= Gaussian limit of a Brownian exit skeleton
{c}
{title2=$B_{\sigma_1^k}/\sqrt k\ \Rightarrow\ N(0,1)$}
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 $\mathbb E\sigma_1=1$.
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