We now turn to the proofs of the three propositions of the previous section. In Sect. 5.1, we will prove Proposition 2. In Sect. 5.2, we will give some bounds involving the function L(h), as well as a key proposition involving coupling of the contact process on \({\tilde{G}}\), Proposition 5. Next, Sect. 5.3 contains the proof of Proposition 3, and Sect. 5.4 contains the proof of Proposition 4.
Proof of Proposition 2
Proof of Proposition 2
We begin with some definitions. For \(0 \le i \le h\), let T(i) denote the set of vertices of \({\mathcal {T}}_h\) at distance i from the root \(\rho \). Using the graphical construction of the contact process with parameter \(\lambda ' \ge \lambda \), we will now define random sets \({\mathcal {Z}}_{\lambda '}(0),\ldots , {\mathcal {Z}}_{\lambda '}({\lfloor h/2\rfloor })\) with \({\mathcal {Z}}_{\lambda '}(i)\subset T(i)\) for each i. We set \({\mathcal {Z}}_{\lambda '}(0) := \{\rho \}\). Assume that \({\mathcal {Z}}_{\lambda '}(i)\) has been defined, let z be a vertex of \(T(i+1)\) and let \(z'\) be the neighbor of z in T(i). We include z in \({\mathcal {Z}}_{\lambda '}({i+1})\) if \(z' \in {\mathcal {Z}}_{\lambda '}(i)\) and, in the time interval \([i,i+1]\), there are no recovery marks on \(z'\) or z, and there is a transmission arrow from \(z'\) to z. Letting \(Z_{\lambda '}(i):= |{\mathcal {Z}}(i)|\) for each i, it is readily seen that \((Z_{\lambda '}(i): 0 \le i \le \lfloor h/2\rfloor )\) is a branching process. Its offspring distribution is equal to the law of \(U \cdot W\), where \(U \sim \text {Bernoulli}(e^{-1})\) and \(W \sim \text {Binomial}(d,e^{-1}\cdot (1-e^{-\lambda '}))\) are independent. The expectation of this distribution is larger than \(m_\lambda > 1\). For this reason, there exists \(\sigma _\lambda > 0\) such that the event
$$\begin{aligned} B_{\lambda '} := \left\{ Z_{\lambda '}(\lfloor h/2 \rfloor ) > (m_\lambda /2)^{\lfloor h/2 \rfloor }\right\} \end{aligned}$$
has
$$\begin{aligned} {\mathbb {P}}\left( B_{\lambda '} \right) > \sigma _\lambda \quad \text {for all}\quad \lambda ' \ge \lambda \quad \text {and}\quad h \in {\mathbb {N}}. \end{aligned}$$
Finally, note that
$$\begin{aligned} B_{\lambda '} \subset \left\{ \xi _{{\mathcal {T}}_h,\lambda ';\lfloor h/2 \rfloor }^{\{\rho \}} \in {\mathcal {A}}_h\right\} . \end{aligned}$$
Now, define \(B_{\lambda '}(0) := B_{\lambda '}\) and, for \(t \in [0,\infty )\), define \(B_{\lambda '}(t)\) as the time translation of \(B_{\lambda '}\), so that time t becomes the time origin (that is, \(B_{\lambda '}(t)\) is defined by using the graphical construction of the contact process on the time intervals \([t,t+1],[t+1,t+2],\ldots ,[t+\lfloor h/2 \rfloor -1, t+ \lfloor h/2 \rfloor ]\)). We evidently have
$$\begin{aligned} {\mathbb {P}}(B_{\lambda '}(t)) = {\mathbb {P}}(B_{\lambda '}) > \sigma _\lambda \quad \text {for any}\;t, \end{aligned}$$
(27)
and moreover,
$$\begin{aligned} \left\{ \rho \in \xi _{{\tilde{G}},\lambda ';t}^{A} \right\} \cap B_{\lambda '}(t) \subset \left\{ \xi _{{\tilde{G}},\lambda ';t + \lfloor h/2\rfloor }^{A} \in {\mathcal {A}}_h \right\} \end{aligned}$$
(28)
for any A. It will be useful to note that, if \(t_1, t_2 \ge 0\) with \(t_2 > t_1 + 2\), then \(B_{\lambda '}(t_1)\) and \(B_{\lambda '}(t_2)\) are independent.
Now, fix \(t>0\) and condition on the event \(\left\{ {\bar{\xi }}_{G,\lambda '}^{A}(o) > t\right\} \) occurs. Note that this event only involves the graphical construction of the contact process on G; in particular, the Poisson processes involving vertices and edges of \({\mathcal {T}}_h\), or the edge \(\{o,\rho \}\), are still unrevealed. Then, by elementary properties of Poisson processes, there exists \(c_\lambda > 0\) (depending only on \(\lambda \)) such that (uniformly on \(\lambda ' \ge \lambda \)) outside probability \(\exp \{-c_\lambda \cdot t\}\), we can find random times \(s_1< \cdots < s_{\lfloor c_\lambda t \rfloor }\) separated from each other by more than two units, and such that for each i, \(o \in \xi _{G,\lambda ';s_i}^{A}\) and there is a transmission arrow from \((o,s_i)\) to \((\rho ,s_i)\). If this is the case, and if \(B_{\lambda '}(s_i)\) also occurs for some i, we then get \(\xi _{{\tilde{G}},\lambda ';s_i + \lfloor h/2\rfloor }^{A} \in {\mathcal {A}}_h\), by (28). The desired result now follows from independence between the events \(B_{\lambda '}(s_i)\), together with (27) and a Chernoff bound. \(\square \)
Preliminary Bounds
In this section we will prove that
$$\begin{aligned} d^{\frac{3h}{4}} \le L(h) \le d^{2h} \end{aligned}$$
for large enough h. These bounds will be instrumental for proving Propositions 3 and 4. We first give an upper bound involving the extinction time of the contact process on \({\tilde{G}}\), in terms of the length L(h).
Lemma 8
We have
$$\begin{aligned} \lim _{h \rightarrow \infty } {\mathbb {P}}\left( \xi _{{\tilde{G}},\lambda ;\exp \{d^{\frac{3}{2}h}\}\cdot (\log L(h))^2}^{\tilde{V}}\ne \varnothing \right) = 0, \end{aligned}$$
that is, the extinction time of the contact process on \({\tilde{G}}\) started from full occupancy is smaller than \(\exp \{d^{\frac{3}{2}h}\}\cdot (\log L(h))^2\) with high probability as \(h \rightarrow \infty \).
Proof
Let \(E_{0}'\) be the event that each vertex in \(V \cup {\mathcal {T}}_h\) has a recovery mark before it sends out any transmission arrow, and before time 1. Since all vertices of \(V \cup {\mathcal {T}}_h\) have degree at most \(d+1\), we have
$$\begin{aligned} {\mathbb {P}}(E_0') \ge \left( (1-e^{-1})\cdot e^{-(d+1)\lambda }\right) ^{|V \cup {\mathcal {T}}_h|} \ge \left( (1-e^{-1})\cdot e^{-(d+1)\lambda }\right) ^{d^{h+2}}, \end{aligned}$$
if h is large enough (since \(|{\mathcal {T}}_h| < d^{h+1}\) and |V| is fixed as \(h \rightarrow \infty \)). Next, let \(E_0''\) denote the event that the contact process on \({\tilde{G}}\) started from \({\mathcal {L}}_h\) infected dies out before time \((\log L(h))^2\), and never infects the root \(\rho \) of \({\mathcal {T}}_h\). That is,
$$\begin{aligned} E_0'' := \left\{ \xi _{{\tilde{G}},\lambda ;(\log L(h))^2}^{{\mathcal {L}}_h} = \varnothing ,\; \rho \notin \xi _{{\tilde{G}},\lambda ;t}^{{\mathcal {L}}_h} \text { for all }t \right\} . \end{aligned}$$
The probability of \(E_0''\) is the same as the probability that a contact process on the line segment \(\{-1,0,\ldots , L_G(h)\}\), with rate \({\bar{\lambda }}\) and initial configuration \(\{0,\ldots , L(h)\}\), dies out before time \((\log L(h))^2\) and never infects vertex \(-1\). Therefore, by Lemmas 2 and 4, we have
$$\begin{aligned} {\mathbb {P}}(E_0'')> \delta > 0 \end{aligned}$$
for all h. Let \(E_0 := E_0' \cap E_0''\); since \(E_0'\) and \(E_0''\) have different supports in the graphical representation, they are independent and hence
$$\begin{aligned} {\mathbb {P}}(E_0) > \delta \left( (1-e^{-1})\cdot e^{-(d+1)\lambda }\right) ^{d^{h+2}}. \end{aligned}$$
For \(i \in \{1,\ldots ,\lfloor \exp \{d^{\frac{3}{2}h}\} \rfloor \}\), let \(E_i\) be the time translation of event \(E_0\) to the graphical construction on the time interval
$$\begin{aligned}{}[i(\log L(h))^2,\;(i+1)(\log L(h))^2]. \end{aligned}$$
Finally, noting that \(E_0,E_1,\ldots \) are independent and
$$\begin{aligned} E_i \subset \left\{ \xi _{{\tilde{G}},\lambda ;(i+1)(\log L(h))^2}^{\tilde{V}} = \varnothing \right\} , \end{aligned}$$
we have
$$\begin{aligned}&{\mathbb {P}}\left( \xi _{{\tilde{G}},\lambda ;\exp \{d^{\frac{3}{2}h}\} \cdot (\log L(h))^2}^{\tilde{V}} \ne \varnothing \right) \\&\quad \le {\mathbb {P}}((E_0)^c)^{\lfloor \exp \{d^{\frac{3}{2}h}\} \rfloor }\\&\quad \le \exp \left\{ -\lfloor \exp \{d^{\frac{3}{2}h}\} \rfloor \cdot \delta \left( (1-e^{-1})\cdot e^{-(d+1)\lambda }\right) ^{d^{h+2}} \right\} \xrightarrow {h \rightarrow \infty }0. \end{aligned}$$
\(\square \)
We now proceed to an upper bound on L(h).
Lemma 9
If h is large enough, we have
$$\begin{aligned} L(h) \le d^{2h} \end{aligned}$$
(29)
Proof
Define
$$\begin{aligned} F_1:= \left\{ \xi _{{\tilde{G}},\lambda ;\exp \{d^{\frac{3}{2}h}\}\cdot (\log L(h))^2}^{V \cup {\mathcal {T}}_h} = \varnothing \right\} . \end{aligned}$$
Recall that \(v_{L(h)-1}\) denotes the neighbor of \({\tilde{o}}\) in \({\mathcal {L}}_h\), and let \(F_2\) be the event that there is no infection path starting from \((\rho ,s)\) for some \(s \le \exp \{d^{\frac{3}{2}h}\}\cdot (\log L(h))^2\), ending at \((v_{L(h)-1},t)\) for some \(t > s\), and entirely contained in \({\mathcal {L}}_h \cup \{\rho \}\). It is easy to see that
$$\begin{aligned} F_1 \cap F_2 \subset \left\{ \xi _{{\tilde{G}},\lambda ;t}^{V \cup {\mathcal {T}}_h}(v_{L(h)-1}) = 0 \; \text {for all }t \ge 0 \right\} . \end{aligned}$$
By Lemma 8 we have \({\displaystyle \lim _{h \rightarrow \infty } {\mathbb {P}}(F_1) = 1}\), and by Corollary 1 we have
$$\begin{aligned} {\mathbb {P}}(F_2) \ge 1 - \left( \exp \{d^{\frac{3}{2}h}\}\cdot (\log L(h))^2 + 1 \right) \cdot \exp \{-c_{\mathbb {L}} \cdot L(h)\}. \end{aligned}$$
This shows that, if we had \(L(h) >d^{2h}\), we would get
$$\begin{aligned} {\mathbb {P}}\left( \xi _{{\tilde{G}},\lambda ;t}^{V \cup {\mathcal {T}}_h}(v_{L(h)-1}) = 0\; \text {for all }t \ge 0\right) \ge {\mathbb {P}}(F_1 \cap F_2) \xrightarrow {h \rightarrow \infty } 1. \end{aligned}$$
On the other hand, the definition of L(h) implies that
a contradiction. \(\square \)
The following guarantees that if the contact process with some initial condition remains active for
time in \({\tilde{G}}\), then it is highly likely to coincide with the process started from full occupancy. This, in turn, will be applied in the proof of Lemma 10 which is an important step toward obtaining lower bounds on L(h).
Proposition 5
If h is large enough, for any \(A \subset {\tilde{V}}\) we have
The proof of this proposition is lengthy and technical, so we postpone it to Appendix.
We are now interested in giving an upper bound for the probability that the infection crosses \({\mathcal {L}}_h\) in a single attempt. For the proof of Proposition 4, it will be important that this bound is given in terms of the extinction time of the infection on \({\tilde{G}}\), starting from full occupancy.
Define
$$\begin{aligned} S(h) := {\mathbb {E}}\left[ \inf \left\{ t: \xi _{{\tilde{G}},\lambda ;t}^{\tilde{V}} = \varnothing \right\} \right] , \end{aligned}$$
that is, S(h) is the expected amount of time it takes for the contact process on \({\tilde{G}}\) with parameter \(\lambda \) started from full occupancy to die out. Also let
$$\begin{aligned} {\mathcal {p}}(\ell ) = {\mathcal {p}}_\lambda (\ell ):= {\mathbb {P}}\left( {\bar{\xi }}_{{\mathbb {N}}_0,\lambda }^{\{0\}}(\ell ) > 0 \right) , \end{aligned}$$
(30)
or equivalently, \({\mathcal {p}}(\ell )\) is the probability that, for the contact process with parameter \(\lambda \) on a line segment of length \(\ell + 1\), an infection starting at one extremity ever reaches the other extremity.
Lemma 10
If h is large enough,
Proof
Recall that \(v_0\) is the vertex of \({\mathcal {L}}_h\) neighboring \(\rho \), the root of \({\mathcal {T}}_h\). Let q(h) denote the probability that there is an infection path starting from \((v_0,0)\), ending at \(({\tilde{o}},t)\) for some
, and entirely contained in \({\mathcal {L}}_h\). Note that \(q(h) \le {\mathcal {p}}(L(h))\) and, by a union bound,
Next, assume that h is large enough that any vertex in V is at distance smaller than h from \(\rho \), the root of \({\mathcal {T}}_h\). With this choice, we claim that for any \(A \subset {\tilde{V}}\), \(A \ne \varnothing \) we have
Indeed, if \(A \cap {\mathcal {L}}_h \ne \varnothing \), then the left-hand side is larger than q(h) by the definition of q(h) and simple monotonicity considerations. If \(A \cap {\mathcal {L}}_h = \varnothing \), then by (9), with probability larger than \(\delta (h):= (e^{-1}(1-e^{-{\lambda }}))^h\), \(\rho \) gets infected within time h, and conditioned on this, with probability q(h), \({\tilde{o}}\) gets infected after at most additional
units of time. Applying (32) and the strong Markov property repeatedly, we have
Now, letting
, we have
where the first inequality follows from the definition of L(h), see (23) and (25). We now claim that
if h is large enough. Plugging this into (34), we obtain
for large enough h, completing the proof.
It remains to prove (35). Noting that
if h is large, we have
By Lemma 1, we have
Next,
Now, the first term on the right-hand side is smaller than
by Lemma 7 (since \(V \cup {\mathcal {T}}_h \in {\mathcal {A}}_h\)), and the second term on the right-hand side is also smaller than
by Proposition 5. Putting things together gives (35) for large enough h. \(\square \)
We end this section with a lower bound on L(h), which again will be important for the proof of Proposition 4.
Lemma 11
If h is large enough,
$$\begin{aligned} L(h) \ge d^{\frac{3h}{4}}. \end{aligned}$$
(36)
Proof
By the simple estimate (9) and Lemma 10, we have
This gives
Recalling that
and noting that
$$\begin{aligned} S(h) \ge {\mathbb {E}}\left[ \inf \{t: \xi _{{\mathcal {T}}_h,\lambda ;t}^{} = \varnothing \} \right] \ge \exp \{c_{\mathbb {T}}\cdot d^h \}, \end{aligned}$$
we obtain
$$\begin{aligned} L(h) \ge \frac{c_{\mathbb {T}} \cdot d^h - 3d^{\sqrt{h}}}{\log (e(1-e^{-\lambda })^{-1})} > d^{\frac{3h}{4}} \end{aligned}$$
if h is large enough. \(\square \)
Proof of Proposition 3
We begin with a simple consequence of Proposition 5.
Lemma 12
If h is large enough, for any \(A \in {\mathcal {A}}_h\) we have
Proof
Since both A and \(V \cup {\mathcal {T}}_h\) belong to \({\mathcal {A}}_h\), Lemma 7 gives
and Proposition 5 gives
The desired statement follows from these four inequalities. \(\square \)
Lemma 13
If h is large enough, we have, for any \(v \in V\),
Proof
Assume h is larger than the graph diameter of G, and fix \(v \in V\). We have, for any \(A \subset {\tilde{V}}\) with \(A \cap {\mathcal {T}}_h \ne \varnothing \),
$$\begin{aligned} {\mathbb {P}}\left( \int _0^{4h}\xi _{{\tilde{G}},\lambda ;t}^{A}(v)\;\mathrm {d}t > h\right) \ge e^{-2h}\cdot (e^{-1}\cdot (1-e^{-\lambda }))^{2h}. \end{aligned}$$
Indeed, by the estimate (9) we have that, with probability at least \((e^{-1}\cdot (1-e^{-\lambda }))^{2h}\), v becomes infected before time 2h, and then, it remains infected for time 2h (by having no recovery marks) with probability \(e^{-2h}\). By iterating this, we obtain
We therefore have
if h is large, which implies the statement of the lemma. \(\square \)
Lemma 14
If h is large enough, we have
Proof
We will separately prove that
$$\begin{aligned} {\mathbb {P}}\left( \int _{0}^\infty \xi _{{\tilde{G}},\lambda ;t}^{V \cup {\mathcal {T}}_h}({\tilde{o}})\,\mathrm {d}t> h \right) > 1 - \frac{1}{4h} \end{aligned}$$
(38)
and
the desired result will then follow.
For (38), let \(u_1 := v_{L(h)-2},u_2:= v_{L(h)-1}\) be such that \(u_1,u_2,{\tilde{o}}\) (in this order) are the three last vertices in \({\mathcal {L}}_h\), as we move away from \({\mathcal {T}}_h\). By Lemma 6 and the definition of L(h) we have
Let \(G'\) denote \({\tilde{G}}\) after removing \(u_2\) and \({\tilde{o}}\). Define the random set of times
$$\begin{aligned} I:= \left\{ t \ge 0: u_1 \in \xi _{G',\lambda ;t}^{V \cup {\mathcal {T}}_h}\right\} . \end{aligned}$$
We have
$$\begin{aligned} {\mathbb {P}}\left( {\bar{\xi }}_{{\tilde{G}},\lambda }^{V \cup {\mathcal {T}}_h}(u_2) = 0\right) = {\mathbb {E}}\big [1- e^{-{\lambda } |I|} \big ], \end{aligned}$$
(41)
where |I| denotes the Lebesgue measure of I. To justify this, note that the first time that \(u_2\) becomes infected in \((\xi ^{V \cup {\mathcal {T}}_h}_{{\tilde{G}},\lambda ;t})_{t \ge 0}\) is necessarily through a transmission from \(u_1\). Hence, one can decide if \(u_2\) is ever infected in this process by inspecting whether there is a point in time at which (1) \(u_1\) is infected in process confined to \(G'\), and (2) there is a transmission arrow from \(u_1\) to \(u_2\). The number of such time instants is a Poisson random variable with parameter \({\lambda } |I|\), justifying (41).
We bound
so
for h large enough.
We next claim that
$$\begin{aligned} {\mathbb {P}}\left( \left. {\bar{\xi }}_{{\tilde{G}},\lambda }^{V \cup {\mathcal {T}}_h}({\tilde{o}}) \le h \right| |I| \ge h^2 \right) < e^{-h}. \end{aligned}$$
(43)
To prove this, we observe that on the event \(\{|I| \ge h^2\}\), we can find an increasing sequence of times \(S_0,\ldots , S_{\lfloor h^2/2 \rfloor } \in I\) with
$$\begin{aligned} |I \cap [S_j+1,S_{j+1}]| \ge 2\quad \text {for each }j. \end{aligned}$$
Next, note that for each interval \([S_j,S_{j+1}]\), with a probability that is positive and depends only on \({\lambda }\), the infection is sent to \({\tilde{o}}\) and remains there for one unit of time. This occurring independently in different time intervals, (43) follows from a simple Chernoff bound. Now, (38) follows from (42) and (43).
We now turn to (39). Note that the event inside the probability there is contained in the event that there is an infection path starting at some time s and ending at some time t with
, connecting the two endpoints of \({\mathcal {L}}_h\). By Corollary 1, the probability that such a path exists is smaller than
if h is large enough. \(\square \)
Proof of Proposition 3
The statements follow readily from Lemmas 12, 13 and 14. \(\square \)
Proof of Proposition 4
Proving Proposition 4 is now just a matter of putting together bounds that were obtained earlier.
Proof of Proposition 4
Fix \(\lambda ' < \lambda \). Let B be the event that, in the graphical construction with parameter \(\lambda '\), there is an infection path starting from \((v_0,s)\) for some
(where \(v_0\) is the vertex of \({\mathcal {L}}_h\) neighboring the root \(\rho \) of \({\mathcal {T}}_h\)), ending at \(({\tilde{o}},t)\) for some \(t > s\), and entirely contained in \({\mathcal {L}}_{h}\). Then, by a union bound,
The first term is bounded using Markov’s inequality and monotonicity:
Next, note that the occurrence of B depends only on the graphical construction of the contact process with parameter \(\lambda '\) on the line segment connecting \(v_0\) and \({\tilde{o}} = v_{L(h)}\). Therefore, using Lemma 3 [and also recalling the definition of \({\mathcal {p}}\) from (30)], we have
Bounding the right-hand side using Lemma 5, we obtain
By using
as in (31) and \(L(h) \ge d^{\frac{3h}{4}}\) as in (36), the right-hand side above is smaller than
which is much smaller than
if h is large enough (depending on \(\lambda \) and \(\lambda '\)). \(\square \)