Rigorous upper bound for the discrete Bak-Sneppen model

Fix some $p\in[0,1]$ and a positive integer $n$. The discrete Bak-Sneppen model is a Markov chain on the space of zero-one sequences of length $n$ with periodic boundary conditions. At each moment of time a minimum element (typically, zero) is chosen with equal probability, and is then replaced together with both its neighbours by independent Bernoulli(p) random variables. Let $\nu^{(n)}(p)$ be the probability that an element of this sequence equals one under the stationary distribution of this Markov chain. It was shown in [Barbay, Kenyon (2001)] that $\nu^{(n)}(p)\to 1$ as $n\to\infty$ when $p>0.54\dots$; the proof there is, alas, not rigorous. The complimentary fact that $\limsup \nu^{(n)}(p)<1$ for $p\in(0,p')$ for some $p'>0$ is much harder; this was eventually shown in [Meester, Znamenski (2002)]. The purpose of this note is to provide a rigorous proof of the result from Barbay et al, as well as to improve it, by showing that $\nu^{(n)}(p)\to 1$ when $p>0.45$. In fact, our method with some finer tuning allows to show this fact even for all $p>0.419533$.


Introduction
The classical Bak-Sneppen model [1,2] is defined as a collection of n individual species located equidistantly on a circumference, each possessing a fitness, which is a number in (0, 1). The process evolves in discrete time as follows. First, one finds the node(s) with the minimal fitness (if there are more than one, each of them is chosen with equal probability), and then this individual is replaced by a new one, with a fitness drawn from a uniform U(0, 1) distribution.
In such formulation, there are no interactions in the model, and it is easy to see that the second highest fitness is always non-decreasing. Consequently, with a little extra work one can show that all but one fitnesses converge to 1 a.s. To make the model interesting, it is also assumed that the "worst" species is replaced together with both its immediate neighbours on the circumference, and each of the three new fitnesses is drawn independently from the same uniform distribution; as a result, the model becomes highly non-trivial. In particular, simulations indicate that as time goes to infinity, for very large n the distribution of each fitness converges to a uniform distribution with parameters [f c , 1] where f c ≈ 0.66. To the best of our knowledge, this has not yet been shown rigorously.
The discrete version of the Bak-Sneppen model, proposed in [4], is defined as follows. Fix a positive integer n ≥ 3 and p ∈ [0, 1] and consider a Markov chain ξ(t), t = 0, 1, 2, . . . , on the state space {0, 1} n with the following transition probabilities. Let ξ(t) = (x 0 (t), x 1 (t), . . . , x n−1 (t)), x i (t) ∈ {0, 1}, and assume that x i (t) are the values assigned at time t to n vertices placed equidistantly on some circumference. Pick uniformly at random a vertex with index i = i t such that x i (t) = 0, and replace it 1 and both of its neighbours (i.e., each of ξ i−1 (t), ξ i (t), and ξ i+1 (t)) by an independent Bernoulli(p) random variable, keeping all the remaining ξ j intact. Throughout the paper we assume periodic boundary conditions, that is, n − 1 and 1 are the neighbours of 0, and n−2 and 0 are the neighbours of n−1, this is equivalent to addition/subtraction modulus n.
For t ≥ 1, let X t is the number of zeroes in ξ(t) and i t be the index of a randomly chosen zero 2 in the configuration ξ(t). The locations of zeros in ξ(t) are denoted by Then we set i t = i ⌊ζ(t+1)Xt⌋ if X t ≥ 1, and i t = i ⌊ζ(t+1)n⌋ otherwise. Now, We define the sigma-algebra F t = σ ζ (1),ζ(2), . . . ,ζ(t) , then ξ(t) is F t -measurable. 1 if all x i (t) = 1, then pick i uniformly amongst {0, 1, . . . , n − 1} 2 or a randomly chosen 1, in case ξ(t) does not contain any zeros Since the Markov chain ξ(t) is irreducible, aperiodic and on a finite state space, it converges to the unique stationary distribution π (n) . Let (c) There exists p c ∈ [0, 1] such that ν(p) = 1 for p > p c and ν(p) < 1 for p < p c .
(d) p c lies strictly between 0 and 1.
None of the parts of the above conjecture are rigorously shown. At the same time [4, Theorem 2.1] claims that lim n→∞ ν (n) (p) = 1 for p > p * = 0.54 . . . . Hence, if p c exists, then p c < p * . At the same time it appears much harder to show that p c > 0, or at least that lim sup n→∞ ν (n) (p) < 1 for small enough p. The latter statement was eventually proven in [8], using a very involved combinatorial avalanche method.
While the statement of [4, Theorem 2.1] is correct, its proof is, unfortunately, not rigorous.
The authors correctly compute the probabilities of bounds of possible changes in D t , which the size of the smallest contiguous area containing all zeros (precisely defined later in our article, see (2)), and show that its drift is negative whenever p > p * . Then they proceed to compare D t with Markov chain X t with some "holding" probabilities α i , and claim that lim t EX t ≥ lim t ED t as long as "alphas are defined to maximize EX t subject to some relevant constraints" (just above Lemma 2.1 in [4]). This cannot in general be true, as, for example, lim t→∞ ED t might not even exist. It is also not clear to us how exactly the authors derived Lemma 2.2 from [10].
The purpose of this short note is twofold. Firstly, we show how one can make the results of [4] rigorous. Secondly, we obtain a better bound on the critical probability by showing that ν(p) = 1 for all p > p ⋄ where p ⋄ = 0.45 · · · < p * . This is done in the next section. The final section contains the required statements for stochastic processes with drift, which are crucial in order to make the proof mathematically rigorous.

Main result
This is the main result of our paper.
Similarly to [4], we will define X t = |{i : ξ i (t) = 0}| as the total number of zeros at time t.
The proof of Theorem 1 will be based on finding some sort of a Lyapunov function M t , which is a function of ξ(t), and is thus F t -measurable, satisfying the following properties: • M t has uniformly bounded up-jumps; • there are some constants C > 0 and ε > 0 such that on the event Once such M t is found (see Lemma 3), Lemma 5 (which itself follows from Lemma 4) would imply that, loosely speaking, EM t remains "on average" bounded by a quantity independent of n, and hence the same holds for X t . Since EX t in fact converges as t → ∞, we will get the desired result. As in [4], let D t be the diameter of the smallest configuration containing all the zeros. Formally, for a given configuration ξ(t) = (x 0 , . . . , x n−1 ) let be a sequential subset of indices of (0, 1, 2, . . . , n−1) with periodic boundary conditions, satisfying the following two properties (a) ξ k (t) = 1 for all k ∈ Z t ; (b) Z t has the smallest number of elements amongst all such subsets.
If there is more than one such subset at time 0, choose any of them arbitrarily 3 . Now we can Also let (l t , r t ) be the pair with the first and the last index of Z t , and set l t = r t = 0 if X t = n or X t = 0, i.e. there are no ones or no zeros amongst ξ i (t). Note that if ξ t = (x 0 , . . . , x n−1 ) and 1 ≤ X t ≤ n − 1 then Suppose that D t ≥ 6, and for some positive constant β ∈ (0, 1/2) define a "corrected" diameter of the configuration that is, M t differs from D t by at most 1. For definiteness, let M t = 0 whenever D t < 6.
Note that if one of the "deeply" internal indices 4 of Z t is chosen, then x lt , x lt+1 , x rt−1 , x rt do not change and Z t+1 = Z t , e.g.
On the other hand, if i t ∈ {l t , l t + 1, r t , r t − 1} then Z t might change; it can increase by at most one point; if i t ∈ {l t + 2, r t − 2} then Z t does not change but M t still can.
The above analysis fails, however, if one can make card(Z t+1 ) even smaller by flipping indices, which can happen when one of the "deeply" internal zeros is chosen, e.g.
Note that if l and r do not flip, then according to the above arguments.
Lemma 1. Suppose that D t ≥ 6, and suppose that the indices l and r flip between times t and t + 1. Then Proof. The first statement follows from the fact that the flip occurs only if card(Z t ) decreases; the second from the fact that 2β < 1.
as the set indices of the three+three points at both ends of the zero area.
Recall that i t denotes the index of the zero, chosen to be replaced with both its neighbours, and note that 1 it∈E t is a function of ξ(t) andζ(t + 1).
Proof. On the event described in the statement, either D t+1 = D t (and there were no changes of the configuration near its endpoints, hence M t+1 = M t ), or the indices l and r flipped, and in this case the result follows from Lemma 1.
Lemma 3. Suppose that p > p ⋄ and β is given by (6). Then Moreover, there is an ε = ε(p) > 0, depending on p only, such that quantity D t . By slightly modifying D t with extra terms in (3), we will be able to obtain a better bound on p c .
Proof of Lemma 3. Throughout the proof we write ξ(t) = x = (x 0 , x 1 , . . . , x n−1 ), l = l t and r = r t . Also, because of Lemma 1 for the rest of the proof we may assume that the flip of the indices l and r does not happen (should that occur, M t decreases by at least 1−2β > 0). Observe also that M t ≥ 8 ensures that D t ≥ 6, and hence (3) holds.
For the moment, assume that D t ≤ n − 2, the remaining two cases will be investigated later.
Due to the symmetry, it is sufficient to study only the left end of the zero configuration; the drift on the right end is identical. Also w.l.o.g. assume l = 2 (and hence x 0 = x 1 = 1, x 2 = 0).
Conditioned on choosing one of the three zeros shown in case (a), each with equal probability, the drift ∆ t+1 is bounded above by In case (d), conditioned on choosing the shown zero, ∆ t+1 is bounded above by In case (b), conditioned on choosing one of the two zeros, ∆ t+1 is bounded above by Finally, in case (c), conditioned on choosing one of the two zeros, ∆ t+1 is bounded above by The reason we have inequalities for the drift, rather than equalities, is that sometimes we do not know whether the new configuration Z t+1 starts with "00" or "01", and we take the worst case scenario (i.e. "01").
While the computations of T ij are quite tedious, for the sake of completeness we will present the detailed calculation of T 10 (case (c)). Suppose w.l.o.g that l t = 2, and thus ξ(t) = 11010?? . . .
where question marks correspond to the unknown values. Here * stands for the unimportant value which is generated in ξ(t + 1). Since i t = 2 or 4 with equal probibility, we get which is the same as (5).
By examining all the cases, we find that the largest value of p for which T (p, β) can be made non-positive, is the real solution of the equation (1); in this case T 00 = T 11 = 0, T 01 = −0.0669 . . . , T 10 = −0.106 . . . .
Proof of Theorem 1. Let M t be defined by (3), and let the sequence of stopping times τ k be defined by τ 0 = 0 and Then chosen with equal probability, the probability of the latter event is bounded below as follows: from (3) since β < 1/2. Hence τ k+1 − τ k is bounded above by a geometric random variable with expectation M τ k + 1.

Now, Lemma 5 implies that
for someR not dependent on n. At the same time M t is a function of a positive recurrent finite-state Markov chain, hence as t → ∞ whereẼ denotes the expectation under the stationary measure π (n) for the chain ξ, and µ (n) is the expectation of M under this measure. Therefore, by Cesàro summation from (7) we obtain that µ (n) ≤R. Since X t ≤ D t ≤ M t + 1, the expected fraction of zeros in the limit is given by The RHS of (8) converges to zero as n → ∞, yielding the statement of the theorem.
Remark 4. By analyzing a larger (but still a finite) set of cases, one can even get a better upper estimate for p c , namely p c ≤ 0.419533, which is closer to the estimated value of 0.36. In order to do this, one can introduce the following more subtle supermartingale (on D t ≥ 8) where α = 0.3764287, β = 0.078811, and γ = 0.423494. For these particular values, we get that ε(p) used in Lemma 3 is at least 3.6 × 10 −8 .
Proof. The idea is borrowed from the proof of [10, Theorem 2.1.7]. Let 0 < h < 1/b. Since Then, taking the expectation of (9), we have on by taking h so small that hb 2 < ε/2. Consequently, ≤ γE e hY t−1 + e h(C+b) .
Hence, since e hy ≥ (hy) m /m! for all y, h ≥ 0, and m = 1, 2, . . . , The proof of the next statement uses some ideas from the renewal theory.
Proof. First, let Y k = M τ k . Because of (a), (b) and (c), Y k satisfies the conditions of Lemma 4.
Let N(t) = max{k : τ k ≤ t} be the counting process. Then, from (c), since N(t) ≤ t. On the other hand, from (d), where R 1 , R 2 < ∞ depend only on C, b, ε by Lemma 4. This, together with (11), implies that lim sup as the Cesàro mean.