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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?

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  • $\begingroup$ 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
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$\def\eps{\varepsilon}$ $\def\P{\mathbb{P}}$ 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".

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  • $\begingroup$ 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

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