Gaussian limit of a Brownian exit skeleton ($B_{{\sigma }_1^k}/\sqrt{k}\Rightarrow N(0,1)$)

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