= Solution
{c}
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> gives
$$
\mathbb P_i\!\left(\sup_{t\leq\tau}X_t>\lambda\right)
=2\mathbb P_i(X_\tau>\lambda)
=\boxed{\frac2\pi\int_\lambda^\infty\frac{dx}{1+x^2}.}
$$
Horizontal symmetry gives the same <probability> for $-\inf X>\lambda$. A union bound therefore yields
$$
\boxed{\mathbb P_i(X^*>\lambda)
\leq\frac4\pi\int_\lambda^\infty\frac{dx}{1+x^2}.}
$$
Also $X^*\geq|X_\tau|$, 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 gives
$$
\mathbb E(X^*)^p=p\int_0^\infty
\lambda^{p-1}\mathbb P(X^*>\lambda)d\lambda.
$$
The large-$\lambda$ integral is finite exactly when $0<p<1$; it diverges for $p\geq1$. Near zero its upper bound by $\int_0^1p\lambda^{p-1}d\lambda$ is finite.
The vertical coordinate remains nonnegative up to exit and starts at one. The strip calculation, now with upper boundary $\lambda\geq1$, gives
$$
\mathbb P(Y^*>\lambda)=\frac1\lambda.
$$
Thus $Y^*$ has density $\lambda^{-2}$ on $[1,\infty)$, and in fact
$$
\boxed{\mathbb E(Y^*)^p=\frac1{1-p}\ (p<1),\qquad
\mathbb E(Y^*)^p=\infty\ (p\geq1).}
$$
For completeness, if the source's $p<1$ includes negative powers, $X^*$ also has all negative moments. Let $\sigma_\eta$ be the first horizontal exit from $(-\eta,\eta)$. Then
$$
\mathbb P(X^*\leq\eta)
\leq\mathbb P(\tau\leq\eta)+\mathbb P(\sigma_\eta\geq\eta)
\leq C e^{-c/\eta}\quad(0<\eta\ll1).
$$
The 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.
\Image[/past-exam-of-the-stochastic-differential-equations-and-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2008-exit-maxima.png]
{title=The vertical maximum has an exact inverse-height tail while the horizontal maximum has comparable Cauchy tails}
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