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 1dimensional 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 onedimensional Boolean percolation.
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Acknowledgements
Sandro Gallo thanks FAPESP 2015/090943 and research fellowships CNPq (312315/20155) and FAPESP (2017/070846). Nancy Garcia thanks FAPESP 2014/264190 and research fellowship CNPq 302598/20146. This article was produced as part of the activities of FAPESP Research, Innovation and Dissemination Center for Neuromathematics (Grant #2013/076990 , São Paulo Research Foundation) and Edital Universal CNPq (480108/20129 and 462064/20140).
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Appendices
A Appendix on renewal sequences
Here we list some basic and wellknown facts concerning renewal sequences that we use in our proofs. Let \({\varvec{\eta }} = (\eta _i)_{i \in \mathbb {N}}\) be a renewal sequence with interarrival time T.
If it starts with \(\eta _0=1\), it is called undelayed. It is wellknown 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
$$\begin{aligned} \mathbb {P}(\eta _n=1\eta _0^{n1}=a_0^{n1})=\mathbb {P}(T= kT\ge k) \end{aligned}$$(8)for any string \(a_0^{n1}\) of symbols of \(\{0,1\}\) such that \(a_{nk}=1\) and \(a_{nk+i}=0\) for \(i=1,\ldots ,k1\). 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,
$$\begin{aligned} \mathbb {E}(\mathbf{1}\{\eta _n=1\}\eta _0^{n1})=\mathbb {E}(\mathbf{1}\{\eta _n=1\}\ell _1(\eta _0^{n1})) \end{aligned}$$where \(\ell _1(\eta _0^{i}):=\inf \{j\ge 0: \,\eta _{ij}=1\}\). 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 \(\zeta ^{(n)}_i:=\eta _{ni},i=0,\ldots ,n\) is that of a renewal process with the same interarrival 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 \(\mathbb {E}\prod _{i=0}^{k1}\alpha _i^{\xi _{i+1}}\).
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 upperbound \(\mathbb {E}(e^{g\mathbb {E}g})\). Let, for \(i=1,\ldots ,k\)
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 \(\mathcal {F}_1^{i1}\subset \mathcal {F}_1^{i}\), we have \(\mathbb {E}(\Delta _i\mathcal {F}_{1}^{i1})= 0\) 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 AzumaHoeffding inequality (see the proof of Lemma 4.1 of McDiarmid [10]) which states
To upper bound \(\sum _id_i^2\), let us compute, for \(i=1,\ldots ,k1\)
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,
where \(D(\xi _{1}^{n})\) is the upper triangular \(k\times k\) matrix defined by
L is the \(k\times 1\) matrix (column vector) with entries \(L_{i,1}=\log \alpha _i,\,i=1,\ldots ,k\) and finally \(\mathcal {Q}((\cdot ,\cdot )1\xi _{1}^{i1}1,1\xi _{1}^{i1}0)\) denotes the law of a coupling between two discrete renewal sequence having the same interarrival distribution and starting from the different configurations \(1\xi _{1}^{i1}1\) and \(1\xi _{1}^{i1}0\).
Coupling and conclusion of the proof of the Lemma. Let \(\ell (1\xi _{1}^{i})\) denote the smaller integer k such that \(\xi ^i_{ik+1}=0^k\), 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 \(1\xi _{1}^{i1}1\)), and the other with delay \(\ell (1\xi _{1}^{i1}0)\ge 1\) (the one starting with \(1\xi _{1}^{i1}0\)), disagree at time j. Recall the coupling that we define before the statement of the lemma. Using this coupling we have the upper bound,
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
and thus,
Recalling that we translated the indexes in the beginning of the proof, what we proved, from (9), is that
and therefore
This concludes the proof of the lemma.
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Gallo, S., Garcia, N.L. Discrete OneDimensional Coverage Process on a Renewal Process. J Stat Phys 173, 381–397 (2018). https://doi.org/10.1007/s1095501821382
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DOI: https://doi.org/10.1007/s1095501821382