Chen--Stein method for the uncovered set of random walk on $\mathbb{Z}_{n}^{d}$ for $d \geq 3$

Let $X$ be a simple random walk on $\mathbb{Z}_n^d$ with $d\geq 3$ and let $t_{\rm{cov}}$ be the expected cover time. We consider the set of points $\mathcal{U}_\alpha$ of $\mathbb{Z}_n^d$ that have not been visited by the walk by time $\alpha t_{\rm{cov}}$ for $\alpha\in (0,1)$. It was shown in [MS17] that there exists $\alpha_1(d)\in (0,1)$ such that for all $\alpha>\alpha_1(d)$ the total variation distance between the law of the set $\mathcal{U}_\alpha$ and an i.i.d.\ sequence of Bernoulli random variables indexed by $\mathbb{Z}_n^d$ with success probability $n^{-\alpha d}$ tends to $0$ as $n \to \infty$. In [MS17] the constant $\alpha_1(d)$ converges to $1$ as $d\to\infty$. In this short note using the Chen--Stein method and a concentration result for Markov chains of Lezaud we greatly simplify the proof of [MS17] and find a constant $\alpha_1(d)$ which converges to $3/4$ as $d\to\infty$.

1. Introduction.Let X be a simple random walk on Z d n with d ≥ 3 started from the stationary distribution.For each x ∈ Z d n we let τ x = min{t ≥ 0 : X(t) = x} be the first time that X visits x.For t ≥ 0 we define the process (U x (t)) x and the uncovered set U (t) respectively by U x (t) = 1(τ x > t) for x ∈ Z d n and U (t) = {x ∈ Z d n : U x (t) = 1}.The expected cover time t cov is given by We recall that the total variation distance between two measures µ and ν is given by For any α > 0 let p α,n ∈ (0, 1) be a parameter to be defined precisely later which satisfies p α,n = n −αd (1 + o(1)).Let t * be a time to be defined precisely later which satisfies Finally let ν α,n be the law of {x ∈ Z d n : Z x = 1} where (Z x ) x is an i.i.d.sequence of Bernoulli random variables with parameter p n,α .The following theorem was shown in [MS17].
The existence of α 1 (d) was the main challenge in [MS17], while the existence of α 0 (d) followed by counting the number of neighbouring points in the uncovered set.In [MS17] they obtained α 0 (d) = (1 + p d )/2, where p d is the return probability to 0 for simple random walk on Z d , while their constant Our contribution in the present paper is to give a much simpler proof of the existence of the constant α 1 (d) and moreover to show that α 1 (d) can be chosen to be bounded away from 1 as d → ∞ as the following theorem shows.
In [MS17] as a corollary of Theorem 1.1 it was shown that the same uniformity statement holds when we wait for the first time the uncovered set contains n d−αd points.Using our improved bound on α 1 (d) one can use exactly the same proof as in [MS17] to obtain the same result for this larger range of α.
Notation.For functions f and g we write . We also write P x to indicate the law of the random walk when started from x.We denote by E x the corresponding expectation.
2. Excursions and hitting probabilities.Let r < R. We write B(x, r) for the closed Euclidean ball centered at x of radius r, i.e.
For a set A we define the boundary ∂A to be the outer boundary, i.e. ∂A = {y ∈ A : ∃ x ∈ A c adjacent to y}.
Definition 2.1.We define the following sequence of stopping times and inductively we set We call a path of the random walk trajectory an excursion if it starts from B(x, R) and it comes back to ∂B(x, R) after hitting B(x, r).
We now define N x (r, R, t) to be the total number of excursions across the annulus B(x, R) \ B(x, r) before time t after the first time that X hits ∂B(x, R), i.e.
We next recall [MS17, Lemma 2.2] proved in the Appendix of [MS17] showing that the mixing time of the exit points of the excursions mix in time of order 1.
Lemma 2.2.Let R ≥ 10r and let Y j be the exit point of the j-th excursion across B(0, R) \ B(0, r).Then (Y j ) j is a finite state space Markov chain.Let π be its stationary distribution.
Then the mixing time of the chain is of order 1, i.e. there exists k 0 < ∞ such that t mix = k 0 and k 0 only depends on d.Then there exists a positive constant c such that for all m and N we have Corollary 2.3.The process (Y i−1 , Y i ) mixes in time of order 1 and its stationary distribution is given by ν(x, y) = π(x)P (x, y), where P is the transition matrix of Y .Moreover, there exist positive constants c 1 and c 2 so that for all (x, y) ∈ ∂B(0, R) × ∂B(0, R) the measure ν satisfies Proof.By the definition of total variation distance it is easy to show that for all times t we have This together with Lemma 2.2 shows that the mixing time of (Y i−1 , Y i ) is of order 1.The second claim follows immediately from the proof of Lemma 2.2 in [MS17].
Definition 2.4.For R ≥ 10r we let i.e.T r,R is the expected length of the excursion when the walk is started on ∂B(0, R) according to the stationary distribution π of the exit points of the excursions across the annulus B(0, R) \ B(0, r) as given in Lemma 2.2.
The following lemma was proved in [MS17].The main idea behind the proof is to allow enough time between excursions so that the walk mixes and this essentially gives an almost i.i.d.sequence of excursion lengths.
Lemma 2.5.For each ψ ∈ (0, 1/2) there exists n 0 ≥ 1 and a positive constant c such that for all n ≥ n 0 the following is true.Suppose that n/4 ≥ R ≥ 10r and t ≍ n d log n.Then for all δ > 0 such that δr d−2 n −ψ−1/2 ≥ 1 and δn ψ ≥ 1, for all x we have where We finally recall another standard result that was proved in the Appendix of [MS17] which shows that conditioning on the entrance and exit points of an excursion does not affect the probability of hitting the centre.The proof is an easy consequence of Harnack's inequality.
Lemma 2.6.There exists a constant C d > 0 depending only on d such that the following is true.Let n/4 ≥ R ≥ 2r such that both r and R tend to infinity as n → ∞.We denote by τ R the first hitting time of ∂B(0, R) and by τ 0 the first hitting time of 0. Then for all x ∈ ∂B(0, r) and all y ∈ ∂B(0, R) we have The constant C d is given by c d /G(0), where c d is the constant from [LL10, Theorem 4.3.1]and G is the Green's function for simple random walk on Z d .
Remark 2.7.To avoid confusion, we emphasize that τ x , τ y and τ z will always refer to hitting times of a point, while τ r and τ R to hitting times of boundaries of balls.
Definition 2.8.We define p d to be the probability that a simple random walk on Z d started from 0 returns to 0.
Remark 2.9.For d = 3, it is well-known (see e.g.[Spi64]) that p 3 ≈ 0.34.It is also easy to see that p d → 0 as d → ∞.Note that p d is equal to the probability that a simple random walk in Z d starting from 0 visits a given neighbour of 0 before escaping to ∞.
Lemma 2.10.Let n/4 ≥ R > 2r → ∞ and x, y ∈ Z d n satisfying x − y = o(r).We denote by τ R the first hitting time of B(x, R) and by τ x (resp.τ y ) the first hitting time of x (resp.y).Then for all a ∈ ∂B(x, r) and all b ∈ ∂B(x, R) we have 3. Total variation distance.In this section we prove Theorem 1.2.The strategy of the proof is to define another set, called U below, that can be coupled with U (αt cov ) with high probability and for which we can apply the Chen-Stein method to show that it is close to the distribution ν α,n .
Recall the definition of the times (ρ i ) and ( ρ i ) from Definition 2.1.Next we define a function i.e. this is the probability that 0 is not hit in an excursion of the walk starting from x and conditioned to exit at y.
Recall the definition of the chain Y from Lemma 2.2 as the sequence of exit points of the excursions and ν stands for its invariant distribution.Let We take δ = r (2−d)/2 n ψ for ψ > 0 sufficiently small and define In particular, Proof.Since r = R 1−ε , it follows from Lemma 2.6 that for all x and y we have Therefore for all x and y we obtain and hence, substituting the value of r proves the first equality of the lemma.
The proof of the second equality follows from [MS17, Lemma 4.1] which is proved in the Appendix of [MS17].The last equality follows from the definition of A from (3.2).
For every x let σ x be the first time that the walk has completed A excursions across the annulus B(x, R) \ B(x, r), i.e.
We also define and consider the set Lemma 3.2.There exists a positive constant c so that for all η > 0 we have In particular, as n → ∞ we have Proof.Since after conditioning on σ((Y i ) i≥1 ) the events that 0 is hit in the i-th excursion become independent, we obtain where the function f was defined in (3.1).Taking logarithms we get Using now [Lez98, Theorem 1.1 and Remark 1] and Corollary 2.3 for a positive constant c and all η ∈ (0, 1) we get This proves the first statement of the lemma.
We turn to prove the second statement.Let F be the event Taking η ≍ n −γ(d−2)/2 log n and using Lemma 3.1 we thus deduce Proof.(a) Let E be the number of excursions across B(x, R) \ B(x, r) that the walk has completed by time σ y .We set Then we have Using Lemmas 2.10 and 3.1 we obtain For the second term on the right hand side of (3.3) we have where the last equality follows from Lemma 2.5.
(b) Define L x to be the number of excursions across B x = B(x, n 2ζ/3 ) \ B(x, n ζ/3 ) that the walk has completed by time σ x and analogously define L y .Set δ = n ψ+(2−d)ζ/6 for ψ sufficiently small and Let F exc be the sigma algebra generated by the entrance and exit points of the first M excursions across B(x, n 2ζ/3 ) \ B(x, n ζ/3 ) and the first M excursions across B(y, n 2ζ/3 ) \ B(y, n ζ/3 ).Then The term P(L x < M ) can be controlled in exactly the same way as in (3.4).To bound the first term appearing on the right hand side above we define the events and F y analogously.Since, after conditioning on entrance and exit points of the excursions, the events of hitting the centres are independent, we obtain where for the last equality we used that which follows from [MS17, Lemma 4.1] proved in the Appendix of [MS17].
We now have all the required ingredients to prove Theorem 1.2.
Proof of Theorem 1.2.By Lemma 3.3 it suffices to show that In order to do so, we are going to use the Chen-Stein method [AGG89, Theorem 3] (see also [Che75,Ste72]).For every x we define its neighbourhood The Chen-Stein method [AGG89] shows that and hence it suffices to prove that b i = o(1) as n → ∞ for all i.
Regarding the quantity b 2 we use Lemma 3.4 to obtain . We finally turn our attention to the quantity b 3 .By transitivity, we have We let F exc be the sigma algebra generated by the exit points of the first A excursions across the annulus B(0, n γ ) \ B(0, n γ(1−ε) ).We write F out for the sigma algebra generated by {Q y , y / ∈ B x }.Then by the tower property we have where for the last equality we used that F out is independent of σ(F exc , σ(Q 0 )) which follows from the fact that the annuli are disjoint by the choice of the radii.Using the same notation as in Lemma 2.2, we let Y i be the exit point of the i-th excursion.Then where f was defined in (3.1).We then get Let η = n −γ(1−ε)(d−2)/2 log n and set F to be the event Then we obtain Substituting the value of γ = 2α − 1 − ε we see that taking α > 3 4 (d − 2 3 )/(d − 1) and ε > 0 sufficiently small gives b 3 = o(1).This concludes the proof.
Acknowledgements.We thank Jason Miller and Ofer Zeitouni for helpful discussions.
Following the notation of[AGG89] we write