= Horizontal and vertical maxima before a half-plane exit
{title2=$\mathbb P(Y^*>\lambda)=1/\lambda\quad(\lambda\geq1)$}
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 $p\geq1$. 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.
Back to article page
