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For a Brownian motion with drift $X$, the process $X_T-X_{T-t}$ on $[0,T]$ has the same law as $X_t$. Thereforehave the same distribution, which equates the corresponding geometric-Brownian lookback payoffs.

The laws on continuous paths of a standard Brownian motion and a Brownian motion with drift $\mu \ne 0$ are mutually singular measures on the whole half-line. The event ${\lim }_{t\to \infty }\omega (t)/t=0$ has full measure for the former and zero measure for the latter by the strong law for Brownian motion. On every fixed finite horizon, their laws are equivalent by the Girsanov theorem. The finite-horizon densities $e^{\mu B_t-{\mu }^{2}t/2}$ tend to zero almost surely while retaining expectation one, so they lack uniform integrability over all time.

First prove the strong law for Brownian motion, $B_t/t\to 0$ with probability one. For each $\epsilon >0$, the Gaussian tail bound givesBy stationary increments and the Brownian reflection principle, followed by the same tail bound,Both bounds are summable in $n$. The First Borel-Cantelli lemma therefore implies that, eventually, both quantities inside these probability events are at most $\epsilon n$. For $n\le t\le n+1$ this gives $|B_t|/t\le 2\epsilon$. Intersect over a sequence of positive rational $\epsilon$ tending to zero to obtain the asserted continuous-time limit.

On this one almost sure event,so $B_t+at\to +\infty$ for $a>0$ and to $-\infty$ for $a<0$. For each real $x$, the path is eventually strictly on the corresponding side of $x$. HenceThe same event works simultaneously for all levels $x$, giving the requested transience of Brownian motion with drift.