a node with a smaller infection eccentricity is more likely to occur.

\[lem:mainlemma\] Consider a tree network 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. For $u,v \in {\cal V}$ such that $(u,v)\in
{\cal E}$, if $t^*_u>t^*_v,$ then $$\begin{aligned}
\Pr({\bf X}^*_u([0,t^*_u])) < \Pr({\bf X}^*_v([0,t^*_v])),\end{aligned}$$ where ${\bf X}^*_u[0, t_u^*]$ is the optimal sample path starting from node $u.$

Recall that $T_{v}$ denotes the tree rooted at $v$ and $T^{-v}_u$ denotes the tree rooted at $u$ but without the branch from $v.$ Furthermore, ${\cal C}(v)$ is the set of children of $v,$ and ${\bf X}([0,t],T_u^{-v})$ is the sample path ${\bf X}[0,t]$ restricted to $T_u^{-v}.$

[**Step 1:**]{} The first step is to show $t^*_u=t^*_v+1.$ First we claim $T^{-u}_v \cap {\cal I}\neq \emptyset.$ Otherwise, all infected node are on $T^{-v}_u.$ Since on a tree, $v$ can only reach nodes in $T^{-v}_u$ through edge $(u,v),$ $t^*_v=t^*_u+1,$ which contradicts $t^*_u>t^*_v.$

If $T^{-v}_u \cap {\cal I}\neq \emptyset,$ $\forall a \in T^{-v}_u
\cap {\cal I}$, we have $$d(u,a)=d(v,a)-1\leq t^*_v-1,$$ and $\forall b \in T^{-u}_v \cap {\cal I}$, $$d(u,b)=d(v,b)+1\leq
t^*_v+1.$$ Hence, $$t^*_u\leq t_v^*+1,$$ which implies that $$t^*_v<t^*_u\leq t^*_v+1,$$ i.e., $t^*_u=t^*_v+1.$

If 