I am currently preparing a seminar on a chapter 3 of the book "An Introduction to Random Interlacements" (by Alexander Drewitz) and I am having some problems proving some of the exercises given by the book.
The first problem I encountered is the following: Assuming that $X_n$ is a simple random walk on the $T_N^d$ and
$H_K$= inf{$n\ge0$ : $X_n(\omega)\in$ K}
$Ĥ_K$=inf{$n\ge1:$$X_n(\omega)\in$ K},
show that for all $N\ge1$, $t\in$$\Bbb{N}$ and $x\in K$,
$P[H_K=t,X_{H_K}=x] = P[X_0=x, Ĥ_K>t]$
As it is written on the book, this assumption can be showed by using the reversibility of the simple random walk. I have no idea on how to solve the problem and I appreciate any help that I can get.
($t>0$ in the sequel.)
Let $\bar K$ be the complement of $K$.
Let us translate in words the relation to be shown. The probability $$P[\ H_K=t,\ X_{H_K}=x\ ]$$ is the probability that we start at time zero the "Brownian motion $X$", and land in $K$ (for the first time) after exactly $t$ steps, and moreover, we land then exactly in $x$. Equivalently, $$X_0, X_1,\dots, X_{t-1}\in \bar K\ ;\ X_t=x\ .$$ If we have to be verbose, the probability can be written as a sum over all relevant paths $$\xi=(x_0\to x_1\to \dots\to x_{t-1}\to x_t)\ ,\qquad x_0,x_1,\dots, x_t\in \bar K;\ x_t=x\in K\ ,$$ with terms $$ P[\ X_0=x_0,\ X_1=x_1,\ \dots\ ,\ X_{t-1}=x_{t-1}\ ,\ X_t=x_t\ ] \ , $$ which depend each only on the transition probabilities for the steps $(x_0\to x_1)$, $(x_1\to x_2)$, $\dots$ , $(x_{t-1}\to x_t)$ in $\xi$.
The other probability, $$P[\ X_0=x\ ,\ \dot H_K > t\ ]$$ is the probability that we start at time zero the "Brownian motion $X$", in $K$, moreover, exactly in $x$, and do not hit $K$ the next $t$ steps. Equivalently, $$X_t, X_{t-1},\dots, X_1\in \bar K\ ;\ X_0=x\in K\ .$$ If we have to be verbose, the probability can be written as a sum over all relevant paths $$\xi'=(x'_t\leftarrow x'_{t-1}\leftarrow \dots\leftarrow x'_1\leftarrow x'_0)\ ,\qquad x'_t,x'_{t-1},\dots, x'_1\in \bar K;\ x'_0=x\in K\ ,$$ with terms $$ P[\ X_t=x'_t,\ X_{t-1}=x'_{t-1},\ \dots\ ,\ X_1=x'_1\ ,\ X_0=x'_0\ ] \ , $$ which depend each only on the transition probabilities for the steps $(x'_t\leftarrow x'_{t-1})$, $(x'_{t-1}\leftarrow x'_{t-2})$, $\dots$ , $(x'_1\leftarrow x'_0)$ in $\xi'$.
Now it is clear how to associate to each $\xi$ as above bijectively a $\xi'$ as above. The corresponding probabilities coincide by "reversibility" of the random walk, since there is a "reversibility" for the one step transitions, since for two "states" $x,y$, the transitions $(x\to y)$ and $(x\leftarrow y)$ have the same probability, since the graph of the torus is symmetric, "as seen from each point in it". (The same symmetry applies for a random walk on any graph with constant adjecent number of vertices. For instance for a torus, for the graph of some regular polyhedra, for graphs associated to groups $G$, where transitions are given by ((left) multiplications with $g\in S$ in) specific systems $S$ of group elements, where the systems are closed w.r.t. taking the inverse, $S=S^{-1}$, etc.)
(I tried to give the story, not the formula for the idea of proof.)