I want to find rate of escape (drift) on free group (with d generators).
From here (page 2): https://arxiv.org/pdf/math/0506129.pdf I know the answer = 1
But I can't fully figure out why
I know there exist next way to compute drift (from Vershik: Dynamic theory of growth in groups: https://www.deepdyve.com/lp/iop/dynamic-theory-of-growth-in-groups-entropy-boundaries-examples-ZH4dbCaiRQ?key=iop-publishing - sorry, I don't have a pdf version to make a screenshot for now):
Let $G = <a_1, ..., a_d, a_1^{-1}, ..., a_d^{-1}> = <S>$, (I will assume d = 2) let $\mu$ - measure with support $S$ (we also assume it is probabilistic and uniform, i.e. $\mu(S) = 1, \mu(g) = \frac{1}{2d}, g \in S$. We then introduce $\mu^{*n}$ - $n^{th}$ convolution ($\mu^{*n}(A) = (\mu \times ... \times \mu){(g_1, ..., g_n): g_i \in S, g_1 \cdot ... \cdot g_n \in A \subset G}$) and the space $(G^{\infty}, \mu^{\infty})$, $G^{\infty}$ - space of trajectories (we can think of it as infinite sequence of elements in S), $\mu^{\infty}$ - measure on that space (it will be nonzero on sets $A_{w_n} = $ {all paths with first n letters fixed} (and it can be easily calculate: $w_n$ as a n-length word, so we can associate in with some element $g_n$ in G, so $\mu^{\infty}(A_{w_n}) = \mu^{*n}(g_n) = $ number of ways represent $g_n$ as a product of n elements from S divided by number of words of length n
Next, drift for G is $l_\mu = \lim_{n \to \infty}\frac{E_{\mu^{*n}}L(g)}{n}$, L(g) - "length" of word g. So, for fixed n, $E_{\mu^{*n}}L(g)$ is "expected reduced length" of n-length word in alphabet S
Vershik in his note has the following lemma: for almost every $\omega = (w_1, w_2, ..., w_n, ...) \in G^{\infty}$, $\lim_{n \to \infty}\frac{L_n(\omega)}{n}$ converge to 1 (by measure), where $L_n$ - length of $w_1 \cdot ... \cdot w_n$
To prove that we introduce $T: G^{\infty} \to G^{\infty}$ - left shift. I can proof that: $T$ - ergodic and $T$ with $L_n$ satisfy the conditions in Kingman's subadditive ergodic theorem, so $\lim_{n \to \infty}\frac{L_n(\omega)}{n}$ converge to some constant, but why is it converge to exactly 1?
Vershik in his notes said that converge by measure follows from almost sure convergence. So I guess $\mu^{\infty}(\omega \in G^{\infty}: lim_{n \to \infty}\frac{L_n(w)}{n} \neq 1) = 0$ must be some obvious statement, but I don't understand why. For fixed n we have $3^n$ reduced (honestly it is $4 \cdot 3^{n-1}$, but it doesn't matter) words and $4^n$ words, so number of reduced words of fixed length n divided by number of words of fixed length tends to 0, so we need observe some set of "almost reduced words" I guess, and show that "this set converge to whole $G^{\infty}$" or something like that
The speed (rate of escape) on the free group with $d$ generators is not 1 but rather $1-1/d$. In each step (when not in the identity) the walk will move forward with probability $(2d-1)/(2d)$ and backward with probability $1/(2d)$. See Chapter 13 in [1] for more on speed of random walks and in particular a formalization of this argument.
[1] https://rdlyons.pages.iu.edu/prbtree/