X}_v\left([0,t^*_v],T^{-v}_u\right)\Big|t^I_u=1\right).\end{aligned}$$

For the subtree $T_v^{-u},$ given ${\bf X}^*_u\left([0,t^*_u],T^{-u}_v\right),$ in which $t_v^I=1$, we construct the partial sample path ${\bf
X}_v\left([0,t^*_v],T^{-u}_v\right)$ to be identical to ${\bf X}^*_u\left([0, t_u^*], T_v^{-u}\right)$ except that all events occur one time slot earlier, i.e., $${\bf X}_v\left([0,t^*_v],T^{-u}_v\right)={\bf X}^*_u\left([1, t_u^*], T_v^{-u}\right).$$ This is feasible because $t_v^*=t_u^*-1.$ Then $$\Pr\left({\bf X}^*_u\left([0,t^*_u],T^{-u}_v\right)\Big|t^I_v=1\right)=\Pr\left({\bf
X}_v\left([0,t^*_v],T^{-u}_v\right)\right).$$

For the subtree $T_u^{-v}$, we construct ${\bf X}_v([0,t^*_v],T_u^{-v})$ such that $$\begin{aligned}
&{\bf X}_v([0,t^*_v],T_u^{-v})\in \\
&{\arg\max}_{\tilde{\bf X}([0,t^*_v],T_u^{-v})\in{\cal
X}(t^*_v,T_u^{-v})}\Pr\left(\tilde{\bf
X}\left([0,t^*_v],T^{-v}_u\right)\Big|t^I_u=1\right).\end{aligned}$$ Based on Lemma \[lem:inequalitytime\], we have $$\begin{aligned}
&\max_{\tilde{\bf X}([0,t^*_v],T_u^{-v})\in{\cal X}(t^*_v,T_u^{-v})}\Pr\left(\tilde{\bf X}\left([0,t^*_v],T^{-v}_u\right)\Big|t^I_u=1\right)=\\
&\max_{\tilde{\bf X}([0,t^*_u-1],T_u^{-v})\in{\cal X}(t^*_u-1,T_u^{-v})}\Pr\left(\tilde{\bf X}\left([0,t^*_u-1],T^{-v}_u\right)\Big|t^I_u=1\right)\\
&> \max_{{\bf X}([0,t^*_u],T_u^{-v})\in{\cal
X}(t^*_u,T_u^{-v})}\Pr\left({\bf
X}\left([0,t^*_u],T^{-v}_u\right)\right).\end{aligned}$$ Therefore, given the optimal sample path rooted at $u$, we have constructed a sample path rooted at $v$ which occurs with a higher probability. The lemma holds.

Next, we give a useful property of the Jordan infection centers in the following lemma.

\[lem:numofcenters\] On a tree network with at least one infected node, there exist at most two Jordan infection centers. When the network has two Jordan infection centers, the two must be neighbors.

First, we claim if there are more than one Jordan infection centers, they must be adjacent. Suppose $v,u\in{\cal V}$ are two Jordan infection centers and $\tilde{e}(v)=\tilde{e}(u)=\lambda.$ Suppose $v$ and $u$ are not adjacent, i.e., $d(v,u)>1$. Then, there 