Crossing Speeds of Random Walks Among “Sparse” or “Spiky” Bernoulli Potentials on Integers

Abstract

We consider a random walk among i.i.d. obstacles on $\mathbb {Z}$ under the condition that the walk starts from the origin and reaches a remote location y. The obstacles are represented by a killing potential, which takes value M>0 with probability p and value 0 with probability 1−p, 0<p≤1, independently at each site of $\mathbb {Z}$. We consider the walk under both quenched and annealed measures. It is known that under either measure the crossing time from 0 to y of such walk, τ _y , grows linearly in y. More precisely, the expectation of τ _y /y converges to a limit as y→∞. The reciprocal of this limit is called the asymptotic speed of the conditioned walk. We study the behavior of the asymptotic speed in two regimes: (1) as p→0 for M fixed (“sparse”), and (2) as M→∞ for p fixed (“spiky”). We observe and quantify a dramatic difference between the quenched and annealed settings.

Appendix:  Proofs of Technical Lemmas

Proof of Proposition 2.1

Let u ₀=u _n =0, and u _k =E ^k(τ _n ;τ _n <τ ₀). Then

Denoting by Δu _k the discrete Laplacian of u _k , we get that

$$-\frac{1}{2}\Delta u_k=\frac{k}{n},\quad k=0,1,2,\dots,n. $$

The continuous analog is −Δu=2x, u(0)=u(1)=0, x∈[0,1], and it is easy to solve: u(x)=x(1−x)(1+x)/3. From the scaling property of the random walk we conclude that u _k =(3n)⁻¹ k(n−k)(n+k). It is also easy to check directly that this expression indeed solves the equation $-\frac{1}{2}\,\Delta u_{k}=k/n,\ k=0,1,2,\dots,n$, and satisfies the boundary conditions. □

Proof of Proposition 2.2

Notice that we can omit from (1.5), (1.6), and (1.16), since on the event {τ _y =∞} the exponential function in the integrand vanishes P-a.s. By the strong Markov property, for every λ≥0 and $y\in \mathbb {N}$ we can write

The ergodic theorem implies that

Next we show that $\mathbb {E}(E^{\omega}(\tau_{1}\mid\tau_{1}<\infty) )<\infty$. Let $\varLambda(t)=-\log \mathbb {E}(e^{-tV(0,\cdot)})$ and . Then, since P ^ω(τ ₁<∞)≥e ^−V(0,ω)/2,

The function α _λ+V (1) is concave and non-decreasing (see [16, p. 272]). Taking the right derivative of α _λ+V (1) at λ=0 we obtain the last statement of proposition. □

Proof of Lemma 2.4

The proof is very simple. If a ₁=0 then a trivial bound is given by the survival probability e ^−M. For $a_{1}\in \mathbb {N}$ we have

It is obvious that the last ratio is bounded by e ^−M but we shall need an improvement of this estimate for the proof of Lemma 2.5. We have

This gives

$$ \frac{P^{\omega,-a_1} (\tau_1<\infty)}{P^{\omega}(\tau_1<\infty)}\leq \frac {e^{-M}}{2a_1(1-e^{-M})+e^{-M}}, $$

(A.1)

and the statement of the lemma follows. □

Proof of Lemma 2.5

Suppose a ₁=0. Recall that τ ₀ is the time of the first return to 0. We just have to bound the expected number of visits to 0 before τ ₁:

We used the following obvious facts: P ^ω(τ ₀<τ ₁)≤(2e ^M)⁻¹ and P ^ω(τ ₁<∞)≥(2e ^M)⁻¹. The last inequality in the multi-line formula above is obtained by considering the Markov chain which starts at 0, gets killed with probability (1−e ^−M) after each visit to 0, otherwise goes with equal probabilities to 1 and 0, and always gets absorbed at 1. For such chain the expected time to hit 1 restricted to the event that it reaches 1 is equal to 2e ^−M/(2−e ^−M)²≤2e ^−M.

Assume now that $a_{1}\in \mathbb {N}$. Since P ^ω(τ ₁<∞)≥1/2, we can replace the conditioning on the event {τ ₁<∞} with the intersection at the cost of factor 2. The time spent in I before hitting 1 is the sum of three terms: (1) $\tau_{-a_{1}}$; (2) the total time spent in excursions to I from −a ₁ that end up in −a ₁; (3) the time needed to get from −a ₁ to 1 without returning to −a ₁.

Term (1) is estimated using Lemma 2.2. We get Ca ₁ times $P^{\omega,-a_{1}}(\tau_{1}<\infty)$. By (A.1) we have the required bound Ce ^−M/(1−e ^−M). Term (3) is bounded by the product of $P^{\omega}(\tau_{-a_{1}}<\nobreak\tau_{1})$, e ^−M, and $1+E^{-a_{1}+1}(\tau_{1};\tau_{1}<\tau_{-a_{1}})$. The latter is again bounded by Ca ₁. Thus, (3) does not exceed Ce ^−M. Term (2) is the sum of a random number of durations of excursions. The expected duration of one such excursion is bounded by Ca ₁ (Lemma 2.2), and the number of them is at most geometric with expectation e ^−M/(1−e ^−M). The total is multiplied by $P^{\omega}(\tau_{-a_{1}}<\tau_{1})$, since to have such excursion the path has to get to −a ₁ before hitting 1. The strong Markov property implies the desired bound. Adding the three terms we get the statement of the lemma. □

Proof of Lemma 2.6

This is a rough bound, which is sufficient for our purposes. Just as in the proof of Lemma 2.5, we replace the conditioning on the event {τ ₁<∞} by the intersection with {τ ₁<∞} at a cost of factor 2e ^M (here we can not exclude the case a ₁=0). The time spent in I _j , $j\in \mathbb {N}$, is equal to the sum of durations of excursions contained in I _j from −a _j+1 and −a _j as well as crossings between the end points of I _j . The total number of such excursions and crossings is again at most geometric with expectation e ^−M/(1−e ^−M) and the expected duration is bounded by C(a _j+1−a _j ) due to Lemma 2.2. But to have a chance to undergo at least one such excursion or crossing the walk has to reach −a _j prior to hitting 1, survive at least one visit to each −a _i , 1≤i<j, and after completing the excursions reach 1 before returning to −a _j surviving the final run through {−a _i ,j≥i≥1}. Thus, the expectation, which we want to estimate is bounded by

$$\frac{C(a_{j+1}-a_j)}{1-e^{-M}}\,\frac{e^{-M(2j-1)}}{(a_j+1)^2}. $$

Replacing 2j−1 with j and a _j +1 with a ₁+1 we obtain the desired upper bound. □

Proof of Lemma 3.1

By simple gambler’s ruin considerations we have

$$u_1=\frac{e^{-V(0,\omega)}}{2r_1}=e^{M-V(0,\omega)}F_M(0,r_1,0), $$

and for n>1

$$ u_n = e^{-M} \biggl(\frac{1}{2r_n} + \frac{u_{n-1}u_n }{2r_{n-1}}+ u_n{ \biggl(1- \frac{1}{2r_n} - \frac{1}{2r_{n-1}} \biggr)} \biggr). $$

Solving the last equation for u _n we get for n>1

$$ u_n = \frac{e^{- M}}{2r_n} \biggl(1- e^{- M} \biggl(1- \frac{1}{2r_n} - \frac{1-u_{n-1}}{2r_{n-1}} \biggr) \biggr )^{-1}=F_M(r_{n-1},r_n,u_{n-1}) $$

as claimed. □

Lemma A.1

For every p∈[0,1), M∈[0,∞), $y\in \mathbb {N}$, and ω∈Ω

$$E_{Q^\omega_{0,y}}\tau_y\leq 1+2y^2. $$

Proof

Let . We need to estimate

$$ E_{Q^\omega_{0,y}}\tau_y=1+\sum _{x=1}^{y-1} E_{Q^\omega_{0,y}}\ell_y(x)=1+ \sum_{x=1}^{y-1} \sum _{m=0}^\infty Q^\omega_{0,y}\bigl( \ell_y(x)>m\bigr). $$

(A.2)

Denote by $\mathcal{G}_{n}$ the sigma-algebra generated by the simple random walk up to time n. Then by the strong Markov property of the simple random walk we have

Substituting this into (A.2) and taking into account that

$$P^x\bigl(\tau_x^{(m)}<\tau_y \wedge\tau_0\bigr) = \biggl(\frac{x-1}{2x}+\frac{y-x-1}{2(y-x)} \biggr)^m= \biggl(1-\frac{y}{2x(y-x)} \biggr)^m, $$

we get

$$E_{Q^\omega_{0,y}}\tau_y\leq 1+\sum_{x=1}^{y-1} \sum_{m=0}^\infty\biggl(1- \frac{y}{2x(y-x)} \biggr)^m= 1+\frac{2}{y} \sum _{x=1}^{y-1} x(y-x)\leq 1+2y^2. $$

 □

Lemma A.2

Fix an arbitrary p∈[0,1), M∈[0,∞), ω∈Ω, and y≥1. Let $\mathcal{O}:=\{x_{1},x_{2},\dots,x_{n-1}\}$ be the set of all occupied sites in (0,y), 0=:x ₀<x ₁<⋯<x _n−1<x _n :=y. Set r _j =x _j −x _j−1, j=1,2,…,n. Then

$$\frac{1}{3}\sum_{j=1}^nr_j^2 \leq E_{Q^\omega_{0,y}}\tau_y\leq \frac{1}{3(1-e^{-M})}\sum _{j=1}^nr_j^2. $$

Proof

If n=1, i.e. $\mathcal{O}=\emptyset$, then the statement follows from the properties of the standard random walk (see Proposition 2.1). Assume that n≥2, and, therefore, y≥n≥2. The proof is based on the decomposition of random walk paths according to the total number of visits, k, to $\mathcal {O}$ before hitting y and to the order, in which the walk visits points of $\mathcal{O}$. For each k≥n−1, such orderings are represented by admissible sequences $(y_{1},y_{2},\dots,y_{k})\in\mathcal {O}^{k}$. For example, for n=4 and k=9 a sequence (x ₁,x ₂,x ₂,x ₃,x ₂,x ₁,x ₂,x ₃,x ₄) is admissible. The set of all admissible sequences of length k will be denoted by $\mathcal {Y}_{k}$ and the set of all random walk paths corresponding to a given admissible $\mathbf{y}_{k}\in\mathcal{Y}_{k}$ by $W_{\mathbf{y}_{k}}$. Define σ ₀=0, σ _i =inf{j>σ _i−1:S _j =y _i }, $i\in \mathbb {N}$. With this notation, we have $\tau_{y}=\sum_{i=1}^{k}(\sigma_{i}-\sigma_{i-1} )$ and

(A.3)

It is easy to see from Proposition 2.1 that

(A.4)

for all $m\in \mathbb {N}$ and ℓ∈{1,…,n−1}. Since the random walk has to cross every interval (x _j−1,x _j ), j=1,…,n, at least once, the strong Markov property of the standard random walk, (A.3), and the first inequality in (A.4) immediately give us the claimed lower bound.

We turn now to the upper bound. Using the strong Markov property and the second inequality in (A.4), we can estimate the right-hand side of (A.3) from above by

where . The last term can be handled in the same way as in the proof of Lemma A.1. The only difference is that each additional visit to an occupied site adds the factor e ^−M, and it is this factor that plays a major role. For every $x\in\mathcal{O}$ we have

This completes the proof. □

Lemma A.3

Let X _i be i.i.d. non-negative random variables such that $P(X_{1} \geq x) \leq e^{- \sqrt{x}}$ for all x≥n ₀. There exist constants C,c∈(0,∞) such that for all n large

$$P \Biggl(\sum_{i=1}^n X_i \geq Cn \Biggr) \leq e ^{-c \sqrt{n}}. $$

This fact is contained in [11, Theorem 1.1, p. 748] but for convenience of the reader we give a short proof.

Proof

Fix an ε∈(0,1) and let , $i\in \mathbb {N}$. Then

$$\Biggl\{ \sum_{i=1}^n X_i \geq Cn \Biggr\}\subset\Biggl(\bigcup_{i=1}^n \{X_i \geq \varepsilon n\} \Biggr)\cup\Biggl\{ \sum _{i=1}^n X^{\varepsilon n}_i \geq Cn \Biggr\}. $$

The probability of the first union is bounded by $n e ^{- \sqrt{\varepsilon n}}$ (for εn>n ₀), which is less than $e ^{- \sqrt{\varepsilon n/2}}$ for all large n. It remains to bound the probability of the last event. By Chebyshev’s inequality

$$P \Biggl( \sum_{i=1}^n X^{\varepsilon n}_i \geq Cn \Biggr) \leq e^{-\frac{C\sqrt{n}}{ 2}} \biggl(E \exp\biggl(\frac{X^{\varepsilon n}_1}{2\sqrt{n}} \biggr) \biggr)^n. $$

We claim that $E\exp(X^{\varepsilon n}_{1}/(2\sqrt{n}) )<1$ for all sufficiently large n. Indeed,

For a fixed ε∈(0,1) and all large n the maximum is attained at the left endpoint. Thus, the last expression converges to $\sqrt{e}-1<1$ as n→∞. □