Thank you!

]]>Yes, I agree.

]]>Thanks. Sounds to me that assuming stochastic continuity is then hardly a restriction at all, but rather serves to exclude some pathological cases.

]]>There are examples using the axiom of choice (i.e., assuming ZFC set theory). There are even deterministic examples, where and X is not continuous (https://en.wikipedia.org/wiki/Cauchy's_functional_equation), but these are all non-measurable (see http://math.stackexchange.com/q/318523/1321) and use the axiom of choice in the construction. There are axiomatizations/models of set theory where all subsets of the reals are Lebesgue measurable (e.g., Solovay model) and then I think that the continuity condition in the definition of Levy processes would be superfluous. However, such axiomatizations do not allow the uncountable axiom of choice.

]]>If we drop the assumption on continuity in probability, what are we left with? Are there processes with independent and stationary increments that do not have a cadlag modification (i.e. a Levy modification)? I have been struggling to come up with an examples of such processes.

]]>Do you know an easy proof of the fact that for two independent Levy processes $X$ and $Y$ the co-variation process $[X,Y]$ is equal to zero? I have a proof of this result but I feel that it is to complicated and I would like to make it shorter. Thank you very much.

Best regards,

Paolo ]]>

Hi. Strictly speaking that’s not true. If *X*_{0} > y then and . You can only conclude that if you assume that *X*_{0} ≤ 0 or, alternatively, if you restrict to . Then, the conclusion holds for any cadlag process, and is nothing specific to Lévy processes. In fact, you have for any right-continuous process and if it has left limits. If it also has no positive jumps then .

Let X be a Levy processes with no positive jumps and then we have

on

Could you explain that why? and does it hold for Levy process with no negative jumps? If X be Hunt process with no positive jumps then does this hold?

Thank you very much!