X_{v_r}(t)=I)\\
&=(1-p)^t\prod_{u\in{\cal C}(v_r)}\Pr({\bf
X}([0,t],T_u^{-v_r})| X_{v_r}(t)=I),\end{aligned}$$ where the last equality holds since $v_r$ is the only infected node in the network at time $t,$ which requires $X_{v_r}(s)=I$ for $0\leq s\leq t.$ Node $u\in {\cal C}(v_r)$ has two possible states $S$ or $R.$

[**Step 1.a**]{} $u$ is susceptible if it was not infected within $t$ time slots. In each time slot, $v_r$ tries to infect $u$ with probability $q.$ The probability that $u$ is susceptible at time slot $t$ is $$(1-q)^t,$$ which implies that $$\begin{aligned}
\Pr({\bf X}([0,t],T_u^{-v_r})| X_{v_r}(t)=I)=(1-q)^t \label{eq: k=0-S}\end{aligned}$$ if $X_u(t)=S.$

[**Step 1.b**]{} If $u$ is in the recovered state, we denote by $t^I_u$ and $t^R_u$ its infection and recovery times, respectively. Then, we have if $X_u(t)=R,$ $$\begin{aligned}
&\Pr({\bf X}([0,t],T_u^{-v_r})| X_{v_r}(t)=I)\\
&=(1-q)^{t^I_u-1}q(1-p)^{t^R_u-t^I_u-1}p\\
&\prod_{w \in {\cal C}(u)}\Pr\left({\bf
X}([0,t],T_w^{-u})|t^I_u,t^R_u\right),\end{aligned}$$ where $(1-q)^{t^I_u-1} q(1-p)^{t^R_u-t^I_u-1}p$ is the probability that node $u$ was infected at time $t^I_u$, and recovered at time $t^R_u.$ Since $T_w^{-u}$ is also an infinite tree, there exists at least one node $\xi \in T_w^{-u}$ such that the node is in the susceptible state but its parent node (say node $\gamma$) is in the recovered state. We denote by $T_w^{-u}\backslash T_{\xi }^{-\gamma}$ the set of nodes that are on subtree $T_w^{-u}$ but not on 