Brownian exit-time skeleton (${\sigma }_a^{k+1}=\inf \{t\ge {\sigma }_a^k:|B_t-B_{{\sigma }_a^k}|=a\}$)

Successive exits from intervals of radius $a$ around the current position embed a simple symmetric random walk into Brownian motion. The Strong Markov property makes the durations independent with common law $a^2{\sigma }_1$; the displacements divided by $a$ are independent symmetric signs. The displacement sequence is independent across exits; no independence between a given displacement and its duration is needed for these conclusions.

The position is the sum of $k$ independent symmetric signs. Its characteristic function after division by $\sqrt{k}$ is $\cos (u/\sqrt{k}{)}^k\to e^{-u^2/2}$. The Lévy continuity theorem gives the normal distribution limit. Scaling the same skeleton to radius $2^{-n}$ and using Brownian exit-skeleton clock convergence identifies its other limit as $B_{\mathbb{E}{\sigma }_1}$, forcing $\mathbb{E}{\sigma }_1=1$.

Unit-radius Brownian exit time has finite variance, since independent unit-time increments give an exponential upper tail. The Brownian exit-time skeleton clock after $2^{2n}$ radius-$2^{-n}$ exits has mean $\mathbb{E}{\sigma }_1$ and variance $2^{-2n}\mathrm{Var}({\sigma }_1)$. Chebyshev's inequality and the Borel-Cantelli first lemma yield the displayed pathwise convergence. Independence between different rows indexed by $n$ is unnecessary.