= Solution
\b[No absolutely continuous probability measure can remove this nonzero drift on the whole half-line.] By the <strong law for Brownian motion>, the event
$$
A=\left\{\lim_{t\to\infty}\frac{B_t}{t}=0\right\}
$$
has $\mathbb P$-probability one. If $\mathbb P^\infty\ll\mathbb P$, <absolute continuity of measures> gives $\mathbb P^\infty(A)=1$. But if $W_t=B_t-\mu t$ is a <Brownian motion> under $\mathbb P^\infty$, the same <strong law for Brownian motion> gives $B_t/t=W_t/t+\mu\to\mu$ with $\mathbb P^\infty$-probability one. Since $\mu>0$, these two limiting events are disjoint, a contradiction.
This is <infinite-horizon singularity of Brownian motion with constant drift>. The finite-horizon densities are consistent on $\mathcal F_t$ but cannot define a globally absolutely continuous measure with the requested property. Indeed,
$$
\boxed{Z_t\longrightarrow0\quad\mathbb P\text{-almost surely},\qquad \mathbb E Z_t=1,}
$$
because $t^{-1}\log Z_t\to-\mu^2/2$. Hence $Z$ lacks <uniform integrability> on $[0,\infty)$. A drifted law does exist on canonical path space, but it is mutually singular with <Wiener measure> on the infinite-horizon path <sigma-algebra>.
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