Two-type annihilating systems on the complete and star graph

TL;DR

This paper investigates the dynamics of two-type annihilating particle systems on complete and star graphs, analyzing extinction times and the impact of asymmetric speeds, extending understanding beyond single-type models.

Contribution

It introduces a comparative analysis of two-type annihilating systems on specific graph structures and explores the effects of asymmetric particle speeds.

Findings

Extinction times are characterized for two-type systems on complete and star graphs.

Asymmetric speeds significantly influence the annihilation dynamics.

Comparison with one-type systems highlights differences in extinction behavior.

Abstract

Red and blue particles are placed in equal proportion through-out either the complete or star graph and iteratively sampled to take simple random walk steps. Mutual annihilation occurs when particles with different colors meet. We compare the time it takes to extinguish every particle to the analogous time in the (simple to analyze) one-type setting. Additionally, we study the effect of asymmetric particle speeds.

Equations

$$2n\log n\le E{T}_p^2(K_{2n})\le 20n(\log n{)}^2/\log \log n$$

$$c\sqrt{n}\le E{T}_{1/2}^2(S_{2n})-E{T}^1(S_{2n})\le C\sqrt{n}\log n$$

$$E{T}^1(K_{2n})-(n\log n+\gamma n)=\Theta (1)$$

$$E{T}^1(S_{2n})-(2n+2\log n)=\Theta (1).$$

$$T_p^2(K_{2n})⪰\sum _{i=1}^{n}X(i/2n)$$

$$E{T}_p^2(K_{2n})-\frac{20n(\log n{)}^2}{\log \log n}=O(1).$$

$$\left(2+\frac{2p-1}{2}\right)n-1\le E{T}_p^2(S_{2n})\le \frac{2}{(1-p)}n.$$

$$E{T}_p^2(S_{2n})-c\log \left(\frac{1}{1-p}\right)n=\Omega (1).$$

$$E{T}_p^2(S_{2n})-C\log \left(\frac{1}{1-p}\right)n=O(1).$$

$$T_1^2(S_{2n})\stackrel{d}{=}2\sum _{i=1}^{n}X(p_i)$$

$$E{T}_1^2(S_{2n})-(4n\log n+4\gamma n)=\Theta (1)$$

$$p_t\sim \{\begin{array}{ll}1/(2\sqrt{\pi t}),& d=1\\ \log t/(2\pi t),& d=2\\ 1/(2{\gamma }_{d}t),& d\ge 3\end{array}$$

$${\rho }_t\approx \{\begin{array}{ll}{t}^{-d/4},& d\le 3\\ t^{-1},& d\ge 4\end{array}.$$

$$E{T}_p^2(S_{2n})\sim 4\log \left(\frac{1}{1-p}\right)n.$$

$$T^1(G)={\tau }_0-{\tau }_n=\sum _{i=1}^n{\tau }_{i-1}-{\tau }_i.$$

$$q_i=\frac{(2i-1)(2n+1)}{4ni}$$

$$E{T}^1(S_{2n})=2\sum _{i=1}^n\frac{1}{q_i}$$

$$=\frac{8n}{2n+1}\sum _{i=1}^n\frac{1}{2}\left(1+\frac{1}{2i-1}\right)$$

$$=2n+\log n+\log 2+o(1).$$

$$\frac{1}{2i}\frac{2i-1}{2n}+\frac{2i-1}{2i}=q_i.$$

$$P(A_{t+1}$$

$$T_p^2(K_{2n})⪰\sum _1^{n}X(p_i).$$

$$E{T}_p^2(K_{2n})\ge E\sum _1^{n}X(p_i)=\sum _1^n\frac{2n}{i}=2n\log n+2\gamma n+O(n^{-1}).$$

$$P(Z_{t+1}=k+1|Z_t=k,A_t=2i)\le \frac{p}{2n},$$

$$P(Z_{t+1}=k-1|Z_t=k,A_t=2i)\ge p\cdot \frac{k}{i}\cdot \left(1-\frac{1}{2n}\right)>\frac{pk}{2n}.$$

$$(pk/2n)/(p/2n)=k,$$

$$\frac{n^3}{(m-1)!}\le n^3\frac{e^m}{m^{m-1}}=n^3\exp \left[-m\log m+m+\log m\right]$$

$$G_t=\{\text{every site has at most m particles through time }t\},$$

$$P(G_t^c\mid A_t=2i)$$

$$\le n^{-1}.$$

$$P(A_{t+1}=2i-2\mid A_t=2i)$$

$$\ge \frac{i/m}{2n}\cdot (1-1/n)$$

Full text

The third author was partially supported by NSF-DMS grant 1855516.

1. Introduction

We introduce a discrete-time annihilating particle system and study the effects of multiple particle types and asymmetric speeds on the time to extinguish every particle. We consider such systems in two geometries: the complete graph on $2n$ vertices, $K_{2n}$, and the star graph with $2n$ leaves and a single non-leaf vertex, called the core, $S_{2n}$. Initially, one particle is placed at every site of $K_{2n}$, or at every leaf of $S_{2n}$. In the one-type system, at each step a particle is chosen uniformly at random and takes one step of a simple random walk. When any two particles meet, they mutually annihilate.

In our two-type system, half of the particles are colored blue and half are colored red. At each step, a blue particle is chosen uniformly at random with probability $p\in[1/2,1]$, or else a red particle is chosen uniformly at random, and the chosen particle takes a random walk step. When two particles with different colors meet, they mutually annihilate; particles of the same color do not interact. Note that the incremental movement of particles corresponds to the embedded jump chain from the analogous process with particles performing continuous time random walks. Increasing $p$ is equivalent to increasing the rate at which blue particles jump. Accordingly, we call the case $p=1/2$ the symmetric speeds case and $p>1/2$ the asymmetric speeds case.

The two-type system belongs to a family of processes that model systems with two compounds in which reactions neutralize both chemicals involved. These dynamics have been rigorously studied on infinite graphs, typically lattices. The main focus is on the asymptotic density of the different particle types. We initiate the study of such systems on finite graphs. Besides being natural for studying reactions with inherently limited space and material, the finite setting also introduces a new quantity: the time to neutralize all reactants. By working with simple geometries—complete and star graphs—we reveal complicated underlying features of the dynamics. For example, clustering of like compounds is a phenomenon that makes the two-type system more challenging to analyze than related one-type systems. Results for these dynamics on more “realistic” finite graphs, such as tori and random networks, or for general topologies would be natural next steps. However, as with the infinite setting, rigorous results appear difficult to obtain. More background and references are provided in Section 1.2.

1.1. Results

Let $T^{1}(G)$ and $T^{2}_{p}(G)$ be the numbers of steps it takes in the one-type and two-type systems for every particle to be annihilated on the graph $G$. We begin with an informal summary of our results. It is straightforward and elementary to compute the distributions of $T^{1}(K_{2n})$ and $T^{1}(S_{2n})$ exactly. This is done in \threfthm:1, which we include for comparison with our quantitative bounds on $ET^{2}_{p}(G)$. In particular, we show that for $G=K_{2n}$ and $G=S_{2n}$ and for all $p\in[1/2,1]$, we have $ET^{2}_{p}(G)$ is asymptotically larger than $ET^{1}(G)$. How much larger depends of course on the particular graph and the value of $p$. For the complete graph we prove that

[TABLE]

for large $n$, and in particular, $\liminf ET^{2}_{p}(K_{2n})/ET^{1}(K_{2n})\geq 2$. Our strongest results are for the star graph. For $p=1/2$, we have

[TABLE]

for large $n$. For $p\in(1/2,1)$, we have that $ET_{p}^{2}(S_{2n})/ET^{1}(S_{2n})$ is bounded away from $1$ and $\infty$ as $n\to\infty$ and diverges like $\log(1/(1-p))$ as $p\uparrow 1$; and for $p=1$, the ratio diverges like $2\log n$.

Throughout this article we let $X(p)$ denote a geometric random variable with distribution $P(X(p)=k)=(1-p)^{k-1}p$ for $k\geq 1$. We write $X\preceq Y$ to denote the usual notion of stochastic dominance $P(X\geq a)\leq P(Y\geq a)$ for all $a\geq 0$. Or, equivalently, that there is a coupling so that $X\leq Y$ almost surely. We say that $X\overset{d}{=}Y$ if $X$ and $Y$ have the same distribution. Our results make use of the standard asymptotic notation:

• •

$f=O(g)$ if $\limsup f/g<\infty$,

• •

$f=\Omega(g)$ if $\liminf f/g>0$, and

• •

$f=\Theta(g)$ if $f=O(g)$ and $g=O(f)$.

• •

We write $f\sim g$ if $\lim f/g=1$.

One can exactly characterize how long it takes to go from having $2i$ to $2(i-1)$ particles in the system in terms of a geometric random variable. Though elementary, this gives us a baseline for comparing to the two-type system.

Proposition 1**.**

\thlabel

thm:1 In both distributional equalities below the geometric random variables being summed are independent.

1. (i)

$T^{1}(K_{2n})\overset{d}{=}\sum_{i=1}^{n}X(p_{i})$* with $p_{i}=(2i-1)/2n$. In particular,*

[TABLE]

where $\gamma=\lim(-\log n+\sum_{1}^{n}i^{-1})$ is the Euler-Mascheroni constant. 2. (ii)

$T^{1}(S_{2n})\overset{d}{=}2\sum_{i=1}^{n}X(q_{i})$* with $q_{i}=1-(\frac{1}{2i})(\frac{2n-2i+1}{2n})$. In particular,*

[TABLE]

Precisely analyzing the two-type system appears to be much more difficult. The issue on the complete graph is that, as the process evolves, like-particles tend to cluster at the same sites. The clustering should not be too extreme. Namely, when there are $\Omega(n)$ particles, red and blue should occupy $\Omega(n)$ distinct sites at all times, and when there are $o(n)$ particles red and blue should be nearly perfectly spread out. However, there is dependence between which particles are removed and the number of particles at each site. This appears to make it difficult to prove that red and blue particles stay sufficiently spread out.

While we do not completely overcome the issues mentioned above, we are able to confirm that the two-type system survives longer than the one-type system. Below we prove that $ET_{p}^{2}(K_{2n})\geq 2ET^{1}(K_{2n})(1-o(1))$. This result should not be all that surprising since the two-type system in some sense has at least twice as many “safe” sites for particles to jump to among the occupied sites as the one-type setting. We also prove an upper bound that differs by a logarithmic factor.

Theorem 2**.**

\thlabel

thm:Kn For all $p\in[1/2,1]$ it holds that

[TABLE]

with the $X(i/2n)$ independent. Thus, $ET_{p}^{2}(K_{2n})-2n\log n=\Omega(1).$ Furthermore, the distributional inequality is an equality when $p=1$, so $ET_{1}^{2}(K_{2n})-2(n\log n+\gamma n)=\Theta(1)$. As for an upper bound, it holds for any fixed $p\in[1/2,1]$ that

[TABLE]

The proof of the lower bound uses a comparison to a process that has red and blue particles take up the maximal amount of space at each time step. Analogous to what occurs in \threfthm:1, we show that $T^{2}_{p}(K_{2n})$ stochastically dominates a sum of geometric random variables. The upper bound goes by showing that it is overwhelmingly unlikely for any site to host more than $C\log n/\log\log n$ particles in the first $n^{3}$ steps of the process. This gives a tractable way to lower bound the probability of a collision, but comes at the cost of the additional logarithmic factor.

We can say more about the two-type system on the star graph. The process in this setting has the same clustering issue at the leaves as what occurs globally on $K_{2n}$. Moreover, the number of like-particles grouped at the core introduces another hub for many like-particles to cluster. While, in principle, one could write down an explicit Markov chain for this process, to do this precisely would require keeping track of the number of particles at the core, as well as the number of red and blue leaves with $1,2,\ldots$ particles at them. Analyzing this Markov chain exactly appears challenging since the state-space is of growing dimension, and there is significant dependence between the transition rate and the population size.

Despite these difficulties, we prove a fairly precise characterization for all $p\in[1/2,1]$. For symmetric speeds we show that the second order term is different for the two-type case. The proof reveals that this is caused by clustering at the core. Recall that \threfthm:1 shows $ET^{1}(S_{2n})$ has a logarithmic second order term. We show that $ET_{1/2}^{2}(S_{2n})$ has a second order term on an order between $\sqrt{n}$ and $\sqrt{n}\log n$. This demonstrates the effect of clustering at the core and, along with \threfthm:1 (ii), also implies that $ET_{1/2}^{2}(S_{2n})-ET^{1}(S_{2n})=\Omega(\sqrt{n}).$

Theorem 3**.**

\thlabel

thm:2 It holds that

1. (i)

$ET_{1/2}^{2}(S_{2n})-(2n+C\sqrt{n})=\Omega(1)$* for any $C<(32\pi)^{-1/2}$, and* 2. (ii)

$ET_{1/2}^{2}(S_{2n})-(2n+c\sqrt{n}\log n)=O(1)$* for some $c>0$.*

The starting point for the lower bound is a “master formula” in \threflem:A_t that equates the number of remaining particles to what has occurred up to that point at the core. We use this to make estimates on the number of particles in the system at time $2n$. This relies on a coupling to the simple random walk which tracks the discrepancy between the number of times red and blue have been sampled. The upper bound again uses the identity in \threflem:A_t, but this time couples to a different random walk to estimate the number of particles clustered at the core as the process evolves. The argument concludes by bounding the probability a particle is sampled at the core.

For asymmetric speeds, we focus on the leading order coefficient and provide universal upper and lower bounds. The lower bound implies that the asymmetric case has a strictly larger leading coefficient than the symmetric case.

Theorem 4**.**

\thlabel

thm:p Fix $p\in(1/2,1)$. It holds for all $n$ that

[TABLE]

The lower bound is proven in a similar manner as \threfthm:2 (i), and the upper bound follows from the observation in \threflem:remy_bound that from any configuration, after two steps, the probability of a collision is at least $1-p$. So $T^{2}_{p}(S_{2n})$ is stochastically dominated by a sum of independent geometric random variables. While these bounds hold for all $p$, they become rather far from the truth for $p$ near $1$. The following theorem addresses what happens in this regime. We provide matching order upper and lower bounds for the rate the leading constant tends to infinity.

Theorem 5**.**

\thlabel

thm:asymptotic Given $c<4$ there exists a value $p_{*}<1$ (which depends on $c$) such that for any fixed $p\in(p_{*},1)$ it holds that

[TABLE]

And, given $C>12$ there exists $p^{*}<1$ (which depends on $C$) such that for any fixed $p\in(p^{*},1)$ it holds that

[TABLE]

These results use stochastic lower and upper bounds that relate $T^{2}_{p}(S_{2n})$ to a coupon collector process. This connection was not so obvious to make, and requires technical estimates on the time to collect a random subset of the coupons, as well as on the number of coupons collected after a random amount of time.

Finally, we handle the case $p=1$, which corresponds to the setting in [DGJ*+*17]. This setting is tractable because the now immobile red particles cannot cluster.

Theorem 6**.**

\thlabel

prop:p=1 It holds that

[TABLE]

with the $X(p_{i})$ independent and $p_{i}=i/2n$. In particular,

[TABLE]

with $\gamma$ the Euler-Mascheroni constant.

1.2. Background

The study of annihilating particle systems dates back to the work of Erdős and Ney [EN74]. They considered a system of continuous time random walks started at each nonzero integer in which collisions cause both particles to annihilate and disappear from the process. In particular, they asked if the origin was visited infinitely often, and, more precisely, they studied the asymptotic decay of $p_{t}$, the probability the origin is occupied at time $t$.

The question of whether or not the origin is visited infinitely often was answered in the affirmative by Lootgieter in [Loo77] in discrete time and by Schwartz in [Sch78] in continuous time. Later, Arratia in [Arr81, Arr83] generalized the process to higher dimensions and more general initial configurations. One of his main findings was that

[TABLE]

where $\gamma_{d}$ is the probability the simple random walk never returns to its starting position in $\mathbb{Z}^{d}$. Due to a parity relation observed by Arratia, $p_{t}$ decays exactly twice as fast as what Bramson and Griffeath in [BG80] proved occurs for coalescing random walk. This is the system in which particles coalesce rather than annihilate upon colliding. The main proof technique in these systems is to analyze a dual process known as the voter model.

Two-type annihilating particle systems first garnered interest in the chemistry and physics literature [OZ78, TW83, KR84, LC95]. Initially particles are assigned to be either of type $A$ or $B$, and only collisions between different particle types result in annihilation. Unlike the one-type annihilating and coalescing systems, the two-type system has no known tractable dual process. Ovchinnikov and Zeldovich and later Toussaint and Wilczek predicted that in low dimensions the density of particles at the origin of $\mathbb{Z}^{d}$ is asymptotically much larger than in the one-type system [OZ78, TW83] due to local clustering of like particles.

In a definitive series of papers, Bramson and Lebowitz [BL88, BL90, BL91, BL01] proved this (and more) for the two-type system on $\mathbb{Z}^{d}$, where initially each site has a $\operatorname{Poi}(\mu_{A})$-distributed number of $A$ particles and a $\operatorname{Poi}(\mu_{B})$-distributed number of $B$ particles. At time [math], pairs of $A$ and $B$ particles at the same site instantly annihilate. Particles then perform continuous time simple random walks at rates $\lambda_{A}$ and $\lambda_{B}$, and annihilate when they meet a particle of opposite type. Since multiple particles can occupy a given site, the main quantity of interest is the expected number of particles at the origin at time $t$, which we denote by $\rho_{t}$. In the critical case, with particle types in balance ($\mu_{A}=\mu_{B}>0$) and symmetric speeds ($\lambda_{A}=\lambda_{B}>0$), Bramson and Lebowitz [BL91] proved that

[TABLE]

Here $f\approx g$ if $0<\liminf f/g\leq\limsup f/g<\infty.$ Note that, in low dimension, this is asymptotically much larger than the formula for $p_{t}$ at (6).

There has been recent interest in extending the results of Bramson and Lebowitz to asymmetric speeds. On lattices, physicists predicted that the asymptotic order of $\rho_{t}$ does not change as the speeds are varied [KR84, LC95]. Cabezas, Rolla, and Sidoravicius in [CRS18] considered the asymmetric speed case on a class of infinite transitive graphs and proved a universal lower bound $\rho_{t}=\Omega(t^{-1})$, and that the root is visited infinitely often when particle types are initially in balance. In a different work [CRS14], Cabezas, Rolla, and Sidoravicius considered the case that red particles move and blue particles are stationary. They proved that there is a phase transition between transience and recurrence when the different particle types are in balance on a broad class of transitive graphs. An Abelian property ensures that the results hold in either discrete or continuous time. More recently Johnson, Junge, Lyu, and Sivakoff proved new upper and lower bounds for the particle density in two-type annihilating systems on lattices and bi-directed regular trees [JJLS20]. Bahl, Barnet, Johnson and Junge further explored how the volatility of the distributions of the initial particle counts impacts the total occupation time of the root [BBJJ21].

Damron, Gravner, Junge, Lyu, and Sivakoff considered a similar problem as [CRS14] in discrete time and proved transience/recurrence results along with more quantitative estimates on the number of visits to the origin when the particle densities are initially out of balance [DGJ*+*17]. Very recently, Przykucki, Roberts, and Scott proved quantitative results in discrete time with $B$-particles stationary on the integers [PRS19]. A slightly different, but related process was studied by Goldschmidt and Przykucki on Galton-Watson trees [GP19]. The papers [DGJ*+*17, PRS19, GP19] refer to the annihilating system as parking since they view $A$-particles as cars and $B$-particles as parking spots. Parking was introduced over fifty years ago in [KW66] and has attracted interest ever since. See [LP16] for an overview.

As for the finite setting, Cooper, Frieze, and Radzik studied similar quantities as us on random regular graphs [CFR09a, CFR09b]. They considered an “explosive” particle system with the same dynamics as our one-type system, but with the modification that all particles move simultaneously. They proved that the time it takes to remove all particles is $O(n\log n)$ when there are sufficiently few particles initially. Additionally, the authors considered a two-type “predator-prey” dynamics, in which predators remove prey on contact, but predators persist, and they studied the expected time to remove all prey. These quantities are closely related to the coalescence time. This is the number of steps needed to reach a single particle when particles coalesce, rather than annihilate, upon colliding. There have been recent results for how this behaves on general finite graphs [CEOR13, KMTS19], as well as a result from Cox concerning the coalescence time on the torus [Cox89]. Note that [Cox89, CEOR13, KMTS19] only considered one-type systems. To the best of our knowledge, the quantity $T^{2}_{p}(G)$ for two-type systems has not been studied on any finite graph.

1.3. Further questions

It would be interesting to find the correct leading order coefficient for $ET_{p}^{2}(K_{2n})$ and for $ET^{2}_{p}(S_{2n})$. Note that currently we do not have a proof that $ET_{p}^{2}(K_{2n})=O(n\log n)$. For the star graph, we conjecture that our asymptotic lower bound in \threfthm:asymptotic is sharp so that, for large enough $p$,

[TABLE]

This is the answer one gets for the simplified model in which one assumes that the core is always occupied by blue particles and that every red step results in a collision. While the connection is difficult to make rigorous, it seems to be a reasonable approximation for large $p$.

We also would like to know the exact second order term for $ET_{1/2}(S_{2n})$. This is a more delicate question, but it would be interesting to decide if the logarithmic factor is needed, and if so, what is causing its appearance. We discuss this a bit more at \threfrem:G. Another future direction is to understand two-type annihilating systems on other finite graphs, such as Erdős-Rényi graphs, tori, and trees.

1.4. Organization

In Section 2 we analyze the one-type system and prove \threfthm:1. In Section 3 we prove our lower and upper bounds for the two-type system on the complete graph from \threfthm:Kn. In Section 4 we analyze the two-type system with symmetric speeds on the star graph by proving the upper and lower bounds in \threfthm:2. Section 5 houses the proofs of Theorems LABEL:thm:p, LABEL:thm:asymptotic and LABEL:prop:p=1 for asymmetric speed two-type systems on the star graph.

2. One-type systems

One-type systems are fairly straightforward to precisely describe because at most one particle can occupy each site. Combining this feature with the simple geometry of the complete and star graphs makes it so the time to annihilate every particle decomposes as a sum of independent geometric random variables.

Proof of \threfthm:1.

We start with a general decomposition then explain how to prove (i) and (ii). Let $\tau_{i}$ be the first time there are $2i$ particles in the system for $0\leq i\leq n$. Notice that $\tau_{0}=T^{1}(G)$ and $\tau_{n}=0$ so that

[TABLE]

To prove (i), notice that on $K_{2n}$ we have $\tau_{i-1}-\tau_{i}\overset{d}{=}X((2i-1)/2n)$. This is true because, when there are $2i$ particles in the system, when a particle is selected there are $2i-1$ out of $2n$ sites with another particle that it could move to and cause a collision.

To prove (ii), for the star graph we claim that $\tau_{i-1}-\tau_{i}\overset{d}{=}2X(q_{i})$ with

[TABLE]

and the $X(q_{i})$ independent. Assuming (9) holds and applying this to (8), we then have

[TABLE]

It remains to justify that $\tau_{i-1}-\tau_{i}\overset{d}{=}2X(q_{i})$ with $q_{i}$ as in (9). First notice that at time $\tau_{i}$ there is no particle at the core, and $2i$ particles at the leaves. This is because the only way an annihilation can occur on the star is by either (a) a particle moving from a leaf to the core, or (b) a particle moving from the core to a leaf. Since only one particle can occupy a given site, both (a) and (b) result in no particles at the core when annihilation occurs. Now, the next step from this configuration will necessarily be a particle from a leaf moving to the core. This particle is destroyed either if it is again selected and then moved to one of the $2i-1$ occupied leaves, or if a particle at a leaf is next selected; these occur with total probability

[TABLE]

To obtain a renewal, notice that if the particle at the core is not destroyed in this second step, then it must move back to an unoccupied leaf. On this event, we have no particle at the core, and $2i$ particles at the leaves, which was the configuration at time $\tau_{i}$. The next two steps once again result in an annihilation with probability $q_{i}$, so we have $\tau_{i-1}-\tau_{i}\overset{d}{=}2X(q_{i}).$ ∎

3. Two-type systems on the complete graph

Let $A_{t}$ be the total number of particles remaining in the system after $t$ steps. The fact that collisions occur in pairs ensures that $0\leq A_{t}\leq 2n$ is even. It is also convenient to let $R_{t}$ and $B_{t}$ be the total number of sites with at least one red or blue particle, respectively. We start by giving the proof of the lower bound at (1), and then give the proof of the upper bound at (2).

Proof of \threfthm:Kn equation (1).

We start by describing the transition probabilities for $A_{t}$ conditional on the number of sites occupied by red and blue particles. Letting $0\leq r,b\leq i\leq n$, we have

[TABLE]

Otherwise $A_{t+1}=A_{t}$. Notice that the above equality does not depend on the value of $A_{t}$. When there are $2i$ particles remaining, the probability of an annihilation occurring on a given step is thus bounded by probability of an annihilation when $R_{t}=i=B_{t}$. This is equal to $p_{i}=i/{2n}.$ Using $p_{i}$ as a bound on the probability of an annihilation is equivalent to comparing to a process that always has red and blue particles occupying the maximal number of sites. For the case $p=1$, the value $p_{i}$ is the actual transition rate, because red particles do not move.

It follows that the number of steps to transition from $2i$ to $2(i-1)$ particles is stochastically larger than $X(p_{i})$. Decomposing $T^{2}_{p}(K_{2n})$ into the time it takes to go from $2n$ to $2(n-1)$ to $2(n-2)$, and so on, we have

[TABLE]

The well-known asymptotic behavior of the harmonic series ensures that

[TABLE]

∎

Proof of \threfthm:Kn equation (2).

We first show that with high probability no site has more than $m:=6\log n/\log\log n$ particles through time $n^{3}$. Let $Z_{t}$ be the number of blue particles at vertex $1$ at time $t$. Note that $Z_{t}$ cannot jump by more than $1$ at any step, so we can dominate it by a birth and death chain. For $k\geq 0$,

[TABLE]

since a blue particle must move to $1$. To decrease the number of blue particles at $1$, it suffices to choose a blue particle at $1$ and move it somewhere else, so for $0<k\leq i$ and $n\geq 2$,

[TABLE]

The ratio between the last two probabilities is at least

[TABLE]

independent of $i$. Therefore, independent of $(A_{t})$, we have that $Z_{t}$ is dominated by a birth and death chain that is $k$ times as likely to move left as right when it is at $k$. Thus, whenever $Z_{t}=1$, the probability that it hits $m$ before hitting [math] is at most $1/(m-1)!$ (see, for instance, Example 5.3.9 in [Dur19]). Using the bound $m!\geq(m/e)^{m}$, the probability that $(Z_{t})$ reaches $m$ by time $n^{3}$ is at most

[TABLE]

independent of $(A_{t})$. We obtain the same bound for the probability that the number of red particles at $1$ hits $m$ by time $n^{3}$ by repeating the same argument as above, replacing the $p$-factor with $1-p$ in (11) and (12).

Define the events

[TABLE]

so the union bound and (13) imply that for all $t\leq n^{3}$ and for all sufficiently large $n$, we have

[TABLE]

On the event $\{A_{t}=2i\}\cap G_{t}$, there are at least $i/m$ sites that contain at least one blue particle, and at least $i/m$ sites that contain at least one red particle. A collision occurs if a particle is selected and moves to one such site containing the opposite type. For $1\leq i\leq n$ and $t\leq n^{3}$, and assuming $n$ is sufficiently large, we have

[TABLE]

Therefore, for $t\leq n^{3}$, we have

[TABLE]

For $t>n^{3}$, letting $S_{n}=X_{1}+\cdots+X_{n}$ be the sum of $n$ i.i.d. Geometric($1/2n$) random variables, we have the trivial upper bound,

[TABLE]

for large $n$. Summing over $t$ and using the last two inequalities for $t\leq n^{3}$ and $t>n^{3}$, respectively, we have

[TABLE]

The first term is equal to

[TABLE]

The second term in (15) is bounded by $8n^{2}e^{-n/4}$ for large $n$, so tends to [math] as $n\to\infty$. We have proved that for large $n$,

[TABLE]

∎

4. The star graph with symmetric speeds

We start by fixing some notation. Again let $A_{t}$ be the total number of particles in the system after $t$ steps. Let $C_{t}$ be the number of particles that are at the core after $t$ steps. Additionally, let $M_{t}$ be the number of times up to time $t$ that a particle at the core is sampled to move, but is not annihilated after taking a step. Let $Z_{t}=1$ if blue is sampled at time $t$, and $-1$ if red is sampled (note that $(Z_{t})_{t\geq 1}$ is an i.i.d. sequence, defined even for $t\geq T^{2}_{p}(S_{2n})$ after there are no particles remaining). Define the quantities

[TABLE]

so that $D_{t}$ has the same law as the displacement of a $p$-biased random walk. When we write $D_{2n}$ it is implicit that this is the value of $D_{t}$ at $t=2n$ for the process on $S_{2n}$.

The quantities $A_{t}$, $C_{t}$ and $M_{t}$ are related by the following identity, which will be useful for proving both lower and upper bounds on $ET^{2}_{p}(S_{2n})$.

Lemma 7**.**

\thlabel

lem:A_t For all $p\in[1/2,1]$ and $t\leq T_{p}^{2}(S_{2n})$ we have

[TABLE]

Moreover, for $t\leq 2n$, we have

[TABLE]

and consequently $T_{p}^{2}(S_{2n})\geq 2n$.

Proof.

Clearly the formula holds for $t=0$. We proceed inductively from here. Suppose that (17) holds through step $t$. At step $t+1$, we sample a particle to move; call this particle $x$. If the step taken by $x$ results in a collision, then by (17),

[TABLE]

Either the collision happens at the core, or $x$ moves from the core to a leaf. In either scenario we have $C_{t+1}=C_{t}-1$ and $M_{t+1}=M_{t}$, so

[TABLE]

This is the desired statement at time $t+1$.

Now, suppose that the step taken by $x$ does not result in a collision. If $x$ moves from a leaf to the core, then $C_{t+1}=C_{t}+1$, and if $x$ moves from the core to a leaf, then $M_{t+1}=M_{t}+1$ and $C_{t+1}=C_{t}-1$. In the first case, we have

[TABLE]

In the second case we have

[TABLE]

Simplifying either case gives $A_{t+1}=2n-(t+1)+C_{t+1}+2M_{t+1}$, as desired.

The second and third statements follow from (17) by observing that $C_{t}\geq 0$ and $M_{t}\geq 0$ for $t=0,1,\ldots,2n-1$. ∎

4.1. A lower bound for the symmetric case

The starting point for our lower bound is a simple observation that relates $T_{p}^{2}(S_{2n})$ to the process stopped at a given time.

Lemma 8**.**

\thlabel

lem:TLB For all $p\in[1/2,1]$ and $t\leq T_{p}^{2}(S_{2n})$ it holds that

[TABLE]

Proof.

At most two particles can be removed from the sytem at each step. Thus, if there are $A_{t}$ particles at time $t$, then it deterministically takes at least $A_{t}/2$ more time steps to remove them all. ∎

We can further bound $M_{2n}$ in terms of $D_{2n}$.

Lemma 9**.**

\thlabel

lem:M_t For any $p\in[1/2,1]$, we have $EM_{2n}\geq(1/8)ED_{2n}-1$.

Proof.

Without loss of generality, suppose that blue is sampled $n+D_{2n}/2$ times through time $2n$. Let $\alpha$ be the number of times through time $2n$ that a blue particle moves from the core to a leaf. By \threflem:A_t we have $T_{p}^{2}(S_{2n})\geq 2n$, so $(n+D_{2n}/2)-\alpha$ is the number of times that a blue particle moves from a leaf to the core. Since we have only $n$ blue particles initially, we must have

[TABLE]

so $\alpha\geq D_{2n}/4$. Since red can occupy at most half of the leaves, each core selection of blue has at least a $1/2$ chance of increasing the count of $M_{2n}$. In particular, $M_{2n}$ stochastically dominates a binomial thinning of $D_{2n}/4$ with success probability $1/2$. The same holds when red is sampled $D_{2n}$ times more than blue. Using the bound $\lfloor ED_{2n}/8\rfloor\geq ED_{2n}/8-1$ gives the claimed inequality. ∎

It is a well-known estimate that $ED_{t}$ grows like $\sqrt{t}$ when $p=1/2$. We give a combinatorial proof of this fact below.

Lemma 10**.**

\thlabel

lem:D_t Suppose $p=1/2$. It holds that $ED_{t}\leq\sqrt{t}$ for all $t\geq 0$. Moreover, as $n\to\infty$ it holds that $ED_{2n}\sim\sqrt{2n/\pi}.$

Proof.

Recalling the definition at (16), it is a standard exercise to show that $W_{t}^{2}-t$ is a martingale, and thus $EW_{t}^{2}=t$. We then have

[TABLE]

Next we prove the asymptotic claim. Observe that for every integer $x>0$, $E[D_{n+1}\mid D_{n}=x]=x$, while $E[D_{n+1}\mid D_{n}=0]=1$. We then have

[TABLE]

Using the parity observation that $D_{2n+1}\neq 0$, and that $ED_{1}=1$, gives the equation

[TABLE]

Stirling’s approximation then yields

[TABLE]

Integrating $1/\sqrt{\pi k}$ from $1$ to $n$ gives the claimed asymptotic formula for $ED_{2n}$. ∎

Proof of \threfthm:2 (i).

Evaluating the formula in \threflem:A_t at $t=2n$ and ignoring the $C_{2n}$ term gives $A_{2n}\geq 2M_{2n}$. \threflem:M_t then tells us that

[TABLE]

which by \threflem:D_t is $(1/8)\sqrt{2n/\pi}+o(\sqrt{n})$. Thus, for any $C^{\prime}<(32\pi)^{-1/2}$ we have

[TABLE]

The result then follows by applying the above bound on $EA_{2n}$ to the inequality $ET^{2}_{1/2}(S_{2n})\geq 2n+EA_{2n}/2$ implied by \threflem:A_t,lem:TLB. ∎

4.2. An upper bound for the symmetric case

We start with a simple observation that provides a stochastic upper bound on $T^{2}_{p}(S_{2n})$ by stopping the process at a given time and then using a worst-case upper bound related to the number of particles still in the system at that time.

Lemma 11**.**

\thlabel

lem:remy_bound For all $p\in[1/2,1)$ we have

[TABLE]

where $(X_{i}(1-p):i\geq 1)$ are i.i.d. Geometric$(1-p)$, and are independent of $A_{t}$; the sum on the right is [math] when $A_{t}=0$.

Proof.

If the core is occupied, then the probability of a collision in the next step is at least $(1-p)\wedge p=1-p$. If the core is not occupied, then after one step it becomes occupied, and the probability of a collision on the next step is at least $1-p$. Therefore, from any configuration of particles, the probability of a collision occurring in the next two steps is always at least $1-p$, and each collision reduces $A_{t}$ by 2. The formula follows. ∎

Keeping in mind the identity in \threflem:A_t, we will require a bound on $C_{2n}$. Because $C_{t}$ always has a slight drift towards zero, it can be dominated by the displacement of a simple random walk.

Lemma 12**.**

\thlabel

lem:C¡D Fix $p=1/2$. Let $D^{\prime}_{t}$ be the displacement from the origin of a simple symmetric random walk on $\mathbb{Z}$ started at [math]. There exists a coupling such that

[TABLE]

for all $t\geq 0$.

Proof.

We explain how to construct $D^{\prime}_{t}$ from $C_{t}$. Notice that the probability $C_{t}$ increases is equal to the probability of picking a particle at a leaf that is the same color as those currently occupying the core, or $1$ if no particles are there. Thus, we define $D^{\prime}_{t+1}=D_{t}^{\prime}+1$ if one of the following occurs:

1. (a)

$D_{t}^{\prime}=0$, 2. (b)

$C_{t}>0$ and $C_{t+1}=C_{t}+1$, 3. (c)

$C_{t}>0$ and $C_{t+1}=C_{t}-1$ because of a particle moving away from the core, 4. (d)

with probability $1/2$ if $C_{t}=0$ and $D_{t}^{\prime}>0$.

Otherwise $D_{t+1}^{\prime}=D_{t}^{\prime}-1.$

It is easy to check that $D_{t}^{\prime}$ is the displacement of a simple random walk, since, when it is nonzero, it transitions up or down with equal probability (the probabilities in (b) and (c) sum to $p=1/2$, the probability of choosing the color at the core). Moreover, $D_{t}^{\prime}$ and $C_{t}$ are coupled so that $D_{t}^{\prime}$ increases whenever $C_{t}$ does with one exception. The only situation in which $C_{t}$ can exceed $D_{t}^{\prime}$ is if $C_{t-1}=0$ and $D_{t-1}^{\prime}=1$ and $D_{t}^{\prime}=0$. When this occurs we have $C_{t}=D_{t}^{\prime}+1$. However, the gap cannot become any larger than this, because while $C_{t}$ is larger than $D_{t}^{\prime}$, case (d) is prohibited, so $D_{t}^{\prime}$ increases whenever $C_{t}$ does. ∎

We will soon require an estimate on a sum via comparison to an integral. We provide the antiderivative and asymptotic behavior of that integral now.

Lemma 13**.**

\thlabel

lem:integral It holds that $\int_{1}^{2n}x^{-1}\sqrt{2n-x}\;dx=O(\sqrt{n}\log n).$

Proof.

This follows immediately from setting $C=2n$ in the equation

[TABLE]

We obtain this formula by computing the indefinite integral $\int x^{-1}\sqrt{C-x}\;dx.$ Start with the substitution $u=\sqrt{C-x}$ so that the integral becomes

[TABLE]

Next, make the substitution $s=iu/\sqrt{C}$ with $i=\sqrt{-1}$ so that the above is equal to

[TABLE]

Substituting back $u$ and then $x$ yields

[TABLE]

We obtain the claimed formula at (19) by applying the identity

[TABLE]

combining logarithmic terms, and then computing the definite integral. ∎

We now put this inequality to work in bounding $EM_{2n}$.

Lemma 14**.**

\thlabel

lem:Mn Fix $p=1/2$. It holds that $EM_{2n}=O(\sqrt{n}\log n)$.

Proof.

Let $G_{t}=\{M_{t}=M_{t-1}+1\}$ be the event that a particle at the core is sampled at time $t$ and the particle is not annihilated after taking a step. This tracks when $M_{t}$ increases; accordingly, at time $2n$ we have

[TABLE]

Notice that $P(G_{t})$ is at most the probability of sampling a particle at the core at time $t$. Given $A_{t}=a$ and $C_{t}=c$, this probability is equal to $c/a$, since $p=1/2$. It follows that

[TABLE]

Using this for $t<2n$ and the bound $P(G_{2n})\leq 1$ as we bring the expectation inside the sum, we obtain

[TABLE]

The second inequality uses the deterministic bound $A_{t}\geq 2n-t$ from \threflem:A_t. Bounding $EC_{t}\leq ED_{t}+1$ via \threflem:C¡D and then bounding $ED_{t}$ with \threflem:D_t, we obtain

[TABLE]

Reindexing with $s=2n-t$ gives

[TABLE]

By comparison to the integral in \threflem:integral, the summation above is $O(\sqrt{n}\log n)$. ∎

Remark 15*.*

\thlabel

rem:G Note that in the previous argument at (20) we made the bound $P(G_{t})\leq E[C_{t}/A_{t}]$. One might suspect that the logarithmic factor comes from this estimate. However, the exact formula is

[TABLE]

where $U_{t}$ is the number of sites occupied by particles of the opposite color from the core at the leaves of $S_{2n}$ at time $t$. Exactly describing the quantity $U_{t}$ is subtle since it depends on clustering at the leaves and on the current particle type occupying the core. Regardless, we have $1/2\leq(2n-U_{t})/2n\leq 1$ for all $t$ since $0\leq U_{t}\leq n$. So, $P(G_{t})\geq(1/2)E[C_{t}/A_{t}]$. Thus, the estimate we make on $P(G_{t})$ is not the source of the logarithmic factor.

In any case, we now we have the necessary ingredients to prove our upper bound.

Proof of \threfthm:2 (ii).

By \threflem:A_t we have

[TABLE]

It follows from \threflem:D_t,lem:C¡D that $EC_{2n}=O(\sqrt{n})$, and from \threflem:Mn we have $EM_{2n}=O(\sqrt{n}\log n)$. Thus, $EA_{2n}=O(\sqrt{n}\log n)$. Applying \threflem:remy_bound with $t=2n$ gives

[TABLE]

It follows from Wald’s lemma and our bound on $EA_{2n}$ that

[TABLE]

∎

5. Asymmetric speeds on the star graph

We break this section into three subsections. The first two subsections contain technical estimates for a modified coupon collector problem, and also describe how these connect back to the two-type system. The third subsection contains the proofs of \threfthm:p,thm:asymptotic,prop:p=1.

5.1. Lemmas for the asymptotic lower bound

The idea behind the lower bound is that after $t\approx-4\log(1-p)n$ steps approximately $r=(1-p)t$ red particles will move from their starting location. To eliminate all of the red particles by time $t$, blue particles must visit all $n-r$ of the sites with red particles that did not move. We identify the sites initially occupied by red particles as coupons, and view each jump from the core by a blue particle as an attempt to collect one of these coupons. So $T_{p}^{2}(S_{2n})$ is lower bounded by the number of steps needed for a coupon collector to collect $n-r$ coupons, which we prove has expected value on the order of $t$.

Lemma 16**.**

\thlabel

lem:coupon1 Fix any $\epsilon\in(0,1)$, let $t_{p}=t_{p}(n)=-n\log(1-p)$ and let $R\overset{d}{=}\operatorname{Bin}(4(1-\epsilon)t_{p},1-p)$. Consider a coupon collector process in which there are $n$ coupons. At each step, with probability $1/2$ no coupon is selected, and otherwise one is picked uniformly at random. Let $T^{\prime}_{p}$ be the number of steps needed to sample $n-R$ distinct coupons. Then there exists $p^{\prime}(\epsilon)<1$ such that for all $p>p^{\prime}(\epsilon)$ we have

[TABLE]

as $n\to\infty$.

Proof.

The time between each new coupon discovery is a geometric random variable, so

[TABLE]

with the convention that the sum is zero if $n-R\leq 0$ and the $X$’s are independent. Let $a=5(1-\epsilon)(1-p)\log(1-p)^{-1}$, and observe that $a\to 0$ as $p\to 1$. Noting that $ER=4an/5$, by a standard Chernoff bound for the binomial distribution, we have

[TABLE]

for some $c=c(\epsilon,p)$. Letting

[TABLE]

then using the bound $\log m\leq\sum_{i=1}^{m}i^{-1}\leq 1+\log m$ we have

[TABLE]

and

[TABLE]

for all $p$ sufficiently close to $1$ and $n$ large enough, depending on $p$. Therefore, by Chebychev’s inequality,

[TABLE]

Combining (23) and (24), we arrive at

[TABLE]

as $n\to\infty$. ∎

Lemma 17**.**

\thlabel

lem:ALB For any $\epsilon\in(0,1)$, there exists $p^{\prime}(\epsilon)<1$ such that for each $p>p^{\prime}(\epsilon)$ and all sufficiently large $n$, we have

[TABLE]

Proof.

Let $t_{p}$, $R$ and $T^{\prime}_{p}$ be as in \threflem:coupon1. Let $\mathtt{Red\_Moved}$ be the number of red particles that have moved from their starting locations through time $4(1-\epsilon)t_{p}$, and let $\mathtt{Red\_Sites\_Visited\_by\_Blue}$ be the number of leaf vertices that were initially occupied by red particles and were visited by at least one blue particle through time $4(1-\epsilon)t_{p}$ (whether or not they are occupied by red at the time of blue’s visit). The number of red particles extinguished through time $4(1-\epsilon)t_{p}$ cannot exceed

[TABLE]

so if all particles are to be removed by time $4(1-\epsilon)t_{p}$, we must have

[TABLE]

Note that $\mathtt{Red\_Moved}$ cannot exceed the number of times that red particles are chosen to move through time $4(1-\epsilon)t_{p}$, which has the same distribution as $R$, so we have $\mathtt{Red\_Moved}\preceq R$. Also, through time $4(1-\epsilon)t_{p}$, a blue particle has moved to a uniformly sampled leaf on at most $2(1-\epsilon)t_{p}$ steps, since each visit to a leaf requires two moves by a blue particle. The random variable $\mathtt{Red\_Sites\_Visited\_by\_Blue}$ is therefore stochastically dominated by the number of distinct coupons collected after $2(1-\epsilon)t_{p}$ steps, where at each step, with probability $1/2$ no coupon is selected, and otherwise one of $n$ coupons is selected uniformly at random. This number can be taken to be independent of $R$, as the number of steps taken by the coupon collector is deterministic, so we are in the setting of \threflem:coupon1, and we have

[TABLE]

as $n\to\infty$. Letting $n$ be sufficiently large so that the probability above is smaller than $\epsilon$ gives the desired lower bound on the expectation. ∎

5.2. Lemmas for the asymptotic upper bound

The idea behind the upper bound is to run the process for $t=-8n\log(1-p)$ steps. At this point, we prove that blue has moved to nearly $pn$ of the sites that were initially red, and at most $r=(1-p)t=(1-p)\log((1-p)^{-8})n$ red particles have moved to avoid a collision. This means that at most $n-pn+r$ red particles have avoided collision through time $t$. We then use the bound at \threflem:remy_bound to show that the expected time to destroy the remaining particles is $O(n)$ with leading constant that does not depend on $p$.

Lemma 18**.**

\thlabel

lem:coupon2 Consider the following coupon collection process like that of \threflem:coupon1. Let $t_{p}=-n\log(1-p)$ and for $r>4$, let $B\overset{d}{=}\operatorname{Bin}(rt_{p},p)$. Let $N=(B-n)/2$, and let $V$ be the number of the $n$ coupons that are not collected through $N$ steps, where we set $V=n$ if $N\leq 0$. For all fixed $p$ sufficiently close to $1$ and for all $n$ sufficiently large,

[TABLE]

Proof.

Note that $B$ is a binomial random variable with mean $rpt_{p}$, so for $p$ sufficiently close to $1$, standard large deviation estimates for the binomial distribution imply that

[TABLE]

for all large enough $n$. Let $a=a(r,p):=\frac{1}{2}(rp\log\frac{1}{1-p}-2)$, which is large for $p$ close to $1$, so for all large enough $n$, we have

[TABLE]

Letting $V^{\prime}$ be the number of coupons not collected through $an$ steps, we have

[TABLE]

where in the last line we use the assumption $r>4$ and take $p$ such that $rp/4>1$.

In anticipation of our variance bound, observe that for $n$ sufficiently large, by Taylor’s theorem, we have

[TABLE]

The probability that coupons labeled $1$ and $2$ (say) are not chosen through $an$ steps is $(1-2/2n)^{an}$, so we have for $p$ close to $1$,

[TABLE]

Finally, noting that $V\leq V^{\prime}\mathbf{1}_{\{N>an\}}+n\mathbf{1}_{\{N\leq an\}}$, we have

[TABLE]

∎

Lemma 19**.**

\thlabel

lem:AUB For any $\epsilon>0$ there exists $p^{\prime}(\epsilon)<1$ such that for each $p>p^{\prime}(\epsilon)$ and all sufficiently large $n$, we have

[TABLE]

Proof.

Let $r>4$ and $t_{p}$ be as in \threflem:coupon2. Let $B\overset{d}{=}\operatorname{Bin}(rt_{p},p)$ be the number of times that blue is chosen to move through time $t=rt_{p}$ (based on the values of $(Z_{s})_{s\leq t}$ defined at the start of Section 4). Like in the proof of \threflem:ALB, we let $\mathtt{Red\_Moved}$ be the number of red particles that have moved from their starting locations through time $t$, and let $\mathtt{Red\_Sites\_Visited\_by\_Blue}$ be the number of leaf vertices that were initially occupied by red particles and were visited by at least one blue particle through time $t$.

The number of times that blue particles are chosen to move from the core to a leaf through time $t$ must be at least $N=(B-n)/2$, provided particles persist through time $t$. To see this, let $C^{\rightarrow}$ be the number of jumps from the core to leaves up to time $t$, and $C^{\leftarrow}$ be the number of jumps from leaves to the core up to time $t$. Adding the equations

[TABLE]

and solving for $C^{\rightarrow}$ gives the claimed inequality $C^{\rightarrow}\geq N$. We now satisfy the hypotheses of \threflem:coupon2. Therefore,

[TABLE]

Moreover, by (25), for $p$ close to $1$ and sufficiently large $n$ we have

[TABLE]

Observe that $A_{t}\leq 2(n-\mathtt{Red\_Sites\_Visited\_by\_Blue}+\mathtt{Red\_Moved})$, since red particles that have not moved and are at sites that are visited by blue particles by time $t$ must be eliminated. Combining this observation with the last two inequalities gives

[TABLE]

Since we have $A_{t}\leq 2n$, we arrive at

[TABLE]

Combining this bound with \threflem:remy_bound applied at time $t=rt_{p}$ and Wald’s equation gives

[TABLE]

Taking $r$ close to $4$, then $p$ close enough to $1$, then $n$ sufficiently large completes the proof. ∎

5.3. Proofs

Proof of \threfthm:p.

The upper bound follows from \threflem:remy_bound by setting $t=0$ and taking expectation. The lower bound follows from similar reasoning as the proof of \threfthm:2 (i). \threflem:A_t,lem:TLB,lem:M_t together imply that

[TABLE]

Recalling the definition $D_{t}=|\sum_{1}^{t}Z_{s}|$ at the start of this section, we have

[TABLE]

Since $EZ_{1}=2p-1$, we then have

[TABLE]

Applying this inequality at (32), it follows that

[TABLE]

∎

Proof of \threfthm:asymptotic.

The lower bound follows from \threflem:ALB and the upper bound from \threflem:AUB. ∎

Proof of \threfprop:p=1 .

Let $M$ denote the value of $M_{t}$ at time $t=T_{1}^{2}(S_{2n})$. Evaluating \threflem:A_t at $t=T^{2}_{1}(S_{2n})$ and rearranging gives

[TABLE]

Let $M(i)$ be the number of times a particle is sampled at the core and moves without being annihilated when there are $2i$ particles in the system. Since red particles do not move, each time a particle at the core is sampled there is an $i/2n$ chance of annihilation, and annihilations cannot occur in any other way. It follows that $M(i)$ has distribution $X(i/2n)-1$ and thus

[TABLE]

In light of (33) and the above equality, we have

[TABLE]

which has expectation $4n\log n+4\gamma n+o(n)$. ∎

Acknowledgements

Thanks to Rick Durrett for helpful conversations, encouragement, and advice.