from the case where the time difference of two sample paths is one, i.e., we will show that $$\begin{aligned}
\max_{{\bf X}[0,t]\in {\cal X}(t)} \Pr({\bf X}[0,t])> \max_{{\bf
X}[0,t+1] \in {\cal X}(t+1)} \Pr({\bf X}[0, t+1]).\label{eqn:3}\end{aligned}$$

We divide all possible infection topologies ${\bf Y}$ into countable subsets $\{{\cal Y}^k\}$ where ${\cal Y}^k$ is the set of infection topologies where the largest distance from $v_r$ to an infected node is $k$. ${\cal Y}^0$ is the topology where there is only one infected node—the root node $v_r$. Note that if no infected node is observed, no algorithm performs better than a random guess. To prove (\[eqn:3\]), we use induction over $k$.

[**Step 1:**]{} First, we consider the case $k=0$. All the sample paths considered in step 1 lead to observation $ {\bf Y}\in{\cal Y}^0.$ We denote by $T_{v_r}$ the tree rooted in $v_r$ and $T^{-v_r}_u$ the tree rooted at $u$ but without the branch from $v_r.$ For example, Figure \[figure:timeinequality\] shows $T_{v_1}^{-v_r},$ $T_{v_2}^{-v_r},$ $T_{v_3}^{-v_r}$ and $T_{v_4}^{-v_r}$. The sample path from time slot $0$ to $t$ restricted to $T_u^{-v_r}$ is denoted by ${\bf
X}([0,t],T_u^{-v_r}).$ Furthermore, denote by ${\cal C}(v)$ the set of children of $v.$ We have $$\begin{aligned}
&\Pr({\bf X}[0,t])\\
&=\Pr(X_{v_r}(s)=I,0\leq s\leq t)\\
&\times\prod_{u\in{\cal C}(v_r)}\Pr({\bf
X}([0,t],T_u^{-v_r})| 