The problem is:Show that the existence of two edge-disjoint self-avoiding paths of open edges starting from 0 has a probability which is either 0 or 1.
My question is: do we require the path to be infinite? otherwise the probability would be $1-(1-p)^{2d}-2dp(1-p)^{2d-1}$ right? If it is infinite, how to prove that?
I try to show for any $x$, we have $\mathbf{P}(A\cap A_x)\approx\mathbf{P}(A)$ and then we can approximate $A$ by $B$ which only depends on finitely many edges, finally pick large enough $x$ to make $B$ and $B_x$ independent, by $\mathbf{P}(B\cap B_x)=\mathbf{P}(B) \mathbf{P}(B_x)\approx \mathbf{P}(A) \mathbf{P}(A_x)=\mathbf{P}(A)^2$, we can prove the statement. However I find it hard to show $\mathbf{P}(A\cap A_x)\approx\mathbf{P}(A)$.
Does my idea make sense? or any other way to show that?
I checked with the author of the lecture notes. The correct form of the exercise is:
Exercise 17 (corrected): Show that the existence a bi-infinite self-avoiding path of open edges has a probability which is either 0 or 1.
This you can easily infer from ergodicity of an array of i.i.d. variables indexed by the edges of the lattice under shifts.
The original formulation of the exercise is wrong for every $p \in (p_c,1)$.