Write $Z=X+iY$, with $Y$ a real Brownian motion started at one. For the nontrivial strip calculation assume $r>1$, so the starting point is in its interior. The first exit time isFor a continuous Brownian path started inside, this equals the first hitting time of $Y=0$ or $Y=r$ almost surely: Brownian motion started at a boundary immediately visits the exterior, so using the closed strip in the definition does not change the infimum.

If $T_S>n$, then $Y_n\in [0,r]$. Its normal density is bounded by $(2\pi n{)}^{-1/2}$, henceThus the exit time is finite almost surely.

The stopped harmonic coordinate $Y_{t\wedge T_S}$ is a martingale bounded between zero and $r$. bounded convergence theorem and optional stopping theorem give $\mathbb{E}{Y}_{T_S}=Y_0=1$. Since the exit coordinate is zero or $r$,If $T$ is the first hit of the real axis, exit through the lower boundary implies $T<\infty$, so $\mathbb{P}(T<\infty )\ge 1-1/r$ for every $r>1$. Letting $r\to \infty$ provesThe strip assumption is worth making explicit: at $r=1$ the initial point is the upper boundary and exit is immediate; for $0<r<1$ it starts outside the strip and the displayed interior probabilities do not apply.

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.