Non-triviality of the vacancy phase transition for the Boolean model

TL;DR

This paper proves that in a spherical Poisson Boolean model with finite $d$-th moment radius distribution, there exists a positive intensity at which the vacant region percolates, highlighting a non-trivial phase transition.

Contribution

It establishes the existence of a non-trivial vacancy phase transition in the Boolean model under finite $d$-th moment conditions.

Findings

Vacant region percolates at positive intensity

Phase transition is non-trivial under finite $d$-th moment

Results apply for Euclidean $d$-space with $d \\geq 2"

Abstract

In the spherical Poisson Boolean model, one takes the union of random balls centred on the points of a Poisson process in Euclidean $d$-space with $d \geq 2$. We prove that whenever the radius distribution has a finite $d$-th moment, there exists a strictly positive value for the intensity such that the vacant region percolates.

Equations

$$Z_{\lambda }={\cup }_{y_k\in {\mathcal{P}}_{\lambda }}B(y_k,{\rho }_k);Z_{\lambda }^{\ast }={\mathbb{R}}^d\setminus Z_{\lambda }.$$

$${\lambda }_c:=\inf \{\lambda :\mathbb{P}[U_{\lambda }]=1\};{\lambda }_c^{\ast }:=\inf \{\lambda :\mathbb{P}[U_{\lambda }^{\ast }]=0\}.$$

$${\lambda }_D:=\inf \{\lambda :\mathbb{E}[D(W_{\lambda })]=\infty \}.$$

$${\lambda }_c^{\ast }\ge {\lambda }_D.$$

$$Z_{\lambda }^A:={\cup }_{\{k:y_k\in {\mathcal{P}}_{\lambda }\cap A\}}B(y_k,{\rho }_k).$$

$$G_{\lambda }(r)=\{Z_{\lambda }^{B(10^{6}r)\setminus S(r{)}_r}\cap S(r)\ne \varnothing \}.$$

$$\mathbb{P}[F_{\lambda }(10r)]\le C_1(\mathbb{P}[F_{\lambda }(r){]}^2+\mathbb{P}[G_{\lambda }(r)]).$$

$$F_{\lambda }(10r)$$

$$\sum _{n\ge 1}\underset{\lambda \in (0,{\lambda }_0]}{\sup }\mathbb{P}[G_{\lambda }(10^n)]<\infty .$$

$$\mathbb{P}[G_{\lambda }(r)]\le \lambda \pi (10^{6}r{)}^2\mathbb{P}[\rho >r]=10^{12}\lambda \pi r^2\mathbb{P}[{\rho }^2>r^2].$$

$$\sum _{n\ge 1}\underset{\lambda \in (0,{\lambda }_0]}{\sup }\mathbb{P}[G_{\lambda }(10^n)]\le 10^{12}{\lambda }_0\pi \sum _{n=1}^{\infty }100^n\mathbb{P}[{\rho }^2>100^n]$$

$$\sum _{n=1}^{\infty }(\mathbb{P}[F_{\lambda }(10^{n}b)]+\mathbb{P}[G_{\lambda }(10^{n}b)])\le 1/2.$$

$$f_{\lambda }(r):=C_1\mathbb{P}[F_{\lambda }(r)];g_{\lambda }(r):=C_1^2\mathbb{P}[G_{\lambda }(r/10)].$$

$$f_{\lambda }(r)\le C_1^2(\mathbb{P}[F_{\lambda }(r/10){]}^2+\mathbb{P}[G_{\lambda }(r/10)])$$

$$=f_{\lambda }(r/10{)}^2+g_{\lambda }(r).$$

$$\sum _{n=1}^{\infty }{g}_{\lambda }(10^{n}b)\le C_2^{-2}.$$

$$f_{\lambda }(10^{n}b)$$

$$\sum _{n\ge 1}{f}_{\lambda }(10^{n}b)\le (C_2^{-2}+g_{\lambda }(10b)+g_{\lambda }(100b)+g_{\lambda }(1000b)+\cdots )$$

$$\times (1+C_2^{-1}+C_2^{-2}+\cdots )$$

$$\sum _{n\ge 1}(\mathbb{P}[F_{\lambda }(10^{n}b)]+\mathbb{P}[G_{\lambda }(10^{n}b)])\le \sum _{n\ge 1}(f_{\lambda }(10^{n}b)+g_{\lambda }(10^{n}b))$$

$$\le 2C_2^{-2}+3\sum _{n\ge 1}{g}_{\lambda }(10^{n}b)\le 5C_2^{-2},$$

$$\mathbb{E}[{\rho }^d]<\infty .$$

$$H_n:=\{S_n\mathrm{is}\mathrm{crossed}\mathrm{by}{Z}_{\lambda }^{S_{n+2}}\mathrm{the}\mathrm{short}\mathrm{way}\};$$

$$J_n:=\{Z_{\lambda }^{S_{n+4}\setminus S_{n+2}}\cap S_n\ne \varnothing \}.$$

$$\mathbb{P}[H_n\cup J_n]\le \mathbb{P}[F_{\lambda }(10^{n}b)]+\mathbb{P}[G_{\lambda }(10^{n}b)].$$

$$1-\mathbb{P}[U_{\lambda }^{\ast }]$$

$$\lambda {\omega }_{d-2}(d-2){\int }_0^{\infty }\mathbb{P}[\rho \ge r]r^{d-3}dr=\lambda {\omega }_{d-2}\mathbb{E}[{\rho }^{d-2}]=:{\lambda }^{\prime },$$

$$\mathbb{E}[D(W_{0,\lambda })]\le \mathbb{E}[D(W_{0,\lambda })|F]=\mathbb{E}[D(W_{\lambda })|F]<\infty ,$$

$$E_{\lambda }:=\left({\cap }_{k=2}^{\infty }\{D(W_{k,\lambda })\le k/2\}\right)\cap \{Z_{\lambda }\cap (I_0\cup I_1)=\varnothing \}.$$

$$\sum _{k\ge 1}\mathbb{P}[D(W_{k,\lambda })>k/2]=\sum _{k\ge 1}\mathbb{P}[D(W_{0,\lambda })>k/2]\le \mathbb{E}[2D(W_{0,\lambda })]<\infty .$$

$$\mathbb{P}[E_{\lambda }]\ge \mathbb{P}\left[\left({\cap }_{k=k_0}^{\infty }\{D(W_{k,\lambda })\le k/2\}\right)\cap \left({\cap }_{k=0}^{k_0}\{Z_{\lambda }\cap I_k=\varnothing \}\right)\right]>0.$$

Full text

1 Introduction

The Boolean model [6, 8] is a classic model of continuum percolation [11, 3] and more general stochastic geometry [9, 4, 14, 10]. In the spherical version of this model, an occupied region in Euclidean $d$-space is defined as a union of balls (sometimes called grains) of fixed or random radius centred on the points of a Poisson process of intensity $\lambda$. One may define a critical value $\lambda_{c}$ of $\lambda$, depending on the radius distribution, above which the occupied region percolates, and a further critical value $\lambda_{c}^{*}$, below which the complementary vacant region percolates. It is a fundamental question whether these critical values are non-trivial, i.e. strictly positive and finite.

For fixed or bounded radii, the non-triviality of $\lambda_{c}$ and $\lambda_{c}^{*}$ for $d\geq 2$ is well known and may be proved using discretization and counting arguments from lattice percolation theory. For unbounded radii, it took some years to fully characterize those radius distributions for which $\lambda_{c}$ is non-trivial [8, 7]. In the present work we carry out a similar task for $\lambda_{c}^{*}$.

We now describe the model in more detail (for yet more details we refer the reader to [11] or [10]). Let $d\in\mathbb{N}$ with $d\geq 2$. Let $\mu$ be a probability measure on $[0,\infty)$ with $\mu(\{0\})<1$. Let $\lambda\in(0,\infty)$. On a suitable probability space $(\Omega,{\cal F},{\mathbb{P}})$ (with associated expectation operator $\mathbb{E}\,$), let ${\cal P}_{\lambda}=\{y_{k}:k\in\mathbb{N}\}$ be a homogeneous Poisson point process in $\mathbb{R}^{d}$ of intensity $\lambda$ (here viewed as a random subset of $\mathbb{R}^{d}$ enumerated in order of increasing distance from the origin), and let $\rho,\rho_{1},\rho_{2},\ldots$ be independent nonnegative random variables with common distribution $\mu$, independent of ${\cal P}_{\lambda}$. For $x\in\mathbb{R}^{d}$ and $r\geq 0$ we let $B(x,r):=\{y\in\mathbb{R}^{d}:\|y-x\|\leq r\}$, where $\|\cdot\|$ is the Euclidean norm. The occupied and vacant regions of the (Poisson, spherical) Boolean model are random sets $Z_{\lambda}\subset\mathbb{R}^{d}$ and $Z^{*}_{\lambda}\subset\mathbb{R}^{d}$, given respectively by

[TABLE]

Let $U_{\lambda}$ be the event that $Z_{\lambda}$ percolates, i.e. has an unbounded connected component, and let $U^{*}_{\lambda}$ be the event that $Z^{*}_{\lambda}$ percolates. By an ergodicity argument (see [11], or [10], Exercise 10.1), $\mathbb{P}[U_{\lambda}]\in\{0,1\}$ and $\mathbb{P}[U^{*}_{\lambda}]\in\{0,1\}$. Also $\mathbb{P}[U_{\lambda}]$ is increasing in $\lambda$, while $\mathbb{P}[U^{*}_{\lambda}]$ is decreasing in $\lambda$. Define the critical values

[TABLE]

It is well known that $\lambda_{c}$ and $\lambda_{c}^{*}$ are finite, and that if $\mathbb{E}\,[\rho^{d}]=\infty$ then $Z_{\lambda}=\mathbb{R}^{d}$ almost surely, for any $\lambda>0$ (see [8], [11] or [10]), so that $\lambda_{c}=\lambda_{c}^{*}=0$. Hence $\mathbb{E}\,[\rho^{d}]<\infty$ is a necessary condition for $\lambda_{c}$ or $\lambda_{c}^{*}$ to be strictly positive. In the case of $\lambda_{c}$, Gouéré [7] has shown that this condition is also sufficient:

Theorem 1**.**

[7]* If $\mathbb{E}\,[\rho^{d}]<\infty$ then $\lambda_{c}>0$.*

We here present a similar result for $\lambda_{c}^{*}$:

Theorem 2**.**

If $\mathbb{E}\,[\rho^{d}]<\infty$ then $\lambda^{*}_{c}>0$.

Theorem 2 says that for the spherical Poisson Boolean model with $\mathbb{E}\,[\rho^{d}]<\infty$, there exists a non-zero value of the intensity $\lambda$ for which the vacant region percolates. In fact we can say more:

Theorem 3**.**

For any $\mu$, if $d=2$ then $\lambda^{*}_{c}=\lambda_{c}$. If $d\geq 3$ then $\lambda_{c}^{*}\geq\lambda_{c}$.

Sarkar [13] has proved the strict inequality $\lambda^{*}_{c}>\lambda_{c}$ for $d\geq 3$ when $\rho$ is deterministic, i.e. when $\mu$ is a Dirac measure.

Theorem 2 could be seen as a trivial corollary of Theorems 1 and 3. However, we would like to prove Theorems 2 and 3 separately, to emphasise that our proof of Theorem 2 is self-contained (and quite short), whereas our proof of Theorem 3 is not, as we now discuss.

In parallel and independent work, Ahlberg, Tassion and Teixeira [2] prove a similar set of results to our Theorems 2 and 3; their proof seems to be completely different from ours. Earlier, in [1] they proved for $d=2$ that (among other things) $\lambda_{c}^{*}=\lambda_{c}$ whenever $\mathbb{E}\,[\rho^{2}\log\rho]<\infty$.

We prove Theorem 2 in the next two sections. The proof of Theorem 3 is given by adapting our proof of Theorem 2 using results in [1], and is therefore heavily reliant on [1]; we give this argument in Section 4.

Finally, we consider the relation between $\lambda_{c}^{*}$ and a different percolation threshold, defined in terms of expected diameter. For non-empty $B\subset\mathbb{R}^{d}$, let $D(B):=\sup_{x,y\in B}(\|x-y\|)$, the Euclidean diameter of $B$, and set $D(\emptyset)=0$. Let $W_{\lambda}$ be the connected component of $Z_{\lambda}$ containing the origin, and set

[TABLE]

It is easy to see that that $\lambda_{D}\leq\lambda_{c}$. Therefore by Theorem 3, for any $\mu$ we have

[TABLE]

In Section 5 we present an alternative, rather simple, direct proof of (1) (not reliant on any other results, either here or in [1]).

A further result in [7] says that $\lambda_{D}>0$, if and only if $\mathbb{E}\,[\rho^{d+1}]<\infty$. Therefore (1) provides an alternative proof that $\lambda_{c}^{*}>0$ under this stronger moment condition. Moreover, it is known in many cases that $\lambda_{D}={\lambda}_{c}$ (see e.g. [11, 15, 5]), and in all such cases our proof of (1) provides another way to show that $\lambda_{c}^{*}\geq\lambda_{c}$.

Our proof of Theorems 2 and 3 for $d=2$ uses a form of multiscale methodology, inspired by [7], which may be of use in other settings. We conclude this section with an outline of the method. At length-scale $r$, we define functions $f(r)$ and $g(r)$. Up to a constant multiple, $f(r)$ is the probability of a ‘local’ event (defined in terms of a box-crossing, using only grains centred near the box) while $g(r)$ is the probability of an ‘outside influence’ event that is still determined at length-scale $r$.

We show that $g(10^{n})$ is summable in $n$ (see Lemma 2 below), and also that $f(10^{n+1})\leq f(10^{n})^{2}+g(10^{n+1})$ (see (2) and (6) below). From this we can deduce that there exists $n_{0}$ such that $\sum_{n\geq n_{0}}(f(10^{n})+g(10^{n}))<1$, if only we can get started by showing $f(n_{0})$ is sufficiently small. This can be done either by taking $\lambda$ small (in the proof of Theorem 2) or for general $\lambda<\lambda_{c}$, by taking $n_{0}$ large and using a result from [1] (in the proof of Theorem 3). Finally, we can take a sequence of boxes of length $10^{n+n_{0}}$, such that if none of these is crossed then $Z_{\lambda}^{*}$ percolates.

We let $o$ denote the origin in $\mathbb{R}^{d}$, and for $r>0$ put $B(r):=B(o,r)$.

2 Preparation for the proof

Throughout this section we assume that $d=2$. We give some definitions and lemmas required for our proof of Theorem 2.

Given $\lambda>0$, for each Borel set $A\subset\mathbb{R}^{2}$ we define the random set

[TABLE]

Also, for $r>0$ set $A_{r}:=\cup_{x\in A}B(x,r),$ the (deterministic) $r$-neighbourhood of $A$.

Given $r>0$, let $S(r):=[-5r,5r]\times[-r/2,r/2]$, the closed $10r\times r$ horizontal rectangle (or ‘strip’) centred at $o$. Note that $S(r)_{r}$ is a $12r\times 3r$ rectangle with its corners smoothed (this smoothing is not important to us).

Let $F_{\lambda}(r)$ be the event that there is a short-way crossing of $S(r)$ by $Z_{\lambda}^{S(r)_{r}}$ (that is, by grains centred within the $r$-neighbourhood of $S(r)$). Also define the event

[TABLE]

Lemma 1**.**

There is a constant $C_{1}\geq 1$ such that for all $\lambda>0$ and $r>0$,

[TABLE]

Proof.

Fix $(\lambda,r)$. Set $S:=S(10r)=[-50r,50r]\times[-5r,5r]$. Let $T:=[-50r,50r]\times[-4.5r,-3.5r]$ and $\tilde{T}:=[-50r,50r]\times[3.5r,4.5r]$, so that $T$ and $\tilde{T}$ are horizontal $100r\times r$ thin strips along $S$ near the bottom and top of $S$, respectively.

We shall now define a collection $R_{1},\ldots,R_{37}$ of horizontal $10r\times r$ and vertical $r\times 10r$ rectangles that knit together in such a way that if there is a long-way vacant crossing of each of $R_{1},\ldots,R_{37}$ then there is a long-way vacant crossing of $T$ (this is a well known technique in these kinds of proof). We shall arrange that they are all contained within the band $\mathbb{R}\times[-12r,-2r]$ and their $r$-neighbourhoods $(R_{1})_{r},\ldots,(R_{37})_{r})$ all lie within the lower half of the region $S_{10r}:=(S(10r))_{10r}$.

Here are the details. Let $R_{1},R_{2},\ldots,R_{19}$ be horizontal $10r\times r$ rectangles centred on $(-45r,-4r),(-40r,-4r),\ldots,(45r,-4r)$ respectively. Let $R_{20},\ldots,R_{37}$ be vertical $r\times 10r$ rectangles centred at $(-42.5r,-7r),(-37.5r,-7r),\ldots,(42.5,-7r)$ respectively.

Similarly, we define a collection $R_{38},\ldots,R_{74}$ of $10r\times r$ and $r\times 10r$ rectangles, such that if each of these has a long-way vacant crossing then there is a long-way vacant crossing of $\tilde{T}$. Each rectangle $R_{37+i}$, $1\leq i\leq 37$, is defined simply as the reflection of $R_{i}$ in the $x$-axis.

For $1\leq i\leq 74$, let $D_{i}$ be the disk of radius $10^{6}r$ with the same centre as $R_{i}$. Let $A_{i}$ denote the event that there exists a grain of the Boolean model that intersects $R_{i}$ and has its centre in the region $D_{i}\setminus(R_{i})_{r}$. Let $B_{i}$ denote the event that the rectangle $R_{i}$ can be crossed the short way in the union of grains that are centred inside $(R_{i})_{r}$. If $R_{i}$ is crossed the short way in the union of grains centred in $D_{i}$, then $A_{i}\cup B_{i}$ must occur.

Suppose $F_{\lambda}(10r)$ occurs, i.e. there is a short-way occupied crossing of $S$, using grains centred in $S_{10r}$. Then there is no long-way vacant crossing of $S$, and hence no long-way vacant crossing either of $T$ or of $\tilde{T}$. Hence

[TABLE]

For $i,j\in\{1,\ldots,37\}$, since $(R_{i})_{r}\cap(R_{j})_{r}=\emptyset$ the events $B_{i}$ and $B_{37+j}$ are independent. Hence by (3) and the union bound we have (2), taking $C_{1}=37^{2}$. ∎

Lemma 2**.**

Suppose $\mathbb{E}\,[\rho^{2}]<\infty$. Let $\lambda_{0}\in(0,\infty)$. Then

[TABLE]

Proof.

Given $\lambda,r>0$, if $G_{\lambda}(r)$ occurs then there exists a point $y_{k}\in{\cal P}_{\lambda}\cap B(10^{6}r)\setminus S(r)_{r}$ with associated radius $\rho_{k}>r$. Therefore by Markov’s inequality $\mathbb{P}[G_{\lambda}(r)]$ is bounded above by the expected number of such points $y_{k}$. Therefore

[TABLE]

Hence,

[TABLE]

which is finite because we assume $\mathbb{E}\,[\rho^{2}]<\infty$. ∎

Lemma 3**.**

Suppose $\mathbb{E}\,[\rho^{2}]<\infty$. Then there exist $b>0$ and $\lambda>0$ such that

[TABLE]

Proof.

Let $C_{1}\geq 1$ be as in Lemma 1. Given $\lambda,r>0$ we define

[TABLE]

Then by (2) we have

[TABLE]

Let $C_{2}=9$. Using (4), we can choose $b$ to be a big enough power of 10 so that for all $\lambda\in(0,1]$, we have

[TABLE]

Now fix this $b$. Choose $\lambda\leq 1$ to be small enough so that $f_{\lambda}(b)\leq C_{2}^{-1}$. Using (6) repeatedly, we have $f_{\lambda}(10^{n}b)\leq C_{2}^{-1}$ for all $n$. Then using (6) repeatedly again, we have for $n\in\mathbb{N}$ that

[TABLE]

and therefore

[TABLE]

so by (7) and the fact that $\sum_{k=0}^{\infty}C_{2}^{-k}\leq 2$, we have

[TABLE]

and hence (5). ∎

3 Proof of Theorem 2

We can now complete the proof of Theorem 2. We assume from now on that

[TABLE]

Consider first the case with $d=2$. Let $b$ and $\lambda$ be as given in Lemma 3.

Let $S_{1},S_{2},S_{3},\ldots$ be a sequence of ‘strips’, i.e. closed rectangles of aspect ratio 10, with successive lengths (the short way) $10b,100b,1000b,\ldots$ alternating between horizontal and vertical strips with each strip $S_{n}$ centred at the origin. Then each strip $S_{n}$ crosses the next one $S_{n+1}$ the short way.

For each $n\in\mathbb{N}$, define the events

[TABLE]

[TABLE]

Lemma 4**.**

If none of the events $H_{1},J_{1},H_{2},J_{2},\ldots$ occurs then $Z^{*}_{\lambda}$ percolates.

Proof.

Suppose none of the events $H_{1},J_{1},H_{2},J_{2},\ldots$ occurs.

We claim for each $n\in\mathbb{N}$ that $Z^{\mathbb{R}^{2}\setminus S_{n+2}}_{\lambda}\cap S_{n}=\emptyset$. Indeed, if $Z^{\mathbb{R}^{2}\setminus S_{n+2}}_{\lambda}\cap S_{n}\neq\emptyset$ then for some integer $m\geq n+2$ with $m-n$ even we have $Z^{S_{m+2}\setminus S_{m}}_{\lambda}\cap S_{n}\neq\emptyset$, and then since $n\leq m$ we also have $S_{n}\subset S_{m}$ so that $Z^{S_{m+2}\setminus S_{m}}_{\lambda}\cap S_{m}\neq\emptyset$, contradicting the assumed non-occurrence of $J_{m-2}$.

For each $n$, by the assumed non-occurrence of $H_{n}$ along with the preceding claim there is no short-way crossing of $S_{n}$ by $Z_{\lambda}$ so there is a long-way crossing of $S_{n}$ by $Z_{\lambda}^{*}$, i.e. a path $\gamma_{n}\subset S_{n}\cap Z_{\lambda}^{*}$ that crosses $S_{n}$ the long way.

Then for each $n$ we have $\gamma_{n}\cap\gamma_{n+1}\neq\emptyset$, so $\cup_{n}\gamma_{n}$ is an unbounded connected set contained in $Z_{\lambda}^{*}$. Therefore $Z_{\lambda}^{*}$ percolates. ∎

Proof of Theorem 2.

Suppose $d=2$. Let $\lambda$ and $b$ be as given in Lemma 3. Recall the definition of events $F_{\lambda}(r)$ and $G_{\lambda}(r)$ at (1). We claim now for each $n$ that

[TABLE]

Indeed, suppose the parity of $n$ is such that $S_{n}$ is horizontal. Then, in terms of earlier notation, $S_{n}=S(10^{n}b)$. Since $S_{n+4}\subset B(10^{6+n}b)$ we have $J_{n}\subset G_{\lambda}(10^{n}b)$ and $H_{n}\subset F_{\lambda}(10^{n}b)\cup G_{\lambda}(10^{n}b)$. Then (9) follows from the union bound.

Using first Lemma 4, then (9), and finally (5), we have

[TABLE]

Therefore by ergodicity $\mathbb{P}[U^{*}_{\lambda}]=1$ so $\lambda\leq\lambda_{c}^{*}$. Hence we have $\lambda_{c}^{*}>0$ as required.

Now suppose $d\geq 3$. Let ${\tilde{Z}}_{\lambda}$, be the intersection of $Z_{\lambda}$ with the two-dimensional subspace $\mathbb{R}^{2}\times\{o^{\prime\prime}\}$ of $\mathbb{R}^{d}$, where $o^{\prime\prime}$ denotes the origin in $\mathbb{R}^{d-2}$.

Let $\omega_{d-2}$ denote the volume of the unit ball in $\mathbb{R}^{d-2}$. It can be seen that ${\tilde{Z}}_{\lambda}$ is a two-dimensional Boolean model with intensity

[TABLE]

which is finite by our assumption (8). Moreover if $\sigma$ denotes a random variable with the radius distribution in this planar Boolean model we claim that $\mathbb{E}\,[\sigma^{2}]<\infty$. This can be demonstrated by a computation, but it is more quickly seen using the fact that, since $\mathbb{P}[o\in Z_{\lambda}]<1$ for the original Boolean model by (8), also $\mathbb{P}[o\in{\tilde{Z}}_{\lambda}]<1$, which would not be the case if $\mathbb{E}\,[\sigma^{2}]$ were infinite.

Therefore by the two-dimensional case already considered, for small enough $\lambda>0$ we have $\lambda^{\prime}$ small enough so that the complement (in the space $\mathbb{R}^{2}\times\{o^{\prime\prime}\}$) of ${\tilde{Z}}_{\lambda}$ percolates. Hence $Z_{\lambda}^{*}$ percolates for small enough $\lambda>0$, so $\lambda^{*}_{c}>0$. ∎

4 Proof of Theorem 3

As mentioned in Section 1, if $\mathbb{E}\,[\rho^{d}]=\infty$ then $\lambda_{c}=\lambda_{c}^{*}=0$, so without loss of generality we assume (8).

First suppose $d=2$. We need to prove that $\lambda_{c}^{*}=\lambda_{c}$.

Suppose $\lambda>\lambda_{c}$. Let $V_{\lambda}^{*}$ be the event that there is an unbounded component of $Z_{\lambda}^{*}$ intersecting with $B(1)$. For $n\in\mathbb{N}$, set $Q(n):=[-n,n]^{2}$. Let $E(n)$ be the event that there exists a path in $Z_{\lambda}^{*}$ from $Q(n)$ to $\mathbb{R}^{2}\setminus Q(3n)$.

The annulus $Q(3n)\setminus Q(n)$ can be written as the union of two $3n\times n$ and two $n\times 3n$ rectangles, and if $Z_{\lambda}$ crosses each of these four rectangles the long way then $Q(n)$ is surrounded by an occupied circuit contained in $Q(3n)$ so $E(n)$ does not occur. Hence by Theorem 1.1 (i) of [1] and the union bound, $\mathbb{P}[E(n)]\to 0$ as $n\to\infty$. Since $V^{*}_{\lambda}\subset\cap_{n=1}^{\infty}E(n),$ we therefore have $\mathbb{P}[V^{*}_{\lambda}]=0$ and hence $\mathbb{P}[U_{\lambda}^{*}]=0$. Hence $\lambda\geq\lambda_{c}^{*}$ so $\lambda_{c}^{*}\leq\lambda_{c}$.

Now suppose $\lambda<\lambda_{c}$. Then by Theorem 1.1(iii) of [1], in the proof of our Lemma 3 we can choose $b$ large enough so that we have both (7), and the inequality $f_{\lambda}(b)<C_{2}^{-1}$. Then the rest of the proof of Lemma 3 carries through for this $(b,\lambda)$, so the conclusion of Lemma 3 holds for this $(b,\lambda)$. Then the proof (for $d=2$) in Section 3 works for this $(b,\lambda)$, showing that $\lambda\leq\lambda^{*}_{c}$ for any $\lambda<\lambda_{c}$ and hence that $\lambda_{c}^{*}\geq\lambda_{c}$. Thus $\lambda_{c}^{*}=\lambda_{c}$ for $d=2$.

Now suppose that $d\geq 3$ and $\lambda<\lambda_{c}$. Then as discussed in Section 3, $Z_{\lambda}\cap(\mathbb{R}^{d-2}\times\{o^{\prime\prime}\})$ is a two-dimensional Boolean model possessing no infinite component, so the radius distribution for this two-dimensional Boolean model has finite second moment and the intensity $\lambda^{\prime}$ of this two-dimensional Boolean model is subcritical (in fact, strictly subcritical since we can repeat the argument for any $\lambda_{1}\in(\lambda,\lambda_{c})$). Therefore by the argument just given for $d=2$, the complement (in $\mathbb{R}^{d-2}\times\{o^{\prime\prime}\}$) of this Boolean model percolates, and therefore the original $Z^{*}_{\lambda}$ also percolates so $\lambda\leq\lambda_{c}^{*}$. Hence $\lambda_{c}^{*}\geq\lambda_{c}$, and the proof is complete.

5 Alternative proof of (1)

We divide the nonnegative $x$-axis into unit intervals $I_{0},I_{1},I_{2},\ldots$ where $I_{k}=[k,k+1)\times\{o^{\prime}\}$ (here $o^{\prime}$ is the origin in $\mathbb{R}^{d-1}$). For each $k\in\mathbb{N}$ let $W_{k,\lambda}$ be the union of $I_{k}$ and all components of $Z_{\lambda}$ which intersect $I_{k}$.

Lemma 5**.**

If $0<\lambda<\lambda_{D}$, then $\mathbb{E}\,[D(W_{0,\lambda})]<\infty$.

Proof.

Fix $\lambda\in(0,\lambda_{D})$. Then $\mathbb{E}\,[D(W_{\lambda})]<\infty$. Let $F$ be the event that $I_{0}\subset Z_{\lambda}$, and set $F^{c}:=\Omega\setminus F$. Then $0<\mathbb{P}[F]<1$. If $F$ occurs then $W_{0,\lambda}=W_{\lambda}$. Hence by the Harris-FKG inequality (see [11] or [10]),

[TABLE]

as required. ∎

Given $\lambda>0$, define the event

[TABLE]

Lemma 6**.**

If $0<\lambda<\lambda_{D}$, then $\mathbb{P}[E_{\lambda}]>0$.

Proof.

Fix $\lambda\in(0,\lambda_{D})$. Then by Lemma 5.

[TABLE]

Choose $k_{0}\in\mathbb{N}$ with $k_{0}>2$, such that $\sum_{k\geq k_{0}}\mathbb{P}[D(W_{k,\lambda})>k/2]<1/2$. Then by the union bound and complementation, $\mathbb{P}[\cap_{k=k_{0}}^{\infty}\{D(W_{k,{\lambda}})\leq k/2\}]\geq 1/2$. Moreover $\mathbb{P}[\cap_{k=0}^{k_{0}}\{Z_{\lambda}\cap I_{k}=\emptyset\}]>0.$ Hence by the Harris-FKG inequality,

[TABLE]

∎

Lemma 7**.**

Suppose that $A\subset\mathbb{R}^{d}$ is closed, connected and unbounded, and that $\mathbb{R}^{d}\setminus A$ has an unbounded connected component. Then $\partial A$, the boundary of $A$, has an unbounded connected component.

Proof.

Let $B$ be an unbounded component of $\mathbb{R}^{d}\setminus A$. Denote the closure of $B$ by $\overline{B}$. Then both $\overline{B}$ and $\mathbb{R}^{d}\setminus B$ are closed and connected. By the unicoherence of $\mathbb{R}^{d}$ [12], the set $\overline{B}\cap A=\overline{B}\cap(\mathbb{R}^{d}\setminus B)$ is connected. Moreover it is unbounded, and contained in $\partial A$. ∎

Given $\varepsilon>0$, let ${\tilde{Z}}_{\lambda,\varepsilon}:=\cup_{k:\rho_{k}>0}B(y_{k},\varepsilon\rho_{k})$ and let ${\tilde{Z}}^{*}_{\lambda,\varepsilon}:=\mathbb{R}^{d}\setminus{\tilde{Z}}_{\lambda,\varepsilon}$. Let $Z_{\lambda}^{0}:=\cup_{\{k:r_{k}=0\}}\{y_{k}\}$, the union of balls of radius zero contributing to our Boolean model (${\tilde{Z}}_{\lambda,\varepsilon}$ is the union of all the other balls, scaled by $\varepsilon$). If $\mathbb{P}[\rho=0]>0$ then $Z_{\lambda}^{0}$ is (almost surely) non-empty but locally finite.

Lemma 8**.**

Let $\varepsilon\in(0,1)$. If $E_{\lambda}$ occurs then ${\tilde{Z}}_{\lambda,\varepsilon}^{*}\cup Z^{0}_{\lambda}$ percolates.

Proof.

Suppose $E_{\lambda}$ occurs. Let $A$ be the union of the half-line $[1,\infty)\times\{o^{\prime}\}$, with all components of $Z_{\lambda}$ intersecting this half-line. Then $A$ is connected, unbounded, and contained in the half-space $[1,\infty)\times\mathbb{R}^{d-1}$ so $o$ lies in an unbounded component of $\mathbb{R}^{d}\setminus A$. Therefore by Lemma 7, $\partial A$ has an unbounded connected component.

No point of $\partial A$ lies in the interior of any of the balls $B(y_{k},\rho_{k})$. Therefore $\partial A\subset{\tilde{Z}}^{*}_{\lambda,\varepsilon}\cup Z_{\lambda}^{0}$. Thus ${\tilde{Z}}^{*}_{\lambda,\varepsilon}\cup Z_{\lambda}^{0}$ has an unbounded connected subset. ∎

Proof of (1).

Assume $\lambda_{D}>0$ (else there is nothing to prove). Suppose $\lambda\in(0,\lambda_{D})$ and $\varepsilon\in(0,1)$. By the last two lemmas, with strictly positive probability the set ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}\cup Z_{\lambda}^{0}$ percolates. Almost surely, ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}$ is open, $Z_{\lambda}^{0}$ is locally finite and all points of $Z_{\lambda}^{0}$ lie either in ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}$ or in the interior of ${\tilde{Z}}_{\lambda,1-\varepsilon}$. Therefore if the set ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}\cup Z_{\lambda}^{0}$ percolates, so does ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}$, and so does ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}\setminus Z_{\lambda}^{0}$. Thus the set ${\tilde{Z}}_{\lambda,1-\varepsilon}^{*}\setminus Z_{\lambda}^{0},$ which is equal to $\mathbb{R}^{d}\setminus\cup_{k}B(y_{k},(1-\varepsilon)\rho_{k})$, percolates with strictly positive probability, and hence by ergodicity, with probability 1. Hence by scaling (see [11]) the set $Z_{(1-\varepsilon)^{d}\lambda}^{*}$ also percolates almost surely, so that $\lambda_{c}^{*}\geq(1-\varepsilon)^{d}\lambda$, and therefore $\lambda_{c}^{*}\geq\lambda_{D}$. ∎

Acknowledgement. I thank the referees for some helpful remarks.