 Solution

Take $\lambda >0$ and let ${\sigma }_{\lambda }$ be the first time $X$ hits $\lambda$. Reflect the horizontal coordinate after ${\sigma }_{\lambda }$, leaving the vertical coordinate unchanged. The Strong Markov property and symmetry preserve the law of planar Brownian motion and leave its real-axis exit time unchanged. On paths with ${\sigma }_{\lambda }<\tau$, the reflection swaps the two terminal events $X_{\tau }<\lambda$ and $X_{\tau }>\lambda$. There is no terminal atom at $\lambda$. Hence the Brownian reflection principle givesHorizontal symmetry gives the same probability for $-\inf X>\lambda$. A union bound therefore yieldsAlso $X^{\ast }\ge |X_{\tau }|$, whose Cauchy distribution is $2{\pi }^{-1}{\int }_{\lambda }^{\infty }(1+x^2{)}^{-1}dx$. Thus the maximum tail is bounded above and below by positive constants times ${\lambda }^{-1}$ for large $\lambda$. The tail integral formula for moments for positive moments givesThe large-$\lambda$ integral is finite exactly when $0<p<1$; it diverges for $p\ge 1$. Near zero its upper bound by ${\int }_0^{1}p{\lambda }^{p-1}d\lambda$ is finite.

The vertical coordinate remains nonnegative up to exit and starts at one. The strip calculation, now with upper boundary $\lambda \ge 1$, givesThus $Y^{\ast }$ has density ${\lambda }^{-2}$ on $[1,\infty )$, and in factFor completeness, if the source's $p<1$ includes negative powers, $X^{\ast }$ also has all negative moments. Let ${\sigma }_{\eta }$ be the first horizontal exit from $(-\eta ,\eta )$. ThenThe first bound follows from the Gaussian reflection bound for the vertical hit of zero; the second follows by splitting time into blocks of length ${\eta }^2$, each having a fixed positive chance of exiting the horizontal interval. This small-value estimate integrates against every negative power. At $p=0$ both moments equal one.