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Assume $X_{t}$ is a 1-dimensional Levy process on a probability $(\Omega, \mathcal{F}, P)$. For a fixed point $x$ in the state space and fixed $t\neq 0$, what's the value of $ P(\omega: X_{t}(\omega)=x)$?

In the case of Brownian motion, $P(\omega: X_{t}(\omega)=x)=0$.

In the case of Pure Jump Levy process, for example $X_{t}$ is a Poisson process, $P(\omega: X_{t}(\omega)=x)\neq 0$.

What confuses me is the case that $X_{t}$ is a Levy process with generating triplets $(\sigma, \gamma, \nu)$. Here $\sigma\neq 0$ is for the Brownian motion part, $\nu\neq 0$ is for the pure jump part. Is $P(\omega: X_{t}(\omega)=x)\neq 0$?

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As long as $\sigma\ne 0$, we get $P(X_t=x)=0$.

By the Levy-Ito decomposition there is a Brownian motion $Y$ and an independent process $Z$ with $X=Y+Z$. Then for $t>0$, $$ P(X_t=x)=P(Y_t=x-Z_t)=0 $$ since $Y$ is independent of $Z$.

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  • $\begingroup$ 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. $\endgroup$
    – Hengyu Zhou
    Commented Mar 9, 2014 at 1:39
  • $\begingroup$ 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. $\endgroup$
    – Hengyu Zhou
    Commented Mar 9, 2014 at 1:48
  • $\begingroup$ Is that a typo, "$P(X_{t-})=0$"? $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Mar 9, 2014 at 2:21
  • $\begingroup$ It's a typo. I mean $P(X_{t-}=0)=0$. $\endgroup$
    – Hengyu Zhou
    Commented Mar 9, 2014 at 23:57

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