I am looking at the speed of a simple random walk on a Galton - Watson (GW) tree in Theorem 3.2, Page 9 here and have a question about the construction of rays on a GW tree $T$ that are used to find the speed.
Theorem: The speed of simple random walk is:
$$l:= \lim_{n\to \infty} \frac{|x_n|}{n} = E\left[ \frac{Z_1-1}{Z_1+1}\right]$$
where $Z_1$ is the number of offspring of the root and $|x|$ is distance from the root to the vertex $x$.
Proof: The set of all rays emanating from the root is called the boundary of $T$ denoted by $\partial T$. We shall calculate the speed as the rate of change of "horodistance" (Busemann function) from a boundary point. Given a boundary point $\xi \in \partial T$ and a vertex $x\in T$, let $[x,\xi]$ denote the ray from $x$ to $\xi$. (More precisely, there is a unique one-to-one correspondence $\xi \to [x, \xi]$ from $\partial T \to \partial \text{MoveRoot}(T, x)$ such that $\xi$ and $[x, ξ]$ have infinitely many vertices in common.) If we change the root of $T$ to a vertex $x ∈ T$, we denote the new rooted tree by MoveRoot$(T, x)$.
So I don't understand the construction of ray $[x,\xi]$ and how $[x,\xi]$ and $\xi$ have infinitely many vertices in common? i.e. for a given a ray $\xi$ and a vertex $x$, may I know how the ray $[x,\xi]$ looks like? Thanks.
$\xi$ is a ray starting at the root; $[x, \xi]$ is a ray that starts at $x$ but is supposed to "go to the same place as" $\xi$.
One way to construct $[x, \xi]$ (in any tree, not necessarily a Galton-Watson tree) is to let $\xi_i$ be the last vertex along $\xi$ which is an ancestor of $x$. (We know that this exists because $\xi_0$ is an ancestor of $x$ - it is the root - but $x$ only has finitely many ancestors.) Then, let $[x, \xi]$ be the ray that walks up the tree from $x$ to $\xi_i$, and then follows $\xi$ down: to $\xi_{i+1}, \xi_{i+2}, \dots$.
All we have to do to check that $[x, \xi]$ is a ray is that it never backtracks. It won't do so along the path from $x$ to $\xi_i$, because it's a path. The step from $\xi_i$ to $\xi_{i+1}$ doesn't backtrack, because $\xi_{i+1}$ is not an ancestor of $x$ and every vertex on the $x, \xi_i$-path is. After this, none of the steps backtrack because $\xi$ is a ray.