Is zero a regular point for a drifted $\alpha$-stable process?

Question

We consider 1-d process of the form $Y_{t} = bt + M_{t}^{\alpha}$, where $M_{t}^{\alpha}$ is $\alpha$-stable process for some $\alpha \in (0,2)$ with its levy symbol $\eta(u) = - |u|^{\alpha}.$, and $b$ is some constant. Let $\tau = \inf\{t>0: Y_{t} >0\}.$

[Q.] For what $(b, \alpha)$ does $\tau = 0$ hold true?

Discussions: The above $\alpha$-stable process $M^{\alpha}$ has its levy measure $$\nu(dy) = dy/ |y|^{1+\alpha}.$$

When $\alpha$ is close to $2$, the behavior of $M^{\alpha}$ resembles Brownian motion. One can show that $bt + W_{t}$ cross zeros infinitely often in any small interval as long as $b$ is a finite constant. So my guess is If $\alpha\ge 1$, then $\tau = 0$ for all $b$ due to unbounded variation?

If $\alpha\in (0,1)$, $M^{\alpha}$ is finite variation, but jumps infinitely often in any small interval symmetrically. In other words, this proves $\tau = 0$ for $b \ge 0$ and $\alpha\in (0,1)$. However, what if $b<0$?

Answer 1

For the case $0<\alpha<1$ : page 84 of Bertoin's book Lévy Processes, you'll find that $\lim_{t\downarrow 0}M^\alpha_t/t=0$. This implies $\tau>0$ almost surely when $b<0$, and $\tau=0$ when $b>0$ (but doesn't give an answer for $b=0$ ...)

Answer 2

For α≥1 you can use scaling plus a 0-1 to show that it is. The 0-1 law says that the probability of immediately going negative is 0 or 1, similarly for positive. Let a be small. $ P \lbrace \inf_{0 < t < a} (bt + M_t)< 0 \rbrace = P \lbrace \inf_{0 < s < 1} (bas + M_{as})< 0 \rbrace = P \lbrace \inf_{0 < s < 1} (bas/a^{1/\alpha} + M_{as}/a^{1/\alpha})< 0 \rbrace $

Note that for α≥1 the drift term, bas/a 1/α is going to 0 (or staying constant in case α=1 ), and the scaled process is always the same process for all a . Therefore the probability is not going to zero, and must be 1.