# Hyperbolic Radial Spanning Tree

Lucas Flammant 111Univ. de Valenciennes, CNRS, EA 4015 - LAMAV, F-59313
Valenciennes Cedex 9, France

###### Abstract

We define and analyze an extention to the $d$-dimensional hyperbolic space of
the Radial Spanning Tree (RST) introduced by Baccelli and Bordenave in the
two-dimensional Euclidean space (2007). In particular, we will focus on the
description of the infinite branches of the tree. The properties shown for the
two-dimensional Euclidean RST are extended to the hyperbolic case in every
dimension: almost surely, every infinite branch admits an asymptotic direction
and each asymptotic direction is reached by at least one infinite branch.
Moreover, the branch converging to any deterministic asymptotic direction is
unique almost surely. Our strategy mainly relies on the two following
ingredients. First, the hyperbolic metric allows us to control fluctuations of
the branches in the hyperbolic DSF without using planarity arguments. Then, we
couple the hyperbolic RST with the hyperbolic DSF introduced and studied in
Flammant 2019.

Key words:  continuum percolation, hyperbolic space, stochastic geometry,
random geometric tree, Radial Spanning Tree, Poisson point processes.

AMS 2010 Subject Classification: Primary 60D05, 60K35, 82B21.

Acknowledgments. This work has been supervised by David Coupier and Viet Chi
Tran, who helped a lot to write this paper. It has been supported by the LAMAV
(Université Polytechnique des Hauts de France) and the Laboratoire P. Painlevé
(Université de Lille). It has also benefitted from the GdR GeoSto 3477, the
Labex CEMPI (ANR-11-LABX-0007-01) and the ANR PPPP (ANR-16-CE40-0016).

###### Contents

  1. 1 Introduction
  2. 2 Definitions, notations and basic properties
    1. 2.1 The hyperbolic space
    2. 2.2 The hyperbolic RST
    3. 2.3 Sketch of proofs
  3. 3 Proof of Theorem 1.1
    1. 3.1 The existence part: proof of (i),(ii),(iv) and (v)
    2. 3.2 The uniqueness part: proof of (iii)
  4. 4 Proof of Proposition 2.5
    1. 4.1 Step 1: a deterministic upper-bound of $\text{MBD}_{r_{0}}^{r_{0}+\delta}(\cdot)$
    2. 4.2 Step 2: a control of the tail decay of $M(z)$
    3. 4.3 Step 3: conclusion
    4. 4.4 Proof of Lemmas 4.1, 4.2, 4.3, 4.4
  5. 5 Proof of Propositions 2.6 and 2.7
  6. 6 Appendix A: proof of Proposition 2.2
  7. 7 Appendix B: computing distances and angles in the half-space model

##  1 Introduction

Geometric random trees are well studied in the literature since it interacts
with many other fields, such as communication networks, particles systems or
population dynamics. Several works have established scaling limits for two-
dimensional radial trees [10, 9] and translation invariant forests [11, 22,
14]. In addition, random spanning trees appear in the context or first passage
percolation [15]. A complete introduction to geometric random graphs is given
in Penrose 2003 [20].

The Radial Spanning Tree (RST) is a random tree whose introduction in the two-
dimensional Euclidean space has been motivated by applications for
communication networks [1]. The set of vertices is given by a homogeneous
Poisson Point Process (PPP) $\mathcal{N}$ of intensity $\lambda$ in
$\mathbb{R}^{2}$. The RST rooted at the origin $0$ is the graph obtained by
connecting each point $z\in\mathcal{N}$ to its parent $A(z)$, defined as the
closest point to $z$ among all points $z^{\prime}\in\mathcal{N}\cup\\{0\\}$
that are closer to the origin than $z$. This defines a random tree rooted at
the origin with a radial structure. An infinite backward path is defined as a
sequence of Poisson points $(z_{n})_{n\geq
0}\in\left(\mathcal{N}\cup\\{0\\}\right)^{\mathbb{N}}$ with $z_{0}=0$ and
$z_{n}=A(z_{n+1})$ for any $n\geq 0$. Given an infinite path, we will say that
the forward direction is towards $0$ and the backward direction is towards
infinity.

The topological properties of the bi-dimensional Euclidean RST are well-
understood. Baccelli and Bordenave showed that almost surely, any infinite
backward path admits an asympotic direction; moreover, a.s., every asymptotic
direction is reached by at least one infinite backward path and there exists
a.s. a unique infinite path in any given deterministic asymptotic direction
[1]. These results on the infinite paths are completed by Baccelli, Coupier &
Tran [2].

For any integer $d\geq 2$, the hyperbolic space $\mathbb{H}^{d}$ is a
$d$-dimensional Riemannian manifold with constant negative curvature, that can
be chosen equal to $-1$ without loss of generality. It admits a set of ideal
boundary points $\partial\mathbb{H}^{d}$, and
$\overline{\mathbb{H}^{d}}:=\mathbb{H}^{d}\cup\partial\mathbb{H}^{d}$ denotes
the hyperbolic space endowed with its boundary. It is a non-amenable space,
i.e. the measure of the boundary of a large subset is not negligible with
respect to its volume. The hyperbolic space is defined in more details in [7]
and [19].

There is a growing interest for the study of random models in a hyperbolic
setting. Benjamini and Schramm establish percolation results on regular
tilings and Voronoï tesselation in the hyperbolic plane [3]. Mean
characteristics of the Poisson-Voronoï tessellation have also been considered
in a general Riemannian manifold by Calka et al. [6]. This interest is
explained by at least two reasons. First, hyperbolic random graphs are well-
fitted to modelize social networks [5]. In addition, strong differences have
been noticed for properties of random models depending whether they are
considered in an Euclidean or hyperbolic setting. Indeed, some hyperbolic
random graphs admits a non-degenerate regime with infinitely many unbounded
components in the hyperbolic space [23, 16], which is generally not the case
in the Euclidean space. In addition, behaviours of non-amenable spaces are
well studied in a discrete context [4, 18, 21].

Thus it is natural to consider and study the hyperbolic RST, which we define
in the same way as the Euclidean RST. A simulation of the two-dimensional
hyperbolic RST is given in Figure 1. In this paper, we extend the results of
Baccelli and his coauthors to hyperbolic geometry in every dimension. Here is
our main result:

######  Theorem 1.1.

For any dimension $d\geq 1$ and any intensity $\lambda$, the following
happens:

  1. (i)

almost surely, any infinite backward path $(z_{n})_{n\in\mathbb{N}}$ admits an
asymptotic direction, i.e. there exists
$z_{\infty}\in\partial\mathbb{H}^{d+1}$ such that
$\lim_{n\to\infty}z_{n}=z_{\infty}$ (in the sense of the topology of
$\overline{\mathbb{H}^{d+1}}$);

  2. (ii)

almost surely, for any $I\in\partial\mathbb{H}^{d+1}$, there exists an
infinite backward path $(z_{n})$ with asymptotic direction $I$ (i.e. such that
$\lim_{n\to\infty}z_{n}=I$);

  3. (iii)

for any deterministic boundary point $I\in\partial\mathbb{H}^{d+1}$, the path
with asymptotic direction $I$ is almost surely unique;

  4. (iv)

the set of boundary points with two infinite backward paths is dense in
$\partial\mathbb{H}^{d+1}$;

  5. (v)

this set is moreover countable in the bi-dimensional case (i.e. $d=1$).

Establishing the results announced in Theorem 1.1 in every dimension
constitutes the main originality of this paper. For the two reasons explained
further, the proofs of Baccelli and Bordenave in the 2D-Euclidean setting [1]
cannot be generalised to higher dimensions.

In both contexts $\mathbb{R}^{2}$ and $\mathbb{H}^{d+1}$, for any $d\geq 1$,
the proofs of (i), (ii), (iv) and (v) of Theorem 1.1 follow the strategy of
Howard and Newman [15], which is to show that the tree is _straight_ , that
is, the descendents subtree of a vertex far from the origin is included in a
thin cone. To prove that the 2D-Euclidean RST is straight, Baccelli and
Bordenave used a translation invariant model derived from the RST: the
Directed Spanning Forest (DSF), which constitutes a local approximation of the
RST far from the origin [1]. They exploit the theory of Markov chains to
upper-bound fluctuations of trajectories in the DSF and then, they deduce the
straightness of the RST via planarity. This strategy cannot be generalised to
higher dimensions. However, in $\mathbb{H}^{d}$, we manage to control the
angular deviations of branches in the RST without resorting to an auxiliary
model, which required planarity in the Euclidean setting. The hyperbolic
metric guarantees that angular deviations decay exponentially fast with the
distance to the origin, which is strong enough to show straightness.

In addition, in the Euclidean context, the uniqueness part (point (iii) in
Theorem 1.1) is only proved in dimension $2$ since it strongly uses planarity
[15, 1], and the strategy of proof cannot be generalised to higher dimensions.
To prove (iii) in $\mathbb{H}^{d}$, our strategy consists in exploiting the
link existing between the hyperbolic RST and another random graph, the
hyperbolic DSF, defined and studied in Flammant 2019 [12], which is the
hyperbolic counterpart of the Euclidean DSF used by Baccelli and Bordenave.
Roughly speaking, the hyperbolic DSF can be defined as the limit of the
hyperbolic RST when the origin point tends to an ideal boundary point.
Similarly to the Euclidean setting, it constitutes a local approximation of
the RST far from the origin. The proof of (iii) exploits the coalescence of
the hyperbolic DSF (i.e. it is almost surely a tree) [12, Theorem 1.1], which
is a non-trivial fact obtained by exploiting the mass-transport principle, and
a local coupling between the two models.

After defining the hyperbolic RST and giving its basic properties, we define
two quantities that encode angular fluctuations along trajectories, the
Cumulative angular Forward Deviations (CFD) and the Maximal Backward
Deviations (MBD). We then establish upper-bounds of these quantities: first,
we upper-bound the Maximal Backward Deviations in a thin annulus of width
$\delta>0$ (Proposition 2.5) and then we deduce a global control of MBD in the
whole space (Proposition 2.6), that roughly says that angular deviations decay
exponentially fast with the distance to the origin. From this upper-bound, we
deduce that the RST is straight in the sense of Howard & Newman (Proposition
2.7). The points (i), (ii), (iv) and (v) in Theorem 1.1 can be deduced from
straightness and the upper-bound of MBD given by Proposition 2.6. The point
(iii) (the uniqueness part) is done by exploiting a local coupling existing
between the RST and the DSF far from the origin.

The rest of paper is organized as follows. In Section 2, we set some reminders
of hyperbolic geometry and we define the hyperbolic RST. Then, we give its
basic properties and a road-map of the proofs. We also announce the upper-
bounds of angular deviations (Propositions 2.5 and 2.6) and the straightness
property (Proposition 2.7). The proof of Theorem 1.1 is done in Section 3.
Proposition 2.5 is proved in Section 4 and the proofs of Propositions 2.6 and
2.7 are done in Section 5.

Figure 1: Simulation of the two-dimensional hyperbolic RST, with $\lambda=30$,
in the Poincaré disc model. The edges are represented by geodesics. The
different connected components of the RST (appart from the root) are
represented with different colors.

##  2 Definitions, notations and basic properties

We denote by $\mathbb{N}$ the set of non-negative integers and by
$\mathbb{N}^{*}$ the set of positive integers. In the rest of the paper, $c$
(resp. $C$) will be some small (resp. large) constant whose value can change
from a line to another.

###  2.1 The hyperbolic space

We refer to [7] or [19] for a complete introduction to hyperbolic geometry.
For $d\in\mathbb{N}^{*}$, the $(d+1)$-dimensional hyperbolic space, denoted by
$\mathbb{H}^{d+1}$, is a $(d+1)$-dimensional Riemannian manifold of constant
negative curvature $-1$ that can be defined by several isometric models. One
of them is the open-ball model consisting in the unit open ball

$\displaystyle
I=\\{(x_{1},...,x_{d+1})\in\mathbb{R}^{d+1},~{}x_{1}^{2}+...+x_{d+1}^{2}<1\\}$
(2.1)

endow with the following metric:

$\displaystyle
ds_{I}^{2}:=4\frac{dx_{1}^{2}+...+dx_{n+1}^{2}}{1-x_{1}^{2}-...-x_{d+1}^{2}}.$
(2.2)

This model is rotation invariant. The metric becomes smaller as we get closer
to the boundary unit sphere $\partial I$, and this boundary is at infinite
distance from the center $0$.

The volume measure on $(I,ds_{I}^{2})$, denoted by $\mbox{Vol}_{I}$, is given
by

$\displaystyle
d\mbox{Vol}_{I}=2^{d+1}\frac{dx_{1}...dx_{d+1}}{1-x_{1}^{2}-...-x_{d+1}^{2}}.$
(2.3)

In this model $(I,ds_{I}^{2})$, the geodesics are of two types: the diameters
of $I$ and the arcs that are perpendicular to the boundary unit sphere
$\partial I$. We refer to discussion [7, P.80] for a proof. Moreover, this
model is conformal, which means that the hyperbolic angle between two
geodesics corresponds to their Euclidean angle in the open-ball
representation.

An important fact about hyperbolic geometry is that all points and all
directions play the same role. More precisely, $\mathbb{H}^{d+1}$ is
homogeneous and isotropic. It means that the group of isometries of
$\mathbb{H}^{d+1}$ acts transitively on the unit tangent bundle of
$\mathbb{H}^{d+1}$: given two points $x,y\in\mathbb{H}^{d+1}$ and two unit
tangent vectors $u\in T_{x}\mathbb{H}^{d+1},~{}v\in T_{y}\mathbb{H}^{d+1}$,
there exists an isometry $g$ of $\mathbb{H}^{d+1}$ such that $g(x)=y$ and that
pushes forward $u$ on $v$. The notations $T_{x}$, $T_{y}$ and the vocabulary
relating to Riemannian geometry are defined in [17]. We refer to [19],
Proposition 1.2.1 p.5, for a proof.

Figure 2: Geodesics in the open ball model

The hyperbolic space $\mathbb{H}^{d+1}$ is naturally equipped with a set of
points at infinity, and the most natural way to identify these points is to
use the open-ball model. In $(I,ds_{I}^{2})$, the set of points at infinity is
identified by the boundary unit sphere $\partial I$. We denote by
$\partial\mathbb{H}^{d+1}$ the boundary set (represented by the boundary unit
sphere in $(I,ds_{I}^{2})$) and by
$\overline{\mathbb{H}^{d+1}}:=\mathbb{H}^{d+1}\cup\partial\mathbb{H}^{d+1}$
the hyperbolic space $\mathbb{H}^{d+1}$ plus the set of points at infinity,
with the topology given by the closed ball. A point
$z_{\infty}\in\partial\mathbb{H}^{d+1}$ is called _ideal point_ or _point at
infinity_.

We denote by $d(\cdot,\cdot)$ the hyperbolic distance in $\mathbb{H}^{d+1}$,
and by $\|\cdot\|$ the Euclidean norm in $\mathbb{R}^{d}$, with the convention
$\|\infty\|=\infty$. Let us denote by Vol the volume measure on
$\mathbb{H}^{d+1}$. For $z_{1},z_{2}\in\overline{\mathbb{H}}^{d+1}$, let us
denote by $[z_{1},z_{2}]$ the geodesic between $z_{1}$ and $z_{2}$. Moreover,
we set the notations:

$[z_{1},z_{2}[:=[z_{1},z_{2}]\backslash\\{z_{2}\\},\quad]z_{1},z_{2}]:=[z_{1},z_{2}]\backslash\\{z_{1}\\},\quad]z_{1},z_{2}[\backslash(\\{z_{2}\\}\cup\\{z_{2}\\}).$

Let us denote by $[z_{1},z_{2})$ (resp. $(z_{1},z_{2}]$) the semi-geodesic
passing threw $z_{2}$ (resp. $z_{1}$) and ending at $z_{1}$ (resp. $z_{2}$).
For $z_{1},z_{2},z_{3}\in\overline{\mathbb{H}}^{d+1}$,
$\widehat{z_{1}z_{2}z_{3}}$ is the measure of the corresponding (non-oriented)
hyperbolic angle. For any subset $B\subset\overline{\mathbb{H}^{d+1}}$,
$\overline{B}$ denotes the closure of $B$ in $\overline{\mathbb{H}^{d+1}}$.
For any point $z\in\overline{\mathbb{H}^{d+1}}$ and $\theta>0$,
$\text{Cone}(z,\theta):=\\{z^{\prime}\in\mathbb{H}^{d+1},~{}\widehat{z0z^{\prime}}\leq\theta\\}$
is defined as the cone of apex $0$ and aperture $\theta$ (if $\theta\geq\pi$
then $\text{Cone}(z,\theta)$ is the hole space $\mathbb{H}^{d+1}$). In
addition, for $r>0$ and $z\in\mathbb{H}^{d+1}$, we define

$\displaystyle B_{S(r)}(z,\theta):=\text{Cone}(z,\theta)\cap S(r).$ (2.4)

Let $0\in\mathbb{H}^{d+1}$ be some arbitrary origin point (it can be thought
as the center of the ball in the open-ball representation), which will plays
the role of the root of the RST. For $z\in\mathbb{H}^{d+1}$ and $r>0$, we
denote by $B(z,r):=\\{z^{\prime}\in H,~{}d(z,z^{\prime})<r\\}$ (resp.
$S(z,r):=\\{z^{\prime}\in H,~{}d(z,z^{\prime})=r\\}$) the hyperbolic ball
(resp. $sphere$) centered at $z$ of radius $r$, and we set $B(r):=B(0,r)$
(resp. $S(r):=s(0,r)$). For $x\in\mathbb{R}^{d}$ and $r>0$, let us also denote
by
$B_{\mathbb{R}^{d}}(x,r):=\\{x^{\prime}\in\mathbb{R}^{d},~{}\|x^{\prime}-x\|<r\\}$
the Euclidean ball centered at $x$ of radius $r$.

Let us denote by $\mathbb{S}^{d}$ the unit Euclidean sphere in
$\mathbb{R}^{d+1}$ and by $\nu$ its $d$-dimensional volume measure. Since the
RST is a rooted graph, a convenient way to represent points in
$\mathbb{H}^{d+1}$ is to use polar coordinates. Recall that $0$ is the origin
point. For any point $z\in\mathbb{H}^{d+1}$, we denote by $z=(r;u)$ its polar
coordinates w.r.t. $0$: $r$ is its distance to $0$ and $u\in
UT_{0}\mathbb{H}^{d+1}\simeq\mathbb{S}^{d}$ is its direction
($UT_{0}\mathbb{H}^{d+1}$ is the unitary tangent space of $0$ in
$\mathbb{H}^{d+1}$). In polar coordinates, the volume measure Vol is given by

$\displaystyle d\mbox{Vol}(r;u)=\sinh(r)^{d}~{}dr~{}d\nu(u).$ (2.5)

A direct consequence of this is that the volume of a ball of radius $r$ is
given by:

$\displaystyle\mbox{Vol}(B(r))=\int_{0}^{r}\sinh(r)^{d}~{}dr~{}d\nu(\mathbb{S}^{d})=\mathcal{S}_{d}\sinh(r)^{d}\asymp
e^{dr}\text{ when }r\to\infty,$ (2.6)

where $\mathcal{S}^{d}$ is the $d$-dimensional volume of $\mathbb{S}^{d}$.

The hyperbolic law of cosines [24, p.13] is a well adapted tool to compute
distances using polar coordinates. Given $z_{1}=(r_{1};u_{1})$,
$z_{2}=(r_{2},u_{2})\in\mathbb{H}^{d+1}$, the hyperbolic law of cosines gives,

$\displaystyle\cosh d(z_{1},z_{2})=\cosh(r_{1})\cosh(r_{2})-\langle
u_{1},u_{2}\rangle\sinh(r_{1})\sinh(r_{2}).$ (2.7)

###  2.2 The hyperbolic RST

In the rest of the paper, the dimension $d$ and the intensity $\lambda>0$ are
fixed. Let $\mathcal{N}$ be a homogeneous PPP of intensity $\lambda$ in
$\mathbb{H}^{d+1}$. The definition of the hyperbolic RST is similar to the
Euclidean case. The set of vertices is $\mathcal{N}\cup\\{0\\}$. Each vertex
$z\in\mathcal{N}$ is connected to the closest Poisson point among those that
are closer to the origin than $z$:

######  Definition 2.1 (Radial Spanning Tree in $\mathbb{H}^{d+1}$).

For any $z=(r;u)\in\mathcal{N}$, the _parent_ of $z$ is defined as

$A(z):=\operatorname*{argmin}_{z^{\prime}\in\mathcal{N}\cap
B(r)}d(z^{\prime},z).$

We call _Radial Spanning Tree (RST)_ in $\mathbb{H}^{d+1}$ rooted at $0$ the
oriented graph $(V,\vec{E})$ where

$V:=\mathcal{N}\cup\\{0\\},\quad\vec{E}:=\\{(z,A(z)),~{}z\in\mathcal{N}\\}.$

It is possible to assume that $\mathcal{N}\cup\\{0\\}$ does not contain
isosceles triangles, since this event has probability $1$. Thus the ancestor
$A(z)$ is well-defined.

For $z\in\mathcal{N}\cup\\{0\\}$ and $k\in\mathbb{N}$, let us define
$A^{(k)}(z)=A\circ..\circ A$ $k$ times and
$A^{(-k)}(z)=\\{z^{\prime}\in\mathcal{N},~{}A^{(k)}(z^{\prime})=z\\}$ (in
particular $A^{(-1)}(z)$ is the set of daughters of $z$). For
$z\in\mathcal{N}$ and $r\geq 0$, let us define

$B^{+}(z,r):=B(z,r)\cap B(0,d(0,z))\text{ and }B^{+}(z):=B^{+}(z,d(z,A(z))).$

By definition of the parent, $B^{+}(z)\cap\mathcal{N}=\emptyset$ for all
$z\in\mathcal{N}$.

Definition 2.1 does not specify the shape of edges, but the results announced
in Theorem 1.1 only concern the graph structure of the hyperbolic RST, so
their veracity does not depend on the geometry of edges. It is more natural to
represent edges with hyperbolic geodesics, but we do another choice which will
appear more convenient for the proofs. Given
$z_{1}=(r_{1};u_{1}),z_{2}=(r_{2};u_{2})\in\mathbb{H}^{d+1}$ such that
$0\notin[z_{1},z_{2}]$, we define a path $[z_{1},z_{2}]^{*}$, in an isotropic
way, verifying the two following conditions:

  1. i)

the distance to the origin $0$ is monotonous along the path
$[z_{1},z_{2}]^{*}$,

  2. ii)

the distance to $z_{1}$ is also monotonous along this path.

It will be necessary for the proofs that the shape of edges satisfy conditions
(i) and (ii), and the geodesic $[z_{1},z_{2}]$ does not verify condition (i)
in general. Since $0\notin[z_{1},z_{2}]$, $u_{1}$ and $u_{2}$ are not
antipodal, thus one can consider the unique geodesic path
$\gamma_{u_{1},u_{2}}:[0,1]\to UT_{0}\mathbb{H}^{d+1}$ on the sphere with
constant speed connecting $u_{1}$ to $u_{2}$. Hence we define the path
$[z_{1},z_{2}]^{*}$ as

$\displaystyle[0,1]$ $\displaystyle\to$ $\displaystyle\mathbb{H}^{d+1}$
$\displaystyle t$ $\displaystyle\mapsto$
$\displaystyle\big{(}(1-t)r_{1}+tr_{2};\gamma_{u_{1},u_{2}}(\phi_{r_{1},r_{2},\widehat{u_{1},u_{2}}}(t))\big{)},$
(2.8)

where $\phi_{r_{1},r_{2},\widehat{u_{1},u_{2}}}:[0,1]\to[0,1]$ is defined as:

$\displaystyle\phi_{r_{1},r_{2},\widehat{u_{1},u_{2}}}(t):=\frac{1}{\widehat{u_{1},u_{2}}}\arccos\left(\frac{(1-t)\sinh(r_{1})+t\cos(\widehat{u_{1}u_{2}})\sinh(r_{2})}{\sinh((1-t)r_{1}+tr_{2})}\right).$

This function $\phi_{r_{1},r_{2},\widehat{u_{1},u_{2}}}$ is built to ensure
that the distance to the origin $z_{1}$ is monotonous along the path
$[z_{1},z_{2}]^{*}$. Indeed, by the hyperbolic law of cosines (2.7),

$\displaystyle\cosh
d\big{(}z_{1},((1-t)r_{1}+tr_{2};\gamma_{u_{1},u_{2}}(\phi(t))\big{)}$
$\displaystyle=\cosh(r_{1})\cosh((1-t)r_{1}+tr_{2})-\cos(\phi(t)(\widehat{u_{1},u_{2}}))\sinh(r_{1})\sinh((1-t)r_{1}+tr_{2})$
$\displaystyle=t\left[\cosh(r_{1})\cosh(r_{2})-\cos(\widehat{u_{1},u_{2}})\sinh(r_{1})\sinh(r_{2})\right]$

is non-decreasing in $t$.

We define $[z_{1},z_{2}[^{*}:=[z_{1},z_{2}]^{*}\backslash\\{z_{2}\\}$ and
$]z_{1},z_{2}]^{*}:=[z_{1},z_{2}]^{*}\backslash\\{z_{1}\\}$. It is possible to
assume that $\mathcal{N}$ does not contain two points $z_{1},z_{2}$ such that
$0\in[z_{1},z_{2}]$ since this event has probability $1$. Let us now define
the random set RST by connecting each point $z\in\mathcal{N}$ to $A(z)$ by the
path $[z,A(z)]^{*}$:

$\text{RST}:=\bigsqcup_{z\in\mathcal{N}}[z,A(z)[^{*}.$

It may exists some points $z$ belonging to several paths
$[z_{1},A(z_{1})[^{*}$, …, $[z_{k},A(z_{k})[^{*}$; in that case, $z$ is
counted with multiplicity $k$ in RST. Formally,
$\text{RST}=\bigcup_{z\in\mathcal{N}}[z,A(z)[^{*}\times\\{z\\}\subset\mathbb{H}^{d+1}\times\mathbb{H}^{d+1}.$,
i.e. an element $z=(z^{\prime},z^{\prime\prime})\in\text{RST}$ is the data of
a point $z\in\mathbb{H}^{d+1}$ and an edge
$[z^{\prime\prime},A(z^{\prime\prime})]^{*}$ containing $z^{\prime}$. For
$z=(z^{\prime},z^{\prime\prime})\in\text{RST}$, we define

$z_{\downarrow}=z^{\prime\prime},\quad z_{\uparrow}=A(z^{\prime\prime}).$

In the following, we will commit an abuse of notations by considering that
$\text{RST}\subset\mathbb{H}^{d+1}$ and identifying an element
$z=(z^{\prime},z^{\prime\prime})\in\text{RST}$ to the corresponding point
$z^{\prime}\in\mathbb{H}^{d+1}$. Given $z\in\text{RST}$, let $n:=\min\\{k\geq
0,~{}A^{(k)}(z_{\uparrow})=0\\}$ be the number of steps required to reach the
origin from $z_{\uparrow}$; we define the _trajectory_ from $z$ as

$\displaystyle\pi(z):=[z,z_{\uparrow}]^{*}\cup\bigcup_{k=0}^{n-1}\left[A^{(k)}(z_{\uparrow}),A^{(k+1)}(z_{\uparrow})\right]^{*}.$

For $r>0$, we define the _level_ $r$ as

$\mathcal{L}_{r}:=\text{RST}\cap S(r).$

For $0<r\leq r^{\prime}$ and for $z^{\prime}\in\mathcal{L}_{r^{\prime}}$, the
_ancestor at level_ $r$ of $z^{\prime}$, denoted by
$\mathcal{A}_{r}^{r^{\prime}}(z^{\prime})$ is the intersection point of
$\pi(z^{\prime})$ and $S(r)$. For $0<r\leq r^{\prime}$ and for
$z\in\mathcal{L}_{r}$, the _set of descendents at level_ $r^{\prime}$ is
defined as
$\mathcal{D}_{r}^{r^{\prime}}(z):=\\{z^{\prime}\in\mathcal{L}_{r^{\prime}},z\in\pi(z^{\prime})\\}$
(we extend the notation for $z\notin\mathcal{L}_{r}$ by setting
$\mathcal{D}_{r}^{r^{\prime}}(z):=\emptyset$). For $z=(r;u)\in\text{RST}$, the
_descendents subtree_ of $z$ is defined as
$\mathcal{D}(z):=\bigcup_{r^{\prime}\geq r}\mathcal{D}_{r}^{r^{\prime}}(z)$.
In addition, we call _infinite backward path_ a sequence
$(z_{i})_{i\in\mathbb{N}}\in\left(\mathbb{H}^{d+1}\right)^{\mathbb{N}}$ such
that $z_{0}=0$ and $z_{i}=A(z_{i+1})$ for all $i\geq 0$.

$0$$r$$r^{\prime}$$z_{2}\in\mathcal{L}_{r}$$z_{1}\in\mathcal{L}_{r^{\prime}}$$A_{r}^{r^{\prime}}(z_{1})$$\mathcal{D}_{r}^{r^{\prime}}(z_{2})$
Figure 3: Representation of levels $r$ and $r^{\prime}$, the ancestor
$A_{r}^{r^{\prime}}(\cdot)$ and the set of descendents
$\mathcal{D}_{r}^{r^{\prime}}(\cdot)$

Let us end this section with basic properties about RST proved in Appendix 6.

######  Proposition 2.2.

The RST is a tree and it has finite degree a.s. Moreover, in the bi-
dimensional case ($d=1$), the representation of the RST obtained by connecting
each vertex $z\in\mathcal{N}$ to its parent $A(z)$ by the geodesic $[z,A(z)]$
(instead of $[z,A(z)]^{*}$) is planar, i.e. their is no two points
$z_{1},z_{2}\in\mathcal{N}$ such that
$[z_{1},A(z_{1})]\cap[z_{2},A(z_{2})]\neq\emptyset$.

###  2.3 Sketch of proofs

In order to prove our main result (Theorem 1.1), the key point is to upper-
bound angular deviations of trajectories. We first introduce two quantities,
the _Cumulative Forward angular Deviations (CFD)_ and _Maximal Backward
Deviations (MBD)_ to quantify those fluctuations.

######  Definition 2.3 (Cumulative Forward angular Deviations).

Let $0<r\leq r^{\prime}$ and $z^{\prime}\in S(r^{\prime})$. If
$z^{\prime}\notin\text{RST}$, we set
$\text{CFD}_{r}^{r^{\prime}}(z^{\prime})=0$ by convention and we now suppose
that $z^{\prime}\in\text{RST}$. Let
$z:=\mathcal{A}_{r}^{r^{\prime}}(z^{\prime})$. We define the _Cumulative
Forward angular Deviations_ of $z^{\prime}$ between levels $r$ and
$r^{\prime}$ as

$\displaystyle\text{CFD}_{r}^{r^{\prime}}(z^{\prime}):=\left\\{\begin{aligned}
&\widehat{z^{\prime}0z}\text{ if }z_{\downarrow}=z^{\prime}_{\downarrow},\\\
&\widehat{z^{\prime}0z^{\prime}_{\uparrow}}+\sum_{k=0}^{n-1}\widehat{A^{(k)}(z^{\prime}_{\uparrow})0A^{(k+1)}(z^{\prime}_{\uparrow})}+\widehat{z_{\downarrow}0z}\text{
else,}\end{aligned}\right.$

where $n$ is the unique non negative integer such that
$A^{(n)}(z^{\prime}_{\uparrow})=z_{\downarrow}$.

######  Definition 2.4 (Maximal Backward angular Deviations).

Let $0<r\leq r^{\prime}$ and $z\in S(r)$. We define the _Maximal Backward
angular Deviations_ between levels $r$ and $r^{\prime}$ as

$\displaystyle\text{MBD}_{r}^{r^{\prime}}(z):=\left\\{\begin{aligned} &0\text{
if }z\notin\text{RST},\\\
&\sup_{r^{\prime\prime}\in[r,r^{\prime}]}\max_{z^{\prime\prime}\in\mathcal{D}_{r}^{r^{\prime\prime}}(z)}\text{CFD}_{r}^{r^{\prime\prime}}(z^{\prime\prime})\text{
if }z\in\text{RST}.\end{aligned}\right.$

We extend the definition to $r^{\prime}=\infty$ by setting:

$\displaystyle\text{MBD}_{r}^{\infty}(z):=\lim_{r^{\prime}\to\infty}\text{MBD}_{r}^{r^{\prime}}(z),$

the limit exists since $r^{\prime}\mapsto\text{MBD}_{r}^{r^{\prime}}(z)$ is
non-decreasing.

These quantities will be upper-bounded in two steps. First, a percolation
argument is used to control angular deviations in any annulus of width
$\delta>0$ for some small $\delta>0$ (Proposition 2.5) and then we deduce a
global control of angular deviations (Proposition 2.6). Recall that
$B_{S(r)}(\cdot,\cdot)$ is defined in (2.4).

######  Proposition 2.5.

There exists $\delta>0$ such that, for any $p\geq 1$, there exists
$C=C(d,p)>0$ such that for any $r>0$, $\theta\geq 0$ and any direction
$u\in\mathbb{S}^{d}$,

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r)}(u,\theta)\cap\text{RST}}\left(\text{MBD}_{r}^{r+\delta}(z)\right)^{p}\right]\leq
C\theta^{d}e^{r(d-p)},$ (2.9)

######  Proposition 2.6.

For any $p$ large enough, there exists some constant $C_{\text{fl}}>0$ such
that, for any $0<r_{0}<\infty$, $A>0$ and any direction $u\in\mathbb{S}^{1}$,

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{\infty}(z)\right)^{p}\right]\leq
C_{\text{fl}}A^{d}e^{-r_{0}p}.$

These controls of angular deviations will be first used to show that the RST
is _straight_ (Proposition 2.7). The straightness property is the key to show
(i), (ii) and (iv) in Theorem 1.1.

######  Proposition 2.7 (straightness property).

Almost surely, the following happens. For any $\varepsilon>0$, there exists
some $R_{0}>0$, such that, for any radius $r_{0}\geq R_{0}$, for any
$z\in\text{RST}$ with $d(0,z)\geq r_{0}$, the descendents subtree
$\mathcal{D}(z)$ is contained in a cone of apex $0$ and aperture
$e^{-(1-\varepsilon)r_{0}}$, i.e. for any
$z^{\prime},z^{\prime\prime}\in\mathcal{D}(z)$,
$\widehat{z^{\prime}0z^{\prime\prime}}\leq e^{-(1-\varepsilon)r_{0}}$.

The proof of (iii) in Theorem 1.1 exploits the controls of angular deviations
(Proposition 2.6) and the link existing between the RST and the hyperbolic
Directed Spanning Forest introduced in [12]: the DSF approximates locally the
RST far from the origin. The unicity of the infinite backward path with some
given deterministic asymptotic direction has been shown for the DSF [13], and
the local coupling existing between the two models permits to show that this
property remains true for the RST.

##  3 Proof of Theorem 1.1

Here we assume that Propositions 2.6 and 2.7 are proved and we show that it
implies Theorem 1.1.

###  3.1 The existence part: proof of (i),(ii),(iv) and (v)

It will be shown in this section that any infinite backward path admits an
asymptotic direction and that any ideal boundary point is the asymptotic
direction of an infinite backward path (point (i) and (ii) in Theorem 1.1. The
strategy consists, in exploiting the straightness property (Proposition 2.7):

Let $(z_{n})$ be an infinite backward path, we prove that $(z_{n})$ admits an
asymptotic direction. For $n\geq 0$, let us decompose $z_{n}$ in polar
coordinates: $z_{n}=(r_{n};u_{n})$. Proposition 2.7 immediately implies that
the sequence $(u_{n})_{n\geq 0}$ is a Cauchy sequence in
$UT_{0}\mathbb{H}^{d+1}\simeq\mathbb{S}^{d}$, thus it converges, so $(z_{n})$
converges to some boundary point $z_{\infty}\in\partial\mathbb{H}^{d+1}$.

Let $\Psi=\\{\lim_{n\to\infty}z_{n},~{}(z_{n})\text{ is an infinite backward
path}\\}\subset\partial\mathbb{H}^{d+1}$ be the set of asymptotic directions
reached by at least one infinite backward path. In order to prove that
$\Psi=\partial\mathbb{H}^{d+1}$, we proceed in two steps: we first show that
$\Psi$ is dense in $\partial\mathbb{H}^{d+1}$, then we show that $\Psi$ is
closed in $\partial\mathbb{H}^{d+1}$.
Since the RST is an infinite tree with finite degree a.s. (Proposition 2.2),
there exists an infinite backward path from $0$ and the corresponding infinite
backward path converges to an ideal boundary point by the previous paragraph,
thus $\Psi\neq\emptyset$ almost surely.

We denote by $\text{Stab}(0)$ the set of isometries that fix $0$, in
particular it contains rotations centered at $0$. Let $B$ be an open subset of
$\partial\mathbb{H}^{d+1}$. Since
$\partial\mathbb{H}^{d+1}\simeq\mathbb{S}^{d}$ is compact, there exists
finitely many isometries $\gamma_{1},...,\gamma_{k}\in\text{Stab}(0)$ such
that $\bigcup_{i=1,...,k}\gamma_{i}B=\partial\mathbb{H}^{d+1}$. The random set
RST is invariant in distribution by $\text{Stab}(0)$, so the events
$\\{\Psi\cap\gamma_{i}B\neq\emptyset\\}$ all have the same probability. Since
$\Psi\neq\emptyset$ almost surely,
$\mathbb{P}\left[\bigcup_{i=1,...,k}\\{\Psi\cap\gamma_{i}B\neq\emptyset\\}\right]=1$
therefore $\mathbb{P}(\Psi\cap B\neq\emptyset)>0$. In addition, for any
neighborhood $\Phi\subset\overline{\mathbb{H}^{d+1}}$ of $B$, the event
$\\{\Psi\cap B\neq\emptyset\\}$ is entirely determined by
$\mathcal{N}\cap\Phi$, therefore it has probability $0$ or $1$. Thus $\Psi\cap
B\neq\emptyset$ almost surely. Since the topology on
$\partial\mathbb{H}^{d+1}$ admits a countable basis, $\Psi$ is almost surely
dense in $\partial\mathbb{H}^{d+1}$.

It remains to show that $\Psi$ is a closed subset of
$\partial\mathbb{H}^{d+1}$. Let $I\in\overline{\Psi}$ (recall that
$\overline{\Psi}$ is the closure of $\Psi$ in $\overline{\mathbb{H}^{d+1}}$).
We construct by induction a sequence $(z_{n})_{n\geq
1}\in\mathcal{N}^{\mathbb{N}^{*}}$ is an infinite backward path such that, for
any $i\in\mathbb{N}$, $I\in\overline{\mathcal{D}(z_{i})}$. Suppose
$0,...,z_{i-1}$ already defined such that $z_{j}=A(z_{j+1})$ for $0\leq j\leq
i-2$ and $I\in\overline{\mathcal{D}(z_{i-1})}$. Since the vertex $z_{i-1}$ has
finitely many daughters, there exists some $z\in A^{(-1)}(z_{i-1})$ such that
$I\in\overline{\mathcal{D}(z)}$. Thus we define $z_{i}$ as such a $z$.

We now use straightness to show that the infinite backward path $(z_{n})$
constructed above converges to $I$ (and thus $I\in\Psi$). This infinite
backward path converges to some $I^{\prime}\in\partial\mathbb{H}^{d+1}$ by
(i). Let $\varepsilon>0$, by Proposition 2.7 there exists some $i\geq 0$ such
that $\mathcal{D}(z_{i})$ (and thus $\overline{\mathcal{D}(z_{i})}$) is
contained in a cone of apex $0$ and aperture at most $\varepsilon$. Since both
$I$ and $I^{\prime}$ belong to $\overline{\mathcal{D}(z_{i})}$,
$\widehat{I0I^{\prime}}\leq\varepsilon$. Thus $I=I^{\prime}$, which achieves
the proof of the existence part.

#### Proof of (iv) and (v)

Let us denote by $\Psi^{\prime}\subset\partial\mathbb{H}^{d+1}$ the set of
asymptotic directions with two infinite backward paths. To show (iv), we first
show that, a.s. $\Psi^{\prime}\neq\emptyset$. For $z\in\text{RST}$, let us
define $\Psi_{z}\subset\partial\mathbb{H}^{d+1}$ as the set of asymptotic
directions of infinite backward paths from $z$. By the same argument as in
Step 2, $\Psi_{z}$ is a closed subset of $\partial\mathbb{H}^{d+1}$. By (ii),
a.s., there exists at least two infinite backward paths, so there exists a.s.
some level $r_{0}>0$ with two points connected to infinity. Thus
$\\{\Psi_{z},~{}z\in\mathcal{L}(r_{0})\\}$ is a covering of
$\partial\mathbb{H}^{d+1}$ by closed subsets, where at least two of them are
nonempty. Since $\partial\mathbb{H}^{d+1}$ is connected, it implies that there
exists $z_{1},z_{2}\in\mathcal{L}_{r_{0}}$ such that
$\Psi_{z_{1}}\cap\Psi_{z_{2}}\neq\emptyset$. Thus $\Psi^{\prime}\neq\emptyset$
a.s.

We use the same argument as in Step 2 to deduce that $\Psi^{\prime}$ is dense.
Let $B$ be an open subset of $\partial\mathbb{H}^{d+1}$. Since
$\partial\mathbb{H}^{d+1}\simeq\mathbb{S}^{d}$ is compact, there exists
finitely many isometries $\gamma_{1},...,\gamma_{k}\in\text{Stab}(0)$ such
that $\bigcup_{i=1,...,k}\gamma_{i}B=\partial\mathbb{H}^{d+1}$. The random set
RST is invariant in distribution by $\text{Stab}(0)$, so the events
$\Gamma_{i}:=\\{\Psi^{\prime}\cap\gamma_{i}B\neq\emptyset\\}$ all have the
same probability. Since $\Psi\neq\emptyset$ almost surely,
$\mathbb{P}\left[\bigcup_{i=1,...,k}\Gamma_{i}\right]=1$ therefore
$\mathbb{P}(\Gamma_{i})>0$. In addition, for any neighborhood
$\Phi\subset\overline{\mathbb{H}^{d+1}}$ of $B$, the event $\Gamma_{i}$ is
entirely determined by $\mathcal{N}\cap\Phi$, therefore it has probability $0$
or $1$. Thus $\Psi^{\prime}\cap B\neq\emptyset$ almost surely. Since the
topology on $\partial\mathbb{H}^{d+1}$ admits a countable basis,
$\Psi^{\prime}$ is almost surely dense in $\partial\mathbb{H}^{d+1}$.

The proof of (v) is done by exploiting the planarity in the bi-dimensional
case (Proposition 2.2). Let us associate to any $z_{\infty}\in\Psi^{\prime}$ a
couple of vertices $P(z_{\infty})=(z_{1},z_{2})\in\mathcal{N}^{2}$ with
$z_{1}\neq z_{2}$ such that $z_{\infty}\in\Psi_{z_{1}}\cap\Psi_{z_{2}}$. By
planarity, such an application $P$ must be injective. Indeed, if
$z_{\infty}\neq z^{\prime}_{\infty}$ are such that
$P(z_{\infty})=P(z^{\prime}_{\infty})=(z_{1},z_{2})$, then there exists four
distinct backward infinite paths joining $z_{\infty}$ to $z_{1}$, $z_{\infty}$
to $z_{2}$, $z^{\prime}_{\infty}$ to $z_{1}$ and $z^{\prime}_{\infty}$ to
$z_{2}$. This implies that two paths among them intersect each other, even if
the representation of edges are replaced by geodesics, which contradicts
planarity. Therefore $\Psi^{\prime}$ is a.s. countable in the case $d=1$.

###  3.2 The uniqueness part: proof of (iii)

The strategy is to exploit the link between the hyperbolic RST and the
hyperbolic DSF. Let us consider the $(d+1)$-dimensional half-space model
$(H,ds^{2})$:

$\displaystyle H=\\{(x_{1},...,x_{d},y)\in\mathbb{R}^{d+1},~{}y>0\\},\quad
ds^{2}=\frac{dx_{1}^{2}+...+dx_{d}^{2}+dy^{2}}{y^{2}}.$

In the following, we will identify the point $(x_{1},...,x_{d+1})\in H$ with
the couple $(x,y)\in\mathbb{R}^{d}\times\mathbb{R}_{+}^{*}$ with

$\displaystyle x:=(x_{1},...,x_{d}),\quad y:=x_{d+1}.$ (3.1)

The coordinate $x$ is referred as the _abscissa_ and $y$ as the _ordinate_.
Let us remind that, in the half-space representation, the boundary set
$\partial\mathbb{H}^{d+1}$ is identifyied as the boundary hyper-plane
$\mathbb{R}^{d}\times\\{0\\}$, plus an additinal point at infinity denoted by
$\infty$, obtained by compactifying the closed half-space
$\mathbb{R}^{d}\times\mathbb{R}_{+}$. Let us define $I_{\infty}$ as the
boundary point represented by $(0,0)$ in $H$.

Let $\mathcal{N}$ be a PPP inside $(H,ds^{2})$, and, for any $h\geq 0$, let us
define $\text{RST}(h)$ as the Radial Spanning Tree of $\mathcal{N}$ with
origin $O(h):=(0,e^{h})$, and let us define DSF as the Directed Spanning
Forest of $\mathcal{N}$ with direction $\infty$ (defined in [13]). Given
$z\in\mathcal{N}$ let $A_{\text{DSF}}(z)$ be the parent of $z$ in DSF, and let
$A_{\text{RST}(h)}(z)$ be the parent of $z$ in $\text{RST}(h)$. We will also
consider, for any given $h\geq 0$, the direction toward $I_{\infty}$ defined
as $u:=(0,...,-1)\in UT_{O(h)}\mathbb{H}^{d+1}\simeq\mathbb{S}^{d}$. The proof
is based on the two following propositions. The next one asserts that the
$\text{RST}(h)$ and the DSF coincide in a given compact set when $h$ is large
enough.

######  Proposition 3.1 (Coupling between RST and DSF).

Let $K\subset\mathbb{H}^{d+1}$ be some compact set. Then

$\displaystyle\lim_{h\to\infty}\mathbb{P}[\forall z\in\mathcal{N}\cap
K,~{}A_{\text{RST}(h)}(z)=A_{\text{DSF}}(z)]=1.$

For $A,a,h\geq 0$, let us define:

$\displaystyle\text{Vois}(A,h):=\text{Cone}_{O(h)}(I_{\infty},Ae^{-h})\backslash
B(O(h),h),$
$\displaystyle\text{Vois}^{\prime}(A,a,h):=(B(O(h),h+a)\cap\text{Cone}_{O(h)}(I_{\infty},Ae^{-h}))\backslash
B(O(h),h),$
$\displaystyle\text{Vois}^{\prime\prime}(A,a,h):=\text{Cone}_{O(h)}(I_{\infty},Ae^{-h-a})\backslash
B(O(h),h+a),$

where $\text{Cone}_{z_{0}}(z,\theta)$ denotes the cone with apex $z_{0}$,
direction $z$ and aperture $\theta$. Let us also define:

$\displaystyle\text{Cyl}(A):=B_{\mathbb{R}^{d}}\left(0,A\right)\times\left(0,\frac{3}{2}\right]\subset
H,$
$\displaystyle\text{Cyl}^{\prime}(A,a):=B_{\mathbb{R}^{d}}\left(0,A\right)\times\left[\frac{1}{2}e^{-a},\frac{3}{2}\right]\subset
H,$
$\displaystyle\text{Cyl}^{\prime\prime}(A,a):=B_{\mathbb{R}^{d}}\left(0,Ae^{-a}\right)\times\left(0,\frac{3}{2}e^{-a}\right]\subset
H.$

The sets
$\text{Vois}(A,h),\text{Vois}^{\prime}(A,a,h),\text{Vois}^{\prime\prime}(A,a,h)$
and $\text{Cyl}(A),\text{Cyl}^{\prime}(A,a),\text{Cyl}^{\prime\prime}(A,a)$
are represented in Figure 4. We will use the following geometrical fact:

######  Lemma 3.2.

For any $A,a\geq 0$, $h$ can be chosen large enough such that

$\displaystyle\text{Vois}(A,h)\subset\text{Cyl}(A),\quad\text{Vois}^{\prime}(A,a,h)\subset\text{Cyl}^{\prime}(A,a),\text{
and }\text{Vois}^{\prime\prime}(A,a,h)\subset\text{Cyl}^{\prime\prime}(A,a).$
(3.2)

For $A,h\geq 0$, let us define the event $E(A,h)$ saying that every infinite
backward path converging to $I_{\infty}$ in $\text{RST}(h)$ restricted to the
annulus $\mathbb{H}^{d+1}\backslash S(O(h),h)$ are contained in
$\text{Vois}(A,h)$:

$\displaystyle E(A,h):=\\{\forall
z\in\text{RST}(h)\cap(\mathbb{H}^{d+1}\backslash
S(O(h),h)),~{}I_{\infty}\in\overline{\mathcal{D}_{\text{RST}(h)}(z)}\implies
z\in\text{Vois}(A,h)\\},$

where $\mathcal{D}_{\text{RST}(h)}(z)$ denotes the descendents subtree of $z$
in $\text{RST}(h)$. Let us also define, for $A,h,a>0$, the event
$E^{\prime}(A,h,a)$ saying that every infinite backward path converging to
$I_{\infty}$ in $\text{RST}(h)$ restricted to the annulus
$\mathbb{H}^{d+1}\backslash S(O(h),h+a)$ are contained in
$\text{Vois}^{\prime\prime}(A,a,h)$:

$\displaystyle E^{\prime}(A,a,h):=\\{\forall
z\in\text{RST}(h)\cap(\mathbb{H}^{d+1}\backslash
S(O(h),h+a)),~{}I_{\infty}\in\overline{\mathcal{D}_{\text{RST}(h)}(z)}\implies
z\in\text{Vois}^{\prime\prime}(A,a,h)\\}.$

The following proposition asserts that, uniformly in $h$, the events $E(A,h)$
and $E(A,a,h)$ occur with high probability when $A$ is large.

######  Proposition 3.3.

We have

$\displaystyle\lim_{A\to\infty}\liminf_{h\to\infty}\mathbb{P}[E(A,h)]=1,\quad\lim_{A\to\infty}\liminf_{h\to\infty}\mathbb{P}[E^{\prime}(A,a,h)]=1\text{
for any }a\geq 0.$

Let us assume Propositions 3.1, 3.3 and Lemma 3.2 for the moment and let us
prove part (iii) of Theorem 1.1. For $h\geq 0$, let us define the event

$\displaystyle U(h):=\\{\text{there is a unique infinite backward path
converging to $I_{\infty}$ in $\text{RST}(h)$}\\}$

By isometries invariance, $\mathbb{P}[U(h)]$ is independent of $h$, and let us
suppose for contradiction that $q:=\mathbb{P}[U(h)^{c}]>0$. For $A,a,h\geq 0$,
let us define the event

$\displaystyle\text{CO}(A,a,h):=\\{\forall
z\in\mathcal{N}\cap\text{Cyl}^{\prime}(A,a),~{}A_{\text{RST}(h)}(z)=A_{\text{DSF}}(z)\\}$

By Proposition 3.3, $A$ can be chosen such that

$\displaystyle\liminf_{h\to\infty}\mathbb{P}[E(A,h)]>1-q/4,\quad\mathbb{P}[E^{\prime}(A,a,h)]>1-q/4.$

Then, by Proposition 3.1 applied to the compact set
$K:=\text{Cyl}^{\prime}(A,a)$ and Lemma 3.2, $h$ can be chosen large enough
such that inclusions (3.2) hold and such that

$\displaystyle\mathbb{P}[\text{CO}(A,a,h)]>1-q/4,\quad\mathbb{P}[E(A,h)]\geq
1-q/4\text{ and }\mathbb{P}[E^{\prime}(A,a,h)]\geq 1-q/4.$

Let us define the event $Z(a)$ as

$\displaystyle Z(a):=U^{c}\cap E(A,h)\cap
E^{\prime}(A,a,h)\cap\text{CO}(A,a,h),$

and define

$\displaystyle Z:=\bigcap_{a_{0}>0}\bigcup_{a\geq a_{0}}Z(a).$

For the choices of $q$, $h$ done before, $\mathbb{P}[Z(a)]\geq q/4$ and so
$\mathbb{P}[Z]\geq q/4>0$. On the event $Z(a)$, and because inclusion (3.2)
holds, there exists two infinite backward paths in $\text{RST}(h)$ whose
restrictions to $\mathbb{R}^{d}\times(0,3/2]$ are contained in
$\text{Cyl}(A,a)$ and converging to $I_{\infty}$, and intersecting
$\text{Cyl}^{\prime\prime}(A,a)$. These two infinite backward paths coincide
with those of DSF inside $\text{Cyl}^{\prime}(A,a)$. Thus, in DSF, there
exists two infinite backward paths contained in $\text{Cyl}(A,a)$ and
intersecting $\text{Cyl}^{\prime\prime}(A,a)$. Therefore, since this is true
for all $a>0$, on the event $Z$, it is possible to construct two infinite
backward paths converging to $I_{\infty}$ in DSF using the fact that DSF is
locally finite (it is true by [13, Proposition 9]). However, by [13, Theorem
3], there almost surely a unique infinite backward path converging to
$I_{\infty}$ in DSF. This leads to a contradiction, which achieves the proof.

Figure 4: Representation of the sets $\text{Vois}(A,h)$,
$\text{Vois}^{\prime}(A,a,h)$, $\text{Vois}^{\prime\prime}(A,a,h)$ and
$\text{Cyl}(A)$, $\text{Cyl}^{\prime}(A,a)$, $\text{Cyl}^{\prime\prime}(A,a)$.
The backward paths of $\text{RST}(h)$ converging to $0$ (in blue) are all
contained in $\text{Vois}^{\prime\prime}(A,a,h)$ up to level $h+a$ and
contained in $\text{Vois}(A,a,h)$ up to level $h$. In the dashed area
($\text{Cyl}^{\prime}(A,a)$), the DSF and $\text{RST}(h)$
coincide.$0$$x\in\mathbb{R}^{d}$$y\in\mathbb{R}$O(h)$\text{Cyl}^{\prime\prime}(A,a)$$\text{Cyl}^{\prime}(A,a)$$\text{Cyl}(A)$$\text{Vois}^{\prime}(A,a,h)$$\text{Vois}(A,h)$$\text{Vois}^{\prime\prime}(A,a,h)$$S(h)$$S(h+a)$

###### Proof of Proposition 3.1.

Let us define, for $z=(x,y)\in\mathcal{N}$ and $h\geq 0$:

$\displaystyle
B^{+}_{\text{DSF}}(z):=B(z,d(z,A(z)))\cap(\mathbb{R}^{d}\times(y,\infty)),\quad
B^{+}_{\text{RST}(h)}(z):=B(z,d(z,A(z)))\cap(y,\infty).$

Let $K\subset\mathbb{H}^{d+1}$ be some compact set. For any given
$z\in\mathcal{N}\cap K$, $h\geq 0$, $A_{\text{DSF}}(z)=A_{\text{RST}(h)}(z)$
if and only if $\mathcal{N}\cap(B^{+}_{\text{DSF}}(z)\Delta
B^{+}_{\text{RST}(h)}(z))=\emptyset$. For any $z\in\mathbb{H}^{d+1}$,
$\mbox{Vol}(B^{+}_{\text{DSF}}(z)\Delta B^{+}_{\text{RST}(h)}(z))\to 0$ as
$h\to\infty$ (recall that Vol is the hyperbolic volume). Campbell formula [8]
gives,

$\displaystyle\mathbb{E}\left[\\#\\{z^{\prime}\in\mathcal{N}\cap
B(z,d(z,A(z))),~{}\mathcal{N}\cap(B^{+}_{\text{DSF}}(z)\Delta
B^{+}_{\text{RST}(h)}(z^{\prime}))\neq\emptyset\\}\right]$
$\displaystyle=\lambda\int_{B(z,d(z,A(z)))}\mathbb{P}\left[\mathcal{N}\cap(B^{+}_{\text{DSF}}(z)\Delta
B^{+}_{\text{RST}(h)}(z^{\prime}))\neq\emptyset\right]~{}d\mbox{Vol}(z)$
$\displaystyle=\lambda\int_{B(z,d(z,A(z)))}1-\exp(-\lambda\mbox{Vol}(B^{+}_{\text{DSF}}(z)\Delta
B^{+}_{\text{RST}(h)}(z)))~{}d\mbox{Vol}(z)\to 0\text{ as }h\to\infty$

by dominated convergence. Proposition 3.1 follows. ∎

###### Proof of Proposition 3.3.

∎

For $n\in\mathbb{N}$ and $h\geq 0$, let us define the event

$\displaystyle F_{n}(h):=\\{\exists z\in
B_{S(h)}(u,2^{n+1}e^{-h}),~{}\text{MBD}_{h}^{\infty}(z)>2^{n}e^{-h}\\}.$

We now show that for $n\in\mathbb{N}$ and $h\geq 0$,
$E(2^{n},h)^{c}\subset\bigcup_{m\geq n}F_{n}(h)$. If $E(2^{n},h)^{c}$ does not
occur, then there exists some $z\in S(h)\backslash B_{S(h)}(u,2^{n}e^{-h})$
such that $I_{\infty}\in\overline{\mathcal{D}_{\text{RST}(h)}(z)}$, so, for
the value of $m\in\mathbb{N}_{\geq n}$ such that
$2^{m}\leq\widehat{z0I_{\infty}}<2^{m+1}$, $z\in B_{S(h)}(u,2^{m+1}e^{-h})$
and $\text{MBD}_{h}^{\infty}(z)>2^{n}e^{-h}\geq\widehat{z0I_{\infty}}\geq
2^{m}$, thus $F_{m}(h)$ occurs.

Therefore

$\displaystyle\mathbb{P}[E(2^{n},h)]\geq 1-\sum_{m\geq
n}\mathbb{P}[F_{n}(h)].$

We now upper-bound $\mathbb{P}[F_{n}(h)]$. On $F_{n}(h)$, the following
occurs:

$\displaystyle\sum_{z\in
B_{S(h)}(u,2^{n+1}e^{-h})\cap\text{RST}}\text{MBD}_{h}^{\infty}(z)^{p}>2^{np}e^{-ph},$
(3.3)

thus, by Markov inequality,

$\displaystyle\mathbb{P}\left[F_{n}(h)\right]\leq
2^{-np}e^{ph}\mathbb{E}\left[\sum_{z\in
B_{S(h)}(u,2^{n+1}e^{-h})\cap\text{RST}}\text{MBD}_{h}^{\infty}(z)^{p}\right]\overset{Prop.\ref{Prop:globalfluct}}{\leq}C_{\text{fl}}2^{n(d-p)}$

for some $C_{\text{fl}}>0$ depending only on $p$. Combining this with (3.3)
for some $p>d$ leads to

$\displaystyle\mathbb{P}[E(2^{n},h)]\geq
1-C_{\text{fl}}\frac{2^{m(d-p)}}{1-2^{d-p}},$

this proves the first part of Proposition 3.3. The second part be deduced from
the first part by applying the dilation $(x,y)\mapsto(e^{a}x,e^{a}y)$ (which
is an isometry of $(H,ds^{2})$).

###### Proof of Lemma 3.2.

Let $A,a,h\geq 0$, and let $z=(x,y)\in\text{Vois}(A,h)$. Considering the
totally geodesic plane containing $I_{\infty},z$ and $O(h)$ (represented by a
half-plane in $H$), it is possible to suppose $d=1$ without loss of
generality. We apply the distance and angle formulas in $(H,ds^{2})$
(Propositions 7.1 and 7.2).

Let $z=(x,y)\in\text{Vois}(A,h)$. On the one hand,
$\widehat{zO(h)I_{\infty}}\leq Ae^{-h}$, so, taking $h$ large enough such that
$Ae^{-h}<\pi/2$,

$\displaystyle\arctan\left|\frac{2xe^{h}}{e^{2h}-x^{2}-y^{2}}\right|\leq
Ae^{-h},$

thus, for $h$ large enough,

$\displaystyle|x|e^{-h}\leq\arctan|2xe^{-h}|\leq\arctan\left|\frac{2xe^{h}}{e^{2h}-x^{2}-y^{2}}\right|\leq
Ae^{-h},$

so $|x|\leq A$.

On the other hand, $d(O(h),z)\geq h$, so

$\displaystyle
2\tanh^{-1}\left(\sqrt{1-\frac{4ye^{h}}{A^{2}+(y+e^{h})^{2}}}\right)$
$\displaystyle\overset{|x|\leq
A}{\geq}2\tanh^{-1}\left(\sqrt{1-\frac{4ye^{-h}}{(xe^{-h})^{2}+(ye^{-h}+1)^{2}}}\right)=d(O(h),z)\geq
h,$

so

$\displaystyle e^{h}$
$\displaystyle\leq\frac{1+\sqrt{1-\frac{4ye^{h}}{A^{2}+(y+e^{h})^{2}}}}{1-\sqrt{1-\frac{4ye^{h}}{A^{2}+(y+e^{h})^{2}}}}=\frac{2}{1-\sqrt{1-\frac{4ye^{h}}{A^{2}+(y+e^{h})^{2}}}}-1=\frac{e^{h}}{y}+o(e^{h})\text{
when }h\to\infty,$

for $h$ large enough this implies $y\leq 3/2$. The two other inclusions are
shown by similar computations. ∎

##  4 Proof of Proposition 2.5

We use a bloc control argument similar to [13, Section 6.3]. Let $\delta>0$
small and $A>0$ large that will be chosen later. For $r_{0}\geq 2$ and $z\in
S(r_{0})$, let us define

$\displaystyle\Psi_{1}(r_{0},z)$ $\displaystyle:=$
$\displaystyle\text{Cone}(z,3Ae^{-r_{0}})\cap(B(r_{0}+\delta)\backslash
B(r_{0})),$ $\displaystyle\Psi_{2}(r_{0},z)$ $\displaystyle:=$
$\displaystyle\text{Cone}(z,Ae^{-r_{0}})\cap(B(r_{0})\backslash B(r_{0}-1)).$
(4.1)

A point $z\in S(r_{0})$ is said to be _good_ if the following event
$G(r_{0},z)$ occurs:

$\displaystyle
G(r_{0},z):=\left\\{\mathcal{N}\cap\Psi_{1}(r_{0},z)=\emptyset\text{ and
}\mathcal{N}\cap\Psi_{2}(r_{0},z)\neq\emptyset\right\\}.$ (4.2)

A good point is represented in Figure 5.

$0$$\in\mathcal{N}$$\emptyset$$\emptyset$$\emptyset$$z$$S(r_{0}-1)$$S(r_{0})$$S(r_{0}+\delta)$$Ae^{-r_{0}}$$\Psi^{1}(r_{0},z)$$\Psi^{2}(r_{0},z)$
Figure 5: The point $z$ is a _good point_ ; the fluctuations of trajectories
crossing $B_{S(r_{0})}(z,Ae^{-r_{0}})$ (in blue) are well controlled.

Let us define the random subsets $\hat{\chi}(r_{0})\subset S(r_{0})$ and
$\chi(r_{0})\subset\mathbb{H}^{d+1}\backslash\\{0\\}$ as

$\displaystyle\hat{\chi}(r_{0}):=\bigcup_{\begin{subarray}{c}z\in S(r_{0}),\\\
G(r_{0},z)\text{
occurs}\end{subarray}}\text{Cone}(z,Ae^{-r_{0}}),\quad\chi(r_{0}):=\bigcup_{z\in\hat{\chi}(r_{0})}]0,z)\subset\mathbb{H}^{d+1}\backslash\\{0\\}.$
(4.3)

The region $\chi(r_{0})$ is the _controlled region_ , where the cumulative
forward deviations in the annulus $S(r_{0}+\delta)\backslash S(r_{0})$ will be
upper-bounded. This control of fluctuations are given by the following lemma:

######  Lemma 4.1.

There exists some deterministic constant $C_{\text{geom}}>0$ such that for any
$r_{0}\geq 2$, $s\in[r_{0},r_{0}+\delta]$, and
$z\in\mathcal{L}_{s}\cap\chi(r_{0})$, $\text{CFD}_{r_{0}}^{s}(z)\leq
C_{\text{geom}}e^{-r_{0}}$.

Lemma 4.1 will be proved in Section 4.4.

For any $r_{0}\geq 2$, we cover the sphere $S(r_{0})$ by balls of angular
radius $e^{-r_{0}}$ such that the number of balls overlapping at a given point
never exceeds some constant $K$. To proceed, we use the following lemma:

######  Lemma 4.2.

There exists $K=K(d,p)\in\mathbb{N}^{*}$ such that, for any $r_{0}\geq 2$,
there exist a non-negative integer $N(r_{0})\geq 0$ and a family of points
$z_{1},...,z_{N(r_{0})}\in S(r_{0})$ such that:

  * •

$\bigcup_{1\leq i\leq N(r_{0})}B_{S(r_{0})}(z_{i},e^{-r_{0}})=S(r_{0})$,

  * •

$\forall z\in S(r_{0})$, $\\#\\{1\leq i\leq N(r_{0}),~{}z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}\leq K$.

Moreover, there exists $C_{\text{ball}}=C_{\text{ball}}(K,d)>0$ such that, for
any $r_{0}\geq 2$, $z\in S(r_{0})$ and $A\geq 1$, the number of balls
intersecting $B_{S(r_{0})}(z,Ae^{-r_{0}})$ is upper-bounded by
$C_{\text{ball}}A^{d}$:

$\displaystyle\exists C_{\text{ball}}>0,~{}\forall r_{0}\geq 2,~{}\forall z\in
S(r_{0}),~{}\forall A\geq 1,$ $\displaystyle\\#\\{1\leq i\leq
N(r_{0}),B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap
B_{S(r_{0})}(z,Ae^{-r_{0}})\neq\emptyset\\}\leq C_{\text{ball}}A^{d}.$

We refer to Section 4.4 for the proof of Lemma 4.2.

For $1\leq i\leq N(r_{0})$, $z_{i}$ is said to be _inhibited_ if the ball
$B_{S(r_{0})}(z_{i},e^{-r_{0}})$ intersects $S(r_{0})\backslash\hat{\chi}$,
and the corresponding event is denoted by $\mbox{Inhib}(i)$. Let
$\Psi(r_{0})\subset\llbracket 1,N(r_{0})\rrbracket$ be the union of all
inhibited balls:

$\displaystyle\Psi(r_{0}):=\bigcup_{\begin{subarray}{c}1\leq i\leq
N(r_{0}),\\\
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\left(S(r_{0})\backslash\hat{\chi}\right)\neq\emptyset\end{subarray}}B_{S(r_{0})}(z_{i},e^{-r_{0}}).$

The region $\Psi$ is the _augmented uncontrolled region_ , that contains
$\left(\mathbb{H}^{d+1}\backslash\\{0\\}\right)\backslash\chi$. For $z\in
S(r_{0})$, let $\hat{\text{Cl}}(z)$ be the cluster of $z$ in $\Psi(r_{0})$ and
let us also define (recall that $]0,z)$ is the semi-geodesic starting at $0$
and containing $z$, without $0$):

$\displaystyle\text{Cl}(z)=\bigcup_{z\in\hat{\text{Cl}}(z)}]0,z)\subset\mathbb{H}^{d+1}\backslash\\{0\\}.$

We define the _angular radius_ of $\text{Cl}(z)$ as
$\text{Rad}(z):=\sup_{z^{\prime}\in\text{Cl}(z)}\widehat{z^{\prime}0z}$.

The next lemma asserts that the connected components of the augmented
uncontrolled region $\Psi$ are small (the radius admits exponential tail
decay):

######  Lemma 4.3.

There exists $\delta>0$ small enough, $A>0$ large enough and some constant
$c_{\text{dec}}>0$ such that, for any $B$ large enough, $r_{0}\geq 2$ and
$z\in S(r_{0})$,

$\displaystyle\mathbb{P}\left[e^{r_{0}}\text{Rad}(z)>B\right]\leq
e^{-c_{\text{dec}}B}.$

Lemma 4.3 is proved in Section 4.4.

In addition, we need a control of the number of points in a given region of a
sphere $S(r_{0})$, which is given by the next lemma:

######  Lemma 4.4.

For any $p\geq 1$, there exists a constant $C=C(d,p)>0$ such that, for any
$r_{0}\geq 0$ and any direction $z\in B_{S(r_{0})}$,

$\displaystyle\mathbb{E}\left[\\#\left(\mathcal{L}_{r_{0}}\cap
B_{S(r_{0})}(z,e^{-r_{0}})\right)^{p}\right]\leq C.$

We refer to Section 4.4 for the proof of Lemma 4.4. Let us choose $A,\delta>0$
as in Lemma 4.3 and $C_{\text{geom}}$ as in Lemma 4.1.

###  4.1 Step 1: a deterministic upper-bound of
$\text{MBD}_{r_{0}}^{r_{0}+\delta}(\cdot)$

For $z\in S(r_{0})$, let us define

$\displaystyle
M(z):=\\#\mathcal{N}\cap\text{Cl}(z)\cap(S(r_{0}+\delta)\backslash S(r_{0})).$
(4.4)

This step is devoted to the proof of the following upper-bound: almost surely,
for any $z\in S(r_{0})$,

$\displaystyle\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\leq
2^{p-1}\big{(}2^{p}\text{Rad}(z)^{p}(M(z)+1)^{p}+C_{\text{geom}}^{p}e^{-pr_{0}}\big{)}.$
(4.5)

The quantity $\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)$ takes into account
backward paths from $z$ that ends (in the backward direction) before level
$r_{0}+\delta$ and backward paths reaching level $r_{0}+\delta$. For $z\in
S(r_{0})$, we define $\text{Stop}(z)$ as the set of ending points of backward
paths from $z$ stopping before level $r_{0}+\delta$ (recall that
$\mathcal{D}(z):=\bigcup_{r\geq r_{0}}\mathcal{D}_{r_{0}}^{r}(z)$):

$\displaystyle\text{Stop}(z):=\\{z^{\prime}=(r^{\prime};u^{\prime})\in\mathcal{N}\cap\mathcal{D}(z),~{}r_{0}\leq
r^{\prime}<r_{0}+\delta,~{}A^{(-1)}(z^{\prime})=\emptyset\\}.$

For any
$z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+\delta}(z)\cup\text{Stop}(z)$, one
the two following cases occur. Either the branch from $z^{\prime}$ to $z$
stays inside $\text{Cl}(z)$, or it crosses $\chi(r_{0})$. Let us define
$\mathcal{C}$ (resp. $\mathcal{C}^{\prime}$) as the set of couples
$(z,z^{\prime})$ such that the branch from $z^{\prime}$ to $z$ crosses
$\chi(r_{0})$ (resp. does not cross $\chi(r_{0})$):

$\displaystyle\mathcal{C}:=\\{(z,z^{\prime}),~{}z\in
S(r_{0}),~{}z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+\delta}(z)\cup\text{Stop}(z),~{}\exists
s\in[r_{0},r_{0}+\delta],~{}\mathcal{A}_{s}^{r_{0}+\delta}(z^{\prime})\in\chi\\},$
$\displaystyle\mathcal{C}^{\prime}:=\\{(z,z^{\prime}),~{}z\in
S(r_{0}),~{}z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+\delta}(z)\cup\text{Stop}(z),~{}\forall
s\in[r_{0},r_{0}+\delta],~{}\mathcal{A}_{s}^{r_{0}+\delta}(z^{\prime})\notin\chi\\}.$

Moreover, for any $(z,z^{\prime})\in\mathcal{C}$, we define
$\text{hit}(z,z^{\prime})$ as the highest level where the branch from
$z^{\prime}$ to $z$ hits $\chi(r_{0})$:

$\displaystyle\text{hit}(z,z^{\prime}):=\max\\{s\in[r_{0},r_{0}+\delta],~{}\mathcal{A}_{s}^{r_{0}+\delta}(z^{\prime})\in\chi\\},$

where the $\max$ is over a set which is finite a.s.

Let $1\leq i\leq N(r_{0})$. Let $z\in B_{S(r_{0})}(z_{i},e^{-r_{0}})$,
$z^{\prime}=(r^{\prime};u^{\prime})\in\mathcal{D}_{r_{0}}^{r_{0}+\delta}(z)\cup\text{Stop}(z)$,
we now upper-bound $\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})$. Let us first
consider the case where $(z,z^{\prime})\in\mathcal{C}^{\prime}$ (the branch
between $z$ and $z^{\prime}$ does not cross $\chi(r_{0})$). Then this branch
stays inside $\text{Cl}(z)$. Thus it crosses at most $M(z)$ points of
$\mathcal{N}$, therefore

$\displaystyle\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})\leq
2\text{Rad}(z)(M(z)+1).$

In the other case, $(z,z^{\prime})\in\mathcal{C}$, let
$z^{\prime\prime}=(r^{\prime\prime};u^{\prime\prime}):=\text{hit}(z,z^{\prime})$.
Let $p\geq 1$. Then, by Jensen inequality,

$\displaystyle\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})^{p}\leq\left(\text{CFD}_{r_{0}}^{r^{\prime\prime}}(z^{\prime\prime})+\text{CFD}_{r^{\prime\prime}}^{r_{0}+\delta}(z^{\prime})\right)^{p}\leq
2^{p-1}\left(\text{CFD}_{r_{0}}^{r^{\prime\prime}}(z^{\prime\prime})^{p}+\text{CFD}_{r^{\prime\prime}}^{r_{0}+\delta}(z^{\prime})^{p}\right)$

By the same argument as in the previous case,

$\displaystyle\text{CFD}_{r_{0}}^{r^{\prime\prime}}(z^{\prime\prime})\leq
2\text{Rad}(z)(M(z)+1),$

and, by Lemma 4.1,
$\text{CFD}_{r^{\prime\prime}}^{r_{0}+\delta}(z^{\prime})\leq
C_{\text{geom}}e^{-r_{0}}$, since, by definition of $z^{\prime\prime}$, the
part of trajectory between $z^{\prime\prime}$ and $z$ is included in
$\text{Cl}(z)$. Thus,

$\displaystyle\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})^{p}\leq
2^{p-1}\left(2^{p}\text{Rad}(z)^{p}(M(z)+1)^{p}+C_{\text{geom}}^{p}e^{-pr_{0}}\right).$
(4.6)

The upper-bound (4.6) holds whatever $(z,z^{\prime})$ belongs to $\mathcal{C}$
or $\mathcal{C}^{\prime}$. It follows that (4.5) holds for any $z\in
S(r_{0})$.

###  4.2 Step 2: a control of the tail decay of $M(z)$

Recall that, for $z\in S(r_{0})$, $M(z)$ is defined in (4.4). In this step, it
is shown that, for any $z\in S(r_{0})$:

$\displaystyle\mathbb{E}\left[M(z)^{4p}\right]\leq C$ (4.7)

for some $C=C(p)>0$.

The quantity $M(z)$ is the number of Poisson points inside a random part of
the thin annulus $S(r_{0}+\delta)\backslash S(r_{0})$, whose diameter admits
exponential tail decay (Lemma 4.3).

Let $z\in S(r_{0})$. For given $R\geq 0$, let us define

$\displaystyle\mbox{Reg}(R):=\text{Cone}(z_{i},Re^{-r_{0}})\cap\big{(}S(r_{0}+\delta)\backslash
S(r_{0})\big{)}.$

For any $R,m\geq 0$,

$\displaystyle\\{M(z)>m\\}\subset\\{\text{Rad}(z)>Re^{-r_{0}}\\}\cup\\{\\#\left(\mathcal{N}\cap\mbox{Reg}(R)\right)>m\\}$

thus

$\displaystyle\mathbb{P}\left[M(z)>m\right]\leq\mathbb{P}\left[\text{Rad}(z)>Re^{-r_{0}}\right]+\mathbb{P}\left[\\#\left(\mathcal{N}\cap\mbox{Reg}(R)\right)>m\right].$
(4.8)

By Lemma 4.3, $\mathbb{P}\left[\text{Rad}(z)>Re^{-r_{0}}\right]\leq
e^{-c_{\text{dec}}R}$. The random variable
$\\#\left(\mathcal{N}\cap\mbox{Reg}(R))\right)$ is distributed according to
the Poisson law with parameter $\lambda\mbox{Vol}(\mbox{Reg}(R))$. Recall that
$\nu$ the $d$-dimensional volume measure on $\mathbb{S}^{d}$. Denoting by $u$
the direction of $z$,

$\displaystyle\lambda\nu(\mbox{Reg}(R))=\lambda\nu(\\{u^{\prime},~{}\langle
u,u^{\prime}\rangle<Re^{-r_{0}}\\})\mbox{Vol}((S(r_{0}+\delta)\backslash
S(r_{0}))\leq CR^{d}.$ (4.9)

for some constant $C>0$ independent of $r_{0},R$, since
$\mbox{Vol}(S(r_{0}+\delta))=O\left(e^{dr_{0}}\right)$ by (2.6). Thus
$\\#(\mathcal{N}\cap\mbox{Reg}(R))\preceq_{st}\mathcal{P}(CR^{d})$. Thus
Chernoff bound for the Poisson distribution [25] leads to:

$\displaystyle\mathbb{P}\left[\\#(\mathcal{N}\cap\mbox{Reg}(R))\geq
m\right]\leq\frac{e^{-CR^{d}}(CeR^{d})^{m}}{m^{m}}$ (4.10)

for any $m\geq CR^{d}$. Let us chose $R=(m/(2eC))^{1/d}$ (thus $m=2eCR^{d}$).
It leads to:

$\displaystyle\mathbb{P}\left[\\#(\mathcal{N}\cap\mbox{Reg}(R))\geq
m\right]\leq\left(\frac{e^{-1/(2e)}}{2}\right)^{m}\leq\left(\frac{1}{2}\right)^{m}.$
(4.11)

Finally, we combine (4.8), Lemma 4.3 and (4.11) to obtain:

$\displaystyle\mathbb{P}\left[\\#\left(\mathcal{N}\cap\text{Cl}(z)\cap(S(r_{0}+\delta)\backslash
S(r_{0}))\right)>m\right]\leq
e^{-c_{\text{dec}}R}+\left(\frac{1}{2}\right)^{m}$
$\displaystyle=\exp\left(-c_{\text{dec}}(m/2eC)^{1/d}\right)+\left(\frac{1}{2}\right)^{m}\leq\exp(-cm^{1/d})$
(4.12)

for some $C>0$. Therefore:

$\displaystyle\mathbb{E}\left[\\#\left(\mathcal{N}\cap\text{Cl}(z)\cap(S(r_{0}+\delta)\backslash
S(r_{0}))\right)^{4p}\right]$
$\displaystyle=\int_{0}^{\infty}\mathbb{P}\left[\\#\left(\mathcal{N}\cap\text{Cl}(z)\cap(S(r_{0}+\delta)\backslash
S(r_{0}))\right)>m^{1/(4p)}\right]~{}dm$
$\displaystyle\overset{(\ref{E:queuenopcl})}{\leq}\int_{0}^{\infty}\exp(-cm^{1/(4dp)})~{}dm<\infty,$

which proves (4.7).

###  4.3 Step 3: conclusion

By (4.5), for any $1\leq i\leq N(r_{0})$,

$\displaystyle\sum_{z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\leq
2^{p-1}\\#\\{\text{RST}\cap B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}$
$\displaystyle\left(2^{p}\text{Rad}(z)^{p}(M(z)+1)^{p}+C_{\text{geom}}^{p}e^{-pr_{0}}\right).$

It follows by Cauchy-Schwartz and Minkowski that

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\right]$
$\displaystyle\leq C\mathbb{E}\left[\\#\\{\text{RST}\cap
B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}^{2}\right]^{1/2}\mathbb{E}\left[\left(\text{Rad}(z_{i})^{p}M(z)^{p}+e^{-pr_{0}}\right)^{2}\right]^{1/2}$
$\displaystyle\leq C\mathbb{E}\left[\\#\\{\text{RST}\cap
B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}^{2}\right]^{1/2}\left(\mathbb{E}\left[\text{Rad}(z_{i})^{2p}M(z)^{2p}\right]^{1/2}+e^{-pr_{0}}\right)$
$\displaystyle\leq C\mathbb{E}\left[\\#\\{\text{RST}\cap
B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}^{2}\right]^{1/2}$
$\displaystyle\times\left(\mathbb{E}\left[\text{Rad}(z_{i})^{4p}\right]^{1/4}\mathbb{E}\left[M(z)^{4p}\right]^{1/4}+e^{-pr_{0}}\right)$
(4.13)

for some $C=C(p)>0$. By Lemma 4.4 applied to $p=2$,

$\displaystyle\mathbb{E}\left[\\#\\{\text{RST}\cap
B_{S(r_{0})}(z_{i},e^{-r_{0}})\\}^{2}\right]\leq C,$ (4.14)

for $C$ independent of $r_{0}$, $z_{i}$. By Lemma 4.3,

$\displaystyle\mathbb{E}\left[\text{Rad}(z_{i})^{4p}\right]$
$\displaystyle=e^{-4pr_{0}}\int_{0}^{\infty}\mathbb{P}\left[e^{4pr_{0}}\text{Rad}(z_{i})^{4p}>B\right]~{}dB$
(4.15)
$\displaystyle=e^{-pr_{0}}\int_{0}^{\infty}\mathbb{P}\left[e^{4r_{0}}\text{Rad}(z_{i})>B^{\frac{1}{4p}}\right]~{}dB$
$\displaystyle\leq
e^{-4pr_{0}}\int_{0}^{\infty}e^{-c_{\text{dec}}B^{\frac{1}{4p}}}~{}dB<\infty.$

Then, by combining (4.3), (4.14), (4.15) and (4.7),

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\right]\leq
Ce^{-pr_{0}},$ (4.16)

for some $C=C(p)>0$.

The final step is to sum over all $i$ such that
$B_{S(r_{0})}(z_{i},e^{-r_{0}})$ intersects $B_{S(r_{0})}(u,\theta)$ for any
given $u\in\mathbb{S}^{d}$ and $\theta>0$. Let $\theta>0$. By Lemma 4.2 it can
be assumed that $B_{S(r_{0})}(z_{i},e^{-r_{0}})$ intersects
$B_{S(r_{0})}(z_{i},\theta)$ for at most $C_{\text{ball}}e^{dr_{0}}\theta^{d}$
values of $i\in\llbracket 1,N(r_{0})\rrbracket$. Therefore

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(u,\theta)\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\right]=\sum_{\begin{subarray}{c}1\leq
i\leq N(r_{0}),\\\ B_{S(r_{0})}(z_{i},e^{-r_{0}})\\\ \cap
B_{S(r_{0})}(u,\theta)\neq\emptyset\end{subarray}}\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+\delta}(z)^{p}\right]$
$\displaystyle\overset{(\ref{E:majorunebouleelem})}{\leq}Ce^{-pr_{0}}\\#\\{1\leq
i\leq N(r_{0}),B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap
B_{S(r_{0})}(u,\theta)\neq\emptyset\\}$ $\displaystyle\leq
Ce^{(d-p)r_{0}}\theta^{d},$

which achieves the proof of Proposition 2.5.

###  4.4 Proof of Lemmas 4.1, 4.2, 4.3, 4.4

###### Proof of Lemma 4.1.

Let $r_{0}\leq s\leq r_{0}+\delta$ and
$z_{1}\in\chi(r_{0})\cap\mathcal{L}(s)$. Since $z_{1}\in\chi(r_{0})$,
$G(z_{1})$ occurs. Let $z_{0}=(r_{0};u_{0}):={z_{1}}_{\downarrow}$,
$z_{2}=(r_{2},u_{2}):=\mathcal{A}_{s}^{r_{0}}(z_{1})$,
$z_{4}=(r_{3},u_{3}):={z_{1}}_{\uparrow}$ and $z_{3}\in[z_{0},z_{4}]^{*}$ is
such that $\widehat{z_{1}0z_{4}}=3Ae^{-r_{0}}$. Since $G(z_{1})$ occurs, there
exists some $z^{\prime}=(r^{\prime};u^{\prime})\in\Psi_{2}(r_{0},z_{1})$. Let
us suppose that $\widehat{z_{1}0z_{2}}\geq 3Ae^{-r_{0}}$. It will be shown
that

$\displaystyle d(z_{0},z^{\prime})\leq d(z_{0},z_{3}).$ (4.17)

It implies by construction of $[z_{0},z_{4}]^{*}$ that
$d(z_{0},z^{\prime})\leq d(z_{0},z_{4})$, which contradicts the fact that
$z_{4}=A(z_{0})$. It follows that $\widehat{z_{1}z^{\prime}z_{2}}\leq
3Ae^{-r_{0}}$, and since $\Psi_{1}(r_{0},z_{1})\cap\mathcal{N}=\emptyset$, it
implies that $\text{CFD}_{s}^{r_{0}}(z_{1})\leq 3Ae^{-r_{0}}$, so Lemma 4.1
holds with $C_{\text{geom}}=3A$.

We move on to show (4.17). By the hyperbolic law of cosines (2.7),

$\displaystyle\cosh d(z_{0},z^{\prime})$
$\displaystyle=\cosh(r_{0})\cosh(r^{\prime})-\cos(\widehat{z_{0}0z^{\prime}})\sinh(r_{0})\sinh(r^{\prime})$
$\displaystyle=1+(1-\cos(\widehat{z_{0}0z^{\prime}}))\sinh(r_{0})\sinh(r^{\prime})$
$\displaystyle\leq
1+\left(1-\cos\left(\widehat{z_{0}0z_{1}}+Ae^{-r_{0}}\right)\right)\sinh(r_{0})\sinh(r^{\prime})$
$\displaystyle\leq
1+\left(1-\cos\left(\widehat{z_{0}0z_{1}}+Ae^{-r_{0}}\right)\right)\sinh(r_{0})^{2}$

and

$\displaystyle\cosh d(z_{0},z_{3})$
$\displaystyle=\cosh(r_{0})\cosh(r_{3})-\cos(\widehat{z_{0}0z_{3}})\sinh(r_{0})\sinh(r-3)$
$\displaystyle=1+(1-\cos(\widehat{z_{0}0z_{3}}))\sinh(r_{0})\sinh(r_{3})$
$\displaystyle=1+\left(1-\cos\left(\widehat{z_{0}0z_{1}}+\widehat{z_{1}0z_{3}}\right)\right)\sinh(r_{0})\sinh(r_{3})$
$\displaystyle\geq
1+\left(1-\cos\left(\widehat{z_{0}0z_{1}}+3Ae^{-r_{0}}\right)\right)\sinh(r_{0})\sinh(r_{3})$
$\displaystyle\geq
1+\left(1-\cos\left(\widehat{z_{0}0z_{1}}+Ae^{-r_{0}}\right)\right)\sinh(r_{0})^{2}.$

This proves (4.17), which achieves the proof of Lemma 4.1. ∎

###### Proof of Lemma 4.2.

Proving the first part of Lemma 4.2 is equivalent to show that there exists
some $K\in\mathbb{N}$ such that, for any $\varepsilon>0$, the Euclidean unit
sphere $\mathbb{S}^{d}$ can be covered by balls of radius $\varepsilon$ such
that the number of balls overlapping some given point $x\in\mathbb{S}^{d}$ is
bounded by $K$, which is a standard fact.

We move on to show the second part, i.e. the existence of $C_{\text{ball}}>0$
such that, for any $r_{0}\geq 2$, $z\in S(r_{0})$ and $A\geq 1$, the number of
balls intersecting $B_{S(r_{0})}(z,Ae^{-r_{0}})$ is upper-bounded by
$C_{\text{ball}}A^{d}$. Let $u\in\mathbb{S}^{d}$ be the direction of $z$ and
let $A\geq 1$. For $i\in\llbracket 1,N(r_{0})\rrbracket$, the ball
$B_{S(r_{0})}(z_{i},e^{-r_{0}})$ intersects $B_{S(r_{0})}(z,Ae^{-r_{0}})$ if
and only if $\widehat{z_{i}0z}\leq(A+1)e^{-r_{0}}$. Thus

$\displaystyle\left(\bigcup_{\begin{subarray}{c}1\leq i\leq N(r_{0}),\\\
B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap\\\
B_{S(r_{0})}(z,Ae^{-r_{0}})\neq\emptyset\end{subarray}}B_{S(r_{0})}(z_{i},e^{-r_{0}})\right)\subset
B_{S(r_{0})}(z,(A+2)e^{-r_{0}}).$

Recall that $\nu$ denotes the $d$-dimensional volume measure on
$\mathbb{S}^{d}$. There exists $C>0$ such that, for any $r_{0}\geq 2$,

$\displaystyle\nu\\{u,~{}\widehat{uu_{0}}\leq e^{-r_{0}}\\}\geq Ce^{-r_{0}d}.$

Moreover,

$\displaystyle\nu\\{u,~{}\widehat{uu_{0}}\leq(A+2)e^{-r_{0}}\\}\leq\left((A+2)e^{-r_{0}}\right)^{d},$

thus the number of balls intersecting $B_{S(r_{0})}(z,Ae^{-r_{0}})$ is upper-
bounded by:

$\displaystyle
K\frac{\nu\\{u,~{}\widehat{uu_{0}}\leq(A+2)e^{-r_{0}}\\}}{\nu\\{u,~{}\widehat{uu_{0}}\leq
e^{-r_{0}}\\}}\leq\frac{K\left((A+2)e^{-r_{0}}\right)^{d}}{ce^{-r_{0}d}}\overset{A\geq
1}{\leq}\frac{3^{d}K}{c}A^{d},$

the conclusion follows. ∎

###### Proof of Lemma 4.3.

Recall that, for $1\leq i\leq N(r_{0})$, $B_{S(r_{0})}(z_{i},e^{-r_{0}})$ is
said to be inhibited if
$B_{S(r_{0})}(z_{i},e^{-r_{0}})\cap(S(r_{0})\backslash\chi)\neq\emptyset$. Let
us first estimate the probability that a given $z_{i}$ is inhibited.

Let $1\leq i\leq N(r_{0})$. Let us consider the following events:

$\displaystyle
E(i):=\\{\mathcal{N}\cap\text{Cone}(z_{i},(3A+1)e^{-r_{0}})\cap(B(r_{0}+\delta)\backslash
B(r_{0}))=\emptyset\\},$ $\displaystyle
E^{\prime}(i):=\\{\mathcal{N}\cap\text{Cone}(z_{i},(A-1)e^{-r_{0}})\cap(B(r_{0})\backslash
B(r_{0}-1))\neq\emptyset\\}.$

We now show that $E(i)\cap E^{\prime}(i)\subset\mbox{Inhib}(i)^{c}$. Let $z\in
B_{S(r_{0})}(z_{i},e^{-r_{0}})$. By triangular inequality,

$\displaystyle\Psi_{1}(r_{0},z)\subset\text{Cone}(z_{i},(3A+1)e^{-r_{0}})\cap(B(r_{0}+\delta)\backslash
B(r_{0}))$ and
$\displaystyle\text{Cone}(z_{i},(A-1)e^{-r_{0}})\cap(B(r_{0}+\delta)\backslash
B(r_{0}))\subset\Psi_{2}(r_{0},z).$

Therefore, on the event $E(i)\cap E^{\prime}(i)$, $z$ is good (i.e.
$G(r_{0},z)$ occurs). Thus, $\mbox{Inhib}(i)^{c}$ occurs, which shows that
$E(i)\cap E^{\prime}(i)\subset\mbox{Inhib}(i)^{c}$. It follows that
$\mathbb{P}[\mbox{Inhib}(i)]\leq\mathbb{P}[E(i)^{c}]+\mathbb{P}[E^{\prime}(i)^{c}]$.
Since

$\displaystyle\mbox{Vol}(\text{Cone}(z_{i},(3A+1)e^{-r_{0}})\cap(B(r_{0}+\delta)\backslash
B(r_{0})))$
$\displaystyle=\int_{r_{0}}^{r_{0}+\delta}\nu\\{u^{\prime},~{}\langle
u,u^{\prime}\rangle\leq(3A+1)e^{-r_{0}}\\}\sinh(r)^{d}~{}dr\leq C\delta A^{d}$

for some $C>0$ independent of $A,r_{0},\delta$. Thus

$\displaystyle\mathbb{P}\left[E(i)^{c}\right]\leq 1-e^{-\lambda c\delta
A^{d}}$

An analogous computation for $E^{\prime}(i)$ leads to:

$\displaystyle\mathbb{P}\left[E^{\prime}(i)^{c}\right]\leq e^{-\lambda
cA^{d}}$

for some $c>0$ independent of $A,r_{0},\delta$.

We move on to show that $A$ and $\delta$ can be chosen such that
$\mathbb{P}[\mbox{Inhib}(i)]\leq 2C_{\text{ball}}^{-1}(6A+4)^{-d}/3$ for any
$1\leq i\leq N(r_{0})$. Indeed, let us first chose $A$ such that $e^{-\lambda
cA^{d}}C_{\text{ball}}(6A+4)^{d}\leq 1/3$. Then let us chose $\delta$ such
that $1-e^{-\lambda c\delta A^{d}}\leq C_{\text{ball}}^{-1}(6A+4)^{-d}/3$.
Hence

$\displaystyle\mathbb{P}[\mbox{Inhib}(z_{i})]\leq\mathbb{P}\left[E(i)^{c}\right]+\mathbb{P}\left[E^{\prime}(i)^{c}\right]\leq
1-e^{-\lambda c\delta A^{d}}+e^{-\lambda
cA^{d}}\leq\frac{2}{3}C_{\text{ball}}^{-1}(6A+4)^{-d}.$

We finally show that, for this choice of $A,\delta$, there exists
$c_{\text{dec}}>0$ such that for any $B$ large enough, $r_{0}\geq 2$ and $z\in
S(r_{0})$,

$\displaystyle\mathbb{P}\left[e^{r_{0}}\text{Rad}(z)>B\right]\leq
e^{-c_{\text{dec}}B}.$

Fix $r_{0}\geq 2$ and $z\in S(r_{0})$. For given $k$, let us denote by $P(k)$
the set of sequences $z_{i_{0}},\cdots,z_{i_{k}}$ among the $\\{z_{i},~{}1\leq
i\leq N(r_{0})\\}$ such that:

  1. 1.

$d(z_{i_{0}},z)\leq e^{-r_{0}}$,

  2. 2.

for any $0\leq j\leq k-1$, $d(z_{i_{j}},z_{i_{j+1}})\leq(6A+4)e^{-r_{0}}$,

  3. 3.

for any $0\leq j,j^{\prime}\leq k$,
$d(z_{i_{j}},z_{i_{j^{\prime}}})\geq(6A+2)e^{-r_{0}}$.

Let us also denote by $\hat{P}(k)\subset P(k)$ the set of sequences
$z_{i_{0}},\cdots,z_{i_{k}}$ verifying 1., 2., 3. and such that:

4\. $z_{i_{0}},\cdots,z_{i_{k}}$ are inhibited.

It can be noticed that $\\{\text{Rad}(z)\geq
k(6A+4)\\}\subset\\{\hat{P}(k)\neq\emptyset\\}$ for any $k\in\mathbb{N}$, thus
it is enough to upper-bound $\\{\hat{P}(k)\neq\emptyset\\}$ for any
$k\in\mathbb{N}$. Let $(z_{i_{0}},\cdots,z_{i_{k}})\in P(k)$. For $0\leq j\leq
k$, the event $\mbox{Inhib}(z_{i_{j}})$ only depends on the Poisson process
$\mathcal{N}$ inside $\text{Cone}(z_{i_{j}},(3A+1)e^{-r_{0}})$, therefore, by
3. the events $\mbox{Inhib}(z_{i_{j}})$ are mutually independent. Thus

$\displaystyle\mathbb{P}[(z_{i_{0}},\cdots
z_{i_{k}})\in\hat{P}(k)]=\mathbb{P}[\mbox{Inhib}(z_{i_{0}})]^{k+1}\leq\left(\frac{2}{3}\right)^{k+1}C_{\text{ball}}^{-(k+1)}(6A+4)^{-d(k+1)}.$

By Lemma 4.2, for any $1\leq i\leq N(r_{0})$, the number of balls intersecting
$B_{S(r_{0})}(z,\theta)$ is upper-bounded by
$C_{\text{ball}}e^{dr_{0}}\theta^{d}$. Thus

$\displaystyle\\#P(k)\leq C_{\text{ball}}^{k+1}(6A+4)^{d(k+1)}.$

It follows that

$\displaystyle\mathbb{P}[\text{Rad}(z)\geq
k(6A+4)]\leq\mathbb{P}[\hat{P}(k)\neq\emptyset]\leq\mathbb{E}[\\#\hat{P}(k)]$
$\displaystyle\leq
C_{\text{ball}}^{k+1}\left(\frac{2}{3}\right)^{k+1}(6A+4)^{d(k+1)}C_{\text{ball}}^{-(k+1)}(6A+4)^{-d(k+1)}\leq\left(\frac{2}{3}\right)^{k+1}.$

Lemma 4.3 follows. ∎

###### Proof of Lemma 4.4.

Let $r_{0},M>0$ and $z\in S(r_{0})$. Let $h\geq 0$ that will be fixed later.
We divide the set
$L=\\{z^{\prime}\in\mathcal{N},~{}[z^{\prime},A(z^{\prime})]^{*}\in
B_{S(r_{0})}(z,e^{-r_{0}})\\}$ into two subsets $L_{\leq h}$ and $L_{>h}$
according to the length of $[z^{\prime},A(z^{\prime})]$:

$\displaystyle L_{\leq h}:=\\{z^{\prime}\in L,~{}d(z^{\prime},z)\leq
h\\},\quad L_{>h}:=\\{z^{\prime}\in L,~{}d(z^{\prime},z)>h\\}.$ (4.18)

Thus $L=L_{\leq h}\cup L_{>h}$, and

$\displaystyle\mathbb{P}\left[\\#L\geq M\right]\leq\mathbb{P}\left[\\#L_{\leq
h}\geq M\right]+\mathbb{P}\left[L_{>h}\neq\emptyset\right].$ (4.19)

We first upper-bound $\mathbb{P}\left[\\#L_{\leq h}\geq M\right]$. Since
$L_{\leq h}\subset B(z,h)$,

$\displaystyle\mathbb{P}\left[\\#L_{\leq h}\geq
M\right]\leq\mathbb{P}\left[\\#\left(\mathcal{N}\cap B(z,h)\right)\geq
M\right].$

By (2.6), $\mbox{Vol}(B(z,h)\leq Ce^{dh}$, for some $C>0$ independent of
$r_{0}$. So the random variable $\\#\left(\mathcal{N}\cap B(z,h)\right)$ is
stochastically dominated by a Poisson law with parameter $C\lambda e^{dh}$,
thus, by the Chernoff bound for the Poisson distribution [25],

$\displaystyle\mathbb{P}\left[\\#L_{\leq h}\geq M\right]\leq\frac{e^{-C\lambda
e^{dh}}(C\lambda e^{dh})^{M}}{M^{M}}.$ (4.20)

The second step is to upper-bound
$\mathbb{P}\left[\\#L_{>h}\neq\emptyset\right]$. Recall that, for $r\geq 0$,

$\displaystyle B^{+}(z,r):=B(z,r)\cap B(0,d(0,z)).$

For any $z^{\prime}\in L_{>h}$, by triangular inequality, denoting by $z^{*}$
the meeting point of $[z^{\prime},A(z^{\prime})]^{*}$ and $S(r_{0})$,
$d(z^{\prime},A(z^{\prime}))\geq d(z^{\prime},z^{*})\geq
d(z^{\prime},z)-d(z^{*},z)$. The hyperbolic law of cosines (2.7) gives,

$\displaystyle d(z^{*},z)$
$\displaystyle=\operatorname*{arcosh}\big{(}\cosh(r_{0})^{2}-\cos(\widehat{z^{*}0z})\sinh(r_{0})^{2}\big{)}$
$\displaystyle\leq\operatorname*{arcosh}\big{(}\cosh(r_{0})^{2}-\cos(e^{-r_{0}})\sinh(r_{0})^{2}\big{)}$
$\displaystyle=\operatorname*{arcosh}\left(1+\left(1-\cos(e^{-r_{0}})\right)\sinh(r_{0})^{2}\right)$
$\displaystyle\leq C_{\text{dis}}$

for some $C_{\text{dis}}>0$ independent of $r_{0}$. Thus
$d(z^{\prime},A(z^{\prime}))\geq d(z^{\prime},z)-C_{\text{dis}}$, so
$B^{+}(z,(d(z^{\prime},z)-C_{\text{dis}})_{+})$ is empty of points. Therefore,
by Campbell formula [8],

$\displaystyle\mathbb{P}\left[L_{>h}\neq\emptyset\right]\leq\mathbb{E}\left[\\#L_{>h}\right]$
$\displaystyle=\lambda\int_{B(r_{0})\backslash
B(h)}\mathbb{P}\left[z^{\prime}\in L_{>h}\right]~{}dz^{\prime}$
$\displaystyle\leq\lambda\int_{B(r_{0})\backslash
B(h)}\mathbb{P}\left[B^{+}(z,(d(z^{\prime},z)-C_{\text{dis}})_{+})\cap\mathcal{N}=\emptyset\right]~{}dz^{\prime}$
$\displaystyle=\lambda\int_{B(r_{0})\backslash
B(h)}\mathbb{P}\left[\exp(-\lambda\mbox{Vol}(B^{+}(z,(d(z^{\prime},z)-C_{\text{dis}})_{+}))\right]~{}dz^{\prime}$
$\displaystyle\overset{(\ref{E:boundbplus})}{\leq}\lambda\int_{B(r_{0})\backslash
B(h)}\mathbb{P}\left[\exp(-\lambda
ce^{-d(d(z^{\prime},z)-C_{\text{dis}})/2})\right]~{}dz^{\prime}$
$\displaystyle\overset{(\ref{E:dvolpolar})}{=}\lambda\mathcal{S}(d)\int_{h}^{r_{0}}\exp(-\lambda
ce^{-rd/2})\sinh(r)^{d}~{}dr,$

where $\mathcal{S}(d)$ denotes the surface area of the Euclidean unit ball
$\mathbb{S}^{d}$.

When $r\to\infty$,

$\exp(-\lambda ce^{-rd/2})\sinh(r)^{d}=o\left(e^{-rd/3}e^{e^{-rd/3}}\right)$

thus

$\lambda\mathcal{S}(d)\int_{h}^{\infty}\exp(-\lambda
ce^{-rd/2})\sinh(r)^{d}~{}dr=o\left(e^{-e^{-rd/3}}\right),$

thus, for $h$ large enough, and for any $r_{0}\geq 0$,

$\displaystyle\mathbb{P}\left[L_{>h}\neq\emptyset\right]\leq e^{-e^{-hd/3}}.$
(4.21)

Finally, combining (4.19), (4.20) and (4.21) with $h=-\ln(M/(2C\lambda))/d$
leads to Lemma 4.4. ∎

##  5 Proof of Propositions 2.6 and 2.7

We first prove Proposition 2.6.

#### Step 1:

Let us fix $p>3d/2$. For any $r_{0}>0$, $n\in\mathbb{N}$, let us define

$\displaystyle S_{n}(A,r_{0}):=\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)^{2p}.$

The strategy of the proof is to construct a family of non-negative random
variables
$(Y_{n}^{M}(A,r_{0}))_{r_{0},A,M\geq 0,n\in\mathbb{N}}$ and
$(Y^{M}(A,r_{0}))_{r_{0},A,M\geq 0}$ such that

  1. (1)

almost surely, $Y_{n}^{M}(A,r_{0})\uparrow Y^{M}(A,r_{0})$ when $n\to\infty$
for any $M,A,r_{0}\geq 0$;

  2. (2)

$\sup_{A,r_{0}}\mathbb{P}\left[Y^{M}(A,r_{0})\geq
M\right]=O\left(M^{-2/3}\right)$ when $M\to\infty$;

  3. (3)

the following implication holds almost surely:

$\displaystyle S_{n}(A,r_{0})\leq(M\land
Y_{n}^{M}(A,r_{0}))A^{d}e^{-2r_{0}p}\implies S_{n+1}(A,r_{0})\leq
Y_{n+1}^{M}(A,r_{0})A^{d}e^{-2r_{0}p}$

.

Let us suppose for the moment that such random variables $Y_{n}^{M}(A,r_{0})$
and $Y^{M}(A,r_{0})$ exist. Let $A,r_{0}\geq 0$ and $M\geq 0$. On the event
$\\{Y^{M}(A,r_{0})\leq M\\}$, it can be shown by induction that
$S_{n}(A,r_{0})\leq MA^{d}e^{-2r_{0}p}$ for any $n\geq 0$. Indeed, $S_{0}=0$,
and if $S_{n}(A,r_{0})\leq Y_{n}^{M}(A,r_{0})A^{d}e^{-2r_{0}p}$, then, using
that

$\displaystyle M\wedge Y_{n}^{M}(A,r_{0})\overset{(1)}{\leq}M\wedge
Y^{M}(A,r_{0})\leq Y^{M}(A,r_{0})\text{ since we are on the event
}\\{Y^{M}(A,r_{0})\leq M\\}.$

So $S_{n+1}(A,r_{0})\leq Y_{n+1}^{M}(A,r_{0})A^{d}e^{-2r_{0}p}$ by (3), which
achieves the induction.

Thus, for any $A,r_{0},M\geq 0$,

$\displaystyle\mathbb{P}[S_{n}(A,r_{0})\geq
MA^{d}e^{-2r_{0}p}]\leq\mathbb{P}[Y^{M}(A,r_{0})\geq M]\leq CM^{-2/3}\text{ by
}(2).$ (5.1)

for $M$ large enough and some constant $C>0$ independent of $A,r_{0},M$. It
follows that

$\displaystyle C^{\prime}:$
$\displaystyle=\sup_{A,r_{0}}\mathbb{E}\left[S_{n}(A,r_{0})^{1/2}A^{-d/2}e^{r_{0}p}\right]=\sup_{A,r_{0}}\int_{0}^{\infty}\mathbb{P}[S_{n}(A,r_{0})^{1/2}A^{-d/2}e^{r_{0}p}\geq
M]~{}dM$ (5.2)
$\displaystyle=\sup_{A,r_{0}}\int_{0}^{\infty}\mathbb{P}\left[S_{n}(A,r_{0})\geq
M^{2}A^{d}e^{-2r_{0}p}\right]\overset{(\ref{E:boundM})}{\leq}\int_{0}^{\infty}CM^{-4/3}~{}dM<\infty.$

Let $K:=\\#\left(\mathcal{N}\cap B_{S(r_{0})}(u,Ae^{-r_{0}})\right)$. Let us
apply Cauchy-Schwartz with the inner product defined by $\langle
X,Y\rangle=\mathbb{E}\left[\sum_{i}X_{i}Y_{i}\right]$,

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)\right)^{p}\right]$
$\displaystyle\leq\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)\right)^{2p}\right]^{1/2}\mathbb{E}\left[\\#B_{S(r_{0})}(u,Ae^{-r_{0}})\right]^{1/2}$
$\displaystyle=\mathbb{E}\left[S_{n}(A,r_{0})^{1/2}\right]\mathbb{E}\left[\\#B_{S(r_{0})}(u,Ae^{-r_{0}})\right]^{1/2}$
$\displaystyle\overset{(\ref{E:definecprime})}{\leq}C^{\prime}A^{d/2}e^{-r_{0}p}\mathbb{E}\left[\\#B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}\right]^{1/2}.$
(5.3)

Let us show that
$\mathbb{E}\left[\\#B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}\right]\leq
CA^{d}$ for some $C>0$ independent of $A,r_{0}$. We use the covering of
$S(r_{0})$ by balls of radius $e^{-r_{0}}$ introduced by Lemma 4.2 in Section
4. For any $1\leq i\leq N(r_{0})$, by Proposition 4.4 applied with $p=1$,
$\mathbb{E}[\\#\text{RST}\cap B_{S(r_{0}}(z_{i},1)]\leq C$ for $C$ independent
of $r_{0}$, $z_{i}$. By Lemma 4.2, the number of balls intersecting
$B_{S(r_{0}}(z_{i},1)$ is bound by $C_{\text{ball}}A^{d}$. It follows that
$\mathbb{E}\left[\\#B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}\right]\leq
CA^{d}$.

Thus, by (5),

$\displaystyle\mathbb{E}\left[\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)\right)^{p}\right]\leq
Ce^{-r_{0}p}.$ (5.4)

Since $r\mapsto\text{MBD}_{r_{0}}^{r}(z)$ is non-decreasing for any $z\in
S(r)$,

$\displaystyle\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{\infty}(z)\right)^{p}=\lim_{n\to\infty}\uparrow\sum_{z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)\right)^{p}.$
(5.5)

Proposition 2.6 follows by (5.4) and by monotone convergence theorem.

#### Step 2:

we build the random variables $Y_{n}^{M}(A,r_{0})$ and $Y^{M}(A,r_{0})$.

Let $A,r_{0}\leq 0$, let $n\in\mathbb{N}$. The strategy is to upper-bound
$S_{n+1}$ in function of $S_{n}$. Fix $z\in B_{S(r_{0})}(u,Ae^{-r_{0}})$. The
quantity $\text{MBD}_{r_{0}}^{r_{0}+n\delta}$ takes into account finite
backwards paths that stop before level $r_{0}+n\delta$ and those (potentially
infinite) that continue after level $r_{0}+n\delta$. Let us define the random
set $\text{Stop}(z)$ as the set of ending points (in the backward direction)
of finite paths from $z$ stopping before level $r_{0}+n\delta$:

$\displaystyle\text{Stop}(z):=\\{z^{\prime}=(r^{\prime};u^{\prime})\in\mathcal{N}\cap\mathcal{D}(z),~{}r_{0}\leq
r^{\prime}\leq
r_{0}+n\delta,~{}A^{-1}(z^{\prime})=\emptyset\\}\subset\mathcal{N}.$

By definition of $\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)$ (resp.
$\text{MBD}_{r_{0}}^{r_{0}+(n+1)\delta}(z)$) (Definition 2.4),

$\displaystyle\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)$ $\displaystyle=$
$\displaystyle\max_{z^{\prime}=(r^{\prime};u^{\prime})\in\text{Stop}(z)}\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})\vee\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\text{CFD}_{r_{0}}^{r_{0}+n\delta}(z^{\prime})$
(5.6)

and

$\displaystyle\text{MBD}_{r_{0}}^{r_{0}+(n+1)\delta}(z)$ (5.7)
$\displaystyle=\max_{z^{\prime}=(r^{\prime};u^{\prime})\in\text{Stop}(z)}\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})\vee\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left(\text{CFD}_{r_{0}}^{r_{0}+n\delta}(z^{\prime})+\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})\right).$

For any $p\geq 1$, $a,b\geq 0$ and $t\in[0,1]$, Jensen inequality gives,

$\displaystyle(a+b)^{p}=\left(t\frac{a}{t}+(1-t)\frac{b}{1-t}\right)^{p}\leq
t\left(\frac{a}{t}\right)^{p}+(1-t)\left(\frac{a}{1-t}\right)^{p}=t^{1-p}a^{p}+(1-t)^{1-p}b^{p}.$
(5.8)

Applying (5.8) with $t=1/n^{2}$ leads to:

$\displaystyle\text{MBD}_{r_{0}}^{r_{0}+(n+1)\delta}(z)^{2p}$
$\displaystyle\overset{(\ref{E:redefmbd2})}{=}\max_{z^{\prime}=(r^{\prime};u^{\prime})\in\text{Stop}(z)}\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})^{2p}\vee\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left(\text{CFD}_{r_{0}}^{r_{0}+n\delta}(z^{\prime})+\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})\right)^{2p}$
$\displaystyle\leq\max_{z^{\prime}=(r^{\prime};u^{\prime})\in\text{Stop}(z)}\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})^{2p}\vee$
$\displaystyle\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left[\left(1-\frac{1}{n^{2}}\right)^{1-{2p}}\left(\text{CFD}_{r_{0}}^{r_{0}+n\delta}(z^{\prime})^{2p}\right)+n^{4p-2}\left(\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}\right)\right]$
$\displaystyle\leq\left(1-\frac{1}{n^{2}}\right)^{1-{2p}}\left[\max_{z^{\prime}=(r^{\prime};u^{\prime})\in\text{Stop}(z)}\left(\text{CFD}_{r_{0}}^{r^{\prime}}(z^{\prime})^{2p}\right)\vee\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left(\text{CFD}_{r_{0}}^{r_{0}+n\delta}(z^{\prime})^{2p}\right)\right]$
$\displaystyle+n^{4p-2}\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left[\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}\right]$
$\displaystyle\overset{(\ref{E:redefmbd1})}{=}\left(1-\frac{1}{n^{2}}\right)^{1-{2p}}\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)^{2p}+n^{4p-2}\max_{z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)}\left[\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}\right].$
(5.9)

Summing (5) over all $z\in B_{S(r_{0})}(u,Ae^{-r_{0}})$ leads to:

$\displaystyle S_{n+1}$
$\displaystyle\leq\left(1-\frac{1}{n^{2}}\right)^{1-2p}S_{n}+n^{4p-2}\left(\sum_{\begin{subarray}{c}z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})\\\
z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)\end{subarray}}\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}\right).$
(5.10)

Let us place on the event $\\{S_{n}\leq MA^{d}e^{-2r_{0}p}\\}$. Then, for any
$z\in B_{S(r_{0})}(u,Ae^{-r_{0}})$,

$\displaystyle\text{MBD}_{r_{0}}^{r_{0}+n\delta}(z)^{2p}\leq S_{n}\leq
MA^{d}e^{-2r_{0}p},$

so, for any $z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)$,
$\widehat{z^{\prime}0z}\leq M^{1/{2p}}A^{d/(2p)}e^{-r_{0}}$, so

$\displaystyle\widehat{z^{\prime}0I_{\infty}}$
$\displaystyle\leq\widehat{z^{\prime}0z}+\widehat{z0I_{\infty}}$
$\displaystyle\leq
M^{\frac{d}{2p}}A^{\frac{d}{2p}}e^{-r_{0}}+Ae^{-r_{0}}\text{ since }z\in
B_{S(r_{0})}(u,Ae^{-r_{0}})$ $\displaystyle\leq
Ae^{-r_{0}}\left(M^{\frac{d}{2p}}+1\right)\text{ since }\frac{d}{2p}\leq 1.$

Therefore, for any $z\in B_{S(r_{0})}(u,Ae^{-r_{0}})$,

$\displaystyle\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)\subset
B_{S(r_{0}+n\delta)}\left(u,Ae^{-r_{0}}\left(M^{\frac{d}{2p}}+1\right)\right).$
(5.11)

Let us define

$\displaystyle Z_{n}(A,r_{0}):=\sum_{z^{\prime}\in
B_{S(r_{0}+(n+1)\delta)}(u,Ae^{-r_{0}}(M^{1/(2p)}+1))\cap\text{RST}}\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}.$
(5.12)

By (5.11),

$\displaystyle\sum_{\begin{subarray}{c}z\in B_{S(r_{0})}(u,Ae^{-r_{0}})\\\
z^{\prime}\in\mathcal{D}_{r_{0}}^{r_{0}+n\delta}(z)\end{subarray}}\text{MBD}_{r_{0}+n\delta}^{r_{0}+(n+1)\delta}(z^{\prime})^{2p}\leq
Z_{n}(A,r_{0}),$ (5.13)

thus, combining (5.10) and (5.13), on the event {$S_{n}\leq
MA^{d}e^{-2r_{0}p}$},

$\displaystyle
S_{n+1}\leq\left(1-\frac{1}{n^{2}}\right)^{1-2p}S_{n}+n^{4p-2}Z_{n}.$ (5.14)

This upper-bound of $S_{n+1}$ suggests the following definition of the random
variables $Y_{n}^{M}(A,r_{0})$. We set $Y_{0}^{M}(A,r_{0}):=0$, and for any
$n\geq 0$,

$\displaystyle
Y_{n+1}^{M}(A,r_{0}):=\left(1-\frac{1}{n^{2}}\right)^{1-2p}Y_{n}^{M}(A,r_{0})+n^{4p-2}A^{-d}e^{2r_{0}p}Z_{n}(A,r_{0}).$

Let us also define $Y^{M}(A,r_{0}):=\lim_{n\to\infty}\uparrow
Y_{n}^{m}(A,r_{0})$ (this is well-defined since $n\to Y_{n}^{m}(A,r_{0})$ is
non-decreasing). We first show that the random variables $Y_{n}^{M}(A,r_{0})$
verify (3): for $n\in\mathbb{N}$, on the event $\\{S_{n}\leq(M\land
Y_{n}^{M}(A,r_{0}))A^{d}e^{-2r_{0}p}\\}$, by (5.14),

$\displaystyle S_{n+1}$
$\displaystyle\leq\left(1-\frac{1}{n^{2}}\right)^{1-2p}S_{n}+n^{4p-2}Z_{n}(A,r_{0})$
$\displaystyle\leq\left(1-\frac{1}{n^{2}}\right)^{1-2p}Y_{n}^{M}(A,r_{0})A^{d}e^{-2r_{0}p}+n^{4p-2}Z_{n}(A,r_{0})$
$\displaystyle=A^{d}e^{-2r_{0}p}\left[\left(1-\frac{1}{n^{2}}\right)^{1-2p}Y_{n}^{M}(A,r_{0})+n^{4p-2}A^{-d}e^{2r_{0}p}Z_{n}(A,r_{0})\right]$
$\displaystyle\leq A^{d}e^{-2r_{0}p}Y_{n+1}^{M}(A,r_{0}).$

Thus the random variables $Y_{n}^{M}(A,r_{0})$ verify (3). We move on to show
that $(Y_{n}^{M}(A,r_{0}))_{n,M,A,r_{0}}$ and $(Y^{M}(A,r_{0}))_{M,A,r_{0}}$
also verify (2). To proceed, we upper-bound $\mathbb{E}[Y_{n}^{M}(A,r_{0})]$
by induction on $n$.

For any $M,A,r_{0},n$, Proposition 2.5 applied for
$\theta=Ae^{-r_{0}}\left(M^{1/(2p)}+1\right)$ gives,

$\displaystyle\mathbb{E}[Z_{n}(A,r_{0})]$ $\displaystyle\leq
C\left(Ae^{-r_{0}}\left(M^{\frac{1}{2p}}+1\right)\right)^{d}e^{(d-2p)(r_{0}+n\delta)}$
(5.15)
$\displaystyle=CA^{d}\left(M^{\frac{1}{2p}}+1\right)^{d}e^{-2pr_{0}+n(d-2p)\delta}.$

Let us define, for any $n\in\mathbb{N}$,

$\displaystyle p(n):=\left(1-\frac{1}{n^{2}}\right)^{1-2p},\quad
q(n):=n^{4p-2}e^{n(d-2p)\delta}$ (5.16)

and

$\displaystyle P(n):=\Pi_{k=0}^{n-1}p(k),\quad Q(n):=\sum_{k=0}^{n-1}q(k).$
(5.17)

with the convention $P(0)=1$ and $Q(0)=0$. It can be noticed that

$\displaystyle\lim_{n\to\infty}P(n)<\infty,\quad\lim_{n\to\infty}Q(n)<\infty\text{
since }d-2p<0.$ (5.18)

Let us show by induction on $n$ that $\mathbb{E}[Y_{n}^{M}(A,r_{0})]\leq
C(M^{1/(2p)}+1)^{d}P(n)Q(n)$ for any $n\in\mathbb{N}$. The assertion is clear
for $n=0$ and, for $n\geq 0$,

$\displaystyle\mathbb{E}[Y_{n+1}^{M}(A,r_{0})]$
$\displaystyle=\left(1-\frac{1}{n^{2}}\right)^{1-2p}\mathbb{E}\left[Y_{n}^{M}(A,r_{0})\right]+n^{4p-2}A^{-d}e^{2r_{0}p}\mathbb{E}\left[Z_{n}(A,r_{0})\right]$
(5.19)
$\displaystyle=p(n)\mathbb{E}[Y_{n}^{M}(A,r_{0})]+A^{-d}e^{2r_{0}p-n(d-2p)\delta}q(n)\mathbb{E}\left[Z_{n}(A,r_{0})\right]$
$\displaystyle\overset{(\ref{E:globalflucteqn6})}{\leq}p(n)\mathbb{E}[Y_{n}^{M}(A,r_{0})]+C\left(M^{\frac{1}{2p}}+1\right)^{d}q(n)$
$\displaystyle\leq
C\left(M^{\frac{1}{2p}}+1\right)^{d}[p(n)P(n)Q(n)+q(n)]\text{ by induction
hypothesis}$
$\displaystyle=C\left(M^{\frac{1}{2p}}+1\right)^{d}[P(n+1)Q(n)+q(n)]$
$\displaystyle\leq C\left(M^{\frac{1}{2p}}+1\right)^{d}P(n+1)Q(n+1)\text{
since }P(n+1)\leq 1,$

which achieves the induction. Thus, there exists some constant $C>0$ such
that, for any $M,A,r_{0}\geq 0$, for any $n\in\mathbb{N}$,
$\mathbb{E}\left[Y_{n}^{M}(A,r_{0})\right]\leq
C\left(M^{1/(2p)}+1\right)^{d}$. By monotone convergence,

$\displaystyle\mathbb{E}\left[Y^{M}(A,r_{0})\right]\leq
C\left(M^{1/(2p)}+1\right)^{d}.$

Thus,for any $M,A,r_{0}\geq 0$, Markov inequality gives,

$\displaystyle\mathbb{P}\left[Y^{M}(A,r_{0})\geq
M\right]\leq\frac{C\left(M^{1/(2p)}+1\right)^{d}}{M}=O(M^{-2/3})$

since $2p>3d$. Thus the family of random variables $Y^{M}(A,r_{0})$ verifies
(2). This achieves the proof.

#### Proof of Proposition 2.7.

This is a direct consequence of Proposition 2.6. Let $\varepsilon>0$ and let
us choose $p$ such that $d/p<\varepsilon$. Applying Proposition 2.6 with
$A=\pi e^{r_{0}}$ gives that, for any $r_{0}\geq 0$,

$\displaystyle\mathbb{E}\left[\sum_{z\in
S(r_{0})\cap\text{RST}}\left(\text{MBD}_{r_{0}}^{\infty}(z)\right)^{p}\right]\leq
Ce^{r_{0}(d-p)}.$ (5.20)

Thus:

$\displaystyle\mathbb{E}\left[\sum_{n\in\mathbb{N}}\left(\max_{z\in\mathcal{L}_{n}}\text{MBD}_{n}^{\infty}(z)e^{(1-\varepsilon)n}\right)^{p}\right]\leq\sum_{n\in\mathbb{N}}Ce^{n((d-p)+(1-\varepsilon)p)}<\infty$

since $\varepsilon<d/p$. Therefore, a.s.,

$\displaystyle\lim_{n\to\infty}e^{(1-\varepsilon)n}\max_{z\in\mathcal{L}_{n}}\text{MBD}_{n}^{\infty}(z)\to
0\text{ as }n\to\infty.$

Moreover,
$r_{0}\mapsto\max_{z\in\mathcal{L}_{r_{0}}}\text{MBD}_{r_{0}}^{\infty}$ is
non-increasing, so for any $n\leq r_{0}<n+1$,

$\displaystyle
e^{(1-\varepsilon)r_{0}}\max_{z\in\mathcal{L}_{r_{0}}}\text{MBD}_{r_{0}}^{\infty}(z)\leq\max_{z\in\mathcal{L}_{n}}\text{MBD}_{n}^{\infty}(z)e^{(1-\varepsilon)r_{0}}\leq
e^{1-\varepsilon}\max_{z\in\mathcal{L}_{n}}\text{MBD}_{n}^{\infty}(z)e^{(1-\varepsilon)n},$
(5.21)

thus

$\displaystyle\lim_{r_{0}\to\infty}\max_{z\in\mathcal{L}_{r_{0}}}\text{MBD}_{r_{0}}^{\infty}(z)e^{(1-\varepsilon)r_{0}}\to
0\text{ as }r_{0}\to\infty.$

Define $R_{0}$ such that, for any $r_{0}\geq R_{0}$,
$e^{(1-\varepsilon)r_{0}}\max_{z\in\mathcal{L}_{r_{0}}}\text{MBD}_{r_{0}}^{\infty}(z)\leq
1/2$. For any $r_{0}\geq R_{0}$, $z\in\mathcal{L}_{r_{0}}$,
$z_{1},z_{2}\in\mathcal{D}(z)$, defining $r_{1}:=d(0,z_{1})$ and
$r_{2}:=d(0,z_{2})$,

$\displaystyle\widehat{z_{1}0z_{2}}$
$\displaystyle\leq\widehat{z_{1}0z}+\widehat{z0z_{2}}\leq\text{CFD}_{r_{0}}^{r_{1}}(z_{1})+\text{CFD}_{r_{0}}^{r_{2}}(z_{2})\leq\text{MBD}_{r_{0}}^{r_{1}}(z)+\text{MBD}_{r_{0}}^{r_{2}}(z)$
(5.22) $\displaystyle\leq 2\text{MBD}_{r_{0}}^{\infty}(z)\leq
2\max_{z^{\prime}\in\mathcal{L}_{r_{0}}}\text{MBD}_{r_{0}}^{\infty}(z^{\prime})\leq
e^{-(1-\varepsilon)r_{0}}.$

This achieves the proof of Proposition 2.7.

##  6 Appendix A: proof of Proposition 2.2

We first show that the RST is a tree. If the RST contains some loop
$z_{0},\cdots,z_{n}$, then the furthest vertex to the origin in the loop, say
$z_{i}$, must have two parents, which contradicts the definition of the RST.
Moreover, for some given vertex $z\in\mathcal{N}$, the sequence
$\left(d\left(A^{(k)}(z),0\right)\right)_{k}$ is decreasing. In addition,
since $\mathcal{N}\cap B(r)$ is finite for any $r\geq 0$, there is no infinite
decreasing sequence $\left(d\left(A^{(k)}(z),0\right)\right)_{k}$. Thus
$A^{(k)}(z)=0$ for some finite $k\geq 0$. Therefore, the RST is a connected
graph, so it is a tree.

We move on to show that the RST is locally finite. Let
$z_{0}=(r_{0};u_{0})\in\mathcal{N}\cup\\{0\\}$. Let us assume for the moment
that, for any $z=(r;u)\in\mathbb{H}^{d+1}$ and $r^{\prime}>0$,

$\displaystyle\mbox{Vol}(B^{+}(z,r^{\prime}))\geq ce^{d(r^{\prime}\wedge
r)/2}.$ (6.1)

for some $c$ independent of $z,r$. For
$z_{0}=(r_{0};u_{0}),z=(r;u)\in\mathbb{H}^{d}$, let us define

$\displaystyle
a(z,z_{0})=\mathbf{1}_{r>r_{0}}\mathbf{1}_{B^{+}(z,d(z,z_{0}))\cap\mathcal{N}=\emptyset}.$

Thus, for any $z_{0}\in\mathcal{N}$, $z_{0}=A(z)$ if and only if
$a(z,z_{0})=1$. By Campbell formula [8],

$\displaystyle\mathbb{E}\left[\\#\\{z_{0}\in\mathcal{N},~{}\\#A^{(-1)}(z_{0})=\infty\\}\right]=\mathbb{E}\left[\sum_{z_{0}\in\mathcal{N}}\mathbf{1}_{\sum_{z\in\mathcal{N}}a(z,z_{0})=\infty}\right]+\mathbb{P}\left[\sum_{z\in\mathcal{N}}a(z,0)=\infty\right]$
$\displaystyle=\mathbb{P}\left[\sum_{z\in\mathcal{N}}a(z,0)=\infty\right]+\lambda\int_{\mathbb{H}^{d+1}}\mathbb{P}\left[\sum_{z\in\mathcal{N}}a(z,z_{0})=\infty\right]~{}dz_{0},$

thus it suffices to show that
$\mathbb{P}\left[\sum_{z\in\mathcal{N}}a(z,z_{0})=\infty\right]=0$ for any
$z_{0}\in\mathbb{H}^{d+1}$. Let $z_{0}=(r_{0};u_{0})\in\mathbb{H}^{d+1}$. Note
that, if $d(z,z_{0})\geq r_{0}$, then $0\in B^{+}(z,d(z,z_{0}))$ so
$a(z,z_{0})=0$. Thus, Campbell formula gives,

$\displaystyle\mathbb{E}\left[\sum_{z\in\mathcal{N}}a(z,z_{0})\right]$
$\displaystyle=\lambda\int_{\mathbb{H}^{d+1}}\mathbb{E}\left[a(z,z_{0})\right]~{}dz\leq\lambda\int_{\mathbb{H}^{d+1}}\mathbf{1}_{d(z,z_{0})<r_{0}}\mathbb{P}[B^{+}(z,d(z,z_{0}))\cap\mathcal{N}=\emptyset]~{}dz$
$\displaystyle=\lambda\int_{\mathbb{H}^{d+1}}\mathbf{1}_{d(z,z_{0})<r_{0}}\mathbb{P}\left[\exp(-\lambda\mbox{Vol}(B^{+}(z,d(z,z_{0})))\right]~{}dz$
$\displaystyle\overset{(\ref{E:boundbplus})}{\leq}\lambda\int_{\mathbb{H}^{d+1}}\mathbb{P}\left[\exp(-\lambda
ce^{-d/2~{}d(z,z_{0})})\right]~{}dz$
$\displaystyle\overset{(\ref{E:dvolpolar})}{=}\lambda\mathcal{S}(d)\int_{0}^{\infty}\exp(-\lambda
ce^{-rd/2})\sinh(r)^{d}~{}dr,$ $\displaystyle<\infty,$

where $\mathcal{S}(d)$ denotes the surface area of the Euclidean unit ball
$\mathbb{S}^{d}$. Thus
$\mathbb{P}\left[\sum_{z\in\mathcal{N}}a(z,z_{0})=\infty\right]=0$.

It remains to show (6.1). Let $\rho\geq 0$ and $0\leq r\leq r_{0}$. Let
$z^{\prime}\in S(r^{\prime})$. Using the hyperbolic law of cosines in the
triangle $0zz^{\prime}$ (2.7),

$\displaystyle z^{\prime}\in B^{+}(z,\rho)$ $\displaystyle\iff
d(z,z^{\prime})\leq\rho$
$\displaystyle\iff\cosh(\rho)\geq\cosh(r_{0})\cosh(r)-\cos(\widehat{z0z^{\prime}})\sinh(r_{0})\sinh(r)$
$\displaystyle\iff\cos(\widehat{z0z^{\prime}})\geq\frac{\cosh(r_{0})\cosh(r)-\cosh(\rho)}{\sinh(r_{0})\sinh(r)}.$

A study of the function
$r\mapsto\frac{\cosh(r_{0})\cosh(r)-\cosh(\rho)}{\sinh(r_{0})\sinh(r)}$ shows
that, if $\rho\geq 1$, then this quantity is non-decreasing when $r_{0}-1\leq
r\leq r_{0}$. Thus, for $\rho\geq 1$ and $1\leq r_{0}-1\leq r\leq r_{0}$,

$\displaystyle\frac{\cosh(r_{0})\cosh(r)-\cosh(\rho)}{\sinh(r_{0})\sinh(r)}\leq\frac{\cosh(r_{0})^{2}-\cosh(\rho)}{\sinh(r_{0})^{2}}=1-\frac{\cosh(\rho)-1}{\sinh(r_{0})^{2}}\leq
1-Ce^{\rho-2r_{0}}$

for some $C>0$ independent of $r_{0},r,\rho$. Thus, there exists $C>0$
independent of $r_{0},r,\rho$ such that, if $\rho\geq 1$ and $1\leq
r_{0}-1\leq r\leq r_{0}$, if $\widehat{z0z^{\prime}}\leq Ce^{\rho/2-r_{0}}$
then $z^{\prime}\in B^{+}(z,\rho)$. Therefore, by (2.5), if $\rho\geq 1$ then
(recall that $\nu$ is the $d$-dimensional volume measure on $\mathbb{S}^{d}$
and $u$ is the direction of $z$),

$\displaystyle\mbox{Vol}(B^{+}(z,\rho))$
$\displaystyle\geq\int_{r_{0}-1}^{r_{0}}\nu\\{u^{\prime},\langle
u^{\prime},u\rangle\leq ce^{\rho/2-r_{0}}\\}\sinh(r)^{d}~{}dr$
$\displaystyle\geq c\int_{r_{0}-1}^{r_{0}}\left[e^{d\rho/2-dr_{0}}\wedge
1\right]\sinh(r)^{d}~{}dr\geq ce^{d(\rho\wedge r_{0})/2}$

Therefore, the RST has finite degree a.s.

It remains to show that the geodesics $[z,A(z)]$ for $z\in\mathcal{N}$ do not
cross a.s. in the bi-dimensional case ($d=1$). Let us suppose that there are
no two points $z_{1}$, $z_{2}$ with $d(0,z_{1})=d(0,z_{2})$ (this happens with
probability $0$). Let $z_{1}=(r_{1};u_{1}),z_{2}=(r_{2};u_{2})\in\mathcal{N}$
and let us set $A(z_{1}):=(r^{\prime}_{1};u^{\prime}_{1})$,
$A(z_{2}):=(r^{\prime}_{2};u^{\prime}_{2})$. Suppose that $[z_{1},A(z_{1})]$
and $[z_{2},A(z_{2})]$ meet at some point $P_{hyp}:=(r_{hyp};u_{hyp})$. We
have $r^{\prime}_{1}<r_{hyp}<r_{2}$, thus by definition of the parent,
$d(z_{2},A(z_{2}))<d(z_{2},A(z_{1}))$. Then

$\displaystyle d(z_{2},P_{hyp})+d(P_{hyp},A(z_{2}))$
$\displaystyle=d(z_{2},A(z_{2}))<d(z_{2},A(z_{1}))$ (6.2) $\displaystyle\leq
d(z_{2},P_{hyp})+d(P_{hyp},A(z_{1})),$

so $d(P_{hyp},A(z_{2}))<d(P_{hyp},A(z_{1}))$. On the other hand, interchanging
$z_{1}$ and $z_{2}$ in the previous calculation leads to
$d(P_{hyp},A(z_{1}))<d(P_{hyp},A(z_{2}))$. This is a contradiction. Therefore
$[z_{1},A(z_{1})]\cap[z_{2},A(z_{2})]=\emptyset$. This achieves the proof of
Proposition 2.2.

##  7 Appendix B: computing distances and angles in the half-space model

######  Proposition 7.1 (Distance formula).

Let $z_{1}=(x_{1},y_{1})\in H$ and $z_{2}=(x_{2},y_{2})\in H$. Let
$\kappa=\|x_{1}-x_{2}\|/y_{1}$ and $v=y_{2}/y_{1}$. Then

$\displaystyle
d(z_{1},z_{2})=2\tanh^{-1}\left(\sqrt{\frac{\kappa^{2}+(v-1)^{2}}{\kappa^{2}+(v+1)^{2}}}\right)=2\tanh^{-1}\left(\sqrt{1-\frac{4v}{\kappa^{2}+(v+1)^{2}}}\right).$
(7.1)

We refer to [13, Proposition 5] for a proof.

######  Proposition 7.2 (Angle formula).

Let $z=(x,y)\in H$ and let $h\geq 0$. Recall that $O(h)=(0,e^{h})$ and
$I_{\infty}=(0,0)$. If $y<e^{h}$, then

$\displaystyle\widehat{I_{\infty}O(h)z}=\arctan\left|\frac{2xe^{h}}{e^{2h}-x^{2}-y^{2}}\right|$

###### Proof of Proposition 7.2.

The proof is done by considering the isometry that sends the half-space model
on the Poincaré disc model [7]. The Poincaré disc model is defined as:

$\displaystyle I=\\{(x,y)\in\mathbb{R}^{2},~{}x^{2}+y^{2}<1\\}$

endowed with the following metric:

$\displaystyle ds_{I}^{2}:=4\frac{dx^{2}+dy^{2}}{1-x^{2}+y^{2}}.$ (7.2)

The application $\phi:H\to I$ defined as:

$\displaystyle(x,y)\mapsto\frac{1}{x^{2}+(y+1)^{2}}\left(x^{2}+y^{2}-1,-2x\right)$

is an isometry sending $(0,1)$ on $(0,0)$ [7]. Let us compose it with the
dilation of factor $e^{-h}$ to build an isometry from $H$ to $I$ sending
$O(h)$ on $(0,0)$. We obtain the application $\phi^{\prime}$ defined as

$\displaystyle\phi^{\prime}(x,y)$
$\displaystyle=\frac{1}{(e^{-h}x)^{2}+(e^{-h}y+1)^{2}}\left((e^{-h}x)^{2}+(e^{-h}y)^{2}-1,-2e^{-h}x\right)$
$\displaystyle=\frac{1}{x^{2}+(y+e^{h})^{2}}\left(x^{2}+y^{2}-e^{2h},-2xe^{h}\right)$

The Poincare disc model is conform (i.e. the hyperbolic angles correspond to
angles in the disc model), and the geodesics containing the origin $(0,0)$ are
represented by straight lines, thus for any two points $z_{1},z_{2}\in I$, the
hyperbolic angle $\widehat{z_{1}0z_{2}}$ coincides with the Euclidean one. For
any $z=(x,y)\in H$,
$\widehat{zO(h)I_{\infty}}=\widehat{\phi^{\prime}(z)0\phi^{\prime}(I_{\infty}})$,
where the second angle is taken in the disc model. Since
$\phi^{\prime}(I_{\infty})=(-1,0)$, if $y<e^{h}$ then
$\widehat{z0(h)I_{\infty}}<\frac{\pi}{2}$, thus

$\displaystyle\widehat{zO(h)I_{\infty}}=\arctan\left|\frac{2xe^{h}}{x^{2}+y^{2}-e^{2h}}\right|.$

∎

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