 Solution

Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.