I know that Lévy process $\{X_t\}_{t\geq 0}$ is a stochastic process that satisfies few conditions:

  • $\mathbb{P}(X_0 = 0) = 1.$
  • $X_t$ has stationary increments and $X_t$ has independent increments.

And in different sources I found different definitions of the last condition. They are as follows:

  1. $X_t$ is a cadlag process (right continuous with left limits)

  2. $X_t$ is continuous in probability, that is: for each $t\geq0$, and for each $\epsilon\geq 0$: $$\lim_{s\rightarrow t}\mathbb{P}(|X_s-X_t|<\epsilon) = 1.$$

My question is: Are both conditions 1. and 2. equivalent? If no, which of these conditions is a correct one?

  • $\begingroup$ Your notion of a Levy process is confused. It isn't stationary. It has stationary increments. See Wikipedia. $\endgroup$ – C Monsour May 24 '18 at 10:40
  • $\begingroup$ Yes I know, but I did not write it is stationary, I just wrote that it has stationary increments. Maybe I could write it a bit more clearly. $\endgroup$ – MathMen May 24 '18 at 10:51
  • $\begingroup$ You did originally write that it was stationary. Thank you for correcting it. $\endgroup$ – C Monsour May 24 '18 at 11:15

$\def\eps{\varepsilon}$ $\def\P{\mathbb{P}}$ Yes 1 and 2 are equivalent. In the one direction it is easy. Let us first assume 1 and show 2. Since the increments are stationary, we have for any $t\ge0$, $\eps>0$ $$ \lim_{s\to0}\P(|X_{t+s}-X_t|>\eps)=\lim_{s\to0}\P(|X_{s}|>\eps) $$ Using the fact that the trajectories are cadlag at $0$, we have that almost surely $X_s$ tends to $X_0=0$ as $s\to0$. Hence $X_s$ converges to $0$ also in probability and $$ \lim_{s\to0}\P(|X_{s}|>\eps)=0. $$ Combined with the first identity, this implies the required stochastic continuity.

The proof in the other direction is much more complicated. I can refer you to Theorem 2.1.8 from the classical Appelbaum's book "Levy processes and Stochastic calculus".

  • $\begingroup$ Your proof in the first direction looks good. Actually, there is no proof of the other direction (the two definitions are not equivalent). I give a counter-example here (see "Edit 2" of the answer): math.stackexchange.com/questions/3269965/… $\endgroup$ – Michael Jun 22 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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