The green process for the forest fire process

Consider the following modification of the forest fire process. Fix an $x\in\{1,2,\ldots\}$ . Assume that there are no fires at all, and wait for the first time when x is reachable from 0 (precisely defined immediately after (1)). This will provide a trivial lower bound for the minimum time necessary for the fire to reach this point.

More formally, let ${\mathcal{P}}_x$ , $x\in \mathbb{Z}_+$ , be a collection of independent Poisson processes, such that ${\mathcal{P}}_x$ has rate ${\lambda}_x$ . Define a $\{0,1\}^{\mathbb{Z}_+}$ -valued continuous-time stochastic process $S^{\color{ForestGreen}{\textbf{G}}}$ as follows. The value $S^{\color{ForestGreen}{\textbf{G}}}(x,t)\in\{0,1\}$ for $x\in \mathbb{Z}+$ , $t\ge 0$ , is the state of site x at time t. We say that x is occupied at time t (resp. vacant) whenever $S^{\color{ForestGreen}{\textbf{G}}}(x,t)=1$ (resp. $S^{\color{ForestGreen}{\textbf{G}}}(x,t)=0$ ). At time $t=0$ all the states are vacant, i.e. $S^{\color{ForestGreen}{\textbf{G}}}(x,0)=0$ for all $x\ge 0$ .

At each arrival time of the Poisson process ${\mathcal{P}}_x$ , the state x becomes 1, regardless of its previous value. Formally,

(1) \begin{align}S^{\color{ForestGreen}{\textbf{G}}}(x,t)=\begin{cases}0 &\text{if $ {\mathcal{P}}_x(t)=0$,}\\1 &\text{if $ {\mathcal{P}}_x(t)\ge 1$.}\end{cases}\end{align}

Let $r\ge 1$ (called the range of the process) be some positive integer; we say that $x\in\{1,2,\ldots\}$ is reachable from 0 at time t if there exists a positive integer n and a sequence of points in $\mathbb{Z}_+$ $0=x_0<x_1<x_2<\dots<x_n= x$ such that $x_i-x_{i-1}\le r$ for $i=1,2,\ldots,n$ and also $S^{\color{ForestGreen}{\textbf{G}}}(x_i,t)=1$ for $i=0,1,2,\ldots,n-1$ (note that we do not require that $S^{\color{ForestGreen}{\textbf{G}}}(x,t)=1$ ). This defines the green process. For the green process, we can define the quantity $N^{\color{ForestGreen}{\textbf{G}}}(t)$ , which is the rightmost reachable vertex at time t; in particular, if $N^{\color{ForestGreen}{\textbf{G}}}(t)=n$ then $S^{\color{ForestGreen}{\textbf{G}}}(n-r,t)=1$ and $S^{\color{ForestGreen}{\textbf{G}}}(n-i,t)=0$ for $i=0,1,\ldots,r-1$ . For definiteness, if $S^{\color{ForestGreen}{\textbf{G}}}(0,t)=0$ let $N^{\color{ForestGreen}{\textbf{G}}}(t)=0$ .

The time when point x becomes reachable for the first time is denoted by

\begin{equation*}\tau^{\color{ForestGreen}{\textbf{G}}}_x=\inf\bigl\{t>0\,:\, N^{\color{ForestGreen}{\textbf{G}}}(t)\ge x\bigr\}.\end{equation*}

It is easy to see that $N^{\color{ForestGreen}{\textbf{G}}}(t)<\infty$ a.s. and that $N^{\color{ForestGreen}{\textbf{G}}}(t)\uparrow\infty$ as $t\to\infty$ a.s.

The forest fire process, denoted by $S^{\color{red}{\textbf{F}}}(x,t)$ , is easily defined on the same probability space generated by the same processes ${\mathcal{P}}_x(t)$ , $x=0,1,2,\ldots,$ as the green process, with the additional rule saying that whenever site 0 becomes occupied, all the sites that are reachable from 0 become vacant. We say that all these sites are burnt at this time (regardless of whether or not they were previously occupied). Throughout the rest of the paper we let N(t) denote the rightmost point burnt by the fire by time t.

The previously studied case (e.g. in [Reference Volkov10]) when $r=1$ is referred to as the ‘short’ range, as opposed to the ‘long’ range when r can be greater than 1.

Remark 1. Suppose that ${\lambda}_x\equiv1$ and $r=1$ as in [Reference Volkov10]. The process $N^{\color{ForestGreen}{\textbf{G}}}$ is then a time-inhomogeneous pure jump Markov process on $\mathbb{Z}_+$ with unit jump rate and with a $\operatorname{Geom}\!({\mathrm{e}}^{-t})$ jump distribution, so $N^{\color{ForestGreen}{\textbf{G}}}(t)\,{\mathrm{e}}^{-t}$ converges in law to an exponential variable, and

\begin{equation*}\mathbb{P}\bigl( N^{\color{ForestGreen}{\textbf{G}}}(t+{\mathrm{d}} t) =N^{\color{ForestGreen}{\textbf{G}}}(t)+k \mid {\text{history up to } \textit{t}}\bigr) =\begin{cases}1-{\mathrm{d}} t+{\mathrm{o}}({\mathrm{d}} t), & k=0, \\ {\mathrm{e}}^{-t}(1-{\mathrm{e}}^{-t})^{k-1} \,{\mathrm{d}} t + {\mathrm{o}}({\mathrm{d}} t),& k \geq 1. \end{cases}\end{equation*}

Indeed, for an individual site the probability of being occupied in time t is $1-{\mathrm{e}}^{-t}$ , and thus the longest stretch of such occupied sites has a geometric distribution. Since at time t we have $N^{\color{ForestGreen}{\textbf{G}}}(t)=n$ , the site $n+1$ must be vacant at time t. If it becomes occupied during time ${\mathrm{d}} t$ , which occurs with probability ${\mathrm{d}} t$ , the process will immediately spread further to the right due to the already occupied sites at $n+1,n+2,\ldots.$

The usefulness of the green process will become clear from Proposition 1, introduced later. At this point we outline our construction. The green and fire processes are naturally coupled: they have the same ‘reach’ at a sequence of times tending to infinity. The green process is a trivial upper bound for the other one, while it can be analysed somewhat explicitly. On the other hand, the fire process has enough of a renewal structure to estimate how much slower than the green process it can be.

Now we rigorously introduce models (i) and (ii) studied in the current paper.

(i) Non-homogeneous, long-range process

Suppose that the rate at which site x becomes occupied depends on the site, and denote it by ${\lambda}_x$ . Throughout the paper we will assume uniform bounds on these rates:

\begin{equation*}c_1\le \lambda_x \le c_2,\quad x=0,1,2\dots\end{equation*}

for some fixed constants $c_2>c_1>0$ . We will assume also that the process is long-range, that is, when the fire hits point x it burns all vertices in $[x,x+1,\ldots,x+r]$ for some fixed positive integer r (‘range’).

In the original model studied in [Reference Volkov10], one has ${\lambda}_x= 1$ for all x, and $r=1$ . The model where all ${\lambda}_x$ are the same will be referred to as the homogeneous model; when $r=1$ we will say that it is a short-range model.

(ii) Continuous tree model on $\mathbb{R}_+$

Suppose that ‘trees’ (or rather their centres) arrive on $\mathbb{R}_+$ as a Poisson process of rate 1. Namely, the number of trees that appeared on [0, x], $x>0$ by time $t>0$ is given by a number of points of two-dimensional Poisson process in a rectangle $[0,x]\times [0,t]$ . Each tree is a closed interval (one-dimensional circle) of a fixed radius 1. See Figure 1. There is a constant source of fire attached to the origin, point 0. Whenever a tree covering the origin appears, it immediately burns down together with the whole connected component of trees (overlapping circles) containing point 0. For the continuous model, we can similarly define N(t) and $N^{\color{ForestGreen}{\textbf{G}}}(t)$ , as we did for the model on $\mathbb{Z}_+$ .

Figure 1. Continuous tree model on $\mathbb{R}_+$ .

Also, to avoid making the paper unnecessarily cumbersome, we restrict our attention to the homogeneous version of model (ii), even though one can think of a more general setup.

Definition 1. Let $\tau_x$ , $x\in \mathbb{R}_+$ , be the time when point x is burnt by fire for the first time in the forest fire model (either (i) or (ii)). Similarly, let $\tau_x^{\color{ForestGreen}{\textbf{G}}}$ be the first time when x is reachable by the green process. More formally,

\begin{align*}\tau_x&=\inf\{t\,:\, N(t)\ge x\},\\ \tau_x^{\color{ForestGreen}{\textbf{G}}}&=\inf\bigl\{t\,:\, N^{\color{ForestGreen}{\textbf{G}}}(t)\ge x\bigr\}.\end{align*}

The following statement applies to both models (i) and (ii).

Figure 2. $\tau_x$ for the forest fire and green processes.

3.1. Short range non-homogeneous process

This is the special case of the process when $r=1$ , which we can study without reference to the green process, using Lemma 3 in [Reference van den Berg and Tóth4] about the invariance of the distribution of $\tau_x$ with respect to permutation of the rates $\lambda_1,\ldots,\lambda_x$ .

It turns out that larger $\lambda$ correspond to smaller $\tau_x$ , hence $\tau_x$ is stochastically bounded below and above by $\tilde\tau_x/c_2$ and $\tilde\tau_x/c_1$ respectively, where $\tilde\tau_x$ is the distribution found in [Reference Volkov10]. In particular, $\mathbb{E} \tau_x={\mathrm{O}}(\!\log x)$ .

To see why this is true, note that by basic coupling, the fire process with rates $\lambda_1,\ldots, \lambda_{x-1}, c_2$ hits point x before $\tau_x$ . Now the permutation invariance result of [Reference van den Berg and Tóth4, Lemma 3] states that the former has the same law as the hitting time of x by the fire process with rates $c_2, \lambda_1,\ldots, \lambda_{x-1}$ . Iterating the argument, we prove our claim.

Remark 5. It is straightforward that for the green process in this case we have

\begin{equation*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t)\ge n\bigr)=\bigl(1-{\mathrm{e}}^{-{\lambda}_1 t}\bigr)\bigl(1-{\mathrm{e}}^{-{\lambda}_2 t}\bigr)\cdots\bigl(1-{\mathrm{e}}^{-{\lambda}_n t}\bigr).\end{equation*}

3.2. Long-range green process

Define $B_i=B_i(t)$ as the event that node i is vacant at time t. Then $B_i$ are independent and $\mathbb{P}(B_i)={\mathrm{e}}^{-{\lambda}_i t}$ . Moreover,

(2) \begin{align}\bigl\{N^{\color{ForestGreen}{\textbf{G}}}(t)<n\bigr\}=\bigcup_{i=0}^{n-r} (B_{i+1} \cap B_{i+2}\cap \dots \cap B_{i+r}),\end{align}

that is, there is a vacant interval of length r somewhere on [0, n]. From equation (2) we have

(3) \begin{array}{*{20}{c}}{\mathbb{P}({N^{\color{ForestGreen}{\mathbf{G}}}}(t) < n) \le \sum\limits_{i = 0}^{n - r} \mathbb{P} (\bigcap\limits_{m = 1}^r {{B_{i + m}}} ) = \sum\limits_{i = 0}^{n - r} {{{\text{e}}^{ - ({\lambda _{i + 1}} + \ldots + {\lambda _{i + r}})t}}} = :{f_n}(t).} \hfill \\ \end{array}

3.2.1. The homogeneous case

Without loss of generality assume that ${\lambda}_x={\lambda}=1$ . The next statement shows that the first time at which $N^{\color{ForestGreen}{\textbf{G}}}(t)\ge n$ is of order $T\,:\!=\, {\log n}/r$ , where r is the range of the process.

Proposition 2. For every $\varepsilon>0$ we have

\begin{align*}\mathbb{P}\bigl( N^{\color{ForestGreen}{\textbf{G}}}((1-\varepsilon)T)>n\bigr)&={\mathrm{o}}(1),\\ \mathbb{P}\bigl( N^{\color{ForestGreen}{\textbf{G}}}((1+\varepsilon)T)<n\bigr)&={\mathrm{o}}(1),\end{align*}

as $n\to\infty$ .

Let $p_n=p_n(t)\,:\!=\, \mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t)\ge n\bigr)$ , that is, the probability that on the set of vertices $\{1,2\ldots,n\}$ there are no ‘holes’ of length of at least r, and site 0 is occupied. Observe that $p_n$ satisfies the following recursion:

\begin{align*}p_n&=\sum_{k=1}^{r} (1-\alpha) \alpha^{k-1} p_{n-k}, \quad n> r,\\p_1&=p_2=\cdots=p_{r-1}=1-{\mathrm{e}}^{-t},\end{align*}

where $\alpha=\alpha_t\,:\!=\, {\mathrm{e}}^{-t}$ . Indeed, in order to reach $n>r$ , one must have an occupied vertex somewhere between 1 and r; the probability that the first such vertex is exactly at k is

\begin{equation*}\underbrace{{\mathrm{e}}^{-t}\cdot {\mathrm{e}}^{-t}\cdot \ldots \cdot {\mathrm{e}}^{-t}}_{\text{$k-1$ times}}\cdot \ (1-{\mathrm{e}}^{-t}),\end{equation*}

and then we need to reach n from k. Therefore the asymptotic behaviour of $p_n$ will depend on the largest solution of the characteristic equation

(4) \begin{align}\xi^r- \sum_{k=0}^{r-1} (1-\alpha)\, \alpha^{(r-1)-k} \, \xi^{k}=0.\end{align}

In particular, when $r=2$ we get $ \xi_{1,2}={(1-\alpha\pm\sqrt{D})}/{2} $ , where $D=1+2\alpha-3\alpha^2$ , resulting in

\begin{equation*}p_n=\dfrac{1}{2}\biggl[\biggl(1+\dfrac{1+\alpha}{\sqrt{D}}\biggr)\cdot\biggl(\dfrac{1-\alpha+\sqrt{D}}{2}\biggr)^n+\biggl(1-\dfrac{1+\alpha}{\sqrt{D}}\biggr)\cdot \biggl(\dfrac{1-\alpha-\sqrt{D}}{2}\biggr)^n\biggr],\end{equation*}

where $n=1,2,\ldots.$

While we cannot solve (4) explicitly except for $r=2$ and $r=3$ , we can still find the critical value of t for a given n, i.e. T. Indeed, from (3) we have

\begin{equation*}\mathbb{P}\bigl( N^{\color{ForestGreen}{\textbf{G}}}(T(1+\varepsilon))<n \bigr)\le n {\mathrm{e}}^{-rT(1+\varepsilon)}=n^{-\varepsilon}\to 0,\end{equation*}

and at the same time, since the unions of the events below are independent,

\begin{align*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t)>n\bigr)&\le \mathbb{P}( (B_1^c\cup \cdots \cup B_r^c) \cap(B_{r+1}^c \dots B_{r+r}^c)\cap \cdots\cap (B_{Mr+1}^c\dots B_{Mr+r}^c))\\&=\prod_{k=0}^{M} \mathbb{P}((B_{kr+1} B_{kr+2} \dots B_{kr+r})^c)\\&= (1-{\mathrm{e}}^{-rt})^{\lfloor n/r\rfloor }\\&\le (1-{\mathrm{e}}^{-rt})^{ n/r-1 } ,\end{align*}

where $M=\lfloor n/r \rfloor -1$ and $\lfloor x\rfloor$ denotes the integer part of $x\in\mathbb{R}$ . Consequently

\begin{equation*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(T(1-\varepsilon))>n\bigr)\le \bigl(1-{\mathrm{e}}^{-rT(1-\varepsilon)}\bigr)^{n/r-1 }= \biggl(1-\dfrac 1{n^{1-\varepsilon}}\biggr)^{n/r-1}\le 2\, {\mathrm{e}}^{-n^{\varepsilon}/r}\to 0.\end{equation*}

Thus Proposition 2 is proved.

3.2.2. The non-homogeneous case

Here we no longer assume that the ${\lambda}_x$ are the same for all x, as we did in Section 3.2.1.

The following statement is trivial and its proof is thus omitted.

Proposition 3. The function $f_n(t)$ defined in (3) satisfies the following properties.

• For a fixed n, $f_n(t)$ is monotonically decreasing with range from $n-r+1$ to $0$ as t goes from $0$ to $+\infty$ .

• For a fixed t, $f_n(t)$ is monotonically increasing in n and moreover $f_{n}(t)-f_{n-1}(t)\le {\mathrm{e}}^{-rc_1 t}$ .

Let us split the sum in $f_n(t)$ into r ‘almost’ equal parts according to the value of the remainder when i is divided by r. Namely, for $j=0,1,\ldots,r-1$ let

\begin{equation*}f_{n,j}(t)=\sum_{k=0}^{\lfloor {(n-j-r)}/{r}\rfloor}{\mathrm{e}}^{-({\lambda}_{rk+j+1}+{\lambda}_{rk+j+2}+\dots+ {\lambda}_{rk+j+r})t} ,\end{equation*}

and therefore

\begin{equation*}f_n(t)=f_{n,0}(t)+f_{n,1}(t)+\dots+f_{n,r-1}(t).\end{equation*}

First we show that

\begin{equation*}1-{\mathrm{e}}^{-{f_n(t)}/r}\le \mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t)<n\bigr)\le f_n(t).\end{equation*}

Indeed, the upper bound follows from (3). To show the lower bound, note that at least one of $f_{n,j}(t)$ , $j=0,1,\ldots,r-1$ , must be larger than $f_n(t)/r$ ; without loss of generality, assume $f_{n,0}(t)\ge f_n(t)/r$ . Then we have

\begin{align*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t) \ge n\bigr)&=\mathbb{P}\Biggl( \bigcap_{i=0}^{n-r} (B_{i+1}B_{i+2} \dots B_{i+r})^c \Biggr)\\&\le\mathbb{P}\Biggl( \bigcap_{k=0}^{\lfloor n/r \rfloor-1}(B_{rk+1} B_{rk+2}\dots B_{rk+r})^c\Biggr)\\&\stackrel{\text{indep.}}{\,=\,}\prod_{k=0}^{\lfloor n/r \rfloor-1}\mathbb{P} ( (B_{rk+1} B_{rk+2}\dots B_{rk+r})^c )\\&=\prod_{k=0}^{\lfloor n/r \rfloor-1}( 1-{\mathrm{e}}^{-({\lambda}_{rk+1}+\dots+{\lambda}_{rk+r})t} )\\&\le\exp\Biggl(-\sum_{k=0}^{\lfloor n/r \rfloor-1}{\mathrm{e}}^{-({\lambda}_{rk+1}+\dots+{\lambda}_{rk+r})t}\Biggr) \\&={\mathrm{e}}^{-f_{n,0}(t)}\\&\le {\mathrm{e}}^{-f_n(t)/r},\end{align*}

hence $P\bigl(N^{\color{ForestGreen}{\textbf{G}}}(t)<n\bigr)\ge 1-{\mathrm{e}}^{-f_n(t)/r}$ .

Let $t_*=t_*(n,\alpha)$ be such that $f_n(t_*)=\alpha\in(0,1)$ ; its existence and uniqueness follow from Proposition 3 and the fact that $f_n({\cdot})$ is strictly decreasing on $[0,\infty)$ and continuous. Now set $\alpha=1/2$ and define $\tilde T=t_*(n,1/2)$ ; later we will establish that $\tilde T={\mathrm{O}}(\!\log n)$ : see (5). Then

\begin{equation*}0<1-{\mathrm{e}}^{-{1}/{(2r)}}\le \mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(\tilde T)<n\bigr)\le \dfrac 12.\end{equation*}

Proposition 4. Fix a small $\varepsilon>0$ . Then, for any $c\in(0,c_1/c_2)$ and all large n,

\begin{align*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(\tilde T-\varepsilon \tilde T\bigr)\ge n)&\le {\mathrm{e}}^{-n^{c\varepsilon}}, \\\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(\tilde T+\varepsilon \tilde T\bigr)<n)&\le n^{-c\varepsilon}.\end{align*}

Proof. Indeed, since

\begin{align*}(n-r+1)\ {\mathrm{e}}^{-rc_2 \tilde T} =\sum_{i=0}^{n-r} {\mathrm{e}}^{-rc_2 \tilde T} \le\sum_{i=0}^{n-r} {\mathrm{e}}^{-({\lambda}_i+\dots+{\lambda}_{i+r})\tilde T} \equiv \dfrac 12\le \sum_{i=0}^{n-r} {\mathrm{e}}^{-rc_1 \tilde T}=(n-r+1)\ {\mathrm{e}}^{-r c_1\, \tilde T},\end{align*}

by summing up the left-hand side, the right-hand side, and $1/2$ of the above chain of inequalities, we immediately get

(5) \begin{align}(2n-2r+2)^{1/c_2}\le {\mathrm{e}}^{r\tilde T}\le (2n-2r+2)^{1/c_1}.\end{align}

Consequently

\begin{equation*}f_n(\tilde T+\varepsilon \tilde T)=\sum_{i=0}^{n-r}\dfrac{{\mathrm{e}}^{-({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\tilde T}}{{\mathrm{e}}^{({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\varepsilon \tilde T}}\le\sum_{i=0}^{n-r} \dfrac{{\mathrm{e}}^{-({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\tilde T}}{{\mathrm{e}}^{r\tilde T \cdot c_1 \varepsilon }}\le \dfrac{1}{2\, (2n-2r+2)^{ {c_1 \varepsilon }/{c_2} }}.\end{equation*}

By the same token,

\begin{align*}f_n(\tilde T-\varepsilon \tilde T)&=\sum_{i=0}^{n-r} {\mathrm{e}}^{-({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\tilde T}\times {\mathrm{e}}^{({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\varepsilon \tilde T}\\&\ge \sum_{i=0}^{n-r} {\mathrm{e}}^{-({\lambda}_{i+1}+\dots+{\lambda}_{i+r})\tilde T}\times {\mathrm{e}}^{r\tilde T \cdot c_1 \varepsilon } \\&\ge \dfrac{1}{2}\, (2n-2r+2)^{ {c_1 \varepsilon }/{c_2} }.\end{align*}

As a result,

\begin{equation*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(\tilde T-\varepsilon \tilde T)<n\bigr)\ge 1-{\mathrm{e}}^{-r^{-1}\,f_n(\tilde T-\varepsilon \tilde T)}\ge 1-\exp\biggl(-\dfrac{(2n-2r+2)^{ {c_1 \varepsilon }/{c_2}}}{4r}\biggr)\end{equation*}

and

\begin{equation*}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}(\tilde T+\varepsilon \tilde T)<n\bigr)\le f_n(\tilde T+\varepsilon \tilde T)\le \dfrac{1}{2\, (2n-2r+2)^{ {c_1 \varepsilon }/{c_2} }},\end{equation*}

which yields the statement of the theorem, since ${c_1\varepsilon}/{c_2}>c\varepsilon$ .

Corollary 1. There exists $C_1>0$ such that

(6) \begin{align}\liminf_{x\to\infty} \dfrac{\tau_x}{\log x}\le C_1\quad\textit{a.s.}\end{align}

Proof. From Proposition 4 we get that for some non-random $C_1>0$

\begin{equation*}\mathbb{P}\bigl(\tau_n^{\color{ForestGreen}{\textbf{G}}}\geq C_1\log n\bigr)\le n^{-c\varepsilon}\quad\text{for all large } {\textit{n}.}\end{equation*}

By choosing a sequence $n_j=j^A$ , $j=1,2,\ldots$ for some large $A>0$ such that $Ac\varepsilon>1$ , by the Borel–Cantelli lemma a.s. for all large enough j, we have

\begin{equation*}\tau_{n_j}^{\color{ForestGreen}{\textbf{G}}} \le C_1 \log\!(j^A).\end{equation*}

Using monotonicity of $\tau_x^{\color{ForestGreen}{\textbf{G}}}$ in x, and the fact that for each x there is a j such that

\begin{equation*}(1-{\mathrm{o}}(1))\, x\le j^A\le x <(j+1)^{A}\le (1+{\mathrm{o}}(1))\, x,\end{equation*}

we conclude that

\begin{equation*}\tau_{x}^{\color{ForestGreen}{\textbf{G}}} \le (C_1+{\mathrm{o}}(1)) \log x\end{equation*}

a.s. for all large x. As a result, by the second part of Proposition 1, which equivalently stated means that $\tau_x=\tau_x^{\color{ForestGreen}{\textbf{G}}}$ for infinitely many x, we obtain (6).

3.3. Coupling of the forest fire and green processes

Fix $\gamma\in (1,2)$ , and let us define the increasing sequence of times $\gamma^k$ , $k=1,2,\ldots.$ Let

\begin{align*}n_k&=\min\{n\in\mathbb{Z}_+\,:\, f_{n}(\gamma^k)\ge 1/2\},\\T_k&=t_*(n_k,1/2).\end{align*}

Proof. The first part of the statement follows immediately from the definition of f. To show the second part, note that from Proposition 3 we have

\begin{equation*}f_{n_k-1}(\gamma^k)<\dfrac 12=f_{n_k}(T_k)\le f_{n_k}(\gamma^k)\ \Longrightarrow \ \gamma^k\le T_k.\end{equation*}

On the other hand, for any small positive $\delta$ ,

\begin{equation*}f_{n_k}(T_k-\delta)\ge f_{n_k}(T_k) {\mathrm{e}}^{r c_1 \delta}=\dfrac{{\mathrm{e}}^{r c_1 \delta} }{2},\end{equation*}

so from Proposition 3 we have

\begin{align*}f_{n_k-1}(T_k-\delta)&=f_{n_k}(T_k-\delta)- [f_{n_k}(T_k-\delta)-f_{n_k-1}(T_k-\delta)]\\& \ge\dfrac{{\mathrm{e}}^{r c_1 \delta} }{2} - {\mathrm{e}}^{-r c_1 (T_k-\delta)}\\&\ge \dfrac 12 +\dfrac{r c_1 \delta}{2} - {\mathrm{e}}^{-r c_1 (T_k-\delta)}.\end{align*}

For k large enough, $\gamma^k$ is also large, and so is $T_k\ge \gamma^k$ , which yields that $ {\mathrm{e}}^{-r c_1 (T_k-\delta)}$ is very small, yielding that the right-hand side of the above expression is larger than $1/2$ , so

\begin{equation*}f_{n_k-1}(T_k-\delta) >\dfrac12> f_{n_k-1}(\gamma^k)\ \Longrightarrow\ T_k-\delta<\gamma^k.\end{equation*}

by monotonicity of $f_{n_k-1}({\cdot})$ . Consequently $\gamma^k\in[T_k-\delta, T_k]$ for all large k.

Corollary 2. For any $\varepsilon>0$ ,

(7) \begin{align}\begin{split}\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}( (1-\varepsilon)\, \gamma^k )\ge n_k\bigr)&\to 0,\\\mathbb{P}\bigl(N^{\color{ForestGreen}{\textbf{G}}}( (1+\varepsilon)\, \gamma^k )< n_k\bigr)&\to 0,\end{split}\end{align}

as $k\to\infty$ .

Now we are ready to present the proof of our main result.

Proof of Theorem Proof of Theorem 1.Fix $k\ge 1$ and define the new ${{\color{blue}{blue}}}$ process $S^{\color{blue}{\textbf{B}}}$ , which takes values in $\{0,1\}^{\mathbb{Z}_+}$ , as follows. Let $\tau_{n_k}^{(1)}$ be the first time when the fire process reaches $n_k$ . Up to this time $S^{\color{blue}{\textbf{B}}}$ completely coincides with the fire process $S^{\color{red}{\textbf{F}}}$ . However, at this hitting time, we set $S^{\color{blue}{\textbf{B}}}\bigl(x,\tau_{n_k}^{(1)}\bigr)=0$ for all x (not just those reachable from 0). After this, the blue process evolves again exactly like the fire process on the same probability space using the same collection of the Poisson process ${\mathcal{P}}_x$ (see the beginning of the paper), until the next time when $n_k$ is burnt, at which point the blue process restarts again with zeros everywhere. Thus $S^{\color{blue}{\textbf{B}}}(x,t)$ is a renewal process with the renewal times $\tau_{n_k}^{(i)}$ , $i=1,2,\ldots,$ where $\tau_{n_k}^{(i)}$ are the consecutive times when the fire process reaches $n_k$ . Moreover, the blue process and the fire process are trivially coupled such that

\begin{alignat*}{3}S^{\color{blue}{\textbf{B}}}(x,t)&=S^{\color{red}{\textbf{F}}}(x,t)\quad&&\text{for $t<\tau_{n_k}^{(1)}$ and all}\, {\textit{x},}\\S^{\color{blue}{\textbf{B}}}(x,t)&=S^{\color{red}{\textbf{F}}}(x,t)\quad&&\text{for all}\, {\textit{t}}\, {\text {and}\, x\le n_k,}\\S^{\color{blue}{\textbf{B}}}(x,t)&\le S^{\color{red}{\textbf{F}}}(x,t)\quad&&\text{for all } {\textit{t}\,} {\text{and} \,x> n_k.}\end{alignat*}

Let $\rho_i$ (resp. $\rho_i^{\color{red}{\textbf{F}}}$ ), $i=1,2,\ldots,$ be the rightmost point burnt by the blue process (resp. the fire process) at time $\tau_{n_k}^{(i)}$ . Define the inter-arrival times $\Delta_{n_k}^{(i)}=\tau_{n_k}^{(i)}-\tau_{n_k}^{(i-1)}$ , $i=2,3,\ldots,$ with the convention that $\Delta_{n_k}^{(1)}=\tau_{n_k}^{(1)}=\tau_{n_k}$ . From the definition of the blue process, it follows that the random variables $\xi_i=\bigl(\Delta_{n_k}^{(i)},\rho_i\bigr)$ , $i=2,3,\ldots$ are i.i.d.

By construction $\rho_1=\rho_1^{\color{red}{\textbf{F}}}$ , that is, the rightmost burnt point is the same for the fire and the blue processes when the fire process reaches $n_k$ for the first time; however, this does not necessarily hold for the times $\tau_{n_k}^{(i)}$ , $i\ge 2$ . At the same time we can claim the following.

Proof of lemma. The first part of the lemma trivially follows from the fact that $S^{\color{blue}{\textbf{B}}}(x,t)\le S^{\color{red}{\textbf{F}}}(x,t)$ for all x and t. To show the second statement, note that at time $s=\tau^{(i-1)}_{n_k}$ we have

\begin{equation*}S^{\color{red}{\textbf{F}}}\bigl(x,\tau_{n_k}^{(i-1)}\bigr)=S^{\color{blue}{\textbf{B}}}\bigl(x,\tau_{n_k}^{(i-1)}\bigr)=0\quad\text{for $x=0,1,\ldots,\rho_{i-1}$.}\end{equation*}

Until time $\tau_{n_k}^{(i)}$ , the fire and the blue processes coincide on $[0,\rho_{i-1}]$ . Since $\rho_{i}\le \rho_{i-1}$ ,

\begin{equation*}S^{\color{blue}{\textbf{B}}}\bigl(x,\tau_{n_k}^{(i)}-0\bigr)=0\quad \text{for $ x=\rho_i,\rho_i-1,\ldots,\rho_i-(r-1)$,}\end{equation*}

where $S^{\color{blue}{\textbf{B}}}(x,t-0)$ denotes the state of x in the infinitesimal moment just before time t. However, since the blue and the fire process coincide on $[0,\rho_{i-1}]$ , this also means that

\begin{equation*}S^{\color{red}{\textbf{F}}}\bigl(x,\tau_{n_k}^{(i)}-0\bigr)=0\quad\text{for $ x=\rho_i,\rho_i-1,\ldots,\rho_i-(r-1)$,}\end{equation*}

and thus the fire cannot reach any point beyond $\rho_i$ .

Now fix an $\varepsilon>0$ so small that

(8) \begin{align}\gamma (1+\varepsilon)<2\, (1-\varepsilon).\end{align}

For each $j=1,2,\ldots,$ let $S^{\color{ForestGreen}{\textbf{G}}}_j$ denote a version of the green process which is reset everywhere to zero at time $\tau_{n_k}^{(j-1)}$ , i.e. we have $S^{\color{ForestGreen}{\textbf{G}}}_j\bigl(x,\tau_{n_k}^{(j-1)}\bigr)=0$ for all $x\in \mathbb{Z}_+$ , with the convention $\tau_{n_k}^{(0)}\equiv 0$ ; thus $S^{\color{ForestGreen}{\textbf{G}}}_j\equiv S^{\color{ForestGreen}{\textbf{G}}}_1$ . Define

\begin{align*}E_{k,j}=\bigl\{&\text{for the green process $S^{\color{ForestGreen}{\textbf{G}}}_j$ there exists a subinterval of } (0, n_{k+1}]\\&\text{of length}\, {\textit{r}}\, \text {that has no Poisson arrivals during the time }\bigl[0,\Delta_{n_k}^{(j)}+\Delta_{n_k}^{(j+1)}\bigr]\bigr\}\end{align*}

as well as $\alpha_k= \mathbb{P}(E_{k,j})$ ; note that this probability does not depend on j since $\Delta_{n_k}^{(j)}$ are i.i.d.

Proof of lemma. Recall $\tau_x^{\color{ForestGreen}{\textbf{G}}}$ from Definition 1 and apply Corollary 2. We have shown that

(9) \begin{align}\mathbb{P}\bigl(|\tau_{n_k}^{\color{ForestGreen}{\textbf{G}}}-\gamma^k|>\varepsilon \gamma^k \bigr)\to 0\quad\text{as $k\to\infty$}\end{align}

and

(10) \begin{align}\mathbb{P}\bigl(\tau_{n_k}<\gamma^k(1-\varepsilon)\bigr)\le \mathbb{P}\bigl(\tau^{\color{ForestGreen}{\textbf{G}}}_{n_k}<\gamma^k(1-\varepsilon)\bigr)={\mathrm{o}}(1),\end{align}

where ${\mathrm{o}}(1)\to 0$ as $k\to\infty$ . From (8), (9), and (10) it follows that

\begin{align*}\mathbb{P}(E_{k,j})& \le \mathbb{P}\bigl(E_{k,j} \cap \bigl\{\Delta_{n_k}^{(j)}+\Delta_{n_k}^{(j+1)}>2(1-\varepsilon)\gamma^k\bigr\}\bigr)+2\,\mathbb{P}(\tau_{n_k}<(1-\varepsilon)\gamma^k)\\ &\le\mathbb{P}\bigl(\tau_{n_{k+1}}^{\color{ForestGreen}{\textbf{G}}}> 2 (1-\varepsilon)\gamma^k\bigr)+{\mathrm{o}}(1)\le\mathbb{P}\bigl(\tau_{n_{k+1}}^{\color{ForestGreen}{\textbf{G}}}> \gamma (1+\varepsilon) \gamma^k \bigr)+{\mathrm{o}}(1)\\&=\mathbb{P}\bigl(\tau_{n_{k+1}}^{\color{ForestGreen}{\textbf{G}}}> (1+\varepsilon) \gamma^{k+1} \bigr)+{\mathrm{o}}(1)\le {\mathrm{o}}(1)+{\mathrm{o}}(1), \nonumber\end{align*}

where ${\mathrm{o}}(1)\to 0$ as $k\to\infty$ .

Corollary 3. For $j=1,2,\ldots $ we have

\begin{equation*}\bigl\{\rho_{j+1}^{\color{red}{\textbf{F}}}\ge \rho_{j}^{\color{red}{\textbf{F}}} \ {and}\ \rho_{j+1}^{\color{red}{\textbf{F}}}<n_{k+1}\bigr\}\subseteq E_{k,j}.\end{equation*}

Proof. The result follows from the construction of $S^{\color{ForestGreen}{\textbf{G}}}_j$ and $S^{\color{red}{\textbf{F}}}$ which are both functionals of the collection of the Poisson processes ${\mathcal{P}}_x$ , $x\in \mathbb{Z}_+$ ; in particular,

\begin{equation*}S^{\color{red}{\textbf{F}}}(x,t)\ge S^{\color{ForestGreen}{\textbf{G}}}_j(x,t)\end{equation*}

for $t\in\bigl[\tau_{n_k}^{(j)},\tau_{n_k}^{(j+1)}\bigr]$ and $x\in \bigl[\rho_j^{\color{red}{\textbf{F}}},n_{k+1}\bigr]$ .

One of the fires that burns $n_k$ will eventually burn $n_{k+1}$ as well, so we can write

\begin{align*}\tau_{n_{k+1}}&=\Delta^{(1)}_{n_k}+\Delta^{(2)}_{n_k} 1_{\{\rho_1^{\color{red}{\textbf{F}}}<n_{k+1}\}}+ \Delta^{(3)}_{n_k} 1_{\{\rho_1^{\color{red}{\textbf{F}}}<n_{k+1},\rho_2^{\color{red}{\textbf{F}}}<n_{k+1}\}}+\dots=\Delta^{(1)}_{n_{k}}+\sum_{i=1}^\infty \Delta^{(i+1)}_{n_{k}} 1_{A_i},\end{align*}

where

\begin{equation*}A_i=A_i^{(k)}=\bigcap_{j=1}^{i}\bigl\{\rho_j^{\color{red}{\textbf{F}}}<n_{k+1}\bigr\}\end{equation*}

is a decreasing sequence of events. In other words, $A_i$ corresponds to the event that during the first i fires at point $n_{k}$ , point $n_{k+1}$ has not yet been burnt. Since $\Delta^{(i)}_{n_k}$ , $i=1,2,\ldots,$ are i.i.d., $\Delta^{(1)}_{n_k}\equiv\tau_{n_k}$ , and $\Delta^{(i+1)}_{n_k}$ is independent of $A_i$ for each i:

(11) \begin{align}\mathbb{E} (\tau_{n_{k+1}})=\mathbb{E} \tau_{n_k}\cdot\Biggl[1+\sum_{i=1}^\infty \mathbb{P}(A_i)\Biggr].\end{align}

Our task now is to estimate $\mathbb{P}(A_i)$ where $i\ge 1$ . Let $y=(y_0,y_1,\ldots,y_i)$ be a sequence of positive numbers of length $i+1$ with the convention that $y_0\equiv+\infty$ . We say that $y_j$ , $j=1,2,\ldots,i-1$ , is a weak local minimum if

\begin{equation*}H_j(y)=\{y_j\le \min\!(y_{j-1},y_{j+1})\}.\end{equation*}

For each such sequence y there exists a non-negative integer $\nu=\nu(y)\ge 0$ , which is a lower bound on the number of local minima, and the increasing sequence of indices $s(y)=(s_1(y),s_2(y),\ldots,s(y)_{\nu(y)})$ with the property

• $s_1(y)=\inf\{j\ge 1\,:\, H_j(y)\text{ occurs}\}$ is the index of the first weak local minimum in the sequence y;

• $s_{m+1}(y)$ is the index of the first local minimum in the sequence y with index at least $s_m(y)+3$ ;

• $s_{\nu}(y)+3>i-1$ , or there is no local minimum with index greater than or equal to $s_{\nu}(y)+3$ .

Set $\nu(y)=0$ if no weak local minima exist in y. (To illustrate this concept, consider the example where $ y=(\infty,3,2,4,1,3,2,5)$ , then $\nu=2$ and $(s_1,s_2)=(2,6)$ . Note that the number of local minima here is $3>\nu$ , since the middle local minimum is too close to the first one.)

Now let

\begin{equation*}y=({+}\infty,\rho_1,\ldots,\rho_i),\quad \sigma=s(y),\end{equation*}

where $\rho_1,\rho_2,\ldots$ are defined at the beginning of the proof of the theorem, and observe that the second condition in the definition of $s_m(y)$ ensures that $\sigma_{m+1}\ge \sigma_m+3$ , so the triples $(\sigma_m-1,\sigma_m,\sigma_m+1)$ are non-overlapping for distinct ms, implying, in turn, that the triples $(\rho_{\sigma_m-1}$ , $\rho_{\sigma_m}$ , $\rho_{\sigma_m+1})$ are independent for distinct ms. As a result,

\begin{equation*}\mathbb{P}(\sigma_{m+1}=\sigma_m+3 \,|\, \sigma_1,\ldots,\sigma_m<i-3)\ge \mathbb{P}(\rho_{\sigma_m+3}\le \min\!(\rho_{\sigma_m+2},\rho_{\sigma_m+4}))\ge \dfrac 13,\end{equation*}

due to the symmetry between $\rho_{\sigma_m+2},\rho_{\sigma_m+3},\rho_{\sigma_m+4}$ . Since the $\rho_i$ are i.i.d., we therefore conclude that $\nu(y)$ is stochastically larger than a $\operatorname{Binomial}(\lfloor (i+1)/3\rfloor,1/3)$ random variable.

By Lemma 1 $\rho_{j+1}^{\color{red}{\textbf{F}}}\ge \rho_{j+1}$ and moreover on the event $H_j(\rho)$ , we have $\rho_j^{\color{red}{\textbf{F}}}=\rho_j$ since $\rho_{j}\le \rho_{j-1}$ . Since also $\bigl(\rho_{j}^{\color{red}{\textbf{F}}}=\bigr)\rho_{j}\le \rho_{j+1} \bigl(\le \rho^{\color{red}{\textbf{F}}}_{j+1}\bigr)$ on $H_j(\rho)$ , by Lemma 1 and Corollary 3,

\begin{equation*}\mathbb{P}\bigl(\rho_{j+1}^{\color{red}{\textbf{F}}} < n_{k+1}, H_j\bigr)\le\mathbb{P}\bigl(\rho_j^{\color{red}{\textbf{F}}} \le \rho_{j+1}^{\color{red}{\textbf{F}}} < n_{k+1}, H_j\bigr)\le\mathbb{P}(E_{k,j}, H_j)\le\mathbb{P}(E_{k,j})\le \alpha_k\to 0.\end{equation*}

By the law of iterated expectations,

\begin{equation*}\mathbb{P}(A_i, \nu(y)=m)=\int_{y\,:\, \nu(y)=m}\mathbb{P}(A_i\,|\, (\rho)_i=y)\,{\mathrm{d}} \mathbb{P}(\rho=y)\le \sup_{y\,:\, \nu(y)=m} \mathbb{P}(A_i\,|\, (\rho)_i=y),\end{equation*}

where $(\rho)_i\,:\!=\, ({+}\infty,\rho_1,\rho_2,\ldots,\rho_i)$ , and the supremum is taken over all sequences y of length $i+1$ with $y_0=+\infty$ and all other elements being positive. However, when $\nu(y)=m$ , for any realization of $s(y)=(\sigma_1,\ldots,\sigma_m)$ we have

(12) \begin{align}\mathbb{P}(A_i \,|\, (\rho)_i=y)&=\mathbb{P}\Biggl( \bigcap_{j=1}^i \bigl\{\rho_{i}^{\color{red}{\textbf{F}}} < n_{k+1}\bigr\} \,|\, (\rho)_i=y \Biggr) \nonumber \\ & \le\mathbb{P}\Biggl( \bigcap_{l=1}^m \bigl[\bigl\{\rho_{\sigma_l+1}^{\color{red}{\textbf{F}}} < n_{k+1}\bigr\}\cap H_{\sigma_l}(\rho)\bigr] \,|\,(\rho)_i=y \Biggr)\le \alpha_k^m\end{align}

due to the independence of the triples $\{\rho_{l-1},\rho_l,\rho_{l+1}\}$ for different $l\in s(y)$ . Consequently

(13) \begin{align}\mathbb{P}(A_i)&=\mathbb{P}\Biggl(A_i \cap \Biggl(\bigcup_{m=0}^\infty\{\nu(y)=m\}\Biggr)\Biggr)\notag \\&=\sum_{m=0}^\infty \mathbb{P}(A_i,\nu(y) =m)\nonumber\\&\le \mathbb{P}(\nu(y)=0)+\sum_{m=1}^{\lfloor i/10\rfloor+1} \mathbb{P}(A_i,\nu(y)=m)+\sum_{m=\lfloor i/10\rfloor+2}^{\infty} \mathbb{P}(A_i,\nu(y)=m)\nonumber\\& \overset{ \text{by (12)} }{\le}\mathbb{P}(\nu(y)=0)+\alpha_k\,\mathbb{P}\biggl(1\le \nu(y)\le \dfrac{i}{10}+1\biggr) +\sum_{m=\lfloor i/10\rfloor+2}^{\infty} \alpha_k^m\nonumber\\&\le \mathbb{P}(\nu(y)=0)+ \alpha_k\,2 {\mathrm{e}}^{-c_4 i} +\dfrac{\alpha_k^{1+i/10}}{1-\alpha_k},\end{align}

since

\begin{equation*}\mathbb{P}\biggl( \nu(y)\le \dfrac{i}{10} +1\biggr)\le \mathbb{P}\biggl({\textrm{Binomial}}\biggl(\biggl\lfloor \dfrac{i+1}3\biggr\rfloor,\dfrac13\biggr) \le \dfrac{i}{10} +1\biggr)\le 2\,{\mathrm{e}}^{-c_4 i}\end{equation*}

for some $c_4>0$ , by the large deviation theory (see e.g. [Reference den Hollander7]). Finally, observe that on $i\ge 2$

\begin{equation*}\mathbb{P}(\nu(y)=0)=\mathbb{P}(\rho_1>\rho_2>\dots >\rho_i)\le \dfrac{1}{i!},\end{equation*}

due to the symmetry of all permutations of $[1,2,\ldots,i]$ .

Trivially $\mathbb{P}(A_1)\le 1$ , so by (13)

\begin{equation*}\sum_{i=1}^\infty \mathbb{P}(A_i)\le 1+\sum_{i=2}^\infty\biggl[\dfrac 1{i!}+\alpha_k\biggl(2 {\mathrm{e}}^{-c_4 i} +\dfrac{\alpha_k^{i/10}}{1-\alpha_k}\biggr) \biggr]=e-1+{\mathrm{o}}(1),\end{equation*}

where ${\mathrm{o}}(1)\to 0$ as $k\to\infty$ , since $\alpha_k\to 0$ . Therefore (11) gives

\begin{align*} \mathbb{E} (\tau_{n_{k+1}})&=\mathbb{E} \tau_{n_k}\cdot\Biggl[1+\sum_{i=1}^\infty \mathbb{P}(A_i)\Biggr]\le(e+{\mathrm{o}}(1))\cdot \mathbb{E} \tau_{n_k},\end{align*}

yielding that for any fixed $\delta>0$

\begin{equation*}\mathbb{E} (\tau_{n_k}) \le (e+\delta)^k\end{equation*}

for all sufficiently large k.

For each $x>0$ one can find a unique k such that $n_{k-1}< x\le n_{k}$ , and by Proposition 5,

\begin{equation*}k= \log_\gamma\biggl(\dfrac{\log x }{r\tilde c_k}\biggr)+{\mathrm{O}}(1)= \log_\gamma\!(\!\log x )+{\mathrm{O}}(1),\end{equation*}

whence

\begin{equation*}\mathbb{E} \tau_x \le \mathbb{E} \tau_{n_k}\le(\!\log x)^{\log_\gamma\! (e+2\delta)},\end{equation*}

where the power can be made arbitrarily close to $(\!\log 2)^{-1}=1.442695\dots $ by choosing $\gamma\uparrow 2$ and $\delta\downarrow 0$ . This proves Theorem 1.