= Solution
Use the <incoming-edge coupling of site and bond percolation>. Start with <independent random variables> indicating open bonds, each with <probability> $q$. For a vertex $v$ of incoming degree $d_v$, let $J_v$ indicate that at least one incoming bond is open. Then
$$
\mathbb P(J_v=1)=1-(1-q)^{d_v}\le r:=1-(1-q)^{\Delta_{\rm in}}.
$$
The indicators $J_v$ are independent: the sets of bonds entering different vertices are disjoint. Using an additional independent coin at each vertex, turn some zeros into ones so that the resulting indicators $S_v$ are independent with common <probability> $r$. Explicitly, when $J_v=0$, set $S_v=1$ with conditional <probability> $(r-\mathbb P(J_v=1))/(1-\mathbb P(J_v=1))$.
Every vertex other than $x$ in an open <bond percolation> out-cluster has an open incoming bond, and is consequently an open site. On $\{S_x=1\}$, the whole bond out-cluster is therefore contained in the <site percolation> out-cluster. The event of infinite bond reachability does not depend on bonds entering $x$: delete cycles from any finite open <directed path> starting at $x$. It is thus independent of $S_x$, including its additional coin. Hence
$$
\theta_s(r)\ge r\theta_b(q).
$$
For every $q>p_H^b$ below $1$, monotonicity and the definition of the survival threshold give $\theta_b(q)>0$. Letting $q\downarrow p_H^b$ yields the \b[quantitative bound]
$$
\boxed{p_H^s(\vec\Lambda;x)\le 1-(1-p_H^b(\vec\Lambda;x))^{\Delta_{\rm in}}.}
$$
In particular, if $p_H^b<1$, the right-hand side is strictly less than $1$. If $p_H^b=1$, no separation from $1$ is implied or generally possible.
Back to article page
