The game behind oriented percolation

The fact that $p_{c}\in(0,1)$ follows from Sections 3 and 6 of [Dur84]. The exponential decay when $p<p_{c}$ is from Section 7 of [Dur84]. The case when $p>p_{c}$ comes from the definition of $p_{c}$ in Section 3 of [Dur84]. For the existence of infinite paths, see Section 3 of [Dur84]. ∎

This is a direct consequence of Theorem 1.2 and 1.3 of [DCTT18], along with the comment just below Theorem 1.3 of [DCTT18]. ∎

Consider $\varepsilon>0$ as in (2.1). Take $1-\varepsilon<\eta<1$ and note that if there is no vertical crossing in $(-n^{\eta},n^{\eta}]\times(-n,n]$, there should be no vertical crossing in any box $(x,0)+(-n^{1-\varepsilon},n^{1-\varepsilon}]\times(-n,n]$ contained within $(-n^{\eta},n^{\eta}]\times(-n,n]$. In particular, denoting $E_{m,n}(z)$ the complement of the event ${\framebox{$|$}}_{m,n}(z)$, we have

We conclude by noting that the events $(E_{n^{1-\varepsilon},n}((3kn^{1-\varepsilon},0))_{k}$ are independent, since the corresponding boxes are disjoint. ∎

Fixing $\delta>0$, one can find $n_{0}$ such that for all $n\geq n_{0}$,

We conclude by coupling the measures $\mathbb{P}_{p}$ and $\mathbb{P}_{p_{c}}$ restricted to the rectangle $(-n^{\sigma},n^{\sigma}]\times(-n,n]$ in such a way that $\mathbb{P}(\omega^{p}\neq\omega^{p_{c}})\leq\delta/2$. ∎

Theorem (1.1) of [MS93] concerns the measurability of the value function in stochastic games with $\limsup$ cost, and our game can be embedded in such a class. Specifically, using the notation from [MS93], we define the following:

• •

  The state space is $X=[0,1]^{E}\times\mathbb{L}\times\mathbb{N}^{*}\times[0,1]$, the action set for Player 1 is $A=\{-1,1\}$, and the action set for Player 2 is $B=\{-1,1\}^{2}$. The functions $F$ and $G$ are constant and equal to $A$ and $B$, respectively.

• •

  The transition function $q:X\times A\times B\mapsto X$ is a (deterministic) function that acts as follows:

  $\displaystyle q((\omega,z,n,v),a,b)=\left(\omega,z+e,n+1,\frac{n}{n+1}v+\frac{1}{n+1}\omega_{z,z+e}\right),$

  where $e$ is $(-1,b_{1})$ or $(1,b_{2})$ if $a$ is $-1$ or $1$, respectively.

• •

  The cost function $u:X\mapsto[0,1]$ is the projection to the last coordinate.

All conditions stated in (1.2) of [MS93] are satisfied, and thus Theorem 1.1 of [MS93] implies that the value function is measurable. ∎

Let $(e_{m})_{m\geq 1}$ be the sequence of edges induced by $\sigma$ and $\tau$, starting from $z$, and $(e^{\prime}_{m})_{m\geq 1}$ be the sequence of edges induced by $\sigma[h]$ and $\tau[h]$, starting from $z^{\prime}$. By definition, we have $e^{\prime}_{m}=e_{m+M-1}$. Hence,

∎

We begin by showing that $v(z)$ does not depend on $z$. To do this, take $\varepsilon>0$ and $z,z^{\prime}\in\mathbb{L}$, without loss of generality, assume $z^{\prime}$ is to the right of $z$. Let $\widehat{T}$ be the strategy of Player 1 that always play $T$, irrespective of the history. Let $\tau$ be an $\varepsilon$-optimal strategy for Player 2 in the game starting from $z$. Let $P_{T}$ be the set of vertices that are visited by the token when $(\widehat{T},\tau)$ is played (see Figure 4).

We, first claim that for all $\hat{z}\in P_{T}$, $v(\hat{z})\geq v(z)-2\varepsilon$. Indeed, let $\hat{z}\in P_{T}$, $M$ be the first instant where $\hat{z}$ is reached under strategies $(\widehat{T},\tau)$, and $h$ be the history at stage $M$. Consider the following strategy $\sigma$ of Player 1: play $T$ until reaching $\hat{z}$, then play an $\varepsilon$-optimal strategy in the game $\Gamma(\hat{z})$⁵⁵5Recall that $\Gamma(\hat{z})$ is the game starting from $\hat{z}$. . By Lemma 4.2, we have $\gamma(z,\sigma,\tau)=\gamma(\hat{z},\sigma[h],\tau[h])$. Because $\sigma[h]$ is $\varepsilon$-optimal in $\Gamma(\hat{z})$, we have $\gamma(\hat{z},\sigma[h],\tau[h])\leq v(\hat{z})+\varepsilon$, hence $\gamma(z,\sigma,\tau)\leq v(\hat{z})+\varepsilon$. Since $\tau$ is $\varepsilon$-optimal in $\Gamma(z)$, we have $\gamma(z,\sigma,\tau)\geq v(z)-\varepsilon$, and we deduce that $v(\hat{z})\geq v(z)-2\varepsilon$ for all $\hat{z}\in P_{T}$.

Analogously, we now construct two other paths: $P_{B}$, the set of vertices visited by the token when the game starts in $z$, Player 1 always play $B$, and Player 2 plays some $\varepsilon$-optimal strategy; and $P^{\prime}$, the set of vertices visited by the token when the game starts in $z^{\prime}$ instead of $z$, Player 2 always plays $L$, and Player 1 plays an $\varepsilon$-optimal strategy (see again Figure 4). The same argument as above shows that for any $\hat{z}\in P_{B}$, $v(\hat{z})\geq v(z)-2\varepsilon$ and $v(\tilde{z})\leq v(z^{\prime})+2\varepsilon$ for all $\tilde{z}\in P^{\prime}$.

We now use the inequalities obtained to conclude. First, set $P=P_{B}\cup P_{T}$ and consider the case where $\emptyset\neq P\cap P^{\prime}\ni z_{0}$. Then, $v(z)-2\varepsilon\leq v(z_{0})\leq v(z^{\prime})+2\varepsilon$, which implies that $v(z)-4\varepsilon<v(z^{\prime})$.

We are left with the case where $P\cap P^{\prime}=\emptyset$. Then, $P^{\prime}$ contains either a finite number of vertices at which Player 1 plays $B$, or a finite number of vertices at which Player 1 plays $T$. Without loss of generality, assume that we are in the first case. Note that in this case, $P_{T}$ has to contain only a finite number of $R$. Hence, by the law of large numbers the total cost along $P_{T}$ and along $P^{\prime}$ is almost surely $p$. We deduce that $v(z)-\varepsilon\leq p$ and $v(z^{\prime})+\varepsilon\geq p$, hence $v(z)-2\varepsilon\leq v(z^{\prime})$.

As $\varepsilon>0$ is arbitrary, the above arguments show that if $z^{\prime}$ is on the right side of $z$, then $v(z)\leq v(z^{\prime})$. By symmetry of the game with respect to the vertical axis, we deduce that for all $z$ and $z^{\prime}$, $v(z)=v(z^{\prime})$.

Once we know that $v$ is independent of $z$, for any $c\in\mathbb{R}$, the event $\left\{v\leq c\right\}$ must have probability $0$ or $1$ as it is invariant by translation⁶⁶6This is a consequence of the ergodicity of translations of i.i.d environments, see for example page 38 of [KL16] or Lemma 2.8 of [DC18].. Thus, $v$ is almost surely deterministic. ∎

We will consider three cases. The first one is included in the second but is presented separately for clarity.

The proof relies crucially on Corollary 2.4. We begin by choosing $0<\delta<1/25$ and $\sigma<1$ such that for all $n\geq n_{0}(\delta)$, there exists $p_{0}<p_{c}$ such that for any $p>p_{0}$:

where $m=\lfloor n^{\sigma}\rfloor$. Throughout this proof, we work exclusively with boxes of the form $x+(-m,m]\times(-n,n]$ with $x\in(m\mathbb{Z})\times(n\mathbb{Z})$. We say a point is “good” if the event ${\framebox{$|$}}_{m,n}(x)$ occurs ⁷⁷7Note that the random variables $({\framebox{$|$}}_{m,n}(x))_{x\in\in(m\mathbb{Z})\times(n\mathbb{Z})}$ are not independent as each rectangle intersects two others, thereby sharing some of the same random variables. However, they are independent as long as they are not vertical neighbours, i.e., as long as they do not intersect. (see Figure 8).

We now establish a recursive strategy for Player 2 that guarantees a total cost close to 1. We start at $x=0$ and depending on whether $x$ is good or not, Player 2 will follow a different strategy (See Figure 9 for a possible result of this strategy):

• •

  If $x$ is good, Player 2 will choose a constant direction that moves the token closer to the nearest vertical crossing of $x+(-m,m]\times(-n,n]$. When the token reaches the crossing, Player 2 will take the necessary actions to remain on the crossing until the token exits the box, either through the top or the bottom (this can be achieved using the strategy explained in Figure 5). Note that in this case, the token spends at least $n/2-2n^{\sigma}$ steps on edges with a cost of $1$, and at most $2n^{\sigma}$ steps on edges with a cost of $0$.

• •

  If $x$ is bad, Player 2 will move the token in the right direction. In this case, the token will spend at most $2n^{\sigma}$ steps on edges with a cost of $0$.

After any of these steps, the token exits the box $x+(-m,m]\times(-n,n]$. This means the token will now be in two different boxes. At least one of these boxes will have a distance from the vertical limits greater than or equal to $n/2$, we call $x$ the center of this box and repeat the procedure described above.

In this strategy, the token never revisits bad boxes represented by the same $x\in(m\mathbb{Z})\times(n\mathbb{Z})$. Assume that it has passed through $N$ different boxes $(x_{i})_{i=1}^{N}$; the cost is then lower bounded by

where time($x_{N})$ denotes the number of turns spent between exiting $x_{N-1}$ and exiting $x_{N}$. We now define the random variable $\tilde{N}=\sum_{i=1}^{N}\mathbf{1}_{x_{i}\text{ is good}}$ and study

Here, an acceptable path is one in $(-m,m]\times[-n,n]$ where each step moves either up, down, right, up-right, or down-right. This path represents a possible path that the players took in the game using the strategy described for Player 2. Note that each time there is a right, up-right or down-right step, it must be from a bad box, and the token never returns to that box again.

Since $\delta<1/25$, and by using Borel-Cantelli, we see that eventually $\tilde{N}>N/2$, leading to the following bound:

We conclude by noting that (6.1) implies $\liminf_{p\nearrow p_{c}}v_{p}(0)\geq 1-n^{\sigma-1}$ for any $n\in\mathbb{N}$.

∎

Player 2 can reach the 0-path in a finite number of stages and then keep the token in it, thereby guaranteeing a total cost of 0 (see Figure 10).

∎

Clearly, $p_{0}<1$ because the value is monotone with respect to $p$, and thus $p_{0}<p_{c}<1$. To see that $p_{0}>0$, we need a Peierls-type argument. We identify squares with their centers ($(x,y)$ such that $x+y$ is odd), and note that the problem of the existence of paths is now a problem of oriented site percolation with $2$-dependence (i.e., random variables that are at a distance greater than $2$ are independent).

If there is no $0$-path starting from $(0,1)$, then there must be some $n\in\mathbb{N}^{*}$ such that there is a dual top-to-bottom crossing of the square $(0,1)+[-n,n]\times[n,n]$ by a non-self-intersecting path⁸⁸8In this context, a dual crossing is a path on faces where at each step, the intersection between the closures of two faces is not trivial, i.e., meaning this path may include “diagonal” edges.. Using the union bound, we see that this probability is upper bounded by

Here, the term $n^{2}5^{l}$ bounds the amount of non-self-intersecting paths of length $l$, $(1-(1-p)^{4})$ is the probability that a given box is not a $0$-square and that each self-avoiding path of length $l$ has at least $l/5$ boxes who do not share any edges. We conclude by taking $p>0$ small enough such that the above upper bound is smaller than $1/2$, in which case we have a positive probability of having a $0$-path. Since the event of having a $0$-path is translation-invariant, we conclude that the probability must be either $0$ or $1$. ∎