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Let ${\xi }_j=B_{{\sigma }_1^j}-B_{{\sigma }_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh Brownian motion gives probability one half to each sign, and the Strong Markov property makes these signs independent across successive exits. Thus $B_{{\sigma }_1^n}={\sum }_{j=1}^n{\xi }_j$.

For the normalized position, the characteristic function issince $\log \cos z=-z^2/2+O(z^4)$ near zero. By the Lévy continuity theorem, this provesThis is the Gaussian limit of a Brownian exit skeleton.

Compare along the subsequence $n=2^{2m}$ with ${\tau }_m={\sigma }_{2^{-m}}^{2^{2m}}$. Joint Brownian scaling of the successive exits givesThe right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{{\tau }_m}\to B_C$ almost surely on the original Brownian probability space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces $C=1$, so finally $\mathbb{E}{\sigma }_a=a^2$. Independence of the position and its random clock was not assumed in this comparison.