Proof of Proposition 4. The reason for constructing the Markov chain is to dominate the sequence of widths $\{L_n\}_{n\geq 0}$ . More precisely, by construction, $M_n\geq L_n$ for all $n\geq 0$ and, in particular, when $M_n=0$ , we have $L_n=0$ . Therefore, almost surely, ${\tau}^M \geq {\tau}$ , which completes the proof.

Proof of Lemma 6. Note that the Markov chain $\{M_n\}_{n\ge 0}$ is irreducible since, for any $m> 0$ ,

(8) \begin{align}\begin{split} \mathbb{P}(M_{1}=m \mid M_0=0 )& \geq \mathbb{P}( \lceil \overline{R}_1 \rceil =m \text{ and } \underline{\Phi}_1\notin T_\theta )>0\quad \text{ and}\\[5pt] \mathbb{P}(M_{1}=0 \mid M_0=m )& \geq \mathbb{P}( \lfloor \underline{R}_1\cos\theta \rfloor >m \text{ and } \underline{\Phi}_1\in T_\theta )>0. \end{split}\end{align}

Now, we demonstrate the recurrence of $\{M_n\}_{n\ge 0}$ by establishing that 0 is a recurrent state. From Lemma 5, we know that, for all $h\geq 2$ ,

\[\mathbb{P}( \lceil \overline{R}_1 \rceil >h)\leq \mathbb{P}( \overline{R}_1 >h/2) = \exp\!(\!-\!\lambda(\pi/2-\theta)h^2/4).\]

Let us define the sequence $\{a_n\}_{n\geq 1}$ as

\[a_n\;:\!=\;\Bigg\lceil \bigg( \frac{8}{\lambda(\pi/2-\theta)}\log n \bigg)^{1/2} \Bigg\rceil.\]

Then, for all sufficiently large n, we have $\mathbb{P}(\lceil \overline{R}_1 \rceil > a_n) \leq n^{-2}$ . This implies that, for all sufficiently large n,

\begin{align*} \mathbb{P}\Big(\max_{i=1}^n \lceil \overline{R}_i \rceil\leq a_n\Big)\geq (1-n^{-2})^n\end{align*}

and, therefore,

(9) \begin{align} \lim_{n\uparrow\infty} \mathbb{P}\Big(\max_{i=1}^n \lceil \overline{R}_i \rceil\leq a_n\Big) =1. \end{align}

Now, let $G_n\;:\!=\; \mathbb{1}{\{\underline{\Phi}_n\in T_\theta \text{ and } \underline{R}_n\cos\theta \geq 1\}} $ and $q\;:\!=\; \mathbb{P}(G_1=1)>0$ . From the definition of $M_n$ , it follows that, whenever $G_n=1$ , we have $M_n\leq (M_{n-1}-1)_+$ . We define $\mathcal{A}_n$ to be the event that the finite sequence $\{G_i\}_{i=1}^n$ has a run of 1s of length at least $a_n $ . Since $\{G_i\}_{i=1}^n$ are i.i.d., we have

\[\mathbb{P}({\mathcal{A}^c_n}) \leq ( 1-q^{ a_n } )^{\left\lfloor n/ a_n \right\rfloor} .\]

Furthermore, since, for all sufficiently large n,

\[ a_n \leq \frac{1}{2(\!-\!\log q)} \log n,\]

we have, for all sufficiently large n,

\[\mathbb{P}(\mathcal{A}^c_n) \leq (1-n^{-1/2})^{\left\lfloor 2(\!-\!\log q)n/\log n \right\rfloor},\]

which tends to 0 as $n\to\infty$ . Therefore, we obtain

(10) \begin{align}\lim_{n\uparrow\infty}\mathbb{P}(\mathcal{A}_n)=1. \end{align}

Now, by construction, we have

\begin{align*} &\mathbb{P}(M_{i}=0\text{ for some $1\leq i\leq n$} \mid M_0=0 ) \\[5pt] &\qquad\geq \mathbb{P}\Big(\max_{i=1}^n M_{i}\leq a_n \text{ and $\{G_{i}\}_{i=1}^n$ has $ a_n $ consecutive 1 s} \,\,\Big|\,\, M_0=0 \Big) \\[5pt] &\qquad\geq \mathbb{P}\Big(\max_{i=1}^n \lceil \overline{R}_i \rceil\leq a_n \text{ and $\{G_{i}\}_{i=1}^n$ has $ a_n $ consecutive 1 s} \,\,\Big| \,\, M_0=0 \Big) \\[5pt] &\qquad= \mathbb{P}\Big(\max_{i=1}^n \lceil \overline{R}_i \rceil\leq a_n \text{ and $\{G_{i}\}_{i=1}^n$ has $ a_n $ consecutive 1 s} \Big).\end{align*}

But this tends to 1 as $n\to\infty$ by (9) and (10). As a result, we established that 0 is a recurrent state, and therefore, in view of (8), the Markov chain $\{M_n\}_{n\ge 0}$ is recurrent.

The rest of the proof is similar to the proof of [Reference Asmussen2, Proposition 5.5], but we include some details for the convenience of the reader. Observe that

(11) \begin{align}\begin{split}&\mathbb{E}[{\operatorname e }^{M_1}\mid M_0=k]\\[5pt] &\qquad=\mathbb{E}\big[\!\exp\big((k-\lfloor \underline{R}_1\cos\theta \rfloor)_+\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} + \max\{ k, \lceil \overline{R}_1 \rceil \} \mathbb{1}{\{\underline{\Phi}_1\notin T_\theta\}}\big)\big]\\[5pt] &\qquad={\operatorname e }^k\mathbb{E}\big[\!\exp\big(\!-\!\min\{k,\lfloor \underline{R}_1\cos\theta \rfloor\}\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} + \max\{ 0, \lceil \overline{R}_1 \rceil -k \} \mathbb{1}{\{\underline{\Phi}_1\notin T_\theta\}}\big) \big] \end{split}\end{align}

and note that

\[\!\exp\big(\!-\!\min\{k,\lfloor \underline{R}_1\cos\theta \rfloor\}\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} + \max\{ 0, \lceil \overline{R}_1 \rceil -k \} \mathbb{1}{\{\underline{\Phi}_1\notin T_\theta\}}\big)\le \exp\!( \lceil \overline{R}_1 \rceil ),\]

which has finite expectation. Therefore, for all $k\geq 0$ ,

(12) \begin{equation}\mathbb{E}[\!\exp\!(M_1)\mid M_0=k] <\infty.\end{equation}

Moreover, by the dominated-convergence theorem, we get

\begin{align*}&\lim_{k\uparrow\infty} \mathbb{E}[\!\exp\!(\!-\!\min\{k,\lfloor \underline{R}_1\cos\theta \rfloor\}\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} + \max\{ 0, \lceil \overline{R}_1 \rceil -k) \} \mathbb{1}{\{\underline{\Phi}_1\notin T_\theta\}}) ]\\[5pt] &\qquad=\mathbb{E}[\!\exp\!(\!-\!\lfloor \underline{R}_1\cos\theta \rfloor\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} ) ]\\[5pt] &\qquad <1\end{align*}

and, thus, there exists $r>1$ and $k_0\geq 0$ such that, for all $k>k_0$ ,

\[\mathbb{E}[\!\exp\!(\!-\!\min\{k,\lfloor \underline{R}_1\cos\theta \rfloor\}\mathbb{1}{\{\underline{\Phi}_1\in T_\theta\}} + \max\{ 0, \lceil \overline{R}_1 \rceil -k) \} \mathbb{1}{\{\underline{\Phi}_1\notin T_\theta\}})] <r^{-1}.\]

But this, together with (11), implies that, for all $k>k_0$ ,

(13) \begin{equation}\mathbb{E}[\!\exp\!(M_1)\mid M_0=k] < r^{-1}\exp\!(k).\end{equation}

Let $\sigma\;:\!=\; \inf\{n>0\colon M_n\le k_0\}$ denote the first time that $M_n$ is below $k_0$ and fix any $l>k_0$ . In view of (13), note that, for $M_0=l$ , the process $\{Y_n\}_{n\geq 0}$ defined as $Y_n\;:\!=\;r^{n\wedge \sigma}\exp\!(M_{n\wedge \sigma})$ is a nonnegative supermartingale. Therefore, by the recurrence of the Markov chain $\{M_n\}_{n\ge 0}$ , almost surely,

\[Y_n\to Y_{\infty}\;:\!=\; r^{ \sigma}\exp\!(M_{ \sigma})\geq r^{\sigma}.\]

This, together with Fatou’s lemma, implies that

\[\mathbb{E}[r^{\sigma} \mid M_0=l] \leq \mathbb{E}[Y_{\infty} \mid M_0=l] \leq \liminf_{n\uparrow\infty} \mathbb{E}[Y_{n} \mid M_0=l] \leq \mathbb{E}[Y_{0} \mid M_0=l]=\exp\!(l).\]

Since $l>k_0$ is arbitrary, we have, for all $l>k_0$ ,

\[\mathbb{E}[r^{\sigma}|M_0=l] \leq \exp\!(l).\]

Therefore, for any $z\leq k_0$ ,

\begin{align*} \mathbb{E}[r^{\sigma} \mid M_0=z] &\leq r+r\sum_{l>k_0} \mathbb{P}(M_1=l \mid M_0=z) \mathbb{E}[r^{\sigma} | M_0=l]\\ & \leq r +r\sum_{l>k_0} \mathbb{P}(M_1=l \mid M_0=z) \exp\!(l) \\ &\leq r +r \mathbb{E}[\!\exp\!(M_1) \mid M_0=z] \\ &<\infty.\end{align*}

Note that, for any $z\leq k_0$ and $l>k_0$ ,

\[\mathbb{P}(M_1=l \mid M_0=z)= \mathbb{P}(\underline{\Phi}_1\notin T_\theta, \lceil \overline{R}_1\rceil =l) = \mathbb{P}(M_1=l \mid M_0\leq k_0),\]

which does not depend on z. Therefore, if we define $\{\sigma_i\}_{i\geq 0}$ as $\sigma_0\;:\!=\;0$ and, for all $i\ge 1$ ,

\[\sigma_{i}\;:\!=\; \inf\{n> \sigma_{i-1}\colon M_{n}\le k_0\} ,\]

then $\{\sigma_{i}-\sigma_{i-1}\}_{i\ge 1}$ are i.i.d. copies of $\sigma$ . Moreover, for all $z\leq k_0$ ,

\[c_1\;:\!=\;\mathbb{E}[r^{\sigma} \mid M_0\le k_0]=\mathbb{E}[r^{\sigma} \mid M_0=z]<\infty.\]

We choose $c_2>0$ such that ${c_1}^{c_2}< r$ . Then, by Markov’s inequality,

\[ \mathbb{P}(\sigma_{\lfloor c_2n\rfloor} \geq n \mid M_0=0) \leq r^{-n}\mathbb{E}[r^{\sigma_{\lfloor c_2n\rfloor}} \mid M_0=0] = r^{-n}\mathbb{E}[r^{\sigma_1} \mid M_0=0]^{\lfloor c_2n\rfloor}\leq (r^{-1}{c_1}^{c_2})^n ,\]

and thus,

\begin{align*} &\mathbb{P}({\tau}^M>n \mid M_0=0\big) \\ &\qquad\le \mathbb{P}({\tau}^M>n, \sigma_{\lfloor c_2n\rfloor} <n \mid M_0=0) + \mathbb{P}(\sigma_{\lfloor c_2n\rfloor} \geq n \mid M_0=0) \\ &\qquad \le \mathbb{P}\Bigg(\bigcap_{i=1}^{\lfloor c_2n\rfloor} \{ \underline{\Phi}_{\sigma_i+1}\notin T_\theta \text{ or } \lfloor \underline{R}_{\sigma_i+1}\cos\theta \rfloor <k_0 \} \Bigg) + \mathbb{P}(\sigma_{\lfloor c_2n\rfloor} \geq n\mid M_0=0) \\ &\qquad \le \mathbb{P}( \underline{\Phi}_{1}\notin T_\theta \text{ or } \lfloor \underline{R}_{1}\cos\theta \rfloor <k_0 )^{\lfloor c_2n\rfloor} + (r^{-1}{c_1}^{c_2})^n \\ &\qquad\le C_1 {\operatorname e }^{-cn}\end{align*}

for some $C_1,c>0$ . This proves the result.

Proof of Lemma 7. From the definition of $H_n$ , we see that, for any $n\geq0$ ,

\[ H_n\subseteq \bigcup_{m=0}^{n-1} B(V_m, R_{m+1}). \]

Now, since, for any $0\le m \le n-1$ , $V_n\in {\mathcal C }_{\theta}(V_{m})$ , it follows that $R_{m+1}\le |V_n-V_m|$ . Therefore,

\[H_n\subseteq \bigcup_{m=0}^{n-1} B(V_m, |V_n-V_m|).\]

Hence, it suffices to show that, for any $0\le m \le n-1$ , ${\mathcal C }_{\pi/2-\theta}(V_{n}) \cap B(V_m, |V_n-V_m|)=\emptyset$ . For this, note that the lines

\begin{align*} L_{n}(\pi/2-\theta) &\;:\!=\;V_n+\{(r, \varphi)\colon r>0 , \varphi=\pi/2-\theta\} \quad\text{ and } \\[5pt] L_{n}(\!-\!\pi/2+\theta)&\;:\!=\;V_n+\{(r, \varphi)\colon r>0 , \varphi=-\pi/2+\theta\}\end{align*}

are the two boundary lines of ${\mathcal C }_{\pi/2-\theta}(V_{n})$ . Hence, it is also adequate to prove that, for any $0\le m \le n-1$ , $(L_{n}(\pi/2-\theta) \cup L_{n}(\!-\!\pi/2+\theta)) \cap B(V_m, |V_n-V_m|)=\emptyset$ .

We now fix $m<n$ and write $V_m$ and $V_n$ in Cartesian coordinates as $V_m=(V_{m,1},V_{m,2} )$ and $V_n=(V_{n,1},V_{n,2} )$ . If $V_{m,2}=V_{n,2}$ then $|V_n-V_m|= V_{n,1}-V_{m,1}$ and since, for any $x=(x_1,x_2)\in L_{n}(\pi/2-\theta) \cup L_{n}(\!-\!\pi/2+\theta)$ , we have $x_1> V_{n,1}> V_{m,1}$ , it follows that

\[|x-V_m|\geq x_1- V_{m,1}> V_{n,1}- V_{m,1} =|V_n-V_m|.\]

Therefore, $(L_{n}(\pi/2-\theta) \cup L_{n}(\!-\!\pi/2+\theta)) \cap B(V_m, |V_n-V_m|)=\emptyset$ .

Now, suppose that $V_{m,2}\neq V_{n,2}$ . Without loss of generality, we assume that $V_{m,2}< V_{n,2}$ . Then, for any $x=(x_1,x_2)\in L_{n}(\pi/2-\theta)$ , we have $x_1> V_{n,1}> V_{m,1}$ and $x_2> V_{n,2} > V_{m,2}$ , which implies that

\[|x-V_m| = \sqrt{(x_1- V_{m,1})^2+(x_2- V_{m,2})^2} > \sqrt{(V_{n,1}- V_{m,1})^2+(V_{n,2}- V_{m,2})^2} =|V_n-V_m|.\]

Therefore, $L_{n}(\pi/2-\theta) \cap B(V_m, |V_n-V_m|)=\emptyset$ . Now, we draw two horizontal line segments $\overline{V_m A}$ and $\overline{V_n B}$ passing through $V_m$ and $V_n$ , respectively, such that $\overline{V_m A}$ intersects $L_{n}(\!-\!\pi/2+\theta)$ at point C; see Figure 2 for an illustration. From the definition of $L_{n}(\!-\!\pi/2+\theta)$ , we know that $\angle B V_n C = \pi/2-\theta$ . Since $\overline{V_m A}$ and $\overline{V_n B}$ are parallel, we also have $\angle V_n C V_m = \pi/2-\theta$ . Furthermore, as $V_n\in {\mathcal C }_{\theta}(V_{m})$ , it follows that $\angle V_n V_m C \leq \theta$ . Now, focusing on the triangle $\triangle V_n V_m C$ , we observe the relationship

\[\angle V_n C V_m + \angle V_n V_m C +\angle V_m V_n C = \pi,\]

which implies that $\angle V_m V_n C \geq \pi/2$ , and this, in turn, implies that $\cos\angle V_m V_n C\leq 0$ . Therefore, for any $x\in L_{n}(\!-\!\pi/2+\theta)$ , by the law of cosines,

\[|x-V_m| = \sqrt{|V_n-V_m|^2+ |x-V_n|^2 - 2\cos\angle V_m V_n C\cdot|V_n-V_m|\cdot |x-V_n| } > |V_n-V_m|,\]

which means that $L_{n}(\pi/2-\theta) \cap B(V_m, |V_n-V_m|)=\emptyset$ . This proves that, for any $0\le m \le n-1$ , ${\mathcal C }_{\pi/2-\theta}(V_{n}) \cap B(V_m, |V_n-V_m|)=\emptyset$ , thereby completing the proof.

Proof of Lemma 5. For $0<\theta\le \pi/4$ , the random variables $\{(R_n,\Phi_n)\}_{n\ge 1}$ are i.i.d. Therefore, Lemma 5 holds trivially by taking $(\underline R_n, \underline\Phi_n)=(\overline R_n, \overline\Phi_n)=(R_n,\Phi_n)$ for all $n\ge1$ . So, we now assume that $\pi/4<\theta<\pi/2$ . We define a sequence of sets $\{D_n\}_{n\geq 1}$ as

\[D_n\;:\!=\; {\mathcal C }_{\theta}(V_{n-1}) \cap \overline{B(V_{n-1}, R_n)} \cap H^c_{n-1}.\]

Essentially, $D_n$ is the previously unexplored set where we have explored at the nth step to find $V_n$ . Clearly, $\{D_n\}_{n\geq 1}$ are pairwise disjoint. Let $\{{\mathcal P }_{\lambda}^{(n)}\}_{n\geq 1}$ be i.i.d. copies of the Poisson point process ${\mathcal P }_{\lambda}$ . We define $\{{\mathcal Q }_{\lambda}^{(n)}\}_{n\geq 1}$ as

\[{\mathcal Q }_{\lambda}^{(n)}\;:\!=\; ({\mathcal P }_{\lambda}\cap D_n)\cup( {\mathcal P }_{\lambda}^{(n)}\cap D^c_n).\]

Since the $D_n$ are disjoint, $\{{\mathcal Q }_{\lambda}^{(n)}\}_{n\geq 1}$ is a sequence of i.i.d. Poisson point processes with intensity $\lambda$ . We write

\[\underline{W}_n \;:\!=\; \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\theta}(V_{n-1}) \cap {\mathcal Q }_{\lambda}^{(n)} \},\]

and define $(\underline{R}_n,\underline{\Phi}_n)$ to be the polar coordinates of $\underline{W}_n-V_{n-1}$ . Similarly, we write

\[\overline{W}_n \;:\!=\; \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\pi/2-\theta}(V_{n-1}) \cap {\mathcal Q }_{\lambda}^{(n)} \},\]

and define $(\overline{R}_n,\overline{\Phi}_n)$ to be the polar coordinates of $\overline{W}_n-V_{n-1}$ . From the definition, it follows that $(\underline{R}_n,\underline{\Phi}_n,\overline{R}_n,\overline{\Phi}_n)$ and $ (\underline{R},\underline{\Phi}, \overline{R},\overline{\Phi})$ are equal in distribution. Furthermore, note that, since $\{{\mathcal Q }_{\lambda}^{(n)}\}_{n\geq 1}$ are i.i.d., $\{(\underline{R}_n, \underline{\Phi}_n, \overline{R}_n, \overline{\Phi}_n)\}_{n\geq 1}$ are also i.i.d. We know that $(R_n,\Phi_n)$ are the polar coordinates of $V_n-V_{n-1}$ , where

\begin{align*}V_n &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\theta}(V_{n-1}) \cap {\mathcal P }_{\lambda} \}\\[5pt] &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\theta}(V_{n-1})\cap \overline{B(V_{n-1}, R_n)} \cap H^c_{n-1} \cap {\mathcal P }_{\lambda} \}\\[5pt] &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in D_n \cap {\mathcal P }_{\lambda} \}\\[5pt] &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in D_n \cap {\mathcal Q }_{\lambda}^{(n)} \}\\[5pt] &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\theta}(V_{n-1})\cap \overline{B(V_{n-1}, R_n)} \cap H^c_{n-1} \cap {\mathcal Q }_{\lambda}^{(n)} \}\\[5pt] &= \textrm{argmin} \{|v-V_{n-1}|\colon v \in {\mathcal C }_{\theta}(V_{n-1}) \cap H^c_{n-1} \cap {\mathcal Q }_{\lambda}^{(n)} \}.\end{align*}

Now, from Lemma 7, we obtain ${\mathcal C }_{\pi/2-\theta}(V_{n-1})\subseteq {\mathcal C }_{\theta}(V_{n-1}) \cap H^c_{n-1}\subseteq {\mathcal C }_{\theta}(V_{n-1})$ , which implies that $\overline{R}_n \geq R_n \geq \underline{R}_n$ . Additionally, if $\underline{\Phi}_n \in T_\theta$ then we have $(\underline{R}_n,\underline{\Phi}_n)=(R_n,\Phi_n)=(\overline{R}_n,\overline{\Phi}_n)$ . This concludes the proof.

Proof of Lemma 2. Using Proposition 4, Lemma 5, and the Cauchy–Schwarz inequality, we get

\begin{align*}\mathbb{E}[{\operatorname e }^{\langle\gamma, U'_1\rangle}]\le \sum_{n\ge 1}\mathbb{E}[{\operatorname e }^{2\langle\gamma, \sum_{j=1}^n U_j\rangle}]^{1/2}\mathbb{P}({\tau}_1^{\theta}\ge n)^{1/2}\le\sum_{n\ge 1}\mathbb{E}[{\operatorname e }^{2(\gamma_1+\gamma_2)\overline{R}}]^{n/2}\sqrt{C}{\operatorname e }^{-cn/2}.\end{align*}

Since $\overline{R}$ has all exponential moments finite, for all sufficiently small $\gamma_1,\gamma_2>0$ , the above sum is also finite.

Proof of Proposition 2. By symmetry and the exponential Markov inequality, for sufficiently small $s>0$ ,

\begin{align*}\mathbb{P}(|Y'_1|\ge n^{1/2+\varepsilon})\le 2{\operatorname e }^{-sn^{1/2+\varepsilon}}\mathbb{E}[{\operatorname e }^{s Y_1'}]<\infty,\end{align*}

where we also used Lemma 2, and thus,

\begin{align*}\limsup_{n\uparrow\infty}n^{-2\varepsilon}\log\mathbb{P}(|Y'_1|\ge n^{1/2+\varepsilon})\le -s\limsup_{n\uparrow\infty}n^{1/2-\varepsilon}=-\infty.\end{align*}

Hence, $\{{\mathcal Y }_{\lfloor t\rfloor}^w\}_{t\ge 0}$ satisfies the conditions of [Reference Eichelsbacher and Löwe9, Theorem 2.2] and, thus, obeys the moderate-deviation principle with rate $x^2/(2w\mathbb{E}[Y_1'^2])$ . But $\{{\mathcal Y }_{\lfloor t\rfloor}^w\}_{t\ge 0}$ and $\{{\mathcal Y }_{t}^w\}_{t\ge 0}$ are exponentially equivalent since, for all $\delta>0$ ,

\begin{align*} \limsup_{t\uparrow\infty} t^{-2\varepsilon}&\log \mathbb{P}(|{\mathcal Y }_{\lfloor t\rfloor}^w-{\mathcal Y }_{t}^w|\geq \delta t^{1/2+\varepsilon})\le \limsup_{t\uparrow\infty} t^{-2\varepsilon}\log \mathbb{P}\Bigg(\sum_{i=1}^{\lceil w+1\rceil}|Y_i'|\geq \delta t^{1/2+\varepsilon}\Bigg),\end{align*}

as $\lfloor tw\rfloor-\lfloor \lfloor t\rfloor w\rfloor\le \lceil w+1\rceil$ . Hence, again using symmetry, independence, and the exponential Markov inequality, for sufficiently small $s>0$ ,

\begin{align*}\mathbb{P}\Bigg(\sum_{i=1}^{\lceil w+1\rceil}|Y_i'|\geq \delta t^{1/2+\varepsilon}\Bigg)\le {\operatorname e }^{-s\delta t^{1/2+\varepsilon}}\mathbb{E}[{\operatorname e }^{s |Y_1'|}]^{\lceil w+1\rceil}\le {\operatorname e }^{-s\delta t^{1/2+\varepsilon}}2^{\lceil w+1\rceil}\mathbb{E}[{\operatorname e }^{s Y_1'}]^{\lceil w+1\rceil}<\infty,\end{align*}

and thus,

\begin{align*}\limsup_{t\uparrow\infty}t^{-2\varepsilon}\log\mathbb{P}(|{\mathcal Y }_{\lfloor t\rfloor}^w-{\mathcal Y }_{t}^w|\geq \delta t^{1/2+\varepsilon})\le -s\delta\limsup_{t\uparrow\infty}t^{1/2-\varepsilon}=-\infty,\end{align*}

as desired.

Proof of Proposition 1. First note that we have ${\mathcal Y }_t=\sum_{i=1}^{K_t}Y_i+Y_{K_t+1}(t-\sum_{i=1}^{K_t}X_i)/X_{K_t+1}$ with $(t-\sum_{i=1}^{K_t}X_i)/X_{K_t+1}\le 1$ , and hence, by Lemma 5,

\begin{align*}\mathbb{P}(|{\mathcal Y }_t-{\mathcal Y }_t'|\geq \delta t^{1/2+\varepsilon})&\le\mathbb{P}\Bigg(\sum_{i=\tau^\theta_{K'_t}+1}^{K_t+1}|Y_i|\geq \delta t^{1/2+\varepsilon}\Bigg)\\[5pt] &\le\mathbb{P}\Bigg(\sum_{i=\tau^\theta_{K'_t}+1}^{\tau^\theta_{K'_t+1}}|Y_i|\geq \delta t^{1/2+\varepsilon}\Bigg)\\[5pt] &\le\sum_{j=0}^{\lfloor t^2\rfloor}\mathbb{P}\Bigg(\sum_{i=\tau^\theta_{j}+1}^{\tau^\theta_{j+1}}|Y_i|\geq \delta t^{1/2+\varepsilon}, K'_t=j \Bigg) + \mathbb{P}(K'_t \ge t^2)\\[5pt] &\le t^2 \mathbb{P}\Bigg(\sum_{i=1}^{\tau^\theta_1}|Y_i|\geq \delta t^{1/2+\varepsilon}\Bigg) + \mathbb{P}(K'_t \ge t^2)\\[5pt] &\le t^2 \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t^{\varepsilon'}\rfloor}\overline{R}_i\geq \delta t^{1/2+\varepsilon}\Bigg) + t^2 \mathbb{P}(\tau^\theta_1 > \lfloor t^{\varepsilon'}\rfloor) + \mathbb{P}\Bigg(\sum_{i=1}^{ \lceil t^2\rceil}X'_i \le t \Bigg)\end{align*}

for some $\varepsilon'\in(2\varepsilon, 1/2+\varepsilon)$ . This, together with Proposition 4 and the exponential Markov inequality, implies that

\begin{align*} \mathbb{P}(|{\mathcal Y }_t-{\mathcal Y }_t'|\geq \delta t^{1/2+\varepsilon})\le t^2{\operatorname e }^{-\delta t^{1/2+\varepsilon} }\mathbb{E}[{\operatorname e }^{\overline{R}}]^{\lfloor t^{\varepsilon'}\rfloor}+ t^2 C{\operatorname e }^{-c\lfloor t^{\varepsilon'}\rfloor} + {\operatorname e }^{t}\mathbb{E}[{\operatorname e }^{-X'_1}]^{\lceil t^2\rceil},\end{align*}

and therefore,

\begin{align*}&\limsup_{t\uparrow\infty} t^{-2\varepsilon}\log \mathbb{P}(|{\mathcal Y }_t-{\mathcal Y }_t'|\geq \delta t^{1/2+\varepsilon})\\[5pt] &\qquad\le -\min\Big\{ \liminf_{t\uparrow\infty} t^{-2\varepsilon}(\delta t^{1/2+\varepsilon}- \lfloor t^{\varepsilon'}\rfloor \log \mathbb{E}[{\operatorname e }^{\overline{R}}] ),\liminf_{t\uparrow\infty} c\lfloor t^{\varepsilon'}\rfloor t^{-2\varepsilon},\\[5pt] &\hskip61pt -\limsup_{t\uparrow\infty}t^{-2\varepsilon}(t+\lceil t^2\rceil\log\mathbb{E}[{\operatorname e }^{-X'_1}] ) \Big\}\\[5pt] &\qquad=-\infty,\end{align*}

as desired.

Proof of Lemma 8. First, note that

\begin{align*} \kappa^{-1}&=\mathbb{E}[X_1'] \\[5pt] &\ge \mathbb{E}[X_1{\mathbb{1}}\{\tau^\theta_1=1\}] \\[5pt] &=\lambda\int_0^\infty\textrm{d} r\; r^2{\operatorname e }^{-\theta r^2}\int_{-\theta}^{\theta}\textrm{d} \varphi\; \cos\varphi{\mathbb{1}}\Bigg\{ \theta-\frac{\pi}{2}\le \varphi\le \frac{\pi}{2}- \theta\Bigg\} \\[5pt] &>0,\end{align*}

since independence after one step is achieved precisely if the angle is in the interval $[\theta-\pi/2,\pi/2- \theta]$ . Using Lemma 5 and Proposition 4 together with the Cauchy–Schwarz inequality, we get

\begin{align*}\kappa^{-1} &=\mathbb{E}[X_1']\\[5pt] &=\mathbb{E}\left[\sum_{i=1}^{\tau}X_i\right]\\[5pt] &\leq \mathbb{E}\left[\sum_{i=1}^{\tau}\overline{R}_i\right]\\[5pt] &\leq \sum_{i=1}^{\infty}\mathbb{E}[\overline{R}_i\mathbb{1}{\{\tau\ge i\}}]\\[5pt] &\le \sum_{i=1}^{\infty}(\mathbb{E}[\overline{R}_i^2]\mathbb{P}(\tau\ge i))^{1/2}\\[5pt] &\le \sum_{i=1}^{\infty} (\mathbb{E}[\overline{R}^2]C)^{1/2} {\operatorname e }^{-ci/2}\\[5pt] &<\infty.\end{align*}

For the second part of the statement, note that we can bound

(15) \begin{align}\mathbb{P}(K'_t/t \notin B_{\varepsilon}( \kappa))&\le\mathbb{P}(K'_t\le t(\kappa-\varepsilon))+\mathbb{P}(K'_t\ge t(\kappa+\varepsilon))\nonumber\\[5pt] &\le \mathbb{P}\Bigg(\sum_{i=1}^{\lceil t(\kappa-\varepsilon)\rceil}X'_i \ge t\Bigg)+\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t(\kappa+\varepsilon)\rfloor}X'_i < t\Bigg), \end{align}

where we used the fact that $K'_t$ , as defined in (2), can also be represented as

\begin{equation*}K'_t=\sup\bigg\{n>0\colon\sum_{i=1}^{n}X'_i < t \bigg\}.\end{equation*}

Now, by the exponential Markov inequality, for any $s>0$ ,

\begin{align*}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\Bigg(\sum_{i=1}^{\lceil t(\kappa-\varepsilon)\rceil}X'_i \ge t\Bigg)\le -s(1-(\kappa-\varepsilon)s^{-1}\log \mathbb{E}[{\operatorname e }^{s X_1'}]).\end{align*}

Using Lemma 5 and Proposition 4 together with the Cauchy–Schwarz inequality, we obtain

\begin{align*} \mathbb{E}[{\operatorname e }^{s X_1'}] &\leq \mathbb{E}[{\operatorname e }^{\sum_{i=1}^{\tau} s\overline R_i}] \\[5pt] &=\sum_{n=1}^{\infty} \mathbb{E}[{\operatorname e }^{\sum_{i=1}^{n} s\overline R_i}\mathbb{1}{\{\tau=n\}}] \\[5pt] &\le \sum_{n=1}^{\infty} (\mathbb{E}[{\operatorname e }^{\sum_{i=1}^{n} 2s\overline R_i}]\mathbb{P}(\tau=n))^{1/2} \\[5pt] &\le \sum_{n=1}^{\infty} (\mathbb{E}[{\operatorname e }^{ 2s\overline R}])^{n/2} \sqrt{C}{\operatorname e }^{-cn/2},\end{align*}

which is finite for all $0<s\leq s_0$ and for some $s_0>0$ , as a consequence of Lemma 5. Hence, we can use dominated convergence to conclude that $\lim_{s\downarrow 0} \mathbb{E} [ {\operatorname e }^{sX'_1}]=1$ . Then, we have

\begin{align*} \lim_{s\downarrow 0} s^{-1}\log \mathbb{E}[{\operatorname e }^{s X_1'}] &= \lim_{s\downarrow 0} s^{-1}\log(1+ ( \mathbb{E}[{\operatorname e }^{s X_1'}]-1)) \\[5pt] &= \lim_{s\downarrow 0} s^{-1}( \mathbb{E}[{\operatorname e }^{s X_1'}]-1) \\[5pt] &= \mathbb{E}[ \lim_{s\downarrow 0} s^{-1}( {\operatorname e }^{s X_1'}-1)] \\[5pt] &= \mathbb{E}[X'_1] \\[5pt] &= \kappa^{-1}.\end{align*}

Here, the third equality also follows from dominated convergence using the bound $s^{-1}( {\operatorname e }^{s X_1'}-1)<s_0^{-1}( {\operatorname e }^{s_0 X_1'}-1)$ for all $0<s\le s_0$ , where $\mathbb{E}[s_0^{-1}( {\operatorname e }^{s_0 X_1'}-1)]<\infty$ . Therefore, there exists $c'_1>0$ such that

(16) \begin{align} \limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\Bigg(\sum_{i=1}^{\lceil t(\kappa-\varepsilon)\rceil}X'_i \ge t\Bigg)<-c'_1. \end{align}

Again, by using the exponential Markov inequality on the other term, we obtain, for all $s>0$ ,

\begin{align*}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t(\kappa+\varepsilon)\rfloor}-X'_i > -t\Bigg)\le s(1+(\kappa+\varepsilon)s^{-1}\log \mathbb{E}[{\operatorname e }^{-s X_1'}]).\end{align*}

Since, by the dominated-convergence theorem, $\lim_{s\downarrow 0} \mathbb{E} [ e^{-sX'_1}]=1$ , we have

\begin{align*} \lim_{s\downarrow 0} s^{-1}\log \mathbb{E}[{\operatorname e }^{-s X_1'}] &= \lim_{s\downarrow 0} s^{-1}\log(1- (1- \mathbb{E}[{\operatorname e }^{-s X_1'}])) \\[2pt] &= \lim_{s\downarrow 0} -s^{-1}(1- \mathbb{E}[{\operatorname e }^{-s X_1'}]) \\[2pt] &= \mathbb{E}[ \lim_{s\downarrow 0} -s^{-1}(1- {\operatorname e }^{-s X_1'})] \\[2pt] &= \mathbb{E}[-X'_1] \\[2pt] &= -\kappa^{-1}.\end{align*}

Here, the third equality follows from the dominated-convergence theorem, using the bound $s^{-1}\big(1- {\operatorname e }^{-s X'_1}\big)<X'_1$ , where $X'_1$ has finite expectation. Therefore, there exists $c'_2>0$ such that

(17) \begin{align} \limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t(\kappa+\varepsilon)\rfloor}-X'_i > -t\Bigg)<-c'_2. \end{align}

Now, combining (15), (16), and (17) proves the lemma.

Proof of Proposition 3. Let us fix $\varepsilon'>0$ and note that

(18) \begin{equation}\begin{split}\mathbb{P}(|{\mathcal Y }^\kappa_t-{\mathcal Y }'_t|\ge \delta t^{1/2+\varepsilon})&\le\mathbb{P}\Bigg(\Bigg\{\Bigg\vert\sum_{i=\lfloor t\kappa\rfloor+1}^{K'_t}Y_i'\Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg\}\cap \Bigg\{\kappa\le \frac{K'_t}{t}\le\kappa+\varepsilon'\Bigg\}\Bigg)\\[2pt] &\hskip10pt+\mathbb{P}\Bigg(\Bigg\{\Bigg\vert\sum_{i=K'_t+1}^{\lfloor t\kappa\rfloor}Y_i'\Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg\}\cap \Bigg\{\kappa-\varepsilon'\le \frac{K'_t}{t}\le \kappa\Bigg\}\Bigg)\\[2pt] &\hskip10pt+\mathbb{P}\Bigg(\frac{K'_t}{t}\not\in B_{\varepsilon'}(\kappa)\Bigg),\end{split}\end{equation}

where, by Lemma 8, $\limsup_{t\uparrow\infty}t^{-2\varepsilon}\log\mathbb{P}(K'_t/t\not\in B_{\varepsilon'}(\kappa))=-\infty$ and, hence, the third summand plays no role, by logarithmic equivalence. For the first summand in (18), since the $Y_i'$ are i.i.d., we have the bound

(19) \begin{align}\sum_{m=\lfloor t\kappa \rfloor+1}^{\lfloor t(\kappa+\varepsilon') \rfloor}\mathbb{P}\Bigg(\Bigg\{\Bigg\vert\sum_{i=\lfloor t\kappa\rfloor+1}^{K'_t}Y_i'\Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg\}\cap \{ K'_t=m\}\Bigg)\le \sum_{m=1}^{r_1(t,\varepsilon')}\mathbb{P}\Bigg(\Bigg\vert\sum_{i=1}^{m}Y_i' \Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg),\end{align}

where the number of summands in the above sum is $r_1(t, \varepsilon')\;:\!=\;\lfloor t(\kappa+\varepsilon')\rfloor-\lfloor t\kappa\rfloor$ . Note that the corresponding upper bound is valid also for the second summand in (18) with $r_1(t,\varepsilon')$ replaced by $r_2(t, \varepsilon')\;:\!=\;\lfloor t\kappa\rfloor-\lfloor t(\kappa-\varepsilon')\rfloor$ . Furthermore, since $\limsup_{t\uparrow \infty}t^{-2\varepsilon}\log r(t,\varepsilon')=0$ , where $r(t,\varepsilon')=r_1(t,\varepsilon')\vee r_2(t,\varepsilon')$ , we have

\begin{align*}\limsup_{t\uparrow \infty}t^{-2\varepsilon}\log\mathbb{P}(|{\mathcal Y }^\kappa_t-{\mathcal Y }'_t|\ge \delta t^{1/2+\varepsilon})&\le \limsup_{t\uparrow \infty}t^{-2\varepsilon}\sup_{0\le \alpha \le \varepsilon'}\log\mathbb{P}\Bigg(\Bigg\vert\sum_{i=1}^{\lfloor t\alpha\rfloor}Y_i'\Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg)\\[2pt] &\le \limsup_{t\uparrow \infty}t^{-2\varepsilon}\log\mathbb{P}\Bigg(\Bigg\vert\sum_{i=1}^{\lfloor t\varepsilon'\rfloor}Y_i'\Bigg\vert\ge \delta t^{1/2+\varepsilon}\Bigg),\end{align*}

where we used the fact that, by symmetry and independence,

\begin{align*}\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t\alpha\rfloor}Y_i'\ge \delta t^{1/2+\varepsilon}\Bigg)&=2 \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t\alpha\rfloor}Y_i'\ge \delta t^{1/2+\varepsilon}, \sum_{i=\lfloor t\alpha\rfloor+1}^{\lfloor t\varepsilon'\rfloor}Y_i'\ge 0\Bigg)\le 2 \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t\varepsilon'\rfloor}Y_i'\ge \delta t^{1/2+\varepsilon}\Bigg),\end{align*}

and similar for $\mathbb{P}(\sum_{i=1}^{\lfloor t\alpha\rfloor}Y_i'\le - \delta t^{1/2+\varepsilon})$ . Hence, using the moderate-deviation principle, Proposition 2, we have

\begin{align*}\limsup_{t\uparrow \infty}t^{-2\varepsilon}\log\mathbb{P}(|{\mathcal Y }^\kappa_t-{\mathcal Y }'_t|\ge \delta t^{1/2+\varepsilon})&\le -I^{\varepsilon'}_{\lambda,\theta}(\delta),\end{align*}

which tends to $-\infty$ as $\varepsilon'$ tends to 0, as desired.

Proof of Lemma 1. The statement follows from an application of the multivariate Cramér theorem for empirical means of sequences of i.i.d. random variables that possess exponential moments; see [Reference Dembo and Zeitouni7, Corollary 6.1.6]. In order to compute the exponential moments, we use polar coordinates, i.e. consider $U_1=(R,\Phi)\in \mathbb{R}_+\times [-\pi,\pi)$ . As already used earlier, the radius follows a Rayleigh distribution, i.e.

$$\mathbb{P}(R>r)=\exp\!(\!-\!\lambda \theta r^2),$$

and we note that, due to the isotropy of the model, $\Phi$ is uniformly distributed in $[-\theta, \theta]$ . Hence,

(24) \begin{equation}\begin{split}\mathbb{E}[\!\exp\!(\langle \gamma, U_1\rangle)]&=\int_0^\infty\textrm{d} r\;\frac{1}{2\theta}\int_{-\theta}^{\theta}\textrm{d} \varphi\; \exp\!(\gamma_1 r\cos\varphi+\gamma_2 r\sin\varphi-\lambda\theta r^2)\lambda 2\theta r\\[5pt] &=\lambda \int_0^\infty\textrm{d} r\; r\exp\!(\!-\!\lambda\theta r^2)\int_{-\theta}^{\theta}\textrm{d} \varphi\;\exp\!(\gamma_1 r\cos\varphi+\gamma_2 r\sin\varphi),\end{split}\end{equation}

as desired. In particular,

\begin{align*}{\operatorname e }^{J_{\lambda, \theta}(\gamma)}\le 2\pi \lambda \int_0^\infty\textrm{d} r\; r\exp\!(\!-\!\lambda\theta r^2)\exp\!(r(|\gamma_1|+|\gamma_2|))<\infty.\end{align*}

Since $J_{\lambda, \theta}$ is strictly convex and differentiable, by [Reference Rockafellar14, Theorem 1], its Legendre transform $\mathcal J_{\lambda, \theta}$ is strictly convex and differentiable on $\{u\in \mathbb{R}^2\colon \mathcal J_{\lambda,\theta}(u)<\infty\}$ .

Further note that, writing $u=(r_u, \varphi_u)$ in polar coordinates,

\begin{align*} {\operatorname e }^{-\mathcal J_{\lambda,\theta}(u)}=\inf_{q\geq 0, \,\alpha\in[-\pi,\pi]} \int_{0}^{\infty}\textrm{d} r\;\int_{-\theta}^{\theta}\textrm{d} \varphi\;\lambda r \exp\!(\!-\!\lambda\theta r^2 + q[r\cos(\varphi-\alpha) - r_u\cos(\varphi_u-\alpha) ] ),\end{align*}

and, if $\varphi_u\notin (\!-\!\theta, \theta)$ , then we can always choose $\alpha_0\in[-\pi,\pi] $ such that, for all $\varphi \in [-\theta,\theta]$ , we have $r\cos(\varphi-\alpha_0) - r_u\cos(\varphi_u-\alpha_0)\leq 0$ . One choice of $\alpha_0$ for example is given by

\begin{align*} \alpha_0=\left\{ \begin{array}{l@{\quad}l} {\pi}/{2}+\theta & \text{ if } \varphi_u\in[\theta, \pi], \\[5pt] -{\pi}/{2}-\theta & \text{ if } \varphi_u\in[-\pi,-\theta]. \end{array} \right.\end{align*}

Then, since

\[\int_{0}^{\infty}\textrm{d} r\;\int_{-\theta}^{\theta}\textrm{d} \varphi\; \lambda r \exp\!(\!-\!\lambda\theta r^2 ) <\infty,\]

by the dominated-convergence theorem, we obtain

\begin{align*}{\operatorname e }^{-\mathcal J_{\lambda,\theta}(u)}&\leq\lim_{q\rightarrow \infty}\int_{0}^{\infty}\textrm{d} r\;\int_{-\theta}^{\theta}\textrm{d}\varphi\; \lambda r \exp\!(\!-\!\lambda\theta r^2 + q[r\cos(\varphi-\alpha_0) - r_u\cos(\varphi_u-\alpha_0) ] )\\[5pt] & = \int_{0}^{\infty}\textrm{d} r\; \int_{-\theta}^{\theta} \textrm{d}\varphi\;\lim_{q\rightarrow \infty} \lambda r \exp\!(\!-\!\lambda\theta r^2 + q[r\cos(\varphi-\alpha_0) - r_u\cos(\varphi_u-\alpha_0) ] )\\[5pt] &= 0.\end{align*}

Therefore, we get $\mathcal J_{\lambda,\theta}(u)= \infty$ whenever $ \varphi_u\notin (\!-\!\theta, \theta)$ . A similar calculation also gives us that $\mathcal J_{\lambda,\theta}(u)= \infty$ for $r_u=0$ . This implies that $\mathcal J_{\lambda,\theta}(u)= \infty$ for $u\notin {\mathcal C }^o_{\theta}$ .

Now, we show that $\mathcal J_{\lambda,\theta}(u)< \infty$ for $u\in{\mathcal C }^o_{\theta}$ . In this case, $r_u>0$ and $\varphi_u\in(\!-\!\theta,\theta)$ . For this, we pick $0<\eta<\pi/16$ small enough such that $[\varphi_u-4\eta, \varphi_u+4\eta]\in(\!-\!\theta,\theta)$ and define two sets $A_{\alpha}$ , $B_{\alpha}$ as

\begin{align*} A_{\alpha}\;:\!=\;\left\{ \begin{array}{l@{\quad}l} (r_u/(\sin\eta), \infty) & \text{ if } |\alpha-\varphi_u|\le \pi/2+ \eta, \\[5pt] (0, r_u \sin\eta) & \text{ if } |\alpha-\varphi_u|> \pi/2+ \eta, \end{array} \right.\end{align*}

and

\begin{align*} B_{\alpha}\;:\!=\;\left\{ \begin{array}{l@{\quad}l} [\varphi_u+2 \eta, \varphi_u+ 3 \eta] & \text{ if } \alpha\in[\varphi_u, \varphi_u+\pi], \\[5pt] [\varphi_u-3 \eta, \varphi_u- 2 \eta] & \text{ if } \alpha\in[\varphi_u-\pi, \varphi_u). \end{array} \right.\end{align*}

Here, we consider the range of $\alpha$ as $[\varphi_u-\pi, \varphi_u+\pi]$ instead of $[-\pi,\pi]$ for simplicity. Now, for any $\alpha\in[\varphi_u-\pi, \varphi_u+\pi]$ , $r\in A_{\alpha}$ , and $\varphi\in B_{\alpha}$ , we have $r\cos(\varphi-\alpha) - r_u\cos(\varphi_u-\alpha)\ge 0$ . Therefore,

\begin{align*} {\operatorname e }^{-\mathcal J_{\lambda,\theta}(u)}&\ge\inf_{\alpha\in[\varphi_u-\pi, \varphi_u+\pi]} \int_{A_{\alpha} }\textrm{d} r\;\int_{B_{\alpha}}\textrm{d} \varphi\;\lambda r \exp\!(\!-\!\lambda\theta r^2 ) \\[5pt] &=\eta\lambda\min\bigg\{\int_{0}^{r_u \sin\eta} \textrm{d} r\;r \exp\!(\!-\!\lambda\theta r^2 ) , \int_{r_u/(\sin\eta)}^{\infty} \textrm{d} r\;r \exp\!(\!-\!\lambda\theta r^2 ) \bigg\} \\[5pt] &>0,\end{align*}

which implies that, for all $u\in{\mathcal C }^o_{\theta}$ , we have $\mathcal J_{\lambda,\theta}(u)< \infty$ . This completes the proof.

Proof of Lemma 10. Using the same arguments as in the proof of Lemma 1, we have

\begin{align*}\mathbb{E}[\!\exp\!(\langle \gamma, \overline U_1\rangle)]&=2\lambda \int_0^\infty\textrm{d} r\;r\exp\!(\!-\!\lambda\theta r^2)\int_0^{\theta}\textrm{d} \varphi\;\exp\!(\gamma_1 r\cos\varphi+\gamma_2 r\sin\varphi)<\infty,\end{align*}

which, together with Cramér’s theorem, gives the result.

Proof of Theorem 2. The key observation is that the sequence of progress variables $\{U_i\}_{i\ge1}$ is i.i.d. for $\theta<\pi/4$ . Indeed, note that, for all $i\ge 1$ , $(C_\theta(V_i)\cap B(V_i,|U_{i+1}|))\cap C_\theta(V_{i+1})=\emptyset$ . Hence, in every step, the navigation discovers a previous undiscovered region in space and the corresponding Poisson point clouds are stochastically independent. However, even though the progress steps are i.i.d., the statement is not a direct application of, e.g. Cramér’s theorem, since $\{{\mathcal Y }_t\}_{t\ge 0}$ only tracks the vertical displacement; however, there is also a random horizontal displacement. Let us write $U_i=(X_i,Y_i)$ , where $X_i$ is the first and $Y_i$ the second Cartesian coordinate of $U_i$ and recall that $K_t\;:\!=\;\sup\{n>0\colon\sum_{i=1}^n X_i< t\}$ denotes the number of steps the navigation takes before reaching t along the x axis.

Step 1: Lower bound for upper tail. Let us start with the lower bound for the upper tail. It suffices to consider $a>0$ . Then, for all $1>b>0, c>0$ , and $\alpha>\beta>0$ ,

\begin{align*}\mathbb{P}({\mathcal Y }_t&> at)\ge \mathbb{P}({\mathcal Y }_t> at,\lfloor \beta t\rfloor\le K_t< \lfloor \alpha t\rfloor)\\[5pt] &\ge \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor} Y_i> at+\sum_{j=\lfloor \beta t\rfloor+1}^{\lfloor \alpha t\rfloor}|Y_j|,bt< \sum_{i=1}^{\lfloor \beta t\rfloor} X_i<t, \sum_{j=\lfloor \beta t\rfloor+1}^{\lfloor \alpha t\rfloor} X_j>(1-b)t\Bigg)\\[5pt] &\ge \mathbb{P}\Bigg(\sum_{j=\lfloor \beta t\rfloor+1}^{\lfloor \alpha t\rfloor}|Y_j|<ct,\sum_{j=\lfloor \beta t\rfloor+1}^{\lfloor \alpha t\rfloor} X_j>(1-b)t\Bigg)\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor} Y_i> (a+c)t,bt<\sum_{i=1}^{\lfloor \beta t\rfloor} X_i< t\Bigg) \end{align*}

by independence. Consequently, for $\delta=\alpha-\beta$ ,

\begin{align*}\liminf_{t\uparrow\infty}t^{-1}\log\mathbb{P}({\mathcal Y }_t> at)&\ge -\delta\inf\{\overline{\mathcal J}_{\theta,\lambda}(x,y)\colon x> (1-b)/\delta, y<c/\delta \}\\[5pt] &\quad-\beta\inf\{\mathcal J_{\theta,\lambda}(x,y)\colon b/\beta<x< 1/\beta, y>(a+c)/\beta\}.\end{align*}

Sending b to 1 and fixing $\delta=\delta(c)$ such that $c/\delta> \mathbb{E}[|Y_1|]$ , the first summand vanishes and we have

\begin{align*}&\liminf_{t\uparrow\infty}t^{-1}\log\mathbb{P}({\mathcal Y }_t> at)\ge -\beta\mathcal J_{\theta,\lambda}(1/\beta,(a+c)/\beta).\end{align*}

Sending c to 0, we arrive at

\begin{align*}&\liminf_{t\uparrow\infty}t^{-1}\log\mathbb{P}({\mathcal Y }_t> at)\ge -\inf\{\beta \mathcal J_{\theta,\lambda}(1/\beta,a/\beta)\colon \beta>0\}.\end{align*}

This expression reflects that the process has to find the optimal compromise between making the right number of steps for the displacement along the y axis to be not too unlikely as well as hitting the time t along the x axis.

Step 2: Upper bound for upper tail. For the upper bound, we can proceed similarly. For all $\alpha>0$ , we can bound

\begin{align*}\mathbb{P}({\mathcal Y }_t\ge at)&\le \mathbb{P}\Bigg(\sum_{i=1}^{K_t}Y_i\ge at-|Y_{K_t+1}|, K_t\le \lfloor \alpha t\rfloor\Bigg)+\mathbb{P}(K_t>\lfloor \alpha t\rfloor)\\[5pt] &=\sum_{m=0}^{\lfloor \alpha t\rfloor} \mathbb{P}\Bigg(\sum_{i=1}^{m}Y_i\ge at-|Y_{m+1}|, K_t=m\Bigg)+\mathbb{P}(K_t>\lfloor \alpha t\rfloor)\\[5pt] &\le\sum_{m=0}^{\lfloor \alpha t\rfloor} \mathbb{P}\Bigg(\sum_{i=1}^{m}Y_i\ge t(a-\varepsilon), t(1-\varepsilon)\le \sum_{i=1}^m X_i <t\Bigg)+\mathbb{P}(\mathfrak B_\varepsilon^c(t))+\mathbb{P}(K_t>\lfloor \alpha t\rfloor),\end{align*}

where $\mathfrak B_\varepsilon(t)=\{X_{m+1}\le \varepsilon t, |Y_{m+1}|\le \varepsilon t\}$ with $\varepsilon>0$ . Note that

\begin{align*}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}(\mathfrak B^c_\varepsilon(t))=-\infty,\end{align*}

since the upper tails of $|U_i|$ are of order $O(\!-\!t^2)$ on the exponential level. Furthermore,

\begin{align*}\limsup_{\alpha\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}(K_t>\lfloor \alpha t\rfloor)=\limsup_{\alpha\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1}\log \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \alpha t\rfloor}X_i< t\Bigg)=-\infty,\end{align*}

and hence, the error terms play no role on the exponential scale with rate t. Now, let $(t_n)_{n\ge 0}$ be a subsequence such that

\begin{align*}\limsup_{t\uparrow\infty} t^{-1}&\log \sum_{m=0}^{\lfloor \alpha t\rfloor} \mathbb{P}\Bigg(\sum_{i=1}^{m}Y_i\ge t(a-\varepsilon), t(1-\varepsilon)\le \sum_{i=1}^m X_i <t\Bigg)\\[5pt] &=\lim_{n\uparrow\infty} t_n^{-1}\log \sum_{m=0}^{\lfloor \alpha t_n\rfloor} \mathbb{P}\Bigg(\sum_{i=1}^{m}Y_i\ge t_n(a-\varepsilon), t_n(1-\varepsilon)\le \sum_{i=1}^m X_i <t_n\Bigg)\end{align*}

and define

$$\beta_{n,\varepsilon}\;:\!=\;{\textrm{argmax}}\Bigg\{\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t_n\rfloor}Y_i\ge t_n(a-\varepsilon), t_n(1-\varepsilon)\le \sum_{i=1}^{\lfloor \beta t_n\rfloor} X_i <t_n\Bigg)\colon 0\le \beta\le \alpha\Bigg\},$$

where we simply take $\beta_{n,\varepsilon}$ to be the smallest solution in case of ambiguity. Then, for a suitable further sub-sequence $(t_{n_k})_{k\ge 0}$ , we have

\begin{align*}&\lim_{n\uparrow\infty} t_n^{-1}\log \sum_{m=0}^{\lfloor \alpha t_n\rfloor} \mathbb{P}\Bigg(\sum_{i=1}^{m}Y_i\ge t_n(a-\varepsilon), t_n(1-\varepsilon)\le \sum_{i=1}^m X_i <t_n\Bigg)\\[5pt] &\qquad\le \limsup_{n\uparrow\infty}t_n^{-1}\log \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor\beta_{n,\varepsilon} t_n\rfloor}Y_i\ge t_n(a-\varepsilon), t_n(1-\varepsilon)\le \sum_{i=1}^{\lfloor \beta_{n,\varepsilon} t_n\rfloor} X_i <t_n\Bigg)\\[5pt] &\qquad= \lim_{k\uparrow\infty}t_{n_k}^{-1}\log \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor\beta_{{n_k},\varepsilon} t_{n_k}\rfloor}Y_i\ge t_{n_k}(a-\varepsilon), t_{n_k}(1-\varepsilon)\le \sum_{i=1}^{\lfloor \beta_{{n_k},\varepsilon} t_{n_k}\rfloor} X_i <t_{n_k}\Bigg),\end{align*}

and we note that $(\beta_{{n_k},\varepsilon})_{k\ge 0}$ must contain a convergent sub-sequence $(\beta_{{n_{k_l}},\varepsilon})_{l\ge 0}$ with limit $0\le \beta^*_\varepsilon\le \alpha$ . Hence,

\begin{align*}\lim_{l\uparrow\infty}t_{n_{k_l}}^{-1}&\log \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor\beta_{{n_{k_l}},\varepsilon} t_{n_{k_l}}\rfloor}Y_i\ge t_{n_{k_l}}(a-\varepsilon), t_{n_{k_l}}(1-\varepsilon)\le \sum_{i=1}^{\lfloor \beta_{{n_{k_l}},\varepsilon} t_{n_{k_l}}\rfloor} X_i <t_{n_{k_l}}\Bigg)\\[5pt] &\le -\inf\{\beta\mathcal J_{\theta,\lambda}\big(1/\beta,a/\beta\big)\colon \beta>0\},\end{align*}

since $\limsup_{t\uparrow\infty}t^{-1}\log \lfloor \alpha t\rfloor=0$ and we could use continuity.

Step 3: General sets. Note that, by symmetry, $\mathbb{P}({\mathcal Y }_t\ge at)=\mathbb{P}({\mathcal Y }_t\le -at)$ and, hence, for $O\subset\mathbb{R}^2$ open, we find that, for all $z=(x,y)\in O$ , there exists $\varepsilon_z>0$ such that $\bar B_{\varepsilon_z}(z)\subset O$ . In particular, for all $z\in O$ ,

\begin{align*}&\mathbb{P}(t^{-1}{\mathcal Y }_t\in O)\ge\mathbb{P}(z-\varepsilon_z< t^{-1}{\mathcal Y }_t< z+\varepsilon_z)=\mathbb{P}(t^{-1}{\mathcal Y }_t> -z-\varepsilon_z)-\mathbb{P}(t^{-1}{\mathcal Y }_t\ge -z+\varepsilon_z),\end{align*}

where the rate to zero is dominated by the first summand. Taking an infimum over $z\in O$ gives the desired result. The upper bound can be proved similarly.

Step 4: Scaling. If we denote by ${\mathcal Y }_{\lambda,t}$ the vertical displacement at time t in the navigation based on a Poisson point process with intensity $\lambda>0$ , then, by scaling both the coordinates by $\sqrt{\lambda}$ , we find that ${\mathcal Y }_{\lambda,t}$ and ${\mathcal Y }_{1,\sqrt{\lambda}t}/\sqrt{\lambda}$ are equal in distribution. Therefore, for any $x>0$ ,

\begin{align*} \mathcal I_{\lambda,\theta}(x)&= -\lim_{t\uparrow\infty}t^{-1}\log\mathbb{P}( t^{-1} {\mathcal Y }_{\lambda,t} >x )\\[5pt] &= - \lim_{t\uparrow\infty} \sqrt{\lambda}(\sqrt{\lambda}t)^{-1}\log\mathbb{P}( (\sqrt{\lambda}t)^{-1} {\mathcal Y }_{1,\sqrt{\lambda}t} >x ) \\[5pt] &= \sqrt{\lambda} \mathcal I_{1,\theta}(x).\end{align*}

By a similar calculation, we also find that, for any $x<0$ , $\mathcal I_{\lambda,\theta}(x)= \sqrt{\lambda} \mathcal I_{1,\theta}(x)$ . This proves the theorem.

Proof of Lemma 3. The proof follows again from an application of the multivariate Cramér theorem for empirical means of sequences of i.i.d. random variables; see [Reference Dembo and Zeitouni7, Corollary 6.1.6]. We need to show existence of $\gamma=(\gamma_1,\gamma_2)\in \mathbb{R}^2\setminus\{0\}$ , such that $J'_{\lambda,\theta}(\gamma)<\infty$ , but this is an immediate consequence of Lemma 2.

Proof of Lemma 4. Observe that if, for all $1\leq m\leq n$ , $\Phi_m\in(\pi/2-\theta,\theta]$ then, for all $1\leq m\leq n$ , $H_m\neq\emptyset$ , which then ensures that ${\tau}^\theta_1>n$ . Therefore,

\[ \mathbb{P}({\tau}^\theta_1>n)\geq \mathbb{P}( \Phi_m\in(\pi/2-\theta,\theta] \text{ for all }1\leq m\leq n). \]

Note that, conditioned on $V_{m-1}$ , $R_m$ and $H_{m-1}$ , $\Phi_m$ is uniformly distributed on the set

\[\{\varphi\in[-\theta,\theta]\colon V_{m-1}+(R_m,\varphi)\notin H_{m-1}\}.\]

Hence, with $\ell$ denoting the Lebesgue measure,

\[\mathbb{P}(\Phi_m\in(\pi/2-\theta,\theta] | V_{m-1},R_m, H_{m-1}) = \frac{\ell (\{\varphi\in(\pi/2-\theta,\theta]\colon V_{m-1}+(R_m,\varphi)\notin H_{m-1}\})}{\ell (\{\varphi\in[-\theta,\theta]\colon V_{m-1}+(R_m,\varphi)\notin H_{m-1}\})}.\]

Also, note that if, for all $1\leq i< m$ , $\Phi_i>0$ then ${\mathcal C }_{\theta}^u(V_{m-1})\cap H_{m-1} =\emptyset$ . Therefore, in this case,

\begin{align*} \frac{\ell (\{\varphi\in(\pi/2-\theta,\theta]\colon V_{m-1}+(R_m,\varphi)\notin H_{m-1}\})}{\ell (\{\varphi\in[-\theta,\theta]\colon V_{m-1}+(R_m,\varphi)\notin H_{m-1}\})} \geq \frac{\ell ((\pi/2-\theta,\theta])}{\ell ([-\theta,\theta])} &= \frac{2\theta-\pi/2}{2\theta} = \frac{4\theta-\pi}{4\theta}.\end{align*}

As a result, we find that

\[\mathbb{P}(\Phi_m\in(\pi/2-\theta,\theta] | \Phi_i\in(\pi/2-\theta,\theta] \text{ for all }1\leq i< m) \geq (4\theta-\pi)/(4\theta),\]

and thus,

\begin{align*} \mathbb{P}(\tau>n)&\geq \mathbb{P}( \Phi_m\in(\pi/2-\theta,\theta] \text{ for all }1\leq m\leq n)\\[5pt] &= \prod_{m=1}^n \mathbb{P}(\Phi_m\in(\pi/2-\theta,\theta] | \Phi_i\in({\pi}/{2}-\theta,\theta] \text{ for all }1\leq i< m)\\[5pt] & \geq ((4\theta-\pi)/(4\theta))^n,\end{align*}

which completes the proof.

Proof of Theorem 3. We proceed via several steps. First, note that, for all $\delta>0$ ,

\begin{align*}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\big(|{\mathcal Y }_t-\hat{\mathcal Y }'_t|\ge \delta t\big)=-\infty,\end{align*}

where $\hat{\mathcal Y }'_t\;:\!=\;\sum_{i=1}^{K_t}Y_i$ , since

\begin{align*} \mathbb{P}\big(|{\mathcal Y }_t-\hat{\mathcal Y }'_t|\ge \delta t\big) &\le \mathbb{P}(|Y_{K_t+1}|>\delta t) \\[5pt] &\leq \mathbb{P}(K_t\geq \lfloor t \rfloor^2) + \mathbb{P}\left(\max_{i=1}^{\lfloor t \rfloor^2} |Y_i|>\delta t\right) \\[5pt] &\leq \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t \rfloor^2} X_i < t\Bigg) + \mathbb{P}\left(\max_{i=1}^{\lfloor t \rfloor^2} |Y_i|>\delta t\right) \\[5pt] &\leq \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor t \rfloor^2} \underline R_i \cos \theta < t\Bigg) + \mathbb{P}\left(\max_{i=1}^{\lfloor t \rfloor^2} \overline R_i>\delta t\right) \\[5pt] &\leq {\operatorname e }^t \mathbb{E}[ {\operatorname e }^{-\underline R \cos \theta}]^{\lfloor t \rfloor^2} + t ^2 {\operatorname e }^{-\lambda (\theta\wedge(\pi/2-\theta)) \delta^2 t^2},\end{align*}

and hence, for the large deviations, we can focus on $\hat{\mathcal Y }'_t$ . For this, as in (3), we can further split $\hat{\mathcal Y }'_t$ into

$$\hat{\mathcal Y }'_t=\sum_{i=1}^{K'_t}Y'_i+\sum_{j=\tau^\theta_{K'_t}+1}^{ K_t}Y_j\;=\!:\;{\mathcal Y }_t'+\hat Y_t.$$

The first summand is a sum of i.i.d. segments, but the second summand is a sum of dependent random variables. The challenge comes from the fact that all the involved random variables $Y'_i, \hat Y_t$ have exponential tails and, therefore, contribute on the large-deviation scale t.

Step 1: Upper bound for upper tail. For all $\alpha>0$ , we can bound

\begin{align*}\mathbb{P}(\hat{\mathcal Y }'_t \ge at)&\le \mathbb{P}\Bigg(\sum_{i=1}^{K'_t}Y'_i+\hat Y_t\ge at, K'_t\le \lfloor \alpha t\rfloor\Bigg)+\mathbb{P}(K'_t>\lfloor \alpha t\rfloor)\\[5pt] &\le \sup\nolimits_{0<\beta\le \alpha}\alpha t\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i+\hat Y_t\ge at, K'_t=\lfloor \beta t\rfloor\Bigg)+\mathbb{P}(K'_t>\lfloor \alpha t\rfloor),\end{align*}

where the second summand plays no role on the exponential scale with rate t since

\begin{align*}\limsup_{\alpha\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}(K'_t>\lfloor \alpha t\rfloor)=\limsup_{\alpha\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1} \log\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \alpha t\rfloor}X'_i< t\Bigg)=-\infty.\end{align*}

Similarly, for all $0<\beta\le \alpha$ ,

\begin{align*}\limsup_{\gamma\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i>\gamma t\Bigg)=-\infty\end{align*}

and

\begin{align*}\limsup_{\delta\uparrow\infty}\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}({\tau}^\theta_1>\delta t)=-\infty,\end{align*}

and hence, it suffices to further bound as

\begin{align*}&\mathbb{P}\Bigg( \sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i+\hat Y_t\ge at, K'_t=\lfloor \beta t\rfloor\Bigg)\\[5pt] &\qquad\le\sup\nolimits_{|b|<\gamma,\ c<1}\gamma t^2\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i{\ \hat =\ } bt, \sum_{i=1}^{\lfloor \beta t\rfloor}X'_i{\ \hat =\ } ct\Bigg)\\[5pt] &\hskip29pt\times\sup\nolimits_{d\le \delta}\delta t\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor d t\rfloor}Y_i\ge (a-b)t, \sum_{i=1}^{\lfloor d t\rfloor}X_i< (1-c)t, \sum_{i=1}^{\lfloor d t\rfloor+1}X_i\ge (1-c)t, {\tau}^\theta_1>d t\Bigg)\\[5pt] &\hskip29pt+\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i>\gamma t\Bigg)+ \mathbb{P}({\tau}^\theta_1>\delta t),\end{align*}

where we write $x{\ \hat =\ } y$ if and only if $\lfloor x\rfloor=\lfloor y\rfloor$ and use independence. Now, we can combine Assumption 1 and Lemma 3 to obtain

\begin{align*}&\limsup_{t\uparrow\infty}t^{-1}\log\mathbb{P}(\hat{\mathcal Y }'_t\ge at)\\[5pt] &\quad\le -\!\!\!\inf_{b\in\mathbb{R},\ c\in(0,1)}\!\big\{\inf\{\beta\mathcal J'_{\lambda,\theta}(c/\beta,b/\beta)\colon \beta>0\}+\inf\{d \mathcal H_{\lambda,\theta}((1-c)/d,(a-b)/d)\colon d>0\}\big\},\end{align*}

as desired.

Step 2: Lower bound for upper tail. Next, we consider the lower bound for the upper tail with $a>0$ . Then, for $b\in \mathbb{R}, 0<c<1$ and $ \beta,d>0$ ,

\begin{align*}\mathbb{P}(\hat{\mathcal Y }'_t> at)&\ge \mathbb{P}\Bigg(\sum_{i=1}^{K'_t}Y'_i>bt,K'_t= \lfloor \beta t\rfloor,\hat Y_t> (a-b)t\Bigg)\\[5pt] &\ge \mathbb{P}\Bigg(\sum_{i=1}^{\lfloor \beta t\rfloor}Y'_i>bt,\sum_{i=1}^{\lfloor \beta t\rfloor}X'_i{\ \hat =\ } ct\Bigg)\\[5pt] &\hskip10pt\times\mathbb{P}\Bigg(\sum_{i=1}^{\lfloor d t\rfloor}Y_i>(a-b)t,\sum_{i=1}^{\lfloor d t\rfloor}X_i< (1-c)t, \sum_{i=1}^{\lfloor d t\rfloor+1}X_i\ge (1-c)t, {\tau}^\theta_{1}>\lfloor d t\rfloor\Bigg), \end{align*}

where we used independence. Consequently,

\begin{align*}\liminf_{t\uparrow\infty}t^{-1}\log\mathbb{P}(\hat{\mathcal Y }'_t> at)\ge -\beta\mathcal{I'_{\lambda,\theta}}(c/\beta,b/\beta)-d \mathcal H_{\lambda,\theta}((1-c)/d,(a-b)/d).\end{align*}

Optimizing first with respect to $\beta$ and d in the individual summands and then with respect to b, c in the joint expression, we arrive at the desired lower bound that matches the upper bound.

Step 3: General sets. Using symmetry and the previous steps, we can follow the exact same arguments as in the independent case, step 4 in the proof of Theorem 2, to arrive at the large-deviation principle.

Step 4: Scaling. A scaling argument similar to that in the proof of Theorem 2 implies that $\mathcal I'_{\lambda,\theta}(x)= \sqrt{\lambda} \mathcal I'_{1,\theta}(x)$ for all $x\in \mathbb{R}$ .