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\title{Inference of a Rumor's Source in the Independent Cascade Model\\(Supplementary Material)}

\begin{document}

\onecolumn
\appendix

\maketitle



\section{Omitted Proofs}

\subsection{Proof of Theorem 1}
Observe that by definition, we have for any $v, w \in V$ that $ \Pr \bc{ \pz = v } = \Pr \bc{ \pz = w }$.
Thus, by Bayes' rule and the law of total probability we get
\begin{align*}
    \Pr & \bc{ \pz = v \mid \activenodes = X} = \frac{ \Pr \bc{ \activenodes = X \mid \pz = v} \Pr \bc{ \pz = v }}{ \Pr \bc{ \activenodes = X } }  \\
    &= \frac{ \Pr \bc{ \activenodes = X \mid \pz = v} \Pr \bc{ \pz = v }}{ \sum\limits_{\omega \in V} \Pr \bc{ \activenodes = X \mid \pz = \omega} \Pr \bc{ \pz = w }} 
    = \frac{ \Pr \bc{ \activenodes = X \mid \pz = v} }{ \sum\limits_{\omega \in V} \Pr \bc{ \activenodes = X \mid \pz = \omega}}.
\end{align*}
As $\sum_{\omega \in V} \Pr \bc{ \activenodes = X \mid \pz = \omega}$ is independent from $v$, we have
\begin{align*}
    \arg\max_{v \in V}~ \Pr \bc{ \pz = v \mid \activenodes = X} &= \arg\max_{v \in V}~ \Pr \bc{ \activenodes = X \mid \pz = v}
\end{align*}
and the theorem follows. \qed


\subsection{Proof of Proposition 8}
\begin{proof}
The recurrence $\bar{x}_t = \exp \bc{- \lambda p (1 - \bar{x}_{t-1})}$ can be easily calculated by the probability generating function of the Poisson distribution. Indeed, let $f_{\Po(\lambda)}(s) = \Erw \brk{ s^{\Po(\lambda)} }$ be the probability generating function of the Poisson distribution. It is well known that
\begin{align*}
    f_{\Po(\lambda)}(s) = \exp \bc{ - \lambda (1 - s) }.
\end{align*}
We refer to \citep{branching_2020} for a detailed explanation of the connection between the probability generating function and the extinction probability of branching processes.

Now, for brevity, suppose that $v = \lca$. Let $\cV_0$ be the event that $v$ has exactly $k \leq \vec d_0 \leq d$ children that get activated by $v$. Similarly as before, $\Pr \bc{ \cV_0 } = \Pr \bc{ \Po( \lambda p ) = \vec d_0 }$ and of course, $\vec d_0$ needs to be at least $k$ as differently, the probability of having $k$ active sub-trees was zero. 

Given $\cV_0$, we again start $\vec d_0$ independent Galton-Watson processes with offspring distribution $\Po( \lambda p )$ in the children. Therefore, the probability of observing exactly $k$ active sub-trees is the probability that exactly $k$ out of $\vec d_0$ of those processes are not extinct after $\distactive{v}$ steps. Of course, the number of such active sub-trees at time $t$ is distributed as $ \Bin \bc{ \vec d_0, \bar{x}_t } $ given $\cV_0$ and the first part of the formula follows.

As in the $d$-regular case, if on contrary $v$ is not the closest candidate but a node further apart from $\activenodes$, we observe that from the originally $1 \leq \vec d_0 \leq d$ Galton-Watson processes originated in the children of $v$, exactly one process needed to survive and $\vec d_0 - 1$ needed to be extinct at time $\distactive{v}$.
\end{proof}

\subsection{Proof of Theorem 3 (i)}
\begin{proof}[Proof of Theorem 3 (i)]
As in the $d$-regular case, the first part of Theorem 3 follows by the first part of Proposition 8. If $\lambda p \leq 1$, the smallest fixed-point of $\bar{x} \mapsto \exp{ - \lambda p (1 - \bar{x}) }$ is $\bar{x} = 1$. Therefore, $\bar{x}_t = 1 - o_t(1)$ describes the probability that the underlying spreading process died out until time-step $t$. More precisely, by the recurrence equation, we find the following. Suppose that $\eps_t = o_t(1)$ denotes the convergence speed towards $1$. Then, by the recurrence equation and a Taylor approximation we have
\begin{align*}
    1 - \eps_t &= 1 - \lambda p \bc{\eps_{t-1} + \frac{\lambda^2 p^2 \eps_{t-1}^2}{2} } + O \bc{ \eps_{t-1}^3 }.
\end{align*}
If $\lambda p < 1$, we directly find that $\eps_t = O \bc{ (\lambda p)^t }$ decays exponentially fast in $t$. If $\lambda p = 1$, this is much more subtle. Indeed, we find
\begin{align*}
    \eps_t & \leq \bc{\eps_{t-1} - \frac{\eps_{t-1}^2}{2} } + O \bc{ \eps_{t-1}^3 }
\end{align*}
and therefore, we only get $\eps_t = O \bc{ t^{-1} }$ in this case.

Since we assume $p$ to be a constant, clearly $\lambda = O(1)$ as well. Unfortunately, the Poisson tails are kind of heavy. More precisely, even if $\lambda$ is a constant, the probability that a $\Po(\lambda)$ variable becomes large is not negligible. We analyze this by a careful application of limits. Recall that we assume that the underlying tree-network is infinite. We model this as follows. Suppose that the tree-network consists of $n$ vertices and we will let $n \to \infty$.

Let $C > 0$, then the probability that the number of neighbors of a specific node $v$ exceeds $C$ is, for large $C$, given by Chernoff bounds as
{
\begin{align*}
    \Pr \bc{ \abs{ \partial v } > C }  \leq \exp \bc{ - \frac{ (C - \lambda)^2 }{ 2 C } } \sim \exp \bc{ - C/2 }. 
\end{align*}
}
As the number of spawned children is independent for all vertices, the number of vertices of degree at least $C$ is stochastically dominated by $\Bin \bc{n, \exp \bc{-C/2}}$. Thus, with probability $1 - o_n(1)$, there are no more than $O( n \sqrt{\log(n)} \exp \bc{ - D } )$ vertices of degree $D > 0$ for a sufficiently large constant $D$ (independent of $n$) if $n \to \infty$.

We denote by $\cD$ the event that this is actually true. Thus, conditioned on $\cD$, there are only $O( n \sqrt{\log(n)} \exp \bc{ - D } )$ vertices of degree at least $D$. Now, we chose $\pz$ uniformly at random out of all vertices. Therefore, given $\cD$, the probability that $\pz$ has small degree is
\begin{align*}
    \Pr \bc{ \abs{ \partial \pz } > D \mid \cD} = 1 - O \bc{ \frac{ \sqrt{\log(n)} }{\exp \bc{ - D }} }.
\end{align*}
Clearly, this becomes only a high probability event if $D = \Omega \bc{ \log \log n}$. In the worst case, we find that a union bound over all activated children of $\pz$ leads only to ultimate extinction of all processes, if 
    $O\bc{ \frac{\log \log n}{t} } = o_t(1),$
or, differently, that $t = \omega(\log \log n)$. As in the theorem, we only claim the assertion in the limit $t \to \infty$ and we assume the underlying tree-network to be infinite. This proves the claim of the theorem. We remark at this point that the assumption that $t$ depends slightly on $n$ does no harm to applications as, on real networks, $\log \log n$ can be seen as a constant.
\end{proof}


\section{Simulation Data}


\begin{table}[b!]
\centering
\caption{Simulation results for random geometric graphs.}
\label{tab:table1}
 \small
\begin{tabular}{lrrrrr}
\toprule
$p$ & \parbox{\widthof{number of}}{\raggedleft number of\newline successes} & $\cc \neq \pz$ & $\activenodes = \emptyset$ & \parbox{\widthof{distance}}{\raggedleft average\newline distance} & \parbox{\widthof{maximum}}{\raggedleft maximum\newline distance}\\
\midrule
0.00 & 0 & 0 & 100 &  & 0 \\
0.05 & 1 & 1 & 98 & 2.25 & 4 \\
0.10 & 5 & 29 & 66 & 2.39 & 5 \\
0.15 & 25 & 59 & 16 & 2.02 & 6 \\
0.20 & 34 & 63 & 3 & 1.68 & 5 \\
0.25 & 51 & 48 & 1 & 1.51 & 4 \\
0.30 & 62 & 36 & 2 & 1.40 & 5 \\
0.35 & 71 & 29 & 0 & 1.24 & 5 \\
0.40 & 86 & 14 & 0 & 1.11 & 4 \\
0.45 & 94 & 6 & 0 & 1.04 & 3 \\
0.50 & 94 & 6 & 0 & 1.13 & 5 \\
0.55 & 95 & 5 & 0 & 1.07 & 5 \\
0.60 & 100 & 0 & 0 & 0.97 & 4 \\
0.65 & 95 & 5 & 0 & 1.03 & 6 \\
0.70 & 99 & 1 & 0 & 0.79 & 3 \\
0.75 & 99 & 1 & 0 & 1.03 & 5 \\
0.80 & 100 & 0 & 0 & 0.97 & 5 \\
0.85 & 100 & 0 & 0 & 0.96 & 6 \\
0.90 & 100 & 0 & 0 & 0.66 & 3 \\
0.95 & 98 & 2 & 0 & 0.87 & 5 \\
1.00 & 100 & 0 & 0 & 0.85 & 6 \\
\bottomrule
\end{tabular}
\end{table}

\begin{table}[p]
\caption{Simulation results for Erdős-Rényi graphs.}
\centering
\label{tab:table2}
 \small
\begin{tabular}{lrrrrr}
\toprule
$p$ & \parbox{\widthof{number of}}{\raggedleft number of\newline successes}& $\cc \neq \pz$ & $\activenodes = \emptyset$ & \parbox{\widthof{distance}}{\raggedleft average\newline distance} & \parbox{\widthof{maximum}}{\raggedleft maximum\newline distance}\\
\midrule
0.00 & 0 & 0 & 100 & - & -\\
0.05 & 0 & 0 & 100 & - & -\\
0.10 & 0 & 0 & 100 & - & -\\
0.15 & 0 & 1 & 99 & 6.00 & 6\\
0.20 & 0 & 6 & 94 & 7.50 & 9\\
0.25 & 0 & 14 & 86 & 6.63 & 8\\
0.30 & 2 & 30 & 68 & 7.34 & 10\\
0.35 & 11 & 35 & 54 & 7.23 & 10\\
0.40 & 21 & 49 & 30 & 5.87 & 9\\
0.45 & 33 & 43 & 24 & 6.14 & 9\\
0.50 & 42 & 33 & 25 & 1.15 & 8\\
0.55 & 54 & 31 & 15 & 0.54 & 3\\
0.60 & 63 & 24 & 13 & 0.36 & 3\\
0.65 & 74 & 19 & 7 & 0.30 & 2\\
0.70 & 78 & 17 & 5 & 0.21 & 2\\
0.75 & 78 & 15 & 7 & 0.17 & 2\\
0.80 & 82 & 12 & 6 & 0.18 & 3\\
0.85 & 81 & 15 & 4 & 0.17 & 2\\
0.90 & 86 & 13 & 1 & 0.17 & 2\\
0.95 & 85 & 10 & 5 & 0.10 & 1\\
1.00 & 92 & 5 & 3 & 0.05 & 1\\
\bottomrule
\end{tabular}
\end{table}


\begin{table}[p]
\centering
\caption{Simulation results for random regular graphs (configuration model).}
\label{tab:table3}
\small
\begin{tabular}{lrrrrr}
\toprule
$p$ & \parbox{\widthof{number of}}{\raggedleft number of\newline successes} & $\cc \neq \pz$ & $\activenodes = \emptyset$ & \parbox{\widthof{distance}}{\raggedleft average\newline distance} & \parbox{\widthof{maximum}}{\raggedleft maximum\newline distance}\\
\midrule
0.00 & 0 & 0 & 100 & - & - \\
0.05 & 0 & 0 & 100 & - & - \\
0.10 & 0 & 0 & 100 & - & - \\
0.15 & 0 & 0 & 100 & - & - \\
0.20 & 0 & 0 & 100 & - & - \\
0.25 & 0 & 5 & 95 & 7.20 & 9 \\
0.30 & 0 & 16 & 84 & 7.53 & 11 \\
0.35 & 2 & 26 & 72 & 6.86 & 12 \\
0.40 & 19 & 43 & 38 & 5.37 & 11 \\
0.45 & 38 & 40 & 22 & 2.70 & 11 \\
0.50 & 43 & 41 & 16 & 2.06 & 9 \\
0.55 & 70 & 23 & 7 & 0.57 & 6 \\
0.60 & 76 & 21 & 3 & 0.34 & 4 \\
0.65 & 86 & 11 & 3 & 0.15 & 3 \\
0.70 & 87 & 10 & 3 & 0.14 & 3 \\
0.75 & 98 & 2 & 0 & 0.02 & 1 \\
0.80 & 97 & 3 & 0 & 0.03 & 1 \\
0.85 & 99 & 1 & 0 & 0.01 & 1 \\
0.90 & 100 & 0 & 0 & 0 & 0 \\
0.95 & 100 & 0 & 0 & 0 & 0 \\
1.00 & 100 & 0 & 0 & 0 & 0 \\
\bottomrule
\end{tabular}
\end{table}




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