Large deviations for transient random walks with asymptotically zero drifts

1.   Introduction

Consider a nearest-neighbor random walk defined as follows. $X = \{X_{n}\}_{n\geq0}$ is a Markov chain on ${{\mathbb N}} = \{0, 1, 2, \cdots\}$ with $X_{0} = 0$ and for $n\geq1,$ the transition probabilities are

where $\alpha, B > 0.$ This random walk $X$ describes the motion of particles that starts at zero, moves on the nonnegative integers, and goes away from $0$ with a larger probability than in the direction of $0$. Obviously, $\frac{B}{i^{\alpha}}$ goes to $0$ as $i\rightarrow \infty$. This means that the state $0$ has a repelling power that decreases as the particle moves away from $0$ (see ^[4]). This is a special case of the so-called Lamperti's random walk ^[7,9].

The transience and recurrence for $X$ are well-known results in the literature (e.g., Chung ^[3]). For $i\geq1$, denote

Theorem A (^[3]) The random walk $X$ defined in (1.1) is transient if and only if

However, this criterion does not explicitly reveal the transient/recurrent classification for sequences of the form $\frac{B}{i^{\alpha}}$. Lamperti (^[7,9]) proved a more general theorem regarding the recurrence and transience of real nonnegative processes. Here we explicitly characterize the types of $\frac{B}{i^{\alpha}}$ sequences and discuss their implications for the transience and recurrence of the random walk $X$. These results also follow straightforwardly from Theorem A.

Transience criterion. For sufficiently large $i$:

● When $0 < \alpha < 1$, the random walk $X$ is transient if $B > 0$ and recurrent if $B < 0$.

● When $\alpha = 1$, the random walk $X$ is transient if $B > \frac{1}{4}$ and recurrent if $B\leq\frac{1}{4}$.

Numerous studies have investigated the limiting behavior of $X$ depending on the sequence $\frac{B}{i^{\alpha}}$. Lamperti ^[8] established the weak convergence of $X$ to a Bessel process. The law of the iterated logarithm for $X$ was provided in ^[1,12]. In ^[13], Voit proved the law of large numbers for $X$, which we restate here in our specific framework.

Theorem B (^[13]) If $E_{i} = \frac{B}{i^{\alpha}}$, where $0 < \alpha < 1$ and $B > 0$, then

There is a minor mistake in ^[13] about the limit value. Hong&Yang ^[6] gave the correct form and provided the limit of $T_{n}$, which is defined as follows.

For $n\geq0$, denote

Theorem C (^[6]) If $0 < \alpha < 1$ and $B > 0$, then

On this basis, we investigate the large deviations for the sequences of hitting times $\{T_{n}, n\geq1\}$. Additionally, for $t\in{{\mathbb N}}$, let

be the maximum of the random walk $X$ up to time $t$. Note that $T_{n}$ defined in (1.2) is the inverse process of $M_{t}.$ Observing the relationship between $M_{t}$ and $T_{n}$, we study the limit theorems and large deviations for $M_{t}$.

The paper is organized as follows. Section 2 presents the main results. Section 3 provides auxiliary results required for the proofs. The proofs of the main theorems are contained in Section 4.

2.   Main results

This section presents the main results of the paper. It is divided into three parts. The first part establishes the law of large numbers for $M_{t}$, and the second part addresses the large deviation principles for $T_{n}$. Finally, we derive the large deviation principles for $M_{t}$.

Theorem 2.1. (Law of Large Numbers for $M_{t}$) When $0 < \alpha < 1$ and $B > 0$, we have

Po-Ning Chen ^[2] established a generalization of the Gärtner–Ellis Theorem for arbitrary random sequences. All subsequent definitions in this section are adapted from reference ^[2].

Definition 2.1. Define

and

The sup-large deviation rate function of $\{T_n\}_{n = 0}^{\infty}$ is defined as

and the inf-large deviation rate function is defined as

Actually, we have $\overline{\Lambda}(\lambda)\geq\underline{\Lambda}(\lambda) > -\infty$ for every $\lambda\in {{\mathbb R}}$ (see Property 2). So the ranges of the supremum operations in (2.1) and (2.2) are always $\{\lambda\in{{\mathbb R}}\}$. Hence, $\overline{\Lambda}^{\ast}(x)$ and $\underline{\Lambda}^{\ast}(x)$ are always defined.

Definition 2.2. Define the sup-Gärter–Ellis set as

where

Define the inf-Gärter–Ellis set as

where

Let us briefly remark on the sup-Gärter–Ellis set defined above. The definitions of $\overline{\mathscr{G}}$ and $\underline{\mathscr{G}}$ are special cases of Definitions 3.4 and 3.5 in ^[2], which only require $h(x) = x$. By Property 2, we can see that $\overline{\Lambda}^{\prime}(\lambda)$ and $\underline{\Lambda}^{\prime}(\lambda)$ exist for $|\lambda| \leq 2B^{2}$. So it can be derived that $\{x = \overline{\Lambda}^{\prime}(\lambda):|\lambda| \leq 2B^{2}\}\subset\overline{\mathscr{G}}$ and $\{x = \underline{\Lambda}^{\prime}(\lambda):|\lambda| \leq 2B^{2}\}\subset\underline{\mathscr{G}}.$

Theorem 2.2. (Large Deviation Principles for $T_{n}$)

(1) For any closed set $F\subset{{\mathbb R}}$, we have

and

(2) For any $(a, b)\subset\overline{\mathscr{G}}$, we have

(3) For any $(a, b)\subset\underline{\mathscr{G}}$, we have

Theorem 2.3. (Large Deviation Principles for $M_{t}$) Define

Then,

(1) for any closed set $F\subset{{\mathbb R}}$,

and

(2) for any open set $G$,

where $\underline{\mathscr{G}}^{o}$ and $\overline{\mathscr{G}}^{o}$ represent the interior of $\underline{\mathscr{G}}$ and $\overline{\mathscr{G}}$ respectively.

3.   Preliminaries

Property 3.1. $\overline{\Lambda}^{\ast}(x)$ and $\underline{\Lambda}^{\ast}(x)$ are the sup- and inf-large deviation rate functions of $\{T_n\}_{n = 0}^{\infty}$ respectively. Denote $\overline{x} = \frac{1}{2B(1+\alpha)}$, then

(1) $\overline{\Lambda}^{\ast}(x)$ and $\underline{\Lambda}^{\ast}(x)$ are always defined;

(2) $\overline{\Lambda}^{\ast}(x)$ and $\underline{\Lambda}^{\ast}(x)$ are both convex rate function;

(3) $\overline{\Lambda}^{\ast}(x)$ is continuous over $\{x\in {{\mathbb R}}:\; \overline{\Lambda}^{\ast}(x) < \infty\}$. Likewise, $\underline{\Lambda}^{\ast}(x)$ is continuous over $\{x\in {{\mathbb R}}:\; \underline{\Lambda}^{\ast}(x) < \infty\}$;

(4) $\overline{\Lambda}^{\ast}(\overline{x}) = \underline{\Lambda}^{\ast}(\overline{x}) = 0.$

Before proving Property 3.1, we provide some preparatory works.

Lemma A (^[11], Lemma 17) Let $p = (p_{i})_{i\in\mathbb{Z}}$ and $q = (q_{i})_{i\in\mathbb{Z}}$ such that

Then we can construct the random walks $\{X_{n}, n\in\mathbb{N}\}$ and $\{Y_{n}, n\in\mathbb{N}\}$, respectively associated with $p$ and $q$, and starting from any common point $d\in\mathbb{Z}$, such that

In ^[14], Ke Zhou provided explicit expressions for the generating functions of hitting times of the skip-free Markov chain on $\mathbb{Z}^{+}$. The chain's upward jumps are restricted to unit size; moreover, it starts at state $0$ and is absorbed by state $d$. The result relevant to our case is stated as follows.

Assume that the random walk defined in 1.1 is absorbed by state $d$ and that $p_{i}\equiv p\in(0, 1)$ for $1\leq i\leq d-1$. We denote this random walk as $X^{*}$. Let $P$ be the transition probability matrix of $X^{*}$, and let $\tau_{d-1, d}$ denote the hitting time of state $d$ when starting from state $d-1$.

Lemma B (^[14]) Denote $f_{d-1}(s)$ as the generating function of $\tau_{d-1, d}$; then we have

where matrix $A_{i}(s)$ is the first $i$ rows and first $i$ lines of $I_{d+1}-P$ for $i = d-1, d$. $I_{d+1}$ is $(d+1)\times (d+1)$ unit matrix.

Actually, Lemma B can be derived from the proof of Theorem 1.1 in ^[14]. Specifically, we have

where $\varphi_{d}(s)$ is a symbol defined in ^[14], representing the generating function of the hitting time from state $0$ to state $d$. We omit further details here.

Lemma 3.1. Let $q = 1-p$; we have

where $\eta, \beta$ are the roots of equation $x^{2}-x+pqs^{2} = 0$, and $g_{2}(s) = 1-s^{2}q, \; g_{3}(s) = 1-s^{2}q-pqs^{2}$.

Proof. We will prove this result using mathematical induction. We can easily calculate that $\det{[A_{2}(s)]} = 1-s^{2}q = g_{2}(s)$ and $\det{[A_{3}(s)]} = 1-s^{2}q-pqs^{2} = g_{3}(s)$. Clearly, (3.1) holds for $d = 3$. Assume (3.1) is also hodes for $d = k$. To calculate $\det{[A_{k+1}(s)]}$, we expand along the last row and obtain

By the quadratic formula, we have

satisfying $\eta\beta = pqs^{2}$ and $\eta+\beta = 1$. Therefore, for $d = k+1$, we have

that is to say, (3.1) is true for $d = k+1$. In conclusion, (3.1) is true for all $d$. This completes the inductive step. □

The assumptions $0 < \alpha < 1, B > 0$ will be used throughout the paper. Recall the definition of $T_{n}$ in (1.2); let

Lemma 3.2. For $\lambda \leq 2B^{2}$, we have

uniform about $1 \leq i \leq n$.

Proof. By Lemma A, let $p^{(1)} = (p_{1}, p_{2}, \cdots, p_{i-1})$ and $p^{(2)} = (p_{i}, \cdots, p_{i})$ be $(i-1)$-dimensional vectors. Obviously, for every $1\leq j \leq i-1$, $p_{j} > p_{i}$. We construct the random walks $\{X^{(1)}_{n}\}_{n\in{{\mathbb N}}}$ and $\{X^{(2)}_{n}\}_{n\in{{\mathbb N}}}$, associated with $p^{(1)}$ and $p^{(2)}$ respectively, starting from $i-1$, reflected at $0$, and absorbed at $i$, such that

For $j = 1, 2$, denote $\tau^{(j)}_{i} = \inf\{n\in{{\mathbb N}}, \; X^{(j)}_{n} = i\}$ as the passage time of $X^{(j)}$ starting from $i-1$ and ending at $i$. Then

implying $Es^{\tau^{(1)}_{i}}\leq Es^{\tau^{(2)}_{i}}$. By Lemma B and Lemma 3.1, the generating function of $\tau^{(2)}_{i}$ is:

where $\eta, \beta$ are the roots of equation $x^{2}-x+p_{i}q_{i}s^{2} = 0$, and

By the definitions of $X_{n}$ in (1.1) and $\tau_{i}$ in (3.2), the random variables $\tau_{i}^{(1)}$ and $\tau_{i}$ share the same distribution. Consequently, their generating functions $Es^{\tau_{i}} = Es^{\tau^{(1)}_{i}}\leq Es^{\tau^{(2)}_{i}}.$ Letting $s = e^{\frac{\lambda}{n^{2\alpha}}}$ and considering sufficiently large $n$ large enough, we employ the representation $\tau_{i}^{(2)}$ from (3.3) to derive that

Hence, to finish the proof of the lemma, we just need to verify that $Ee^{\frac{\lambda}{n^{2\alpha}}\tau_{n}^{(2)}}\rightarrow 1$, as $n\rightarrow \infty$. To ensure the existence of $\eta, \beta$, we require

where $p_{n} = 1-q_{n} = \frac{1}{2}+\frac{B}{n^{\alpha}}$. In other words,

Now,

so $\lim_{n\rightarrow \infty}\gamma^{n-3} = 0$.

Additionally, note that $as\; n\rightarrow \infty$,

From this, we can conclude that

Similarly, we have

Hence, for $\lambda\leq2B^{2}$,

□

Lemma 3.3. When $|\lambda|\leq2B^{2}$, the series

are convergent.

Proof. Indeed,

Therefore, when $|\frac{\lambda}{4B^{2}}| < \frac{1}{2}$, i.e., $|\lambda| < 2B^{2}$, the series

converges.

When $\lambda = 2B^{2}$,

By the label discriminant, we can conclude that the series $\sum\limits_{j = 1}^{\infty}\frac{(2B^{2})^{j}}{j!}\frac{(2j-3)!!}{(4B^{2})^{\frac{2j-1}{2}}}$ is convergent.

Similarly, when $\lambda = -2B^{2}$, the series $\sum\limits_{j = 1}^{\infty}\frac{(-2B^{2})^{j}}{j!}\frac{(2j-3)!!}{(4B^{2})^{\frac{2j-1}{2}}}$ is also convergent. □

Lemma 3.4. Suppose $0 < \alpha < 1$; then we have

for $|\lambda| \leq 2B^{2}$, where

Proof. Recalling that the random variables $\tau_i, \; i\geq1$, defined in (3.2), are independent, we observe that for any fixed $\lambda\leq 2B^{2}$, the following holds:

Hence,

where,

Furthermore, by (3.3), we obtain

Consequently, for all $1\leq i\leq n$, it follows that

Hence, $L(j)\leq U(j)\leq\frac{(2j-3)!!}{(4B^{2})^{\frac{2j-1}{2}}}$.

By Lemma 3.3, we have $-\infty < \underline{\Lambda}(\lambda)\leq\overline{\Lambda}(\lambda) < \infty$ for $|\lambda|\leq2B^{2}$. □

Proposition 3.1. For $j = 1, 2$, $L(j) = U(j).$

Proof. For $j = 1$, we have

so, $L(1) = U(1) = \frac{1}{2B(1+\alpha)}$.

For $j = 2$, we have

By Proposition 2.1 ^[6],

Hence,

According to Theorem1.2 ^[6],

The combination of (3.4, 3.5, and 3.6) yields:

Hence, $L(2) = U(2) = \frac{1}{4B^{3}(1+3\alpha)}$. □

Property 3.2. $\overline{\Lambda}^{\prime}(\lambda)$ and $\underline{\Lambda}^{\prime}(\lambda)$ exist for $|\lambda| \leq 2B^{2}$.

Proof. By Lemma 3.4, $\overline{\Lambda}(\lambda)$ and $\underline{\Lambda}(\lambda)$ can be expressed as power series when $|\lambda|\leq 2B^{2}$. According to the properties of power series, it follows that $\overline{\Lambda}^{\prime}(\lambda)$ and $\underline{\Lambda}^{\prime}(\lambda)$ exist for $|\lambda| \leq 2B^{2}$. □

Proposition 3.2. For every $\lambda\in {{\mathbb R}}$, $\overline{\Lambda}(\lambda)\geq\underline{\Lambda}(\lambda) > -\infty$.

Proof. Since for every $n\geq 0$, $T_{n}$ is non-negative, we have $\underline{\Lambda}(\lambda) > -\infty$ for every $\lambda\in {{\mathbb R}}^{+}$. Let $\mathscr{D}_{\underline{\Lambda}}\stackrel{\triangle}{ = }\{\lambda:\underline{\Lambda}(\lambda) > -\infty\}$, and let $\mathscr{D}_{\underline{\Lambda}}^{o}$ be the interior of $\mathscr{D}_{\underline{\Lambda}}$. By Lemma 3.4, we know that $0 \in \mathscr{D}_{\underline{\Lambda}}^{o}$, so there exists $\lambda_{0} < 0$ such that $\underline{\Lambda}(\lambda_{0}) > -\infty$. Hence, for every $\lambda < \lambda_{0}$, by Jensen's inequality we have the following:

So,

Hence, we have $\overline{\Lambda}(\lambda)\geq\underline{\Lambda}(\lambda) > -\infty$ for every $\lambda\in {{\mathbb R}}$. □

The proof of Property 3.1

By Proposition 3.2, it follows that $\overline{\Lambda}^{\ast}(x)$ and $\underline{\Lambda}^{\ast}(x)$ are always defined. Properties (2) and (3) follow from the results in ^[2]. Specifically, $\overline{x} = \frac{1}{2B(1+\alpha)}$ is the limiting value of $\frac{T_{n}}{n^{1+\alpha}}$ as stated in Theorem C. The idea of the proof of (4) comes from Lemma2.2.5 in ^[5]. For all $\lambda\in {{\mathbb R}}$, by Jensen's inequality,

Since $\overline{\Lambda}(0) = 0$, we obtain $\overline{\Lambda}^{\ast}(\overline{x}) = \sup\limits_{\{\lambda\in{{\mathbb R}}\}} \{\lambda \overline{x}-\overline{\Lambda}(\lambda)\} = 0$. Similarly, we also have $\underline{\Lambda}^{\ast}(\overline{x}) = 0$. □

Proposition 3.3. For all $x\geq\overline{x}$

is a non-decreasing function. Similarly, for all $x\leq \overline{x},$

is a non-increasing function.

Proof. For every $x\geq\overline{x}$ and every $\lambda < 0$,

so $\overline{\Lambda}^{\ast}(x) = \sup\limits_{\{\lambda\geq0\}}\{\lambda x-\overline{\Lambda}(\lambda)\}.$ This also implies the monotonicity of $\overline{\Lambda}^{\ast}(x)$ on $(\overline{x}, \infty)$, since for every $\lambda\geq 0$, the function $\lambda x-\overline{\Lambda}(\lambda)$ is nondecreasing as a function of $x$.

For every $x\leq\overline{x}$ and every $\lambda > 0$,

so $\overline{\Lambda}^{\ast}(x) = \sup\limits_{\{\lambda\leq0\}}\{\lambda x-\overline{\Lambda}(\lambda)\}.$ This also implies the monotonicity of $\overline{\Lambda}^{\ast}(x)$ on $(-\infty, \overline{x})$, since for every $\lambda\leq 0$, the function $\lambda x-\overline{\Lambda}(\lambda)$ is nonincreasing as a function of $x$.

Similarly, we can also obtain analogous properties for $\underline{\Lambda}^{\ast}(x).$ □

4.   Proof of main results

Proof of Theorem 2.1

Let $k_{n}$ be the unique (random) integers such that

Since $T_{k_{n}}$ represents the time when the random walk first reaches position $k_{n}$, and $T_{k_{n}}\leq n$, the maximum position $M_{n}$ must satisfy $M_{n}\geq k_{n}$. Similarly, because $n\leq T_{k_{n}+1}$ and the definition of $T_{k_{n}+1}$, it follows that $M_{n}\leq k_{n}+1$. Therefore, we conclude $k_{n}\leq M_{n} \leq k_{n}+1$. Hence,

Note that Theorem C states that $\lim_{n\rightarrow \infty}\frac{T_{k_{n}}}{(k_{n})^{1+\alpha}} = \lim_{n\rightarrow \infty}\frac{T_{k_{n}+1}}{(k_{n})^{1+\alpha}} = \frac{1}{2B(1+\alpha)}$. As a consequence, dividing both sides of inequality (4.1) by $(k_{n})^{1+\alpha}$ yields $\lim_{n\rightarrow \infty}\frac{n}{(k_{n})^{1+\alpha}} = \frac{1}{2B(1+\alpha)}$. Thus,

The proof is completed.

Proof of Theorem 2.2

The idea originates from the proof of Cramér's Theorem in ^[5] and Theorem 3 in ^[10]. Let $F$ be a non-empty closed set. Note that (2.3) holds trivially if

Assume instead that

Since $\overline{\Lambda}^{\ast}(\overline{x}) = 0$ (see Property 3.1), $\overline{x}$ must lie in the open set $F^{c}$. Let $(x_{-}, x_{+})$ denote the union of all open intervals $(a, b)\subset F^{c}$ containing $\overline{x}.$ Observe that $x_{-} < x_{+}$ and at least one of $x_{-}$ or $x_{+}$ must be finite (since $F$ is non-empty).

(1) If $x_{-}$ is finite and $x_{+} = +\infty$, then $x_{-}\in F\subset (-\infty, x_{-})$. Consequently,

For every $\lambda\leq0$,

Observe that the random variable $\exp\{n^{1-\alpha}\lambda(\frac{T_{n}}{n^{1+\alpha}}-x_{-})\}\geq1$ on the set $\{\frac{T_{n}}{n^{1+\alpha}}-x_{-}\leq0\}, for\; \lambda\leq0$, we used Chebyshev's inequality in the last inequality above. Next, by taking the logarithm on both sides of the above inequality and considering the upper limit with proper scaling, we obtain

The above inequality holds for all $\lambda\leq0$; consequently,

where the last equality follows from Proposition 3.3. The final inequality holds trivially since $x_{-}\in F$.

By a similar argument, if $x_{+}$ is finite and $x_{-} = -\infty$, then

(2) If $x_{-}, x_{+}$ are all finite, then $x_{-}, x_{+}\in F$, and $x_{-} < \overline{x} < x_{+}$.

Combining (4.3) with (4.4), we obtain

In summary, we have completed the proof of (2.3). The proof of (2.4) follows by taking the limit inferior in (4.2), (4.3), (4.4) and (4.5).

The large deviations lower bound of $T_{n}$ (statements (2) and (3) in Theorem 2.2) follow directly from Theorem3.5-3.6 in ^[2] with $h(x) = x$; therefore, we omit the details here.

Proof of Theorem 2.3

Let $v_{0} = (2B(1+\alpha))^{\frac{1}{1+\alpha}}$, which is the limit value of $M_{n}/n^{\frac{1}{1+\alpha}}$ in Theorem 2.1. For every $v > v_{0}$, i.e. $\frac{1}{v^{1+\alpha}} < \frac{1}{v_{0}^{1+\alpha}} = \frac{1}{2B(1+\alpha)} = \overline{x}$, note that

The event $M_{n}\geq n^{\frac{1}{1+\alpha}} v$ implies that the random walk has reached $n^{\frac{1}{1+\alpha}} v$ before time $n$, and thus

where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$, and $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. Taking logarithms on both sides of the inequality and taking the upper limit with proper scaling yields

where the last inequality follows from (4.3), noting that $\lim_{n\rightarrow \infty}\frac{n}{(\lfloor n^{\frac{1}{1+\alpha}}v \rfloor)^{1+\alpha}} = \frac{1}{v^{1+\alpha}}$ and the continuity of $-\overline{\Lambda}^{\ast}(x)$ (see Property 3.1). Multiplying both sides by $v^{1-\alpha}$ gives

Next, we derive an upper bound for $P(M_{n}/n^{\frac{1}{1+\alpha}}\leq v)$, where $v < v_{0}$ (i.e., $\frac{1}{v^{1+\alpha}} > \frac{1}{v_{0}^{1+\alpha}} = \overline{x}$). Observe that

Thus, (4.4) implies

Consequently,

The remainder of the proof follows similarly to the large deviations upper bound for $T_{n}$ (Theorem 2.2). We have completed the proof of part (1) in Theorem 2.3.

Next, we consider the large deviation lower bound for $M_{n}$. For $v < v_{0}$ (i.e., $\frac{1}{v^{1+\alpha}} > \overline{x}$) satisfying $\frac{1}{v^{1+\alpha}}\in \overline{\mathscr{G}}^{o}$, there exists a neighborhood $(\frac{1}{v^{1+\alpha}}-\delta, \frac{1}{v^{1+\alpha}}+\delta)\subset\overline{\mathscr{G}}^{o}$. For any $0 < \epsilon < \delta/2$, we have

for sufficiently large $n$. By Theorem 2.2, we obtain

Since $\frac{1}{v^{1+\alpha}}\in \overline{\mathscr{G}}^{o}$, $\overline{\Lambda}^{\ast}(\frac{1}{v^{1+\alpha}}) < \infty$, and $\overline{\Lambda}^{\ast}(x)$ is continuous at $\frac{1}{v^{1+\alpha}}$ (see (3) in Property 3.1), letting $\epsilon\rightarrow 0$ yields

Combining this with (4.6), we conclude

Similarly, for $v < v_{0}$, satisfying $\frac{1}{v^{1+\alpha}}\in \underline{\mathscr{G}}^{o}$, we have

Hence, for any $v$ with $\frac{1}{v^{1+\alpha}} \in \overline{\mathscr{G}}^{o}\cap \underline{\mathscr{G}}^{o}$, there exists a neighborhood $(\frac{1}{v^{1+\alpha}}-\delta, \frac{1}{v^{1+\alpha}}+\delta)\subset\overline{\mathscr{G}}^{o}\cap \underline{\mathscr{G}}^{o}$. Assume $v < v_{0}$ (the case $v > v_{0}$ is analogous). Choose $\delta_{1} < \delta$ and $\delta_{2} < \delta$, combining (4.7) with (4.8), we obtain

Hence, for any $\epsilon^\prime > 0$ and sufficiently large $n$,

Thus,

Since $\overline{I}(v-\delta_{1}) > \underline{I}(v+\delta_{2})$ (by (4.9)); choosing $\epsilon^\prime, \delta_{1}$, and $\delta_{2}$ such that $\overline{I}(v-\delta_{1})-\underline{I}(v+\delta_{2})-2\epsilon^\prime > 0$ yields

By the continuity $\underline{I}(v)$, for any $\epsilon > 0$, we can choose $\delta_{2}$ sufficiently small such that $\underline{I}(v+\delta_{2}) < \underline{I}(v)+\epsilon$. Consequently,

Since $\epsilon$ and $\epsilon^\prime$ are arbitrary, we conclude

5.   Conclusions

In recent years, random walks with asymptotic perturbations in transition probabilities have received widespread attention from scholars. These perturbations bring many new phenomena to random walks. For transient near-critical random walks, Voit ^[13] established the law of large numbers for the random walks, showing that the escape velocity of such processes is significantly slower than that of simple random walks. Building on this foundation, we derive large deviation principles for such random walks and demonstrate that their velocity order is substantially sublinear in $n$. This result further indicates that asymptotic perturbations reduce the wandering speed of random walks.

Use of Generative-AI tools declaration

The author declares she has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author thanks the anonymous referees for their meticulous reading of the manuscript. This work was partly supported by the National Natural Science Foundation of China (Grant Nos. 12001558, 71971228, 72371261).

Conflict of interest

The author declares no conflicts of interest in this paper.