3.1 Simulations
In this section, we evaluate the performance of the reverse infection algorithm for the heterogeneous SIR model on different networks including tree networks and realworld networks.
We first describe the heterogeneous SIR model we used in the simulation. Each edge e∈ is assigned with a weight q_{
e
} which is uniformly distributed over (0,1). The infection time over each edge e∈ is geometrically distributed with mean 1/q_{
e
}. Similarly, each node v∈ is assigned with a weight p_{
v
} generated by a uniform distribution over (0,1), and the recovery time is geometrically distributed with mean 1/p_{
v
}. The information source is randomly selected. The total number of infected and recovered nodes in each infection graph is within the range of [ 100,300]. Each infected node v in the infection graph reports with probability σ, independently. The snapshots used in the simulations have at least one infected node. We changed σ and evaluated the performance on different networks.
We briefly introduce the three main algorithms which were used to compare with the reverse infection algorithm (RI).

1.
Closeness centrality algorithm (CC): The closeness centrality algorithm selects the node with the maximum infection closeness as the information source.

2.
Weighted reverse infection algorithm (wRI): The weighted reverse infection algorithm selects the node with the minimum weighted infection eccentricity as the information source where the weighted infection eccentricity is similar to the infection eccentricity except that the length of a path is defined to be the sum of the link weights instead of the number of hops, and the link weight is the average time it takes to spread the information over the link, i.e., ⌊1/q _{
e
}⌋ on edge e.

3.
Weighted closeness centrality algorithm (wCC): The weighted closeness centrality algorithm selects the node with the maximum weighted infection closeness as the information source.
3.1.1 Tree networks
We first evaluated the performance of the RI algorithm on tree networks.
Regular trees
A gregular tree is a tree where each node has g neighbors. We set the degree g=5 in our simulations.
We varied the sample probability σ from 0.01 to 0.1. The simulation results are summarized in Figure 5a, which shows the average distance between the estimator and the actual information source versus the sampling probability. When the sample probability increases, the performance of all algorithms improves. When the sample probability is larger than 6%, the average distance becomes stable which means that a small number of infected nodes is enough to obtain a good estimator. We also notice that the average distance of RI is smaller than all other algorithms and is less than one hop when σ≥0.04. wRI has a similar performance with RI when the sample probability is small (=0.01) but becomes much worse when the sample probability increases.
Binomial trees
We further evaluated the performance of RI and other algorithms on binomial trees T(ξ,β) where the number of children of each node follows a binomial distribution such that ξ is the number of trials and β is the success probability of each trial. In the simulations, we selected ξ=10 and β=0.4. Again, we varied σ from 0.01 to 0.1. The results are shown in Figure 5b. Similar to the regular trees, the performance of RI dominates CC, wRI, and wCC, and the difference in terms of the average number of hops is approximately 1 when σ≥0.03.
3.1.2 Realworld networks
In this section, we conducted experiments on two realworld networks: the Internet autonomous systems (IAS) network which is available at http://snap.stanford.edu/data/index.html and the power grid (PG) network which is available at http://wwwpersonal.umich.edu/~mejn/netdata/.
The power grid network
The power grid network has 4,941 nodes and 6,594 edges. On average, each node has 1.33 edges. So the power grid network is a sparse network. The simulation results are shown in Figure 5c. In the power grid network, we can see that RI and wRI have similar performance, and both outperform CC and wCC by at least one hop when σ≥0.04.
The internet autonomous systems network
The Internet autonomous systems network is the data collected on 31 March 2001. There are 10,670 nodes and 22,002 edges in the network. The simulation results are shown in Figure 5d. wRI and wCC always perform worse than RI. Although RI and CC have similar performance when the sample probability is large, RI outperforms CC when σ≤0.03.
3.1.3 RI versus DMP
We finally compared the performance of RI and DMP. We conducted the simulation on the power grid network and fixed the sample probability to be 10%. Under this setting, the complexity of DMP is very high since the DMP computation needs to be repeated for every node in the network. Since nodes far away from the observed infected nodes are not likely to be the information source, we ran DMP over a small subset of nodes close to the Jordan infection centers (roughly 10%) to reduce the complexity of the algorithm.
We tested the speed of RI and DMP on a machine with 1.8 GB memory, 4 cores 2.4 GHz Intel i5 CPU and Ubuntu 12.10. The algorithms are implemented in Python 2.7. On average, it took RI 0.57 s to locate the estimator for one snapshot and took DMP 229.12 s. So RI is much faster than DMP.
Figure 6 shows the cumulative distribution function (CDF) of the distance from the estimator to the actual source under DMP and RI. We can see that RI dominates DMP; in particular, 71% of the estimators under RI are no more than seven hops from the actual source compared to 57% under DMP. Therefore, RI outperforms DMP in terms of both speed and accuracy. We remark that we did not compare the performance of RI and DMP on the IAS network because the complexity of running DMP on a largesized network like the IAS network is prohibitively high.
3.2 Proofs
In this section, we present the proofs of the main results.
3.2.1 Proof of Theorem 1
Denote by {\mathcal{I}}_{\mathbf{\text{Y}}}=\left\{v\right{Y}_{v}=1\} the set of observed infected nodes and {\mathcal{\mathscr{H}}}_{\mathbf{\text{Y}}}=\left\{v\right{Y}_{v}=0\} the set of unobserved nodes. Given a node v, define the optimal time {t}_{v}^{\ast} to be
{t}_{v}^{\ast}\triangleq \underset{t}{arg}\underset{t,\mathbf{\text{X}}[0,t]\in \mathcal{X}\left(t\right)}{max}Pr\left(\mathbf{\text{X}}[0,t]v\phantom{\rule{1em}{0ex}}\text{is information source}\right),
i.e., it is the duration of the optimal sample path with node v as the information source.
Lemma 1 (Time Inequality).
Consider an infinite tree rooted at v_{
r
}. Assume that v_{
r
} is the information source and the observed snapshot Y contains at least one infected node. If \stackrel{~}{e}({v}_{r},{\mathcal{I}}_{\mathbf{\text{Y}}})\le {t}_{1}<{t}_{2}, the following inequality holds:
\underset{\mathbf{\text{X}}[0,{t}_{1}]\in \stackrel{~}{\mathcal{X}}\left({t}_{1}\right)}{max}Pr\left(\mathbf{\text{X}}\right[\phantom{\rule{0.3em}{0ex}}0,{t}_{1}\left]\phantom{\rule{0.3em}{0ex}}\right)>\underset{\mathbf{\text{X}}[0,{t}_{2}]\in \stackrel{~}{\mathcal{X}}\left({t}_{2}\right)}{max}Pr\left(\mathbf{\text{X}}\right[\phantom{\rule{0.3em}{0ex}}0,{t}_{2}\left]\phantom{\rule{0.3em}{0ex}}\right),
where \stackrel{~}{\mathcal{X}}\left(t\right)=\left\{\mathbf{\text{X}}\right[0,t\left]\right\mathbf{\text{Y}}=\mathbf{\text{F}}\left(\mathbf{\text{X}}\right(t\left)\right)\}. In addition,
{t}_{{v}_{r}}^{\ast}=\stackrel{~}{e}({v}_{r},{\mathcal{I}}_{\mathbf{\text{Y}}})=\underset{u\in {\mathcal{I}}_{\mathbf{\text{Y}}}}{max}d({v}_{r},u),
i.e., {t}_{{v}_{r}}^{\ast} is equal to the observed infection eccentricity of v_{
r
} with respect to {\mathcal{I}}_{\mathbf{\text{Y}}}.
Proof.
We adopt the notations defined in [6], which are listed below:
\mathcal{C}\left(v\right) is the set of children of v.
black ϕ(v) is the parent of node v.
{\mathcal{Y}}^{k} is the set of infection topologies where the maximum distance from v_{
r
} to an infected node is k. All possible infection topologies are then partitioned into countable subsets \left\{{\mathcal{Y}}^{k}\right\}.
T_{
v
} is the tree rooted in v.
{T}_{v}^{u} is the tree rooted in v without the branch from its neighbor u.
\mathbf{\text{X}}\left(\right[\phantom{\rule{0.3em}{0ex}}0,t],{T}_{v}^{u}) is the sample path restricted to topology {T}_{v}^{u}.
{t}_{v}^{I},{t}_{v}^{R} are the infection time and recovery time of node v.
Considering the case where the time difference of two sample paths is 1, we will show that
\begin{array}{l}\underset{\mathbf{\text{X}}[0,t]\in \stackrel{~}{\mathcal{X}}\left(t\right)}{max}Pr\left(\mathbf{\text{X}}\right[\phantom{\rule{0.3em}{0ex}}0,t\left]\phantom{\rule{0.3em}{0ex}}\right)>\underset{\mathbf{\text{X}}[0,t+1]\in \stackrel{~}{\mathcal{X}}(t+1)}{max}Pr\left(\mathbf{\text{X}}\right[\phantom{\rule{0.3em}{0ex}}0,t+1\left]\phantom{\rule{0.3em}{0ex}}\right).\end{array}
Next, we use induction over {\mathcal{Y}}^{k}.
Step 1 k=0 v_{
r
} is the only observed infected node in this case. Given a sample path \mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t+1]\in \stackrel{~}{\mathcal{X}}(t+1), the probability of the sample path can be written as
Pr\left(\mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t+1]\phantom{\rule{0.3em}{0ex}}\right)=Pr\left(\mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t]\phantom{\rule{0.3em}{0ex}}\right)Pr\left(\mathbf{\text{X}}\right(t+1\left)\right\mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t]\phantom{\rule{0.3em}{0ex}}).
Since v_{
r
} is the only observed infected node and all other nodes’ states are unknown, we assign {\mathbf{\text{X}}}^{\prime}[\phantom{\rule{0.3em}{0ex}}0,t]\in \stackrel{~}{\mathcal{X}}\left(t\right) to be same as the first t time slots in X[ 0,t+1], i.e., X^{′}[ 0,t]=X[ 0,t]. Hence, we obtain that
Pr\left({\mathbf{\text{X}}}^{\prime}[\phantom{\rule{0.3em}{0ex}}0,t]\phantom{\rule{0.3em}{0ex}}\right)=Pr\left(\mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t]\phantom{\rule{0.3em}{0ex}}\right)>Pr\left(\mathbf{\text{X}}[\phantom{\rule{0.3em}{0ex}}0,t+1]\phantom{\rule{0.3em}{0ex}}\right).
Therefore, the case k=0 is proved.
Step 2 Assume the inequality holds for k≤n and consider k=n+1, i.e., \mathbf{\text{Y}}\in {\mathcal{Y}}^{n+1}. Clearly, t≥n+1≥1 for each X[ 0,t]. Furthermore, the set of subtrees \mathcal{T}=\left\{{T}_{u}^{{v}_{r}}\rightu\in \mathcal{C}\left({v}_{r}\right)\} are divided into two subsets:
{\mathcal{T}}^{h}=\left\{{T}_{u}^{{v}_{r}}\rightu\in \mathcal{C}\left({v}_{r}\right),{T}_{u}^{{v}_{r}}\cap {\mathcal{I}}_{\mathbf{\text{Y}}}=\varnothing \}
and
{\mathcal{T}}^{i}=\mathcal{T}\setminus {\mathcal{T}}^{h}.
Given {t}_{{v}_{r}}^{R}, the infection processes on the subtrees are mutually independent.
We construct X^{′}[ 0,t] which occurs more likely than X^{∗}[ 0,t+1] according to the following steps, where {\mathbf{\text{X}}}^{\ast}[\phantom{\rule{0.3em}{0ex}}0,t+1]=arg\underset{\mathbf{\text{X}}[0,t+1]\in \stackrel{~}{\mathcal{X}}(t+1)}{max}Pr\left(\mathbf{\text{X}}\right[\phantom{\rule{0.3em}{0ex}}0,t+1\left]\phantom{\rule{0.3em}{0ex}}\right).
Part 1{\mathcal{T}}^{i}. For a subtree in {\mathcal{T}}^{i} the proof follows Step 2.b and Step 2.c of Lemma 1 in [6]. The intuition is as follows: Consider a subtree and a sample path on it with duration t+1. If u is not infected at the first time slot, we can construct a sample path with duration t by moving the events one time slot earlier. The new sample path (with duration t) has a higher probability to occur than the original one. If u is infected in the first time slot, we can invoke the induction assumption to the subtree rooted at u, which belongs to {\mathcal{Y}}^{n}.
Part 2 v_{
r
}. In this part, we have the freedom to assign the unobserved node as infected or healthy. In part 1, the infection time of each root u in subtrees {\mathcal{T}}^{i} of X^{′}[ 0,t] is either the same as or one time slot earlier than its infection time in X^{∗}[ 0,t+1]. Therefore, if {t}_{{v}_{r}}^{R}\le t, the recovery time of the source v_{
r
} in X^{′}[ 0,t] can be assigned the same as that in X^{∗}[ 0,t+1].
If {t}_{{v}_{r}}^{R}=t+1, the source v_{
r
} recovers at time slot t+1 which means v_{
r
} is not observed since the observation set only contains infected nodes. Therefore, in X^{′}[ 0,t] we assign the source to be in state I at time t, which is the same as the state of v_{
r
} at time t in X^{∗}[ 0,t+1].
If {t}_{{v}_{r}}^{R}>t+1,v_{
r
} remains infected in the sample path X^{∗}[0,t+1]. We assign the source to be in state I in X^{′}[0,t].
As a summary, according to the assignment above, the states of the source v_{
r
} in X^{′}[ 0,t] are the same as those of the first t time slots in X^{∗}[ 0,t+1].
Part 3{\mathcal{T}}^{\mathbf{h}}. Based on the conclusion of part 2, the subtrees belonging to {\mathcal{T}}^{h} in X^{′}[ 0,t] mimic the behaviors of the first t time slots in X^{∗}[ 0,t+1].
Since X^{∗}[ 0,t+1] has one extra time slot during which some extra events occur, X^{′}[ 0,t] occurs with a higher probability on the subtrees in {\mathcal{T}}^{h}.
According to the discussion above, we conclude that time inequality holds for k=n+1 and hence for any k according to the principle of induction. Therefore, the lemma holds. □
Lemma 2 (Adjacent nodes inequality).
Consider an infinite tree with partial observation Y which contains at least one infected node. For u,v\in \mathcal{V} such that (u,v)\in \mathcal{E}, if {t}_{u}^{\ast}>{t}_{v}^{\ast}
\begin{array}{l}Pr\left(\underset{u}{\overset{\ast}{\mathbf{\text{X}}}}\right[\phantom{\rule{0.3em}{0ex}}0,\underset{u}{\overset{\ast}{t}}\left]\phantom{\rule{0.3em}{0ex}}\right)<Pr\left(\underset{v}{\overset{\ast}{\mathbf{\text{X}}}}\right[\phantom{\rule{0.3em}{0ex}}0,\underset{v}{\overset{\ast}{t}}\left]\phantom{\rule{0.3em}{0ex}}\right),\end{array}
where {\mathbf{\text{X}}}_{u}^{\ast}[\phantom{\rule{0.3em}{0ex}}0,{t}_{u}^{\ast}] is the optimal sample path associated with root u.
Proof.
The proof of the lemma follows the proof of Lemma 2 in [6]. The key idea is to construct a sample path rooted at v, which has a higher probability than the optimal sample path rooted at u. It is not hard to see that {t}_{u}^{\ast}={t}_{v}^{\ast}+1 based on the definition of the infection eccentricity. The graph is partitioned into {T}_{v}^{u} and {T}_{u}^{v} which are mutually independent after the infection of v and u. With this observation, we construct {\stackrel{~}{\mathbf{\text{X}}}}_{v}[\phantom{\rule{0.3em}{0ex}}0,{t}_{v}^{\ast}] which infects u at the first time slot. {\stackrel{~}{\mathbf{\text{X}}}}_{v}\left([\phantom{\rule{0.3em}{0ex}}0,{t}_{v}^{\ast}],{T}_{v}^{u}\right) then mimics the behavior of {\mathbf{\text{X}}}_{u}^{\ast}\left([\phantom{\rule{0.3em}{0ex}}0,{t}_{u}^{\ast}],{T}_{v}^{u}\right), and {\stackrel{~}{\mathbf{\text{X}}}}_{v}\left([\phantom{\rule{0.3em}{0ex}}0,{t}_{v}^{\ast}1],{T}_{u}^{v}\right) has a higher probability than {\mathbf{\text{X}}}_{u}^{\ast}\left([\phantom{\rule{0.3em}{0ex}}0,{t}_{u}^{\ast}],{T}_{u}^{v}\right) based on Lemma 1. □
The adjacent nodes inequality results in partial orders in the tree and makes it possible to compare the likelihood of optimal sample paths associated with adjacent nodes without knowing the actual probability of the optimal sample path. Following the proof of Theorem 4 in [6], it can be shown that in tree networks, from any node, there exists a path from the node to a Jordan infection center such that the observed infection eccentricity strictly decreases along the path. By repeatedly using Lemma 2, we can then prove that the source of the optimal sample path must be a Jordan infection center.
3.2.2 Proof of Theorem 2
In this subsection, we present the proof that shows that the sample path estimator is within a constant distance from the actual source independent of the size of the infected subnetwork. Given a tree rooted in v^{∗} where the information starts from v^{∗} following the general SIR model, we define the following three branching processes:

1.
{\mathcal{Z}}_{l}\left({T}_{{v}^{\ast}}\right) denotes the set of nodes which are in infected or recovered states at level l on tree {T}_{{v}^{\ast}}. Let {Z}_{l}\left({T}_{{v}^{\ast}}\right) denote the cardinality of {\mathcal{Z}}_{l}\left({T}_{{v}^{\ast}}\right). Note that {\mathcal{Z}}_{0}\left({T}_{{v}^{\ast}}\right)=\left\{{v}^{\ast}\right\}. We call this process the original infection process.

2.
{\mathcal{Z}}_{l}^{\tau}\left({T}_{{v}^{\ast}}\right) denotes the set of infected and recovered nodes at level l whose parents are in set {\mathcal{Z}}_{l1}^{\tau}\left({T}_{{v}^{\ast}}\right) and who were infected within τ time slots after their parents were infected. This process adds a deadline τ on infection. If a node is not infected within τ time slots after its parent is infected, it is not included in this branching process. This process is called τ deadline infection process. From the definition, if u,v\in {\mathcal{Z}}_{l}^{\tau}\left({T}_{{v}^{\ast}}\right), then
{t}_{u}^{I}{t}_{v}^{I}\le l(\tau 1).
For τ=1, we call {\mathcal{Z}}_{l}^{1}\left({T}_{{v}^{\ast}}\right) the onetimeslot infection process. The extinction probability of a branching process is the probability that there is no offspring at a certain level of the branching process, i.e., {Z}_{l}^{1}\left({T}_{{v}^{\ast}}\right)=0 for some l. Denote by ρ_{
v
} the extinction probability of {Z}_{l}^{1}\left({T}_{v}^{\varphi \left(v\right)}\right).

3.
We define the binomial branching process as a branching process whose offspring distribution follows binomial distribution B(g,φ) where g is the number of trials and φ is the success probability. Denote by ρ the extinction probability of the binomial branching process.
The following notations will be used in later analysis:
v^{†} denotes the optimal sample path estimator.
g_{min} is the lower bound on the number of children, i.e.,
\underset{v}{min}\left\mathcal{C}\right(v\left)\right\ge {g}_{min},\forall v\in \mathcal{V.}
q_{min} is the lower bound on the infection probability, i.e.,
{q}_{min}=\underset{e}{min}{q}_{e},\forall e\in \mathcal{E.}
{\sigma}_{v}^{\tau} is the probability that a node v infects at least one of its children within τ time slot after v is infected.
Given n_{0}>0 and τ>0, define l^{†}= minl^{′} where {Z}_{{l}^{\prime}}^{\tau}\left({T}_{{v}^{\ast}}\right)>{n}_{0}, i.e., l^{†} is the first level where the τdeadline infection process has more than n_{0} offsprings.
Given τ and level L≥2, we consider the following two events:
Event 1: {Z}_{L}\left({T}_{{v}^{\ast}}\right)=0.
Event 2: l^{†}≤L and at least two onetimeslot infection processes starting from level l^{†} survive, i.e., \exists u,v\in {\mathcal{Z}}_{{l}^{\u2020}}^{\tau}\left({T}_{{v}^{\ast}}\right) such that ∀l,{Z}_{l}^{1}\left({T}_{u}^{\varphi \left(u\right)}\right)\ne 0 and {Z}_{l}^{1}\left({T}_{v}^{\varphi \left(v\right)}\right)\ne 0. In addition, at least one infected node at the bottom of each survived onetimeslot infection process is observed.
For event 1, no node at level L gets infected and the infection process terminates at level L−1. So the infection eccentricity of v^{∗} is at most L−1, and the minimum infection eccentricity of the network is at most L−1. Therefore, the distance between v^{∗} and v^{†} is no more than 2(L−1).
Considering event 2, we assume that the information propagates for t time slots. The deadline property of the τdeadline infection process indicates {t}_{{u}_{1}}^{I}\le \tau {l}^{\u2020} and {t}_{{u}_{2}}^{I}\le \tau {l}^{\u2020}. Given a node \stackrel{~}{v} at level (τ+1)l^{†}−1 where \stackrel{~}{v}\in {T}_{{u}_{2}}^{\varphi \left({u}_{2}\right)} and a node {v}^{\prime}\in {T}_{{u}_{1}}^{\varphi \left({u}_{1}\right)} which is an observed infected node at the bottom of the infection tree, from Figure 7, we obtain
\begin{array}{ll}d\left(\stackrel{~}{v},{v}^{\prime}\right)& =t{t}_{{u}_{1}}^{I}+\tau {l}^{\u2020}+1\ge t+1.\phantom{\rule{2em}{0ex}}\end{array}
Note that \forall u\in \mathcal{I},
d\left({v}^{\ast},u\right)\le t<d\left(\stackrel{~}{v},{v}^{\prime}\right).
Since l^{†}≤L, any node at or below level L(τ+1)−1 has an infection eccentricity larger than that of v^{∗}. Hence, v^{†} cannot be at or below level L(τ+1)−1. Therefore,
d\left({v}^{\u2020},{v}^{\ast}\right)<(\tau +1)L1.
Next, we prove the probability that either event 1 or event 2 happens goes asymptotically to 1. Denote by {K}_{{l}^{\u2020}} the number of onetimeslot infection processes which start from level l^{†} and survive. Denote by E the event that a survived onetimeslot infection process has at least one observed infected node at its lowest level.
According to the discussion above, the probability that the distance between the estimator and the actual source is no more than (τ+1)L−1 is at least
\begin{array}{cc}Pr\left({Z}_{L}\phantom{\rule{0.3em}{0ex}}\right.& \left(\right)close=")">\left({T}_{{v}^{\ast}}\right)=0& +Pr\left({K}_{{l}^{\u2020}}\ge 2,{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}\end{array}\ge Pr\left({Z}_{L}\left({T}_{{v}^{\ast}}\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}0\right)\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}Pr\left({l}^{\u2020}\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}L\right)Pr\left({K}_{{l}^{\u2020}}\phantom{\rule{0.3em}{0ex}}\ge \phantom{\rule{0.3em}{0ex}}2\left\right.{l}^{\u2020}\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}L\right)\phantom{\rule{0.3em}{0ex}}Pr{\left(E\right)}^{2}\\ =Pr\left({Z}_{L}\left({T}_{{v}^{\ast}}\right)=0\right)+Pr\left(\bigcup _{i=1}^{L}\underset{i}{\overset{\tau}{Z}}{n}_{0}\right)\\ \phantom{\rule{1em}{0ex}}\times Pr({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}\le L)Pr{\left(E\right)}^{2}\\ =\left(1\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}Pr\left(\bigcap _{i=1}^{L}0\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}Pr\left(\bigcup _{i=1}^{L}\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}0\right)\right)\\ \phantom{\rule{1em}{0ex}}\times Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}+Pr\left({Z}_{L}\left({T}_{{v}^{\ast}}\right)=0\right).\n
In addition, we have
\begin{array}{ll}Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}\le L\right)& =\sum _{l=1}^{L}Pr\left({K}_{{l}^{\u2020}}\ge 2,{l}^{\u2020}=l{l}^{\u2020}\le L\right)\phantom{\rule{2em}{0ex}}\end{array}
(3)
\begin{array}{l}=\sum _{l=1}^{L}Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}=l\right)Pr\left({l}^{\u2020}=l{l}^{\u2020}\le L\right).\phantom{\rule{2em}{0ex}}\end{array}
(4)
In Lemma 3, we prove that the extinction probability of each branching process from level l^{†} is upper bounded by the extinction probability ρ of the binomial infection process B(g_{min},q_{min}). Therefore, at level l^{†}, we have n_{0} i.i.d onetime infection processes whose extinction probabilities are upper bounded by ρ. The probability that at least two of them survive goes asymptotically to 1 when n_{0} increases. Therefore, ∀ε_{1}>0, we have enough large n_{0}, such that
Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}=l\right)\ge 1{\epsilon}_{1}.
Therefore, Equation 4 becomes
\begin{array}{ll}Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}\le L\right)& \ge (1{\epsilon}_{1})\sum _{l=1}^{L}Pr\left({l}^{\u2020}=l{l}^{\u2020}\le L\right)\phantom{\rule{2em}{0ex}}\\ =(1{\epsilon}_{1}).\phantom{\rule{2em}{0ex}}\end{array}
We show in Lemma 4 that Pr(E)≥1−ε_{2} given ε_{2}>0. If n_{0} and t are sufficiently large, we have
Pr\left({K}_{{l}^{\u2020}}\ge 2{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}\ge (1{\epsilon}_{1}){\left(1{\epsilon}_{2}\right)}^{2}.
Therefore,
\begin{array}{cc}Pr({Z}_{L}& \left({T}_{{v}^{\ast}}\right)=0)+Pr\left({K}_{{l}^{\u2020}}\ge 2,{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}\\ \ge \left(1Pr\left(\bigcap _{i=1}^{L}0<\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)\right)(1{\epsilon}_{1}){(1{\epsilon}_{2})}^{2}\\ \phantom{\rule{1em}{0ex}}Pr\left(\bigcup _{i=1}^{L}\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)=0\right)+Pr\left({Z}_{L}\right({T}_{{v}^{\ast}})=0)\\ =\underset{\mathrm{Part1}}{\underset{\#}{\left(1Pr\left(\bigcap _{i=1}^{L}0<\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)\right)}}(1{\epsilon}_{1}){(1{\epsilon}_{2})}^{2}\\ \phantom{\rule{1em}{0ex}}+\underset{\mathrm{Part2}}{\underset{\#}{Pr\left({Z}_{L}\right({T}_{{v}^{\ast}})=0)Pr\left(\underset{L}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)=0\right)}},\end{array}
(5)
where Equation 5 holds since {Z}_{l}^{\tau}\left({T}_{{v}^{\ast}}\right)=0 implies that {Z}_{l}^{\tau}\left({T}_{{v}^{\ast}}\right)=0 for l≤L.
For part 1 in Equation 5, we prove in Lemma 4, given ε_{3}>0, when τ and L are sufficiently large,
1Pr\left(\bigcap _{i=1}^{L}0<\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)>1{\epsilon}_{3}.
For part 2 in Equation 5, we have
{lim}_{\tau \to \infty}Pr\left(\underset{L}{\overset{\tau}{Z}}\right({T}_{{v}^{\ast}})=0)=Pr\left({Z}_{L}\right({T}_{{v}^{\ast}})=0).
Therefore, given ε_{4}>0, when τ is sufficiently large,
Pr\left({Z}_{L}\right({T}_{{v}^{\ast}})=0)Pr\left(\underset{L}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)=0\right)\ge {\epsilon}_{4}.
Hence, we have
\begin{array}{ll}Pr\left({Z}_{L}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right.& \left(\right)close=")">\left({T}_{{v}^{\ast}}\right)=0& +Pr\left({K}_{{l}^{\u2020}}\ge 2,{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}\phantom{\rule{2em}{0ex}}\end{array}\n \n \n \u2265\n (\n 1\n \u2212\n \n \n \epsilon \n \n \n 1\n \n \n )\n \n \n \n \n 1\n \u2212\n \n \n \epsilon \n \n \n 2\n \n \n \n \n \n \n 2\n \n \n \n \n 1\n \u2212\n \n \n \epsilon \n \n \n 3\n \n \n \n \n \u2212\n \n \n \epsilon \n \n \n 4\n \n \n .\n \n \n \n
Now choosing ε_{1}=ε_{2}=ε_{3}=ε_{4}=ε_{5}/5 for some ε_{4}>0, we have
\begin{array}{l}Pr\left({Z}_{L}\left({T}_{{v}^{\ast}}\right)=0\right)+Pr\left({K}_{{l}^{\u2020}}\ge 2,{l}^{\u2020}\le L\right)Pr{\left(E\right)}^{2}\ge 1{\epsilon}_{5}.\phantom{\rule{2em}{0ex}}\end{array}
Now let Y denote the number of infected nodes in the observation Y. Define events E_{1}={Z_{
L
}=0} and E_{2}={K_{
l
}≥2f o r s o m e l≤L}, and E_{3} is the event that two of the survived onetimeslot infection processes have at least one observed infected node each at their bottoms. We have
\begin{array}{l}Pr\left({E}_{1}\right\left\mathbf{\text{Y}}\right\ge 1)+Pr\left({E}_{2}\cap {E}_{3}\left\right\mathbf{\text{Y}}\ge 1\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{Pr\left(\right\mathbf{\text{Y}}\ge 1)}\left(Pr({E}_{1}\cap \{\left\mathbf{\text{Y}}\right\ge 1\left\}\right)\right.\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\right)close=")">+Pr\left({E}_{2}\cap {E}_{3}\cap \left\{\right\mathbf{\text{Y}}\ge 1\}\right)& .\phantom{\rule{2em}{0ex}}\end{array}\n
Since E_{2}∩E_{3} implies that Y≥1, we have
\begin{array}{cc}Pr\left({E}_{1}\right& \mathbf{\text{Y}}\ge 1)+Pr\left({E}_{2}\cap {E}_{3}\left\right\mathbf{\text{Y}}\ge 1\right)\\ =\frac{1}{Pr\left(\right\mathbf{\text{Y}}\ge 1)}\left(Pr({E}_{1}\cap \{\left\mathbf{\text{Y}}\right\ge 1\left\}\right)+Pr\left({E}_{2}\cap {E}_{3}\right)\right)\\ =\frac{1}{Pr\left(\right\mathbf{\text{Y}}\ge 1)}\left(Pr\left({E}_{1}\right)Pr({E}_{1}\cap \{\left\mathbf{\text{Y}}\right=0\left\}\right)\right.\\ \left(\right)close=")">\phantom{\rule{1em}{0ex}}+Pr\left({E}_{2}\cap {E}_{3}\right)\end{array}\n \n \n \u2265\n \n \n 1\n \n \n Pr\n (\n \n Y\n \n \u2265\n 1\n )\n \n \n \n \n Pr\n (\n \n \n E\n \n \n 1\n \n \n )\n \u2212\n Pr\n (\n {\n \n Y\n \n =\n 0\n }\n )\n +\n Pr\n \n \n \n \n E\n \n \n 2\n \n \n \u2229\n \n \n E\n \n \n 3\n \n \n \n \n \n \n \n \n \n \n \u2265\n \n \n 1\n \n \n Pr\n (\n \n Y\n \n \u2265\n 1\n )\n \n \n \n \n Pr\n (\n {\n \n Y\n \n \u2265\n 1\n }\n )\n \u2212\n \n \n \epsilon \n \n \n 5\n \n \n \n \n =\n 1\n \u2212\n \n \n \n \n \epsilon \n \n \n 5\n \n \n \n \n Pr\n (\n \n Y\n \n \u2265\n 1\n )\n \n \n .\n \n \n
(6)
Note that Pr(Y≥1) is a positive constant since blackthe onetimeslot infection process starting from the information source survives with nonzero probability. The theorem holds by choosing ε_{5}=ε Pr(Y≥1).
Lemma 3.
The extinction probability of a onetimeslot infection process is smaller than the extinction probability of a binomial branching process B(g_{min},q_{min}), i.e., \forall v\in \mathcal{V},
{\rho}_{v}<\mathrm{\rho .}
Proof.
As shown in Figure 8, we construct a virtual source process{Z}_{l}^{\left(\mathit{\text{vs}}\right)}\left({T}_{v}^{\varphi \left(v\right)}\right) and a mininfection process{Z}_{l}^{\left(\mathit{\text{mi}}\right)}\left({T}_{v}^{\varphi \left(v\right)}\right) as auxiliary processes over the same tree topology where {Y}_{v}^{\left(\mathit{\text{vs}}\right)} and {Y}_{v}^{\left(\mathit{\text{mi}}\right)} are the binary numbers indicating whether node v has been infected. Denote by {\rho}_{v}^{\left(\mathit{\text{vs}}\right)} and {\rho}_{v}^{\left(\mathit{\text{mi}}\right)} the extinction probabilities, respectively.
In the mininfection process, infection spreads over edges with probability q_{min}. In the virtual source process, the probability that a node gets infected is
\begin{array}{ll}Pr\left(\underset{v}{\overset{\left(\mathit{\text{vs}}\right)}{Y}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}1\right)& =Pr\left(\underset{v}{\overset{\left(\mathit{\text{mi}}\right)}{Y}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}1\right)\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}Pr\left(\underset{v}{\overset{\left(\mathit{\text{mi}}\right)}{Y}}=0\right)\phantom{\rule{0.3em}{0ex}}\xb7\phantom{\rule{0.3em}{0ex}}\frac{{q}_{\mathit{\text{uv}}}{q}_{min}}{1{q}_{min}}={q}_{\mathit{\text{uv}}},\phantom{\rule{2em}{0ex}}\end{array}
i.e., for each node u\in \mathcal{C}\left(v\right),v tries to infect u with probability q_{min}. If v fails to infect u, a virtual source v^{′} tries to infect u with probability \frac{{q}_{\mathit{\text{vu}}}{q}_{min}}{1{q}_{min}}. Therefore, the virtual source process has the same distribution with the onetimeslot infection process.
We now couple the mininfection process and the virtual source infection process as follows:
If {Y}_{v}^{\left(\mathit{\text{mi}}\right)}=1, then {Y}_{v}^{\left(\mathit{\text{vs}}\right)}=1.
If {Y}_{v}^{\left(\mathit{\text{mi}}\right)}=0, then {Y}_{v}^{\left(\mathit{\text{vs}}\right)}=1 with probability \frac{{q}_{\mathit{\text{uv}}}{q}_{min}}{1{q}_{min}}.
Since a node is more likely to get infected in the virtual source infection process, we obtain
{\rho}_{v}^{\left(\mathit{\text{vs}}\right)}\le {\rho}_{v}^{\left(\mathit{\text{mi}}\right)}.
Recalling that the onetimeslot infection process has the same distribution with the virtual source branching process, we obtain {\rho}_{v}\le {\rho}_{v}^{\left(\mathit{\text{mi}}\right)},\forall \mathrm{v.}
In addition, the mininfection process has more children than the binomial branching process with the same infection probability for each child. It is obvious that the binomial branching process is more likely to die out, i.e., {\rho}_{v}^{\left(\mathit{\text{mi}}\right)}<\mathrm{\rho .}
As a summary, we prove
{\rho}_{v}<\mathrm{\rho .}
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Lemma 4.
Assume ∃ξ>0 such that {\sigma}_{v}^{\tau}<1\xi ,\forall v\in \mathcal{V.} Given any ε>0, there exists a constant L^{′} such that for any L≥L^{′},
Pr\left(\bigcap _{i=1}^{L}0<\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)\le \epsilon
Proof.
Follows the same argument of Lemma 7 in [6], and by choosing
{L}^{\prime}=\u2308\frac{log\epsilon}{log\left(1{\xi}^{{n}_{0}}\right)}\u2309,
we obtain for any L≥L^{′},ε>0
Pr\left(\bigcap _{i=1}^{L}0<\underset{i}{\overset{\tau}{Z}}\left({T}_{{v}^{\ast}}\right)\le {n}_{0}\right)\le \mathrm{\epsilon .}
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Lemma 5.
For any ε>0, there exists a sufficiently large t such that
Pr\left(E\right)\ge 1\mathrm{\epsilon .}
Proof.
Note that the binomial branching process B(g_{min},q_{min}) is a GaltonWatson (GW) process [12] which requires each node to have an i.i.d offspring distribution. The previous result about the instability of the GaltonWatson process in Theorem 6.2 in [12] proves that the GW process either goes to infinity or goes to 0. If the GW process survives, the number of offsprings goes to infinity as the level increases. Therefore, for a sufficiently long time, the survived binomial branching process will have a sufficiently large number of offsprings at the lowest level. Since the onetimeslot infection process always has at least the same number of children as the binomial branching process, the survived onetimeslot infection process will have enough number of infected nodes at the lowest level as time increases. According to the unbiased property of the partial observation, after a sufficiently long time, the probability that at least one infected node in the lowest level is observed goes to 1 asymptotically, i.e.,
Pr\left(E\right)\ge 1\mathrm{\epsilon .}
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