1
$\begingroup$

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 $b<0$?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $b=0$ ...)

$\endgroup$
1
  • $\begingroup$ Thanks for the partial answer. I've looked over the book, and find that the result is for the subordinator, which is increasing process. In my post, $M^\alpha$ with given levy symbol is a symmetric process in the sense that positive and negative jumps have the same chance. In other words, it's not monotone in any subinterval almost surely. $\endgroup$
    – kenneth
    Aug 7, 2015 at 2:17
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Why is that scaled process always the same process for all a? $\endgroup$
    – kenneth
    Aug 29, 2015 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.