Proof. Let $(\Omega, \mathcal{F}, {\mathbb{P}})$ be a probability space on which $\boldsymbol{\Delta}=(\Delta_{x,l,j})_{x \in {\mathbb{Z}}, l,j \in {\mathbb{N}}}$ is a collection of i.i.d. random variables satisfying ${\mathbb{P}}(\Delta_{x,l,j}=1)={\mathbb{P}}(\Delta_{x,l,j}=-1)=1/2$ , and $(U_{x,l,k})_{x \in {\mathbb{Z}}, l,k \in {\mathbb{Z}}_+}$ be independent uniform random variables on [0, 1], independent of $\boldsymbol{\Delta}$ . Given $0<p\le p'\le 1$ , $\boldsymbol{\eta}\in {\mathbb{Z}}^{{\mathbb{Z}}_+}$ , and $S\subset {\mathbb{Z}}$ , let $X(p,\boldsymbol{\eta},S)$ denote the frog process with $\eta(x)$ frogs starting at x, with sites in S activated at time 0, and such that, once activated, the frog (x, l) with $1\le l\le \eta(x)$ conducts a lazy walk, stepping k time units after it was activated ( $k\ge 1$ ) if $U_{x,l,k}\le p$ . Its jth step is given by $\Delta_{x,l,j}$ . Also let $X'(p',\boldsymbol{\eta'},S')$ denote the frog process with $\eta'(x)$ frogs starting at x, with sites in S $^{\prime}$ activated at time 0, and such that, once activated, the frog (x, l) with $l\le \eta'(x)$ conducts a lazy walk, stepping k time units after it was activated ( $k\ge 1$ ) if $U_{x,l,k}\le p'$ . Its jth step is given by $\Delta_{x,l,j}$ .

By the construction on this space we have the following trivial facts:

• For fixed p, the range (set of vertices visited) of any particle/frog $\beta=(x,l)$ at time n is increasing in its time since activation.

• Each frog (x, l) has the same sequence of moves in the X process and the X $^{\prime}$ process (given by the $\Delta_{x,l,j}$ variables).

• If $U_{x,l,k}\le p$ then $U_{x,l,k}\le p'$ , so the range of a single p frog is dominated by that of the corresponding p $^{\prime}$ frog, relative to their times of activation.

We can thus conclude that all frogs in the X process are activated no earlier than in the X $^{\prime}$ process, and their ranges trail those in X $^{\prime}$ from that time forward.

Proof. By Lemma 1 it suffices to prove the result in the case $p=1$ , since the range is stochastically increasing in p.

Let $T_x$ denote the first time that an active frog reaches x. We use a (non-lazy) simple symmetric random walk $(S_k)_{k \ge 0}$ defined under a measure ${\mathbb{P}}'$ . Then, for $n \in {\mathbb{N}}$ the following holds:

\begin{align*}&{\mathbb{P}}^*\Big( T_{-n}\le n^{3/2}\Big)\\&= \sum_{y=0}^\infty \sum_{\ell=0}^\infty\sum_{k=0}^\infty\sum_{j=1}^{\infty}{\mathbb{P}}^*\big(T_{-n}\le n^{3/2},P(\!-n)=y,\eta(y)=\ell,\eta(y+1)=k,\eta(0)=j\big)\\&\le \sum_{y=0}^\infty \sum_{\ell\ge 0}\sum_{k\ge 0}\sum_{j\ge 1} \nu(\{\ell\})\nu(\{k\}){\mathbb{P}}^*(\eta(0)=j)\cdot(\ell+k+j) \cdot{\mathbb{P}}'\Big(\max_{1\le k\le n^{3/2}}S_k\ge n+y\Big)\\&\le \frac{3\mu}{(1-\nu(\{0\}))}\sum_{y=0}^\infty {\mathbb{P}}'\Big(\max_{1\le k\le n^{3/2}}S_k\ge n+y\Big),\end{align*}

where the first inequality is obtained by conditioning and using a union bound. Using a union bound, Chernoff’s bound, and $\cosh\!(t)\le \mathrm{e}^{t^2/2}$ yields, for any $t>0$ ,

\begin{align*}{\mathbb{P}}^*\!\Big( T_{-n}\le n^{3/2}\Big)&\le\frac{3\mu}{(1-\nu(\{0\}))}\sum_{y=0}^\infty \sum_{k\le n^{3/2}} \frac{(\!\cosh\!(t))^k}{\mathrm{e}^{t(n+y)}}\le \frac{3\mu}{(1-\nu(\{0\}))}n^{3/2}\sum_{y=0}^\infty \frac{\mathrm{e}^{\frac12 t^2n^{3/2}}}{\mathrm{e}^{t(n+y)}}.\end{align*}

Now take $t=(n+y)n^{-3/2}$ and $c=3\mu (1-\nu(\{0\}))^{-1}$ to see that

\begin{align*}{\mathbb{P}}^*\!\Big( T_{-n}\le n^{3/2}\Big)\le c\cdot n^{3/2}\cdot \sum_{y=0}^\infty\exp\!\bigg\{{-}\dfrac{(n+y)^2}{2n^{3/2}}\bigg\}.\end{align*}

By comparing the sum to an integral and using standard Gaussian tail bounds, we obtain

\begin{align}{\mathbb{P}}^*\!\Big( T_{-n}\le n^{3/2}\Big)&\le c n^{3/2}\left(\exp\!\bigg\{{-}\dfrac{n^2}{2n^{3/2}}\bigg\}+\int_0^\infty\exp\!\bigg\{{-}\dfrac{(n+y)^2}{2n^{3/2}}\bigg\}\mathrm{d}y\right)\nonumber\\&=cn^{3/2}\!\left(\exp\!\big\{{-}\sqrt{n}/2\big\}+\int_n^\infty\exp\!\bigg\{{-}\dfrac{y^2}{2n^{3/2}}\bigg\}\mathrm{d}y\right)\nonumber\\&\le cn^{3/2}\!\left( \exp\!\big\{{-}\sqrt{n}/2\big\}+\int_n^\infty\frac{y}{n}\exp\!\bigg\{{-}\dfrac{y^2}{2n^{3/2}}\bigg\}\mathrm{d}y\right)\nonumber\\&=cn^{3/2}\!\left(\exp\!\big\{{-}\sqrt{n}/2\big\}+\sqrt{n}\exp\!\big\{{-}n^2/(2n^{3/2})\big\}\right) \nonumber\\&\le 2cn^2\exp\!\big\{{-}\sqrt{n}/2\big\}.\nonumber\end{align}

By the Borel–Cantelli lemma, $T_{-n}>n^{3/2}$ for all but finitely many n, ${\mathbb{P}}^+$ -a.s., which implies that $n/T_{-n}$ converges to 0 ${\mathbb{P}}^+$ -almost surely, which implies that ${\mathbb{P}}^+(A_-(p)=0)=1$ for every p.

Let us prove the second statement. Since $\liminf_{n\to \infty}n^{-1}L_n=0$ , for any $\delta,\varepsilon>0$ there exists $m_0\in {\mathbb{N}}$ such that, for all $m\ge m_0$ ,

\[\mathbb{P}^*\!\left(\inf_{n\ge m_0}{\frac{L_n}{n}}<-\varepsilon\right)<\delta.\]

As we work in discrete time, for all $0\le n\le m$ , $L_n+m\ge0$ almost surely, which proves the second statement.

Proof. Let $\bar{A}_+(p)=\limsup_{n \to \infty}n^{-1}R_n(p)$ denote the $\limsup$ of the rightmost visited site up to time n. By Lemma 2 and the definition of ${\mathbb{P}}^{*-}$ we have ${\mathbb{P}}^{*-}(\bar{A}_+(p)=0)=1$ . By Lemma 1 this implies that for the one-sided model with $\eta(x)$ frogs per site ( $x\le 0$ ) and with all frogs activated at time zero, also $\bar{A}_+(p)=0$ a.s.

Now let $R^{(\!-)}_n$ denote the rightmost point reached by frogs originating from sites $x<0$ up to time n. For the two-sided model under ${\mathbb{P}}$ , this quantity is stochastically dominated by the same quantity when all such frogs are activated at time 0. But the latter grows sublinearly as above ( $\bar{A}_+(p)=0$ a.s.). Since ${\mathbb{P}}(A_+(p)=A(p))=1$ and $A(p)>0$ , this shows that there exists an a.s. finite random variable K such that ${\mathbb{P}}(P(x)\ge 0 \mathrm{\ for\ all\ } x\ge K)=1$ . The claim now follows since ${\mathbb{P}}(K>k_\delta)<\delta$ for $k_\delta$ sufficiently large.

Proof. Let $\delta>0$ . From Corollary 2 there exists $k_{\delta} \in {\mathbb{N}}$ such that ${\mathbb{P}}(P(x)\ge 0 \mathrm{\ for\ all\ } x\ge k_{\delta})\ge 1-\delta$ . Since the activation time $T^+_{k_{\delta}}$ of $k_{\delta}$ in the one-sided model is a.s. finite, there exists $t_{\delta}>0$ such that ${\mathbb{P}}^+(T^+_{k_{\delta}}\le t_\delta)\ge 1-\delta$ . Both of these statements are conditional on $\eta(0)\ge 1$ , with (only) the origin activated at time 0.

A natural coupling $\overline{{\mathbb{P}}}$ of these two processes is as follows. Under $\overline{\mathbb{P}}$ , the initial configuration for X is $(\eta(x))_{ x\in\mathbb{Z}}$ which are $\mathrm{i.i.d.} \sim \nu$ , with $\eta(0)$ conditioned to be strictly positive. The initial configuration $(\eta^+(x))_{x\in\mathbb{Z}}$ for the one-sided process is defined as $\eta^+(x)=0$ if $x<0$ and $\eta^+(x)=\eta(x)$ otherwise. On top of this structure define two frog processes X and $X^+$ from an i.i.d. collection of steps $\boldsymbol{\Delta}$ as in Lemma 1, with $S=S^+=\{0\}$ , so that X dominates $X^+$ . On this space we therefore have

(1) \begin{equation}\overline{\mathbb{P}}\Bigg(\bigcap_{x\ge k_{\delta}}\{P(x)\ge 0\}, \, T^+_{k_{\delta}}\le t_\delta\Bigg)\ge 1-2\delta.\end{equation}

Now note that $\bigcap_{x\ge k_{\delta}}\{P(x)\ge 0\}\cap\Big\{ T^+_{k_{\delta}}\le t_\delta\Big\}\subset \bigcap_{x\ge k_{\delta}}\{T_x\le T^+_x\le T_x+t_{\delta}\}$ , where $T_x$ denotes the activation time of x by the two-sided process. Since $R_{T_n}/T_n=n/T_n\to A(p)$ a.s., this shows that $T^+_n/n\to A(p)^{-1}$ a.s. on the event in (1), and therefore also $R_m^+/m \to A(p)$ a.s. on this event (consider $m\in [T^+_n,T^+_{n+1})$ ). This shows that ${\mathbb{P}}^+(A^+(p)=A(p))=\overline{\mathbb{P}}(A^+(p)=A(p))\ge 1-2\delta$ . Since $\delta$ was arbitrary, this completes the proof.

Proof. On the almost sure (by Lemma 3) event $\{A_+(p)=A(p)\}$ , $\inf_{n\ge m}R_n/n\to A(p)$ almost surely as m goes to infinity. Thus, there exists $m_1<\infty$ such that, for all $m\ge m_1$ ,

\[ \mathbb{P}^+\bigg(A_+(p)=A(p), \inf_{n\ge m} \frac{R_n}{n}>\frac{A(p)}{2}\bigg)>1-\delta. \]

Now, for all $0\le n\le m$ , $(R_n+m)/n\ge m/n\ge1$ , which gives the conclusion.

Proof of Theorem 1. We choose to provide a detailed proof of our main theorem, but the idea is fairly simple and easily obtained from Lemma 2, Corollary 3 and Remark 1.

Let $p_1,p_2>0$ be given. Under the measure ${\mathbb{P}}_{p_1,p_2}$ , let the following collection of random variables all be mutually independent:

• $(\eta(x))_{x\in {\mathbb{Z}}\setminus \{0\}}$ , all $\sim \nu$ and with $\eta_1\!(0)$ , $\eta_2(0)$ having distribution $\nu$ conditional on being strictly positive;

• $\boldsymbol{\Delta}=(\Delta_{x,l,j})_{x \in {\mathbb{Z}}, l,j \in {\mathbb{N}}}$ satisfying ${\mathbb{P}}_{p_1,p_2}(\Delta_{x,l,j}=1)={\mathbb{P}}_{p_1,p_2}(\Delta_{x,l,j}=-1)=1/2$ ;

• $(U_{x,l,k})_{x \in {\mathbb{Z}}, l,k \in {\mathbb{Z}}_+}$ are independent standard uniform random variables.

Define

\[\eta_1(x)=\begin{cases}\eta(x) & \text{ if }x>0 , \\ \\[-8pt] 0 & \text{ if }x<0,\end{cases}\qquad \eta_2(x)=\begin{cases}0 & \text{ if }x>0 , \\ \\[-8pt] \eta(x) & \text{ if }x<0.\end{cases}\]

We will define three processes on this space:

• $X^1$ denotes a one-sided one-type frog process with initial configuration $\boldsymbol{\eta_1}$ (with only the frogs at 0 activated at time 0) such that, once activated, the frog (x, l) with $1\le l\le \eta_1(x)$ conducts a lazy walk, stepping on the kth attempt ( $k\ge 1$ ) if $U_{x,l,k}\le p_1$ . Its jth step is given by $\Delta_{x,l,j}$ .

• $X^2$ denotes a one-sided one-type frog process with initial configuration $\boldsymbol{\eta_2}$ (with only the frogs at 0 activated at time 0, and where the $\eta_2(0)$ frogs started at 0 are labelled $(0,\eta_1(0)+1), \dots ,(0,\eta_1(0)+\eta_2(0)$ ), such that, once activated, the frog (x, l) (with $x\ne 0$ and $1\le l\le \eta_1(x)$ , or with $x=0$ and $l=\eta_1(0)+1,\dots, \eta_1(0)+\eta_2(0)$ ) conducts a lazy walk, stepping on the kth attempt ( $k\ge 1$ ) if $U_{x,l,k}\le p_2$ . Its jth step is given by $\Delta_{x,l,j}$ .

• X denotes a two-sided two-type frog process with initial configuration $\boldsymbol{\eta}$ (with only the frogs at 0 activated at time 0), where the frogs labelled $(0,1),\dots, (0,\eta_1(0))$ are type 1, and those labelled $(0,\eta_1(0)+1),\dots, (0,\eta_1(0)+\eta_2(0))$ are type 2. Once activated, if activated by a frog of type $i\in \{1,2\}$ , the frog (x, l) $1\le l\le \eta(x)$ conducts a lazy walk, stepping on the kth attempt ( $k\ge 1$ ) if $U_{x,l,k}\le p_i$ . If $k_i$ frogs of type i land on a previously unvisited site y at the same time, we use the random variable $U_{y,0,0}$ to choose the activator—all dormant frogs at y become type 2 if $U_{y,0,0}\le k_2/(k_1+k_2)$ , and otherwise they all become type 1.

Let $\tau$ be the first time (possibly infinite) that in the process X a frog of type 2 activates a site in ${\mathbb{N}}$ (i.e. a type 2 frog arrives at a positive site before any type 1 frog has), or a frog of type 1 activates a site in $-{\mathbb{N}}$ . On the event $\{\tau=\infty\}$ for each $i \in \{1,2\}$ we have that the moves of all the frogs in the frog process $X^i$ follow exactly those of the frogs of type i in the frog process X.

Let $R^{i}_n$ and $L^{i}_n$ be the rightmost and leftmost positions visited by active frogs of type $i\in \{1,2\}$ up to time n for the $X^i$ process. We will show that

(2) \begin{equation}{\mathbb{P}}_{p_1,p_2}\bigg(\sup_{n\in {\mathbb{N}}}\Big(R_n^{1}\wedge -L_n^{2}\Big)=\infty ,\tau=\infty\bigg)>0.\end{equation}

On the event in (2) we have coexistence because type 1 frogs activate every positive site, and frogs of type 2 every negative site. Clearly the first event in (2) has probability 1 since simple random walk has infinite range almost surely (as a Markov chain on an infinite irreducible class, $\mathbb{Z}$ ), so we need only show that ${\mathbb{P}}_{p_1,p_2}(\tau=\infty)>0$ .

Let $\ell_{\min}=\min\{\ell\ge0 \colon \nu(\{\ell\})>0\}$ and $\ell^+_{\min}=\min\{\ell>0 \colon \nu(\{\ell\})>0\}$ . Define $\varepsilon=\min\{A(p_1), A(p_2)\}/4$ and let us fix $m\in{\mathbb{N}}$ so that $m= m_0(\varepsilon,1/5)+m_1(p_1,\nu(\{\ell^+_{\min}\})/5)+m_1(p_2,\nu(\{\ell^+_{\min}\})/5)$ , where $m_0$ and $m_1$ are given by Lemma 2 and Corollary 3.

Now let $A^1_{m}$ be the event that $\eta_1(0)=\ell^+_{\min}$ , $\eta(x)=\ell_{\min}$ for all $1\le x< 2m$ , and, in the first 2m time units for the process X, the frogs of type 1 at the origin all take 2m steps to the right while all the activated frogs from the region $[1,2m-1]$ take alternate steps left and right (in that order). Let $A^2_{m}$ be the event that $\eta_2(0)=\ell^+_{\min}$ , $\eta(x)=\ell_{\min}$ for all $-2m< x\le -1$ , and, in the first 2m time units, the frogs of type 2 at the origin all take 2m steps to the left while all the activated frogs from the region $[\!-(2m-1),-1]$ alternate stepping right and left (in that order). Let $A_{m}=A^1_{m}\cap A^2_{m}$ (see Fig. 2).

Figure 2.

A depiction of the event $A_{{m}}$ when $m=2$ and $\ell_{\min}=1$ . The first (resp. second) diagram shows an example of an initial configuration $\boldsymbol{\eta}_1$ (resp. $\boldsymbol{\eta}_2$ ) on the event $A_{{m}}$ . The third diagram shows the corresponding configuration of frogs at time $2m=4$ , in particular, the original type 1 (resp. type 2) frog at the origin at time 0 is now at 2m (resp. $-2m$ ).

Recall that $\varepsilon=\min\{A(p_1), A(p_2)\}/4$ , and define

\begin{alignat*}{2} B_m^{1,\mathrm{l}} & = \left\{\inf_{n\ge0}{\frac{L_{n+2m}^{1}+m}{n}}\ge-\varepsilon\right\} , & B_m^{1,\mathrm{r}} & = \left\{ \inf_{n\ge0} \frac{R_{n+2m}^{1}-m}{n}>\frac{A(p_1)}{2}\right\} , \\ B_m^{2,\mathrm{l}} & = \left\{ \sup_{n\ge0} \frac{L_{n+2m}^{2}+m}{n}<-\frac{A(p_2)}{2}\right\} , \qquad & B_m^{2,\mathrm{r}} & = \left\{\sup_{n\ge0}{\frac{R_{n+2m}^{2}-m}{n}}\le\varepsilon\right\}.\end{alignat*}

Also define $B_m^i=B_m^{i,\mathrm{l}}\cap B_m^{i,\mathrm{r}}$ and $B_m=B_m^1\cap B_m^2$ . Note that $A_m\subset \{\tau>2m\}$ , while on $B_m$ we have, for any $n\ge 0$ , $L_{n+2m}^{1}\ge -n\varepsilon-m> -nA(p_2)/2-m\ge L_{n+2m}^{2}$ and similarly $R_{n+2m}^{1}> R_{n+2m}^{2}$ , so $\tau\ne n+2m$ for any n on $B_m$ . This shows that $A_m\cap B_m\subset \{\tau=\infty\}$ .

It therefore remains to show that ${\mathbb{P}}_{p_1,p_2}(A_m\cap B_m)>0$ . The event $A_m$ does not depend on $\eta(x)=\eta_i(x)$ for $|x|\ge 2m$ . On the other hand, we have that ${\mathbb{P}}_{p_1,p_2}(A_m)>0$ for all $m\ge1$ , since $A_m$ only prescribes the value of the environment in a finite box and requires a bounded number of frogs to perform a fixed finite set of moves. We say that a ${\mathbb{Z}}$ -valued process $\boldsymbol{W}=(W_n)_{n \ge 0}$ stochastically dominates $\boldsymbol{Y}=(Y_n)_{n \ge 0}$ if there exists a probability space with processes $\boldsymbol{W}'\sim \boldsymbol{W}$ and $\boldsymbol{W}'\sim \boldsymbol{Y}$ on which $W'_n\ge Y'_n$ for every n a.s. Let us briefly justify that, conditional on $A_m$ , the following dominations hold:

• $\left(R^{1}_{n+2m}-2m\right)_{n\ge0}$ stochastically dominates the rightmost position of a one-sided frog process $\left(R_{n}\right)_{n\ge0}$ under $\mathbb{P}^+_{p_1}$ , conditioned on $\Big\{\eta(0)=\ell^+_{\min}\Big\}$ .

• $\left(L^{1}_{n+2m}\right)_{n\ge 0}$ stochastically dominates $\left(L_{n}\right)_{n\ge0}$ under $\mathbb{P}^*_{p_1}$ .

• $\left(\!-L^{2}_{n+2m}-2m\right)_{n\ge0}$ stochastically dominates the rightmost position of a one-sided frog process $\left(R_{n}\right)_{n\ge0}$ under $\mathbb{P}^+_{p_2}$ , conditioned on $\{\eta(0)=\ell^+_{\min}\}$ .

• $\left(\!-R^{2}_{n+2m}\right)_{n\ge0}$ stochastically dominates the leftmost position of a one-sided frog process $\left(L_{n}\right)_{n\ge0}$ under $\mathbb{P}^*_{p_2}$ .

We will prove the first two items, as the other two follow by symmetry. For this purpose, let us denote by $N_1(x)$ the number of active frogs on site x for the $X^1$ process at time 2m; see Fig. 2.

Note that, on $A_m$ , at time 2m, $N_1(x)=0$ for $x\le-1$ and x odd, $N_1(0)=\ell_{\min}$ , $N_1(x)=2\ell_{\min}$ for all even $x\in [1,2m-1]$ , and finally $N_1(2m)=\eta_1(2m)+\ell_{\min}^+$ . On all $x\ge 2m+1$ , there are $\eta_1(x)$ dormant frogs.

The first item follows by the monotonicity provided by Lemma 1 and using that $\eta_1(x)$ has the same law as $\eta_1(x-2m)$ for all $x>2m$ , as there are at least $\ell^+_{\min}$ active frogs on 2m. To prove the second item, note that, for every $x\ge 0$ , $N_1(x)\le \eta_1(0)+\eta_1(x)+\eta_1(x+1)$ , and use the definition of $\ell^+_{\min}$ together with Lemma 1.

Thus, by our choice of $\varepsilon$ and m, using Lemma 2 we obtain

\begin{align*}{\mathbb{P}}_{p_1,p_2}\big(\big(B^{1,\mathrm{l}}_m\big)^\mathrm{c} \mid A_m\big)\le {\mathbb{P}}_{p_1}^*\!\bigg(\inf_{n\ge 0}\frac{L_n+m}{n}<-\varepsilon\bigg)\le \frac15.\end{align*}

Similarly, by Corollary 3 together with Remark 1, we obtain

\begin{align*}{\mathbb{P}}_{p_1,p_2}\big(\big(B^{1,\mathrm{r}}_m\big)^\mathrm{c} \mid A_m\big) & = {\mathbb{P}}_{p_1,p_2}\bigg(\inf_{n\ge 0}\frac{R^{+1}_{n+2m}-2m+m}{n}\le \frac{A(p_1)}{2} \, \Big| \, A_m\bigg) \\&\le {\mathbb{P}}^+_{p_1}\bigg(\inf_{n\ge 0}\frac{R_n+m}{n}\le \frac{A(p_1)}{2} \, \Big| \, \eta(0)=\ell^+_{\min}\bigg)\le \frac15.\end{align*}

Similarly, ${\mathbb{P}}_{p_1,p_2}\big(\big(B^{2,\mathrm{r}}_m\big)^\mathrm{c}\mid A_m\big) \le 1/5$ and ${\mathbb{P}}_{p_1,p_2}\big(\big(B^{2,\mathrm{l}}_m\big)^\mathrm{c}\mid A_m\big)\le 1/5$ , and thus ${\mathbb{P}}_{p_1,p_2}(B_m \mid A_m)\ge1/5$ . This proves that ${\mathbb{P}}_{p_1,p_2}\left(A_m\cap B_m\right)>0$ as claimed.

Proof of Proposition 1. Recall that $\nu$ is the probability measure on ${\mathbb{Z}}_+$ that is the law of $\eta(x)$ . Let $\mu={\mathbb{E}}[\eta(x)]<\infty$ .

Consider a branching random walk where each individual (independently of other particles) has:

1. (i) exactly one offspring, with displacement 0, with probability $1-p$ ;

2. (ii) exactly $1+k$ offspring, all at displacement $+1$ , with probability $\nu(\{k\}){p}/{2}$ ;

3. (iii) exactly $1+k$ offspring at displacement ${-}1$ , with probability $\nu(\{k\}){p}/{2}$ ,

and then the parent particle dies immediately. The number of offspring is always at least 1, with mean $1+p\mu$ . It is easy to couple this branching random walk (BRW) with our lazy frog model such that the set of visited points in the BRW contains that of the frog model. To see this, note that in the lazy frog model an active frog at x either doesn’t move (with probability $1-p$ ), and then it doesn’t activate any new frogs, or it moves left or right with probability $p/2$ each, and then it activates at most $\eta(x+1)$ or $\eta(x-1)$ new frogs.

It is therefore sufficient to show that the speed of the front (the maximal site visited) for the BRW model goes to zero as $p\downarrow 0$ . We will verify this by applying small modifications to standard results (in particular [Reference Shi8, Lemma 1.5]) that are typically stated for the position of the rightmost particle at time n (so this differs slightly from the quantity we are after).

First, note that since $R_n/n$ converges almost surely and is bounded above by 1 for every n, we have, by dominated convergence, that the limiting speed A(p) for the lazy frog model with parameter p is equal to $\limsup_{n \to \infty}{\mathbb{E}}_p[n^{-1}R_n]$ . By the above coupling it is therefore enough for us to show that $\limsup_{n \to \infty}{\mathbb{E}}_p[n^{-1}M^*_n]\to 0$ as $p \to 0$ , where $M^*_n$ is the largest site visited by the BRW up to (and including) time n. Let $|x|$ denote the generation of a particle x in the BRW, and V(x) denote the location of x. Then

\begin{align*} \frac{1}{n}{\mathbb{E}}_p[M^*_n] & = \frac{1}{n}{\mathbb{E}}_p\Big[\max_{x:|x|\le n} V(x)\Big]\le \frac{1}{n}\log\!\Big({\mathbb{E}}_p\Big[\mathrm{e}^{\max_{x:|x|\le n} V(x)}\Big]\Big) \\&\le \frac{1}{n}\log\!\Bigg({\mathbb{E}}_p\Bigg[\sum_{x:|x|\le n} \mathrm{e}^{V(x)}\Bigg]\Bigg) = \frac{1}{n}\log\!\Bigg(\sum_{k=0}^n {\mathbb{E}}_p\Bigg[\sum_{x:|x|=k} \mathrm{e}^{V(x)}\Bigg]\Bigg).\end{align*}

By conditioning on generation $k-1$ of the branching process we have

\begin{equation*}{\mathbb{E}}_p\Bigg[\sum_{x:|x|=k} \mathrm{e}^{V(x)}\Bigg] = {\mathbb{E}}_p\Bigg[\sum_{x:|x|=k-1} \mathrm{e}^{V(x)}\Bigg] \mathrm{e}^{\psi_p(1)},\end{equation*}

where, for $t>0$ ,

\begin{equation*}\psi_p(t)=\log {\mathbb{E}}_p\Bigg[\sum_{x:|x|=1} \mathrm{e}^{V(x)}\Bigg] = \log\!\left( (1-p) +\frac{p}{2}\sum_{k=0}^\infty \nu(\{k\})(k+1)(\mathrm{e}^{-t} + \mathrm{e}^{t})\right).\end{equation*}

Thus, by induction, we obtain ${\mathbb{E}}_p\big[\sum_{x:|x|=k} \mathrm{e}^{V(x)}\big]=\mathrm{e}^{k \psi_p(1)}$ . Alternatively, we could have applied the many-to-one lemma (see [Reference Shi8, Theorem 1.1]).

Thus, we have $\frac{1}{n}{\mathbb{E}}_p[M^*_n] \le \frac{1}{n}\log\!\big(\sum_{k=0}^n \mathrm{e}^{k\psi_p(1)}\big)$ , where

\begin{align*}\psi_p(1)=\log\!\left(1+p\bigg[\frac{1}{2}(\mathrm{e}^{-1}+\mathrm{e})(1+\mu)-1\bigg]\right)\ge 0.\end{align*}

Thus,

\begin{align*}\frac{1}{n}{\mathbb{E}}_p[M^*_n]&\le \frac{1}{n}\log\!\big((n+1)\mathrm{e}^{n\psi_p(1)}\big)=\frac{\log(n+1)}{n}+\psi_p(1).\end{align*}

Since $\log\!(1+x)\le x$ , we see that $\psi_p(1)$ can be made arbitrarily small by making p arbitrarily small, which completes the proof.