Timeline for The probability of Levy process staying at a point
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2014 at 23:57 | comment | added | Hengyu Zhou | It's a typo. I mean $P(X_{t-}=0)=0$. | |
| Mar 9, 2014 at 2:21 | comment | added | Bjørn Kjos-Hanssen | Is that a typo, "$P(X_{t-})=0$"? | |
| Mar 9, 2014 at 1:48 | comment | added | Hengyu Zhou | The motivation of this question is as follows. Assume $X_{t}$ is a Levy process. For any fixed $t$, is $P(X_{t-})=0$? Maybe you already answer my questions. But I am still not sure what's happening when $X_{t}$ is a pure jump process. | |
| Mar 9, 2014 at 1:39 | comment | added | Hengyu Zhou | Yes, you are right. If the generating triplet of $X_{t}$ is $(\sigma\neq 0, \gamma, \nu)$, $P(X_{t}=x)=0$. If the generating triplet of $X_{t}$ is $(0, 0, \nu)$, $P(X_{t}=x)$ is dependent on the measure $\nu$. If $\nu$ is absolute with respect to Lebesgue measure, $P(X_{t}=x)$ is still equal to 0. | |
| Mar 9, 2014 at 1:39 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
deleted 20 characters in body
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| Mar 9, 2014 at 1:31 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |