let $U=(U_k)_{k \in \mathbb {N}}$ be a sequence of independent random variables (at least one of which has a non-degenerate distribution). Let $m_k=\mathbb {E}(U_k)$ and $\sigma _k^2 =\mathbb {V}\mathrm {ar}(U_k)$. Define

$$\begin{aligned} \chi _n = \sum _{k=1}^n \sigma _k^2. \end{aligned}$$

Then the Lindeberg condition

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{\chi _n} \sum _{k=1}^n \mathbb {E}\left( (U_k -m_k)^2 \,1_{\left| U_k-m_k\right| \ge \epsilon \sqrt{\chi _n}}\right) = 0 \end{aligned}$$

implies that

$$\begin{aligned} w-\lim _{n\rightarrow \infty } \frac{1}{\sqrt{\chi _n}} \sum _{k=1}^n (U_k-m_k) = \mathcal {N}(0,1). \end{aligned}$$

Moreover, the Lindeberg condition is implied by the Lyapunov condition

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{\chi _n^{p/2}} \sum _{k=1}^n \mathbb {E}\left( |U_k -m_k|^p\right) = 0 \quad \text { for some } p > 2. \end{aligned}$$

C RWRE: $L^p$ Convergence Under Recurrence

The authors are grateful to Zhan Shi for suggesting the proof of Proposition 4 below.

We begin by observing that all the moments of the limiting random variable V in Theorem 2 are finite.

Lemma 3

Let V be the random variable with density function (5). Let P denote its law. Then $E\left( V^p\right) <\infty$ for all $p>0$ with $E\left( V^{2k}\right) =0$ for $k\in \mathbb {N}$.

Proof

For $k\in \mathbb {N}$, it follows from (5) that $E(V^{2k})=0$. For arbitrary $p>0$, compute

$$\begin{aligned} E\left( |V|^p\right) = \frac{4}{\pi }\sum _{k\in \mathbb {N}_0} \frac{(-1)^k}{2k+1} \int _0^\infty x^p\,\exp \left( -\frac{(2k+1)^2\pi ^2}{8}x \right) \mathrm {d}x. \end{aligned}$$

(32)

Since $b^q \int _0^\infty x^{q-1}\,\text {e}^{-bx}\,\mathrm {d}x = \varGamma (q)$, the integral in (32) equals

$$\begin{aligned} \frac{8^{p+1}\varGamma (p+1)}{(2k+1)^{2(p+1)}\pi ^{2(p+1)}}. \end{aligned}$$

Therefore

$$\begin{aligned} E\left( |V|^p\right) = \frac{4\varGamma (p+1)8^{p+1}}{\pi ^{2p+3}} \sum _{k\in \mathbb {N}_0} \frac{(-1)^k}{(2k+1)^{2p+3}}, \end{aligned}$$

which is finite for all $p>0$.   $\square$

Proposition 4

The convergence in Proposition 2 holds in $L^p$ for all $p>0$.

Proof

The proof comes in 3 Steps.

• 1. As shown by Sinai [11],

  $$\begin{aligned} w-\lim _{n\rightarrow \infty } \frac{Z_n - b_n}{\log ^2 n} = 0 \quad \text {under the law } \mathbb {P}, \end{aligned}$$

  where $b_n$ is the bottom of the valley of height $\log n$ containing the origin for the potential process $(U(x))_{x\in \mathbb {Z}}$ given by

  $$\begin{aligned} U(x) = \left\{ \begin{array}{ll} \sum \nolimits _{y=1}^x \log \rho (y), &{}x \in \mathbb {N},\\ 0, &{}x=0,\\ - \sum \nolimits _{y=x}^{-1} \log \rho (y), &{}x \in -\mathbb {N}, \end{array} \right. \end{aligned}$$

  with $\rho (y) = (1-\bar{\omega }(y))/\bar{\omega }(y)$. This process depends on the environment $\omega$ only, and

  $$\begin{aligned} w-\lim _{n\rightarrow \infty } \frac{b_n}{\log ^2 n} = V \quad \text {under the law } \alpha ^\mathbb {Z}. \end{aligned}$$

  We will prove the claim by showing that, for all $p>0$,

  $$\begin{aligned} \sup _{n \ge 3} \,\mathbb {E}_{\alpha ^\mathbb {Z}} \Big (\Big |\frac{b_n}{\log ^2 n}\Big |^p\Big )< \infty , \qquad \sup _{n \ge 3} \,\mathbb {E}\Big (\Big |\frac{Z_n}{\log ^2 n}\Big |^p\Big ) < \infty . \end{aligned}$$

  (33)

  To simplify the proof we may assume that there is a reflecting barrier at the origin, in which case $b_n$ and $Z_n$ take values in $\mathbb {N}_0$. This restriction is harmless because without reflecting barrier we can estimate $|b_n| \le \max \{b_n^+,-b_n^-\}$ and $|Z_n| \le \max \{Z_n^+,-Z_n^-\}$ in distribution for two independent copies of $b_n$ and $Z_n$ with reflecting barrier to the right, respectively, to the left.

• 2. To prove the first half of (33) with reflecting barrier, define

  $$\begin{aligned} H(r) = \inf \{x \in \mathbb {N}_0:\, |U(x)| \ge r\}, \qquad r \ge 0. \end{aligned}$$

  Then

  $$\begin{aligned} b_n \le H(\log n). \end{aligned}$$

  (34)

  We have

  $$\begin{aligned} \mathbb {E}_{\alpha ^\mathbb {Z}}\Big (\Big |\frac{H(\log n)}{\log ^2 n}\Big |^p\Big ) = \int _0^\infty p \lambda ^{p-1}\,\mathbb {P}\big (H(\log n) > \lambda \log ^2 n\big )\,\mathrm {d}\lambda . \end{aligned}$$

  Since $\int _0^1 p \lambda ^{p-1} \mathrm {d}\lambda = 1$, we need only care about $\lambda \ge 1$. To that end, note that

  $$\begin{aligned} \big \{H(\log n) > \lambda \log ^2 n\big \} = \Big \{\max _{0 \le x \le \lambda \log ^2 n} |U(x)| < \log n\Big \} \end{aligned}$$

  and

  $$\begin{aligned} \alpha ^\mathbb {Z}\left( \max _{0 \le x \le \lambda \log ^2 n} |U(x)|< \log n\right) \le \hat{P}\left( \max _{0 \le t \le \lambda N} \sigma |W(t)| < \sqrt{N}\,\right) , \ N = \log ^2 n, \end{aligned}$$

  where $\sigma ^2$ is the variance of $\rho (0)$ and $(W(t))_{t\ge 0}$ is standard Brownian motion on $\mathbb {R}$ with law $\hat{P}$. But there exists a $c>0$ (depending on $\sigma$) such that

  $$\begin{aligned} \hat{P}\Big (\max _{0 \le t \le \lambda N} \sigma |W(t)|< \sqrt{N} \Big ) = \hat{P}\Big (\max _{0 \le t \le \lambda } \sigma |W(t)| < 1 \Big ) \le \text {e}^{-c\lambda }, \ \lambda \ge 1. \end{aligned}$$

  (35)

  Combining (34)–(35), we get the first half of (33).

• 3. To prove the second half of (33), write

  $$\begin{aligned} \mathbb {E}\Big (\Big |\frac{Z_n}{\log ^2 n}\Big |^p\Big ) = \int _0^\infty p \lambda ^{p-1}\, \mathbb {P}\big (Z_n > \lambda \log ^2 n\big ) \,\mathrm {d}\lambda . \end{aligned}$$

  Again we need only care about $\lambda \ge 1$. As in Step 2, we have

  $$\begin{aligned} \sup _{n\ge 3} \,\int _1^\infty p \lambda ^{p-1}\, \alpha ^\mathbb {Z}\big (H(\lambda ^{1/3} \log n) > \lambda \log ^2 n\big ) \,\mathrm {d}\lambda < \infty . \end{aligned}$$

It therefore remains to check that

$$\begin{aligned} \sup _{n \ge 3} \, \int _1^\infty p \lambda ^{p-1}\,\mathbb {P}(\mathcal {E}_{\lambda ,n})\,\mathrm {d}\lambda < \infty \end{aligned}$$

(36)

with

$$\begin{aligned} \mathcal {E}_{\lambda ,n} = \Big \{Z_n > \lambda \log ^2 n,\, H(\lambda ^{1/3} \log n) \le \lambda \log ^2 n\Big \}. \end{aligned}$$

To that end, for $x\in \mathbb {N}_0$, let $T(x) = \inf \{n\in \mathbb {N}_0:\,Z_n =x\}$. On the event $\mathcal {E}_{\lambda ,n}$ we have $T(H(\lambda ^{1/3} \log n)) \le n$. Therefore, by Golosov [5, Lemma 7],

$$\begin{aligned} \mathbb {P}(T(x) \le n \mid \omega ) \le n \,\exp \Big (- \max _{0 \le y<x} [U(x-1)-U(y)]\Big ), \ x \in \mathbb {N},\,n \in \mathbb {N}, \end{aligned}$$

which is bounded from above by $n\,\text {e}^{-U(x-1)}$. Picking $x = H(\lambda ^{1/3} \log n)$, we obtain

$$\begin{aligned} \mathbb {P}(\mathcal {E}_{\lambda ,n} \mid \omega ) \le n \,\text {e}^{-U(H(\lambda ^{1/3} \log n)-1)}, \end{aligned}$$

which is approximately $n\,\text {e}^{-\lambda ^{1/3} \log n}$ because U(H(x)) is approximately x. (The undershoot at x can be neglected because it has finite first moment, by our assumption that $\sigma <\infty$.) Taking the expectation over $\omega$, we get

$$\begin{aligned} \mathbb {P}(\mathcal {E}_{\lambda ,n}) \le n^{-(\lambda ^{1/3}-1)}. \end{aligned}$$

This implies (36), and hence we have proved the second half of (33).   $\square$