Horizontal and vertical maxima before a half-plane exit ($\mathbb{P}(Y^{\ast }>\lambda )=1/\lambda (\lambda \ge 1)$)

Start planar Brownian motion at $i$ and stop at its first hit of the real axis. The vertical maximum has the displayed tail by optional stopping theorem in a strip, so its $p$th moment is $1/(1-p)$ for $p<1$ and infinite for $p\ge 1$. Reflection of the horizontal coordinate at a first level hit gives $\mathbb{P}(\sup X>\lambda )=2{\pi }^{-1}{\int }_{\lambda }^{\infty }(1+x^2{)}^{-1}dx$. The absolute horizontal maximum has upper and lower bounds of this same inverse-level order, hence the same positive-moment threshold. Its small-value probability is exponentially small in the reciprocal level, so all negative moments also exist.