For a standard linear Brownian motion $\{B(t)\mid\ 0\le t\le 1\}$, for natural $n\ge 0$ and natural $1\le k\le 2^n$, let $d(n,k)=B\left(k2^{-n}\right)-B\left((k-1)2^{-n}\right)$ be the differences of height of the Brownian motion along the $k$'th $n$-diadic interval.
Denote the (random) sequence of averages of such differences as follows: $$\mu_n = 2^{-n}\sum_{k=1}^{2^n}d(n,k)$$ and the sequence of standard deviations of such differences as follows: $$\sigma_n = \sqrt{2^{-n}\sum_{k=1}^{2^n} \left(d(n,k) - \mu_n\right)^2}$$ and the sequence of volatilities as follows: $$v_n = \sigma_n \cdot 2^{n/2}$$
First question: does it follow that $v_n$ converges almost surely?
Second question: does it follow that $v_n$ converges almost surely to 1?
Third question: if so, what can I say about the rate of convergence?
Some motivation / intuition: the story here is diadic sampling of a Brownian motion. The idea is that such a sampling should behave "properly" as the mesh goes to 0 (there's nothing special about "diadic" here). By properly I mean - having "annualized" standard deviation 1. To form such "annualizing", I remove the normalizing factor ($2^{-n}$) from the variance calculation.
The origin of the question about "rate of convergence" is in order to figure out what is a "proper mesh" of sampling, that will guarantee me behaviour which is "similar enough" to that of Brownian motion.
This is all done for a computer simulation, but the questions are here also for proper understanding of the subject.