2. Results for the ERW in the triangular array setting

Before we state our results, we give a short synopsis of the results for the standard ERW $\{W_n\}$ [Reference Baur and Bertoin3, Reference Bercu4, Reference Coletti, Gava and Schütz6–Reference Guérin, Laulin and Raschel8, Reference Kubota and Takei11, Reference Qin14]. Let $\alpha\,:\!=\,2p-1$.

• • For $\alpha \in ({-}1,1)$ i.e. $p \in (0,1)$,

  (2.1) \begin{align} \lim_{n \to \infty} \dfrac{W_n}{n} = 0\quad\text{almost surely (a.s.) and in}\ L^2. \end{align}

• • For $\alpha \in \big({-}1,\frac12\big)$, i.e. $p \in \big(0,\frac34\big)$,

  (2.2) \begin{align} & \frac{W_n}{\sqrt{n}} \stackrel{\text{d}}{\to} N\bigg(0,\frac{1}{1-2\alpha}\bigg) \quad \text{as}\ n \to \infty, \\[-10pt] \nonumber \end{align}
  (2.3) \begin{align} & \limsup_{n\to\infty} \pm\frac{W_n}{\sqrt{2n\log\log n}\,} = \frac{1}{\sqrt{1-2\alpha}\,} \quad \text{a.s.} \\[6pt] \nonumber \end{align}

• • For $\alpha=\frac12$, i.e. $p = \frac34$,

  (2.4) \begin{align} & \frac{W_n}{\sqrt{n \log n}\,} \stackrel{\text{d}}{\to} N(0,1) \quad \text{as}\ n \to \infty, \\[-10pt] \nonumber\end{align}
  (2.5) \begin{align} & \limsup_{n\to\infty} \pm\frac{W_n}{\sqrt{2n\log n\log\log\log n}\,} = 1 \quad \text{a.s.} \\[6pt] \nonumber \end{align}

• • For $\alpha \in \big(\frac12,1\big)$, i.e. $p \in \big(\frac34,1\big)$, there exists a random variable M such that

  (2.6) \begin{align} \lim_{n \to \infty}\frac{W_n}{n^{\alpha}} = M \quad \text{a.s. and in}\ L^2, \end{align}
  where $\mathbb{E}[M] = {\beta}/{\Gamma(\alpha+1)}$, $\mathbb{E}[M^2] \gt 0$, $\mathbb{P}(M \neq 0)=1$, and
  (2.7) \begin{align} \frac{W_n-Mn^{\alpha}}{\sqrt{n}} \stackrel{\text{d}}{\to} N\bigg(0,\frac{1}{2\alpha-1}\bigg) \quad \text{as}\ n \to \infty. \end{align}

Our first result improves and extends [Reference Gut and Stadtmüller10, Theorem 3.1].

Theorem 2.1. Let $p \in (0,1)$ and $\alpha=2p-1$. Assume that $\{m_n \colon n \in \mathbb{N}\}$ satisfies (1.3), (1.7), and

(2.8) \begin{align} \gamma_n \,:\!=\, \frac{m_n}{n}, \qquad \lim_{n \to \infty} \gamma_n = \gamma \in [0,1]. \end{align}

1. (i) If $\alpha \in \big({-}1,\frac12\big)$, i.e. $p \in \big(0,\frac34\big)$, then

   (2.9) \begin{align} \frac{\gamma_n T_n}{\sqrt{m_n}\,} \stackrel{\mathrm{d}}{\to} N\bigg(0,\frac{\{\gamma+\alpha(1-\gamma)\}^2}{1-2\alpha}+\gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty. \end{align}

2. (ii) If $\alpha = \frac12$, i.e. $p=\frac34$, then

   (2.10) \begin{align} \frac{\gamma_n T_n}{\sqrt{m_n\log m_n}\,} \stackrel{\text{d}}{\to} N\bigg(0,\frac{(1+\gamma)^2}{4}\bigg) \quad \text{as}\ n \to \infty. \end{align}

3. (iii) If $\alpha \in \big(\frac12,1\big)$, i.e. $p \in \big(\frac34,1\big)$, then

   (2.11) \begin{align} \lim_{n \to \infty} \frac{\gamma_n T_n}{(m_n)^{\alpha}} =\{\gamma+\alpha(1-\gamma)\}M \quad \text{a.s. and in}\ L^2, \end{align}
   where M is the random variable in (2.6). Moreover,
   (2.12) \begin{align} \frac{\gamma_n T_n - M\{\gamma_n + \alpha(1-\gamma_n)\}(m_n)^{\alpha}}{\sqrt{m_n}} \stackrel{\mathrm{d}}{\to} N\bigg(0,\frac{\{\gamma+\alpha(1-\gamma)\}^2}{2\alpha-1}+\gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty. \end{align}

The next theorem concerns the strong law of large numbers and its refinement.

Theorem 2.2 Let $p \in (0,1)$ and $\alpha=2p-1$. Assume that $\{m_n \colon n \in \mathbb{N}\}$ satisfies (1.3), (1.7), and (2.8). Then

(2.13) \begin{align} \lim_{n \to \infty} \frac{T_n}{n} = 0 \quad \text{a.s.} \end{align}

Actually, we obtain the following sharper result: If $c\in\big(\max\big\{\alpha,\frac12\big\},1\big)$ then

(2.14) \begin{align} \lim_{n \to \infty}\frac{\gamma_n T_n}{(m_n)^c} = 0 \quad \text{a.s.} \end{align}

3. Proof of Theorem 2.1

Throughout this section we assume that (1.3), (1.7), and (2.8) hold.

Proof. Let $\mathcal{F}_n$ be the $\sigma$-algebra generated by $X_1,\ldots,X_n$. For $n \in \mathbb{N}$, the conditional distribution of $X_{n+1}$ given the history up to time n is

\begin{align*} \mathbb{P}(X_{n+1}= \pm 1 \mid \mathcal{F}_n) & = \frac{\#\{k=1,\ldots,n \colon X_k=\pm 1\}}{n} \cdot p + \frac{\#\{k=1,\ldots,n \colon X_k= \mp 1\}}{n} \cdot (1-p) \\ & = \frac{1}{2}\bigg(1 \pm \alpha \cdot \frac{W_n}{n} \bigg). \end{align*}

For each $n \in \mathbb{N}$, let $\mathcal{G}_{m_n}^{(n)} = \mathcal{F}_{\infty} \,:\!=\, \sigma(\{X_i \colon i \in \mathbb{N} \}) = \sigma(\{X_1\} \cup \{ U_i \colon i \in \mathbb{N} \})$ and

\begin{equation*} \mathcal{G}_k^{(n)} \,:\!=\, \sigma( \{X_i \colon i \in \mathbb{N} \} \cup \{ X_i^{(n)} \colon m_n \lt i \leq k\}) = \sigma(\{X_1\} \cup \{ U_i \colon i \in \mathbb{N} \} \cup \{ U_{i,n} \colon m_n \lt i \leq k\}) \end{equation*}

for $k \in (m_n,n] \cap \mathbb{N}$. From (1.5), we can see that the conditional distribution of $X_k^{(n)}$ for $k \in (m_n,n] \cap \mathbb{N}$ is given by

\begin{equation*} \mathbb{P} (X_k^{(n)} = \pm 1 \mid \mathcal{G}_{k-1}^{(n)}) = \frac{1}{2} \bigg( 1 \pm \alpha \cdot \frac{W_{m_n}}{m_n}\bigg). \end{equation*}

(This corresponds to [Reference Gut and Stadtmüller10, (2.2)].) From this we have that

(3.1) \begin{align} \mathbb{E}[ X_k^{(n)} \mid \mathcal{F}_{\infty}] = \alpha \cdot \frac{W_{m_n}}{m_n} \end{align}

for each $k \in (m_n,n] \cap \mathbb{N}$, and

(3.2) \begin{align} \mathbb{E}[T_n-W_{m_n}\mid\mathcal{F}_{\infty}] = \sum_{k=m_n+1}^{n}\mathbb{E}[X_k^{(n)}\mid\mathcal{F}_{\infty}] = \alpha(n-m_n)\cdot\frac{W_{m_n}}{m_n}. \end{align}

We introduce

(3.3) \begin{align} A_n \,:\!=\, \mathbb{E}[T_n \mid \mathcal{F}_{\infty}], \qquad B_n \,:\!=\, T_n-A_n. \end{align}

Noting that

(3.4) \begin{align} A_n = W_{m_n} + \mathbb{E}[T_n-W_{m_n} \mid \mathcal{F}_{\infty}] = \frac{W_{m_n}}{\gamma_n} \cdot \{ \gamma_n + \alpha(1-\gamma_n) \}, \end{align}

we have

(3.5) \begin{align} \gamma_n T_n = \gamma_n (A_n + B_n) = c_n W_{m_n}+\gamma_n B_n, \end{align}

where $c_n=c_n(\alpha)\,:\!=\,\gamma_n+\alpha(1-\gamma_n)$.

First, we prove Theorem 2.1(i). Assume that $\alpha \in \big({-}1,\frac12\big)$. By (3.4) and (2.2),

\begin{equation*} \frac{\gamma_n A_n}{\sqrt{m_n}\,} = \frac{c_n W_{m_n}}{\sqrt{m_n}} \stackrel{\text{d}}{\to} \{\gamma + \alpha(1-\gamma)\} \cdot N\bigg(0,\frac{1}{1-2\alpha}\bigg) \quad \text{as}\ n \to \infty. \end{equation*}

In terms of characteristic functions, this is equivalent to, for $\xi \in \mathbb{R}$,

(3.6) \begin{align} \mathbb{E}\bigg[\exp\bigg(\frac{\mathrm{i}\xi\gamma_n A_n}{\sqrt{m_n}}\bigg)\bigg] \to \exp\bigg({-}\frac{\xi^2}{2} \cdot \frac{\{\gamma+\alpha(1-\gamma)\}^2}{1-2\alpha}\bigg) \quad \text{as}\ n \to \infty. \end{align}

Now we turn to $\{B_n\}$. Unless specified otherwise, all the results on $\{B_n\}$ hold for all $\alpha \in ({-}1,1)$. Since, for each $n \in \mathbb{N}$, $X_k^{(n)}$ for $k \in (m_n,n] \cap \mathbb{N}$ are independent and identically distributed under $\mathbb{P}(\,\cdot\mid \mathcal{F}_{\infty})$, so

(3.7) \begin{align} B_n=\sum_{k=m_n+1}^n\{ X_k^{(n)}-\mathbb{E}[X_k^{(n)}\mid \mathcal{F}_{\infty}]\} \end{align}

is a sum of centered i.i.d. random variables. The conditional variance of $X_k^{(n)}$ for

(3.8) \begin{align} \mathbb{V}[X_k^{(n)} \mid \mathcal{F}_{\infty}] & = \mathbb{E}[(X_k^{(n)})^2 \mid \mathcal{F}_{\infty}] - (\mathbb{E}[X_k^{(n)} \mid \mathcal{F}_{\infty}])^2 \notag \\ & = \begin{cases} 0 & \text{for}\ k \in [1,m_n] \cap \mathbb{N}, \\ 1 - \alpha^2 \cdot \bigg(\dfrac{W_{m_n}}{m_n}\bigg)^2 & \text{for}\ k \in (m_n,n] \cap \mathbb{N}. \end{cases} \end{align}

We have

(3.9) \begin{align} \mathbb{V}\bigg[\frac{\gamma_n B_n}{\sqrt{m_n}\,} \mid \mathcal{F}_{\infty}\bigg] = \frac{(\gamma_n)^2}{m_n}\cdot(n-m_n)\cdot\bigg\{1-\alpha^2\cdot\bigg(\frac{W_{m_n}}{m_n}\bigg)^2\bigg\} = \gamma_n (1-\gamma_n) \cdot \bigg\{1 - \alpha^2 \cdot \bigg(\frac{W_{m_n}}{m_n}\bigg)^2\bigg\}, \end{align}

which converges to $\gamma(1-\gamma)$ as $n \to \infty$ a.s. by (2.1). Based on this observation, we prove the following result.

Lemma 3.1. For $\gamma \in [0,1]$,

(3.10) \begin{align} \mathbb{E}\bigg[\exp\bigg(\frac{\mathrm{i}\xi\gamma_nB_n}{\sqrt{m_n}}\bigg) \mid \mathcal{F}_{\infty}\bigg] \to \exp\bigg({-}\frac{\xi^2}{2} \cdot \gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty\ \text{a.s.} \end{align}

Proof. Because $B_n$ is the sum (3.7) of centered i.i.d. random variables under $\mathbb{P}(\,\cdot\mid \mathcal{F}_{\infty})$, by (3.8) we have

\begin{equation*} \mathbb{E}\bigg[\exp\bigg(\frac{i\xi\gamma_nB_n}{\sqrt{m_n}}\bigg) \mid \mathcal{F}_{\infty}\bigg] = \bigg[1 - \frac{\xi^2\gamma_n}{2n} \cdot \bigg\{1 - \alpha^2\cdot\bigg(\frac{W_{m_n}}{m_n}\bigg)^2\bigg\} + o\bigg(\frac{\gamma_n}{n}\bigg)\bigg]^{n-m_n} \quad\!\! \text{as}\ n \to \infty\ \text{a.s.} \end{equation*}

Note that ${\gamma_n}/{n} \to 0$ and $({\gamma_n}/{n})\cdot (n-m_n) =\gamma_n(1-\gamma_n) \to \gamma(1-\gamma)$ as $n \to \infty$. Now (3.10) follows from this together with (2.1).

From (3.5), (3.6), and (3.10), together with the bounded convergence theorem, we can see that

\begin{equation*} \mathbb{E}\bigg[\exp\bigg(\frac{i\xi\gamma_nT_n}{\sqrt{m_n}\,}\bigg)\bigg] = \mathbb{E}\bigg[\exp\bigg(\frac{i\xi\gamma_nA_n}{\sqrt{m_n}\,}\bigg) \cdot \mathbb{E}\bigg[\exp\bigg(\frac{i\xi\gamma_nB_n}{\sqrt{m_n}\,}\bigg) \mid \mathcal{F}_{\infty}\bigg]\bigg] \end{equation*}

converges to

\begin{equation*} \exp\bigg({-}\frac{\xi^2}{2} \cdot \frac{\{\gamma+\alpha(1-\gamma)\}^2}{1-2\alpha}\bigg) \cdot \exp\bigg({-}\frac{\xi^2}{2} \cdot \gamma(1-\gamma)\bigg) \end{equation*}

as $n \to \infty$. This gives (2.9).

The proof of Theorem 2.1(ii) is along the same lines as that of (i), and is actually simpler. Assume that $\alpha =\frac12$. As $c_n\big(\frac12\big)=(1+\gamma_n)/2$, from (3.4) and (2.4) we have

\begin{equation*} \frac{\gamma_nA_n}{\sqrt{m_n\log m_n}\,} = \frac{c_n W_{m_n}}{\sqrt{m_n \log m_n}\,} \stackrel{\text{d}}{\to} \frac{1+\gamma}{2} \cdot N(0,1) \quad \text{as}\ n \to \infty. \end{equation*}

Also, from (3.9) and (2.1), we have

(3.11) \begin{align} \mathbb{E}\bigg[\bigg(\frac{\gamma_nB_n}{\sqrt{m_n}\,}\bigg)^2 \bigg] = \gamma_n(1-\gamma_n)\bigg\{1 - \alpha^2\cdot\mathbb{E}\bigg[\bigg(\frac{W_{m_n}}{m_n}\bigg)^2\bigg]\bigg\} \to \gamma(1-\gamma) \quad \text{as}\ n \to \infty. \end{align}

This implies that $\gamma_n B_n/\sqrt{m_n \log m_n} \to 0$ as $n \to \infty$ in $L^2$. By Slutsky’s lemma, we obtain (2.10).

Finally, we prove Theorem 2.1(iii). Assume that $\alpha \in \big(\frac12,1\big)$. By (3.4) and (2.6),

(3.12) \begin{align} \frac{\gamma_nA_n}{(m_n)^{\alpha}} = \frac{c_nW_{m_n}}{(m_n)^{\alpha}} \to \{\gamma+\alpha(1-\gamma)\}M \end{align}

as $n \to \infty$ a.s. and in $L^2$. Noting that $\gamma_n B_n/(m_n)^{\alpha} \to 0$ as $n \to \infty$ in $L^2$, by (3.11) we obtain the $L^2$-convergence in (2.11). From (4.2), which will be proved in the next section, the almost-sure convergence in (2.11) follows. To prove (2.12), by (3.5) we have

(3.13) \begin{align} \gamma_n T_n-c_n \cdot M \cdot (m_n)^{\alpha} = c_n \{W_{m_n}-M \cdot (m_n)^{\alpha}\} + \gamma_n B_n. \end{align}

Note that M is $\mathcal{F}_{\infty}$-measurable. Using (2.7), (3.10), and (3.13), we obtain (2.12) similarly to the proof of (2.9).

4. Proof of Theorem 2.2

In this section we assume that (1.3), (1.7), and (2.8) hold.

Lemma 4.1 For any $\alpha \in ({-}1,1)$ and $\gamma \in [0,1]$,

(4.1) \begin{align} \limsup_{n \to \infty} \frac{\gamma_n B_n}{\sqrt{2\gamma_n(1-\gamma_n)m_n \log n}\,} \leq 1 \quad \text{a.s.} \end{align}

In particular, for any $c \in \big(\frac12,1\big)$,

(4.2) \begin{align} \lim_{n \to \infty} \frac{\gamma_n B_n}{(m_n)^c} = 0 \quad \text{a.s.} \end{align}

Proof. Note that $|X_k^{(n)}-\mathbb{E}[X_k^{(n)}\mid\mathcal{F}_{\infty}]| \leq 1$ for each $1\leq k\leq n$. For $\lambda \in \mathbb{R}$, it follows from Azuma’s inequality [Reference Azuma2] that

\begin{align*} \mathbb{E}[\exp(\lambda\gamma_nB_n) \mid \mathcal{F}_{\infty}] & = \mathbb{E}\Bigg[\exp\Bigg(\lambda\gamma_n\sum_{k=m_n+1}^n \big\{X_k^{(n)}-\mathbb{E}[X_k^{(n)}\mid\mathcal{F}_{\infty}]\big\}\Bigg) \mid \mathcal{F}_{\infty}\Bigg] \\ & \leq \exp((\lambda \gamma_n)^2 (n-m_n)/2), \end{align*}

and

\begin{equation*} \mathbb{P}(|\gamma_n B_n| \geq x ) \leq 2\exp\bigg({-}\frac{x^2}{2\gamma_n(1-\gamma_n)m_n}\bigg) \quad \text{for}\ x \gt 0. \end{equation*}

Taking $x=\sqrt{2(1+\varepsilon) \gamma_n (1-\gamma_n)m_n \log n}$ for some $\varepsilon \gt 0$, we have

\begin{equation*} \sum_{n=1}^{\infty}\mathbb{P}(|\gamma_nB_n| \geq \sqrt{2(1+\varepsilon)\gamma_n(1-\gamma_n)m_n\log n}) \leq \sum_{n=1}^{\infty}\frac{2}{n^{1+\varepsilon}} . \end{equation*}

This, together with the Borel–Cantelli lemma, implies (4.1). To obtain (4.2) note that, for $c \in \big(\frac12,1\big)$,

\begin{equation*} \frac{2\gamma_n(1-\gamma_n)m_n \log n}{(m_n)^{2c}} = \frac{2(1-\gamma_n) \log n}{n(m_n)^{2c-2}} \leq \frac{2(1-\gamma_n)\log n}{n^{2c-1}} \to 0 \quad \text{as}\ n \to \infty, \end{equation*}

where we used $2c-2 \lt 0 \lt 2c-1$ and $m_n \leq n$.

We now prove (2.14) in Theorem 2.2. Equation (2.13) is readily derived from (2.14). For the case $\alpha \in \big({-}1,\frac12\big)$, from (2.3) and (3.4) we have

\begin{equation*} \limsup_{n \to \infty}\frac{\gamma_n A_n}{\sqrt{2m_n \log \log m_n}\,} \leq \frac{\gamma + \alpha(1-\gamma)}{\sqrt{1-2\alpha}\,} \quad \text{a.s.} \end{equation*}

For the case $\alpha=\frac12$, from (2.5) and (3.4) we have

\begin{equation*} \limsup_{n \to \infty}\frac{\gamma_n A_n}{\sqrt{2m_n\log m_n \log \log \log m_n}\,} \leq \frac{1+\gamma}{2} \quad \text{a.s.} \end{equation*}

By (4.2), if $\alpha \in \big({-}1,\frac12\big]$ then (2.14) holds for any $c \in \big(\frac12,1\big)$. As for the case $\alpha \in \big(\frac12,1\big)$, almost-sure convergence in (2.11) follows from (3.12) and (4.2). Thus, (2.14) holds for any $c \in (\alpha,1)$.

5. The ERW with stops in the triangular array setting

Let $s \in [0,1]$, and assume that $p,q,r \in [0,1)$ satisfy $p+q+r=1$. We consider a sequence $X_1,X_2,\ldots$ of random variables taking values in $\{+1, 0, -1\}$ given by

(5.1) \begin{align} X_1 = \begin{cases} +1 & \text{with probability } s, \\ -1 & \text{with probability } 1-s; \end{cases} \end{align}

$\{U_n\colon n \geq 1\}$ a sequence of independent random variables, independent of $X_1$, with $U_n$ having a uniform distribution over $\{1, \ldots, n\}$; and, for $n \in \mathbb{N}$,

(5.2) \begin{align} X_{n+1} = \begin{cases} X_{U_n} & \text{with probability}\ p, \\ -X_{U_n} & \text{with probability}\ q, \\ 0 & \text{with probability}\ r. \\ \end{cases} \end{align}

The ERW with stops $\{W_n\}$ is defined by $W_n= \sum_{k=1}^n X_k$ for $n\in \mathbb{N}$. Note that if $r=0$ then it is the standard ERW defined in Section 1. Hereafter we assume that $r \in (0,1)$.

The ERW with stops was introduced in [Reference Kumar, Harbola and Lindenberg12]. To describe the limit theorems obtained in [Reference Bercu5], it is convenient to introduce the new parameters $\alpha\,:\!=\,p-q$ and $\beta\,:\!=\,1-r$, where $\beta \in (0,1)$ and $\alpha \in [-\beta,\beta]$. Let $\Sigma_n$ be the number of moves up to time n, i.e.

\begin{equation*} \Sigma_n \,:\!=\, \sum_{k=1}^n \mathbf{1}_{\{X_k \neq 0\}} = \sum_{k=1}^n X_k^2 \quad \text{for}\ n\in \mathbb{N}.\end{equation*}

It is shown in [Reference Bercu5] that

(5.3) \begin{align} \lim_{n \to \infty}\frac{\Sigma_n}{n^{\beta}} = \Sigma \gt 0 \quad \text{a.s. and in}\ L^2, \end{align}

where $\Sigma$ has a Mittag–Leffler distribution with parameter $\beta$. We turn to the central limit theorem for $\{W_n\}$ in [Reference Bercu5].

• • For $\alpha \in [-\beta,\beta/2)$,

  (5.4) \begin{align} \frac{W_n}{\sqrt{\Sigma_n}\,} \stackrel{\text{d}}{\to} N\bigg(0,\frac{\beta}{\beta-2\alpha}\bigg) \quad \text{as}\ n \to \infty. \end{align}

• • For $\alpha=\beta/2$,

  (5.5) \begin{align} \frac{W_n}{\sqrt{\Sigma_n \log \Sigma_n}\,} \stackrel{\text{d}}{\to} N(0,1) \quad \text{as}\ n \to \infty. \end{align}

• • For $\alpha \in (\beta/2,\beta]$, there exists a random variable M such that

  (5.6) \begin{align} \lim_{n \to \infty} \frac{W_n}{n^{\alpha}} = M \quad \text{a.s. and in}\ L^2 \end{align}
  and
  (5.7) \begin{align} \frac{W_n-Mn^{\alpha}}{\sqrt{\Sigma_n}} \stackrel{\text{d}}{\to} N\bigg(0,\frac{\beta}{2\alpha-\beta}\bigg) \quad \text{as}\ n \to \infty, \end{align}
  where $\mathbb{P}(M \gt 0) \gt 0$.

Next, we define the sequence $\{T_n\}$ as in (1.6); however, $Y_k^{(n)}$ and $X_k^{(n)}$ of (1.4) and (1.5) are defined with $\{X_i\}$ as in (5.1) and (5.2) instead of (1.1) and (1.2). We call this the ERW with stops in the triangular array setting.

Our first result of this section is an extension of [Reference Gut and Stadtmüller10, Theorem 4.1]. We note here that [Reference Gut and Stadtmüller10] allows $X_1$ to take value 0 with probability r, unlike this paper. As such they have an extra $\delta_0$ in their results for the case $\gamma=0$.

1. (i) If $\alpha \in [-\beta,\beta/2)$ then

   (5.8) \begin{align} \frac{\gamma_n T_n}{\sqrt{\Sigma_{m_n}}\,} \stackrel{\mathrm{d}}{\to} N\bigg(0,\frac{\beta\{\gamma+\alpha(1-\gamma)\}^2}{\beta-2\alpha}+\beta\gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty. \end{align}

2. (ii) If $\alpha = \beta/2$ then

   (5.9) \begin{align} \frac{\gamma_n T_n}{\sqrt{\Sigma_{m_n} \log \Sigma_{m_n}}\,} \stackrel{\mathrm{d}}{\to} N(0,\{\gamma+\beta(1-\gamma)/2\}^2) \quad \text{as}\ n \to \infty. \end{align}

3. (iii) If $\alpha \in (\beta/2,\beta]$ then

   (5.10) \begin{align} \lim_{n \to \infty} \frac{\gamma_n T_n}{(m_n)^{\alpha}} =\{\gamma+\alpha(1-\gamma)\}M \quad \text{in}\ L^2, \end{align}
   where M is the random variable in (5.6). Moreover,
   (5.11) \begin{align} & \frac{\gamma_nT_n- M\cdot\{\gamma_n + \alpha(1-\gamma_n)\}\cdot(m_n)^{\alpha}}{\sqrt{\Sigma_{m_n}}} \notag \\ & \qquad \stackrel{\mathrm{d}}{\to} N\bigg(0,\frac{\beta\{\gamma+\alpha(1-\gamma)\}^2}{2\alpha-\beta}+\beta\gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty. \end{align}

We also consider the process $\{\Xi_n \colon n \in \mathbb{N} \}$ defined by $\Xi_n\,:\!=\, \sum_{k=1}^n \{X_k^{(n)}\}^2$ for $n \in \mathbb{N}$. The next theorem is an improvement of [Reference Gut and Stadtmüller10, Theorem 4.2].

Theorem 5.2. Under the same conditions as in Theorem 5.1, we have

(5.12) \begin{align} \lim_{n\to\infty}\frac{\gamma_n\Xi_n}{(m_n)^{\beta}} = \{\gamma + \beta(1-\gamma)\}\Sigma \quad \text{in}\ L^2, \end{align}

where $\Sigma$ is defined in (5.3).

The strong law of large numbers and its refinement can also be obtained for the ERW with stops.

6. Proof of Theorem 5.1

Proof. Noting that $p=(\beta+\alpha)/2$ and $q=(\beta-\alpha)/2$, for $n \in \mathbb{N}$ we have

\begin{align*} \mathbb{P}(X_{n+1}= \pm 1 \mid \mathcal{F}_n) & = \frac{\#\{k=1,\ldots,n \colon X_k=\pm 1\}}{n} \cdot p + \frac{\#\{k=1,\ldots,n \colon X_k= \mp 1\}}{n} \cdot q \\ & = \frac{1}{2}\bigg(\beta\cdot\frac{\Sigma_n}{n} \pm \alpha\cdot\frac{W_n}{n}\bigg). \end{align*}

For $k \in (m_n,n] \cap \mathbb{N}$, we have

(6.1) \begin{align} \mathbb{P}\big(X_k^{(n)}=\pm 1 \mid \mathcal{G}_{k-1}^{(n)}\big) & = \frac{1}{2}\bigg(\beta\cdot\frac{\Sigma_{m_n}}{m_n} \pm \alpha\cdot\frac{W_{m_n}}{m_n}\bigg), \end{align}

(6.2) \begin{align} \mathbb{P}\big(\{X_k^{(n)}\}^2 = 1 \mid \mathcal{G}_{k-1}^{(n)}\big) & = \beta \cdot \dfrac{\Sigma_{m_n}}{m_n}. \end{align}

From (6.1), we see that (3.1) and (3.2) continue to hold in this setting. Defining $\{A_n\}$ and $\{B_n\}$ by (3.3), we note that they satisfy (3.4) and (3.5).

We prepare a lemma about $\{B_n\}$.

1. (i) For $\alpha \in [-\beta,\beta]$ and $\xi \in \mathbb{R}$,

   (6.3) \begin{align} \mathbb{E}\bigg[\exp\bigg(\frac{\mathrm{i}\xi\gamma_nB_n}{\sqrt{\Sigma_{m_n}}\,}\bigg) \mid \mathcal{F}_{\infty}\bigg] & \to \exp\bigg({-}\frac{\xi^2}{2}\cdot\beta\gamma(1-\gamma)\bigg) \quad \text{as}\ n \to \infty\ \text{a.s.,} \end{align}
   (6.4) \begin{align} \mathbb{E}\bigg[\exp\bigg(\frac{\mathrm{i}\xi\gamma_nB_n}{\sqrt{\Sigma_{m_n}\log\Sigma_{m_n}}\,}\bigg) \mid \mathcal{F}_{\infty}\bigg] & \to 1 \quad \text{as}\ n \to \infty\ \text{a.s.} \end{align}

2. (ii) If $\alpha \in (\beta/2,\beta]$ then $\gamma_n B_n/(m_n)^{\alpha} \to 0$ as $n \to \infty$ in $L^2$.

Proof. Note that $\mathbb{E}[ B_n \mid \mathcal{F}_{\infty}] = 0$. By (6.1) and (6.2),

\begin{equation*} \mathbb{V}[X_k^{(n)} \mid \mathcal{F}_{\infty}] = \beta \cdot \frac{\Sigma_{m_n}}{m_n}- \alpha^2 \cdot \bigg(\frac{W_{m_n}}{m_n}\bigg)^2 \end{equation*}

for $k \in (m_n,n] \cap \mathbb{N}$. As in (3.9), we have

(6.5) \begin{align} \mathbb{V}[B_n \mid \mathcal{F}_{\infty}] = (n-m_n)\cdot\bigg\{\beta\cdot\frac{\Sigma_{m_n}}{m_n}-\alpha^2\cdot\bigg(\frac{W_{m_n}}{m_n}\bigg)^2\bigg\} = \frac{1-\gamma_n}{\gamma_n}\cdot\bigg\{\beta\Sigma_{m_n}-\alpha^2\cdot\frac{W_{m_n}^2}{m_n}\bigg\}. \quad \end{align}

From this,

(6.6) \begin{align} \mathbb{V}\bigg[\frac{\gamma_nB_n}{\sqrt{\Sigma_{m_n}}\,} \mid \mathcal{F}_{\infty}\bigg] = \gamma_n(1-\gamma_n)\cdot\bigg\{\beta - \alpha^2\cdot\frac{(m_n)^{\beta}}{\Sigma_{m_n}} \cdot \bigg(\frac{W_{m_n}}{(m_n)^{(1+\beta)/2}}\bigg)^2\bigg\}. \end{align}

For any $\beta \in (0,1)$ and $\alpha \in [-\beta,\beta]$, we show that

(6.7) \begin{align} \lim_{n \to \infty} \frac{W_{m_n}}{(m_n)^{(1+\beta)/2}} = 0 \quad \text{a.s.} \end{align}

Indeed, if $\alpha \in [-\beta,\beta/2)$ then

(6.8) \begin{align} \limsup_{n \to \infty}\frac{W_n}{\sqrt{2n^{\beta}\log\log n}\,} = \sqrt{\frac{\beta\Sigma}{\beta-2\alpha}} \quad \text{a.s.} \end{align}

by [Reference Bercu5, (3.5)]. If $\alpha=\beta/2$ then

(6.9) \begin{align} \limsup_{n \to \infty}\frac{W_n}{\sqrt{2n^{\beta}\log n\log\log\log n}\,} = \sqrt{\beta\Sigma} \quad \text{a.s.} \end{align}

by [Reference Bercu5, (3.13)]. If $\alpha \in (\beta/2,\beta]$ then $W_{m_n}/(m_n)^{\alpha} \to M$ as $n \to \infty$ a.s. by (5.6), and $(1+\beta)/2 \gt \alpha$ since $2\alpha - \beta\leq \beta \lt 1$. In any case we have (6.7). Since $(m_n)^{\beta}/\Sigma_{m_n}\to 1/\Sigma$ as $n \to \infty$ a.s. by (5.3), we see that (6.6) converges to $\beta \gamma(1-\gamma)$ as $n \to \infty$ a.s., and

\begin{equation*} \mathbb{V}\bigg[\frac{\gamma_nB_n}{\sqrt{\Sigma_{m_n}\log\Sigma_{m_n}}\,} \mid \mathcal{F}_{\infty}\bigg] \to 0 \quad \text{as}\ n \to \infty\ \text{a.s.} \end{equation*}

By a similar computation to Lemma 4.1, we obtain (6.3) and (6.4) in (i).

Next, we consider (ii). By (6.5),

\begin{equation*} \mathbb{E}\bigg[\bigg(\frac{\gamma_nB_n}{(m_n)^{\alpha}}\bigg)^2\bigg] = \gamma_n(1-\gamma_n)\cdot\bigg\{\beta\cdot\frac{\mathbb{E}[\Sigma_{m_n}]}{(m_n)^{2\alpha}} - \alpha^2\cdot\frac{\mathbb{E}[(W_{m_n})^2]}{(m_n)^{1+2\alpha}}\bigg\}. \end{equation*}

From [Reference Bercu5, (A.6)], $\mathbb{E}[(W_n)^2] \sim n^{2\alpha}/\{(2\alpha-\beta)\Gamma(2\alpha)\}$ as $n \to \infty$. On the other hand, from [Reference Bercu5, (4.4)] we can see that $\mathbb{E}[\Sigma_n] \sim n^{\beta}/\Gamma(1+\beta)$ as $n \to \infty$. Noting that $\beta \lt 2\alpha$, we have (ii).

Assume that $\alpha \in [-\beta,\beta/2)$. By (3.4) and (5.4), we have

\begin{equation*} \frac{\gamma_nA_n}{\sqrt{\Sigma_{m_n}}\,} = \frac{c_nW_{m_n}}{\sqrt{\Sigma_{m_n}}\,} \stackrel{\text{d}}{\to} \{\gamma + \alpha(1-\gamma)\} \cdot N\bigg(0,\frac{\beta}{\beta-2\alpha}\bigg) \quad \text{as}\ n \to \infty. \end{equation*}

Combining this and (6.3), we can prove (5.8) by the same method as for (2.9). Next, we consider the case $\alpha=\beta/2$. By (3.4) and (5.5), we have

\begin{equation*} \frac{\gamma_nA_n}{\sqrt{\Sigma_{m_n}\log\Sigma_{m_n}}\,} = \frac{c_nW_{m_n}}{\sqrt{\Sigma_{m_n}\log\Sigma_{m_n}}\,} \stackrel{\text{d}}{\to} \{\gamma + \alpha(1-\gamma)\} \cdot N(0,1) \quad \text{as}\ n \to \infty. \end{equation*}

This together with (6.4) gives (5.9). As for the case $\alpha \in (\beta/2,\beta]$, by (3.4) and (5.6),

\begin{equation*} \frac{\gamma_nA_n}{(m_n)^{\alpha}} = \frac{c_nW_{m_n}}{(m_n)^{\alpha}} \to \{\gamma+\alpha(1-\gamma)\}M \quad \text{as}\ n \to \infty\ \text{a.s.\ and\ in}\ L^2. \end{equation*}

Now (5.10) follows from Lemma 6.1(ii). The proof of (5.11) is almost identical to that of (2.12): use (3.5), (5.7), and (6.3).

7. Proof of Theorem 5.2

Proof. Put $A^{\prime}_n\,:\!=\,\mathbb{E}[\Xi_n \mid \mathcal{F}_{\infty}]$ and $B^{\prime}_n\,:\!=\, \Xi_n - A^{\prime}_n$. Using (6.2), we can see that $\gamma_n A^{\prime}_n = c_n(\beta) \Sigma_{m_n}$, which together with (5.3) imply

\begin{equation*} \frac{\gamma_n A^{\prime}_n}{(m_n)^{\beta}} = \frac{c_n(\beta)\Sigma_{m_n}}{(m_n)^{\beta}} \to \{\gamma+\beta(1-\gamma)\} \cdot \Sigma \quad \text{as}\ n \to \infty\ \text{a.s.\ and in}\ L^2. \end{equation*}

As for $B^{\prime}_n$, again by (6.2) we can see that

\begin{align*} \mathbb{V}\bigg[\bigg(\frac{\gamma_nB^{\prime}_n}{(m_n)^{\beta}}\bigg)^2 \mid \mathcal{F}_{\infty}\bigg] & = \frac{(\gamma_n)^2}{(m_n)^{2\beta}}\cdot\sum_{k=m_n+1}^n\mathbb{V}[\{X_k^{(n)}\}^2 \mid \mathcal{F}_{\infty}] \\ & = \frac{(\gamma_n)^2}{(m_n)^{2\beta}}\cdot(n-m_n)\cdot\beta\cdot\frac{\Sigma_{m_n}}{m_n}\cdot \bigg(1-\beta\cdot\frac{\Sigma_{m_n}}{m_n}\bigg), \\ \mathbb{E}\bigg[\bigg(\frac{\gamma_nB^{\prime}_n}{(m_n)^{\beta}}\bigg)^2\bigg] & = \frac{\beta\gamma_n(1-\gamma_n)}{(m_n)^{\beta}}\cdot \mathbb{E}\bigg[\frac{\Sigma_{m_n}}{(m_n)^{\beta}}\cdot\bigg(1-\beta\cdot\frac{\Sigma_{m_n}}{m_n}\bigg)\bigg]. \end{align*}

Since $\beta \lt 1$ and $\Sigma_{m_n}/(m_n)^{\beta}$ converges to $\Sigma$ in $L^2$ by (5.3), we have

\begin{equation*} \mathbb{E}\bigg[\frac{\Sigma_{m_n}}{(m_n)^{\beta}}\cdot\bigg(1-\beta\cdot\frac{\Sigma_{m_n}}{m_n}\bigg)\bigg] = \mathbb{E}\bigg[\frac{\Sigma_{m_n}}{(m_n)^{\beta}}\bigg] - \frac{\beta}{(m_n)^{1-\beta}}\cdot\mathbb{E}\bigg[\bigg(\frac{\Sigma_{m_n}}{(m_n)^{\beta}}\bigg)^2\bigg] \to \mathbb{E}[\Sigma] \quad \text{as}\ n \to \infty. \end{equation*}

Noting that $\beta \gt 0$, this shows that $\gamma_n B^{\prime}_n/(m_n)^{\beta} \to 0$ as $n \to \infty$ in $L^2$, which completes the proof.

8. Proof of Theorem 5.3