Discrete One-Dimensional Coverage Process on a Renewal Process

Abstract

Consider the following coverage model on $\mathbb {N}$, for each site $i \in \mathbb {N}$ associate a pair $(\xi _i, R_i)$ where $(\xi _i)_{i \ge 0}$ is a 1-dimensional undelayed discrete renewal point process and $(R_i)_{i \ge 0}$ is an i.i.d. sequence of $\mathbb {N}$-valued random variables. At each site where $\xi _i=1$ start an interval of length $R_i$. Coverage occurs if every site of $\mathbb {N}$ is covered by some interval. We obtain sharp conditions for both, positive and null probability of coverage. As corollaries, we extend results of the literature of rumor processes and discrete one-dimensional Boolean percolation.

Appendices

A Appendix on renewal sequences

Here we list some basic and well-known facts concerning renewal sequences that we use in our proofs. Let ${\varvec{\eta }} = (\eta _i)_{i \in \mathbb {N}}$ be a renewal sequence with inter-arrival time T.

If it starts with $\eta _0=1$, it is called undelayed. It is well-known that this sequence is positive recurrent if, and only if, $\mathbb {E}T<\infty$, and, as far as T has a proper distribution, the Renewal Theorem states that $\lim _n \mathbb {P}(\eta _n=1)=1/\mathbb {E}T$ (use $1/\infty =0$).

We can also suppose that ${\varvec{\eta }}$ is delayed, that is, there is a $\mathbb {N}$-valued random variable $T_0$ which gives the time elapsed until the first occurrence of a 1. That is, for $i\ge 0$, $\eta _i=1$ if there exists $n\ge 1$ such that $\sum _{k=0}^nT_k=i$, and we recover an undelayed sequence by putting $T_0\equiv 0$. As long as $T_0$ is a proper random variable, the Renewal Theorem also holds, and the limit is the same, no matter the delay, that is $\lim _n \mathbb {P}(\eta _n=1)=1/\mathbb {E}T$.

There are two other important observations concerning renewal sequences, delayed or not, that we use in the present paper.

• (i) For $k\in \{1,\ldots ,n\}$, ${\varvec{\eta }}$ has conditional probabilities

  $$\begin{aligned} \mathbb {P}(\eta _n=1|\eta _0^{n-1}=a_0^{n-1})=\mathbb {P}(T= k|T\ge k) \end{aligned}$$

  (8)

  for any string $a_0^{n-1}$ of symbols of $\{0,1\}$ such that $a_{n-k}=1$ and $a_{n-k+i}=0$ for $i=1,\ldots ,k-1$. In other words, the conditional probability of time n given the whole past depends only on the distance since the last occurrence of a 1 in the realisation, that is,

  $$\begin{aligned} \mathbb {E}(\mathbf{1}\{\eta _n=1\}|\eta _0^{n-1})=\mathbb {E}(\mathbf{1}\{\eta _n=1\}|\ell _1(\eta _0^{n-1})) \end{aligned}$$

  where $\ell _1(\eta _0^{i}):=\inf \{j\ge 0: \,\eta _{i-j}=1\}$. This is a property that characterises 1 as a renewal event for the sequence. Naturally, the sequence ${\varvec{\eta }}$ can be constructed by applying recursively the conditional probabilities (8), beginning with $\eta _0=1$ if it is undelayed, and from $T_0$ if it is delayed. So, together with the distribution of $T_0$, these conditional probabilities define the process univocally.

• (ii) ${\varvec{\eta }}$ has the following reversibility property: for any $n\ge 1$, the law of the sequence $\zeta ^{(n)}_i:=\eta _{n-i},i=0,\ldots ,n$ is that of a renewal process with the same inter-arrival distribution. Moreover, if ${\varvec{\eta }}$ is undelayed, on the event that $\eta _n=1$, the sequence $\zeta ^{(n)}_i,i=0,\ldots ,n$ has the same distribution as $\eta _i,i=0,\ldots ,n$. These property are easy to obtain when we observe that these sequences are constructed via a concatenation of blocks of the form $0\ldots 01$, of independent size equally distributed with T.

B Proof of Lemma 1

The proof will be done in three parts. First we explain the martingale difference method used to obtain concentration inequalities. At some point, we will need to estimate the probability of discrepancy between coupled delayed renewal processes, and thus, the second part of the proof is the explicit construction of a coupling of these processes. The proof of the lemma is concluded in a third step.

Notation alert B.1

For notational simplicity, we will translate the indexes of one unit along the present section, to study $\mathbb {E}\prod _{i=1}^{k}\alpha _i^{\xi _i}$ instead of $\mathbb {E}\prod _{i=0}^{k-1}\alpha _i^{\xi _{i+1}}$.

The method of martingale difference. We refer to Sect. 4 of McDiarmid [10] for what we now present. Let $g(\xi _1,\ldots ,\xi _{k}):=\sum _{i=1}^{k} \xi _i\log \alpha _i$. We will upper-bound $\mathbb {E}(e^{g-\mathbb {E}g})$. Let, for $i=1,\ldots ,k$

$$\begin{aligned} \Delta _i:=\mathbb {E}(g|\mathcal {F}_{1}^{i})-\mathbb {E}(g|\mathcal {F}_{1}^{i-1}), \end{aligned}$$

where, for $i\ge 1$, $\mathcal {F}_1^i$ is the $\sigma$-algebra generated by $\xi _j,j=1,\ldots ,i$ and $\mathcal {F}_1^0$ is the trivial $\sigma$-algebra. These quantities sum telescopically in i to $\sum _{i=1}^{k}\Delta _i(\xi _{1}^{i})=g(\xi _{1}^{k})-\mathbb {E}g(\xi _1^k)$. Since $\mathcal {F}_1^{i-1}\subset \mathcal {F}_1^{i}$, we have $\mathbb {E}(\Delta _i|\mathcal {F}_{1}^{i-1})= 0$ which means that $\Delta _i$, $i\ge 1$ forms a martingale difference sequence. If there exists, for any $i\ge 1$, a finite real number $d_i$ such that $|\Delta _i|\le d_i$ a.s., we can use the Azuma-Hoeffding inequality (see the proof of Lemma 4.1 of McDiarmid [10]) which states

$$\begin{aligned} \mathbb {E}e^{g-\mathbb {E}g}\le e^{\frac{1}{8}\sum _{i=1}^kd_i^2}. \end{aligned}$$

(9)

To upper bound $\sum _id_i^2$, let us compute, for $i=1,\ldots ,k-1$

$$\begin{aligned} \Delta _i(\xi _{1}^{i})=&\sum _{u_{i+1}^{k}\in \{0,1\}^{k-i}}\mathbb {P}(u_{i+1}^{k}|\xi _{1}^{i})g(\xi _1^iu_{i+1}^{k})-\sum _{u_{i}^{k}\in \{0,1\}^{k-i+1}}\mathbb {P}(u_{i}^{k}|\xi _{1}^{i-1})g(\xi _1^{i-1}u_{i}^{k})\\ \le&|\sum _{u_{i+1}^{k}}\mathbb {P}(u_{i+1}^{k}|\xi _{1}^{i-1}1)g(\xi _1^{i-1}1u_{i+1}^{k})-\sum _{u_{i+1}^{k}}\mathbb {P}(u_{i+1}^{k}|\xi _{1}^{i-1}0)g(\xi _1^{i-1}0u_{i+1}^{k})| \end{aligned}$$

where we used the convention that $\xi _1^0=\emptyset$ and the notation of concatenation between strings $a_i^jb_k^l=(a_i,\ldots ,a_j,b_k,\ldots ,b_l)$. A similar computation yields $\Delta _k(\xi _{1}^{k})\le \left| \log \alpha _k\right|$.

Therefore,

$$\begin{aligned} \Delta _i(\xi _{1}^{i})\le \sum _{j=0}^{k-i}D_{i,i+j}(\xi _{1}^{i})|\log \alpha _{i+j}|=(D(\xi _1^k)L)_i\,,\,\,i=1,\ldots ,k \end{aligned}$$

where $D(\xi _{1}^{n})$ is the upper triangular $k\times k$ matrix defined by

$$\begin{aligned} D_{i,i+j}(\xi _{1}^{i})&:=\sum _{u_{i+1}^{k},v_{i+1}^{k}\in \{0,1\}^{k-i}}\mathcal {Q}(u_{i+1}^{k},v_{i+1}^{k}|1\xi _{1}^{i-1}1,1\xi _{1}^{i-1}0)\mathbf{1}\{u_{i+j}\ne v_{i+j}\}\\&=\mathcal {Q}(u_{i+j}\ne v_{i+j}|1\xi _{1}^{i-1}1,1\xi _{1}^{i-1}0), \quad \text{ for } j=0,\ldots ,k-i, \end{aligned}$$

L is the $k\times 1$ matrix (column vector) with entries $L_{i,1}=|\log \alpha _i|,\,i=1,\ldots ,k$ and finally $\mathcal {Q}((\cdot ,\cdot )|1\xi _{1}^{i-1}1,1\xi _{1}^{i-1}0)$ denotes the law of a coupling between two discrete renewal sequence having the same inter-arrival distribution and starting from the different configurations $1\xi _{1}^{i-1}1$ and $1\xi _{1}^{i-1}0$.

Coupling and conclusion of the proof of the Lemma. Let $\ell (1\xi _{1}^{i})$ denote the smaller integer k such that $\xi ^i_{i-k+1}=0^k$, where $0^k=(0,\ldots ,0)$ denotes the string of k consecutive 0’s. Observe that $\ell (1\xi _{1}^{i})\le \ell (10^i)=i$ (this is one of the main differences with the paper of Chazottes et al. [3]). Then, for $i\ge 1$, $D_{i,i+j}(\xi _{1}^{i})$ is equal to the probability that two coupled renewal processes, one undelayed (the one starting with $1\xi _{1}^{i-1}1$), and the other with delay $\ell (1\xi _{1}^{i-1}0)\ge 1$ (the one starting with $1\xi _{1}^{i-1}0$), disagree at time j. Recall the coupling that we define before the statement of the lemma. Using this coupling we have the upper bound,

$$\begin{aligned} D_{i,i+j}(\xi _{1}^{i})=\mathbb {P}_\mathbf{U}\left( {\tilde{\xi }}^{(0)}_j\ne {\tilde{\xi }}^{(\ell (\xi _{1}^{i}))}_j\right) \le \mathbb {P}_\mathbf{U}(\tau _{0,\ell (\xi _{1}^{i})}\ge j) \le \sup _{\ell =1,\ldots ,i}\mathbb {P}_\mathbf{U}(\tau _{0,\ell }\ge j) \end{aligned}$$

where the last inequality follows taking the supremum over all the possible values of $\ell (\xi _{1}^{i})$ for any $i=1,\ldots ,k$. In view of (9), we now take

$$\begin{aligned} d_i:=( DL)_i=\sum _{j=i}^k\sup _{\ell =1,\ldots ,i}\mathbb {P}_\mathbf{U}(\tau _{0,\ell }\ge j)|\log \alpha _j| \end{aligned}$$

and thus,

$$\begin{aligned} \sum _{i=1}^kd_i^2\le \sum _{i=1}^k\left( \sum _{j=i}^k\sup _{\ell =1,\ldots ,i}\mathbb {P}_\mathbf{U}(\tau _{0,\ell }\ge j)|\log \alpha _j|\right) ^2. \end{aligned}$$

Recalling that we translated the indexes in the beginning of the proof, what we proved, from (9), is that

$$\begin{aligned} \mathbb {E}\left( e^{\sum _{i=0}^{k-1} \xi _i\log \alpha _i-\mathbb {E}\sum _{i=0}^{k-1} \xi _i\log \alpha _i}\right) \le e^{\frac{1}{8}\sum _{i=1}^k\left( \sum _{j=i}^k\sup _{\ell =1,\ldots ,i}\mathbb {P}_\mathbf{U}(\tau _{0,\ell }\ge j)|\log \alpha _{j-1}|\right) ^2} \end{aligned}$$

and therefore

$$\begin{aligned} \mathbb {E}\prod _{i=0}^{k-1}\alpha _i^{\xi _i}\le \prod _{i=0}^{k-1}\alpha _i^{\mathbb {P}(\xi _i=1)}e^{\frac{1}{8}\sum _{i=1}^k\left( \sum _{j=i}^k\sup _{\ell =1,\ldots ,i}\mathbb {P}_\mathbf{U}(\tau _{0,\ell }\ge j)|\log \alpha _{j-1}|\right) ^2}. \end{aligned}$$

(10)

This concludes the proof of the lemma.