= Solution
The given limit inferior says that almost surely there is a random $\delta>0$ such that $t^{-a}W_t>-1/2$ whenever $0<t\leq\delta$. Therefore
$$
S_t=t^a\bigl(1+t^{-a}W_t\bigr)>\frac12t^a>0
$$
on that interval. The first entrance time
$$
\tau=\inf\{t>0:S_t\leq\tfrac12t^a\}\wedge T
$$
is a <stopping time> by continuity and satisfies $\tau>0$ almost surely. Before $\tau$ one has $S_t>t^a/2$, and continuity gives $S_\tau\geq\tau^a/2>0$. Hence $S_t>0$ for every $0<t\leq\tau$.
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