Proof. The stochastic domination in (20) is the consequence of a standard coupling argument: construct the graphical construction of ${\mathrm {CP}}_{f_2,\lambda }(G,{\underline {\xi }}_0)$ , that is, of the process with higher infection rates $(\lambda /f_2(u,v))_{u,v}$ . Then, independently for different pairs $uv$ , on $\mathrm {PPP}_{uv}$ , keep every infection event (point) with probability $(\lambda /f_1(u,v)) / (\lambda /f_2(u,v) ) = f_2(u,v)/f_1(u,v)$ , independently across points. The thinned PPP has rate $\lambda f_1(u,v)$ , hence we obtain a graphical construction of ${\mathrm {CP}}_{f_1,\lambda }(G,{\underline {\xi }}_0)$ . This joint realization of the two processes gives a coupling of ${\mathrm {CP}}_{f_1,\lambda }(G,{\underline {\xi }}_0)$ and ${\mathrm {CP}}_{f_2,\lambda }(G,{\underline {\xi }}_0)$ , so that every infection event in the former process is also an infection event in the latter process. This finishes the proof of (20).

Proof. Proof of part (a). We argue by induction in $m = \mathfrak {l}(\pi )$ . In case $m=0$ , we have $\pi = (u_0)$ , and the process $(y_s(\pi ))_{s \ge 0}$ is a continuous-time Markov chain that starts at $y_0(\pi ) = x_0(u_0)$ at time 0 and can only decrease, doing so with rate $y_s(\pi )$ at any time $s \ge 0$ . If we interpret the state of this chain as a number of particles, where each particle dies with rate 1 (and no particles are born), then the probability that a particle is still alive at time t is $\mathrm {e}^{-t}$ , so the expected number of living particles at time t is $x_0(u_0)\mathrm {e}^{-t}$ , as desired.

Now assume that $m \ge 1$ and the statement in (25) holds for all $\pi '\in \mathscr {T}$ with $\mathfrak {l}(\pi ')\le m-1$ . Let

$$\begin{align*}\pi_0 = (u_0),\quad \pi_1 = (u_0,u_1),\quad \ldots, \quad \pi_m = \pi = (u_0,\ldots,u_m),\end{align*}$$

and let $\mathcal {F}$ be the $\sigma $ -algebra generated by

$$\begin{align*}\{y_s(\pi_i): 1 \le i \le m-1,\; s \ge 0\}.\end{align*}$$

Conditioned on $\mathcal {F}$ , the process $(y_s(\pi ))_{s \ge 0}$ is an ${\mathbb {N}}$ -valued (time-inhomogeneous) Markov process that starts at $0$ at time $0$ and, at any time $s \ge 0$ , increases by $1$ with rate

$$\begin{align*}y_s(\pi_{m-1})r(\mathfrak{s}(\mathfrak{p}(\pi)),\mathfrak{s}(\pi)) = y_s(\pi_{m-1})r(u_{m-1},u_m),\end{align*}$$

and decreases by $1$ with rate $y_s(\pi )$ , by (23) and (22). Again seeing this process as counting particles (which as before die with rate 1, but now can also be born with a time-dependent rate), the conditional expectation of the number of particles at time t is

$$\begin{align*}\mathbb{E}[y_t(\pi)\mid \mathcal{F}] = r(u_{m-1},u_m)\cdot \int_0^t y_s(\pi_{m-1})\cdot \mathrm{e}^{-(t-s)}\;\mathrm{d}s.\end{align*}$$

Taking expectation and using Tonelli’s theorem, this gives

$$\begin{align*}\mathbb{E}[y_t(\pi)] = r(u_{m-1},u_m)\cdot \int_0^t \mathbb{E}[y_s(\pi_{m-1})]\cdot \mathrm{e}^{-(t-s)}\;\mathrm{d}s.\end{align*}$$

Using this recursively m times, and then using the base induction case $\mathbb {E}[y_s(\pi _0)] = x_0(u_0)\mathrm {e}^{-s}$ , we obtain

$$ \begin{align*} &\mathbb{E}[y_t(\pi)] \\& \quad = x_0(u_0) \prod_{i=0}^{m-1}r(u_{i-1},u_i) \int_0^t \int_{s_1}^t \cdots \int_{s_{m-1}}^t \mathrm{e}^{-s_1} \mathrm{e}^{-(s_2-s_1)}\cdots \mathrm{e}^{-(s_m-s_{m-1})} \mathrm{e}^{-(t-s_m)}\;\mathrm{d}s_m\cdots \mathrm{d}s_1\\& \quad = x_0(u_0) \prod_{i=0}^{m-1}r(u_{i-1},u_i)\cdot \mathrm{e}^{-t}\cdot \frac{t^m}{m!}. \end{align*} $$

Proof of part (b). In case $\mathfrak {l}(\pi ) = 0$ , the statement is obvious, using the fact that $y_0((v)) = x_0(v)$ for all v. Assume that $\mathfrak {l}(\pi ) = m \ge 1$ , and write $\pi = (u_0,\ldots ,u_{m})$ . Since the transition occurs with rate $\underline {y}(\mathfrak {p}(\pi ))\cdot r(u_{m-1},u_{m})$ by (23), we have

$$ \begin{align*} \mathbb{E}[Z(\pi)] &= r(u_{m-1},u_{m})\cdot \mathbb{E}\left[\int_0^\infty y_t(\mathfrak{p}(\pi))\;\mathrm{d}t\right]\\ &= r(u_{m-1},u_{m})\cdot \mathbb{E}\left[\int_0^\infty y_t((u_0,\ldots,u_{m-1}))\;\mathrm{d}t\right]. \end{align*} $$

Using Tonelli’s theorem and (25) on the right-hand side, we obtain

$$ \begin{align*} \mathbb{E}[Z(\pi)] = r(u_{m-1},u_{m}) \cdot \left(x_0(u_0) \prod_{i=0}^{m-2}r(u_i,u_{i+1}) \right) \cdot \underbrace{\int_0^\infty \frac{t^{m-1}}{(m-1)!}\mathrm{e}^{-t}\;\mathrm{d}t}_{=1}, \end{align*} $$

as desired.

Proof. Let

$$\begin{align*}\unicode{x3c4}:= \inf\{s \ge t:\;y_s(\pi)> 0\},\end{align*}$$

so that the left-hand side of (28) equals $\mathbb {P}(\unicode{x3c4} < \infty )$ . Next, define the event

$$\begin{align*}\mathcal{A}:= \{\unicode{x3c4} < \infty,\; y_s(\pi)> 0 \text{ for all } s \in [\unicode{x3c4},\unicode{x3c4}+1]\}.\end{align*}$$

It is easy to check that

(29) $$ \begin{align} \mathbb{P}(\mathcal{A}) \ge \mathbb{P}(\unicode{x3c4} < \infty) \cdot \mathrm{e}^{-1}. \end{align} $$

Using first Tonelli’s theorem and then the fact that $y_s(\pi )$ is integer-valued, we have

Now, the right-hand side is bounded from below by

where the equality follows from the definition of $\mathcal {A}$ . Also using (29), we have thus obtained

$$\begin{align*}\mathbb{P}(y_s(\pi)> 0 \text{ for some } s \ge t)=\mathbb{P}(\unicode{x3c4} < \infty) \le \mathrm{e} \int_t^\infty \mathbb{E}[y_s(\pi)]\;\mathrm{d}s.\end{align*}$$

The desired bound in (28) now follows from (25) in Lemma 3.6 (a).

Claim 4.1 (Supermartingale for global extinction).

Let $f(x,y)=a x^\mu y^\nu $ for some $a> 0$ and $\mu ,\nu \ge 0$ such that $\mu +\nu \ge 1$ , and let $G=(V,E)$ be an arbitrary finite graph. Consider the process $(x_t(v))_{t\ge 0, v\in V}={\mathrm {BRW}}_{f,\lambda }(G, {\underline {x}}_0)$ for $\lambda>0$ on a finite graph G, starting from a given state $\underline {x}_0\in {\mathbb {N}}^{V}$ . Define, for any $\alpha \in [1-\mu ,\nu ]$ ,

$$\begin{align*}M_t:=\sum_{v\in V}x_t(v)d_v^\alpha.\end{align*}$$

Then, whenever $\sum _{v\in V}x_0(v)d_v^\alpha <\infty $ , the process $(M_t)_{t\ge 0}$ is a supermartingale with respect to the filtration $\mathcal {F}_t=\sigma \big ((x_s(v))_{v\in V(G),s\le t}\big )$ for all $\lambda \in (0,a]$ and a strict supermartingale when $\lambda \in (0,a)$ .

Proof. We start by observing that the interval $[1-\mu ,\nu ]$ is nonempty since $\mu +\nu \ge 1$ . To prove the supermartingale property we analyze the expected increments of $(M_t)_{t\ge 0}$ , using the definition of ${\mathrm {BRW}}_{f,\lambda }$ in Def. 1.2. Here we use the transition rates for computations instead of the construction in Definition 3.5. This is justified since we assume that the graph is finite. The change in $M_t$ may come from either a particle disappearing at v due to a death event, or from a new particle appearing at v due to reproduction events from neighboring particles. We obtain, using the rates in (3) and (4) with $r(u,v)=\lambda e(u,v)/f(d_u,d_v)$ , that

$$ \begin{align*} \mathbb{E}[M_{t+\mathrm{d}t}-M_t \mid \mathcal{F}_t]&=\mathbb{E}\left[\left.\sum_{v\in V(G)} (x_{t+\mathrm{d}t}(v)-x_t(v))d_v^\alpha \;\right|\; \mathcal{F}_t\right]\\ &=\sum_{v\in V(G)}\Bigg(-x_t(v)+\sum_{u\in N(v)} x_t(u)r(u,v)\Bigg)\mathrm{d}t\cdot d_v^\alpha\\ &=-M_t\mathrm{d}t+\sum_{v\in V(G)}\sum_{u\in N(v)}x_t(u)\cdot [\lambda e(u,v)/f(d_u,d_v)]\cdot d_v^\alpha\mathrm{d}t. \end{align*} $$

We substitute $f(d_u,d_v)=a d_u^{\mu }d_v^{\nu }$ in the last line above, and use that $d_v^{\alpha -\nu }\le 1$ by the assumption that $\alpha \le \nu $ to obtain that:

$$ \begin{align*} \mathbb{E}[M_{t+\mathrm{d}t}-M_t \mid \mathcal{F}_t] &=-M_t\mathrm{d}t+(\lambda/a)\cdot\sum_{v\in V(G)}\sum_{u\in N(v)}x_t(u) e(u,v)d_u^{-\mu}d_v^{\alpha-\nu}\mathrm{d}t\\ &\le-M_t\mathrm{d}t+(\lambda/a)\cdot\sum_{v\in V(G)}\sum_{u\in N(v)}x_t(u) e(u,v)d_u^{-\mu}\mathrm{d}t. \end{align*} $$

Exchanging the sums and using that $\sum _{v\in V}e(u,v)=d_u$ (see Notation in Section 1), we obtain

$$ \begin{align*} \mathbb{E}[M_{t+\mathrm{d}t}-M_t \mid \mathcal{F}_t]\le-M_t\mathrm{d}t+(\lambda/a)\cdot\sum_{u\in V(G)} x_t(u) d_u^{1-\mu}\mathrm{d}t. \end{align*} $$

Finally, since $d_u\ge 0$ is an integer, $d_u^{1-\mu }\le d_u^{\alpha }$ holds by the assumption $1-\mu \le \alpha $ . Hence,

(31) $$ \begin{align} \mathbb{E}[M_{t+\mathrm{d}t}-M_t \mid \mathcal{F}_t]\le -M_t\mathrm{d}t+(\lambda/a)\cdot\sum_{u\in V(G)} x_t(u) d_u^\alpha\mathrm{d}t=[(\lambda/a)-1]\cdot M_t\mathrm{d}t. \end{align} $$

Since $M_t\ge 0$ , for $\lambda \le a$ we obtain the supermartingale property, as $[(\lambda /a)-1]\cdot M_t\mathrm {d}t\le 0$ , with strict inequality when $\lambda <a$ . The finiteness of the initial state $M_0$ is ensured by the assumption that $M_0=\sum _{v\in V} x_0(v)d_v^\alpha <\infty $ . This finishes the proof.

Proof of Theorem 2.2.

Let G be a finite graph. Without loss of generality we may assume that all vertices in G have degree at least $1$ . Indeed, if G contained a (finite) number of vertices with degree $0$ , the contact process on those, starting from any ${\underline {\xi }}_0$ with finitely many infected vertices, reduces to a pure death process where each particle dies at rate $1$ . This is because infection cannot happen to and from these vertices. This process goes almost surely extinct. Hence we assume wlog that $d_v\ge 1$ for all $v\in V$ .

By Lemma 3.8, it is sufficient to prove the almost sure extinction of ${\mathrm {BRW}}_{f,\lambda }(G, {\underline {\xi }}_0)$ for any $\xi _0$ that is almost surely finite, that is, $\sum _{v\in V} \xi _0(v)<\infty $ almost surely. Fix now any such realization of the initial state. Then, since only finitely many coordinates are nonzero, $\sum _{v\in V} \xi _0(v) d_v^\alpha =\sum _{v\in V} x_0(v) d_v^\alpha <\infty $ also holds for any $\alpha>0$ . We assumed $\lambda \in (0,1)$ also in Theorem 2.2. Hence, the conditions of Claim 4.1 are satisfied with $\nu =\mu \ge 1/2$ , and we can set $\alpha =1-\mu $ there to obtain the non-negative (strict) supermartingale $(M_t)_{t\ge 0}$ , (i.e., not a martingale).

Apply Doob’s martingale convergence theorem for the non-negative supermartingale $(M_t)_{t\ge 0}$ . Since $(M_t)_{t\ge 0}\ge 0$ cannot take values in $(0,1)$ , (as $d_v\ge 1$ and $x_t(v)\in {\mathbb {N}}$ for all v) its almost sure limit can only be $0$ . Therefore, almost surely $M_t=0$ for large enough t. By the coupling between ${\mathrm {CP}}_{f,\lambda }$ and ${\mathrm {BRW}}_{f,\lambda }$ and that $d_v^{1-\mu }\ge 1$ whenever $d_v\ge 1$ , we obtain that

$$\begin{align*}\sum_{v\in V} \xi_t(v) \le \sum_{v\in V} x_t(v)d_v^{1-\mu} =M_t \ {\buildrel a.s. \over \longrightarrow}\ 0 \end{align*}$$

implying global extinction. To compute the extinction time, by definition, $T_{\mathrm {ext}}(G, {\underline {\xi }}_0)\ge t$ implies the existence of at least one infected particle at time t. Since $1-\mu>0$ and $d_v\ge 1$ for all v, the existence of at least one infected particle at time t in turn implies $M_t\ge 1$ . By Markov’s inequality, and since $M_0=|{\underline {\xi }}_0|$ , taking expectation of (31) and solving the resulting differential equation for $\mathbb E[M_t \mid G, {\underline {\xi }}_0]$ yields for all $\lambda \le 1$ :

$$ \begin{align*} \mathbb{P}(T_{\mathrm{ext}}(G, {\underline{\xi}}_0) \ge t \mid G, {\underline{\xi}}_0) &\le \mathbb{P}(M_t\ge 1\mid G, {\underline{\xi}}_0) \le \mathbb E[M_t\mid G, {\underline{\xi}}_0]\\ &\le \Big(\sum_{v\in V}\xi_0(v)d_v^{1-\mu}\Big)\exp(-(1-\lambda) t). \end{align*} $$

Hence,

$$ \begin{align*} \mathbb{E}[T_{\mathrm{ext}}(G, {\underline{\xi}}_0) \mid G, {\underline{\xi}}_0]&\le \int_{0}^{\infty} \Big(\sum_{v\in V}\xi_0(v)d_v^{1-\mu}\Big)\exp(-(1-\lambda) t) \mathrm dt\\ &= \Big(\sum_{v\in V}\xi_0(v)d_v^{1-\mu}\Big)/(1-\lambda). \end{align*} $$

This finishes the proof for finite graphs. To extend the result to infinite graphs, we can take an exhausting sequence of sets $V_n$ (increasing finite subgraphs whose union is the whole graph) and work with the transition rates inherited from the original infinite graph and take a monotone limit; we omit the details.

The extensions in Remark 2.4 follow immediately from the stochastic domination in (20) and then the martingale argument applied to the monomial obtained.

Proof of Theorem 2.3(b).

The bound in (17) in Theorem 2.2 applied to the configuration model $G_n$ yields that $\mathbb E[T^{\mathrm {cp}}_{\mathrm {ext}}(G_n, \underline 1_{G_n})| (d_i)_{i\le n}]\le \sum _{i=1}^n d_i^{1-\mu }/(1-\lambda )$ . Fast extinction now follows by using the assumption that $\sum _{i=1}^n d_i^{1-\mu }=O_{\mathbb {P}}(\mathrm {poly}(n))$ . Assumption 1.10 implies that $\max _{i\le n} d_i=o(n)$ , so then this condition is automatically satisfied, but it holds even in a much larger class of degree sequences $(\underline d_n)$ that do not grow superpolynomially.

Proof of Theorem 2.6(b).

For all $\mu \ge 0$ ,

$$\begin{align*}f_1(x,y):=\max(x,y)^\mu \ge x^{\mu/2}y^{\mu/2}=:f_2(x,y)\end{align*}$$

holds for all $x,y\ge 1$ . Hence, the stochastic domination in (20) applies and ${\mathrm {CP}}_{f_1,\lambda }\ {\buildrel d \over \le }\ {\mathrm {CP}}_{f_2,\lambda }$ . Since the exponent in $f_2$ is $\mu /2\ge 1/2$ by the assumption that $\mu \ge 1$ , Theorem 2.2 applies for ${\mathrm {CP}}_{f_2,\lambda }$ , and the process goes extinct for all $\lambda <1$ . Hence, so does ${\mathrm {CP}}_{f_1,\lambda }$ .

Proof of Theorem 2.9,(b).

The stochastic domination between the product and max-penalties discussed in the proof of Theorem 2.6(b) above implies the result from Theorem 2.3(b).

Claim 4.3 (Removal of one backtracking step).

Let $G=(V,E)$ be a graph, $\lambda> 0$ and $\mu \ge 1/2$ . Let $z(\cdot )$ be as in (33). For any $\pi \in \mathscr {T}$ and any index $a \in {\mathbb {N}}$ , we have

(36) $$ \begin{align} \sum_{ \pi' \in g^{-1}(\pi):\ \tau(\pi') = a}z(\pi') \le \lambda^2\cdot z(\pi) \max_{v: v\neq \pi_{a-2}}e(\pi_{a-2},v). \end{align} $$

Proof. Fix $\pi $ and a as in the statement of the lemma. Write $\pi = (\pi _0,\ldots ,\pi _m)$ , where $m = \mathfrak {l}(\pi )$ . We assume that the set $\{\pi ' \in g^{-1}(\pi ):\; \tau (\pi ') = a\}$ is nonempty, as the desired inequality is trivial otherwise. By (34) we then have $a \in \{2,\ldots , m+2\}$ , and any $\pi ' \in g^{-1}(\pi )$ with $\tau (\pi ') = a$ and $\pi $ are of the form

(37) $$ \begin{align} \pi' = (\pi_0,\ldots, \pi_{a-3}, u,v,u,\pi_{a-1}, \ldots, \pi_m),\qquad \pi=(\pi_0, \ldots, \pi_{a-3}, u, \pi_{a-1}, \ldots, \pi_m), \end{align} $$

where $u = \pi _{a-2}=\pi {a-2}=\pi ^{\prime }_a$ and v is a neighbor of u (with $v \neq u$ ). (We obtain that the next vertex on the path $\pi $ has index $a-1$ by the erasure of the $(a-2)$ nd (u) and $(a-1)$ th vertex (v) on $\pi '$ ). Then, by (33),

$$ \begin{align*} z(\pi') = r(u,v)^2\cdot z(\pi) = \left(\frac{\lambda\cdot \mathrm{e}(u,v)}{\max(d_u,d_v)^\mu}\right)^2 \cdot z(\pi) \le \frac{\lambda^2\cdot \mathrm{e}(u,v)^{2}}{(d_u)^{2\mu} } \cdot z(\pi), \end{align*} $$

and

$$ \begin{align*} \sum_{\pi' \in g^{-1}(\pi):\ \tau(\pi') = a}z(\pi') &\le \frac{\lambda^2\cdot z(\pi)}{(d_u)^{2\mu}} \sum_{v:v\neq u} \mathrm{e}(u,v)^{2}\le \frac{\lambda^2\cdot z(\pi)}{(d_u)^{2\mu}} \cdot d_u \cdot \max_{v: v\neq u}e(u,v) \\ &= \lambda^2\cdot z(\pi)\cdot (d_u)^{1-2\mu} \cdot \max_{v: v\neq u}e(u,v)\le \lambda^2\cdot z(\pi) \cdot \max_{v: v\neq \pi_{a-2}}e(\pi_{a-2},v) , \end{align*} $$

where the last inequality follows from $d_u \ge 1$ and $\mu \ge 1/2$ , and that $u=\pi _{a-2}$ .

Proof. The proof is by induction on k, the case $k=1$ being Claim 4.3. Assume the statement has been proved for k, and fix $\pi \in \mathscr {T}$ and a sequence $(a_1,\ldots ,a_{k+1})$ . Then, since $\tau $ gives the location of the first backtracking step,

(38) $$ \begin{align} &\sum\left\{ z(\pi'): \;\pi' \in (g^{(k+1)})^{-1}(\pi),\; \tau(\pi') = a_1,\; \ldots,\; \tau(g^{(k)}(\pi')) = a_{k+1} \right\}\nonumber\\& \quad = \sum\left\{\sum\left\{ z(\pi'): \begin{array}{l} \pi' \in g^{-1}(\pi"),\\[.1cm]\tau(\pi') = a_1 \end{array} \right\}: \begin{array}{l} \pi" \in (g^{(k)})^{-1}(\pi),\\[.1cm] \tau(\pi") = a_2,\; \ldots,\; \tau(g^{(k-1)}(\pi")) = a_{k+1} \end{array}\right\}. \end{align} $$

By Claim 4.3, for each $\pi "$ , the inner sum above is smaller than

$$\begin{align*}\lambda^2 z(\pi") \max_{v: v\neq \pi^{\prime\prime}_{a_1-2}}e(\pi^{\prime\prime}_{a_1-2},v)\le \lambda^2 z(\pi") \max_{u,v \in G: v\neq u}e(u,v),\end{align*}$$

so the double sum in (38) is smaller than

$$ \begin{align*} \max_{u,v \in G: v\neq u}e(u,v)\lambda^2 \cdot \sum \left\{z(\pi"): \pi" \in (g^{(k)})^{-1}(\pi),\; \tau(\pi") = a_2,\; \ldots,\; \tau(g^{k-1}(\pi")) = a_{k+1} \right\}. \end{align*} $$

Using the induction hypothesis, this is smaller than $\big (\max _{u,v \in G: v\neq u}\big )^{k+1}\lambda ^{2(k+1)}z(\pi )$ , as required.

Claim 4.5. Let G be a graph and $\pi \in \mathscr {T}$ be such that $\tau (\pi ) < \infty $ and $\tau (g(\pi )) < \infty $ . Then,

$$\begin{align*}\tau(g(\pi)) \ge \tau(\pi) - 1.\end{align*}$$

Proof. This follows from the observation that the sub-path $(\pi _0,\ldots ,\pi _{\tau -2})$ remains intact after applying g to $\pi $ , and this sub-path contains no backtracking steps by the minimality of $\tau (\pi )$ .

Proof. Fix $\pi $ and k as in the statement. Define

$$\begin{align*}\mathcal{A}:= \{(\tau(\pi'),\tau(g(\pi')),\ldots, \tau(g^{(k-1)}(\pi'))):\; \pi' \in (g^{(k)})^{-1}(\pi)\}.\end{align*}$$

That is, for a single $\pi ' \in (g^{(k)})^{-1}(\pi )$ , the sequence $(\tau (\pi '),\tau (g(\pi ')),\ldots , \tau (g^{(k-1)}(\pi ')))$ gives the locations – that is, not the vertex but its index on the “current” path – of loop erasure when we sequentially apply g, k times, on the path $\pi '$ . $\mathcal {A}$ is then the set of all sequences of length k that can be obtained by taking $\pi ' \in (g^{(k)})^{-1}(\pi )$ and applying $\tau $ , $\tau \circ g$ , $\ldots $ , $\tau \circ g^{(k-1)}$ to $\pi '$ . By Lemma 4.4, the left-hand side of (39) is smaller than

$$\begin{align*}\sum_{ \pi' \in (g^{(k)})^{-1}(\pi)} z(\pi')\le\Big(\max_{u,v \in G: v\neq u}e(u,v)\Big)^{k}\lambda^{2k}z(\pi)\cdot |\mathcal{A}|.\end{align*}$$

The desired bound will then follow from the inequality $|\mathcal {A}| \le 2^{\mathfrak {l}(\pi )+2k}$ , which we now prove.

For each $\pi ' \in (g^{(k)})^{-1}(\pi )$ , we add $2(i-1)$ to the location of the ith erasure in the sequential application of loop erasure g on $\pi '$ , which, by Claim 4.5 leads to a a strictly increasing sequence of numbers, that is, we define

$$\begin{align*}c_i(\pi'):= \tau(g^{(i-1)}(\pi')) + 2(i-1),\quad i \in \{1,\ldots, k\}\end{align*}$$

(with $g^{(0)}(\pi ') =\pi '$ ). Note that

$$ \begin{align*} c_k(\pi') &= \tau(g^{(k-1)}(\pi')) + 2(k-1) \le \mathfrak{l}(g^{(k-1)}(\pi')) + 2(k-1) \\ &= \mathfrak{l}(\pi)+2 + 2(k-1) = \mathfrak{l}(\pi) + 2k. \end{align*} $$

Moreover, for $i \in \{1, \ldots , k-1\}$ ,

$$\begin{align*}c_{i+1}(\pi') - c_i(\pi') = \tau(g^{(i)}(\pi')) - \tau(g^{(i-1)}(\pi')) + 2,\end{align*}$$

which is positive by Claim 4.5. These considerations show that $(c_1(\pi '),\ldots ,c_k(\pi '))$ is an increasing sequence in $\{1,\ldots ,\mathfrak {l}(\pi )+2k\}$ . Therefore, $\mathcal {A}$ can be mapped injectively into the set of increasing sequences with k elements in $\{1,\ldots ,\mathfrak {l}(\pi ) +2k\}$ . It is a combinatorial exercise to show that the number of such sequences is $\binom {\mathfrak {l}(\pi )+2k} {k} \le 2^{\mathfrak {l}(\pi )+2k}$ .

Proof of Theorem 2.6(a).

By Lemma 3.8 it is enough to prove the result for the branching random walk. Assume that $\mu \ge 1/2$ and $\lambda < 1/2$ . Let $\mathcal {T}$ be a tree with a root $\varnothing $ . For each vertex u of $\mathcal {T}$ , let $\pi _{\downarrow u}$ denote the geodesic path from $\varnothing $ to u. Consider the branching random walk on $\mathcal {T}$ with penalty function $f(x,y) = \max (x,y)^\mu $ , birth rate $\mu $ and initial configuration consisting of a single particle, located at the root. For this process, let $Z(\cdot )$ be as in (26) and $z(\cdot )$ be as in (33); note that by (27), we have $\mathbb {E}[Z(\pi )] = z(\pi )$ for any $\pi \in \mathscr {T}$ . Further, let $\mathscr {T}_0=\{\pi \in \mathscr {T}:\ \pi _0=\varnothing \}$ denote the set of paths in $\mathcal {T}$ that start at the root. Then, since $\mathcal {T}$ is a tree and $e(u,v)\in \{0,1\}$ for all pairs $u,v\in \mathcal {T}$ ,

(40) $$ \begin{align} \sum_{\pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u}\mathbb{E} \left[Z(\pi) \right]&= \sum_{\pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u}z(\pi) = \sum_{k=0}^\infty \; \;\sum_{\pi \in (g^{(k)})^{-1}(\pi_{\downarrow u})} z(\pi) \nonumber\\& \le z(\pi_{\downarrow u}) \cdot 2^{\mathfrak{l}(\pi_{\downarrow u})}\cdot \sum_{k=0}^\infty (4\lambda^2)^k = \frac{2^{\mathfrak{l}(\pi_{\downarrow u})}}{1-4\lambda^2}\cdot z(\pi_{\downarrow u}), \end{align} $$

where the inequality follows from Corollary 4.6. Since the right-hand side above is finite, we see that the expectation of the number of particles ever born at u is finite, so this number is almost surely finite. This proves local extinction for the initial configuration in which there is a single particle at the root. As already observed, this implies local extinction for the branching random walk, and also the contact process, started from any finite initial configuration.

To prove the exponential decay of the local extinction time, we will use Corollary 3.7 to write, for any $t>0$ ,

(41)

where $X_{m}$ is a Gamma( $1,m$ ) variable for any $m>0$ . Let $\alpha \in (0,1)$ be a constant specified later. Further bounding the right-hand side of (41), we write

(42)

First, we bound the first sum on the right-hand side of (42). Noting that $X_{\lfloor \alpha t\rfloor }$ stochastically dominates $X_{\mathfrak {l}(\pi )+1}$ when $\mathfrak {l}(\pi )<\lfloor \alpha t\rfloor $ , and using Corollary 4.6 we get

(43) $$ \begin{align} \sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)<\lfloor\alpha t\rfloor}}e\cdot z(\pi)&\cdot\mathbb{P}(X_{\mathfrak{l}(\pi)+1}\ge t)\le e\cdot \mathbb{P}(X_{\lfloor\alpha t\rfloor}\ge t)\cdot\sum_{r=\mathfrak{l}(\pi_{\downarrow}(u))}^{\lfloor\alpha t\rfloor-1}\sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)=r}}z(\pi)\nonumber\\ &\le e\cdot \mathbb{P}(X_{\lfloor\alpha t\rfloor}\ge t)\cdot z(\pi_{\downarrow}(u))\cdot2^{\mathfrak{l}(\pi_{\downarrow}(u))}\sum_{k=0}^{(\lfloor\alpha t\rfloor-1-\mathfrak{l}(\pi_{\downarrow}(u)))/2}(4\lambda^2)^k. \end{align} $$

Since $\lambda <1/2$ , the sum on the right-hand side of (43) is bounded by $1/(1-4\lambda ^2)$ . By (33), we have

(44) $$ \begin{align} z(\pi_{\downarrow}(u))=\lambda^{\mathfrak{l}(\pi_{\downarrow}(u))}\prod_{i=0}^{\mathfrak{l}(\pi_{\downarrow}(u))-1}(\max(d_{\pi_i},d_{\pi_{i+1}}))^{-\mu}\le(2^{-\mu}\lambda)^{\mathfrak{l}(\pi_{\downarrow}(u))}. \end{align} $$

Combining (43) and (44) to further upper bound the right-hand side of (43) yields

(45) $$ \begin{align} \sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)<\lfloor\alpha t\rfloor}}e\cdot z(\pi)\cdot\mathbb{P}(X_{\mathfrak{l}(\pi)+1}\ge t)\le \frac{e}{1-4\lambda^2}\cdot \mathbb{P}(X_{\lfloor\alpha t\rfloor}\ge t)\cdot (2^{1-\mu}\lambda)^{\mathfrak{l}(\pi_{\downarrow}(u))}. \end{align} $$

To bound the probabilistic term on the right-hand side of (45), we use the large deviation principle for Gamma variables to write

$$ \begin{align*} \mathbb{P}(X_{\lfloor\alpha t\rfloor}\ge t)\le e^{-\lfloor\alpha t\rfloor I_{\mathrm{exp}}(1/\alpha)}, \end{align*} $$

where $I_{\mathrm {exp}}$ is the large deviation rate function of the exponential distribution with parameter $1$ , defined as

(46) $$ \begin{align} I_{\mathrm{exp}}(a)=a-1+\log(1/a) \end{align} $$

for $a>1$ . As a result, we get

(47) $$ \begin{align} \sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)<\lfloor\alpha t\rfloor}}e\cdot z(\pi)\cdot\mathbb{P}(X_{\mathfrak{l}(\pi)+1}\ge t)\le \frac{e\cdot (2^{1-\mu}\lambda)^{\mathfrak{l}(\pi_{\downarrow}(u))}}{1-4\lambda^2}\cdot e^{-\lfloor\alpha t\rfloor I_{\mathrm{exp}}(1/\alpha)}. \end{align} $$

Next, we bound the second sum on the right-hand side of (42). Similarly to (43), again using Corollary 4.6, we get

(48) $$ \begin{align} \sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)\ge\lfloor\alpha t\rfloor}}e\cdot z(\pi)\le e\cdot z(\pi_{\downarrow}(u))\cdot2^{\mathfrak{l}(\pi_{\downarrow}(u))}\sum_{k=(\lfloor\alpha t\rfloor-\mathfrak{l}(\pi_{\downarrow}(u)))/2}^{\infty}(4\lambda^2)^k. \end{align} $$

Bounding $z(\pi _{\downarrow }(u))$ as in (44), and evaluating the geometric sum in (48) yields

(49) $$ \begin{align} \sum_{\substack{ \pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u, \\ \mathfrak{l}(\pi)\ge\lfloor\alpha t\rfloor}}e\cdot z(\pi)&\le e\cdot (2^{1-\mu}\lambda)^{\mathfrak{l}(\pi_{\downarrow}(u))}\cdot\frac{(4\lambda^2)^{(\lfloor\alpha t\rfloor-\mathfrak{l}(\pi_{\downarrow}(u)))/2}}{1-4\lambda^2}\nonumber\\ &=\frac{e\cdot 2^{-\mu\mathfrak{l}(\pi_{\downarrow}(u))}}{1-4\lambda^2}\cdot (2\lambda)^{\lfloor\alpha t\rfloor}. \end{align} $$

Substituting the bounds (47) and (49) into (42) yields

(50)

For $\lambda <1/2$ , (50) shows the exponential decay of the local extinction time at u. Since the first term on the right-hand side is increasing in $\alpha $ , whereas the second term is decreasing, the optimized bound is given by $\alpha =\alpha ^\star $ , where $\alpha ^\star $ is the solution of

(51) $$ \begin{align} e^{-\lfloor\alpha^\star t\rfloor I_{\mathrm{exp}}(1/\alpha^\star)}=(2\lambda)^{\lfloor\alpha^\star t\rfloor}. \end{align} $$

Using (46), (51) simplifies to

(52) $$ \begin{align} 1/\alpha^\star-1+\log(\alpha^\star)=-\log(2\lambda). \end{align} $$

Since the left-hand side of (52) is strictly decreasing from $\infty $ to $0$ as $\alpha ^\star $ increases from $0$ to $1$ , there is exactly one solution $\alpha ^\star \in (0,1)$ for any given $\lambda <1/2$ . This finishes the proof for .

To extend the argument to any starting state ${\underline {x}}_0$ with $|{\underline {x}}_0|<\infty $ , we make two observations. First, since the above argument is valid for any tree $\mathcal {T}$ with any fixed root $\varnothing $ , (by rerooting the tree) this implies that

(53)

for any $u,v\in \mathcal {T}$ and $t>0$ (for $\lambda <1/2$ ). Here, the constant $c_1(v)$ further depends on $u,\lambda ,\mu $ , while $c_2$ depends on $\lambda $ , but, importantly, not on v. Second, when $({\underline {x}}_t)_{t\ge 0}={\mathrm {BRW}}_{f,\lambda }(\mathcal {T},{\underline {x}}_0)$ , then by the independent behavior of the particles in BRW, we have that

(54) $$ \begin{align} ({\underline{x}}_t)_{t\ge 0}\stackrel{d}{=}\left(\sum_{v:x_0(v)>0}\sum_{i=1}^{x_0(v)}{\underline{x}}^{(v,i)}_t\right)_{t\ge 0}, \end{align} $$

where $({\underline {x}}^{(v,i)}_t)_{v,i}$ are independent realizations of the processes . Hence, if $T_{\mathrm {ext}}^{(v,i,u)}$ denotes the local extinction time of $({\underline {x}}^{(v,i)}_t)_{t\ge 0}$ at u, then a union bound combined with (53) gives

$$ \begin{align*} \mathbb{P}\left(T_{\mathrm{ext}}^{\mathrm{brw}}(\mathcal{T},{\underline{x}}_0,u)>t\right)&=\mathbb{P}\left(\max_{v,i} T_{\mathrm{ext}}^{(v,i,u)}>t\right)\\&\le\sum_{v:x_0(v)>0}\sum_{i=1}^{x_0(v)}\mathbb{P}\left(T_{\mathrm{ext}}^{(v,i,u)}>t\right)\le\sum_{v:x_0(v)>0}\sum_{i=1}^{x_0(v)} c_1(v)e^{-c_2 t}, \end{align*} $$

that is, exponential decay of the distribution of the local extinction time (with the same constant in the exponent for any $|{\underline {x}}_0|$ ). This finishes the proof.

Proof. We continue using the notation $\mathscr {T}_0=\{\pi \in \mathscr {T}:\ \pi _0=\varnothing \}$ for the set of paths in $\mathcal {T}$ that start at the root. Repeating the estimate in (40) and using (56), for any $N \in {\mathbb {N}}$ and any vertex $u \in \mathrm {Gen}_N(\mathcal {T})$ we have $\mathfrak {l}(\pi _{\downarrow u})=N$ , so summing over all infection paths ending at u gives

$$ \begin{align*} \sum_{\pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u}\mathbb{E}[Z(\pi)] =\sum_{\pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u} z(\pi) \le \frac{(2\lambda)^N}{1-4\lambda^2}\cdot \zeta(u). \end{align*} $$

Then, when summing over all infection paths in the tree, we have

$$ \begin{align*} \sum_{\pi \in \mathscr{T}_0} \mathbb{E}[Z(\pi)] &= \sum_{N=0}^\infty\; \sum_{u \in \mathrm{Gen}_N(\mathcal{T})}\; \sum_{\pi \in \mathscr{T}_0:\ \mathfrak{s}(\pi) = u}\mathbb{E}[Z(\pi)]\\ &\le \frac{1}{1-4\lambda^2} \sum_{N=0}^\infty (2\lambda)^N \sum_{u \in \mathrm{Gen}_N(\mathcal{T})} \zeta(u)< \infty \end{align*} $$

by the assumption. This shows that, starting from a single particle at the root, the expected number of particles ever born (overall in $\mathcal {T}$ ) is finite, so this number is finite almost surely. This implies global extinction.

Proof of Theorem 2.5(c).

We assume that the offspring distribution of the Galton-Watson tree satisfies $\mathbb {E}[D^{1-\mu }] < \infty $ . We claim that, for any $N \ge 1$ ,

(60) $$ \begin{align} \mathbb{E}\left[\sum_{u \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(u)\right] = (\mathbb{E}[D^{1-\mu}])^N. \end{align} $$

This is obvious in case $N=1$ . Assume that it has been proved for N. Recalling that $d_v\ge 2$ for all v except possibly the root, for the induction step, by (59), we note that

(61) $$ \begin{align} \sum_{u \in \mathrm{Gen}_{N+1}(\mathcal{T})} \tilde{\zeta}(u) &= \sum_{v \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(v) \sum_{\substack{u \in \mathrm{Gen}_{N+1}(\mathcal{T}):\\u \sim v}} {(d_v-1)^{-\mu}} \nonumber \\&=\sum_{v \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(v)\cdot (d_v - 1)\cdot {(d_v-1)^{-\mu}} = \sum_{v \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(v)\cdot (d_v-1)^{1-\mu}. \end{align} $$

Let $\mathcal {T}_N$ denote the truncation of $\mathcal {T}$ at generation N, that is, $\mathcal {T}_N$ is the subgraph of $\mathcal {T}$ induced by the set of vertices at graph distance at most N from $\varnothing $ . Note that $\mathcal {T}_N$ does not include information about the offsprings of vertices in generation N, and conditioned on $\mathcal {T}_N$ , the sizes of these offsprings are iid, with same law as D. Taking expectations in (61), we have

$$ \begin{align*} \mathbb{E}\left[\sum_{u \in \mathrm{Gen}_{N+1}(\mathcal{T})} \tilde{\zeta}(u) \right]&= \mathbb{E}\left[ \mathbb{E}\left[\left.\sum_{v \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(v)\cdot (d_v-1)^{1-\mu}\right| \mathcal{T}_N\right]\right]\\&=\mathbb{E}\left[ \sum_{v \in \mathrm{Gen}_N(\mathcal{T})}\tilde{\zeta}(v)\cdot\mathbb{E}\left[ \left. (d_v-1)^{1-\mu}\right| \mathcal{T}_N\right]\right]\\&=\mathbb{E}\left[ \sum_{v \in \mathrm{Gen}_N(\mathcal{T})}\tilde{\zeta}(v)\right]\cdot \mathbb{E}[D^{1-\mu}] = (\mathbb{E}[D^{1-\mu}])^{N+1}, \end{align*} $$

where the last equality follows from the induction hypothesis. This completes the proof of (60).

Now, if $\lambda < (2\mathbb {E}[D^{1-\mu }])^{-1}$ , then

$$\begin{align*}\mathbb{E}\left[ \sum_{N=1}^\infty (2\lambda)^N \cdot \sum_{u \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(u) \right] = \sum_{N=1}^\infty (2\lambda \cdot \mathbb{E}[D^{1-\mu}])^N < \infty.\end{align*}$$

Hence, $\sum _{N=1}^\infty (2\lambda )^N \cdot \sum _{u \in \mathrm {Gen}_N(\mathcal {T})} \tilde {\zeta }(u)$ is finite for almost all realizations of $\mathcal {T}$ . It then follows from Lemma 4.7 (and the observation following its proof) that there is global extinction of the penalized branching random walk for almost every realization of $\mathcal {T}$ .

Proof of Corollary 2.7.

The case of trees with finite upper branching number b follows from verifying condition (57) with the simple bound $\zeta (u) \le 1$ for all u. For the case of spherically symmetric trees, we can verify condition (58) directly instead of working with the branching number. Note that, for any $N \ge 1$ , we have

$$\begin{align*}\tilde{\zeta}(u) = (d_0)^{-\mu} \prod_{i=1}^{N-1} (d_i-1)^{-\mu} \quad \text{for any } u \in\mathrm{Gen}_N(\mathcal{T}),\end{align*}$$

so

$$\begin{align*}\sum_{u \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(u) = (d_0)^{-\mu} \prod_{i=1}^{N-1} (d_i-1)^{-\mu} \cdot |\mathrm{Gen}_N(\mathcal{T})| = (d_0)^{1-\mu}\prod_{i=1}^{N-1}(d_i -1)^{1-\mu},\end{align*}$$

and then

$$ \begin{align*} &(2\lambda)^N\sum_{u \in \mathrm{Gen}_N(\mathcal{T})} \tilde{\zeta}(u)\\& \quad = \exp\left\{N \left(\log(2) + \log(\lambda) + \frac{(1-\mu)(\log d_0)}{N} + \frac{1-\mu}{N} \sum_{i=1}^{N-1} \log(d_i-1) \right) \right\}. \end{align*} $$

Now, it is easy to check that $\limsup 1/N\cdot \sum _{i=1}^{N-1} \log (d_i-1)\le \log \overline {\mathrm {br}}(\mathcal {T})$ , so if $\lambda < \mathrm {e}^{-(1-\mu )\log \overline {\mathrm {br}}(\mathcal {T})}/2$ , then there exists $c < 0$ such that the expression inside parentheses above is smaller than c for N large enough. It readily follows that (58) is satisfied, so global extinction follows from Lemma 4.7.

Proof. Let $(\underline {y}_t)_{t \ge 0}$ be the genealogic branching random walk corresponding to $(\underline {x}_t)_{t \ge 0}$ as in Lemma 3.5; in particular, ${y}_0((\bar {v})) = 1$ and ${y}_0(\pi ) = 0$ for any $\pi \neq \bar {v}$ . We note that

$$ \begin{align*} &\left\{ (\underline{x}_t) \text{ is alive at time } CN, \text{ or reaches some vertex at distance } N \text{ from } \bar{v} \right\} \\ &\subset \{y_{CN}(\pi)> 0 \text{ for some } \pi \in \mathscr{T} \text{ with } \pi_0 = \bar{v},\; \mathfrak{l}(\pi) <N \}\\ &\quad\cup \{y_t(\pi)>0 \text{ for some } \pi \in \mathscr{T} \text{ with } \pi_0 = \bar{v},\; \mathfrak{l}(\pi) = N \text{ and some } t > 0\}. \end{align*} $$

Using a union bound and the inequalities $\mathbb {P}(y_{CN}(\pi )> 0) \le \mathbb {E}[y_{CN}(\pi )]$ and $\mathbb {P}(y_t(\pi )> 0 \text { for some } t) \le \mathrm {e} \cdot \mathbb {E}[Z(\pi )] = \mathrm e \cdot z(\pi )$ from Corollary 3.7, we have

(64) $$ \begin{align} \mathbb{P}\left(\begin{array}{l} (\underline{x}_t) \text{ is alive at time } CN, \text{ or reaches }\\ \text{some vertex at distance } N \text{ from } \bar{v} \end{array}\right) \le \sum_{\substack{\pi \in \mathscr{T}:\\\pi_0 = \bar{v},\\\mathfrak{l}(\pi) < N }} \mathbb{E}[y_{CN}(\pi)] + \mathrm{e}\cdot \sum_{\substack{\pi \in \mathscr{T}:\\ \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = N}}z(\pi). \end{align} $$

We bound the two sums in the rhs separately. Using (33) the following bound holds for any path:

(65) $$ \begin{align} z(\pi) \le (\ell\lambda)^{\mathfrak{l}(\pi)},\end{align} $$

which follows from $\max (d_u,d_v)^\mu \ge 1$ and the assumption that $\mathrm {e}(u,v) \le \ell $ .

We first deal with the second sum in (64). Recall that if $\pi ' \in (g^{(k)})^{-1}(\pi )$ , then $\mathfrak {l}(\pi ') = \mathfrak {l}(\pi )+2k$ . Then, we break the sum as follows:

$$ \begin{align*} \sum_{\substack{\pi : \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = N}} z(\pi) &= \sum_{\substack{(m,k):\\ m+2k = N}} \; \sum_{\substack{ \pi: \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = m,\\ \tau(\pi) = \infty}} \;\sum_{\pi' \in (g^{(k)})^{-1}(\pi)}\; z(\pi'). \end{align*} $$

Using (39) in Corollary 4.6, the right-hand side is at most

$$\begin{align*}\sum_{\substack{\pi : \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = N}} z(\pi)\le \sum_{\substack{(m,k):\\ m+2k = N}} \; \sum_{\substack{ \pi: \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = m,\\ \tau(\pi) = \infty}} \;2^m\cdot (4\lambda^2\ell)^k \cdot z(\pi).\end{align*}$$

Using (65) and $b_N$ from (62), this is at most

$$ \begin{align*}\sum_{\substack{(m,k):\\ m+2k = N}} \; &\sum_{\substack{ \pi: \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = m,\\ \tau(\pi) = \infty}} \;2^m\cdot (4\ell\lambda^2)^k \cdot (\ell\lambda)^m\\ &= \sum_{\substack{(m,k):\\ m+2k = N}} (4\ell)^{m+k}\cdot \lambda^{m+2k} \cdot |\{\pi:\; \pi_0 = \bar{v},\; \mathfrak{l}(\pi) = m,\; \tau(\pi) = \infty\}|\\[.2cm] &\le b_N\cdot \sum_{\substack{(m,k):\\ m+2k = N}} (4\ell)^{m+k}\cdot \lambda^{m+2k}. \end{align*} $$

Using that $m+2k=N$ implies that $m+k=(N+m)/2$ for each $m\in {0, \dots , N}$ , the above sum is at most

(66) $$ \begin{align} \sum_{\substack{\pi : \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) = N}} z(\pi)&\le b_N\cdot \sum_{m=0}^N (4\ell)^{(N+m)/2}\cdot \lambda^N = (2\ell^{1/2}\lambda)^N\cdot b_N\cdot \sum_{m=0}^N (2\ell^{1/2})^m \nonumber \\ &\le 2\ell^{1/2}(4\ell\lambda)^N\cdot b_N. \end{align} $$

We now turn to the first term in (64). Using (25), we have

(67) $$ \begin{align} \sum_{\substack{\pi: \pi_0 = \bar{v},\\\mathfrak{l}(\pi) < N}} \mathbb{E}[y_{CN}(\pi)] \le \left( \max_{0 \le m < N} \frac{(CN)^m}{m!} \mathrm{e}^{-CN}\right) \cdot \sum_{\substack{\pi: \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) < N}} z(\pi). \end{align} $$

Let us bound the sum in the right-hand side using (39) with $\max e(u,v)\le \ell $ and then (65) as

$$ \begin{align*} \sum_{\substack{\pi: \pi_0 = \bar{v},\\ \mathfrak{l}(\pi) < N}} z(\pi) &\le \sum_{\substack{\pi: \pi_0= \bar{v},\\ \mathfrak{l}(\pi) <N,\\ \tau(\pi) = \infty}} \;\sum_{k=0}^\infty \;\sum_{\pi' \in (g^{(k)})^{-1}(\pi)} z(\pi')\\ &\stackrel{({39})}{\le} \frac{1}{1-4\ell\lambda^2}\sum_{\substack{\pi: \pi_0= \bar{v},\\ \mathfrak{l}(\pi) <N,\\ \tau(\pi) = \infty}}2^{\mathfrak{l}(\pi)} z(\pi) \stackrel{(65)}{\le} \frac{1}{1-4\ell\lambda^2} \sum_{\substack{\pi: \pi_0= \bar{v},\\ \mathfrak{l}(\pi) <N,\\ \tau(\pi) = \infty}} (2\ell\lambda)^{\mathfrak{l}(\pi)}. \end{align*} $$

Since $\lambda < 1/(4\ell )$ with $\ell \ge 1$ , we have $\frac {1}{1-4\ell \lambda ^2} < 2$ and $2\ell \lambda < 1/2 < 1$ , so the last factor in (67) is smaller than

(68) $$ \begin{align} 2|\{\pi: \pi_0 = \bar{v},\; \mathfrak{l}(\pi) < N,\;\tau(\pi) = \infty\}| \le 2b_N.\end{align} $$

Next, the expression inside the maximum in (67) equals $\mathbb {P}(W = m)$ for W having the $\mathrm {Poisson}(CN)$ distribution. We bound

$$\begin{align*}\max_{0 \le m < N} \mathbb{P}(W = m) \le \mathbb{P}(W \le N).\end{align*}$$

We use a Chernoff bound for Poisson random variables: for $X \sim \mathrm {Poisson}(\nu )$ we have $\mathbb {P}(X \le \nu - t) \le \mathrm {e}^{-t^2/(2\nu )}$ , see [Reference van der Hofstad67, Exercise 2.21]. This gives

$$\begin{align*}\mathbb{P}(W \le N) \le \exp\left\{- \frac{(CN-N)^2}{2CN} \right\} = \exp\left\{-\frac{(C-1)^2}{2C}\cdot N\right\}.\end{align*}$$

Combining this with (68) in (67) and (66) completes the proof of (63).