= Solution
{c}
For a <continuous local martingale> $M$ starting at zero, <Burkholder-Davis-Gundy inequalities> compare its maximal size with its quadratic-variation clock:
$$
c_p\,\mathbb E\langle M\rangle_T^{p/2}
\leq\mathbb E\sup_{t\leq T}|M_t|^p
\leq C_p\,\mathbb E\langle M\rangle_T^{p/2},\qquad p>0.
$$
They remain valid with infinite values, by stopping and passage to the limit. Their significance here is that the horizontal and vertical <martingales> share the same stopping-time clock, so one coordinate's maximum can determine moments of the other and of the clock. For a process started away from zero, first subtract its initial value.
Starting at $i$, the vertical coordinate starts at one and is stopped on first hitting zero. For $\lambda>1$, it reaches level $\lambda$ before zero with probability $1/\lambda$, by <optional stopping theorem> of one-dimensional <Brownian motion> on $[0,\lambda]$. Thus
$$
\boxed{\rho(\lambda)=
\begin{cases}1,&0<\lambda\leq1,\\1/\lambda,&\lambda>1.\end{cases}}
$$
The tail <integral> formula gives
$$
\mathbb E(Y^*)^p=p\int_0^\infty \lambda^{p-1}\rho(\lambda)d\lambda
=\frac1{1-p}\quad(0<p<1),
$$
and it diverges for $p\geq1$.
The stopped <martingale> $Y_{t\wedge T_{\mathbb H}}-1$ has <quadratic variation> $t\wedge T_{\mathbb H}$. Its maximum in absolute value is $\max(1,Y^*-1)$, between $Y^*/2$ and $Y^*$, so it has precisely the same finite positive moments as $Y^*$. BDG therefore gives $\mathbb E T_{\mathbb H}^{p/2}<\infty$ exactly for $p<1$. Applying BDG to $X_{t\wedge T_{\mathbb H}}$, which starts at zero and has the same clock, yields
$$
\boxed{\mathbb E_i(X^*_{T_{\mathbb H}})^p<\infty\iff0<p<1,\qquad
\mathbb E_iT_{\mathbb H}^q<\infty\iff0<q<\tfrac12.}
$$
Taking the displayed $1/p$ or $1/q$ powers does not change finiteness. In particular the exit time is almost surely finite while its mean is infinite.
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