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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 $\mathbb{E}{\sigma }_1=1$.

At each finite stopping time ${\sigma }_a^k$, the Strong Markov property says that $B_{{\sigma }_a^k+t}-B_{{\sigma }_a^k}$ is a standard Brownian motion independent of the past sigma-algebra. Its exit time from $(-a,a)$ is exactly ${\sigma }_a^{k+1}-{\sigma }_a^k$. By induction the stopping times are finite almost surely, and these durations are independent with common law ${\sigma }_a$, not merely identically distributed. This is the Brownian exit-time skeleton.

Put $v=\mathrm{Var}({\sigma }_1)<\infty$, which follows from the moment estimate. Brownian scaling gives $\mathrm{Var}({\sigma }_a)=a^{4}v$. Forthere are $2^{2n}$ independent durations of mean $C{2}^{-2n}$ and variance $v{2}^{-4n}$. ConsequentlyFor each $\epsilon >0$, Chebyshev's inequality givesApply the Borel-Cantelli first lemma for a countable sequence of positive errors tending to zero. We obtain the Brownian exit-skeleton clock convergence:No independence between the clocks for different values of $n$ is required.