at the third step, we will show that given any two nodes $u$ and $v,$ if $v$ has the minimum infection eccentricity and $u$ has a larger infection eccentricity, then there exists a path from $u$ to $v$ along which the infection eccentricity monotonically decreases, which implies that the source of the optimal sample path must be a Jordan infection center. For example, in Figure \[figure:infectioneccentricity\], node $v_4$ has a larger infection eccentricity than $v_1$ and $v_4\rightarrow v_3
\rightarrow v_2 \rightarrow v_1$ is the path along which the infection eccentricity monotonically decreases from $5$ to $2.$

The Optimal Time
----------------

\[lem:inequalitytime\] Consider a tree network rooted at $v_r$ and with infinitely many levels. Assume the information source is the root, and the observed infection topology is ${\bf Y}$ which contains at least one infected node. If $\tilde{e}(v_r)\leq t_1<t_2,$ then the following inequality holds $$\max_{{\bf X}[0,t_1]\in {\cal X}(t_1)} \Pr({\bf X}[0,t_1])> \max_{{\bf X}[0,t_2] \in {\cal X}(t_2)} \Pr({\bf X}[0,t_2]),$$ where ${\cal X}(t)=\{{\bf X}[0,t]|{\bf F}({\bf
X}(t))={\bf Y}\}.$ In addition, $$t_{v_r}^*=\tilde{e}(v_r)=\max_{u\in{\cal I}}d(v_r,u),$$ where $d(v_r,u)$ is the length of the shortest path between $v_r$ and $u$ and also called the distance between $v_r$ and $u$, and ${\cal I}$ is the set of infected nodes.

We start 