= Maximum occupation is not a hitting probability
{title2=$\sup_{t>0}\mathbb P(Y_t>\ell)\ne\mathbb P(\sup_{t\geq0}Y_t\geq\ell)$}
For $Y_t=-vt+\sqrt{2D}\,B_t$ with $v,D,\ell>0$ and standard <Brownian motion>, the one-time occupation probability is $\tfrac12\operatorname{erfc}[(\ell+vt)/\sqrt{4Dt}]$. It is maximized at $t=\ell/v$, giving $\tfrac12\operatorname{erfc}(\sqrt{v\ell/D})$. The actual <hitting probability> is $e^{-v\ell/D}$. To derive it, solve $Dh''-vh'=0$ on $(-A,\ell)$ with $h(-A)=0$, $h(\ell)=1$, obtaining $h(0)=(1-e^{-vA/D})/(e^{v\ell/D}-e^{-vA/D})$, and let $A\to\infty$. A path that reached the level and returned contributes to the latter event but not to a given one-time tail.
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