Proof. By the Markov inequality, since $\mathbb{E}{\mathrm{e}}^{uZ}={\mathrm{e}}^{\mu({\mathrm{e}}^u-1)}$ , we have, for $u>0$ ,

\begin{equation*} \mathbb{P}\bigg(Z\ge\frac{3\mu}2\bigg) \le \mathbb{P}({\mathrm{e}}^{uZ}\ge{\mathrm{e}}^{{3\mu u}/2}) \le {\mathrm{e}}^{-{3\mu u}/2}\mathbb{E}\,{\mathrm{e}}^{uZ} = \exp\!\bigg\{{-}\mu\bigg[\frac{3u}{2}-{\mathrm{e}}^u+1\bigg]\bigg\}. \end{equation*}

Setting $u=\ln(3/2)$ yields the first inequality in the claim.

For the second inequality, we use

\begin{equation*} \mathbb{P}\bigg(Z\le \frac{\mu}2\bigg) \le \mathbb{P}({\mathrm{e}}^{-uZ}\ge{\mathrm{e}}^{-{\mu u}/2}) \le {\mathrm{e}}^{{\mu u}/2}\mathbb{E}\,{\mathrm{e}}^{-uZ} = \exp\!\bigg\{{-}\mu\bigg[{-}\frac{u}2-{\mathrm{e}}^{-u}+1\bigg]\bigg\}. \end{equation*}

Now let $u=\ln 2$ .

Proof. Recall that N(s) denotes the number of points of the PPP by time s. Then $N(s)\sim\mathrm{Poi}(\Lambda(s))$ , and $\mathbb{P}(\tau_k\le s)=\mathbb{P}(N(s)\ge k)$ . Let $T_n=\Lambda^{(-1)}(n)=\sqrt[1-\alpha]{(1-\alpha)n}$ . Noting that $N(T_n)\sim\mathrm{Poi}(n)$ for all n, we have

\begin{align*} \mathbb{P}(\tau_k\le T_{2k/3}) = \mathbb{P}(N(T_{2k/3})\ge k) = \mathbb{P}\bigg(Z\ge\frac32\mu\bigg) \le {\mathrm{e}}^{-0.072\,13\dots k} \end{align*}

by Claim 2.1, with $Z\sim\mathrm{Poi}(\mu)$ where $\mu=2k/3$ . Similarly,

\begin{align*} \mathbb{P}(\tau_k\ge T_{2k}) = \mathbb{P}(N(T_{2k})\le k) \le \mathbb{P}\bigg(Z\le\frac12\mu\bigg) \le {\mathrm{e}}^{-0.306\,85\dots k} \end{align*}

by Claim 2.1, with $Z\sim\mathrm{Poi}(\mu)$ where $\mu=2k$ . Note that $0.072\,13>\frac1{15}$ , $0.306\,85>\frac1{15}$ . Now the statement follows with $c_0=\sqrt[1-\alpha]{2(1-\alpha)/3}$ and $c_1=\sqrt[1-\alpha]{2(1-\alpha)}$ .

Proof. Assume without loss of generality that n is even. We will show that for any $\rho>0$ the walk $W_n$ visits $[-\rho,\rho]$ finitely often a.s. With probabilities close to 1, both the events

(3.1) \begin{equation} \mathcal{E}_1 = \big\{\tau_{n/2} \ge c_0(n/2)^{{1}/({1-\alpha})}\big\}, \qquad \mathcal{E}_2 = \big\{\tau_{n} \le c_1 n^{{1}/({1-\alpha})}\big\} \end{equation}

occur; indeed,

(3.2) \begin{equation} \mathbb{P}(\mathcal{E}_1^\mathrm{c}) \le {\mathrm{e}}^{-n/30}, \qquad \mathbb{P}(\mathcal{E}_2^\mathrm{c}) \le {\mathrm{e}}^{-n/15} \end{equation}

by Lemma 2.2.

Since the rate $\lambda(t)=t^{-\alpha}$ of the Poisson process is monotonically decreasing, the random variables $\tau_i - \tau_{i-1}$ , $n/2\le i\le n$ , under the condition stated in the event $\mathcal{E}_1$ , are stochastically larger than i.i.d. exponential random variables $\zeta_i$ with rates equal to $[c_0(n/2)^{1/{1-\alpha}}]^\alpha=\tilde c_0 n^{-{\alpha}/({1-\alpha})}$ for some $\tilde c_0>0$ . For $\zeta$ , there exists $\beta=\beta(c_0,\alpha)>0$ such that

\begin{align*} \mathbb{P}\big(\zeta_i>\beta n^{{\alpha}/({1-\alpha})}\big) = \exp\!\big({-}\tilde c_0 n^{-{\alpha}/({1-\alpha})}\cdot\beta n^{{\alpha}/({1-\alpha})}\big) = {\mathrm{e}}^{-\tilde c_0 \beta} = \frac{2}{3}. \end{align*}

Let $I_n=\big\{i\in[n/2,n]\colon\;i\;\mathrm{is\;even},\,\tau_i-\tau_{i-2}>\beta n^{{\alpha}/({1-\alpha})}\big\}$ , and note that $\mathrm{card}(I_n)\le n/4$ . Then, since $\tau_i-\tau_{i-2}>\tau_i-\tau_{i-1}$ , by stochastic monotonicity and Hoeffding’s inequality (2.1) with $m=n/4$ , $p=\frac23$ , and $\varepsilon=\frac23-\frac12=\frac16$ ,

(3.3) \begin{align} \mathbb{P}\bigg(\!\mathrm{card}(I_n) <\frac12\cdot\frac{n}{4}\mid\mathcal{E}_1\bigg) & \le \mathbb{P}\bigg(\!\mathrm{card}\big(\big\{i\in[n/2,n]\colon\;\textit{i}\;\mathrm{is\;even},\, \zeta_i>\beta n^{{\alpha}/({1-\alpha})}\big\}\big)< \frac{n}{8}\bigg) \nonumber \\[5pt] & \le 2{\mathrm{e}}^{-n/72}. \end{align}

Let $J_n=\{i\in I_n\colon\textbf{f}_{i-1}=-\textbf{f}_{i-2}\}$ . Since the $\textbf{f}_i$ are i.i.d. and independent of $\{\tau_1,\tau_2,\dots\}$ , and $\mathbb{P}(\textbf{f}_{i-1}={-}\textbf{f}_i)=\frac12$ , on the event $\{\mathrm{card}(I_n)\ge n/8\}$ we have, by (2.1) with $m=\mathrm{card}(I_n)$ , $p=\frac12$ , and $\varepsilon=\frac{1}{34}$ ,

(3.4) \begin{align} \mathbb{P}\bigg(\!\mathrm{card}(J_n)<\frac{n}{17}\mid\mathrm{card}(I_n)\ge\frac{n}{8}\bigg) & \le \mathbb{P}\bigg(\bigg|\mathrm{card}(J_n)-\frac{\mathrm{card}(I_n)}2\bigg| > \frac{\mathrm{card}(I_n)}{34}\mid\mathrm{card}(I_n)\ge\frac{n}{8}\bigg) \nonumber \\[5pt] & \le 2\exp\!\bigg\{{-}2\cdot\frac{1}{34^2}\cdot\frac{n}{8}\bigg\} = 2{\mathrm{e}}^{-c_* n}, \end{align}

where $c_*=1/4624$ .

Our proof will rely on conditioning over the even stopping times, i.e. on the event $\mathcal{D}=\mathcal{D}_0\cap\mathcal{D}_1\cap\mathcal{E}_2$ , where

\begin{equation*} \mathcal{D}_0 = \big\{\tau_{n/2}=t_{n/2},\tau_{n/2+2}=t_{n/2+2},\dots,\tau_n=t_n\big\}, \qquad \mathcal{D}_1 = \{\mathrm{card}(J_n)\ge n/17\}\cap\mathcal{E}_1 \end{equation*}

for some strictly increasing sequence $0<t_{n/2}<t_{n/2+2}<\dots<t_n$ . Note that

(3.5) \begin{align} \mathbb{P}(\mathcal{D}_1^\mathrm{c}) & \le \mathbb{P}(\mathcal{E}_1^\mathrm{c}) + \mathbb{P}\bigg(\!\mathrm{card}(J_n)<\frac{n}{17}\mid\mathcal{E}_1\bigg) \nonumber \\[5pt] & \le \mathbb{P}(\mathcal{E}_1^\mathrm{c}) + \mathbb{P}\bigg(\!\mathrm{card}(J_n)<\frac{n}{17}\mid\mathrm{card}(I_n)\ge\frac{n}{8},\mathcal{E}_1\bigg) + \mathbb{P}\bigg(\!\mathrm{card}(I_n)<\frac{n}{8}\mid\mathcal{E}_1\bigg) \nonumber \\[5pt] & \le {\mathrm{e}}^{-n/30} + 2{\mathrm{e}}^{-c_* n} + 2{\mathrm{e}}^{-n/72} \le 5{\mathrm{e}}^{-c_*n} \end{align}

by (3.2), (3.3), and (3.4).

We denote the ith step of the embedded walk by $X_i=W_i-W_{i-1}$ , $i=1,2,\dots$ , and

\begin{align*} \xi_{i}\;:\!=\;X_{i-1}+X_{i}=(\tau_{i-1}-\tau_{i-2})\textbf{f}_{i-2}+(\tau_{i}-\tau_{i-1})\textbf{f}_{i-1}. \end{align*}

Conditioned on $\mathcal{D}$ , the random variables $\xi_2,\xi_4,\xi_6,\dots$ are then independent.

Proof of Lemma 3.1. Given $\tau_{i-2}=t_{i-2}$ and $\tau_i=t_i$ , $\tau_{i-1}$ has the distribution of the only point of the PPP on $[t_{i-2},t_i]$ with rate $\lambda(t)$ conditioned on the fact that there is exactly one point in this interval. Hence, the conditional density of $\tau_{i-1}$ is given by

\begin{align*} f_{\tau_{i-1}\mid\mathcal{D}}(x) = \begin{cases} \dfrac{\lambda(t)}{\int_{t_{i-2}}^{t_i}\lambda(u)\,\mathrm{d}u} & \mathrm{if }\;x\in[t_{i-2},t_i], \\[5pt] 0 & \mathrm{otherwise.} \end{cases} \end{align*}

Assume without loss of generality that $X_{i-1}>0>X_i$ (recall that $i\in J_n$ ). Then

\begin{align*} \xi_i=[\tau_{i-1}-\tau_{i-2}]-[\tau_{i}-\tau_{i-1}]=2\tau_{i-1}-(t_{i-2}+t_i) \end{align*}

so that the maximum of the conditional density of $\xi_i$ equals one-half of the maximum of the conditional density of $\tau_{i-1}$ . At the same time, since $\lambda$ is a decreasing function,

\begin{align*} \sup_{x\in\mathbb{R}}f_{\tau_{i-1}\mid\mathcal{D}}(x) & = \frac{\lambda(t_{i-2})}{\int_{t_{i-2}}^{t_i}\lambda(u)\,\mathrm{d}u} \le \frac{\lambda(t_{i-2})}{(t_i-t_{i-2})\lambda(t_{i})} \le \frac{\lambda(t_{i-2})}{\beta n^{{\alpha}/({1-\alpha})}\lambda(t_{i})} = \frac{1}{\beta n^{{\alpha}/({1-\alpha})}} \cdot \bigg(\frac{t_i}{t_{i-2}}\bigg)^\alpha \\[5pt] & \le \frac{1}{\beta n^{{\alpha}/({1-\alpha})}} \cdot \bigg(\frac{\tau_n}{\tau_{n/2}}\bigg)^\alpha \le \frac{1}{\beta n^{{\alpha}/({1-\alpha})}} \cdot \bigg(\frac{c_1}{c_0 2^{-{1}/({1-\alpha})}}\bigg)^\alpha \end{align*}

since the events $\mathcal{E}_1$ and $\mathcal{E}_2$ occur. This implies the stated result with $c=\beta^{-1}(c_0^{-1}c_1\sqrt[1-\alpha]{2})^\alpha$ .

Proof of Lemma 3.2. The result follows immediately from Lemma 2.1 with $a_i\equiv 2\rho=L$ , using Lemma 3.1 and the fact that $\mathrm{card}(J_n)\ge n/17$ .

So,

\begin{align*} \mathbb{P}(|W_n|\le\rho\mid\mathcal{D}) & = \mathbb{P}(A+B\in[-\rho,\rho]\mid\mathcal{D}) = \int\mathbb{P}(A+b\in[-\rho,\rho]\mid\mathcal{D})f_{B|\mathcal{D}}(b)\,\mathrm{d}b \\[5pt] & \le \int\sup_x\mathbb{P}(A\in[x-\rho,x+\rho]\mid\mathcal{D})f_{B|\mathcal{D}}(b)\,\mathrm{d}b \le \frac{C}{n^{{\alpha}/({1-\alpha})+1/2}}, \end{align*}

where $f_{B|\mathcal{D}}(\!\cdot\!)$ is the density of B conditional on $\mathcal{D}$ .

Finally, using (3.2) and (3.5),

\begin{align*} \mathbb{P}(W_n\in[-\rho,\rho]) \le \mathbb{P}(W_n\in[-\rho,\rho]\mid\mathcal{D}) + \mathbb{P}(\mathcal{D}_1^\mathrm{c}) + \mathbb{P}(\mathcal{E}_2^\mathrm{c}) \le \frac{C}{n^{{\alpha}/({1-\alpha})+1/2}} + 5{\mathrm{e}}^{-c_* n} + {\mathrm{e}}^{-n/15}, \end{align*}

which is summable over n, so we can apply the Borel–Cantelli lemma to show that $\{|W_n|\le\rho\}$ occurs finitely often a.s.

Proof of Theorem 3.2. We provide the proof only for the case $d=2$ and $\rho=1$ ; it can be easily generalized for all $d\ge 3$ and $\rho>0$ . Denote the coordinates of the embedded walk by $X_n$ and $Y_n$ ; thus $W_n=(X_n,Y_n)\in\mathbb{R}^2$ . Fix some small $\varepsilon>0$ and consider the event $\mathcal{R}_n=\{Z(t)\in[-1,1]^2\;\mathrm{ for\;some }\;t\in(\tau_n,\tau_{n+1}]\}$ . We will show that events $\mathcal{R}_n$ occur finitely often a.s., thus ensuring the transience of Z(t).

For the walk Z(t) to hit $[-\rho,\rho]^2$ between times $\tau_n$ and $\tau_{n+1}$ , we have to have either

• • $|X_n|\le\rho$ , $\textbf{f}_n=-\mathrm{sign}(Y_n)\textbf{e}_2$ , and $\tau_{n+1}-\tau_n\ge|Y_n|-1$ , or

• • $|Y_n|\le\rho$ , $\textbf{f}_n=-\mathrm{sign}(X_n)\textbf{e}_1$ , and $\tau_{n+1}-\tau_n\ge|X_n|-1$

(see Figure 1.)

Figure 1.

Model A: Recurrence of conservative random walk on $\mathbb{R}^2$ .

Using the fact that the $\textbf{f}_n$ are independent of anything, we get

\begin{align*} \mathbb{P}(\mathcal{R}_n) \le \frac14\mathbb{P}(|X_n|\le\rho,\,\tau_{n+1}-\tau_n\ge|Y_n|-1) + \frac14\mathbb{P}(|Y_n|\le\rho,\,\tau_{n+1}-\tau_n\ge|X_n|-1). \end{align*}

We will show that $\mathbb{P}(|X_n|\le\rho,\,|Y_n|\le\tau_{n+1}-\tau_n+1)$ is summable in n (and the same holds for the other summand by symmetry), hence by the Borel–Cantelli lemma that $\mathcal{R}_n$ occurs finitely often a.s.

Proof of Lemma 3.3. For $s,t\ge 0$ ,

\begin{align*} \mathbb{P}(\tau_{n+1}-\tau_n\ge s\mid \tau_n=t) = {\mathrm{e}}^{-[\Lambda(t+s)-\Lambda(t)]} = \exp\!\bigg\{{-}\frac{(t+s)^{1-\alpha}-t^{1-\alpha}}{1-\alpha}\bigg\}, \end{align*}

so if $t<c_1 n^{1/({1-\alpha})}$ , where $c_1$ is the constant from Lemma 2.2, and $s=n^{{\alpha}/({1-\alpha})+\varepsilon}-1 = o(n^{1/({1-\alpha})})$ , we have

\begin{align*} \mathbb{P}(\tau_{n+1}-\tau_n \ge s \mid \tau_n=t) & \le \exp\!\bigg\{{-}\frac{(c_1 n^{1/({1-\alpha})}+s)^{1-\alpha}-c_1^{1-\alpha} n}{1-\alpha}\bigg\} \\[5pt] & \le \exp\!\bigg\{{-}\frac{nc_1^{1-\alpha}}{1-\alpha}\bigg[\bigg(1+\frac{n^{{\alpha}/({1-\alpha})+\varepsilon}-1} {c_1 n^{1/({1-\alpha})}}\bigg)^{1-\alpha}-1\bigg]\bigg\} \\[5pt] & = \exp\!\bigg\{{-}\frac{n^\varepsilon(1+o(1))}{c_1^{\alpha}}\bigg\}. \end{align*}

Consequently, using Lemma 2.2,

(3.6) \begin{align} \mathbb{P}\big(\tau_{n+1}-\tau_n \ge n^{{\alpha}/({1-\alpha})+\varepsilon}-1\big) & \le \mathbb{P}\big(\tau_{n+1}-\tau_n \ge n^{{\alpha}/({1-\alpha})+\varepsilon}-1\mid\tau_n<c_1 n^{1/({1-\alpha})}\big) \nonumber \\[5pt] & \quad + \mathbb{P}\big(\tau_n \ge c_1 n^{1/({1-\alpha})}\big) \nonumber \\[5pt] & \le {\mathrm{e}}^{-c_1^* n^\varepsilon} + {\mathrm{e}}^{-n/15} \le 2{\mathrm{e}}^{-c_1^* n^\varepsilon} \end{align}

for some $c_1^*\in\big(0,\frac{1}{15}\big)$ .

Now we will modify the proof of Theorem 3.1 slightly to adapt to our needs. Let the events $\mathcal{E}_1$ and $\mathcal{E}_2$ be the same as in the proof of Theorem 3.1. We will also use the set $I_n$ , but instead of $J_n$ we introduce the sets

\begin{align*} J_n^1 & = \{i\in I_n\colon\textbf{f}_{i-1}=-\textbf{f}_{i-2},\, \textbf{f}_{i-1}\in\{\textbf{e}_1,-\textbf{e}_1\}\}, \\[5pt] J_n^2 & = \{i\in I_n\colon\textbf{f}_{i-1}=-\textbf{f}_{i-2},\,\textbf{f}_{i-1}\in\{\textbf{e}_2,-\textbf{e}_2\}\}. \end{align*}

Similarly to the proof of Theorem 3.1, inequality (3.5), we immediately obtain that

(3.7) \begin{equation} \begin{split} \mathbb{P}(\mathrm{card}(J_n^1) \le n/34) & \le \mathbb{P}(\mathrm{card}(J_n^1) \le n/34\mid\mathcal{E}_1) + \mathbb{P}(\mathcal{E}_1^\mathrm{c}) \le 6{\mathrm{e}}^{-c'_{\!\!\ast}n}, \\[5pt] \mathbb{P}(\mathrm{card}(J_n^2) \le n/34) & \le \mathbb{P}(\mathrm{card}(J_n^2) \le n/34\mid\mathcal{E}_1) + \mathbb{P}(\mathcal{E}_1^\mathrm{c}) \le 6{\mathrm{e}}^{-c'_{\!\!\ast}n} \end{split} \end{equation}

for some $c'_{\!\!\ast}>0$ .

Fix a deterministic sequence of unit vectors $\textbf{g}_1,\dots,\textbf{g}_n$ such that each $\textbf{g}_i\in\{\pm\textbf{e}_1,\pm\textbf{e}_2\}$ . We now also define the event

\begin{align*}\mathcal{D}&=\{\textbf{f}_1=\textbf{g}_1,\dots,\textbf{f}_n=\textbf{g}_n\}\cap\{\tau_k=t_k\text{ for all } k\le n\colon k\text{ is even or }\textbf{f}_k\perp \textbf{f}_{k+1}\} \cap \\& \bigg\{\mathrm{card}({J^1_n})\ge \frac{n}{34}\bigg\}\cap \bigg\{\mathrm{card}({J^2_n})\ge \frac{n}{34}\bigg\}\cap \mathcal{E}_1 \cap \mathcal{E}_2.\end{align*}

Therefore, repeating the previous arguments for each of the horizontal and vertical components of $W_n$ , we immediately obtain

\begin{equation*} \mathbb{P}(|X_n|\le1\mid\mathcal{D}) \le \frac{C}{n^{{\alpha}/({1-\alpha})+1/2}}, \qquad \sup_{x\in\mathbb{R}}\mathbb{P}(|Y_n-x|\le1\mid\mathcal{D}) \le \frac{C}{n^{{\alpha}/({1-\alpha})+1/2}}. \end{equation*}

The second inequality implies that

\begin{align*} \mathbb{P}(|Y_n| \le n^{{\alpha}/({1-\alpha})+\varepsilon} \mid \mathcal{D}) \le n^{{\alpha}/({1-\alpha})+\varepsilon}\cdot\frac{C}{n^{{\alpha}/({1-\alpha})+1/2}} = \frac{C}{n^{1/2-\varepsilon}}. \end{align*}

Now, by Lemma 3.3,

\begin{align*} \mathbb{P}(|X_n|\le 1,\, & |Y_n|\le \tau_{n+1}-\tau_n+1) \\[5pt] & \le \mathbb{P}\big(|X_n|\le1,|Y_n|\le\tau_{n+1}-\tau_n+1 \mid \tau_{n+1}-\tau_n+1\le n^{{\alpha}/({1-\alpha})+\varepsilon}\big) \\[5pt] & \quad + \mathbb{P}\big(\tau_{n+1}-\tau_n+1\ge n^{{\alpha}/({1-\alpha})+\varepsilon}\big) \\[5pt] & \le \mathbb{P}\big(|X_n|\le1,|Y_n|\le n^{{\alpha}/({1-\alpha})+\varepsilon}\big) + 3{\mathrm{e}}^{-c_1^*n^\varepsilon}. \end{align*}

Observing that $X_n$ and $Y_n$ are actually independent given $\mathcal{D}$ , we conclude that

\begin{align*} \mathbb{P}\big(|X_n|\le1,|Y_n|\le n^{{\alpha}/({1-\alpha})+\varepsilon}\big) & \le \mathbb{P}\big(|X_n|\le1,|Y_n|\le n^{{\alpha}/({1-\alpha})+\varepsilon}\mid\mathcal{D}\big) \\[5pt] & \quad + \mathbb{P}\bigg(\!\mathrm{card}(J_n^1)<\frac{n}{34}\;\mathrm{ or }\; \mathrm{card}(J_n^1)<\frac{n}{34}\bigg) + \mathbb{P}(\mathcal{E}_1^\mathrm{c}) + \mathbb{P}(\mathcal{E}_2^\mathrm{c}) \\[5pt] & \le \frac{C}{n^{{\alpha}/({1-\alpha})+1/2}}\cdot\frac{C}{n^{1/2-\varepsilon}} + 12{\mathrm{e}}^{-c'_{\!\!\ast} n} + {\mathrm{e}}^{-n/15} + {\mathrm{e}}^{-n/15} \\[5pt] & = \frac{C^2+o(1)}{n^{1+[{\alpha}/({1-\alpha})-\varepsilon]}} \end{align*}

by (3.2) and (3.7). Assuming $\varepsilon\in(0,{\alpha}/({1-\alpha}))$ , the right-hand side is summable, and thus $\mathbb{P}(|X_n|\le\rho,|Y_n|-1\le\tau_{n+1}-\tau_n)$ is also summable in n.

Proof. The proof is analogous to the proof of Theorem 3.2; we will only indicate how that proof should be modified for this case. As before, we assume that $\rho=1$ and $d=2$ , without loss of generality. First, we prove the following lemma.

Proof of Lemma 3.4. Fix an $\varepsilon\in(0,1)$ . Then

\begin{align*} \Lambda(T) = \int_{\mathrm{e}}^T\frac{\mathrm{d}s}{(\!\ln s)^\beta} = \int_{\mathrm{e}}^{T^{1-\varepsilon}}\frac{\mathrm{d}s}{(\!\ln s)^\beta} + \int_{T^{1-\varepsilon}}^T\frac{\mathrm{d}s}{(\!\ln s)^\beta}< T^{1-\varepsilon} + \frac{T}{(\!\ln T)^\beta(1-\varepsilon)^\beta}. \end{align*}

At the same time, trivially, $\Lambda(T)>({T-{\mathrm{e}}})/{(\!\ln T)^\beta}$ . Now, the limit of the ratio of the upper and the lower bounds of ${\Lambda(T)}$ can be made arbitrarily close to 1 by choosing a small enough $\varepsilon$ . Hence the statement of the lemma follows.

The rest of the proof goes along the same lines as that of Theorem 3.2. First, note that for some $c>0$ the event $\{\tau_n \leq cn(\!\ln{n})^{\beta}\}$ occurs finitely often almost surely. Indeed,

\begin{align*} \tilde\Lambda &= \Lambda(cn(\!\ln{n})^{\beta}) = \frac{cn(\!\ln{n})^{\beta}}{[\ln(cn(\!\ln{n})^{\beta})]^{\beta}}(1+o(1)) = \frac{(\!\ln n)^\beta}{(1+o(1))(\!\ln n)^\beta}cn(1+o(1)) \\[5pt] &= (1+o(1))cn \end{align*}

by Lemma 3.4, and thus

\begin{align*} \mathbb{P}(\tau_n \leq cn(\!\ln{n})^{\beta}) & = \mathbb{P}(N(cn(\!\ln{n})^{\beta}) \ge n) \\[5pt] & = {\mathrm{e}}^{-\tilde\Lambda}\bigg(\frac{\tilde\Lambda^n}{n!} + \frac{\tilde\Lambda^{n+1}}{(n+1)!} + \frac{\tilde\Lambda^{n+2}}{(n+2)!} + \cdots\bigg) \\[5pt] & = {\mathrm{e}}^{-\tilde\Lambda}\frac{\tilde\Lambda^n}{n!}\bigg(1 + \frac{\tilde\Lambda}{(n+1)} + \frac{\tilde\Lambda^2}{(n+1)(n+2)} + \cdots\bigg) \\[5pt] & \le {\mathrm{e}}^{-\tilde\Lambda}\frac{\tilde\Lambda^n}{n!}\bigg(1 + \frac{\tilde\Lambda}{1!} + \frac{\tilde\Lambda^2}{2!} + \cdots\bigg) \\[5pt] & = \frac{\tilde\Lambda^n}{n!} = \frac{[(1+o(1))cn]^{n}}{n!} = \mathcal{O}\bigg(\frac{[(1+o(1))c]^n n^{n}}{n^n{\mathrm{e}}^{-n}\sqrt{n}}\bigg). \end{align*}

This quantity is summable as long as $c{\mathrm{e}}<1$ , and the statement follows from the Borel–Cantelli lemma.

Second, for any positive $\varepsilon$ , the event $\{\tau_n \geq c^*n(\!\ln{n})^{\beta}\}$ , where $c^*=1+\varepsilon$ , occurs finitely often, almost surely. Indeed,

\begin{align*} \bar\Lambda \;:\!=\; \Lambda( c^*n(\!\ln{n})^{\beta}) = \frac{c^*n(\!\ln{n})^{\beta}}{[\ln(c^*n(\!\ln{n})^{\beta})]^\beta}(1+o(1)) = c^* n(1+o(1)) \ge (1+\varepsilon/2)n \end{align*}

for large enough n by Lemma 3.4. Hence, since $\bar\Lambda>n$ ,

\begin{align*} \mathbb{P}(\tau_n \geq c^*n(\!\ln{n})^{\beta}) & = \mathbb{P}(N(c^*n(\!\ln{n})^\beta) \le n) \\[5pt] & = {\mathrm{e}}^{-\bar\Lambda}\bigg(1+\bar\Lambda+\frac{{\bar\Lambda}^2}{2!}+\dots+\frac{{\bar\Lambda}^n}{n!}\bigg) \\[5pt] & \le {\mathrm{e}}^{-\bar\Lambda}(n+1)\frac{{\bar\Lambda}^n}{n!} \\[5pt] & = {\mathrm{e}}^{-\bar\Lambda}(n+1)\frac{{\bar\Lambda}^n}{n^n{\mathrm{e}}^{-n}\sqrt{2\pi n}}(1+o(1)) \\[5pt] & = (1+o(1))\sqrt{\frac{n}{2\pi}} \exp\!\bigg\{{-}n\bigg[\frac{\bar\Lambda}{n}-1-\ln\frac{\bar\Lambda}{n}\bigg]\bigg\}, \end{align*}

which is summable in n, as $c^*=1+\varepsilon$ implies that the expression in square brackets is strictly positive (this follows from the easy fact that $c-1-\ln c>0$ for $c>1$ ). Hence, $\{\tau_n \geq c^*n(\!\ln{n})^{\beta}\}$ happens finitely often, almost surely, as stated.

Third, the event $\{\tau_{n+1}-\tau_{n} \geq 2(\!\ln{n})^{1+\beta}\}$ occurs finitely often, almost surely. This holds because, for all sufficiently large n, $\tau_{n+1}\le c^*(n+1)(\!\ln(n+1))^\beta$ , and hence, for $t\le\tau_{n+1}$ ,

\begin{align*} \lambda(t) \ge \frac{1}{[\ln(c(n+1)(\!\ln(n+1))^\beta)]^\beta} = \frac{1+o(1)}{(\!\ln n)^\beta}, \end{align*}

and thus $(\tau_{n+1}-\tau_{n})$ is stochastically smaller than an exponential random variable $\mathcal{E}$ with parameter $({1+o(1)})/{(\!\ln{n})^\beta}$ . So, for all sufficiently large n,

(3.9) \begin{align} \mathbb{P}(\tau_{n+1}-\tau_n\geq2(\!\ln{n})^{1+\beta}) \leq \mathbb{P}(\mathcal{E}\geq2(\!\ln{n})^{1+\beta}) = \frac{1}{n^{2-o(1)}}, \end{align}

which is summable in n, and we can apply the Borel–Cantelli lemma.

Using the previous arguments for horizontal and vertical components, we obtain

\begin{equation*} \mathbb{P}(|X_n|\leq1) \leq \frac{C}{\sqrt{n}(\!\ln{n})^{\beta}}, \qquad \mathbb{P}(|Y_n|\leq(\!\ln{n})^{1+\beta}) \leq (\!\ln{n})^{1+\beta}\cdot\frac{C}{\sqrt{n}(\!\ln{n})^{\beta}} = \frac{C\ln{n}}{\sqrt{n}}. \end{equation*}

Again, similar to Theorem 3.2, the event $\{|X_n|\leq\rho,\,|Y_n|\leq2(\!\ln{n})^{1+\beta}\}$ has to happen infinitely often almost surely for Z(t) to be recurrent.

Thus, it follows that

\begin{align*} \mathbb{P}(Z(t)\;\mathrm{ visits }\;[-1,1]^2\;\mathrm{ for }\;t\in[\tau_n,\tau_{n+1}]) & \leq \mathbb{P}(|X_n|\leq1,\,|Y_n|\leq2(\!\ln{n})^{1+\beta}) + f(n) \\[5pt] & \leq \frac{C^2}{n(\!\ln{n})^{\beta-1}} + f(n) + g(n), \end{align*}

where f and g are two summable functions over n, similar to Theorem 3.2. The right-hand side is summable over n as long as $\beta>2$ . Hence, Z(t) visits $[-1, 1]^2$ finitely often, almost surely.

Proof of Lemma 4.1. We use the following well-known representation (see, e.g., [Reference Blum, Hopcroft and Kannan2, Section 2.5]) of $\textbf{f}$ :

\begin{align*} \textbf{f} = \bigg(\frac{\eta_1}{\Vert\eta\Vert},\dots,\frac{\eta_d}{\Vert\eta\Vert}\bigg), \end{align*}

where $\eta_i$ , $i=1,2,\dots$ , are i.i.d. standard normal and $\Vert\eta\Vert=\sqrt{\eta_1^2+\dots+\eta_d^2}$ . For some large enough $A>0$ ,

\begin{align*} \mathbb{P}\Big(\max_{i=3,\dots,d}|\eta_i|<A\Big) = (\Phi(A)-\Phi(-A))^{d-2} \ge \sqrt{\frac23}, \end{align*}

where $\Phi(\!\cdot\!)$ is the distribution function of the standard normal random variable. Also, for some small enough $a\in(0,A)$ , $\mathbb{P}(\max_{i=1,2}|\eta_i|>a)\ge\sqrt{\frac23}$ . On the intersection of these two independent events we have

\begin{equation*} f_1^2+f_2^2 = \frac{\eta_1^2+\eta_2^2}{(\eta_1^2+\eta_2^2)+(\eta_3^2+\dots+\eta_d^2)} \ge \frac{a^2}{a^2+(d-2)A^2} \;=\!:\; \gamma^2. \end{equation*}

Proof. Assume that the event $\mathcal{E}_1$ defined by (3.1) has occurred. We can write

\begin{align*} \hat W_n = (X_n,Y_n) = \sum_{k=1}^n(\tau_k-\tau_{k-1})\tilde{\textbf{f}}_k \end{align*}

where $\tilde{\textbf{f}}_k=\ell_k[\textbf{e}_{\textbf{1}}\cos(\phi_k)+\textbf{e}_{\textbf{2}}\sin(\phi_k)]$ , $\phi_k$ , $k=1,2,\dots$ , are uniformly distributed on $[\!-\!\pi,\pi]$ , and $\ell_k$ is the length of the projection $\tilde{\textbf{f}}_k$ of $\textbf{f}_k$ on the two-dimensional plane. Note that the elements of the set $\{\ell_1,\eta_2,\eta_3,\dots,\phi_1,\phi_2,\phi_3,\dots\}$ are all independent. Also, define

\begin{align*} \mathcal{D}_2 = \{\mathrm{for\;at\;least\;half\;of\;the\;integers }\; i\in[n/2,n]\;\mathrm{ we\;have }\;\ell_i\ge\gamma\}, \end{align*}

where $\gamma$ is the constant from Lemma 4.1. Then $\mathbb{P}(\mathcal{D}_2^\mathrm{c})\le2{\mathrm{e}}^{-n/36}$ by (2.1) and Lemma 4.1.

Let $\tilde W_n$ be the distribution of $\hat W_n\in\mathbb{R}^2$ conditioned on $\mathcal{E}_1 \cap \mathcal{D}_2$ , and

\begin{align*} \varphi_{\tilde W_n}(t) = \mathbb{E}\,{\mathrm{e}}^{\mathrm{i}t\cdot\tilde W_n} = \mathbb{E}\exp\Bigg\{i\sum_{k=1}^n t\cdot\tilde{\textbf{f}}_{k}(\tau_k-\tau_{k-1})\Bigg\} \end{align*}

be its characteristic function (here, $t\cdot\tilde{\textbf{f}}_k=\ell_k(t_1\cos(\phi_k)+t_2\sin(\phi_k))$ ). We use the Lévy inversion formula, which allows us to compute the density of $\tilde W_n$ , provided $|\varphi_{\tilde W_n}(t)|$ is integrable:

(4.1) \begin{equation} f_{\tilde W_n}(x,y) = \frac{1}{(2\pi)^2} \iint_{\mathbb{R}^2}{\mathrm{e}}^{-\mathrm{i}(t_1 x+t_2 y)}\varphi_{\tilde W_n}(t)\,\mathrm{d}t_1\,\mathrm{d}t_2 \le \frac{1}{(2\pi)^2}\iint_{\mathbb{R}^2}\big|\varphi_{\tilde W_n}(t)\big|\,\mathrm{d}t_1\,\mathrm{d}t_2. \end{equation}

Let $\Delta_k=\tau_k-\tau_{k-1}$ . Since $\phi_1,\dots,\phi_n$ are i.i.d. Uniform $[-\pi,\pi]$ and independent of anything, we have

\begin{align*} & \varphi_{\tilde W_n}(t) = \mathbb{E}\Bigg[\prod_{k=1}^n\mathbb{E}\big( {\mathrm{e}}^{\mathrm{i}\Delta_k\ell_k(t_1\cos\phi_k+t_2\sin\phi_k)}\mid\tau_1,\dots,\tau_n;\;\ell_1,\dots,\ell_n\big)\Bigg] = \mathbb{E}\Bigg[\prod_{k=1}^n J_0(\Vert t\Vert\Delta_k\ell_k)\Bigg] \\[5pt] & \Longrightarrow \\[5pt] & |\varphi_{\tilde W_n}(t)| \le \mathbb{E}\Bigg[\prod_{k=1}^n|J_0(\Vert t\Vert \Delta_k\ell_k)|\Bigg], \end{align*}

where $\Vert t\Vert=\sqrt{t_1^2+t_2^2}$ , and $J_0(x)=\sum_{m=0}^\infty{(-x^2/4)^{m}}/{m!^2}$ is the Bessel $J_0$ function. Indeed, for any $x\in\mathbb{R}$ , setting $\tilde x=x\sqrt{t_1^2+t_2^2}$ and $\beta=\arctan({t_2}/{t_1})$ , we get

\begin{align*} & \mathbb{E}\big[{\mathrm{e}}^{\mathrm{i}x(t_1\cos\phi_k+t_2\sin\phi_k)}\big] = \frac{1}{2\pi}\int_0^{2\pi}{\mathrm{e}}^{\mathrm{i}x(t_1\cos\phi+t_2\sin\phi)}\,\mathrm{d}\phi = \frac{1}{2\pi}\int_0^{2\pi}{\mathrm{e}}^{\mathrm{i}\tilde x(\cos\beta\cos\phi+\sin\beta\sin\phi)}\, \\[5pt] &\mathrm{d}\phi = \frac{1}{2\pi}\int_0^{2\pi}{\mathrm{e}}^{\mathrm{i}\tilde x\cos(\phi+\beta)}\,\mathrm{d}\phi = \frac{1}{2\pi}\int_0^{2\pi}{\mathrm{e}}^{\mathrm{i}\tilde x\cos(\phi)}\,\mathrm{d}\phi = \frac{1}{2\pi}\int_0^{2\pi}\cos(\tilde x\cos(\phi))\,\mathrm{d}\phi = J_0(\tilde x) \end{align*}

due to periodicity of $\cos(\!\cdot\!)$ , the fact that the function $\sin(\!\cdot\!)$ is odd, and [Reference Abramowitz and Stegun1, (9.1.18)].

Let $\xi_k$ be the random variable with the distribution of $\Delta_k$ given $\tau_{k-1}$ . Then, recalling that we are on the event $\mathcal{E}_1$ , we get $\xi_k\succ\tilde\xi^{(1)}$ for $\hat{W}^{(1)}_n$ and $\xi_k\succ\tilde\xi^{(2)}$ for $\hat{W}^{(2)}_n$ , where $\tilde\xi^{(1)}$ ( $\tilde\xi^{(2)}$ respectively) is an exponential random variable with rate $1/{a_1}$ ( $1/a_2$ respectively) with $a_1=\tilde c_1 n^{{\alpha}/({1-\alpha})}$ and $a_2=\tilde c_2(\!\ln{n})^\beta$ for some constants $\tilde c_1>0$ and $\tilde c_2>0$ , and where ‘ $\zeta_a\succ\zeta_b$ ’ denotes that random variable $\zeta_a$ is stochastically larger than random variable $\zeta_b$ . Consequently, setting $a=a_1$ , $\tilde\xi=\tilde\xi^{(1)}$ or $a=a_2$ , $\tilde\xi=\tilde\xi^{(2)}$ , depending on which of the two models we are talking about, and $\mathcal{F}_k=\sigma(\tau_1,\dots,\tau_k)$ , we get

(4.2) \begin{align} \mathbb{E}[|J_0(\Vert t\Vert\Delta_k\ell_k)| \mid \mathcal{F}_{k-1},\ell_k=\ell] & = \mathbb{E}\,|J_0(\Vert t\Vert \xi_k\ell)| \nonumber \\[5pt] & \le \mathbb{E}\,G(\Vert t\Vert\xi_k\ell) \nonumber \\[5pt] & \le \mathbb{E}\,G(\Vert t\Vert\tilde\xi\ell) \nonumber \\[5pt] & = \int_0^\infty a^{-1}{\mathrm{e}}^{-{y}/{a}}G(\Vert t\Vert y\ell)\,\mathrm{d}y \nonumber \\[5pt] & = \int_0^\infty{\mathrm{e}}^{-u}G(su)\,\mathrm{d}u \nonumber \\[5pt] & = \int_0^\infty\frac{{\mathrm{e}}^{-u}}{\sqrt[4]{1+s^2 u^2}\,}\,\mathrm{d}u \;=\!:\; h(s,\ell), \end{align}

where $s=a\ell\Vert t\Vert$ , since $|J_0(x)|\le G(x)$ by (A.1) and the facts that $G(\!\cdot\!)$ is a decreasing function and $\xi_k\succ\tilde\xi$ .

We now estimate the function $h(t,\ell)$ . Since $|J_0(x)|\le 1$ , we trivially get $0\le h(s,\ell)\le 1$ . Additionally, for all $s>0$ ,

\begin{align*} h(s,\ell) \le \int_{0}^\infty\frac{{\mathrm{e}}^{-u}}{\sqrt{su}\,}\,\mathrm{d}u = \sqrt{\frac{\pi}{s}} = \sqrt{\frac{\pi}{a\ell\Vert t\Vert}}. \end{align*}

Let $n/2\le j_1<j_2<\dots<j_m\le n$ be the indices $i\in[n/2,n]$ for which $\ell_i\ge\gamma$ . Since $|J_0(x)|\le 1$ , we have, from (4.2),

\begin{align*} \mathbb{E}\Bigg[\prod_{k=1}^n|J_0(\Vert t\Vert\Delta_k\ell_k)|\Bigg] & \le \mathbb{E}\Bigg[\prod_{i=1}^m\big|J_0\big(\Vert t\Vert\Delta_{j_i}\ell_{j_i}\big)\|\Bigg] \\[5pt] & = \mathbb{E}\Bigg[\mathbb{E}\big(\big|J_0\big(\Vert t\Vert\Delta_{j_m}\ell_{j_m}\big)\big| \mid \mathcal{F}_{j_m-1},\ell_{j_m}\big) \prod_{i=1}^{m-1}\big|J_0\big(\Vert t\Vert\Delta_{j_i}\ell_{j_i}\big)\big|\Bigg] \\[5pt] & \le \sqrt{\frac{\pi}{a\gamma\Vert t\Vert}} \cdot \mathbb{E}\Bigg[\prod_{i=1}^{m-1}\big|J_0\big(\Vert t\Vert\Delta_{j_i}\ell_{j_i}\big)\big|\Bigg]. \end{align*}

By iterating this argument for $i=j_{m-1},j_{m-2},\dots,j_1$ , we get

\begin{align*} \big|\varphi_{\tilde W_n}(t)\big| \le \mathbb{E}\Bigg[\prod_{k=1}^n|J_0(\Vert t\Vert\Delta_k\ell_k)|\Bigg] \le \bigg(\sqrt{\frac{\pi}{a\gamma\Vert t\Vert}}\bigg)^{n/4} \end{align*}

(recall that $m\ge n/4$ on $\mathcal{D}_1$ ).

Now consider two cases. For $\Vert t\Vert\ge\frac{2\pi}{a\gamma}$ , part of the inversion formula gives

(4.3) \begin{align} \iint_{\Vert t\Vert>{2\pi}/{a\gamma}}\big|\varphi_{\tilde W_n}(t)\big|\,\mathrm{d}t_1\,\mathrm{d}t_2 & \le \iint_{\Vert t\Vert\ge{2\pi}/{a\gamma}} \bigg(\frac{\pi}{\gamma\Vert t\Vert a}\bigg)^{n/8}\,\mathrm{d}t_1\,\mathrm{d}t_2 \nonumber \\[5pt] & = \bigg(\frac{\pi}{a\gamma}\bigg)^2\int_0^{2\pi}\,\mathrm{d}\theta\int_2^\infty\frac{r\,\mathrm{d}r}{r^{n/8}} \nonumber \\[5pt] & = \frac{2\pi^3}{a^2\gamma^2}\cdot\frac{r^{2-n/8}}{2-n/8}\bigg|_2^\infty = o(2^{-n/8}) \end{align}

by changing the variables $t_1=({\pi}/{a\gamma})r\cos\theta$ , $t_2=({\pi}/{a\gamma})r\sin\theta$ .

On the other hand, for $\Vert t\Vert\le{2\pi}/{a\gamma}$ , when $\ell\ge\gamma$ and thus $\sigma\;:\!=\;a\gamma\Vert t\Vert\le s$ ,

\begin{align*} h(s,\ell) & = \int_0^\infty\frac{{\mathrm{e}}^{-u}\,\mathrm{d}u}{\sqrt[4]{1+s^2u^2}\,} \\[5pt] & \le \int_0^\infty\frac{{\mathrm{e}}^{-u}\,\mathrm{d}u}{\sqrt[4]{1+\sigma ^2u^2}\,} \\[5pt] & \le \int_0^1\frac{{\mathrm{e}}^{-u}\,\mathrm{d}u}{\sqrt[4]{1+\sigma^2u^2}\,} + \int_1^\infty\frac{{\mathrm{e}}^{-u}\,\mathrm{d}u}{\sqrt[4]{1+\sigma^2u^2}\,} \\[5pt] & \le \int_0^1\bigg(1-\frac{\sigma^2u^2}{50}\bigg){\mathrm{e}}^{-u}\,\mathrm{d}u + \int_1^\infty{\mathrm{e}}^{-u}\,\mathrm{d}u \\[5pt] & = -\int_0^1\frac{\sigma^2u^2}{50}{\mathrm{e}}^{-u}\,\mathrm{d}u + \int_{0}^\infty{\mathrm{e}}^{-u}\,\mathrm{d}u \\[5pt] & = 1 - 0.0016\ldots\sigma^2 \le \exp\!\bigg\{{-}\frac{\sigma^2}{700}\bigg\} = \exp\!\bigg\{{-}\frac{\gamma^2\Vert t\Vert^2 a^2}{700}\bigg\}, \end{align*}

since $(1+x^2)^{-1/4}\le 1-x^2/50$ for $0\le x\le 7$ , and $\sigma\le 2\pi<7$ by assumption. By iterating the same argument as before, we get

\begin{align*} \varphi_{\tilde W_n} \le h(s,\ell)^{n/4} \le \exp\!\bigg\{{-}\frac{n\gamma^2\Vert t\Vert^2 a^2}{2800}\bigg\}. \end{align*}

Consequently,

(4.4) \begin{align} \iint_{\Vert t\Vert\le{2\pi}/{a\gamma}}\big|\varphi_{\tilde W_n}(t)\big|\,\mathrm{d}t_1\,\mathrm{d}t_2 & \le \iint_{\Vert t\Vert\le{2\pi}/{a\gamma}}\exp\!\bigg\{{-}\frac{n\gamma^2\Vert t\Vert^2 a^2}{2800}\bigg\}\, \mathrm{d}t_1\mathrm{d}t_2 \nonumber \\[5pt] & \le \frac{1400}{a^2\gamma^2}\int_0^{2\pi}\mathrm{d}\theta \int_0^\infty\exp\!\bigg\{{-}\frac{nr^2}2\bigg\}r\,\mathrm{d}r = \frac{2800\pi}{na^2\gamma^2} \end{align}

by changing the variables $t_1=({10\sqrt{14}}/{a\gamma})r\cos\theta$ , $t_2=({10\sqrt{14}}/{a\gamma})r\sin\theta$ .

Finally, recalling that $a_1\sim n^{{\alpha}/({1-\alpha})}$ for $\hat W_t^{(1)}$ and $a_2\sim(\!\ln{n})^\beta$ for $\hat W_t^{(2)}$ , from (4.1), (4.3), and (4.4) we obtain

\begin{equation*} f_{\tilde W_n^{(1)}}(x,y) \le O\bigg(\frac{1}{n^{1+({2\alpha}/({1-\alpha}))}}\bigg), \qquad f_{\tilde W_n^{(2)}}(x,y) \le O\bigg(\frac{1}{n(\!\ln{n})^{2\beta}}\bigg). \end{equation*}

Hence,

\begin{align*} \mathbb{P}(\hat W_n\in R_k) & \le \mathbb{P}(\mathcal{E}_1^\mathrm{c}) + \mathbb{P}(\mathcal{D}_2^\mathrm{c}) + \mathbb{P}(\tilde W_n\in R_k) \\[5pt] & = 4{\mathrm{e}}^{-n/36} + \iint_{(x,y)\in R_k}f_{\tilde W_n}(x,y)\,\mathrm{d}x\,\mathrm{d}y \\[5pt] & = 2\pi k\times \begin{cases} O\bigg(\dfrac{1}{n^{1+({2\alpha}/({1-\alpha}))}}\bigg) & \mathrm{for }\;W_n^{(1)}, \\[5pt] O\bigg(\frac{1}{n(\!\ln{n})^{2\beta}}\bigg) & \mathrm{for }\;W_n^{(2)}, \end{cases} \end{align*}

since the area of $R_k$ is $\pi(2k+1)$ .

Proof. First, note that $\hat W_n$ lies outside the unit circle on $\mathbb{R}^2$ . The projection $\tilde{\textbf{f}}_{n+1}$ of $\textbf{f}_{n+1}$ on the first two coordinates’ plane has an angle $\phi$ uniformly distributed over $[0,2\pi]$ . The probability in the statement of the lemma is monotone increasing in $\tau_{n+1}$ , and hence is bounded above by

\begin{align*} \mathbb{P}(\mathrm{the\;infinite\;ray\;from }\;\hat W_n\;\mathrm{ in\;the\;direction }\;\tilde{\textbf{f}}_{n+1} \;\mathrm{ passes\;through\;the\;unit\;circle}) = \frac{2\arcsin(1/r)}{2\pi} \end{align*}

(see Figure 3.) Finally, $\arcsin(x)\le {\pi x}/{2}$ for $0\le x\le 1$ .

Proof. As mentioned before, we only show that the walk is not $\rho$ -recurrent for $\rho=1$ . To do that, it will suffice that, a.s., there will be only finitely many n such that the event $A_n=\{\Vert\hat{W}(t)\Vert\le1\;\mathrm{ for\;some }\;t\in[\tau_n,\tau_{n+1}]\}$ occurs. Indeed, fix a positive integer n. At this time, $\hat W_n\in R_k$ for some $k\in \{0,1,2,\dots\}$ . According to Lemma 4.3,

\begin{align*} \mathbb{P}(A_n\mid\hat W_n\in R_k) \le \begin{cases} 1 & \mathrm{if }\; k=0, \\[5pt] \dfrac{1}{2k} & \mathrm{if }\; k\ge 1. \end{cases} \end{align*}

Fix some very small $\varepsilon>0$ . Then

\begin{align*} \mathbb{P}(A_n) & \le \sum_{k=1}^{n^{{\alpha}/({1-\alpha})+\varepsilon}} \mathbb{P}\big(A_n\mid\hat W_n^{(1)}\in R_k\big)\mathbb{P}\big(\hat W_n^{(1)}\in R_k\big) + \mathbb{P}\big(A_n\mid\big\Vert\hat W_n^{(1)}\big\Vert\ge n^{{\alpha}/({1-\alpha})+\varepsilon}\big) \\[5pt] & \le \frac{Cn^{{\alpha}/({1-\alpha})+\varepsilon}}{n^{1+({2\alpha}/({1-\alpha}))}} + \mathbb{P}\big(\tau_{n+1}-\tau_n\ge n^{{\alpha}/({1-\alpha})+\varepsilon}-1\big) \le \frac{C}{n^{1+({\alpha}/({1-\alpha}))-\varepsilon}} + 3{\mathrm{e}}^{-c_1^* n^\varepsilon} \end{align*}

by (3.6) and Lemma 4.2. Hence, $\sum_n\mathbb{P}(A_n)<\infty$ and the result follows from the Borel–Cantelli lemma.

Similarly, in the other case,

\begin{align*} \mathbb{P}(A_n) & \le \sum_{k=1}^{2(\!\ln{n})^{1+\beta}}\mathbb{P}\big(A_n,\hat W_n^{(2)}\in R_k\big) + \mathbb{P}\big(A_n\mid\big\Vert\hat W_n^{(2)}\big\Vert\ge{2(\!\ln{n})^{1+\beta}}\big) \\[5pt] & \le \frac{2C(\!\ln{n})^{1+\beta}}{n(\!\ln{n})^{2\beta}} + \mathbb{P}(\tau_{n+1}-\tau_n\ge2(\!\ln{n})^{1+\beta}-1) \le \frac{2C}{n(\!\ln{n})^{\beta-1}} + \frac{1}{n^2} \end{align*}

for all sufficiently large n by (3.9) and Lemma 4.2. Hence, $\sum_n\mathbb{P}(A_n)<\infty$ for $\beta>2$ , and the result again follows from the Borel–Cantelli lemma.

Proof. From [Reference Krasikov15, Theorem 1 and (1)] we get (with $\nu=0$ , $\mu=3$ )

\begin{equation*} J_0^2(x) \le \frac{4(4x^2-5)}{\pi( (4x^2-3)^{3/2}-3 )}, \quad x \ge 1.13. \end{equation*}

This inequality also implies that

\begin{align*} |J_0(x)| \le \frac{1}{\sqrt[4]{1+x^2}\,}, \quad x \ge 1.13. \end{align*}

At the same time, for $0\le x\le 2$ (using [1, 9.1.14]),

\begin{align*} J_0^2(x) = \sum_{k=0}^\infty(-1)^k\frac{(2k)!}{(k!)^4}\bigg(\frac{x}{2}\bigg)^{2k} \le 1 - 2\bigg(\frac{x}{2}\bigg)^2 + \frac32\bigg(\frac{x}{2}\bigg)^4 - \frac59\bigg(\frac{x}{2}\bigg)^6 + \frac{35}{288}\bigg(\frac{x}{2}\bigg)^8, \end{align*}

and at the same time

\begin{align*} \frac{1}{\sqrt{1+x^2}\,} \ge 1 - 2\bigg(\frac{x}{2}\bigg)^2 + \frac32\bigg(\frac{x}{2}\bigg)^4 - \frac59\bigg(\frac{x}{2}\bigg)^6 + \frac{35}{288}\bigg(\frac{x}{2}\bigg)^8, \end{align*}

yielding (A.1).