 Solution

For a continuous local martingale $M$ starting at zero, Burkholder-Davis-Gundy inequalities compare its maximal size with its quadratic-variation clock: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 $[0,\lambda ]$. ThusThe tail integral formula givesand it diverges for $p\ge 1$.

The stopped martingale $Y_{t\wedge T_{\mathbb{H}}}-1$ has quadratic variation $t\wedge T_{\mathbb{H}}$. Its maximum in absolute value is $\max (1,Y^{\ast }-1)$, between $Y^{\ast }/2$ and $Y^{\ast }$, so it has precisely the same finite positive moments as $Y^{\ast }$. 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, yieldsTaking 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.