key: cord-0649575-80ebtfwr
authors: Sun, Hongyan; Wang, Hua-Ming
title: On a maximum of nearest-neighbor random walk with asymptotically zero drift on lattice of positive half line
date: 2020-04-26
journal: nan
DOI: nan
sha: e689ad15ee92179dccb3d7f0b702bd4bf6ff967a
doc_id: 649575
cord_uid: 80ebtfwr

Consider a nearest-neighbor random walk with certain asymptotically zero drift on the positive half line. Let $M$ be the maximum of an excursion starting from $1$ and ending at $0.$ We study the distribution of $M$ and characterize its asymptotics, which is quite different from those of simple random walks.

Let X = {X k } k≥0 be a Markov chain on Z + := {0, 1, 2, ...} starting from some x 0 ∈ Z + and with transition probabilities P (X k+1 = 1|X k = 0) = p 0 = 1 P (X k+1 = n + 1|X k = n) = p n ∈ (0, 1), P (X k+1 = n − 1|X k = n) = q n := 1 − p n , n ≥ 1.

Unless otherwise stated, we always assume x 0 = 1. For i ≥ 1 set

which we will work with throughout.

Next we define the so-called maximum we concern for the chain X. Let

where and throughout, we use the convention inf φ = ∞. Clearly, D is the time that the walk X hits X 0 − 1 for the fist time and M is the maximum that the walk has ever reached in the time interval [0, D).

Our aim is to compute the exact distribution of M and study its asymptotics. If for n ≥ 1, p n ≡ p ∈ (0, 1), then X is just a simple random walk and in this case for all i ≥ 1, ρ i ≡ ρ := q/p. In this situation, the asymptotical distribution of M on the event {D < ∞} is as the following proposition.

Proposition 1. Suppose that p i ≡ p ∈ (0, 1), i ≥ 1 and let ρ := 1−p p . Then

Proposition 1 has nothing new and can be easily deduced from the following Corollary 1. Clearly, if X is recurrent, then P (D < ∞) = 1. Taking the wellknown recurrence criterion for simple random walk into account, we come to the conclusion that, as n → ∞, (a) if ρ = 1, X is null recurrent and P (M = n) decays polynomially;

(b) if ρ < 1, X is transient and P (M = n, D < ∞) decays exponentially;

(c) if ρ > 1, X is positive recurrent and P (M = n) decays exponentially.

One asks naturally that, besides the above polynomial and exponential speeds, are there any situations, under which P (M = n, D < ∞) decays with other speed, for example, n −s , s > 0 or (n log β n) −1 , β > 1, et al?

The answer is affirmative. Adding some perturbation on the recurrent random walk, we get a near-recurrent random walk, known as Lamperti random walk which dates back to Harris [4] and Lamperti [7] and is extensively studied by many other authors, refer for example, to [1, 2, 5, 6, 8, 10] etc. To introduce Lamperti random walk, we borrow the perturbations from [1] .

For fixed B ∈ R and K = 1, 2, ... set

The following fact, parts of which can be found in [1] , page 208, is a corollary of Lemma 2 and Lemma 3 below.

In the rest of the paper, unless otherwise stated, c is a strictly positive constant which may change from line to line. (2), we get (3) and vice versa. II) We see that the asymptotics of M are quite different from those of the simple random walk in (1). For example, fix K = 1 and let p i = 1/2 + r i , i ≥ 1. Let us consider the null recurrent case. That is −1 ≤ B ≤ 1. If B = 1, P (M = n) decays with speed c n log 2 n , but if −1 ≤ B < 1, P (M = n) decays with speed c n 2−B . We conclude that even for null recurrent case, the decay speeds are quite different among different B. While, for simple null-recurrent random walk, P (M = n) always decays with polynomial speed cn −2 .

The rest of the paper is arranged as follows. In Section 2, firstly, we describe the distribution of M by the escape probability, which can be written as the function of ρ 1 · · · ρ j , j ≥ 1. Secondly, we study the limit behaviors of ρ 1 · · · ρ n as n → ∞, which are essential for proving Theorem 1 and the recurrence criterion. Finally, Theroem 1 is proved in Section 3.

It is easily seen that

By Markov property, we have for a < k < b,

Solving the homonic system (5) with the boundary condition (4), we get the following lemma, which is very standard and can found in many documents, see for example [9, 11] .

Here and throughout the paper, we always assume that empty product equals 1 and empty sum equals 0.

On the event {D < ∞}, the distribution of M can be deduced from Lemma 1 directly. Corollary 1. For the chain X, we have P (M = n,D < ∞) = (1 − P 1 (0, n, −))P n (0, n + 1, −)

Both the asymptotics of M and the recurrent criterion of X rely on the limit behavior of ρ 1 · · · ρ n , as n → ∞, which we will give in the next lemma. Lemma 2. Fix K = 1, 2, .. and B ∈ R. (a) If p i = 1 2 + r i , i ≥ 1, then ρ 1 · · · ρ n ∼ c n log n log log n · · · log K−2 n(log K−1 n) B , as n → ∞.

(b) If p i = 1 2 − r i , i ≥ 1, then ρ 1 · · · ρ n ∼ cn log n log log n · · · log K−2 n(log K−1 n) B , as n → ∞.

Before proving Lemma 2, let us study the recurrent criterion of X at first. Define another Markov chain X ′ = {X ′ k } k≥0 , which starts from some x ′ 0 ∈ Z + , with transition probabilities

In literatures, if q ′ n ≡ p n for all n ≥ 1, X ′ is called the adjoint chain of X and vise versa. We have the following recurrence criterion, which can be found in Derriennic [3] , page 204-205. It is easily seen from (6) that X is transient or recurrent according as ∞ j=1 ρ 1 · · · ρ j < ∞ or = ∞ respectively. Thus, with Lemma 3 in hands, the recurrence criterion for X in Section 1 follows from (7) since ∞ j=1 ρ 1 · · · ρ j converges if and only if B > 1.

Proof of Lemma 2. Case 1: Assume p i = 1/2 + r i , i ≥ 1. Then, ρ n = q n p n = 1 2 − r n 1 2 + r n = 1 − 4r n + 8r 2 n + o(r 2 n ), as n → ∞.

By the mean value theorem, there exists θ n between 0 and ρ n − 1 such that log(ρ n ) = (ρ n − 1) − 1 2(1 + θ n ) 2 (ρ n − 1) 2 .

Since ρ n − 1 + 4r n = 8r 2 n + o(r 2 n ) and r 2 n ∼ 1 16n 2 as n → ∞,

Let f (x) := 1 x + 1

x log x + ... + 

When K = 1, B < 0, let f (x) := − B x , a similar argument as above also yields (11) . From (8), (9) and (11), we get

converges to some constant as n → ∞. But n M 1

x

x log x · · · log K−1 x dx = log n + log log n + ... + log K−1 n + B log K n − (log M + log log M + ... + log K−1 M + B log K M).

Therefore, ρ 1 ...ρ n ∼ c n log n... log K−2 n(log K−1 n) B , as n → ∞.

The result can be deduced directly from Case 1. 

Note that when K > 1 or K = 1, B ≥ 0, x → 1

x log x... log K−2 x(log K−1 x) B is a decreasing function in [n 0 , ∞), for some integer n 0 ≥ i 0 depending on K and B.

Therefore, by a similar argument as the proof of (11), we get

which coincides with (13) whenever K = 1, B < 0.

From Corollary 1, (12), (13) and (14), it follows that for K ≥ 1 and B ≤ 1, On the other hand,

Note that If K = 1 and B > −1,

Note that x → x log x... log K−2 x(log K−1 x) B is monotone in [k 0 , ∞), for some integer k 0 ≥ i 0 depending on K and B. Using the same argument as the proofs of (13) and (14), we see that for K > 1 or K = 1, B ≥ −1, as n → ∞. Case 2 is proved and so is Theorem 1.

The programme is carried out during the outbreak of COVID 2019 epidemic in the world. We would like to thank all people, especially those medical workers, over the world who fight against the virus. Without their sacrifice and dedication, we cannot be so safe. Also, we are very grateful to Prof. W.M. Hong who introduces to us the Lamperti problem.

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