My question concerns the comparison of durations of two Ito integrals. By duration I mean time elapsed between the origin and the first time that a process changes in absolute value by a specified threshold. For details, please see below.
For any given process $Z$, we put $Z^\star_t = \sup_{s \leq t} |Z_s|$ as the running supremum of $Z$.
In the sequel we let $W$ denote a standard Wiener process. Define $X=\sigma_1W$ and $Y=\sigma_2W$, where $\sigma_1 > \sigma_2$ are strictly positive nonrandom real numbers. It's easy to see $X^\star > Y^\star$ (up to an evanescent set). If we define two stopping times $S = \inf\{t:|X_t|>\alpha\}$ and $T = \inf\{t:|Y_t|>\alpha\}$, where $\alpha$ is a positive real number, then for any given $t>0$, $P(S \leq t) \geq P(T \leq t)$ .
Now comes my question: say $\sigma_1$ and $\sigma_2$ are no longer two constants, but two stochastic processes. Define $X = \sigma_1 \cdot W$ and $Y = \sigma_2 \cdot W$, namely they are Ito integrals. Suppose $\sigma_1 > \sigma_2$ except on an evanescent set (that is, on most all paths, $\sigma_{1,t} > \sigma_{2,t}$ for all $t$). In this case, can we still conclude that $X^\star > Y^\star$? If not, can we deduce the weaker conclusion, that is, with $S$ and $T$ defined as above, for any $t>0$, $P(S \leq t) \geq P(T \leq t)$?
Any help will be greatly appreciated!