Let consider a finite graph $G$ with a fixed root $v_0$. Let $v_1,\ldots,v_n$ be a labelling of the vertices of $G$. Suppose we apply Wilson's algorithm following the labbelling $v_1,\ldots, v_n$ and sample the LERW by sampling SRW and erasing their loops in chronological order. Let $K$ denote the total number of steps of all these SRW. Show that the law of $K$ does not depend on the labelling $v_1,\ldots,v_n$ but that it may depend on $v_0$.
I'm not sure of what I did and if it is sufficent to conclude.
Let define $A_j = G \setminus \{ v_0, v_1,\ldots, v_j\} $, and $\partial A_j = \{v_0,v_1,\dots,v_j \} $. We have that $$ K = \sum_{0 \leq j \leq n-1} \tau_j $$ where $ \tau_j := \{ n \geq 0 : X_n \in \partial A_j \} $, i.e. the stopping time of the SRW starting at $v_{j+1}$. Now suppose that we consider another labelling of the vertices $v_1',v_2',\ldots,v_n' $. Define $A_j' = G \setminus \{ v_0, v_1',\ldots, v_j'\} $ and note that $v_0$ is the same, similarly $\partial A_j' = \{v_0,v_1',\dots,v_j' \} $ and similarly define $\tau_j'$ as the stopping time of the SRW starting at $v_{j+1}'$. We want to prove that if we define $$ K' = \sum_{0 \leq j \leq n-1 } \tau_j' $$ then the law of $K$ and $K'$ are the same. We use the fact that for every $j$ we have that $ \mathbb{E}[\tau_j] $ is the unique solution of
$$ \left\{\begin{matrix} \Delta f(x) =-1 & \text{ on } A_j \\ f(x)=0 & \text{ on } \partial A_j \end{matrix}\right.$$
Now we prove the result inductively. First of all we want to prove that $\mathbb{E}[\tau_0] = \mathbb{E}[\tau_0']$, indeed we have that both solve this equation $$ \left\{\begin{matrix} \Delta f(x) =-1 & \text{ on } A_0 = A_0' \\ f(x)=0 & \text{ on } \{v_0\} \end{matrix}\right.$$ hence the difference $\mathbb{E}[\tau_0']-\mathbb{E}[\tau_0] $ is harmonic and possess maximum (and the minimum ) on the boundary, which is $0$. Hence $\mathbb{E}[\tau_0']=\mathbb{E}[\tau_0] $. Now $ \mathbb{E}[\tau_0' + \tau_1'] $ and $ \mathbb{E}[\tau_0 + \tau_1] $ both solve this equation
$$ \left\{\begin{matrix} \Delta f(x) =-1 & \text{ on } G \setminus \{ v_0, v_1, v_1 '\} \\ f(x)=0 & \text{ on } \{v_0\} \\ f(x)= \mathbb{E}[\tau_0] + \mathbb{E}[\tau_1] + \mathbb{E}[\tau_1']& \text{ on } \{v_1,v_1'\} \end{matrix}\right.$$ And again we conclude by harmonicity of the difference and by the maximum principle. Now inductively we have that $$ \mathbb{E}[\tau_0+ \ldots + \tau_k] = \mathbb{E}[\tau_0'+\ldots + \tau_k'] $$
since both solve the discrete PDE $$ \left\{\begin{matrix} \Delta f(x) =-1 & \text{ on } G \setminus \{ v_0, v_1, \ldots, v_k, v_1',\ldots,v_k' \} \\ f(x)=0 & \text{ on } \{v_0\} \\ f(x)= \mathbb{E}[\tau_0+\ldots+\tau_{k-1}] + \mathbb{E}[\tau_k] + \mathbb{E}[\tau_k']& \text{ on } \{v_1, \ldots, v_k, v_1',\ldots,v_k'\} \end{matrix}\right.$$ and we deduce that $\mathbb{E}[K] = \mathbb{E}[K']$
this suffices to conclude that $K$ doesn't depend on the order of $v_1,\ldots,v_n $ ?
How argue that it still may depend on $v_0$ ? I mean I think the problem is that $A_0 \neq A_0' $ if we relabel $v_0$ into $v_0'$ but suffices to conclude?